Software Verification
User Manual: Software Verification
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Software Verification Examples
Software Verification Examples
For ETABS® 2016
ISO ETA122815M6 Rev. 1
Proudly developed in the United States of America
January 2017
Copyright
Copyright Computers & Structures, Inc., 1978-2017
All rights reserved.
The CSI Logo® and ETABS®are registered trademarks of Computers & Structures, Inc.
The computer program ETABS® and all associated documentation are proprietary and
copyrighted products. Worldwide rights of ownership rest with Computers & Structures,
Inc. Unlicensed use of these programs or reproduction of documentation in any form,
without prior written authorization from Computers & Structures, Inc., is explicitly
prohibited.
No part of this publication may be reproduced or distributed in any form or by any means, or
stored in a database or retrieval system, without the prior explicit written permission of the
publisher.
Further information and copies of this documentation may be obtained from:
Computers & Structures, Inc.
www.csiamerica.com
info@csiamerica.com (for general information)
support@csiamerica.com (for technical support)
DISCLAIMER
CONSIDERABLE TIME, EFFORT AND EXPENSE HAVE GONE INTO THE DEVELOPMENT
AND DOCUMENTATION OF THIS SOFTWARE. HOWEVER, THE USER ACCEPTS AND
UNDERSTANDS THAT NO WARRANTY IS EXPRESSED OR IMPLIED BY THE DEVELOPERS
OR THE DISTRIBUTORS ON THE ACCURACY OR THE RELIABILITY OF THIS PRODUCT.
THIS PRODUCT IS A PRACTICAL AND POWERFUL TOOL FOR STRUCTURAL DESIGN.
HOWEVER, THE USER MUST EXPLICITLY UNDERSTAND THE BASIC ASSUMPTIONS OF
THE SOFTWARE MODELING, ANALYSIS, AND DESIGN ALGORITHMS AND COMPENSATE
FOR THE ASPECTS THAT ARE NOT ADDRESSED.
THE INFORMATION PRODUCED BY THE SOFTWARE MUST BE CHECKED BY A
QUALIFIED AND EXPERIENCED ENGINEER. THE ENGINEER MUST
INDEPENDENTLY VERIFY THE RESULTS AND TAKE PROFESSIONAL
RESPONSIBILITY FOR THE INFORMATION THAT IS USED.
Contents
Introduction
Methodology
Conclusions
Problems
Analysis Problems
1
Plane Frame with Beam Span Loads, Static Gravity Load Analysis
2
Three-Story Plane Frame, Dynamic Response Spectrum Analysis
3
Three-Story Plane Frame, Code-Specific Static Lateral Load Analysis
4
Single Story Three-Dimensional Frame, Dynamic Response Spectrum
Analysis
5
Three-Story Three-Dimensional Braced Frame, Dynamic Response
Spectrum Analysis
6
Nine-Story Ten-Bay Plane Frame, Eigenvalue Analysis
7
Seven-Story Plane Frame, Gravity and Lateral Loads Analysis
8
Two-Story Three-Dimensional Frame, Dynamic Response Spectrum
Analysis
9
Two-Story Three-Dimensional Unsymmetrical Building Frame, Dynamic
Response Spectrum Analysis
10
Three-Story Plane Frame with ADAS Elements, Nonlinear Time History
Analysis
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11
Three-Story Plane Frame with Viscous Damper Elements, Nonlinear Time
History Analysis
12
Pounding of Two Planar Frames, Nonlinear Time History Analysis
13
Base Isolated Two-Story 3D Frame, Nonlinear Time History Analysis
14
Friction Pendulum Base-Isolated 3D Frame, Nonlinear Time History
Analysis
15
Wall Object Behavior, Static Lateral Loads Analysis
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Design Examples
Steel Frame
AISC 360-05 Example 001
Wide Flange Member Under Bending
AISC 360-05 Example 002
Build-up Wide Flange Member Under
Compression
AISC 360-10 Example 001
Wide Flange Member Under Bending
AISC 360-10 Example 002
Build-up Wide Flange Member Under
Compression
AISC ASD-89 Example 001
Wide Flange Member Under Bending
AISC ASD-89 Example 002
Wide Flange Member Under Compression
AISC LRFD-93 Example 001
Wide Flange Member Under Bending
AISC LRFD-93 Example 002
Wide Flange Member Under Combined
Compression & Biaxial Bending
AS 4100-1998 Example 001
Wide Flange Member Under Compression
AS 4100-1998 Example 002
Wide Flange Member Under Bending
AS 4100-1998 Example 003
Wide Flange Member Under Combined
Compression & Bending
BS 5950-2000 Example 001
Wide Flange Member Under Bending
BS 5950-2000 Example 002
Square Tube Member Under Compression &
Bending
CSA S16-09 Example 001
Wide Flange Member Under Compression &
Bending
CSA S16-09 Example 002
Wide Flange Member Under Compression &
Bending
CSA S16-14 Example 001
Wide Flange Member Under Compression &
Bending
CSA S16-14 Example 002
Wide Flange Member Under Compression &
Bending
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EC 3-2005 Example 001
Wide Flange Member Under Combined
Compression & Bending
EN 3-2005 Example 002
Wide Flange Section Under Bending
EN 3-2005 Example 003
Wide Flange Section Under Combined
Compression & Bending
IS 800-2007 Example 001
Wide Flange Member Under Compression
IS 800-2007 Example 002
Wide Flange Member Under Bending
IS 800-2007 Example 003
Wide Flange Member Under Combined
Compression & Biaxial Bending
KBC 2009 Example 001
Wide Flange Member Under Bending
KBC 2009 Example 002
Build-up Wide Flange Member Under
Compression
NTC 2008 Example 001
Wide Flange Section Under Combined
Compression & Bending
NTC 2008 Example 002
Wide Flange Section Under Combined
Compression & Bending
NZS 3404-1997 Example 001
Wide Flange Member Under Compression
NZS 3404-1997 Example 002
Wide Flange Member Under Bending
NZS 3404-1997 Example 003
Wide Flange Member Under Combined
Compression & Bending
Concrete Frame
ACI 318-08 Example 001
Beam Shear & Flexural Reinforcing
ACI 318-08 Example 002
ACI 318-11 Example 001
P-M Interaction Check for Rectangular Column
Beam Shear & Flexural Reinforcing
ACI 318-11 Example 002
ACI 318-14 Example 001
P-M Interaction Check for Rectangular Column
Beam Shear & Flexural Reinforcing
ACI 318-14 Example 002
P-M Interaction Check for Rectangular Column
AS 3600-2009 Example 001
Beam Shear & Flexural Reinforcing
AS 3600-2009 Example 002
P-M Interaction Check for Rectangular Column
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BS 8110-1997 Example 001
Beam Shear & Flexural Reinforcing
BS 8110-1997 Example 002
P-M Interaction Check for Rectangular Column
CSA A23.3-04 Example 001
Beam Shear & Flexural Reinforcing
CSA A23.3-04 Example 002
P-M Interaction Check for Rectangular Column
CSA A23.3-14 Example 001
Beam Shear & Flexural Reinforcing
CSA A23.3-14 Example 002
P-M Interaction Check for Rectangular Column
EN 2-2004 Example 001
Beam Shear & Flexural Reinforcing
EN 2-2004 Example 002
P-M Interaction Check for Rectangular Column
IS 456-2000 Example 001
Beam Shear & Flexural Reinforcing
IS 456-2000 Example 002
P-M Interaction Check for Rectangular Column
NTC 2008 Example 001
Beam Shear & Flexural Reinforcing
NTC 2008 Example 002 KBC
P-M Interaction Check for Rectangular Column
2009 Example 001
Beam Shear & Flexural Reinforcing
KBC 2009 Example 002
P-M Interaction Check for Rectangular Column
RCDF 2004 Example 001
Beam Moment Strength Using Equivalent
Rectangular Stress Distribution
RCDF 2004 Example 002
P-M Interaction Check for Rectangular Column
NZS 3101-2006 Example 001
Beam Shear & Flexural Reinforcing
NZS 3101-2006 Example 002
P-M Interaction Check for Rectangular Column
SS CP 65-1999 Example 001
Beam Shear & Flexural Reinforcing
SS CP 65-1999 Example 002
P-M Interaction Check for Rectangular Column
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TS 500-2000 Example 001
Beam Shear & Flexural Reinforcing
TS 500-2000 Example 002
P-M Interaction Check for Rectangular Column
Shear Wall
ACI 318-08 WALL-001
P-M Interaction Check for Wall
ACI 318-08 WALL-002
P-M Interaction Check for Wall
ACI 318-11 WALL-001
P-M Interaction Check for Wall
ACI 318-11 WALL-002
P-M Interaction Check for Wall
ACI 318-14 WALL-001
P-M Interaction Check for Wall
ACI 318-14 WALL-002
P-M Interaction Check for Wall
ACI 530-11 Masonry–WALL-001
P-M Interaction Check for Wall
ACI 530-11 Masonry–WALL-002
P-M Interaction Check for Wall
AS 360-09 WALL-001
P-M Interaction Check for a Wall
AS 360-09 WALL-002
P-M Interaction Check for a Wall
BS 8110-97 WALL-001
P-M Interaction Check for a Wall
BS 8110-97 WALL-002
P-M Interaction Check for a Wall
CSA A23.3-04 WALL-001
P-M Interaction Check for a Wall
CSA A23.3-04 WALL-002
P-M Interaction Check for a Wall
CSA A23.3-14 WALL-001
P-M Interaction Check for a Wall
CSA A23.3-14 WALL-002
P-M Interaction Check for a Wall
EC 2-2004 WALL-001
P-M Interaction Check for a Wall
EC 2-2004 WALL-002
P-M Interaction Check for a Wall
Hong Kong CP-04 WALL-001
P-M Interaction Check for a Wall
Hong Kong CP-04 WALL-002
P-M Interaction Check for a Wall
Indian IS 456-2000 WALL-001
P-M Interaction Check for a Wall
Indian IS 456-2000 WALL-002
P-M Interaction Check for a Wall
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KBC 2009 WALL-001
P-M Interaction Check for Wall
KBC 2009 WALL-002
P-M Interaction Check for Wall
Mexican RCDF-04 WALL-001
P-M Interaction Check for a Wall
Mexican RCDF-04 WALL-002
P-M Interaction Check for a Wall
NZS-3103-2006 WALL-001
P-M Interaction Check for a Wall
NZS-3103-2006 WALL-002
P-M Interaction Check for a Wall
Singapore CP65-99-001
P-M Interaction Check for a Wall
Singapore CP65-99-002
P-M Interaction Check for a Wall
Turkish TS 500-2000 WALL-001
P-M Interaction Check for a Wall
Turkish TS 500-2000 WALL-002
P-M Interaction Check for a Wall
Composite Beam
AISC 360-05 Example 001
Composite Girder Design
AISC 360-10 Example 001
Composite Girder Design
AISC 360-10 Example 002
Composite Girder Design
BS 5950-90 Example 001
Steel Designers Manual Sixth Edition – Design of
Simply Supported Composite Beam
CSA S16-09 Example 001
Handbook of Steel Construction Tenth Edition –
Composite Beam
EC 4-2004 Example 001
Steel Designers Manual Seventh Edition –
Design of Simply Supported Composite Beam
Composite Column
AISC 360-10 Example 001
Composite Column Design
AISC 360-10 Example 002
Composite Column Design
AISC 360-10 Example 003
Composite Column Design
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Slab
ACI 318-08 PT-SL Ex001
Post-Tensioned Slab Design
ACI 318-08 RC-PN Ex001
Slab Punching Shear Design
ACI 318-08 RC-SL Ex001
Slab Flexural Design
ACI 318-11 PT-SL Ex001
Post-Tensioned Slab Design
ACI 318-11 RC-PN Ex001
Slab Punching Shear Design
ACI 318-11 RC-SL Ex001
Slab Flexural Design
ACI 318-14 PT-SL Ex001
Post-Tensioned Slab Design
ACI 318-14 RC-PN Ex001
Slab Punching Shear Design
ACI 318-14 RC-SL Ex001
Slab Flexural Design
AS 3600-2001 PT-SL Ex001
Post-Tensioned Slab Design
AS 3600-2001 RC-PN Ex001
Slab Punching Shear Design
AS 3600-2001 RC-SL Ex001
Slab Flexural Design
AS 3600-2009 PT-SL Ex001
Post-Tensioned Slab Design
AS 3600-2009 RC-PN Ex001
Slab Punching Shear Design
AS 3600-2009 RC-SL Ex001
Slab Flexural Design
BS 8110-1997 PT-SL Ex001
Post-Tensioned Slab Design
BS 8110-1997 RC-PN Ex001
Slab Punching Shear Design
BS 8110-1997 RC-SL Ex001
Slab Flexural Design
CSA A23.3-04 PT-SL Ex001
Post-Tensioned Slab Design
CSA A23.3-04 RC-PN Ex001
Slab Punching Shear Design
CSA A23.3-04 RC-SL Ex001
Slab Flexural Design
CSA A23.3-14 PT-SL Ex001
Post-Tensioned Slab Design
CSA A23.3-14 RC-PN Ex001
Slab Punching Shear Design
CSA A23.3-14 RC-SL Ex001
Slab Flexural Design
EN 2-2004 PT-SL Ex001
Post-Tensioned Slab Design
EN 2-2004 RC-PN Ex001
Slab Punching Shear Design
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EN 2-2004 RC-SL Ex001
Slab Flexural Design
HK CP-2004 PT-SL Ex001
Post-Tensioned Slab Design
HK CP-2004 RC-PN Ex001
Slab Punching Shear Design
HK CP-2004 RC-SL Ex001
Slab Flexural Design
HK CP-2013 PT-SL Ex001
Post-Tensioned Slab Design
HK CP-2013 RC-PN Ex001
Slab Punching Shear Design
HK CP-2013 RC-SL Ex001
Slab Flexural Design
IS 456-2000 PT-SL Ex001
Post-Tensioned Slab Design
IS 456-2000 RC-PN Ex001
Slab Punching Shear Design
IS 456-2000 RC-SL Ex001
Slab Flexural Design
NTC 2008 PT-SL Ex001
Post-Tensioned Slab Design
NTC 2008 RC-PN Ex001
Slab Punching Shear Design
NTC 2008 RC-SL Ex001
Slab Flexural Design
NZS 3101-2006 PT-SL Ex001
Post-Tensioned Slab Design
NZS 3101-2006 RC-PN Ex001
Slab Punching Shear Design
NZS 3101-2006 RC-SL Ex001
Slab Flexural Design
SS CP 65-1999 PT-SL Ex001
Post-Tensioned Slab Design
SS CP 65-1999 RC-PN Ex001
Slab Punching Shear Design
SS CP 65-1999 RC-SL Ex001
Slab Flexural Design
TS 500-2000 PT-SL Ex001
Post-Tensioned Slab Design
TS 500-2000 RC-PN Ex001
Slab Punching Shear Design
TS 500-2000 RC-SL Ex001
Slab Flexural Design
ETABS
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References
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ETABS Software Verification Log
Revision
Number
Date
0
19 Apr 2013
Description
Initial release of ETABS, Version 13.0.0
1
9 July 2013
2
11 Apr 2014
Minor documentation errors in the Verification manuals have
been corrected
Minor improvements have been made to some of the
examples, and some example file names have been changed
for consistency. The design results produced and reported by
ETABS are correct. The reported results are not changed
except where the model has been changed.
Three new examples have been added for steel frame design.
Analysis model EX8.EDB - The response-spectrum function
damping was incorrect and did not match the responsespectrum load case damping, hence the results produced did
not match the documented value. After correction, the example
produces the expected and documented results. No change was
made to the Verification manual.
Analysis Example 03 - The name of code IBC2000 was
changed to ASCE 7-02, as actually used in ETABS (IBC2000
was used in v9.7.4). In addition, the Verification manual was
corrected for the actual values produced by ETABS. These
values have not changed since v13.0.0. The documented
values were for ETABS v9.7.4 and some changed in v13.0.0
due to the use of a different solver. The change has no
engineering significance.
Analysis Example 06 and Example 07 - The Verification
manual was corrected for the actual values produced by
ETABS. These values have not changed since v13.0.0. The
documented values were for ETABS v9.7.4 and some changed
in v13.0.0 due to the use of a different solver. The change has
no engineering significance.
Analysis Example 15 - The Verification manual was corrected
for the actual values produced by ETABS. These values have
not changed since v13.0.0. The documented values were for
ETABS v9.7.4 and some changed in v13.0.0 due to the use of
a different solver, and due to the difference in how wall
elements are connected to beams. The change due to the solver
has no engineering significance. The change for wall elements
was an enhancement.
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ETABS Software Verification Log
Revision
Number
Date
3
3 Nov 2014
Description
Concrete Frame Design EN 2-2004 Example 001, Concrete
Frame Design NTC 2008 Example 002 - The values produced
by ETABS 2014 were updated in the Verification manual for a
change in v13.1.3 under Incident 59154 (Ticket 23901) where
the coefficients Alpha_CC and Alpha_LCC were not taken
into account in certain cases.
Concrete Frame Design AS 3600-2009 Example 002, Shear
Wall Design AS 3600-2009 WALL-002 - The values produced
by ETABS were updated in the Verification manual for a
change in v13.1.4 under Incident 59973 where the phi factor
was incorrectly computed.
Analysis Example 14 – Minor changes have been made to the
results as the result of an enhancement made under Incident
67283 to improve the convergence behavior of nonlinear static
and nonlinear direct-integration time history analysis.
Composite Beam Design AISC-360-05 Example 001 was
updated to reflect the fact that, under Incident 59912 it is now
possible to specify that the shear stud strength is to be
computed assuming the weak stud position. A typo in the
version number of the referenced Design Guide example was
corrected. A slight error in the hand-calculation for the partial
composite action Mn was corrected, resulting in perfect
agreement with the value produced by ETABS.
Composite Beam Design AISC-360-10 Example 001 was
updated to reflect the fact that, under Incident 59912 it is
possible to specify that the shear stud strength is to be
computed assuming the weak stud position. The handcalculation for the partial composite action Mn was revised
to account for a lower percentage of composite action caused
by an increase in the number of shear studs per deck rib in
places, and a corresponding decrease in shear stud strength.
Composite Beam Design BS-5950-90 Example 001- The
hand-calculations in the Verification manual were updated to
reflect the actual section area of a UKB457x191x167, which
differs from the value in the reference example, and to reflect
that the maximum number of shear studs that can be placed on
the beam is 78 studs and not the 80 the reference example calls
for. Also the value of the live load deflection produced by
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ETABS Software Verification Log
Revision
Number
Date
Description
ETABS was updated for a change in v13.2.0 under Incident
56782.
Composite Beam Design CSA-S16-09 Example 001. The
values produced by ETABS for the shear stud capacity were
updated in the Verification manual for a change in v13.2.0
under Incident 71303. This change in turn affects the value of
the partial composite moment capacity Mc but has no
engineering significance. A typo affecting the value of precomposite deflection in the Results Comparison table was
corrected.
Composite Beam Design EC-4-2004 Example 001. The handcalculation of the construction moment capacity, Ma,pl,Rd was
updated to reflect a more accurate value of the section Wpl and
typos affecting the pre-composite deflection and beam camber
were corrected. None of the values computed by ETABS
changed.
Initial release of ETABS 2015, Version 15.0.0
4
7 Jan 2015
Shear Wall Design example Eurocode 2-2004 Wall-002 has
been updated due to changes previously reported under
Incident #56569.
Shear Wall Design example AS 3600-09 Wall-001 has been
updated due to changes previously reported under Incident
#56113.
Shear Wall Design example CSA A23.3-04 Example 001 has
been updated due to changes previously reported under
Incident #71922.
Concrete Frame Design example CSA A23.3-04 Example 002
has been updated due to changes previously reported under
Incident #71922.
New steel frame design examples have been added for CSA
S16-14 and KBC 2009.
New concrete frame design examples have been added for
ACI 318-14, CSA A23.3-14, and KBC 2009.
New shear wall design examples have been added for ACI
318-14, CSA A23.3-14, and KBC 2009.
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ETABS Software Verification Log
Revision
Number
Date
5
7 July 2016
INTRODUCTION
Description
Initial release of ETABS 2016, Version 16.0.0
Added SAFE design verification examples for slab
design, punching shear design, and post-tension design
for all codes supported in both SAFE and ETABS.
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INTRODUCTION
This manual provides example problems used to test various features and capabilities of the
ETABS program. Users should supplement these examples as necessary for verifying their
particular application of the software.
METHODOLOGY
A series of test problems, or examples, designed to test the various elements and analysis
features of the program were created. For each example, this manual contains a short
description of the problem; a list of significant ETABS options tested; and a comparison of
key results with theoretical results or results from other computer programs. The comparison
of the ETABS results with results obtained from independent sources is provided in tabular
form as part of each example.
To validate and verify ETABS results, the test problems were run on a PC platform that was
a Dell machine with a Pentium III processor and 512 MB of RAM operating on a Windows
XP operating system.
Acceptance Criteria
The comparison of the ETABS validation and verification example results with independent
results is typically characterized in one of the following three ways.
Exact: There is no difference between the ETABS results and the independent results within
the larger of the accuracy of the typical ETABS output and the accuracy of the independent
result.
Acceptable: For force, moment and displacement values, the difference between the
ETABS results and the independent results does not exceed five percent (5%). For internal
force and stress values, the difference between the ETABS results and the independent
results does not exceed ten percent (10%). For experimental values, the difference between
the ETABS results and the independent results does not exceed twenty five percent (25%).
Unacceptable: For force, moment and displacement values, the difference between the
ETABS results and the independent results exceeds five percent (5%). For internal force and
stress values, the difference between the ETABS results and the independent results exceeds
ten percent (10%). For experimental values, the difference between the ETABS results and
the independent results exceeds twenty five percent (25%).
The percentage difference between results is typically calculated using the following
formula:
ETABS Result
Percent Difference 100
1
Independent Result
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For examples with multiple versions of meshing density of area elements, only the models
with the finest meshing density are expected to fall within Exact or Acceptable limits.
Summary of Examples
The example problems addressed plane frame, three-dimensional, and wall structures as
well as shear wall and floor objects. The analyses completed included dynamic response
spectrum, eigenvalue, nonlinear time history, and static gravity and lateral load.
Other program features tested include treatment of automatic generation of seismic and wind
loads, automatic story mass calculation, biaxial friction pendulum and biaxial hysteretic
elements, brace and column members with no bending stiffness, column pinned end
connections, multiple diaphragms, non-rigid joint offsets on beams and columns, panel
zones, point assignments, rigid joint offsets, section properties automatically recovered from
the database, uniaxial damper element, uniaxial gap elements, vertical beam span loading
and user specified lateral loads and section properties.
Slab design examples verify the design algorithms used in ETABS for flexural, shear design
of beam; flexural and punching shear of reinforced concrete slab; and flexural design and
serviceability stress checks of post-tensioned slab by comparing ETABS results with hand
calculations.
Analysis: Of the fifteen Analysis problems, eight showed exact agreement while the
remaining seven showed acceptable agreement between ETABS and the cited independent
sources.
Design – Steel Frame: All 30 Steel Frame Design problems showed acceptable agreement
between ETABS and the cited independent sources.
Design – Concrete Frame: All 34 Concrete Frame Design problems showed acceptable
agreement between ETABS and the cited independent sources.
Design – Shear Wall: All 32 of the Shear Wall Design problems showed acceptable
agreement between ETABS and the cited independent sources.
Design – Composite Beam: The 6 Composite Beam Design problems showed acceptable
agreement between ETABS and the cited independent sources.
Design – Composite Column: The 3 Composite Column Design problems showed
acceptable agreement between ETABS and cited independent sources.
Design – Slab: The 48 Slab Design problems showed acceptable agreement between ETABS
and cited independent sources.
Summary of Examples
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CONCLUSIONS
ETABS is the latest release of the ETABS series of computer programs. Since development,
ETABS has been used widely for structural analysis. The ongoing usage of the program
coupled with continuing program upgrades are strong indicators that most program bugs
have been identified and corrected.
Additionally, the verification process conducted as described in this document demonstrates
that the program features tested are operating reliably and with accuracy consistent with
current computer technology capabilities.
MESHING OF AREA ELEMENTS
It is important to adequately mesh area elements to obtain satisfactory results. The art of
creating area element models includes determining what constitutes an adequate mesh. In
general, meshes should always be two or more elements wide. Rectangular elements give
the best results and the aspect ratio should not be excessive. A tighter mesh may be needed
in areas where the stress is high or the stress is changing quickly.
When reviewing results, the following process can help determine if the mesh is adequate.
Pick a joint in a high stress area that has several different area elements connected to it.
Review the stress reported for that joint for each of the area elements. If the stresses are
similar, the mesh likely is adequate. Otherwise, additional meshing is required. If you choose
to view the stresses graphically when using this process, be sure to turn off the stress
averaging feature when displaying the stresses.
CONCLUSIONS
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EXAMPLE 1
Plane Frame with Beam Span Loads - Static Gravity Load Analysis
Problem Description
This is a one-story, two-dimensional frame subjected to vertical static loading.
To be able to compare ETABS results with theoretical results using prismatic members and
elementary beam theory, rigid joint offsets on columns and beams are not modeled, and axial
and shear deformations are neglected. Thus, the automatic property generation feature of
ETABS is not used; instead, the axial area and moment of inertia for each member are explicitly input.
Geometry, Properties and Loading
The frame is a three-column line, two-bay system. Kip-inch-second units are used. The
modulus of elasticity is 3000 ksi. All columns are 12"x24"; all beams are 12"x30".
The frame geometry and loading patterns are shown in Figure 1-1.
50k
Eq.
100k
Eq.
100k
Eq.
100k
Eq.
50k
Case 1
Case 2
10k/ft
Pinned
Connection
10’
Origin
Figure 1-1 Plane Frame with Beam Span Loads
Plane Frame with Beam Span Loads - Static Gravity Load Analysis
1-1
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Technical Features of ETABS Tested
Two-dimensional frame analysis
Vertical beam span loading
No rigid joint offsets on beams and columns
Column pinned end connections
Results Comparison
The theoretical results for bending moments and shear forces on beams B1 and B2 are easily
obtained from tabulated values for propped cantilevers (American Institute of Steel Construction 1989). These values for beam B1 are compared with ETABS results in Table 1-1.
Table 1-1 Comparison of Results for Beam B1 – Case 1
Load Case I
(Concentrated Load)
Quantity
Bending Moments
Shear Forces
Location
End I
ETABS
Theoretical
0.00
0.00
¼ Point
1,687.50
1,687.50
½ point
3,375.00
3,375.00
¾ point
-337.50
-337.50
End J
-4,050.00
-4,050.00
End I
-31.25
-31.25
¼ Point
-31.25
-31.25
½ point
68.75
68.75
¾ point
68.75
68.75
End J
68.75
68.75
Table 1-1 Comparison of Results for Beam B1 – Case II
Load Case II
(Uniformly Distributed Load)
Quantity
Bending Moments
Location
End I
ETABS
Theoretical
0.00
0.00
¼ Point
2,430.00
2,430.00
½ point
2,430.00
2,430.00
¾ point
0.00
0.00
Plane Frame with Beam Span Loads - Static Gravity Load Analysis
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Table 1-1 Comparison of Results for Beam B1 – Case II
Load Case II
(Uniformly Distributed Load)
Quantity
Shear Forces
Location
ETABS
Theoretical
End J
-4,860.00
-4,860.00
End I
-67.50
-67.50
¼ Point
-22.50
-22.50
½ point
22.50
22.50
¾ point
67.50
67.50
End J
112.50
112.50
Computer File
The input data file for this example is Example 01.EDB. This file is provided as part of the
ETABS installation.
Conclusion
The comparison of results shows an exact match between the ETABS results and the theoretical data.
Plane Frame with Beam Span Loads - Static Gravity Load Analysis
1-3
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EXAMPLE 2
Three-Story Plane Frame - Dynamic Response Spectrum Analysis
Problem Description
This is a three-story plane frame subjected to the El Centro 1940 seismic response spectra, N-S
component, 5 percent damping.
Assuming the beams to be rigid and a rigid offset at the column top ends of 24 inches (i.e.,
equal to the depth of the beams), and neglecting both shear deformations and axial deformations, the story lateral stiffness for this example can be calculated (Przemieniecki 1968).
The example then reduces to a three-spring, three-mass system with equal stiffnesses and
masses. This can be analyzed using any exact method (Paz 1985) to obtain the three natural
periods and mass normalized mode shapes of the system.
The spectral accelerations at the three natural periods can then be linearly interpolated from
the response spectrum used.
The spectral accelerations can in turn be used with the mode shapes and story mass information to obtain the modal responses (Paz 1985). The modal responses for story displacements and column moments can then be combined using the complete quadratic combination
procedure (Wilson, et al. 1981).
Geometry, Properties and Loading
The frame is modeled as a two-column line, single bay system. Kip-inch-second units are
used. Other parameters associated with the structure are as follows:
All columns are W14X90
All beams are infinitely rigid and 24" deep
Modulus of elasticity
= 29500 ksi
Typical story mass
= 0.4 kip-sec2/in
The column is modeled to have infinite axial area, so that axial deformation is neglected. Also, zero column shear area is input to trigger the ETABS option of neglecting shear deformations. These deformations are neglected to be consistent with the hand-calculated model
with which the results are compared.
The frame geometry is shown in Figure 2-1.
Three-Story Plane Frame - Dynamic Response Spectrum Analysis
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Figure 2-1 Three-Story Plane Frame
Technical Features in ETABS Tested
Two-dimensional frame analysis
Rigid joint offsets on beams and columns automatically calculated
Dynamic response spectrum analysis
Results Comparison
The three theoretical natural periods and mass normalized mode shapes are compared in Table 2-1 with ETABS results.
Table 2-1 Comparison of Results for Periods and Mode Shapes
Mode
1
2
3
Period, secs.
0.4414
0.1575
0.1090
Mode Shape
ETABS
Theoretical
Roof Level
1.165
1.165
2nd Level
0.934
0.934
1st Level
0.519
0.519
Roof Level
0.934
0.934
2nd Level
-0.519
-0.519
1st Level
-1.165
-1.165
Roof Level
0.519
0.519
Three-Story Plane Frame - Dynamic Response Spectrum Analysis
2-2
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Table 2-1 Comparison of Results for Periods and Mode Shapes
Mode
Period, secs.
Mode Shape
ETABS
Theoretical
2nd Level
-1.165
-1.165
1st Level
0.934
0.934
The story displacements and column moments thus obtained are compared in Table 2-2 with
ETABS results. The results are identical.
Table 2-2 Comparison of Displacements and Column Moments
Quantity
ETABS
Theoretical
Displacement at
Roof
2.139
2.139
2nd
1.716
1.716
1st
0.955
0.955
11,730
11,730
Moment, Column C1, at Base
Computer Files
The input data file for this example is Example 02.EDB. The response spectrum file is
ELCN-RS1. These files are provided as part of the ETABS installation.
Conclusion
The result comparison shows an exact match between the ETABS results and the theoretical
data.
Three-Story Plane Frame - Dynamic Response Spectrum Analysis
2-3
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EXAMPLE 3
Three-Story Plane Frame, Code-Specified Static Lateral Load Analysis
Problem Description
The frame is modeled as a two-column line, single bay system. This three-story plane frame
is subjected to the following three code-specified lateral load cases:
UBC 1997 specified seismic loads (International Conference of Building Officials 1997)
ASCE 7-02 specified seismic loads (American Society of Civil Engineers 2002)
UBC 1997 specified wind loads (International Conference of Building Officials 1997)
Geometry, Properties and Loads
Kip-inch-second units are used. Other parameters associated with the structure are as follows:
All columns are W14X90
All beams are infinitely rigid and 24" deep
Modulus of elasticity
= 29500 ksi
Poisson's ratio
= 0.3
Typical story mass
= 0.4 kip-sec2/in
The frame geometry is shown in Figure 3-1.
Figure 3-1 Three-Story Plane Frame
Three-Story Plane Frame, Code-Specified Static Lateral Load Analysis
3-1
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For the UBC97 seismic load analysis, the code parameters associated with the analysis are as
follows:
UBC Seismic zone factor, Z
UBC Soil Profile Type
UBC Importance factor, I
UBC Overstrength Factor
UBC coefficient Ct
UBC Seismic Source Type
Distance to Source
= 0.40
= SC
= 1.25
= 8.5
= 0.035
=B
= 15 km
For the ASCE 7-02 seismic load analysis, the code parameters associated with the analysis
are as follows:
Site Class
Response Accel, Ss
Response Accel, S1
Response Modification, R
Coefficient Ct
Seismic Group
=C
=1
= 0.4
=8
= 0.035
=I
For the UBC97 wind load analysis, the exposure and code parameters associated with the
analysis are as follows:
Width of structure supported by frame
UBC Basic wind speed
UBC Exposure type
UBC Importance factor, I
UBC Windward coefficient, Cq
UBC Leeward coefficient, Cq
= 20 ft
= 100 mph
=B
=1
= 0.8
= 0.5
Technical Features in ETABS Tested
Two-dimensional frame analysis
Section properties automatically recovered from AISC database
Automatic generation of UBC 1997 seismic loads
Automatic generation of ASCE 7-02 seismic loads
Automatic generation of UBC 1997 wind loads
Three-Story Plane Frame, Code-Specified Static Lateral Load Analysis
3-2
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Results Comparison
For each of the static lateral load analyses, the story shears can be computed using the
formulae given in the applicable references. For the seismic loads, the fundamental period
computed by ETABS can be used in the formulae. From ETABS results, this fundamental
period is 0.5204 second. (Note the difference between the calculated fundamental period for
this example and Example 2, which neglects shear and axial deformations.)
Hand-calculated story shears are compared with story shears produced by the ETABS
program in Table 3-1 for UBC seismic loads, Table 3-2 for ASCE 7-02 seismic loads and
Table 3-3 for UBC wind loads.
Table 3-1 Comparison of Results for Story Shears - UBC 1997 Seismic
Level
ETABS (kips)
Theoretical (kips)
Roof
34.07
34.09
2nd
56.78
56.82
1st
68.13
68.19
Table 3-2 Comparison of Results for Story Shears - ASCE 7-02 Seismic
Level
ETABS (kips)
Theoretical (kips)
Roof
19.37
19.38
2nd
32.23
32.25
1st
38.61
38.64
Table 3-3 Comparison of Results for Story Shears - UBC 1997 Wind
Level
ETABS (kips)
Theoretical (kips)
Roof
3.30
3.30
2nd
9.49
9.49
1st
15.21
15.21
Computer File
The input data file for this example is Example 03.EDB. This file is provided as part of the
ETABS installation.
Three-Story Plane Frame, Code-Specified Static Lateral Load Analysis
3-3
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Conclusion
The results comparison shows an exact match between the ETABS results and the theoretical
data.
Three-Story Plane Frame, Code-Specified Static Lateral Load Analysis
3-4
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EXAMPLE 4
Single-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
Problem Description
This is a one-story, four-bay, three-dimensional frame. The frame is subjected to the El Centro 1940 N-S component seismic response spectrum, for 5 percent damping, in two orthogonal directions. The columns are modeled to neglect shear and axial deformations to be consistent with the assumptions of hand calculations with which the results are compared.
The example is a three-degree-of-freedom system. From the individual column lateral stiffnesses, assuming rigid beams and rigid offsets at column top ends equal to 36 inches (i.e., the
depth of the beams) and neglecting both shear deformations and column axial deformations,
the structural stiffness matrix can be assembled (Przemieniecki 1968).
Geometry, Properties and Loads
The frame geometry is shown in Figure 4-1.
Figure 4-1 Single-Story Three-Dimensional Frame
The structure is modeled as a single frame with four column lines and four bays. Kip-inchsecond units are used. Other parameters associated with the structure are as follows:
Columns on lines C1 and C2: 24" x 24"
Single-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
4 -1
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Columns on lines C3 and C4: 18" x 18"
All beams infinitely rigid and 36" deep
Modulus of elasticity = 3000 ksi
Story weight
= 150 psf
Technical Features of ETABS Tested
Three-dimensional frame analysis
Automatic story mass calculation
Dynamic response spectrum analysis
Results Comparison
From the stiffness and mass matrices of the system, the three natural periods and mass normalized mode shapes of the system can be obtained (Paz 1985). These are compared in Table
4-1 with ETABS results.
Table 4-1 Comparison of Results for Periods and Mode Shapes
Mode
1
Quantity
Period, sec.
ETABS
Theoretical
0.1389
0.1389
X-translation
-1.6244
-1.6244
Y-translation
0.0000
0.000
Z-rotation
0.0032
0.0032
0.1254
0.1254
X-translation
0.000
0.000
Y-translation
1.6918
1.6918
Z-rotation
0.000
0.000
0.0702
0.070
X-translation
0.4728
0.4728
Y-translation
0.000
0.000
Z-rotation
0.0111
0.0111
Mode Shape
2
Period, sec.
Mode Shape
3
Period,sec.
Mode Shape
Single-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
4 -2
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Computer File
The input data file for this example is Example 04.EDB. This file is provided as part of the
ETABS installation.
Conclusion
The results comparison shows an exact match between the ETABS results and the theoretical
data.
Single-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
4 -3
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EXAMPLE 5
Three-Story, Three-Dimensional Braced Frame - Dynamic Response Spectrum
Analysis
Problem Description
This is an L-shaped building structure with four identical braced frames. All members (columns and braces) carry only axial loads.
The structure is subject to the El Centro 1940 N-S component seismic response spectrum in
the X-direction. The structural damping is 5 percent. The structure is modeled by appropriately placing four identical planar frames. Each frame is modeled using three column lines. Kipinch-second units are used.
Geometry, Properties and Loading
The modulus of elasticity is taken as 29500 ksi and the typical member axial area as 6 in2. A
story mass of 1.242 kip-sec2/in and a mass moment of inertia of 174,907.4 kip-sec2-in are
used.
The geometry of the structure and a typical frame are shown in Figure 5-1.
Technical Features of ETABS Tested
Three-dimensional structure analysis using planar frames
Brace (diagonal) and column members with no bending stiffness
Dynamic response spectrum analysis
Results Comparison
This example has been solved in Wilson and Habibullah (1992) and Peterson (1981). A
comparison of ETABS results for natural periods and key member forces for one frame
with these references is given in Table 5-1.
Three-Story, Three-Dimensional Braced Frame - Dynamic Response Spectrum Analysis
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D4
D1
D2
D3
D1
D2
D3
D4
D1
D2
D3
D4
Figure 5-1 Three-Story, Three-Dimensional Braced Frame Building
Three-Story, Three-Dimensional Braced Frame - Dynamic Response Spectrum Analysis
5 -2
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Table 5-1 Comparison of Results
ETABS
Wilson and
Habibullah
Peterson
Period, Mode 1
0.32686
0.32689
0.32689
Period, Mode 2
0.32061
0.32064
0.32064
Axial Force
Column C1, Story 1
279.39
279.47
279.48
Axial Force
Brace D1, Story 1
194.44
194.51
194.50
Axial Force
Brace D3, Story 1
120.49
120.53
120.52
Quantity
Computer File
The input data file is Example 05.EDB. This file is provided as part of the ETABS installation.
Conclusions
The results comparison reflects acceptable agreement between the ETABS results and reference data.
Three-Story, Three-Dimensional Braced Frame - Dynamic Response Spectrum Analysis
5 -3
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EXAMPLE 6
Nine-Story, Ten-Bay Plane Frame - Eigenvalue Analysis
Problem Description
An eigenvalue analysis is completed.
Geometry, Properties and Loads
The frame is modeled with eleven column lines and ten bays. Kip-ft-second units are used. A
modulus of elasticity of 432,000 ksf is used. A typical member axial area of 3ft2 and moment
of inertia of 1ft4 are used. A mass of 3kip-sec2/ft/ft of member length is converted to story
mass using tributary lengths and used for the analysis.
This is a nine-story, ten-bay plane frame, as shown in Figure 6-1.
Figure 6-1 Nine-Story, Ten-Bay Plane Frame
Nine-Story, Ten-Bay Plane Frame - Eigenvalue Analysis
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Technical Features of ETABS Tested
Two-dimensional frame analysis
Eigenvalue analysis
Results Comparison
This example is also analyzed in Wilson and Habibullah (1992) and Bathe and Wilson
(1972). There are two differences between the ETABS analysis and the analyses of the
references. The models of the references assign vertical and horizontal mass degrees of
freedom to each joint in the structure. However, the ETABS model only assigns horizontal
masses and additionally, only one horizontal mass is assigned for all the joints associated
with any one floor level.
The eigenvalues obtained from ETABS are compared in Table 6-1 with results from Wilson
and Habibullah (1992) and Bathe and Wilson (1972).
Table 6-1 Comparison of Results for Eigenvalues
Quantity
ETABS
Wilson and
Habibullah
Bathe and
Wilson
1
0.58965
0.58954
0.58954
2
5.53196
5.52696
5.52695
3
16.5962
16.5879
16.5878
Computer File
The input data filename for this example is Example 06.EDB. This file is provided as part of
the ETABS installation.
Conclusions
Considering the differences in modeling enumerated herein, the results comparison between
ETABS and the references is acceptable.
Nine-Story, Ten-Bay Plane Frame - Eigenvalue Analysis
6 -2
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EXAMPLE 7
Seven-Story, Plane Frame - Gravity and Lateral Loads Analysis
Problem Description
This is a seven-story plane frame. The frame is modeled with three column lines and two
bays. Kip-inch-second units are used. Because the wide flange members used in the frame
are older sections, their properties are not available in the AISC section property database
included with the ETABS program, and the required properties therefore need to be explicitly provided in the input data.
The example frame is analyzed in Wilson and Habibullah (1992) for gravity loads, static
lateral loads and dynamic response spectrum loads. DYNAMIC/EASE2 analyzes the example frame under static lateral loads and dynamic response spectrum and time history
loads. A comparison of key ETABS results with Wilson and Habibullah (1992) and DYNAMIC/EASE2 results is presented in Tables 7-1, 7-2, 7-3 and 7-4. Note the difference in
modal combination techniques between ETABS and Wilson and Habibullah, which uses
complete quadratic combination (CQC), and DYNAMIC/EASE2, which uses square root
of the sum of the squares combination (SRSS).
Geometry, Properties and Loads
The gravity loads and the geometry of the frame are shown in Figure 7-1.
The frame is subjected to the following lateral loads:
Static lateral loads, shown in Figure 7-1
Lateral loads resulting from the El Centro 1940 N-S component seismic response spectra, 5 percent damping
Lateral loads resulting from the El Centro 1940 N-S component acceleration time history
Seven-Story, Plane Frame - Gravity and Lateral Loads Analysis
7 -1
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Vertical Loading,
typical for all
levels
Global Reference Point
All columns are W14s
All beams are W24s
Member weights are indicated
Typical story mass = 0.49 kip-sec 2/in
Figure 7-1 Seven-Story Plane Frame
Seven-Story, Plane Frame - Gravity and Lateral Loads Analysis
7 -2
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Technical Features of ETABS Tested
Two-dimensional frame analysis
User-specified section properties
User-specified lateral loads
Dynamic response spectrum analysis
Dynamic time history analysis
Results Comparison
The comparison of the results for all three analyses is excellent.
Table 7-1 Comparison of Results for Static Lateral Loads
ETABS
Wilson and
Habibullah
DYNAMIC/EASE2
Lateral Displacement
at Roof
1.4508
1.4508
1.4508
Axial Force
Column C1, at ground
69.99
69.99
69.99
Moment
Column C1, at ground
2324.68
2324.68
2324.68
Quantity
Table 7-2 Comparison of Results for Periods of Vibration
Mode
ETABS
Wilson and
Habibullah
DYNAMIC/EASE2
1
1.27321
1.27321
1.27321
2
0.43128
0.43128
0.43128
3
0.24205
0.24204
0.24204
4
0.16018
0.16018
0.16018
5
0.11899
0.11899
0.11899
6
0.09506
0.09506
0.09506
7
0.07952
0.07951
0.07951
Seven-Story, Plane Frame - Gravity and Lateral Loads Analysis
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Table 7-3 Comparison of Results for Response Spectrum Analysis
Wilson and
ETABS
Habibullah
DYNAMIC/EASE2
CQC
SRSS
CQC
Quantity
Combination
Combination
Combination
Lateral Displacement
at Roof
5.4314
5.4314
5.4378
Axial Force
Column C1 at ground
261.52
261.50
261.76
Moment
Column C1 at ground
9916.12
9916.11
9868.25
Table 7-4 Comparison of Results for Time History Analysis
ETABS
Wilson and
Habibullah
Maximum Roof Displacement
5.49
5.48
Maximum Base Shear
285
284
Maximum Axial Force, Column C1 at ground
263
258
Maximum Moment, Column C1 at ground
9104
8740
Quantity
Computer Files
The input data file is Example 07.EDB. The input history is ELCN-THU. Time history results are obtained for the first eight seconds of the excitation. This is consistent with DYNAMIC/EASE2, with which the results are compared. These computer files are provided
as part of the ETABS installation.
Conclusions
Noting the difference in modal combination techniques between ETABS and Wilson and
Habibullah, which uses complete quadratic combination (CQC), and DYNAMIC/EASE2,
which uses square root of the sum of the squares combination (SRSS), the results of the
testing are acceptable.
Seven-Story, Plane Frame - Gravity and Lateral Loads Analysis
7 -4
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EXAMPLE 8
Two-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
Problem Description
This is a two-story, three-dimensional building frame subjected to a response spectrum of
constant amplitude. The three-dimensional structure is modeled as a single frame with nine
column lines and twelve bays. Kip-foot-second units are used.
For consistency with the models documented in other computer programs with which the
ETABS results are compared (see Table 8-1), no story mass moments of inertia are assigned in the ETABS model.
Geometry, Properties and Loads
The geometry of the structure is shown in Figure 8-1.
B5
B6
B10
B8
B12
B3
13'
B4
B7
B9
B1
B2
B11
13'
C8
C7
C4
Z
C1
C5
C9
C6
25'
Y
C3
C2
X
35'
GLOBAL
AND FRAME
REFERENCE POINT
25'
35'
STORY 1 CENTER OF MASS AT (38,27,13)
STORY 2 CENTER OF MASS AT (38,27,26)
TYPICAL STORY MASS = 6.212 kip-sec 2 /ft
Figure 8-1 Two-Story Three-Dimensional Frame
Two-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
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A response spectrum with a constant value of 0.4g is used. Other parameters associated
with the structure are as follows:
Columns
4 ft2
1.25 ft4
1.25 ft4
350000 ksf
Axial area
Minor moment of inertia
Major moment of inertia
Modulus of elasticity
Beams
5 ft2
1.67 ft4
2.61 ft4
500000 ksf
Technical Features of ETABS Tested
Three-dimensional frame analysis
User-specified section properties
Dynamic response spectrum analysis
Comparison of Results
This example is also analyzed in Wilson and Habibullah (1992) and Peterson (1981). A
comparison of the key ETABS results with Wilson and Habibullah (Reference 1) and Peterson (Reference 2) is shown in Table 8-1.
Table 8-1 Comparison of Results
Quantity
ETABS
Reference 1
Reference 2
Period, Mode 1
0.22708
0.22706
0.22706
Period, Mode 2
0.21565
0.21563
0.21563
Period, Mode 3
0.07335
0.07335
0.07335
Period, Mode 4
0.07201
0.07201
0.07201
X-Displacement
Center of mass, 2nd Story
0.0201
0.0201
0.0201
Computer File
The input data file is Example 08.EDB. This file is provided as part of the ETABS installation.
Conclusion
The results comparison shows acceptable agreement between ETABS and the references.
Two-Story, Three-Dimensional Frame - Dynamic Response Spectrum Analysis
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EXAMPLE 9
Two-Story, 3D Unsymmetrical Building Frame - Dynamic Response Spectrum
Analysis
Problem Description
This is a two-story three-dimensional unsymmetrical building frame. The structure is subjected to a seismic response spectrum along two horizontal axes that are at a 30-degree angle to the building axes. The seismic excitation is identical to the one used in Wilson and
Habibullah (1992).
Geometry, Properties and Loads
The geometry of the structure is shown in Figure 9-1. The three-dimensional structure is
modeled as a single frame with six column lines and five bays. Kip-foot-second units are
used. Typical columns are 18"x18" and beams are 12"x24". The modulus of elasticity is
taken as 432,000 ksf.
Figure 9-1 Two-Story Three-Dimensional Unsymmetrical Building Frame
Two-Story, 3D Unsymmetrical Building Frame - Dynamic Response Spectrum Analysis
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Technical Features of ETABS Tested
Three-dimensional frame analysis
Dynamic response spectrum analysis
Results Comparison
The structure is also analyzed in Wilson and Habibullah (1992). Key ETABS results are
compared in Table 9-1.
Table 9-1 Comparison of Results
ETABS
Wilson and
Habibullah
Period, Mode 1
0.4146
0.4146
Period, Mode 2
0.3753
0.3753
Period, Mode 3
0.2436
0.2436
Period, Mode 4
0.1148
0.1148
Period, Mode 5
0.1103
0.1103
Period, Mode 6
0.0729
0.0729
Seismic at 30° to X
0.1062
0.1062
Seismic at 120° to X
0.0617
0.0617
Quantity
X- Displacement
Center of Mass at 2nd Story for:
Computer File
The input data file is Example 09.EDB. This file is provided as part of the ETABS installation.
Conclusions
The results comparison shows exact agreement between ETABS and the reference material.
Two-Story, 3D Unsymmetrical Building Frame - Dynamic Response Spectrum Analysis
9 -2
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EXAMPLE 10
Three-Story Plane Frame with ADAS Elements - Nonlinear Time History Analysis
Problem Description
This is a single bay three-story plane frame subjected to ground motion, as shown in Figure
10-1. The El Centro 1940 (N-S) record is used in the nonlinear time history analysis. Three
elements that absorb energy through hysteresis (ADAS elements as described in Scholl 1993
and Tsai, et al. 1993) are used to connect the chevron braces to the frame. Two models are
investigated. In the first model, the ADAS elements are intended to produce about 5% damping in the fundamental mode. In the second model, damping is increased to 25%. The manufacturer supplied the properties of the ADAS elements.
The ADAS elements are modeled in ETABS by assigning a panel zone with a nonlinear link
property to the mid-span point object where the chevrons intersect the beams at each story.
The link properties use the uniaxial hysteretic spring property (PLASTIC1) and provide
beam-brace connectivity with nonlinear behavior in the U2 (shear in the 1-2 plane) direction.
Under this arrangement, displacements are transferred between the chevrons and the frame
via the link elements undergoing shear deformation.
Geometry, Properties and Loads
The frame is modeled as a two-column line, one-bay system. Kip-inch-second units are used.
The modulus of elasticity is taken as 29000 ksi. Column, beam and brace section properties
are user-defined.
A single rigid diaphragm is allocated to each story level and connects all three point objects
(two column points and one mid-span point) at each story. Because of the rigid diaphragms,
no axial force will occur in the beam members. All members are assigned a rigid zone factor
of 1.
In both models the value of post yield stiffness ratio is taken as 5% and the time increment
for output sampling is specified as 0.02 second.
Three-Story Plane Frame with ADAS Elements - Nonlinear Time History Analysis
10 - 1
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1
ETABS
0
2
D1
D2
D1
D2
D1
D2
Figure 10-1 Planar Frame with ADAS Elements
Technical Features of ETABS Tested
Two-dimensional frame analysis
Panel zones
Point assignments
Nonlinear time history analysis
Ritz vectors
Three-Story Plane Frame with ADAS Elements - Nonlinear Time History Analysis
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Results Comparison
Sample results are compared in Table 10-1 with results from the nonlinear analysis program
DRAIN-2DX (Prakash, et al. 1993) for both 5% and 25% damping cases.
Table 10-1 Results Comparison
5% Damping
Level
ETABS
DRAIN-2DX
25% Damping
ETABS
DRAIN-2DX
Comparison of Maximum Story Deflections
3rd
4.57
4.57
2.10
1.92
2nd
3.48
3.51
1.68
1.55
1st
1.82
1.82
0.92
0.86
Comparison of Maximum Link Shear Force
3rd
7.29
7.31
17.75
17.40
2nd
13.97
13.92
36.70
36.20
1st
17.98
18.00
47.79
47.10
Comparison of Maximum Brace Axial Force
3rd
5.16
5.17
12.55
12.30
2nd
9.88
9.84
25.95
25.60
1st
12.71
12.70
33.79
33.28
Computer Files
The input data files for this example are Example 10A.EDB (5% damping) and Example
10B.EDB (25% damping). The time history file is ELCN-THE. These files are provided as
part of the ETABS installation.
Conclusions
The results comparison show acceptable to exact agreement between ETABS and DRAIN2DX.
Three-Story Plane Frame with ADAS Elements - Nonlinear Time History Analysis
10 - 3
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Example 11
Three-Story Plane Frame with Viscous Damper Elements - Nonlinear Time History
Analysis
Problem Description
The El Centro 1940 (N-S) record is used in the nonlinear time history analysis. Three viscous
damper elements of the type described in Hanson (1993) are used to connect the chevron
braces to the frame. Two models are investigated. In the first model, the damper elements are
intended to produce about 5% damping in the fundamental mode. In the second model, damping is increased to 25%.
The ETABS viscous damper element (DAMPER) is a uniaxial damping device with a linear
or nonlinear force-velocity relationship given by F = CVα.
The damper elements are modeled in ETABS by assigning a panel zone with a nonlinear link
property to the mid-span point object where the chevrons intersect the beams at each story.
The link properties use the uniaxial damper property (DAMPER) and provide beam-brace
connectivity with nonlinear behavior in the U2 (shear in the 1-2 plane) direction. Under this
arrangement, displacements are transferred between the chevrons and the frame via the link
elements (dampers) undergoing shear deformation.
The time increment for output sampling is specified as 0.02 second.
Geometry, Properties and Loads
This is a single-bay, three-story plane frame subjected to ground motion, as shown in Figure
11-1. The frame is modeled as a two-column line, one-bay system. Kip-inch-second units are
used. The modulus of elasticity is taken as 29000 ksi. Column, beam and brace section properties are user defined.
A single rigid diaphragm is allocated to each story level and connects all three point objects
(two column points and one mid-span point) at each story. Because of the rigid diaphragms,
no axial force will occur in the beam members. All members are assigned a rigid zone factor
of 1.
Three-Story Plane Frame with Viscous Damper Elements - Nonlinear Time History Analysis
11 - 1
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Figure 11-1 Planar Frame with Damper Elements
Technical Features of ETABS Tested
Two-dimensional frame analysis
Use of panel zones
Use of uniaxial damper elements
Point assignments
Nonlinear time history analysis
Ritz vectors
Results Comparison
Sample results for α = 1 are compared in Table 11-1 with results from the nonlinear analysis program DRAIN-2DX (Prakash, et al. 1993) for both 5% and 25% damping cases.
Three-Story Plane Frame with Viscous Damper Elements - Nonlinear Time History Analysis
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Table 11-1 Results Comparison
5% Damping
Level
ETABS
25% Damping
DRAIN-2DX
ETABS
DRAIN-2DX
Comparison of Maximum Story Deflections
3rd
4.09
4.11
2.26
2.24
2nd
3.13
3.14
1.75
1.71
1st
1.63
1.63
0.89
0.87
Comparison of Maximum Link Shear Force
3rd
6.16
5.98
14.75
14.75
2nd
10.79
10.80
32.82
32.84
1st
15.15
15.02
44.90
44.97
Comparison of Maximum Brace Axial Force
3rd
4.36
4.23
10.43
10.43
2nd
7.63
7.63
23.21
23.22
1st
10.71
10.62
31.75
31.80
Computer File
The input data files for this example are Example 11A.EDB (5% damping) and Example
11B.EDB (25% damping). The time history file is ELCN-THE. These files are provided
as part of the ETABS installation.
Conclusions
The comparison of results shows acceptable agreement between ETABS and DRAIN-2DX.
Three-Story Plane Frame with Viscous Damper Elements - Nonlinear Time History Analysis
11 - 3
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EXAMPLE 12
Pounding of Two Planar Frames, Nonlinear Time History Analysis
Problem Description
A two-bay, seven-story plane frame is linked to a one-bay four-story plane frame using
ETABS GAP elements. The structure experiences pounding because of ground motion.
The El Centro 1940 (N-S) record is used in the nonlinear time history analysis.
This example illustrates the use of gap elements to model pounding between buildings.
Geometry, Properties and Loads
The geometry of the structure is shown in Figure 12-1.
The combined structure is modeled as a single frame with five column lines and three beam
bays. Kip-inch-second units are used. The modulus of elasticity is taken as 29500 ksi. Column and beam section properties are user defined.
Through the joint assignment option, Column lines 4 and 5 are connected to Diaphragm 2.
Column lines 1 to 3 remain connected to Diaphragm 1 by default. This arrangement physically divides the structure into two parts. The interaction is provided via the gap elements,
which are used as links spanning Column lines 3 and 4. The local axis 1 of the links is in
the global X-direction.
Technical Features of ETABS Tested
Two-dimensional frame analysis
Use of uniaxial gap elements
Point assignments
Nonlinear time history analysis
Use of multiple diaphragms
Results Comparison
The example frame analyzed using ETABS is also analyzed using SAP2000 (Computers
and Structures 2002) for time history loads (SAP2000 has been verified independently). A
comparison of key ETABS results with SAP2000 is presented in Table 12-1.
Pounding of Two Planar Frames, Nonlinear Time History Analysis
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Figure 12-1 Planar Frame with Gap Elements
Pounding of Two Planar Frames, Nonlinear Time History Analysis
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Table 12-1 Comparison of Results for Time History Analysis
Quantity
ETABS
SAP2000
Maximum Lateral Displacement at Roof
5.5521
5.5521
Maximum Axial Force, Column C1 at ground
266.89
266.88
A typical output produced by the program is shown in Figure 12-2. It shows the variations
of the displacement of Column lines 3 and 4 and the link force at Story 4. It is clearly evident that the link force is generated whenever the two column lines move in phase and their
separation is less than the specified initial opening or if they move towards each other out
of phase. For display purposes, the link forces are scaled down by a factor of 0.01.
Figure 12-2 Variations of Displacement of Column Lines 3 and 4
and Link Force at Story 4
Computer Files
The input data for this example is Example 12.EDB. The time history file is ELCN-THU.
Both of the files are provided as part of the ETABS installation.
Pounding of Two Planar Frames, Nonlinear Time History Analysis
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Conclusions
The results comparison shows essentially exact agreement between ETABS and SAP2000.
Pounding of Two Planar Frames, Nonlinear Time History Analysis
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EXAMPLE 13
Base-Isolated, Two-Story, 3D Frame - Nonlinear Time History Analysis
Problem Description
This is a two-story, three-dimensional frame with base isolation. The structure is subjected
to earthquake motion in two perpendicular directions using the Loma Prieta acceleration
records.
Hysteretic base isolators of the type described in Nagarajaiah et al. (1991) are modeled using the ETABS ISOLATOR1 elements, which show biaxial hysteretic characteristics.
Geometry, Properties and Loads
The structure is modeled as a single reinforced concrete frame with nine column lines and
twelve bays. The floor slab is taken to be 8 inches thick, covering all of the specified floor
bays at the base and the 1st story level. At the second story level the corner column as well
as the two edge beams are eliminated, together with the floor slab, to render this particular
level unsymmetric, as depicted in Figure 13-1.
A modulus of elasticity of 3000 ksi is used. The self-weight of concrete is taken as 150 pcf.
Kip-inch-second units are used.
The geometry of the structure is shown in Figure 13-1.
Technical Features of ETABS Tested
Three-dimensional frame analysis
Use of area (floor) objects
Use of biaxial hysteretic elements
Point assignments
Nonlinear time history analysis using ritz vectors
Base-Isolated, Two-Story, 3D Frame - Nonlinear Time History Analysis
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Figure 13-1 Base-Isolated Three-Dimensional Frame
Results Comparison
The example frame analyzed using ETABS is also analyzed using SAP2000 (Computers
and Structures 2002) for time history loads (SAP2000 has been verified independently). A
comparison of key ETABS results with SAP200 is presented in Table 13-1.
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Table 13-1 Comparison of Results for Time History Analysis
Quantity
ETABS
SAP2000
Maximum Uy Displacement, Column C9 at 2nd Floor
3.4735
3.4736
Maximum Axial Force, Column C1 at base
13.56
13.55
A typical output produced by the program is shown in Figure 13-2. It shows the loaddeformation relationship in the major direction for a typical isolator member.
Figure 13-2 Load Deformation Diagram
Computer Files
The input data file for this example is Example 13.EDB. The time history files are LP-TH0
and LP-TH90. All of these files are provided as part of the ETABS installation.
Conclusion
The results comparison shows essentially exact agreement between ETABS and SAP2000.
Base-Isolated, Two-Story, 3D Frame - Nonlinear Time History Analysis
13 - 3
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EXAMPLE 14
Friction Pendulum Base-Isolated 3D Frame - Nonlinear Time History Analysis
Problem Description
This is a two-story, three-dimensional frame with base isolation using friction pendulum
base isolators. The structure is subjected to earthquake motion in two perpendicular directions using the Loma Prieta acceleration records.
Friction pendulum type base isolators of the type described in Zayas and Low (1990) are
modeled using the ETABS ISOLATOR2 elements.
It is important for these isolator elements that the axial load from other loads be modeled
before starting the nonlinear analysis. This is achieved by using a factor of unity on the
dead load (self weight) on the structure in the nonlinear analysis initial conditions data.
Geometry, Properties and Loads
The structure is modeled as a single reinforced concrete frame with nine column lines and
twelve bays. The floor slab is taken to be 8 inches thick, covering all of the specified floor
bays at the base and the 1st story level. At the second story level, the corner column and the
two edge beams are eliminated, together with the floor slab, to render this particular level
anti-symmetric, as depicted in Figure 14-1.
The isolator properties are defined as follows:
Stiffness in direction 1
Stiffness in directions 2 and 3
Coefficient of friction at fast speed
Coefficient of friction at slow speed
Parameter determining the variation
of the coefficient of friction with velocity
Radius of contact surface in directions 2 and 3
1E3
1E2
.04
.03
20
60
A modulus of elasticity of 3000 ksi is used. The self-weight of concrete is taken as 150 pcf.
Kip-inch-second units are used.
The geometry of the structure is shown in Figure 14-1.
Friction Pendulum Base-Isolated 3D Frame - Nonlinear Time History Analysis
14 - 1
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Figure 14-1 Base-Isolated Three-Dimensional Frame
Technical Features of ETABS Tested
Three-dimensional frame analysis
Use of area (floor) objects
Use of biaxial friction pendulum elements
Point assignments
Nonlinear time history analysis using ritz vectors
Friction Pendulum Base-Isolated 3D Frame - Nonlinear Time History Analysis
14 - 2
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3
Results Comparison
The example frame analyzed using ETABS is also analyzed using SAP2000 (Computers and
Structures 2002) for time history loads (SAP2000 has been verified independently). A comparison of key ETABS results with SAP2000 is presented in Table 14-1.
Table 14-1 Comparison of Result for Time History Analysis
Quantity
ETABS
SAP2000
Maximum Uy Displacement, Column C9 at 2nd Floor
4.2039
4.2069
Maximum Axial Force, Column C1 at base
37.54
38.25
A typical output produced by the program is shown in Figure 14-2. It shows the variation of
the displacement of the second story at column line 1.
Figure 14-2 Variation of Displacement
Computer Files
The input data file for this example is Example 14.EDB. The time history files are LP-TH0
and LP-TH90. All of the files are provided as part of the ETABS installation.
Friction Pendulum Base-Isolated 3D Frame - Nonlinear Time History Analysis
14 - 3
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Conclusion
The results comparison shows acceptable agreement between ETABS and SAP2000.
Friction Pendulum Base-Isolated 3D Frame - Nonlinear Time History Analysis
14 - 4
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ETABS
2
EXAMPLE 15
Wall Object Behavior - Static Lateral Loads Analysis
Problem Description
This example analyzes a series of wall configurations to evaluate the behavior of the
ETABS shell object with wall section assignments. All walls are subjected to a static lateral
load applied at the top of the wall.
The following walls are included:
Planar shear wall, shown in Figure 15-1
Wall supported on columns, shown in Figure 15-2
Wall-spandrel system, shown in Figure 15-3
C-shaped wall section, shown in Figure 15-4
Wall with edges thickened, shown in Figure 15-5
E-shaped wall section, shown in Figure 15-6
Geometry, Properties and Loads
A modulus of elasticity of 3000 ksi and a Poisson's ratio of 0.2 are used for all walls. Kipinch-second units are used throughout. The following sections describe the models for the
different walls.
Planar Shear Wall , Example 15a
This shear wall is modeled with one panel per story. Three different wall lengths of 120",
360" and 720" are analyzed. Also, one-story and three-story walls are analyzed, together
with the six-story wall shown in Figure 15-1. A wall thickness of 12" is used.
Wall Area Object Behavior - Static Lateral Loads Analysis
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Figure 15-1 Planar Shear Wall, Example 15a
Wall Supported on Columns, Example 15b
This wall is modeled with two column lines. Columns are used for the first story, and the
top two stories have a single shell object with end piers, as shown in Figure 15-2. End
piers are 40" by 12" in cross section and panels are 12" thick. Columns are 40" by 20" in
cross section.
Wall Area Object Behavior - Static Lateral Loads Analysis
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Figure 15-2 Wall Supported on Columns, Example 15b
Wall Area Object Behavior - Static Lateral Loads Analysis
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Wall-Spandrel System, Example 15c
This wall is modeled with four column lines. The spandrels are modeled as beams. Two
different spandrel lengths of 60" and 240" are analyzed. Each wall is modeled with two
shell objects per story. Three-story walls are also analyzed together with the six-story wall
shown in Figure 15-3. A wall and spandrel thickness of 12" is used.
Figure 15-3 Wall-Spandrel System, Example 15c
Wall Area Object Behavior - Static Lateral Loads Analysis
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Shaped Wall Section, Example 15d
This wall is modeled with six column lines and five shell objects per story, to model the
shape of the wall. A three-story wall was also analyzed together with the six-story wall, as
shown in Figure 15-4. A wall thickness of 6" is used.
POINT OF LOAD
APPLICATION
TH
TH
TH
RD
ND
ST
ELEVATION
GLOBAL
REFERENCE
POINT
Y
100k
X
C3
C2
C4
C5
C6
C1
80”
100k
120”
80”
40”
80”
PLAN
Figure 15-4 C-Shaped Wall Section, Example 15d
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Wall with Edges Thickened, Example 15e
This wall is modeled with two column lines and one shell object, with end piers, per story
as shown in Figure 15-5. A three-story wall was also analyzed together with the six-story
wall shown in Figure 15-5.
TH
TH
TH
RD
ND
ST
30”
30”
C1
C2
210”
Y
8”
18”
X
Global
Reference
Point
Figure 15-5 Wall with Thickened Edges, Example 15e
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E-Shaped Wall Section, Example 15f
This wall is modeled with six column lines and five shell objects per story to model the
shape of the wall. A three-story wall was also analyzed together with the six-story wall, as
shown in Figure 15-6. A wall thickness of 6" is used.
POINT OF LOAD
APPLICATION
6TH
120”
5 TH
120”
4 TH
120”
3 RD
120”
2 ND
120”
1
ST
120”
BASELINE
ELEVATION
GLOBAL
REFERENCE
POINT
Y
100k
100k C3
C1
C2
X
120”
C4
C5
C6
120”
120”
PLAN
Figure 15-6 E-Shaped Wall Section, Example 15f
Wall Area Object Behavior - Static Lateral Loads Analysis
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Technical Features of ETABS Tested
Use
of area objects
Two-dimensional
Static
and three-dimensional shear wall systems
lateral loads analysis
Results Comparison
All walls analyzed in this example using ETABS were also analyzed using the general
structural analysis program SAP2000 (Computers and Structure 2002), using refined meshes of the membrane/shell element of that program. The SAP2000 meshes used are shown in
Figures 15-7, 15-8, 15-9, 15-10, 15-11 and 15-12. For the SAP2000 analysis, the rigid diaphragms at the floor levels were modeled by constraining all wall nodes at the floor to have
the same lateral displacement for planar walls, or by adding rigid members in the plane of
the floor for three-dimensional walls.
Figure 15-7 SAP2000 Mesh, Example 15a
Wall Area Object Behavior - Static Lateral Loads Analysis
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Figure 15-8 SAP2000 Mesh, Example 15b
Figure 15-9 SAP2000 Mesh, Example 15c
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Figure 15-10 SAP2000 Mesh, Example 15d
Figure 15-11 SAP2000 Mesh, Example 15e
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Figure 15-12 SAP2000 Mesh, Example 15f
The lateral displacements from the ETABS and SAP2000 analyses are compared in Tables
15-1, 15-2, 15-3, 15-4, 15-5 and 15-6 for the various walls.
Table 15-1 Results Comparison for Top Displacements (Inches), Example 15a
Number
Wall Height
Wall Length
of Stories
(inches)
(inches)
ETABS
SAP2000
6
720
120
2.3921
2.4287
360
0.0986
0.1031
720
0.0172
0.0186
3
360
120
0.3071
0.3205
360
0.0170
0.0187
720
0.0046
0.0052
1
120
120
0.0145
0.0185
360
0.0025
0.0029
720
0.0011
0.0013
Table 15-2 Results Comparison for Displacements (Inches), Example 15b
Location
ETABS
SAP2000
Story 3
0.0691
0.0671
Story 2
0.0524
0.0530
Story 1
0.0390
0.0412
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Table 15-3 Results Comparison for Top Displacements (inches)
Example 15c (1-4)
Number of Stories
Beam Length (inches)
ETABS
60
0.0844
6
240
0.1456
60
0.0188
3
240
0.0313
ETABS
2
SAP2000
0.0869
0.1505
0.0200
0.0332
Table 15-4 Results Comparison for Top Displacements (Inches) at Load
Application Point, Example 15d (1-2)
Number of
Load
Displacement
Stories
Direction
Direction
ETABS
SAP2000
X
X
0.8637
0.8936
6
X
Z-Rotation
0.0185
0.0191
Y
Y
1.1447
1.1882
X
X
0.1249
0.1337
3
X
Z-Rotation
0.0024
0.0025
Y
Y
0.1623
0.1733
Table 15-5 Results Comparison for Top Displacements (Inches),
Example 15e(1-2)
Number of Stories
ETABS
6
0.2822
3
0.0464
SAP2000
0.2899
0.0480
Table 15-6 Results Comparison for Displacements at Load Application,
Example 15f (1-2)
Number of
Load
Displacement
Stories
Direction
Direction
ETABS
SAP2000
X
X
0.3707
0.3655
6
X
Z-Rotation
0.0042
0.0039
Y
Y
0.7295
0.7490
X
X
0.0602
0.0628
3
X
Z-Rotation
0.0005
0.0005
Y
Y
0.0993
0.1058
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Computer Files
The input data files for the planar shear walls are included as files Example 15A1.EDB
through Example 15A9.EDB. These and the following input data files are provided as part
of the ETABS installation.
The input data for the wall supported on columns is Example 15B.EDB.
The input data files for the wall-spandrel system are Example C1.EDB through Example
C4.EDB.
The input data files for the shaped wall section are included as files Example 15D1.EDB
and Example 15D2.EDB.
The input data for the wall with thickened edges are included as files Example 15E1.EDB
and Example 15E2.EDB.
The input data for the E-shaped wall section are included as files Example 15F1.EDB and
Example 15F2.EDB.
Conclusion
The results comparison show acceptable agreement between ETABS and SAP2000. In
general, the comparisons become better as the number of stories increases.
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AISC 360-05 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The design flexural strengths are checked for the beam shown below. The beam
is loaded with a uniform load of 0.45 klf (D) and 0.75 klf (L). The flexural
moment capacity is checked for three unsupported lengths in the weak direction,
Lb = 5 ft, 11.667 ft and 35 ft.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W18X50
E = 29000 ksi
Fy = 50 ksi
Loading
w = 0.45 klf (D)
w = 0.75 klf (L)
Geometry
Span, L = 35 ft
TECHNICAL FEATURES TESTED
Section Compactness Check (Bending)
Member Bending Capacities
Unsupported length factors
AISC 360-05 Example 001 - 1
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RESULTS COMPARISON
Independent results are comparing with the results of Example F.1-2a from the
AISC Design Examples, Volume 13 on the application of the 2005 AISC
Specification for Structural Steel Buildings (ANSI/AISC 360-05).
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
Cb ( Lb =5ft)
1.004
1.002
0.20%
378.750
378.750
0.00%
1.015
1.014
0.10%
307.124
306.657
0.15%
Cb ( Lb =35ft)
1.138
1.136
0.18%
φb M n ( Lb =35ft) (k-ft)
94.377
94.218
0.17%
Output Parameter
φb M n ( Lb =5ft) (k-ft)
Cb ( Lb =11.67ft)
φb M n ( Lb =11.67ft) (k-ft)
COMPUTER FILE: AISC 360-05 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
AISC 360-05 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi
Section: W18x50
bf = 7.5 in, tf = 0.57 in, d = 18 in, tw = 0.355 in
h = d − 2t f = 18 − 2 • 0.57 = 16.86 in
h0 = d − t f =18 − 0.57 =17.43 in
S33 = 88.9 in3, Z33 = 101 in3
Iy =40.1 in4, ry = 1.652 in, Cw = 3045.644 in6, J = 1.240 in4
rts
=
40.1 • 3045.644
= 1.98 in
88.889
I y Cw
=
S33
Rm = 1.0 for doubly-symmetric sections
Other:
c = 1.0
L = 35 ft
Loadings:
wu = (1.2wd + 1.6wl) = 1.2(0.45) + 1.6(0.75) = 1.74 k/ft
Mu =
∙
wu L2
= 1.74 352/8 = 266.4375 k-ft
8
Section Compactness:
Localized Buckling for Flange:
λ=
bf
2t f
=
7.50
= 6.579
2 • 0.57
AISC 360-05 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
λ p = 0.38
ETABS
0
E
29000
= 0.38
= 9.152
Fy
50
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
λ
=
h 16.86
=
= 47.49
tw 0.355
λ p = 3.76
29000
E
= 3.76
= 90.553
50
Fy
λ < λ p , No localized web buckling
Web is Compact.
Section is Compact.
Section Bending Capacity:
M p =Fy Z 33 =50 • 101 =5050 k − in
Lateral-Torsional Buckling Parameters:
Critical Lengths:
E
29000
Lp =
1.76 ry
=
1.76 • 1.652
=
70.022 in =
5.835 ft
Fy
50
E
=
Lr 1.95rts
0.7 Fy
Lr = 1.95 • 1.98
0.7 Fy S33 ho
Jc
1 + 1 + 6.76
S33 ho
Jc
E
2
29000 1.240 • 1.0
0.7 • 50 88.9 • 17.43
1 + 1 + 6.76
0.7 • 50 88.9 • 17.43
29000 1.240 • 1.0
2
AISC 360-05 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Lr = 16.966 ft
Non-Uniform Moment Magnification Factor:
For the lateral-torsional buckling limit state, the non-uniform moment magnification factor is
calculated using the following equation:
Cb =
2.5M max
12.5M max
Rm ≤ 3.0
+ 3M A + 4 M B + 3M C
Eqn. 1
Where MA = first quarter-span moment, MB = mid-span moment, MC = second quarter-span
moment.
The required moments for Eqn. 1 can be calculated as a percentage of the maximum mid-span
moment. Since the loading is uniform and the resulting moment is symmetric:
1 L
M A = MC = 1− b
4 L
2
Member Bending Capacity for Lb = 5 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 5
1− b =
1− =
0.995
MA =
MC =
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.995 ) + 4 (1.00 ) + 3 ( 0.995 )
Cb = 1.002
Lb < L p , Lateral-Torsional buckling capacity is as follows:
M
=
M
=
5050 k − in
n
p
ϕb M=
0.9 • 5050 /12
n
=
ϕb M n 378.75 k − ft
AISC 360-05 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Bending Capacity for Lb = 11.667 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 11.667
1− b =
1−
0.972
MA =
MC =
=
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.972 ) + 4 (1.00 ) + 3 ( 0.972 )
Cb = 1.014
L p < Lb < Lr , Lateral-Torsional buckling capacity is as follows:
Lb − L p
≤ M p
M n = C b M p − (M p − 0.7 Fy S 33 )
−
L
L
p
r
11.667 − 5.835
=
=
M n 1.014 5050 − ( 5050 − 0.7 • 50 • 88.889 )
4088.733 k − in
16.966 − 5.835
ϕb M=
0.9 • 4088.733 /12
n
=
ϕb M n 306.657 k − ft
Member Bending Capacity for Lb = 35 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 35
1− b =
1− =
0.750 .
MA =
MC =
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.750 ) + 4 (1.00 ) + 3 ( 0.750 )
(1.00 )
Cb = 1.136
Lb > Lr , Lateral-Torsional buckling capacity is as follows:
AISC 360-05 Example 001 - 6
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Fcr =
Cbπ 2 E
Lb
r
ts
2
Jc
1 + 0.078
S 33 ho
L
b
rts
2
1.136 • π 2 • 29000
1.24 • 1 420
1 + 0.078
14.133 ksi
Fcr =
=
2
88.889 • 17.4 1.983
420
1.983
2
M n = Fcr S 33 ≤ M p
M n= 14.133 • 88.9= 1256.245 k − in
ϕb M=
0.9 • 1256.245 /12
n
=
ϕb M n 94.218 k − ft
AISC 360-05 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC 360-05 Example 002
BUILT UP WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
A demand capacity ratio is calculated for the built-up, ASTM A572 grade 50,
column shown below. An axial load of 70 kips (D) and 210 kips (L) is applied to
a simply supported column with a height of 15 ft.
GEOMETRY, PROPERTIES AND LOADING
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Warping constant calculation, Cw
Member compression capacity with slenderness reduction
AISC 360-05 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
Example E.2 AISC Design Examples, Volume 13.0 on the application of the 2005
AISC Specification for Structural Steel Buildings (ANSI/AISC 360-05).
Output Parameter
Compactness
φcPn (kips)
ETABS
Independent
Percent
Difference
Slender
Slender
0.00%
506.1
506.1
0.00 %
COMPUTER FILE: AISC 360-05 EX002
CONCLUSION
The results show an exact comparison with the independent results.
AISC 360-05 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50
E = 29,000 ksi, Fy = 50 ksi
Section: Built-Up Wide Flange
d = 17.0 in, bf = 8.00 in, tf = 1.00 in, h = 15.0 in, tw = 0.250 in.
Ignoring fillet welds:
A = 2(8.00)(1.00) + (15.0)(0.250) = 19.75 in2
2(1.0)(8.0)3 (15.0)(0.25)3
Iy =
+
=85.35 in 3
12
12
Iy
85.4
ry =
=
= 2.08 in.
A
19.8
I x = ∑ Ad 2 + ∑ I x
(0.250)(15.0)3 2(8.0)(1.0)3
+
= 1095.65 in 4
12
12
t +t
1+1
d ' =−
d 1 2 =
17 −
=
16 in
2
2
Iy • d '2 (85.35)(16.0) 2
=
Cw =
= 5462.583 in 4
4
4
bt 3 2(8.0)(1.0)3 + (15.0)(0.250)3
=
J ∑
=
= 5.41 in 4
3
3
Member:
K = 1.0 for a pinned-pinned condition
L = 15 ft
I x = 2(8.0)(8.0) 2 +
Loadings:
Pu = 1.2(70.0) + 1.6(210) = 420 kips
AISC 360-05 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
Check for slender elements using Specification Section E7
Localized Buckling for Flange:
b 4.0
=
= 4.0
t 1.0
E
29000
=
= 0.38 = 9.152
λ p 0.38
Fy
50
λ=
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
h 15.0
=
= 60.0 ,
t 0.250
E
29000
=
= 1.49 = 35.9
λr 1.49
Fy
50
λ=
λ > λr , Localized web buckling
Web is Slender.
Section is Slender
Member Compression Capacity:
Elastic Flexural Buckling Stress
Since the unbraced length is the same for both axes, the y-y axis will govern by
inspection.
KL y
ry
Fe =
=
1.0(15 • 12 )
= 86.6
2.08
π 2E
KL
r
2
=
π 2 • 29000
(86.6)2
= 38.18 ksi
AISC 360-05 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Elastic Critical Torsional Buckling Stress
Note: Torsional buckling will not govern if KLy > KLz, however, the check is included
here to illustrate the calculation.
π 2 EC w
1
+
Fe =
GJ
2
(K z L )
Ix + Iy
π 2 • 29000 • 5462.4
1
= 91.8 ksi > 38.18 ksi
=
Fe
+
11200
•
5.41
2
1100 + 85.4
(180 )
Therefore, the flexural buckling limit state controls.
Fe = 38.18 ksi
Section Reduction Factors
Since the flange is not slender,
Qs = 1.0
Since the web is slender,
For equation E7-17, take f as Fcr with Q = 1.0
4.71
KLy
E
29000
=4.71
=113 >
=86.6
1.0 ( 50 )
QFy
ry
So
QFy
1.0( 50 )
f = Fcr = Q 0.658 Fe Fy = 1.0 0.658 38.2 • 50 = 28.9 ksi
0.34 E
1 −
≤ b, where b = h
b
t
f
(
)
29000
0.34
29000
be = 1.92 ( 0.250 )
1 −
≤ 15.0in
28.9 (15.0 0.250 ) 28.9
be 12.5in ≤ 15.0in
=
be = 1.92t
E
f
AISC 360-05 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
therefore compute Aeff with reduced effective web width.
Aeff =
betw + 2b f t f =
(12.5)( 0.250 ) + 2 (8.0 )(1.0 ) =19.1 in 2
where Aeff is effective area based on the reduced effective width of the web, be.
Aeff
19.1
=
= 0.968
A 19.75
Q Q=
=
=
(1.00 )( 0.968
) 0.968
s Qa
=
Qa
Critical Buckling Stress
Determine whether Specification Equation E7-2 or E7-3 applies
4.71
KLy
E
29000
= 4.71
= 115.4 >
= 86.6
QFy
ry
0.966 ( 50 )
Therefore, Specification Equation E7-2 applies.
When 4.71
E
KL
≥
QFy
r
QFy
1.0( 50 )
Fe
38.18
Fy 0.966 0.658
Fcr Q 0.658
=
=
=
• 50 28.47 ksi
Nominal Compressive Strength
Pn =Fcr Ag =28.5 • 19.75 =562.3kips
φc =0.90
φc P=
Fcr Ag= 0.90 ( 562.3=) 506.1kips > 420 kips
n
φc Pn =
506.1kips
AISC 360-05 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC 360-10 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The design flexural strengths are checked for the beam shown below. The beam
is loaded with a uniform load of 0.45 klf (D) and 0.75 klf (L). The flexural
moment capacity is checked for three unsupported lengths in the weak direction,
Lb = 5 ft, 11.667 ft and 35 ft.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W18X50
E = 29000 ksi
Fy = 50 ksi
Loading
w = 0.45 klf (D)
w = 0.75 klf (L)
Geometry
Span, L = 35 ft
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacities
Unsupported length factors
AISC 360-10 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are comparing with the results of Example F.1-2a from the
AISC Design Examples, Volume 13 on the application of the 2005 AISC
Specification for Structural Steel Buildings (ANSI/AISC 360-10).
ETABS
Independent
Percent
Difference
Compact
Compact
0.00%
1.004
1.002
0.20%
378.750
378.750
0.00%
1.015
1.014
0.10%
307.124
306.657
0.15%
Cb ( Lb =35ft)
1.138
1.136
0.18%
φb M n ( Lb =35ft) (k-ft)
94.377
94.218
0.17%
Output Parameter
Compactness
Cb ( Lb =5ft)
φb M n ( Lb =5ft) (k-ft)
Cb ( Lb =11.67ft)
φb M n ( Lb =11.67ft) (k-ft)
COMPUTER FILE: AISC 360-10 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
AISC 360-10 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi
Section: W18x50
bf = 7.5 in, tf = 0.57 in, d = 18 in, tw = 0.355 in
h = d − 2t f = 18 − 2 • 0.57 = 16.86 in
h0 = d − t f =18 − 0.57 =17.43 in
S33 = 88.9 in3, Z33 = 101 in3
Iy =40.1 in4, ry = 1.652 in, Cw = 3045.644 in6, J = 1.240 in4
=
rts
40.1 • 3045.644
= 1.98in
88.889
I y Cw
=
S33
Rm = 1.0 for doubly-symmetric sections
Other:
c = 1.0
L = 35 ft
Loadings:
wu = (1.2wd + 1.6wl) = 1.2(0.45) + 1.6(0.75) = 1.74 k/ft
Mu =
∙
wu L2
= 1.74 352/8 = 266.4375 k-ft
8
Section Compactness:
Localized Buckling for Flange:
λ=
bf
2t f
=
7.50
= 6.579
2 • 0.57
AISC 360-10 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
λ p = 0.38
ETABS
0
E
29000
= 0.38
= 9.152
Fy
50
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
λ
=
h 16.86
=
= 47.49
tw 0.355
λ p = 3.76
E
29000
= 3.76
= 90.553
Fy
50
λ < λ p , No localized web buckling
Web is Compact.
Section is Compact.
Section Bending Capacity:
M p =Fy Z 33 =50 • 101 =5050 k-in
Lateral-Torsional Buckling Parameters:
Critical Lengths:
E
29000
Lp =
1.76 ry
=
1.76 • 1.652
=
70.022 in =
5.835ft
Fy
50
E
=
Lr 1.95rts
0.7 Fy
Lr = 1.95 • 1.98
0.7 Fy S33 ho
Jc
1 + 1 + 6.76
S33 ho
Jc
E
2
29000 1.240 • 1.0
0.7 • 50 88.9 • 17.43
1 + 1 + 6.76
0.7 • 50 88.9 • 17.43
29000 1.240 • 1.0
2
Lr = 16.966 ft
AISC 360-10 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Non-Uniform Moment Magnification Factor:
For the lateral-torsional buckling limit state, the non-uniform moment magnification
factor is calculated using the following equation:
Cb =
2.5M max
12.5M max
Rm ≤ 3.0
+ 3M A + 4 M B + 3M C
Eqn. 1
where MA = first quarter-span moment, MB = mid-span moment, MC = second quarterspan moment.
The required moments for Eqn. 1 can be calculated as a percentage of the maximum
mid-span moment. Since the loading is uniform and the resulting moment is symmetric:
1 L
M A = MC = 1− b
4 L
2
Member Bending Capacity for Lb = 5 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 5
1− b =
1− =
0.995
MA =
MC =
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.995 ) + 4 (1.00 ) + 3 ( 0.995 )
Cb = 1.002
Lb < L p , Lateral-Torsional buckling capacity is as follows:
M
=
M
=
5050 k-in
n
p
φb M n =0.9 • 5050 /12
φb M n =
378.75 k-ft
Member Bending Capacity for Lb = 11.667 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 11.667
MA =
MC =
1− b =
1−
0.972
=
4 L
4 35
AISC 360-10 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
Cb =
ETABS
0
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.972 ) + 4 (1.00 ) + 3 ( 0.972 )
Cb = 1.014
L p < Lb < Lr , Lateral-Torsional buckling capacity is as follows:
Lb − L p
≤Mp
M n = C b M p − (M p − 0.7 Fy S 33 )
L − L
r
p
11.667 − 5.835
=
M n 1.014 5050 − ( 5050 − 0.7 • 50 • 88.889 )
=
4088.733 k-in
16.966 − 5.835
φb M n =
0.9 • 4088.733 /12
φb M n =
306.657 k-ft
Member Bending Capacity for Lb = 35 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 35
1− b =
1− =
0.750 .
MA =
MC =
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.750 ) + 4 (1.00 ) + 3 ( 0.750 )
(1.00 )
Cb = 1.136
Lb > Lr , Lateral-Torsional buckling capacity is as follows:
Fcr =
Cbπ 2 E
Lb
rts
2
Jc
1 + 0.078
S 33 ho
Lb
rts
2
1.136 • π2 • 29000
1.24 • 1 420
1 + 0.078
14.133ksi
Fcr =
=
2
88.889 • 17.4 1.983
420
1.983
2
AISC 360-10 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
M n = Fcr S 33 ≤ M p
M n= 14.133 • 88.9= 1256.245 k-in
φb M n =
0.9 •1256.245 /12
φb M n =
94.218 k-ft
AISC 360-10 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC 360-10 Example 002
BUILT UP WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
A demand capacity ratio is calculated for the built-up, ASTM A572 grade 50,
column shown below. An axial load of 70 kips (D) and 210 kips (L) is applied to
a simply supported column with a height of 15 ft.
GEOMETRY, PROPERTIES AND LOADING
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Warping constant calculation, Cw
Member compression capacity with slenderness reduction
AISC 360-10 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
Example E.2 AISC Design Examples, Volume 13.0 on the application of the 2005
AISC Specification for Structural Steel Buildings (ANSI/AISC 360-10).
Output Parameter
Compactness
φcPn (kips)
ETABS
Independent
Percent
Difference
Slender
Slender
0.00%
506.1
506.1
0.00 %
COMPUTER FILE: AISC 360-10 EX002
CONCLUSION
The results show an exact comparison with the independent results.
AISC 360-10 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50
E = 29,000 ksi, Fy = 50 ksi
Section: Built-Up Wide Flange
d = 17.0 in, bf = 8.00 in, tf = 1.00 in, h = 15.0 in, tw = 0.250 in.
Ignoring fillet welds:
A = 2(8.00)(1.00) + (15.0)(0.250) = 19.75 in2
2(1.0)(8.0)3 (15.0)(0.25)3
Iy =
+
=85.35 in3
12
12
Iy
85.4
ry =
=
= 2.08 in.
19.8
A
I x = ∑ Ad 2 + ∑ I x
(0.250)(15.0)3 2(8.0)(1.0)3
+
= 1095.65 in 4
12
12
t1 + t2
1+1
d ' =−
d
=
17 −
=
16 in
2
2
Iy • d '2 (85.35)(16.0) 2
=
Cw =
= 5462.583 in 4
4
4
3
bt
2(8.0)(1.0) 3 + (15.0)(0.250) 3
J =∑
=
= 5.41 in 4
3
3
Member:
K = 1.0 for a pinned-pinned condition
L = 15 ft
I x = 2(8.0)(8.0) 2 +
Loadings:
Pu = 1.2(70.0) + 1.6(210) = 420 kips
Section Compactness:
Check for slender elements using Specification Section E7
AISC 360-10 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Localized Buckling for Flange:
b 4.0
=
= 4.0
t 1.0
E
29000
=
= 0.38 = 9.152
λ p 0.38
Fy
50
λ=
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
h 15.0
=
= 60.0 ,
t 0.250
E
29000
=
= 1.49 = 35.9
λr 1.49
Fy
50
λ=
λ > λr , Localized web buckling
Web is Slender.
Section is Slender
Member Compression Capacity:
Elastic Flexural Buckling Stress
Since the unbraced length is the same for both axes, the y-y axis will govern by
inspection.
KL y
ry
Fe
=
=
1.0(15 • 12 )
= 86.6
2.08
π2 E
π2 • 29000
= 38.18 ksi
=
2
2
(86.6 )
KL
r
Elastic Critical Torsional Buckling Stress
Note: Torsional buckling will not govern if KLy > KLz, however, the check is included
here to illustrate the calculation.
AISC 360-10 Example 002 - 4
Software Verification
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π2 EC
1
w
Fe
GJ
=
+
2
( K z L )
I x + I y
π2 • 29000 • 5462.4
1
= 91.8 ksi > 38.18 ksi
Fe
=
+ 11200 • 5.41
2
1100 + 85.4
(180 )
Therefore, the flexural buckling limit state controls.
Fe = 38.18 ksi
Section Reduction Factors
Since the flange is not slender,
Qs = 1.0
Since the web is slender,
For equation E7-17, take f as Fcr with Q = 1.0
4.71
KLy
E
29000
=4.71
=113 >
=86.6
QFy
1.0 ( 50 )
ry
So
QFy
1.0( 50 )
Fe
f = Fcr = Q 0.658 Fy = 1.0 0.658 38.2 • 50 = 28.9 ksi
0.34 E
1 −
≤ b, where b = h
(b t ) f
29000
0.34
29000
be = 1.92 ( 0.250 )
1 −
≤ 15.0in
28.9 (15.0 0.250 ) 28.9
=
be 12.5in ≤ 15.0in
be = 1.92t
E
f
therefore compute Aeff with reduced effective web width.
Aeff =
betw + 2b f t f =
(12.5)( 0.250 ) + 2 (8.0 )(1.0 ) =19.1 in 2
where Aeff is effective area based on the reduced effective width of the web, be.
AISC 360-10 Example 002 - 5
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Aeff
19.1
=
= 0.968
A 19.75
=
Q Q=
=
) 0.968
(1.00 )( 0.968
s Qa
=
Qa
Critical Buckling Stress
Determine whether Specification Equation E7-2 or E7-3 applies
4.71
KLy
E
29000
= 4.71
= 115.4 >
= 86.6
QFy
ry
0.966 ( 50 )
Therefore, Specification Equation E7-2 applies.
When 4.71
E
KL
≥
QFy
r
QFy
1.0( 50 )
Fe
38.18
Fy 0.966 0.658
Fcr Q 0.658
=
=
=
• 50 28.47 ksi
Nominal Compressive Strength
Pn =Fcr Ag =28.5 • 19.75 =562.3kips
φc =0.90
φc P=
Fcr Ag= 0.90 ( 562.3=) 506.1kips > 420 kips
n
φc Pn =
506.1kips
AISC 360-10 Example 002 - 6
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AISC ASD-89 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The beam below is subjected to a bending moment of 20 kip-ft. The compression
flange is braced at 3.0 ft intervals. The selected member is non-compact due to
flange criteria.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W6X12, M10X9,
W8X10
E = 29000 ksi
Loading
w = 1.0 klf
Geometry
Span, L = 12.65 ft
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacity
AISC ASD-89 Example 001 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are taken from Allowable Stress Design Manual of Steel
Construction, Ninth Edition, 1989, Example 3, Page 2-6.
ETABS
Independent
Percent
Difference
Non-Compact
Non-Compact
0.00%
Design Bending Stress, fb
(ksi)
30.74
30.74
0.00%
Allowable Bending Stress,
Fb (ksi)
32.70
32.70
0.00 %
Output Parameter
Compactness
COMPUTER FILE: AISC ASD-89 EX001
CONCLUSION
The results show an exact match with the independent results.
AISC ASD-89 Example 001 - 2
Software Verification
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HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi
Section: W8x10
bf = 3.94 in, tf = 0.205 in, d = 7.98 in, tw = 0.17 in
h = h − 2t f = 7.89 − 2 • 0.205 = 7.48 in
Member:
L = 12.65 ft
lb = 3 ft
Loadings:
w = 1.0 k/ft
M=
∙
wL2
= 1.0 12.652/8 = 20.0 k-ft
8
Design Bending Stress
f=
M / S33= 20 • 12 / 7.8074
b
fb = 30.74 ksi
Section Compactness:
Localized Buckling for Flange:
=
λ
bf
3.94
=
= 9.610
2t f 2 • 0.205
=
λp
λr =
65
=
Fy
95
Fy
65
= 9.192
50
=
95
50
= 13.435
AISC ASD-89 Example 001 - 3
Software Verification
PROGRAM NAME:
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λ > λ p , Localized flange buckling is present.
λ < λr ,
Flange is Non-Compact.
Localized Buckling for Web:
λ
=
d 7.89
=
= 46.412
tw 0.17
f
P
= 0 and a = 0 ≤ 0.16,so
A
Fy
No axial force is present, so f a=
λ=
p
640
Fy
f 640
0
1 − 3.74 a =
1 − 3.74 • =
90.510
50
F
50
y
λ < λ p , No localized web buckling
Web is Compact.
Section is Non-Compact.
Section Bending Capacity
Allowable Bending Stress
Since section is Non-Compact
bf
Fb 33 0.79 − 0.002
=
2t f
Fb 33 =
Fy Fy
( 0.79 − 0.002 • 9.61•
)
50 50
Fb 33 = 32.70 ksi
AISC ASD-89 Example 001 - 4
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Member Bending Capacity for Lb = 3.0 ft:
Critical Length, lc:
76b f 20,000 A f
l c = min
,
dFy
Fy
76 • 3.94 20, 000 • 3.94 • 0.205
,
lc = min
7.89 • 50
50
lc = min {42.347, 40.948}
lc = 40.948 in
l22 =lb =3 • 12 =36 in
l 22 < l c , section capacity is as follows:
Fb 33 = 32.70 ksi
AISC ASD-89 Example 001 - 5
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AISC ASD-89 Example 002
WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
The column design features for the AISC ASD-89 code are checked for the frame
shown below. This frame is presented in the Allowable Stress Design Manual of
Steel Construction, Ninth Edition, 1989, Example 3, Pages 3-6 and 3-7. The
column K factors were overwritten to a value of 2.13 to match the example. The
transverse direction was assumed to be continuously supported. Two point loads
of 560 kips are applied at the tops of each column. The ratio of allow axial stress,
Fa, to the actual, fa, was checked and compared to the referenced design code.
GEOMETRY, PROPERTIES AND LOADING
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Member compression capacity
AISC ASD-89 Example 002 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are taken from Allowable Stress Design Manual of Steel
Construction, Ninth Edition, 1989, Example 3, Pages 3-6 and 3-7.
ETABS
Independent
Percent
Difference
Compact
Compact
0.00%
Design Axial Stress, fa (ksi)
15.86
15.86
0.00%
Allowable Axial Stress,
16.47
16.47
0.00%
Output Parameter
Compactness
Fa (ksi)
COMPUTER FILE: AISC ASD-89 EX002
CONCLUSION
The results show an exact comparison with the independent results.
AISC ASD-89 Example 002 - 2
Software Verification
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HAND CALCULATION
Properties:
Material: A36 Steel
E = 29,000 ksi, Fy = 36 ksi
Section: W12x120:
bf = 12.32 in, tf = 1.105 in, d =13.12 in, tw=0.71 in
A = 35.3 in2
rx=5.5056 in
Member:
K = 2.13
L = 15 ft
Loadings:
P = 560 kips
Design Axial Stress:
f=
a
P 560
=
A 35.3
f a = 15.86 ksi
Compactness:
Localized Buckling for Flange:
=
λ
=
λp
bf
12.32
=
= 5.575
2t f 2 • 1.105
65
=
Fy
65
= 10.83
36
λ < λ p , No localized flange buckling
Flange is Compact.
AISC ASD-89 Example 002 - 3
Software Verification
PROGRAM NAME:
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Localized Buckling for Web:
f a 15.86
=
= 0.44
Fy
36
λ
=
d 13.12
=
= 18.48
tw 0.71
fa
0.44 > 0.16
Since =
Fy
=
λp
257 257
= = 42.83
Fy
36
λ < λ p , No localized web buckling
Web is Compact.
Section is Compact.
Member Compression Capacity
KL x 2.13 • (15 • 12 )
=
= 69.638
5.5056
rx
Cc
=
2π2 E
=
Fy
2π2 • 29000
= 126.099
36
KL x
rx
69.638
=
= 0.552
126.099
Cc
KL x
< Cc
rx
AISC ASD-89 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
Fa =
1 KL x rx
1.0 −
2 Cc
5 3 KL x rx
+
3 8 C c
2
ETABS
0
Fy
1 KL r
− x x
8 Cc
3
1
2
1.0 − (0.552 ) • 36
2
Fa =
5 3
1
3
+ (0.552 ) − (0.552 )
3 8
8
Fa = 16.47 ksi
AISC ASD-89 Example 002 - 5
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PROGRAM NAME:
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AISC LRFD-93 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The design flexural strengths are checked for the beam shown below. The beam
is loaded with an ultimate uniform load of 1.6 klf. The flexural moment capacity
is checked for three unsupported lengths in the weak direction, Lb = 4.375 ft,
11.667 ft and 35 ft.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W18X40
E = 29000 ksi
Fy = 50 ksi
Loading
wu = 1.6 klf
Geometry
Span, L = 35 ft
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacity
Unsupported length factors
AISC LRFD-93 Example 001 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are comparing with the results of Example 5.1 in the 2nd
Edition, LRFD Manual of Steel Construction, pages 5-12 to 5-15.
Output Parameter
Compactness
ETABS
Independent
Percent
Difference
Compact
Compact
0.00%
1.003
1.002
0.10%
294.000
294.000
0.00%
1.015
1.014
0.10%
213.0319
212.703
0.15%
1.138
1.136
0.18%
50.6845
50.599
0.17%
Cb ( Lb =4.375ft)
φb M n ( Lb =4.375 ft) (k-ft)
Cb ( Lb =11.67 ft)
φb M n ( Lb = 11.67ft) (k-ft)
Cb ( Lb = 35ft)
φb M n ( Lb = 35ft) (k-ft)
COMPUTER FILE: AISC LRFD-93 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
AISC LRFD-93 Example 001 - 2
Software Verification
PROGRAM NAME:
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HAND CALCULATION
Properties:
Material: ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi
Fr = 10 ksi (for rolled shapes)
FL = Fy − Fr = 50 − 10 = 40 ksi
Section: W18x40
bf = 6.02 in, tf = 0.525 in, d = 17.9 in, tw = 0.315 in
hc = d − 2t f = 17.9 − 2 • 0.525 = 16.85 in
A = 11.8 in2
S33 = 68.3799 in3, Z33 = 78.4 in3
Iy = 19.1 in4, ry = 1.2723 in
Cw = 1441.528 in6, J = 0.81 in4
Other:
L = 35 ft
φb = 0.9
Loadings:
wu = 1.6 k/ft
Mu =
∙
wu L2
= 1.6 352/8 = 245.0 k-ft
8
Section Compactness:
Localized Buckling for Flange:
=
λ
=
λp
bf
6.02
=
= 5.733
2t f 2 • 0.525
65
=
Fy
65
= 9.192
50
AISC LRFD-93 Example 001 - 3
Software Verification
PROGRAM NAME:
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λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
=
λ
=
λp
hc 16.85
=
= 53.492
tw 0.315
640 640
= = 90.510
50
Fy
λ < λ p , No localized web buckling
Web is Compact.
Section is Compact.
Section Bending Capacity
Mp =
Fy Z 33 =
50 • 78.4 =
3920 k-in
Lateral-Torsional Buckling Parameters:
Critical Lengths:
=
X1
29000 • 11153.85 • 0.81 • 11.8
π EGJA
π
=
= 1806 ksi
2
68.3799
2
S33
2
2
Cw S33
1441.528
68.3799
4
0.0173in 4
=
=
X 2 4=
19.1 11153.85 • 0.81
I 22 GJ
=
Lp
300 r22 300 • 1.2723
=
= 53.979in
= 4.498ft
50
Fy
=
Lr r22
Lr
=
X1
1 + 1 + X 2 FL 2
FL
1.27 • 1810
2
1 + 1 + 0.0172 • 40
=
144.8in
= 12.069 ft
40
AISC LRFD-93 Example 001 - 4
Software Verification
PROGRAM NAME:
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Non-Uniform Moment Magnification Factor:
For the lateral-torsional buckling limit state, the non-uniform moment magnification
factor is calculated using the following equation:
Cb =
2.5M max
12.5M max
R m ≤ 3. 0
+ 3M A + 4 M B + 3M C
Eqn. 1
where MA = first quarter-span moment, MB = mid-span moment, MC = second quarterspan moment.
The required moments for Eqn. 1 can be calculated as a percentage of the maximum
mid-span moment. Since the loading is uniform and the resulting moment is symmetric:
1 L
M A = MC = 1− b
4 L
2
Member Bending Capacity for Lb = 4.375 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 4.375
1− b =
1−
0.996
MA =
MC =
=
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.996 ) + 4 (1.00 ) + 3 ( 0.996 )
Cb = 1.002
Lb < L p , Lateral-Torsional buckling capacity is as follows:
M n = M p = Fy Z 33 = 50 • 78.4 = 3920 < 1.5S33 Fy = 1.5 • 68.3799 • 50 = 5128.493k-in
φb M n =0.9 • 3920 /12
φb M n =
294.0 k-ft
AISC LRFD-93 Example 001 - 5
Software Verification
PROGRAM NAME:
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Member Bending Capacity for Lb = 11.667 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 11.667
1− b =
1−
0.972
MA =
MC =
=
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.972 ) + 4 (1.00 ) + 3 ( 0.972 )
Cb = 1.014
L p < Lb < Lr , Lateral-Torsional buckling capacity is as follows:
Lb − Lp
−
−
M=
C
M
M
F
S
(
)
n
b
p
p
L 33
Lr − L p
≤ M p
11.667 − 4.486
=
=
M n 1.01 3920 − ( 3920 − 40 • 68.4 )
2836.042 k-in
12.06 − 4.486
φb M n =0.9 • 2836.042 /12
φb M n =
212.7031 k-ft
Member Bending Capacity for Lb = 35 ft:
M=
M
=
1.00
max
B
2
2
1 L
1 35
1− b =
1− =
0.750 .
MA =
MC =
4 L
4 35
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.750 ) + 4 (1.00 ) + 3 ( 0.750 )
(1.00 )
Cb = 1.136
Lb > Lr , Lateral-Torsional buckling capacity is as follows:
M n = Fcr S 33 ≤ M p
AISC LRFD-93 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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2
πE
Cb π
=
M cr
EI 22GJ +
I 22CW
Lb
Lb
1.136 • π
π • 29000
M cr
=
29000 • 19.1 • 11153.85 • 0.81 +
19.1 • 1441.528
35 • 12
35 • 12
2
M
=
M
=
674.655 k-in
n
cr
φb M n =0.9 • 674.655 /12
φb M n =
50.599 k-ft
AISC LRFD-93 Example 001 - 7
Software Verification
PROGRAM NAME:
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AISC LRFD-93 Example 002
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BIAXIAL BENDING
EXAMPLE DESCRIPTION
A check of the column adequacy is checked for combined axial compression and
flexural loads. The column is 14 feet tall and loaded with an axial load,
Pu = 1400 kips and bending, M ux , M uy = 200k-ft and 70k-ft, respectively. It is
assumed that there is reverse-curvature bending with equal end moments about
both axes and no loads along the member. The column demand/capacity ratio is
checked against the results of Example 6.2 in the 3rd Edition, LRFD Manual of
Steel Construction, pages 6-6 to 6-8.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W14X176
E = 29000 ksi
Fy = 50 ksi
Loading
Pu = 1,400 kips
Mux = 200 kip-ft
Muy = 70 kip-ft
Geometry
H = 14.0 ft
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Member compression capacity
Member bending capacity
Demand/capacity ratio, D/C
AISC LRFD-93 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
Example 6.2 in the 3rd Edition, LRFD Manual of Steel Construction, pages 6-6 to
6-8.
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
φc Pn (kips)
1937.84
1937.84
0.00%
φb M nx (k-ft)
1200
1200
0.00%
φb M ny (k-ft)
600.478
600.478
0.00%
0.974
0.974
0.00%
Output Parameter
D/C
COMPUTER FILE: AISC LRFD-93 EX002
CONCLUSION
The results show an exact comparison with the independent results.
AISC LRFD-93 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
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HAND CALCULATION
Properties:
Material: ASTM A992 Grade 50 Steel
Fy = 50 ksi, E = 29,000 ksi
Section: W14x176
A = 51.8 in2,
bf = 15.7 in, tf = 1.31 in, d = 15.2 in, tw = 0.83 in
hc = d − 2t f = 15.2 − 2 • 1.31 = 12.58 in
Ix = 2,140 in4, Iy = 838 in4, rx = 6.4275 in, ry = 4.0221 in
Sx = 281.579 in3, Sy = 106.7516 in3, Zx = 320.0 in3, Zy = 163.0 in3.
Member:
Kx = Ky = 1.0
L = Lb = 14 ft
Other
φc =0.85
φb =0.9
Loadings:
Pu = 1400 kips
Mux = 200 k-ft
Muy = 70 k-ft
Section Compactness:
Localized Buckling for Flange:
/ 2)
( b=
(15.7
f / 2)
=
λ
= 5.99
tf
1.31
λ=
p
65
=
Fy
65
= 9.19
50
λ < λ p , No localized flange buckling
Flange is Compact.
AISC LRFD-93 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Localized Buckling for Web:
h 12.58
λ= c=
= 15.16
tw 0.83
φb Py =
φb Ag Fy =
0.9 • 51.8 • 50 =
2331 kips
Pu
1400
=
= 0.601
φb Py 2331
Pu
Since =
0.601 > 0.125
φb Py
=
λp
191
Fy
P
2.33 − u
φb Py
253
≥
Fy
191
253
=
( 2.33 − 0.601
) 46.714 ≥ = 35.780
50
50
λ < λ p , No localized web buckling
Web is Compact.
=
λp
Section is Compact.
Member Compression Capacity:
For braced frames, K = 1.0 and KxLx = KyLy = 14.0 ft, From AISC Table 4-2,
φc Pn =
1940 kips
Or by hand,
=
λc
K y L Fy 1.0 • 14 • 12
50
=
= 0.552
ry π E
4.022 • π 29000
Since λ c < 1.5,
(
Fcr =
Fy 0.658λc
2
50 • 0.658
)=
0.5522
=
44.012 ksi
φc Pn =
φc Fcr Ag =
0.85 • 44.012 • 51.8
φc Pn =
1937.84 kips
AISC LRFD-93 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
From LRFD Specification Section H1.2,
Pu
1400
=
= 0.722 > 0.2
φc Pn 1937.84
Therefore, LRFD Specification Equation H1-1a governs.
Section Bending Capacity
50 • 310
M
=
F=
= 1333.333 k-ft
px
yZx
12
M py = Fy Z y
Zy
163
However,=
= 1.527 > 1.5,
S y 106.7516
So
Zy =
1.5 S y =
1.5 • 106.7516 =
160.1274in 3
=
M py
50 • 160.1274
= 667.198 k-ft
12
Member Bending Capacity
From LRFD Specification Equation F1-4,
L p = 1.76ry
E
Fyf
1.76 • 4.02
L=
p
29000 1
• = 14.2 ft > L=
14 ft
b
12
50
φb M nx =
φb M px
φb M nx =0.9 • 1333.333
φb M nx =
1200 k-ft
φb M ny =
φb M py
φb M ny =
0.9 • 667.198
φb M ny =
600.478 k-ft
AISC LRFD-93 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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Interaction Capacity: Compression & Bending
From LRFD Specification section C1.2, for a braced frame, Mlt = 0.
M ux = B1x M ntx , where M ntx = 200 kip-ft; and
M uy = B1 y M nty , where M nty = 70 kip-ft
B1 =
Cm
P
1 − u
Pe1
≥1
For reverse curvature bending and equal end moments:
M1
= +1.0
M2
M
C m = 0.6 − 0.4 1
M2
C m = 0.6 − 0.4(1.0 ) = 0.2
pe1 =
=
pe1x
π2 EI
( KL )
2
π2 • 29000 • 2140
=
21, 702 kips
2
(14.0 • 12 )
π2 • 29000 • 838
=
pe1 y =
8, 498
2
(14.0 • 12 )
B1x =
C mx
≥1
Pu
1 −
Pe1x
0.2
=
B1x
= 0.214 ≥ 1
1400
1 −
21702
AISC LRFD-93 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
B1x = 1
C my
≥1
P
u
1 −
P
e1 y
0.2
B1 y
=
= 0.239 ≥ 1
1400
1 −
8498
B1 y = 1
B1 y =
M ux = 1.0 • 200 = 200 kip-ft;
and
M uy = 1.0 • 70 = 70 kip-ft
From LRFD Specification Equation H1-1a,
1400 8 200
70
+
+
0.974 < 1.0 , OK
=
1940 9 1200 600.478
D
= 0.974
C
AISC LRFD-93 Example 002 - 7
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AS 4100-1998 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous column is subjected to factored load N = 200 kN. This example
was tested using the AS4100-1998 steel frame design code. The design capacities
are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
N
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
N =
200 kN
Design Properties
fy = 250 MPa
Section: 350WC197
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Section compression capacity
Member compression capacity
AS 4100-1998 Example 001 - 1
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RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-AS-4100-1998.pdf”, which is
also available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
6275
6275
0.00%
4385
4385
0.00%
Section Axial Capacity,
Ns (kN)
Member Axial Capacity,
Nc (kN)
COMPUTER FILE: AS 4100-1998 EX001
CONCLUSION
The results show an exact comparison with the independent results.
AS 4100-1998 Example 001 - 2
Software Verification
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HAND CALCULATION
Properties:
Material:
fy = 250 MPa
Section: 350WC197
Ag = An = 25100 mm2
bf = 350 mm, tf = 28 mm, h = 331 mm, tw = 20 mm
r33 = 139.15 mm, r22 = 89.264 mm
Member:
le33 = le22 = 6000 mm (unbraced length)
Considered to be a braced frame
Loadings:
N * = 200 kN
Section Compactness:
Localized Buckling for Flange:
=
λe
(b f − tw ) f y
350 − 20 250
= = 5.89
2•tf
250
2 • 28
250
Flange is under uniform compression, so:
=
λep 9,=
λey 16,=
λew 90
λe = 5.89 < λep = 9 , No localized flange buckling
Flange is compact
Localized Buckling for Web:
=
λe
fy
h
331 250
=
= 16.55
tw 250 20 250
AS 4100-1998 Example 001 - 3
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Web is under uniform compression, so:
=
λep 30,
=
λey 45,
=
λew 180
λe= 16.55 < λep= 30 , No localized web buckling
Web is compact.
Section is Compact.
Section Compression Capacity:
Section is not Slender, so Kf = 1.0
Ns =
K f An f y =
1 • 25,100 • 250 /103
N s = 6275kN
Member Weak-Axis Compression Capacity:
Frame is considered a braced frame in both directions, so k=
k=
1
e 22
e 33
le 22
le 33
6000
6000
=
= 67.216 and
=
= 43.119
r22 89.264
r33 139.15
Buckling will occur on the 22-axis.
λ=
n 22
=
α a 22
le 22
r22
K f fy
6000
=
•
250
89.264
(1 • 250=
)
250
67.216
2100(λ n 22 − 13.5)
= 20.363
λ n 22 2 − 15.3λ n 22 + 2050
α b 22 = 0.5 since cross-section is not a UB or UC section
λ 22 = λ n 22 + α a 22 α b 22 = 67.216 + 20.363 • 0.5 = 77.398
=
η22 0.00326(λ 22 − 13.5)
= 0.2083 ≥ 0
AS 4100-1998 Example 001 - 4
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λ 22
77.398
+ 1 + η22
+ 1 + 0.2083
90
90
=
ξ 22
=
=
1.317
2
2
λ 22
77.398
2
2
90
90
2
α c 22 = ξ 22 1 −
2
90 2
1 −
ξ 22 λ22
1 −
α c=
1.317
22
2
90
0.6988
1 −
=
•
1.317
77.398
N c 22 =
α c 22 N s ≤ N s
N c 22 = 0.6988 • 6275= 4385 kN
AS 4100-1998 Example 001 - 5
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AS 4100-1998 Example 002
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The frame object bending strengths are tested in this example.
A continuous column is subjected to factored moment Mx = 1000 kN-m. This
example was tested using the AS 4100-1998 steel frame design code. The design
capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
Mx
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
Mx =
1000 kN-m
Design Properties
fy = 250 MPa
Section: 350WC197
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Section bending capacity
Member bending capacity
AS 4100-1998 Example 002 - 1
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RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-AS-4100-1998.pdf,” which is
also available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
837.5
837.5
0.00%
837.5
837.5
0.00%
Section Bending Capacity,
Ms,major (kN-m)
Member Bending Capacity,
Mb (kN-m)
COMPUTER FILE: AS 4100-1998 EX002
CONCLUSION
The results show an exact comparison with the independent results.
AS 4100-1998 Example 002 - 2
Software Verification
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HAND CALCULATION
Properties:
Material:
fy = 250 MPa
Section: 350WC197
bf = 350 mm, tf = 28 mm, h = 331 mm, tw = 20 mm
I22 = 200,000,000 mm4
Z33 = 2,936,555.891 mm2
S33 = 3,350,000 mm2
J
= 5,750,000 mm4
Iw = 4,590,000,000,000 mm6
Member:
le22 = 6000 mm (unbraced length)
Considered to be a braced frame
Loadings:
M m * = 1000 kN-m
This leads to:
M 2 * = 250 kN-m
M 3 * = 500 kN-m
M 4 * = 750 kN-m
Section Compactness:
Localized Buckling for Flange:
=
λe
(b f − tw ) f y
350 − 20 250
=
= 5.89
2•tf
250
2 • 28
250
Flange is under uniform compression, so:
λ ep = 9, λ ey = 16, λ ew = 90
AS 4100-1998 Example 002 - 3
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PROGRAM NAME:
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=
λ e 5.89 < λ=
9 , No localized flange buckling
ep
Flange is compact
Localized Buckling for Web:
=
λe
fy
h
331 250
=
= 16.55
tw 250 20 250
Web is under bending, so:
λ ep = 82, λ ey = 115, λ ew = 180
=
λ e 16.55 <=
λ ep 30 , No localized web buckling
Web is compact.
Section is Compact.
Section Bending Capacity:
Z=
Z=
min( S ,1.5Z ) for compact sections
e
c
3,350, 000 mm 2
Z=
Z=
e 33
c 33
250 • 3,350, 000 /10002
M=
M s ,major
= f y Z=
s 33
e 33
M s 33 M
=
=
837.5 kN-m
s ,major
Member Bending Capacity:
kt = 1 (Program default)
kl = 1.4 (Program default)
kr = 1 (Program default)
lLTB = le22 = 6000 mm
le = kt kl kr lLTB = 1 • 1.4 • 1 • 6000 = 8400 mm 2
AS 4100-1998 Example 002 - 4
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PROGRAM NAME:
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π2 EI 22
π2 EI w
GJ +
l 2
le 2
e
M=
M
=
oa
o
π2 • 2 • 105 • 2 • 108
π2 • 2 • 105 • 4.59 • 1012
76,923.08
•
5,
750,
000
+
M oa =
Mo =
8, 4002
8, 4002
M=
M
=
1786.938 kN-m
oa
o
2
2
M
Ms
837.5
837.5
s
=
α s 0.6
+
=
3
−
0.6
+
3
−
1786.938
M oa
M oa
1786.938
α s =0.7954
=
αm
α=
m
1.7 M m *
( M 2 *) + ( M 3 *) + ( M 4 *)
2
2
2
≤ 2.5
1.7 • 1000
= 1.817 ≤ 2.5
2
( 250 ) + ( 500 ) + ( 750 )
2
2
M b = α mα s M s = 0.7954 • 1.817 • 837.5 ≤ M s
=
M b 1210.64 kN-m ≤ 837.5 kN-m
AS 4100-1998 Example 002 - 5
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AS 4100-1998 Example 003
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object interacting axial and bending strengths are tested in this
example.
A continuous column is subjected to factored loads and moments N = 200 kN;
Mx = 1000 kN-m. This example was tested using the AS4100-1998 steel frame
design code. The design capacities are compared with independent hand
calculated results.
GEOMETRY, PROPERTIES AND LOADING
Mx
N
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
N =
Mx =
200 kN
1000 kN-m
Design Properties
fy = 250 MPa
Section: 350WC197
AS 4100-1998 Example 003 - 1
Software Verification
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TECHNICAL FEATURES TESTED
Section compactness check (bending, compression)
Section bending capacity with compression reduction
Member bending capacity with in-plane compression reduction
Member bending capacity with out-of-plane compression reduction
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-AS-4100-1998.pdf,” which is
available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
837.5
837.5
0.00%
823.1
823.1
0.00%
837.5
837.5
0.00%
Reduced Section Bending Capacity,
Mr33 (kN-m)
Reduced In-Plane Member Bending
Capacity,
Mi33 (kN-m)
Reduced Out-of-Plane Member
Bending Capacity, Mo (kN-m)
COMPUTER FILE: AS 4100-1998 EX003
CONCLUSION
The results show an exact comparison with the independent results.
AS 4100-1998 Example 003 - 2
Software Verification
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HAND CALCULATION
Properties:
Section: 350WC197
Ag = An = 25100 mm2
I22 = 200,000,000 mm4
I33 = 486,000,000 mm4
J
= 5,750,000 mm4
Iw = 4,590,000,000,000 mm6
Member:
lz=le33 = le22 = 6000 mm (unbraced length)
Considered to be a braced frame
φ =0.9
Loadings:
N * = 200 kN
=
M m * 1000 kN − m
Section Compactness:
From example SFD – IN-01-1, section is Compact in Compression
From example SFD – IN-01-2, section is Compact in Bending
Section Compression Capacity:
From example SFD – IN-01-1, N s = 6275kN
Member Compression Capacity:
From example SFD – IN-01-1, N c 22 = 4385 kN
Section Bending Capacity:
From example SFD – IN-01-2,=
M s 33 M
=
837.5 kN-m
s ,major
AS 4100-1998 Example 003 - 3
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PROGRAM NAME:
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Section Interaction: Bending & Compression Capacity:
N*
200
M r 33 =1.18M s 33 1 −
=1.18 • 837.5 1 −
≤ M s 33 =837.5
0.9 • 6275
φN s
=
M r 33 953.252 ≤ 837.5
M r 33 = 837.5kN-m
Member Strong-Axis Compression Capacity:
Strong-axis buckling strength needs to be calculated:
Frame is considered a braced frame in both directions, so ke 33 = 1
le 33
6000
=
= 43.119
r33 139.15
l
r33
e 33
λn 33 =
=
α a 33
K f fy
6000
=
•
250 139.15
(1 • 250 ) =
43.119
250
2100(λ n 33 − 13.5)
= 19.141
λ n 332 − 15.3λ n 33 + 2050
α b 33 = 0.5 since cross-section is not a UB or UC section
λ 33 = λ n 33 + α a 33α b 33 = 43.119 + 19.141 • 0.5 = 52.690
=
η33 0.00326(λ 33 − 13.5)
= 0.1278 ≥ 0
λ 33
52.690
90 + 1 + η33
+ 1 + 0.1278
90
=
ξ33
=
=
2.145
2
2
λ 33
52.690
2
2
90
90
2
2
AS 4100-1998 Example 003 - 4
Software Verification
PROGRAM NAME:
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α c 33
=
ξ33 1 −
ETABS 2013
0
90 2
1 −
ξ33λ 33
α
=
2.145 1 −
c 33
2
90
0.8474
1 −
=
2.145 • 50.690
N c 33 =
α c 33 N s ≤ N s
=
N c 33 0.8474 • 6275
N c 33 = 5318 kN
Member Interaction: In-Plane Bending and Compression Capacity:
β=
m
M min
0
=
= 0
M max 1000
Since the section is compact,
3
1 + β 3
N*
N*
1 + βm
m
M i = M s 33 1 −
+ 1.18
1−
1 −
2 φN c 33
φN c 33
2
3
1 + 0 3
200
200
1+ 0
−
+
Mi =
837.5 1 −
1
1.18
1−
2 0.9 • 5318
•
2
0.9
5318
M i = 823.11 kN-m
Member Interaction: Out-of-Plane Bending and Compression Capacity:
1
α bc =
3
1 − βm 1 − βm
N *
0.4
0.23
+
−
φN c 22
2
2
AS 4100-1998 Example 003 - 5
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PROGRAM NAME:
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0
1
α bc =
3
1− 0 1− 0
200
0.4
0.23
+
−
0.9 • 4385
2
2
α bc =
4.120
π2 EI w
π2 • 2 • 106 • 4.59 • 1012
2
lz
60002
N oz = GJ +
= 76923.08 • 5.75 • 106 +
I 33 + I 22
( 4.86 + 2 ) •108
Ag
25100
=
N oz 4.423 • 1011 kN
M b 33o = α mα s M sx w/ an assumed uniform moment such that α m =1.0
M b 33o =
1.0 • 0.7954 • 837.5 =
666.145 kN-m
N *
N*
α bc M b 33o 1 −
M o 33 =
1 −
φN c 22 φN oz
≤ M r 33
200
200
4.12 • 666.145 1 −
2674 ≤ 837.5
M o 33 =
=
1 −
11
0.9 • 4385 0.9 • 4.423 • 10
M o 33 = 837.5 kN-m
AS 4100-1998 Example 003 - 6
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BS 5950-2000 Example 001
WIDE FLANGE SECTION UNDER BENDING
EXAMPLE DESCRIPTION
The frame object moment and shear strength is tested in this example.
A simply supported beam is laterally restrained along its full length and is
subjected to a uniform factored load of 69 kN/m and a factored point load at the
mid-span of 136 kN. This example was tested using the BS 5950-2000 steel
frame design code. The moment and shear strengths are compared with
independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
L=6.5 m
Material Properties
E = 205000 MPa
v = 0.3
G = 78846.15 MPa
Loading
W = 69 kN/m
P = 136 kN
Design Properties
Ys = 275 MPa
Section: UB533x210x92
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Section bending capacity
Section shear capacity
BS 5950-2000 Example 001 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are hand calculated based on the methods in Example 2 on
page 5 of the SCI Publication P326, Steelwork Design Guide to BS5950-1:2000
Volume 2: Worked Examples by M.D. Heywood & J.B. Lim.
ETABS
Independent
Percent
Difference
Class 1
Class 1
0.00%
Design Moment,
M33 (kN-m)
585.4
585.4
0.00%
Design Shear, Fv (kN)
292.25
292.25
0.00%
Moment Capacity,
Mc (kN-m)
649.0
649
0.00%
Shear Capacity, Pv (kN)
888.4
888.4
0.00%
Output Parameter
Compactness
COMPUTER FILE: BS 5950-2000 EX001
CONCLUSION
The results show an exact comparison with the independent results.
BS 5950-2000 Example 001 - 2
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HAND CALCULATION
Properties:
Material:
E = 205000 MPa
Ys = 275 MPa
ρ y = 1.0 • Ys = 275 MPa
Section: UB533x210x92
Ag = 11,700 mm2
D = 533.1 mm, b = 104.65 mm
t = 10.1 mm, T = 15.6 mm
d = D − 2t = 533.1 − 2 • 10.1 = 501.9 mm
Z33 = 2,072,031.5 mm3
S33 = 2,360,000 mm3
Loadings:
Paxial = 0
wu = (1.4wd + 1.6wl) = 1.4(15) + 1.6(30) = 69 kN/m
Pu = (1.4Pd + 1.6Pl) = 1.4(40) + 1.6(50) = 136 kN
wu l 2 Pu l 69 • 6.52 136 • 6.5
Mu =
+
=
+
8
4
8
4
M u = 585.4 kN-m
=
Fv
wu l + Pu 69 • 6.5 + 136
=
2
2
Fv = 292.25 kN
BS 5950-2000 Example 001 - 3
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Section Compactness:
r1
=
P
= 0 (since there is no axial force)
dt ρ y
=
r2
P
= 0 (since there is no axial force)
Ag ρ y
=
ε
275
=
ρy
275
= 1
275
Localized Buckling for Flange:
λ=
b 104.65
=
= 6.71
T
15.6
λ ep = 9 ε = 9
=
λ 6.71 < =
λ p 9 , No localized flange buckling
Flange is Class 1.
Localized Buckling for Web:
λ=
d 501.9
=
= 49.69
10.1
t
Since r1 = r2 = 0 and there is no axial compression:
λ p= 80ε= 80
=
λ 49.69 <=
λ p 80 , No localized web buckling
Web is Class 1.
Section is Class 1.
BS 5950-2000 Example 001 - 4
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Section Shear Capacity:
Av 2 =Dt =533.1 • 10.1 =5,384.31mm 2
Pv 2 = 0.6 ρ y Av 2 = 0.6 • 275 • 5384.31
Pv 2 = 888.4 kN
Section Bending Capacity:
With Shear Reduction:
0.6 • P
=
533 kN > =
Fv 292.3kN
v2
So no shear reduction is needed in calculating the bending capacity.
M c = ρ y S33 ≤ 1.2 ρ y Z 33 = 275 • 2,360, 000 ≤ 1.2 • 275 • 2, 072, 031.5
=
M c 649 kN-m ≤ 683.77 kN-m
M c = 649 kN-m
BS 5950-2000 Example 001 - 5
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BS 5950-2000 Example 002
SQUARE TUBE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object axial and moment strengths are tested in this example.
A continuous column is subjected to factored loads and moments N = 640 kN;
Mx = 10.5 kN-m; My = 0.66 kN-m. The moment on the column is caused by
eccentric beam connections. This example was tested using the BS 5950-2000
steel frame design code. The design capacities are compared with independent
hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
N
Mx
My
H
A
A
Section A-A
H=5m
Material Properties
E = 205000 MPa
v = 0.3
G = 78846.15 MPa
Loading
N =
Mx =
My =
640 kN
10.5 kN-m
0.66 kN-m
Design Properties
Ys = 355 MPa
TECHNICAL FEATURES TESTED
Section compactness check (compression & bending)
Member compression capacity
Section bending capacity
BS 5950-2000 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
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0
RESULTS COMPARISON
Independent results are taken from Example 15 on page 83 of the SCI Publication
P326, Steelwork Design Guide to BS5950-1:2000 Volume 2: Worked Examples
by M.D. Heywood & J.B. Lim.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 1
Class 1
0.00%
773.2
773.2
0.00%
68.3
68.3
0.00%
Axial Capacity,
Nc (kN)
Bending Capacity,
Mc (kN-m)
COMPUTER FILE: BS 5950-2000 EX002
CONCLUSION
The results show an exact comparison with the independent results.
BS 5950-2000 Example 002 - 2
Software Verification
PROGRAM NAME:
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HAND CALCULATION
Properties:
Material:
E = 205000 MPa
G = 78846.15 MPa
Ys = 355 MPa
ρ y = 1.0 • Ys = 355 MPa
Section: RHS 150x150x6.3:
Ag = 3580 mm2
D = B = 150 mm, T=t = 6.3 mm
b = B − 3 • t = d = D − 3 • T = 150 − 2 • 6.3 = 131.1mm
r33 = 58.4483 mm
Z33 = 163,066.7 mm3
S33 = 192,301.5 mm3
Loadings:
N = 640 kN
Mx = 10.5 kN-m
My = 0.66 kN-m
Fv33 = Mx/H = 10.5 / 5 = 2.1 kN
Section Compactness:
=
r1
=
ε
P
640
=
= 0.002183
dt ρ y 131 • 6.3 • 355
275
=
ρy
275
= 0.880
355
BS 5950-2000 Example 002 - 3
Software Verification
PROGRAM NAME:
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Localized Buckling for Flange:
λ=
b 131.1
=
= 20.81
6.3
T
131.1
d
λ p= 28 ε < 80ε − = 28 • 0.880 < 80 • 0.880 −
t
6.3
λ=
24.6 < 49.6
p
=
λ 20.81 <=
λ p 24.6 , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
λ=
=
λp
d 131.1
=
= 20.81 :
t
6.3
64ε
64 • 0.88
< 40
=
ε
< 40 • 0.88
= 56.3 > 35.2
1 + 0.6r1
1 + 0.6 • 0.002183
So λ p =
35.2
=
λ 20.81 <=
λ p 35.2 , No localized web buckling
Web is compact.
Section is Compact.
Member Compression Capacity:
l
K l
1.0 • 5000
λ 22 =λ 33 =e 33 = 33 33 =
=
85.546
r33
r33
58.4483
=
λ max {λ 22 , λ=
85.546
33 }
π2 E
π2 • 205000
=
λ o 0.2= 0.2
= 15.1
ρy
355
Robertson Constant: a = 2.0 (from Table VIII-3 for Rolled Box Section in CSI
code documentation)
=
η 0.001a ( λ −=
λ 0 ) 0.001 • 2 ( 85.546 − 15.1
=
Perry Factor:
) 0.141
BS 5950-2000 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
π2 E π2 • 205000
Euler Strength: ρ=
=
= 276.5 MPa
E
λ2
85.5462
=
φ
=
ρc
ρ y + ( η + 1) ρ E 355 + ( 0.141 + 1) • 276.5
=
= 355.215 MPa
2
2
ρE ρ y
276.5 • 355
=
= 215.967 MPa
φ + φ2 − ρ E ρ y 335.215 + 335.2152 − 276.5 • 355
3580 • 215.967
=
N c A=
g ρc
N c = 773.2 kN
Section Shear Capacity:
ρ y = 1.0 • Ys = 275 MPa
D
150
3580 •
1790 mm 2
Av =
Ag
=
=
+
+
150
150
D
B
Pv = 0.6 ρ y Av 2 = 0.6 • 355 • 1790
Pv = 381.3kN
Section Bending Capacity:
With Shear Reduction
0.6=
• Pv 228.8 kN=
> Fv 2.1kN
So no shear reduction is needed in calculating the bending capacity.
Mc =
ρ y S33 ≤ 1.2ρ y Z 33 =
355 • 192,301.5 ≤ 1.2 • 355 • 163, 066.7
=
M c 68.3kN-m ≤ 69.5 kN-m
M c = 68.3kN-m
With LTB Reduction
Not considered since the section is square.
BS 5950-2000 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA S16-09 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object moment and shear strength is tested in this example.
A simply supported beam is (a) laterally restrained along its full length, (b)
laterally restrained along its quarter points, at mid-span, and at the ends (c)
laterally restrained along mid-span, and is subjected to a uniform factored load of
DL = 7 kN/m and LL = 15 kN/m. This example was tested using the CSA S1609 steel frame design code. The moment and shear strengths are compared with
Handbook of Steel construction (9th Edition) results.
GEOMETRY, PROPERTIES AND LOADING
L = 8.0 m
Material Properties
E=
Fy =
2x108 kN/m2
350 kN/m2
Design Properties
Loading
WD =
WL =
7 kN/m
15 kN/m
ASTM A992
CSA G40.21 350W
W410X46
W410X60
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacity, Mr (fully restrained)
Member bending capacity, Mr (buckling)
Member bending capacity, Mr (LTB)
CSA S16-09 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULT COMPARISON
Independent results are taken from Examples 1, 2 and 3 on pages 5-84 and 5-85
of the Hand Book of Steel Construction to CSA S16-01 published by Canadian
Institute of Steel Construction.
ETABS
Independent
Percent
Difference
Class 1
Class 1
0.00%
250.0
250.0
0.00%
(a) Moment Capacity, Mr33 of
W410X46 (kN-m) w/ lb = 0 m
278.775
278.775
0.00%
(b) Moment Capacity, Mr33 of
W410X46 (kN-m) w/ lb = 2 m
268.97
268.83
0.05%
(c) Moment Capacity, Mr33 of
W410X60 (kN-m) w/ lb = 4 m
292.10
292.05
0.02%
Output Parameter
Compactness
Design Moment, Mf (kN-m)
COMPUTER FILE: CSA S16-09 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
CSA S16-09 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: CSA G40.21 Grade 350W
fy = 350 MPa
E = 200,000 MPa
G = 76923 MPa
Section: W410x46
bf = 140 mm, tf = 11.2 mm, d = 404 mm, tw = 7 mm
h = d − 2t f = 404 − 2 • 11.2 = 381.6 mm
Ag = 5890 mm2
I22 = 5,140,000 mm4
Z33 = 885,000 mm3
J
= 192,000 mm4
=
Cw 1.976 • 1011 mm 6
Section: W410x60
bf = 178 mm, tf = 12.8 mm, d = 408 mm, tw = 7.7 mm
h = d − 2t f = 408 − 2 • 12.8 = 382.4 mm
Ag = 7580 mm2
I22 = 12,000,000 mm4
Z33= 1,190,000 mm3
J
= 328,000 mm4
=
Cw 4.698 • 1011 mm 6
Member:
L=8m
Ф = 0.9
CSA S16-09 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Loadings:
wf = (1.25wd + 1.5wl) = 1.25(7) + 1.5(15) = 31.25 kN/m
=
Mf
w f L2 31.25 • 82
=
8
8
M f = 250 kN-m
Section Compactness:
Localized Buckling for Flange:
λCl .1
=
145
=
Fy
145
= 7.75
350
W410x46
=
λ
bf
140
=
= 6.25
2t f 2 • 11.2
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
W410x60
=
λ
bf
178
=
= 6.95
2t f 2 • 12.8
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
Localized Buckling for Web:
Cf
1100
1 − 0.39
Cy
Fy
λCl .1 =
1100
0
=
1 − 0.39
=58.8
5890
350
•
350
W410x46
=
λ
h 381.6
=
= 54.51
tw
7
CSA S16-09 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
W410x60
λ
=
h 382.4
=
= 49.66
tw
7.7
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
Calculation of ω2:
ω2 is calculated from the moment profile so is independent of cross section and is
calculated as:
ω2 =
4 • M max
M max 2 + 4 M a 2 + 7 M b 2 + 4 M c 2
where: Mmax = maximum moment
Ma = moment at ¼ unrestrained span
Mb = moment at ½ unrestrained span
Mc = moment at ¾ unrestrained span
Section Bending Capacity for W410x46:
M p = Fy Z 33 = 350 • 885,000 / 10 6 = 309.75 kN-m
φM p = 0.9 • 309.75 = 278.775 kN-m
Member Bending Capacity for Lb = 0 mm (Fully Restrained):
Lb = 0, so Mmax = Ma = Mb = Mc = Mu = 250 kN-m and ω2 = 1.000
CSA S16-09 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ω2 π
πE
Mu
EI 22GJ +
=
I 22Cw → ∞ as L → 0
L
L
2
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
0.28
M p 33
Mu
→ 0 as M u → ∞
leading to M =
1.15 φ M p 33 > φ M p 33
r 33
So
278.775 kN-m
φ M p 33 =
M r 33 =
Member Bending Capacity for Lb = 2000 mm:
M a @ x=
a
Ma =
L − Lb Lb 8 − 2 2
+ =
+ = 3.5 m
2
4
2
4
ω f Lxa
2
−
ω f xa 2
2
31.25 • 8 • 3.5 31.25 • 3.52
=
−
= 246.094 kN-m
2
2
Ma = Mc = 246.094 kN-m @ 3500 mm and 4500 mm
Mmax = Mb = 250 kN-m @ 4000 mm
ω2
4 • 250
= 1.008
2
250 + 4 • 246.0942 + 7 • 2502 + 4 • 246.0942
ω2 = 1.008
ω2 π
πE
EI 22GJ +
I 22Cw
L
L
2
=
Mu
CSA S16-09 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
( 2 •10 ) • ( 5.14 •10 ) • 76923 • (192 •10 )
π ( 2 • 10 )
( 5.14 • 10 )(197.6 • 10 )
+
2000
5
Mu =
1.008 • π
2000
ETABS
0
6
3
2
5
6
9
=
M u 537.82 • 106 N-mm = 537.82 kN-m
0.67 M p = 0.67 • 309.75 = 208 < M u = 537.82 kN-m, so
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
309.75
M r 33 = 1.15 • 0.9 • 309.75 1 − 0.28
= 268.89 kN-m < 278.775 kN-m
537.82
M r 33 = 268.89 kN-m
Section Capacity for W410x60:
M p = Fy Z 33 = 350 • 1190,000 / 10 6 = 416.5 kN-m
φM p = 0.9 • 416.5 = 374.85 kN-m
Member Bending Capacity for Lb = 4000 m:
M a @ x=
a
Ma =
L − Lb Lb 8 − 4 4
+ =
+ = 3m
2
4
2
4
ω f Lxa
2
−
ω f xa 2
2
31.25 • 8 • 3 31.25 • 32
=
−
= 234.375 kN-m
2
2
Ma = Mc = 234.375 kN-m @ 3500 mm and 4500 mm
Mmax = Mb = 250 kN-m @ 4000 mm
ω2
4 • 250
= 1.032
2
250 + 4 • 234.3752 + 7 • 2502 + 4 • 234.3752
ω2 = 1.032
CSA S16-09 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ω2 π
πE
Mu
EI 22GJ +
=
I 22Cw
Ly
L
2
( 2 •10 ) • (12 •10 ) • 76923 • ( 328 •10 )
π ( 2 • 10 )
(12 • 10 )( 469.8 • 10 )
+
4000
5
Mu =
1.032 • π
4000
6
3
2
5
6
9
=
M u 362.06 • 106 N-mm = 362.06 kN-m
0.67 M p = 0.67 • 309.75 = 279 < M u = 362.06 kN-m, so
M p 33
M r=
M
φ
−
1.15
1
0.28
≤ φM p 33
p 33
33
Mu
416.5
M r 33 = 1.15 • 0.9 • 416.5 1 − 0.28
362.06
M r 33 = 292.23kN-m
CSA S16-09 Example 001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA S16-09 Example 002
WIDE FLANGE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object axial and moment strengths are tested in this example.
A continuous column is subjected to factored loads and moments Cf = 2000 kN;
Mfx-top= 200 kN-m; Mfx-bottom= 300 kN-m. This example was tested using the CSA
S16-09 steel frame design code. The design capacities are compared with
Handbook of Steel Construction (9th Edition) results.
GEOMETRY, PROPERTIES AND LOADING
2000 kN
Mxf = 200 kN-m
3.7 m
A
A
W310x118
Mxf = 300 kN-m
Material Properties
E=
ν=
G=
200,000 MPa
0.3
76,923.08 MPa
Section A-A
Loading
Design Properties
Cf
= 2000 kN
Mfx-top
= 200 kN-m
Mfx-bottom = 300 kN-m
fy = 345 MPa
TECHNICAL FEATURES TESTED
Section compactness check (compression & bending)
Member compression capacity
Member bending capacity with no mid-span loading
CSA S16-09 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from Example 1 on page 4-114 of the Hand Book
of Steel Construction to CSA S16-01 published by the Canadian Institute of Steel
Construction.
ETABS
Independent
Percent
Difference
Compactness
Class 2
Class 2
0.00%
Axial Capacity, Cr (kN)
3849.5
3849.5
0.00%
Bending Capacity, Mr33
(kN-m)
605.5
605.5
0.00%
Output Parameter
COMPUTER FILE: CSA S16-09 EX002
CONCLUSION
The results show an exact comparison with the independent results.
CSA S16-09 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material:
fy = 345 MPa
E = 200,000 MPa
G = 76923.08 MPa
Section: W310x118
Ag = 15000 mm2
r33 = 135.4006 mm, r22 = 77.5457 mm
I22 = 90,200,000 mm4
Z33 = 1,950,000 mm3
J = 1,600,000 mm4
=
Cw 1.966 • 1012 mm 6
ro 2 = xo 2 + yo 2 + r22 2 + r332 = 02 + 02 + 77.5457 2 + 135.40062
ro 2 = 24346.658 mm 2
Member:
lz= le33 = le22 = 3700 mm (unbraced length)
kz=k33 = k22 =1.0
φ =0.9
Loadings:
C f = 2000 kN
=
Ma M
=
200 kN-m
xf ,top
=
Mb M
=
300 kN-m
xf ,bottom
CSA S16-09 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
Localized Buckling for Flange:
λ Cl .1=
145
=
Fy
145
= 7.81
345
λ Cl .2=
170
=
Fy
170
= 9.15
345
=
λ
bf
307
=
= 8.21
2t f 2 • 18.7
λ Cl .1 < λ < λ Cl .2 ,
Flange is Class 2.
Localized Buckling for Web:
=
C y f=
y Ag
λ=
Cl .1
λ=
345 • 15000
= 5175 kN
1000
C f 1100
1100
2000
=
1 − 0.39 =
1 − 0.39
50.30
5175
Cy
345
Fy
h 276.6
=
= 23.24
tw 11.9
λ < λ Cl .1 ,
Web is Class 1.
Section is Class 2
Member Compression Capacity:
Flexural Buckling
n = 1.34 (wide flange section)
CSA S16-09 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
λ = max(λ 22 , λ 33 ) = λ 22 =
(
Cr = φAg Fy 1 + λ
2n
)
−
1
n
k22l22
r22 π
fy
E
=
ETABS
0
1.0 • 3700
345
= 0.6308
77.5457 200000
(
= 0.9 • 15000 • 345 • 1 + 0.6308
2•1.34
)
−
1
1.34
Cr = 3489.5 kN
Torsional & Lateral-Torsional Buckling
=
Fex
π2 E
π2 2 • 105
=
=
2643MPa
2
2
k33l33
1 • 3700
135.4006
r33
π2 E
π2 2 • 105
=
Fey =
=
867 MPa
2
2
k22l22
1 • 3700
77.5457
r22
π2 EC
1
w
=
Fez
+ GJ
2
(k l )
Ag ro 2
zz
π2 • 2 • 105 • 1.966 • 1012
1
=
+ 76923.08 • 1.6 • 106
Fez
2
15000 • 24347
(1 • 3700 )
FeZ = 1113.222 MPa
F
=
min ( Fex , Fey , Fez=
867 MPa
) F=
e
ey
(
Cr = φAg Fe 1 + λ 2 n
)
−
1
n
(
= 0.9 • 15000 • 867 • 1 + 0.63082•1.34
)
−
1
1.34
Cr = 9674.5 kN (does not govern)
Section Bending Capacity:
M p 33= Z 33 Fy = 1,950, 000 • 345= 672.75 kN-m
CSA S16-09 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Bending Capacity:
2
M
M
ω=
1.75 + 1.05 a + 0.3 a ≤ 2.5
2
Mb
Mb
2
200
200
1.75 + 1.05
ω=
2
+ 0.3
= 2.583 ≤ 2.5
300
300
So ω2 =2.5
2
πE
ω2 π
=
Mu
EI 22GJ +
I 22Cw
l22
l22
=
Mu
2.5 • π
3700
2
π • 2 • 105
7
12
2 • 105 • 9.02 • 107 • 76923.08 • 1.6 • 106 +
9.02 • 10 • 1.966 • 10
3700
M u = 3163.117 kN-m
Since M u > 0.67 • M p 33
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
672.75
M r 33 = 1.15 • 0.9 • 672.75 1 − 0.28
≤ 0.9 • 672.75
3163.117
=
M r 33 654.830 ≤ 605.475
M r 33 = 605.5 kN-m
CSA S16-09 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA S16-14 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object moment and shear strength is tested in this example.
A simply supported beam is (a) laterally restrained along its full length, (b)
laterally restrained along its quarter points, at mid-span, and at the ends (c)
laterally restrained along mid-span, and is subjected to a uniform factored load of
DL = 7 kN/m and LL = 15 kN/m. This example was tested using the CSA S1614 steel frame design code. The moment and shear strengths are compared with
Handbook of Steel construction (9th Edition) results.
GEOMETRY, PROPERTIES AND LOADING
L = 8.0 m
Material Properties
E=
Fy =
2x108 kN/m2
350 kN/m2
Design Properties
Loading
WD =
WL =
7 kN/m
15 kN/m
ASTM A992
CSA G40.21 350W
W410X46
W410X60
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Member bending capacity, Mr (fully restrained)
Member bending capacity, Mr (buckling)
Member bending capacity, Mr (LTB)
CSA S16-14 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULT COMPARISON
Independent results are taken from Examples 1, 2 and 3 on pages 5-84 and 5-85
of the Hand Book of Steel Construction to CSA S16-01 published by Canadian
Institute of Steel Construction.
ETABS
Independent
Percent
Difference
Class 1
Class 1
0.00%
250.0
250.0
0.00%
(a) Moment Capacity, Mr33 of
W410X46 (kN-m) w/ lb = 0 m
278.775
278.775
0.00%
(b) Moment Capacity, Mr33 of
W410X46 (kN-m) w/ lb = 2 m
268.97
268.83
0.05%
(c) Moment Capacity, Mr33 of
W410X60 (kN-m) w/ lb = 4 m
292.10
292.05
0.02%
Output Parameter
Compactness
Design Moment, Mf (kN-m)
COMPUTER FILE: CSA S16-14 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
CSA S16-14 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: CSA G40.21 Grade 350W
fy = 350 MPa
E = 200,000 MPa
G = 76923 MPa
Section: W410x46
bf = 140 mm, tf = 11.2 mm, d = 404 mm, tw = 7 mm
h = d − 2t f = 404 − 2 • 11.2 = 381.6 mm
Ag = 5890 mm2
I22 = 5,140,000 mm4
Z33 = 885,000 mm3
J
= 192,000 mm4
=
Cw 1.976 • 1011 mm 6
Section: W410x60
bf = 178 mm, tf = 12.8 mm, d = 408 mm, tw = 7.7 mm
h = d − 2t f = 408 − 2 • 12.8 = 382.4 mm
Ag = 7580 mm2
I22 = 12,000,000 mm4
Z33= 1,190,000 mm3
J
= 328,000 mm4
=
Cw 4.698 • 1011 mm 6
Member:
L=8m
Ф = 0.9
CSA S16-14 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Loadings:
wf = (1.25wd + 1.5wl) = 1.25(7) + 1.5(15) = 31.25 kN/m
=
Mf
w f L2 31.25 • 82
=
8
8
M f = 250 kN-m
Section Compactness:
Localized Buckling for Flange:
λCl .1
=
145
=
Fy
145
= 7.75
350
W410x46
=
λ
bf
140
=
= 6.25
2t f 2 • 11.2
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
W410x60
=
λ
bf
178
=
= 6.95
2t f 2 • 12.8
λ < λCl .1 , No localized flange buckling
Flange is Class 1.
Localized Buckling for Web:
Cf
1100
1 − 0.39
Cy
Fy
λCl .1 =
1100
0
=
1 − 0.39
=58.8
5890
350
•
350
W410x46
=
λ
h 381.6
=
= 54.51
tw
7
CSA S16-14 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
W410x60
λ
=
h 382.4
=
= 49.66
tw
7.7
λ < λCl .1 , No localized web buckling
Web is Class 1.
Section is Class 1
Calculation of ω2:
ω2 is calculated from the moment profile so is independent of cross section and is
calculated as:
ω2 =
4 • M max
M max 2 + 4 M a 2 + 7 M b 2 + 4 M c 2
where: Mmax = maximum moment
Ma = moment at ¼ unrestrained span
Mb = moment at ½ unrestrained span
Mc = moment at ¾ unrestrained span
Section Bending Capacity for W410x46:
M p = Fy Z 33 = 350 • 885,000 / 10 6 = 309.75 kN-m
φM p = 0.9 • 309.75 = 278.775 kN-m
Member Bending Capacity for Lb = 0 mm (Fully Restrained):
Lb = 0, so Mmax = Ma = Mb = Mc = Mu = 250 kN-m and ω2 = 1.000
CSA S16-14 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ω2 π
πE
Mu
EI 22GJ +
=
I 22Cw → ∞ as L → 0
L
L
2
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
0.28
M p 33
Mu
→ 0 as M u → ∞
leading to M =
1.15 φ M p 33 > φ M p 33
r 33
So
278.775 kN-m
φ M p 33 =
M r 33 =
Member Bending Capacity for Lb = 2000 mm:
M a @ x=
a
Ma =
L − Lb Lb 8 − 2 2
+ =
+ = 3.5 m
2
4
2
4
ω f Lxa
2
−
ω f xa 2
2
31.25 • 8 • 3.5 31.25 • 3.52
=
−
= 246.094 kN-m
2
2
Ma = Mc = 246.094 kN-m @ 3500 mm and 4500 mm
Mmax = Mb = 250 kN-m @ 4000 mm
ω2
4 • 250
= 1.008
2
250 + 4 • 246.0942 + 7 • 2502 + 4 • 246.0942
ω2 = 1.008
ω2 π
πE
EI 22GJ +
I 22Cw
L
L
2
=
Mu
CSA S16-14 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
( 2 •10 ) • ( 5.14 •10 ) • 76923 • (192 •10 )
π ( 2 • 10 )
( 5.14 • 10 )(197.6 • 10 )
+
2000
5
Mu =
1.008 • π
2000
ETABS
0
6
3
2
5
6
9
=
M u 537.82 • 106 N-mm = 537.82 kN-m
0.67 M p = 0.67 • 309.75 = 208 < M u = 537.82 kN-m, so
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
309.75
M r 33 = 1.15 • 0.9 • 309.75 1 − 0.28
= 268.89 kN-m < 278.775 kN-m
537.82
M r 33 = 268.89 kN-m
Section Capacity for W410x60:
M p = Fy Z 33 = 350 • 1190,000 / 10 6 = 416.5 kN-m
φM p = 0.9 • 416.5 = 374.85 kN-m
Member Bending Capacity for Lb = 4000 m:
M a @ x=
a
Ma =
L − Lb Lb 8 − 4 4
+ =
+ = 3m
2
4
2
4
ω f Lxa
2
−
ω f xa 2
2
31.25 • 8 • 3 31.25 • 32
=
−
= 234.375 kN-m
2
2
Ma = Mc = 234.375 kN-m @ 3500 mm and 4500 mm
Mmax = Mb = 250 kN-m @ 4000 mm
ω2
4 • 250
= 1.032
2
250 + 4 • 234.3752 + 7 • 2502 + 4 • 234.3752
ω2 = 1.032
CSA S16-14 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ω2 π
πE
Mu
EI 22GJ +
=
I 22Cw
Ly
L
2
( 2 •10 ) • (12 •10 ) • 76923 • ( 328 •10 )
π ( 2 • 10 )
(12 • 10 )( 469.8 • 10 )
+
4000
5
Mu =
1.032 • π
4000
6
3
2
5
6
9
=
M u 362.06 • 106 N-mm = 362.06 kN-m
0.67 M p = 0.67 • 309.75 = 279 < M u = 362.06 kN-m, so
M p 33
M r=
M
φ
−
1.15
1
0.28
≤ φM p 33
p 33
33
Mu
416.5
M r 33 = 1.15 • 0.9 • 416.5 1 − 0.28
362.06
M r 33 = 292.23kN-m
CSA S16-14 Example 001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA S16-14 Example 002
WIDE FLANGE MEMBER UNDER COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object axial and moment strengths are tested in this example.
A continuous column is subjected to factored loads and moments Cf = 2000 kN;
Mfx-top= 200 kN-m; Mfx-bottom= 300 kN-m. This example was tested using the CSA
S16-14 steel frame design code. The design capacities are compared with
Handbook of Steel Construction (9th Edition) results.
GEOMETRY, PROPERTIES AND LOADING
2000 kN
Mxf = 200 kN-m
3.7 m
A
A
W310x118
Mxf = 300 kN-m
Material Properties
E=
ν=
G=
200,000 MPa
0.3
76,923.08 MPa
Section A-A
Loading
Design Properties
Cf
= 2000 kN
Mfx-top
= 200 kN-m
Mfx-bottom = 300 kN-m
fy = 345 MPa
TECHNICAL FEATURES TESTED
Section compactness check (compression & bending)
Member compression capacity
Member bending capacity with no mid-span loading
CSA S16-14 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from Example 1 on page 4-114 of the Hand Book
of Steel Construction to CSA S16-01 published by the Canadian Institute of Steel
Construction.
ETABS
Independent
Percent
Difference
Compactness
Class 2
Class 2
0.00%
Axial Capacity, Cr (kN)
3849.5
3849.5
0.00%
Bending Capacity, Mr33
(kN-m)
605.5
605.5
0.00%
Output Parameter
COMPUTER FILE: CSA S16-14 EX002
CONCLUSION
The results show an exact comparison with the independent results.
CSA S16-14 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material:
fy = 345 MPa
E = 200,000 MPa
G = 76923.08 MPa
Section: W310x118
Ag = 15000 mm2
r33 = 135.4006 mm, r22 = 77.5457 mm
I22 = 90,200,000 mm4
Z33 = 1,950,000 mm3
J = 1,600,000 mm4
=
Cw 1.966 • 1012 mm 6
ro 2 = xo 2 + yo 2 + r22 2 + r332 = 02 + 02 + 77.5457 2 + 135.40062
ro 2 = 24346.658 mm 2
Member:
lz= le33 = le22 = 3700 mm (unbraced length)
kz=k33 = k22 =1.0
φ =0.9
Loadings:
C f = 2000 kN
=
Ma M
=
200 kN-m
xf ,top
=
Mb M
=
300 kN-m
xf ,bottom
CSA S16-14 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
Localized Buckling for Flange:
λ Cl .1=
145
=
Fy
145
= 7.81
345
λ Cl .2=
170
=
Fy
170
= 9.15
345
=
λ
bf
307
=
= 8.21
2t f 2 • 18.7
λ Cl .1 < λ < λ Cl .2 ,
Flange is Class 2.
Localized Buckling for Web:
=
C y f=
y Ag
λ=
Cl .1
λ=
345 • 15000
= 5175 kN
1000
C f 1100
1100
2000
=
1 − 0.39 =
1 − 0.39
50.30
5175
Cy
345
Fy
h 276.6
=
= 23.24
tw 11.9
λ < λ Cl .1 ,
Web is Class 1.
Section is Class 2
Member Compression Capacity:
Flexural Buckling
n = 1.34 (wide flange section)
CSA S16-14 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
λ = max(λ 22 , λ 33 ) = λ 22 =
(
Cr = φAg Fy 1 + λ
2n
)
−
1
n
k22l22
r22 π
fy
E
=
ETABS
0
1.0 • 3700
345
= 0.6308
77.5457 200000
(
= 0.9 • 15000 • 345 • 1 + 0.6308
2•1.34
)
−
1
1.34
Cr = 3489.5 kN
Torsional & Lateral-Torsional Buckling
=
Fex
π2 E
π2 2 • 105
=
=
2643MPa
2
2
k33l33
1 • 3700
135.4006
r33
π2 E
π2 2 • 105
=
Fey =
=
867 MPa
2
2
k22l22
1 • 3700
77.5457
r22
π2 EC
1
w
=
Fez
+ GJ
2
(k l )
Ag ro 2
zz
π2 • 2 • 105 • 1.966 • 1012
1
=
+ 76923.08 • 1.6 • 106
Fez
2
15000 • 24347
(1 • 3700 )
FeZ = 1113.222 MPa
F
=
min ( Fex , Fey , Fez=
867 MPa
) F=
e
ey
(
Cr = φAg Fe 1 + λ 2 n
)
−
1
n
(
= 0.9 • 15000 • 867 • 1 + 0.63082•1.34
)
−
1
1.34
Cr = 9674.5 kN (does not govern)
Section Bending Capacity:
M p 33= Z 33 Fy = 1,950, 000 • 345= 672.75 kN-m
CSA S16-14 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Bending Capacity:
2
M
M
ω=
1.75 + 1.05 a + 0.3 a ≤ 2.5
2
Mb
Mb
2
200
200
1.75 + 1.05
ω=
2
+ 0.3
= 2.583 ≤ 2.5
300
300
So ω2 =2.5
2
πE
ω2 π
=
Mu
EI 22GJ +
I 22Cw
l22
l22
=
Mu
2.5 • π
3700
2
π • 2 • 105
7
12
2 • 105 • 9.02 • 107 • 76923.08 • 1.6 • 106 +
9.02 • 10 • 1.966 • 10
3700
M u = 3163.117 kN-m
Since M u > 0.67 • M p 33
M p 33
M r=
1.15φM p 33 1 − 0.28
≤ φM p 33
33
Mu
672.75
M r 33 = 1.15 • 0.9 • 672.75 1 − 0.28
≤ 0.9 • 672.75
3163.117
=
M r 33 654.830 ≤ 605.475
M r 33 = 605.5 kN-m
CSA S16-14 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EN 3-2005 Example 001
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example considering in-plane
behavior only.
A continuous column is subjected to factored load N = 210 kN and My,Ed = 43
kN-m. This example was tested using the Eurocode 3-2005 steel frame design
code. The design capacities are compared with independent hand calculated
results.
GEOMETRY, PROPERTIES AND LOADING
NEd
My,Ed
L
A
A
Section A-A
L = 3.5 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80770 MPa
Loading
N = 210 kN
My,Ed = 43 kN-m
Design Properties
fy = 235 MPa
Section: IPE 200
TECHNICAL FEATURES TESTED
Section compactness check (beam-column)
Member interaction capacities, D/C, Method 1
EN 3-2005 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-EC-3-2005.pdf,” which is
available through the program “Help” menu. This example was taken from "New
design rules in EN 1993-1-1 for member stability," Worked example 1 in section
5.2.1, page 151.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 1
Class 1
0.00%
D/CAxial
0.334
0.334
0.00%
D/CBending
0.649
0.646
0.46%
COMPUTER FILE: EN 3-2005 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
EN 3-2005 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: S235
fy = 235 MPa
E = 210,000 MPa
G = 80,770 MPa
Section: IPE 200
A = 2848 mm2
h = 200 mm, bf = 100 mm, tf = 8.5 mm, tw = 5.6 mm, r = 12 mm
hw = h − 2t f = 200 − 2 • 85 = 183mm
=
c
Iyy
b f − tw − 2r 100 − 5.6 − 2 • 12
=
= 35.2 mm
2
2
= 19,430,000 mm4
Wel,y = 194,300 mm3
Wpl,y = 220,600 mm3
Member:
Lyy = Lzz = 3,500 mm (unbraced length)
γM 0 =
1
γM1 =
1
αy = 0.21
Loadings:
N Ed = 210, 000 N
M Ed , y ,Left = 0 N-m
M Ed , y ,Right = 43000 N-m
EN 3-2005 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
235
=
fy
=
ε
−1 ≤=
α
=
α
235
= 1
235
N Ed
1
1 −
2 2htw f y
≤ 1
1
210, 000
=
1 −
0.6737
2 2 • 200 • 5.6 • 235
Localized Buckling for Flange:
For the tip in compression under combined bending and compression
λ cl .1 =
λe =
9ε
9 •1
=
= 13.36
α 0.6737
c 35.2
=
= 4.14
tf
8.5
=
λ e 4.14 < λ=
13.36
cl .1
So Flange is Class 1 in combined bending and compression
Localized Buckling for Web:
α > 0.5, so
λ=
cl .1
λe =
396ε
396 • 1
=
= 51.05 for combined bending & compression
13α − 1 13 • 0.6737 − 1
d 183
=
= 28.39
tw 5.6
=
λ e 32.68 < =
λ cl .1 51.05
So Web is Class 1 in combined bending and compression
Since Flange and Web are Class 1, Section is Class 1.
EN 3-2005 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compression Capacity:
N c , Rk = Af y = 2.848 • 10−3 • 235 • 106 = 669 kN
Member Compression Capacity:
=
N cr ,22
π2 EI 22 π2 • 210000 • 106 • 19.43 • 10−6
=
= 3287 kN
3.52
L2
Section Bending Capacity:
M pl , y , Rk= W pl , y f y= 220.6 • 10−6 • 235 • 106= 51.8 kN-m
Interaction Capacities - Method 1:
Member Bending & Compression Capacity with Buckling
Compression Buckling Factors
=
λy
Af y
=
N cr , y
2.858 • 10−3 • 235 • 106
= 0.451
3287 • 103
2
2
0.5 1 + 0.21 • ( 0.451 − 0.2 ) + 0.451
0.628
=
φ y 0.5 1 + α y ( λ y − 0.2 ) + λ=
=
y
1
1
=
χy
=
= 0.939 ≤ 1
φ + φ 2 − λ 2 0.628 + 0.6282 − 0.4512
y
y
y
(
)
(
)
Auxiliary Terms
N Ed
210
1−
N cr , y
3287
0.996
=
µy
=
=
210
N Ed
1 − 0.939
1− χy
3287
N cr , y
1−
wy
=
W pl , y 220.6 • 10−6
=
= 1.135 ≤ 1.5
Wel , y 194.3 • 10−6
EN 3-2005 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Cmo Factor
=
ψy
M Ed , y ,right
0
0
=
=
M Ed , y ,left 43 • 103
0.79 + 0.21ψ y + 0.36 ( ψ y − 0.33)
Cmy=
,0
0.79 + 0.21 • 0 + 0.36 ( 0 − 0.33)
Cmy =
,0
N Ed
N cr , y
210
= 0.782
3287
0.782 because no LTB is likely to occur.
Cmy C=
=
my ,0
Elastic-Plastic Bending Resistance
Because LTB is prevented, bLT = 0 so aLT = 0
1.6
N
1.6
C yy =
1 + ( wy − 1) 2 −
Cmy 2 λ 22 −
Cmy 2 λ y 2 Ed − bLT
N c , Rk
wy
wy
γM1
3
1.6
1.6
210 • 10
C yy =
1 + (1.135 − 1) 2 −
0
• 0.7822 • 0.451 −
• 0.7822 • 0.4512
−
3
1.135
669 • 10
1.135
1.0
C yy = 1.061 ≥
D
=
/ CAxial
Wel , y
W pl , y
=
N Ed
=
N c , Rk
χy
γ M1
194.3 • 10−6
= 0.881
220.6 • 10−6
210 • 103
669 • 103
0.939
1
D / CAxial = 0.334
EN 3-2005 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
D / CBending
Cmy M Ed , y ,right
=
µy
1 − N Ed C M pl , y , Rk
N cr , y yy γ M 1
ETABS
0
3
0.782 • 43 • 10
=
0.996
3
3
210 • 10
51.8 • 10
1.061
1 −
3
3287 • 10
1
D / CBending = 0.646
D / CTotal = 0.980
EN 3-2005 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EN 3-2005 Example 002
WIDE FLANGE SECTION UNDER BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A beam is subjected to factored load N = 1050 kN. This example was tested
using the Eurocode 3-2005 steel frame design code. The design capacities are
compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
NEd
A
L/2
L/2
A
Section A-A
L = 1.4 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80770 MPa
Loading
N
=
1050 kN
Design Properties
fy = 275 MPa
Section: 406x178x74 UB
TECHNICAL FEATURES TESTED
Section compactness (beam)
Section shear capacity
Section bending capacity with shear reduction
EN 3-2005 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-EC-3-2005.pdf,” which is
available through the program “Help” menu. Examples were taken from Example
6.5 on pp. 53-55 from the book “Designers’ Guide to EN1993-1-1” by R.S.
Narayanan & A. Beeby.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 1
Class 1
0.00%
689.2
689.2
0.00%
412.8
412.8
0.00%
386.8
386.8
0.00%
Section Shear Resistance,
Vpl,Rd (kN)
Section Bending Resistance,
Mc,y,Rd (kN-m)
Section Shear-Reduced Bending
Resistance, Mv,y,Rd (kN-m)
COMPUTER FILE: EN 3-2005 EX002
CONCLUSION
The results show an exact comparison with the independent results.
EN 3-2005 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: S275 Steel
fy = 275 MPa
E = 210000 MPa
Section: 406x178x74 UB
A = 9450 mm2
b = 179.5 mm, tf = 16 mm, h = 412.8 mm, tw = 9.5 mm, r = 10.2 mm
hw = h − 2t f = 412.8 − 2 • 16 = 380.8 mm
d = h − 2 ( t f + r ) = 412.8 − 2 • (16 + 10.2 ) = 360.4 mm
=
c
b − tw − 2r 179.5 − 9.5 − 2 • 10.2
=
= 74.8 mm
2
2
Wpl,y = 501,000 mm3
Other:
γ M 0 = 1.0
η = 1.2
Loadings:
N Ed = 0 kN
N = 1050 kN @ mid-span
Results in the following internal forces:
VEd = 525 kN
M Ed = 367.5 kN-m
EN 3-2005 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
=
ε
235
=
fy
235
= 0.924
275
Localized Buckling for Flange:
λ cl .1 = 9ε = 9 • 0.924 = 8.32 for pure compression
λe =
c 74.8
=
= 4.68
tf
16
=
λ e 4.68 < λ=
8.32
cl .1
So Flange is Class 1 in pure compression
Localized Buckling for Web:
λ cl .1= 72ε= 72 • 0.924= 66.56 for pure bending
λe =
d 360.4
=
= 37.94
tw
9.5
=
λ e 37.94 < =
λ cl .1 66.56
So Web is Class 1 in pure bending
Since Flange & Web are Class 1, Section is Class 1.
Section Shear Capacity
Av − min = η h wtw = 1.2 • 380.8 • 9.5 = 4341mm 2
Av = A − 2bt f + (tw + 2r )t f = 9450 − 2 • 179.5 • 16 + ( 9.5 + 2 • 10.2 ) • 16
=
Av 4021.2 mm 2 < Av − min
So Av = 4341mm 2
EN 3-2005 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
=
V pl , Rd
ETABS
0
Av f y 4341 275
=
= 689, 245 N
γ M 0 3 1.0 3
V pl , Rd = 689.2 kN
Section Bending Capacity
M
=
c , y , Rd
W pl , y f y 1501, 000 • 275
=
= 412, 775, 000 N-mm
γM0
1
M c , y , Rd = 412.8 kN-m
With Shear Reduction:
2
2VEd
2 • 525 2
=
ρ
− 1=
− 1= 0.27
V pl , Rd
689.2
Aw = hwtw = 380.8 • 9.5 = 3617.6 mm 2
=
M v , y , Rd
fy
ρ Aw 2 275
0.27 • 3617.62
−
=
−
1,501,
000
W
4 • 9.5
γ M 0 pl , y 4tw 1.0
M v , y , Rd = 386,829, 246 N-mm
M v , y , Rd = 386.8 kN-m
EN 3-2005 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EN 3-2005 Example 003
WIDE FLANGE SECTION UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous beam-column is subjected to factored axial load P = 1400 kN and
major-axis bending moment M = 200 kN-m. This example was tested using the
Eurocode 3-2005 steel frame design code. The design capacities are compared
with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
A
M
P
A
L
Section A-A
L = 0.4 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80769 MPa
Loading
Design Properties
fy = 235 MPa
P = 1400 kN
Section: 457x191x98 UB
M = 200 kN-m
TECHNICAL FEATURES TESTED
Section compactness check (beam-column)
Section compression capacity
Section bending capacity with compression reduction
EN 3-2005 Example 003 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-EC-3-2005.pdf”, which is also
available through the program “Help” menu. Examples were taken from Example
6.6 on pp. 57-59 from the book “Designers’ Guide to EN1993-1-1” by R.S.
Narayanan & A. Beeby.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 2
Class 2
0.00%
2937.5
2937.5
0.00%
524.1
524.5
-0.08%
341.9
342.2
Section Compression Resistance,
Npl,Rd (kN)
Section Plastic Bending Resistance,
Mpl,y,Rd (kN-m)
Section Reduced Bending Resistance,
Mn,y,Rd (kN-m)
-0.09%
COMPUTER FILE: EN 3-2005 EX003
CONCLUSION
The results show an acceptable comparison with the independent results.
EN 3-2005 Example 003 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: S275 Steel
E = 210000 MPa
fy = 235 MPa
Section: 457x191x98 UB
A = 12,500 mm2
b = 192.8 mm, tf = 19.6 mm, h = 467.2 mm, tw = 11.4 mm, r = 10.2 mm
hw = h − 2t f = 467.2 − 2 • 19.6 = 428 mm
d = h − 2 ( t f + r ) = 467.2 − 2 • (19.6 + 10.2 ) = 407.6 mm
=
c
b − tw − 2r 192.8 − 11.4 − 2 • 10.2
=
= 80.5 mm
2
2
Wpl,y = 2,232,000 mm3
Other:
γM 0 =
1.0
Loadings:
P = 1400 kN axial load
Results in the following internal forces:
N Ed = 1400 kN
M = 200 kN-m
Section Compactness:
=
ε
235
=
fy
235
= 1
235
EN 3-2005 Example 003 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
−1 ≤ α=
=
α
N Ed
1
1 −
2 2htw f y
ETABS
0
≤ 1
1
1, 400, 000
=
1 −
2.7818 > 1, so
2 2 • 467.2 • 11.4 • 235
α =1.0
Localized Buckling for Flange:
For the tip in compression under combined bending & compression
λ cl .1 =
λe =
9ε 9 • 1
=
= 9
α
1
c 80.5
=
= 4.11
t f 19.6
=
λ e 4.11 < λ=
9
cl .1
So Flange is Class 1 in combined bending and compression
Localized Buckling for Web:
α > 0.5, so
λ=
cl .1
λe =
396ε
396 • 1
=
= 33.00 for combined bending & compression
13α − 1 13 • 1 − 1
d 407.6
=
= 35.75
tw 11.4
=
λ e 35.75 > =
λ cl .1 33.00
λ=
cl .2
456ε
456 • 1
=
= 38.00
13α − 1 13 • 1 − 1
=
λ e 35.75 < λ
=
38.00
cl .2
So Web is Class 2 in combined bending & compression.
EN 3-2005 Example 003 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Since Web is Class 2, Section is Class 2 in combined bending & compression.
Section Compression Capacity
N pl=
, Rd
Af y 12,500 • 235
=
γM 0
1
N pl , Rd = 2937.5 kN
Section Bending Capacity
M=
pl , y , Rd
W pl , y f y 2, 232, 000 • 235
=
γM 0
1
M pl , y , Rd = 524.5 kN-m
Axial Reduction
1400kN > 0.25 N pl , Rd =
0.25 • 2937.5 =
734.4 kN
N Ed =
N Ed =
1400kN > 0.5
hwtw f y
γM 0
428 • 11.4 • 235
=
0.5 •
=
573.3kN
1
So moment resistance must be reduced.
=
n
N Ed
1400
=
= 0.48
N pl , Rd 2937.5
A − 2bt f 12,500 − 2 • 192.8 • 19.6
=
= 0.40
A
12,500
1− n
1 − 0.48
M N=
M pl , y , Rd
= 524.5 •
, y , Rd
1 − 0.5a
1 − 0.5 • 0.4
=
a
M N , y , Rd = 342.2 kN-m
EN 3-2005 Example 003 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
IS 800-2007 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous column is subjected to factored load N = 1 kN. This example was
tested using the Indian IS 800:2007 steel frame design code. The design
capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
NEd
L
A
A
Section A-A
L = 3m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923 MPa
Loading
N = 1 kN
Design Properties
fy = 250 MPa
fu = 410 MPa
Section: ISMB 350
TECHNICAL FEATURES TESTED
Section compactness check (column)
Member compression capacity
IS 800-2007 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-IS-800-2007.pdf,” which is
available through the program “Help” menu. The example was taken from
Example 9.2 on pp. 765-766 in “Design of Steel Structures” by N. Subramanian.
ETABS
Independent
Percent
Difference
Compactness
Plastic
Plastic
0.00%
Design Axial Strength, Ncrd
733.85
734.07
-0.03%
Output Parameter
COMPUTER FILE: IS 800-2007 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
IS 800-2007 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: Fe 250
E = 200,000 MPa
fy = 250 MPa
Section: ISMB 350
A = 6670 mm2
b = 140 mm, tf = 14.2 mm, d = 350 mm, tw = 8.1 mm, r = 1.8 mm
h =d − 2 ( t f + r ) =350 − 2 (14.2 + 1.8 ) =318 mm
ry = 28.4 mm, rz = 143 mm
Member:
KLy = KLz = 3,000 mm (unbraced length)
γM 0 =
1.1
Loadings:
N Ed = 1 kN
Section Compactness:
=
ε
250
=
fy
250
= 1
250
Localized Buckling for Flange:
λ=
8.4ε= 8.4 •=
1 8.4
p
λe =
b
70
=
= 4.93
t f 14.2
=
λ e 4.93 < λ
=
8.40
p
So Flange is Plastic in compression
IS 800-2007 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Localized Buckling for Web:
λ p= N / A & λ s= 42ε= 42 for compression
λe =
d 318
=
= 39.26
tw 8.1
=
λ e 39.26 < =
λ s 42
So Web is Plastic in compression
Since Flange & Web are Plastic, Section is Plastic.
Member Compression Capacity:
Non-Dimensional Slenderness Ratio:
h 350
= = 2.5 > 1.2
b f 140
and
=
t f 14.2 mm < 40 mm
So we should use the Buckling Curve ‘a’ for the z-z axis and Buckling Curve ‘b’
for the y-y axis (IS 7.1.1, 7.1.2.1, Table 7).
Z-Z Axis Parameters:
For buckling curve a, α =0.21 (IS 7.1.1, 7.1.2.1, Table 7)
Euler Buckling Stress:
=
f cc
=
λz
fy
=
f cc
π2 E
π2 200, 000
=
=
4485 MPa
2
2
K z Lz
3, 000
143
rz
250
= 0.2361
4485
IS 800-2007 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
=
φ 0.5 1 + α ( λ − 0.2 ) + =
λ 2 0.5 1 + 0.21( 0.2361 − 0.2 ) + 0.23612
φ =0.532
Stress Reduction
Factor: χ
=
1
1
=
= 0.9920
2
2
φ+ φ −λ
0.532 + 0.5322 − 0.23612
fy
250
f cd , z =
χ
=
0.992 •
=
255.5 MPa
γM 0
1.1
Y-Y Axis Parameters:
For buckling curve b, α =0.34 (IS 7.1.1, 7.1.2.1, Table 7)
Euler Buckling Stress:
=
f cc
λ
=
y
fy
=
f cc
π2 E
π2 200, 000
=
=
177 MPa
2
2
K z Lz
3, 000
28.4
rz
250
= 1.189
177
=
φ 0.5 1 + α ( λ − 0.2 ) + =
λ 2 0.5 1 + 0.34 (1.189 − 0.2 ) + 1.1892
φ =1.375
Stress Reduction
Factor: χ
=
1
1
=
= 0.4842
2
2
φ+ φ −λ
1.375 + 1.3752 − 1.1892
fy
250
f cd , y =
χ
=
=
0.4842 •
110.1MPa Governs
γM 0
1.1
=
Pd Af=
6670 • 110.1
cd , y
Pd = 734.07 kN
IS 800-2007 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
IS 800-2007 Example 002
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous beam is subjected to factored distributed load w = 48.74 kN/m.
This example was tested using the Indian IS 800:2007 steel frame design code.
The design capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
Section A-A
L1
L2
L3
A
w
A
L1 = 4.9 m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923 MPa
L2 = 6 m
Loading
w = 48.74 kN/m
L3 = 4.9 m
Design Properties
fy = 250 MPa
Section: ISLB 500
IS 800-2007 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES TESTED
Section compactness check (beam)
Section shear capacity
Member bending capacity
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-IS-800-2007.pdf,” which is
available through the program “Help” menu. The example was taken from
Example 10.8 on pp. 897-901 in “Design of Steel Structures” by N. Subramanian.
The torsional constant, It, is calculated by the program as a slightly different
value, which accounts for the percent different in section bending resistance.
Output Parameter
Compactness
Section Bending Resistance,
Md(LTB) (kN-m)
Section Shear Resistance,
Vd (kN)
ETABS
Independent
Percent
Difference
Plastic
Plastic
0.00%
157.70
157.93
0.14%
603.59
603.59
0.00%
COMPUTER FILE: IS 800-2007 EX002
CONCLUSION
The results show an acceptable comparison with the independent results.
IS 800-2007 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: Fe 250
E = 200,000 MPa
G = 76,923 MPa
fy = 250 MPa
Section: ISLB 500
(Note: In ETABS, the section is not available with original example properties,
including fillet properties. A similar cross-section with fillet r = 0 was used, with
similar results, shown below.)
A = 9550 mm2
h = 500 mm, bf = 180 mm, tf = 14.1 mm, tw = 9.2 mm
=
b
b f 180
=
= 90 mm
2
2
d =h − 2 ( t f + r ) =500 − 2 (14.1 + 0 ) =471.8 mm
Iz = 385,790,000 mm4, Iy = 10,639,000.2 mm4
Zez = 1,543,160 mm3, Zpz = 1,543,200 mm3
ry = 33.4 mm
Member:
Lleft = 4.9 m
Lcenter = 6 m (governs)
Lright = 4.9 m
KLy = KLz = 6,000 mm (unbraced length)
γM 0 =
1.1
IS 800-2007 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Loadings:
N Ed = 0 kN
ω = 48.75 kN/m
Section Compactness:
=
ε
250
=
fy
250
= 1
250
r1 = 0 since there is no axial force
Localized Buckling for Flange:
λ=
9.4ε= 9.4 •=
1 9.4
p
λe =
b
90
=
= 6.38
t f 14.1
9.40
=
λ e 6.38 < λ
=
p
So Flange is Plastic in pure bending
Localized Buckling for Web:
=
λp
λe =
84ε
84 • 1
=
= 84
(1 + r1 ) (1 + 0)
d 471.8
=
= 51.28
tw
9.2
=
λ e 51.28 <=
λ p 84.00
So Web is Plastic in pure bending
Since Flange & Web are Class 1, Section is Plastic.
IS 800-2007 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Shear Capacity:
Vd=
fy
γM 0 3
htw=
250
• 500 • 9.2
1.1 3
Vd = 603.59 kN
Member Bending Capacity
h f =h − t f =500 − 14.1 =485.9
3
bi ti 3 2b f t f
d t 3 2 • 180 • 14.13 485.9 • 9.23
I t =∑
=
+ iw =
+
=4.63 • 105 mm 4
3
3
3
3
3
From Roark & Young, 5th Ed., 1975, Table 21, Item 7, pg.302
t1= t2= t f and b1= b2= b f
for symmetric sections
h f 2t1t2b13b23
485.92 • 14.1 • 14.1 • 1803 • 1803
=
Iw
=
=
8.089 • 1011 mm 6
3
3
3
3
12 t1b1 + t2b2
12 • 14.2 • 180 + 14.2 • 180
(
)
(
)
C1 = 1.0 (Assumed in example and specified in ETABS)
π2 EI y
π2 EI w
GI
+
t
2
2
( KL )
( KL )
=
M cr C1
M cr 1.0
=
π2 • 200, 000 • 10, 639, 000.2
π2 • 200, 000 • 8.089 • 1011
76,923
462,508
•
+
2
2
( 6, 000 )
( 6, 000 )
M cr = 215,936,919.3 N-mm
α LT =
0.21
βb =
1.0
=
λ LT
βb Z pz f y
=
M cr
1 • 1,543, 200 • 250
= 1.337
215,936,919.3
IS 800-2007 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
2
0.5 1 + 0.21 • (1.337 − 0.2 ) + 1.337 2
=
φ LT 0.5 1 + α LT ( λ LT − 0.2 ) + λ LT
=
φ LT =
1.5127
=
χ LT
χ LT
=
=
fbd
1
φ LT + φ LT 2 − λ LT 2
≤ 1.0
1
= 0.450 ≤ 1.0
1.5127 + 1.5127 2 − 1.337 2
χ LT f y 0.450 • 250
=
= 102.3MPa
γM 0
1.1
M d , LTB
= Z pz fbd= 1543.2 • 103 • 102.3
= 157,925, 037.7 N-mm
M d , LTB = 157.93kN-m
IS 800-2007 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
IS 800-2007 Example 003
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BIAXIAL BENDING
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
In this example a beam-column is subjected factored distributed load N = 2500
kN, Mz = 350 kN-m, and My = 100 kN-m. The element is moment-resisting in
the z-direction and pinned in the y-direction. This example was tested using the
Indian IS 800:2007 steel frame design code. The design capacities are compared
with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
Y-Axis
Z-Axis
Y-Y
My,top
Mz,top
Z-Z
L
N
A
A
Mz,bot
My,bot
Section A-A
L=4m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
N = 2500 kN
Mz,top = 350 kN-m
Mz,bot = -350 kN-m
My,top = 100 kN-m
My,bot = 0
Design Properties
fy = 250 MPa
Section: W310x310x226
IS 800-2007 Example 003 - 1
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
TECHNICAL FEATURES TESTED
Section Compactness Check (Beam-Column)
Section Compression Capacity
Section Shear Capacity for Major & Minor Axes
Section Bending Capacity for Major & Minor Axes
Member Compression Capacity for Major & Minor Axes
Member Bending Capacity for Major & Minor Axes
Interaction Capacity, D/C, for Major & Minor Axes
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-IS-800-2007.pdf”, which is also
available through the program “Help” menu. The example was taken from
Example 13.2 on pp. 1101-1106 in “Design of Steel Structures” by N.
Subramanian.
Output Parameter
ETABS
Independent
Percent
Difference
Plastic
Plastic
0.00%
6520
6520
0.00%
6511
6511
0.00%
Compactness
Plastic Compression Resistance,
Nd (kN)
Buckling Resistance in Compression,
Pdz (kN)
Buckling Resistance in Compression,
Pdy (kN)
Section Bending Resistance,
Mdz (kN-m)
Section Bending Resistance,
Mdy (kN-m)
Buckling Resistance in Bending,
MdLTB (kN-m)
Section Shear Resistance,
Vdy (kN)
Section Shear Resistance,
Vdz (kN)
5295
5295
0.00%
897.46
897.46
0.00%
325.65
325.65
0.00%
886.84
886.84
0.00%
1009.2
1009.2
0.00%
2961.6
2961.6
0.00%
Interaction Capacity, D/C
1.050
1.050
0.00%
IS 800-2007 Example 003 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
COMPUTER FILE: IS 800-2007 EX003
CONCLUSION
The results show an acceptable comparison with the independent results.
IS 800-2007 Example 003 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: Fe 410
E = 200,000 MPa
G = 76,923.08 MPa
fy = 250 MPa
Section: W310x310x226
A = 28,687.7 mm2
bf = 317 mm, tf = 35.6 mm, h = 348 mm, tw = 22.1 mm, r = 0 mm
=
b
b f 317
=
= 158.5 mm ,
2
2
d =h − 2 ( t f + r ) =348 − 2 ( 35.6 + 0 ) =276.8 mm
Iz = 592,124,221 mm4, Iy = 189,255,388.9 mm4
rz = 143.668 mm, ry = 81.222 mm
Zez = 3,403,012. 8 mm3, Zey = 1,194,040.3 mm3
Zpz = 3,948,812.3 mm3, Zpy = 1,822,502.2 mm3
It = 10,658,941.4 mm6, Iw = 4.611 • 1012 mm6
Member:
Ly = Lz = 4,000 mm (unbraced length)
γ M 0 = 1.1
Loadings:
P = 2500 kN
Vz = 25 kN
Vy = 175 kN
IS 800-2007 Example 003 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
=
M z −1 350 kN − m
M z −2 =
−350 kN − m
=
M y −1 100 kN − m
M=
0 kN − m
y −2
Section Compactness:
=
ε
=
r1
fy
=
250
250
= 1
250
2,500, 000
P
=
= 2.01676
2.5
fy
246.8 • 22.1 •
dtw
1.1
γ mo
Localized Buckling for Flange:
λ p= 9.4ε= 9.4 • 1= 9.4
λ=
e
b 158.5
=
= 4.45
tf
35.6
λe = 4.45 < λ p = 9.40
So Flange is Plastic in pure bending
Localized Buckling for Web:
=
λp
λ=
e
84ε
84 • 1
=
= 27.84
(1 + r1 ) (1 + 2.01676)
d 246.8
=
= 11.20
tw
22.1
λe= 11.20 < λ p= 27.84
So Web is Plastic in bending & compression
Section is Plastic.
IS 800-2007 Example 003 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compression Capacity:
Ag f y 28687.7 • 250
=
γM0
1.1
=
Nd
N d = 6520 kN
Section Shear Capacity:
For major z-z axis
Avz =htw =348 • 22.1 =7690.8 mm 2
=
VPz
fy
γM0
250
=
• 7690.8
Avz
3
1.1 3
VPz = 1009.2 kN
For minor y-y axis
Avy =
2b f t f =•
2 317 • 35.6 =
22,570.4 mm 2
VPy
=
fy
γM0
250
Avy
=
• 22570.4
3
1.1 3
VPy = 2961.6 kN
Section Bending Capacity:
For major z-z axis
M dz =
βb Z pz f y 1 • 3,948,812.3 • 250 1.2Z ez f y 1.2 • 3, 403, 012.8 • 250
=
≤
=
1.1
1.1
γM0
γM0
=
M dz 897.46 kN − m ≤ 933.54 kN − m
=
M dz 897.46 kN − m
IS 800-2007 Example 003 - 6
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
For minor y-y axis
M dy =
βb Z py f y 1 • 1,822,502.2 • 250 1.2 Z ey f y 1.2 • 1,194, 040.3 • 250
=
≤
=
γM0
γM0
1.1
1.1
=
M dy 414.21 kN − m ≤ 325.65 kN − m
=
M dy 325.65 kN − m
With Shear Reduction:
For major z-z axis
Vz =
25 kN < 0.6VPz =
0.6 • 1009.2 =
605.5 kN No shear reduction is needed.
For minor y-y axis
Vy =
175 kN < 0.6VPy =
0.6 • 2961.6 =
1777 kN No shear reduction is needed.
With Compression Reduction:
=
n
P 2500
=
= 0.383
N d 6520
For major z-z axis
M ndz = 1.11M dz (1 − n ) = 1.11 • 897.46 (1 − 0.383) ≤ M dz
=
M ndz 614.2 kN − m < 897.46 kN − m
For minor y-y axis, since n > 0.2
M ndy = 1.56 M dy (1 − n )( n + 0.6 ) = 1.56 • 325.65 (1 − 0.383)( 0.383 + 0.6 )
=
M ndy 308.0 kN − m
IS 800-2007 Example 003 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Compression Capacity:
Non-Dimensional Slenderness Ratio:
h 348
=
= 1.1 < 1.2
b f 317
and
=
t f 35.6 mm < 40 mm
So we should use the Buckling Curve ‘b’ for the z-z axis and Buckling Curve ‘c’ for the
y-y axis (IS 7.1.1, 7.1.2.1, Table 7).
Z-Z Axis Parameters:
For buckling curve b, α = 0.34 (IS 7.1.1, 7.1.2.1, Table 7)
K z = 0.65
K z Lz =0.65 • 4000 =2600 mm,
Euler Buckling Stress:
=
f cr , z
=
λz
fy
=
f cr , z
K z Lz
2600
=
=18.097
rz
143.668
π 2E
π 2 • 200, 000
=
=
6027 MPa
2
2
K z Lz
(18.097 )
rz
250
= 0.2037
6022
φz = 0.5 1 + α z ( λz − 0.2 ) + λz 2 = 0.5 1 + 0.34 ( 0.2037 − 0.2 ) + 0.2037 2
φz = 0.5214
Stress Reduction
Factor: χ z
=
f cd , z= χ
fy
γM0
= 0.9987 •
1
1
=
= 0.9987
2
2
φ z + φ z − λz
0.5214 + 0.52142 − 0.2037 2
250
= 226.978 MPa
1.1
=
Pdz f=
226.978 • 28, 687.7
cd , z Ag
IS 800-2007 Example 003 - 8
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Pdz = 6511 kN
Y-Y Axis Parameters:
For buckling curve c, α = 0.49 (IS 7.1.1, 7.1.2.1, Table 7)
K y = 1.00
K y Ly =
1 • 4000 =
4000 mm,
Euler Buckling Stress:
=
f cr , y
=
λy
fy
=
f cr , y
K y Ly
ry
4000
=
=
49.25
81.222
π 2E
=
2
K y Ly
ry
π 2 • 200, 000
=
813.88 MPa
2
( 49.25)
250
= 0.5542
813.88
φ y = 0.5 1 + α y ( λ y − 0.2 ) + λ y 2 = 0.5 1 + 0.49 ( 0.5542 − 0.2 ) + 0.55422
φ y = 0.7404
Stress Reduction
Factor: χ y
=
f cd , y= χ
fy
γM0
= 0.8122 •
1
1
=
= 0.8122
2
2
φ y + φ y − λy
0.7404 + 0.74042 − 0.55422
250
= 184.584 MPa
1.1
=
Pdy f=
184.584 • 28, 687.7
cd , y Ag
Pdy = 5295 kN
IS 800-2007 Example 003 - 9
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Bending Capacity:
C1 = 2.7 (Program Calculation from AISC equation, where C1 < 2.7 )
π 2 EI y
π 2 EI w
+
GI
t
2
2
( KL )
( KL )
=
M cr C1
=
M cr 2.7
π 2 • 200, 000 • 189,300, 000
( 4, 000 )
76,923.08 • 10, 658,941.4 +
2
π 2 • 200, 000 • 4.611 • 1012
( 4, 000 )
2
=
M cr 15,374, 789,309 N − mm
α LT = 0.21
βb = 1.0
=
λLT
βb Z pz f y
=
M cr
1 • 3,948,812.3 • 250
= 0.2534
15,374, 789,309
φLT = 0.5 1 + α LT ( λLT − 0.2 ) + λLT 2 = 0.5 1 + 0.21 • ( 0.2534 − 0.2 ) + 0.25342
φLT = 0.5377
χ LT
=
1
φLT + φLT 2 + λLT 2
≤ 1.0
1
=
χ LT
= 0.9882 ≤ 1.0
0.5377 + 0.5377 2 + 0.25342
=
fbd
χ LT f y 0.9882 • 250
=
= 224.58 MPa
γM0
1.1
M dLTB = Z pz fbd = 3,948,812.3 • 224.58 = 886,839, 489 N − mm
=
M dLTB 886.84 kN − m
IS 800-2007 Example 003 - 10
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Interaction Capacity: Compression & Bending
Member Bending & Compression Capacity with Buckling
Z-Z Axis
=
nz
P 2500
=
= 0.3839
Pdz 6511
K z = 1 + ( λz − 0.2 ) nz = 1 + ( 0.2037 − 0.2 ) • 0.3839 ≤ 1 + 0.8nz = 1 + 0.8 • ( 0.3839 )
=
K z 1.0014 ≤ 1.3072 so K z = 1.0014
ψz =
M 2 −350
=
= −1
350
M1
Cmz
= 0.6 + 0.4ψ
= 0.6 + 0.4 • −=
1 0.2 > 0.4 so Cmz
= 0.4
Y-Y Axis
=
ny
P 2500
=
= 0.4721
Pdy 5295
K y = 1 + ( λ y − 0.2 ) n y = 1 + ( 0.554 − 0.2 ) • 0.4721 ≤ 1 + 0.8n y = 1 + 0.8 • ( 0.4721)
=
K y 1.167 ≤ 1.378 so K y = 1.167
ψ
=
y
M2
0
= = 0
M 1 100
Cmy = 0.6 + 0.4ψ = 0.6 + 0.4 • 0= 0.6 > 0.4 so Cmy = 0.6
Lateral-Torsional Buckling
CmLT = 0.4
K LT = 1 −
0.1λLT n y
CmLT − 0.25
≥ 1−
0.1n y
CmLT − 0.25
IS 800-2007 Example 003 - 11
Software Verification
PROGRAM NAME:
REVISION NO.:
K LT = 1 −
ETABS
0
0.1 • 0.2534 • 0.4721
0.1 • 0.4721
= 0.920 ≥ 1 −
= 0.831
0.4 − 0.25
0.4 − 0.25
K LT = 0.920
Formula IS 9.3.2.2 (a)
D P K y Cmy M y K LT M z 2500 1.167 • 0.6 • 100 0.920 • 350
=+
+
= +
+
C Pdy
M dy
M dLTB
5295
325.65
886.84
D
= 0.472 + 0.215 + 0.363
C
D
= 1.050 (Governs)
C
Formula IS 9.3.2.2 (b)
D P 0.6 K y Cmy M y K z Cmz M z 2500 0.6 • 1.167 • 0.6 • 100 1.0014 • 0.4 • 350
=
+
+
=+
+
6511
325.65
886.84
C Pdz
M dy
M dLTB
D
= 0.384 + 0.129 + 0.158
C
D
= 0.671
C
IS 800-2007 Example 003 - 12
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
KBC 2009 Example 001
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The design flexural strengths are checked for the beam shown below. The beam
is loaded with a uniform load of 6.5 kN/m (D) and 11 kN/m (L). The flexural
moment capacity is checked for three unsupported lengths in the weak direction,
Lb = 1.75 m, 4 m and 12 m.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W460x74
E = 205,000 MPa
Fy = 345 MPa
Loading
w = 6.5 kN/m (D)
w = 11.0 kN/m (L)
Geometry
Span, L = 12m
TECHNICAL FEATURES TESTED
Section Compactness Check (Bending)
Member Bending Capacities
Unsupported length factors
KBC 2009 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are comparing with the results of ETABS.
ETABS
Independent
Percent
Difference
Compact
Compact
0.00%
Cb ( Lb =1.75m)
1.004
1.002
0.20%
φb M n ( Lb =1.75m) (kN-m)
515.43
515.43
0.00%
Cb ( Lb =4m)
1.015
1.014
0.10%
φb M n ( Lb =4m) (kN-m)
394.8
394.2
0.15%
Cb ( Lb =12m)
1.136
1.136
0.00%
φb M n ( Lb =12m) (kN-m)
113.48
113.45
0.03%
Output Parameter
Compactness
COMPUTER FILE: KBC 2009 EX001
CONCLUSION
The results show an acceptable comparison with the independent results.
KBC 2009 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material:
E = 205,000 MPa, Fy = 345 MPa
Section: W460x74
bf = 191 mm, tf = 14.5 mm, d = 457 mm, tw = 9 mm
h = d − 2t f = 457 − 2 • 14.5 = 428 mm
h0 = d − t f = 457 − 14.5 = 442.5 mm
S33 = 1457.3 cm3, Z33 = 1660 cm3
Iy =1670 cm4, ry = 42 mm, Cw = 824296.4 cm6, J = 51.6 cm4
=
rts
1670 • 824296.4
= 50.45 mm
1457.3
I y Cw
=
S33
Rm = 1.0 for doubly-symmetric sections
Other:
c = 1.0
L = 12 m
Loadings:
wu = (1.2wd + 1.6wl) = 1.2(6.5) + 1.6(11) = 25.4 kN/m
Mu =
∙
wu L2
= 25.4 122/8 = 457.2 kN-m
8
Section Compactness:
Localized Buckling for Flange:
=
λ
bf
191
=
= 6.586
2t f 2 • 14.5
KBC 2009 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
E
205, 000
=
λ p 0.38
= 0.38 = 9.263
Fy
345
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
λ
=
h 428
=
= 47.56
9
tw
E
205, 000
=
= 3.76 = 91.654
λ p 3.76
Fy
345
λ < λ p , No localized web buckling
Web is Compact.
Section is Compact.
Section Bending Capacity:
M p =Fy Z 33 =345 • 1660 =572.7 kN-m
Lateral-Torsional Buckling Parameters:
Critical Lengths:
E
205, 000
Lp =
1.76 ry
=
1.76 • 42
=
1801.9 mm =
1.8 m
Fy
345
E
=
Lr 1.95rts
0.7 Fy
0.7 Fy S33 ho
Jc
1 + 1 + 6.76
S33 ho
Jc
E
2
205, 000
51.6 • 1
0.7 • 345 1457.3 • 44.8
Lr = 1.95 • 50.45
1 + 1 + 6.76
0.7 • 345 1457.3 • 44.25
205, 000 51.6 • 1
2
KBC 2009 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Lr = 5.25 m
Non-Uniform Moment Magnification Factor:
For the lateral-torsional buckling limit state, the non-uniform moment magnification factor is
calculated using the following equation:
Cb =
2.5M max
12.5M max
Rm ≤ 3.0
+ 3M A + 4 M B + 3M C
Eqn. 1
Where MA = first quarter-span moment, MB = mid-span moment, MC = second quarter-span
moment.
The required moments for Eqn. 1 can be calculated as a percentage of the maximum mid-span
moment. Since the loading is uniform and the resulting moment is symmetric:
1 L
M A = MC = 1− b
4 L
2
Member Bending Capacity for Lb = 1.75 m:
M=
M
=
1.00
max
B
2
2
1 L
1 1.75
1− b =
1−
0.995
MA =
MC =
=
4 L
4 12
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.995 ) + 4 (1.00 ) + 3 ( 0.995 )
Cb = 1.002
Lb < L p , Lateral-Torsional buckling capacity is as follows:
M
=
M
=
572.7 kN-m
n
p
φb M=
0.9 • 572.7
n
φb M n = 515.43 kN-m
KBC 2009 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Member Bending Capacity for Lb = 4 m:
M=
M
=
1.00
max
B
2
2
1 L
1 4
MA =
MC =
1− b =
1− =
0.972
4 L
4 12
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.972 ) + 4 (1.00 ) + 3 ( 0.972 )
Cb = 1.014
L p < Lb < Lr , Lateral-Torsional buckling capacity is as follows:
Lb − L p
≤Mp
M n = C b M p − (M p − 0.7 Fy S 33 )
L − L
r
p
4.00 − 1.80
=
=
M n 1.014 572.7 − ( 572.7 − 0.7 • 0.345 • 1457.3)
437.97 kN-m
5.25 − 1.80
φb M=
0.9 • 437.97
n
φb M n = 394.2 kN-m
Member Bending Capacity for Lb = 12 m:
M=
M
=
1.00
max
B
2
2
1 L
1 12
MA =
MC =
1− b =
1− =
0.750 .
4 L
4 12
Cb =
12.5 (1.00 )
2.5 (1.00 ) + 3 ( 0.750 ) + 4 (1.00 ) + 3 ( 0.750 )
(1.00 )
Cb = 1.136
Lb > Lr , Lateral-Torsional buckling capacity is as follows:
KBC 2009 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
Fcr
=
Cbπ 2 E
Lb
rts
2
Jc
1 + 0.078
S33 ho
Lb
rts
ETABS
0
2
1.136 • π 2 • 205, 000
51.6 • 1
12000
1 + 0.078
86.5 MPa
Fcr =
=
2
1457.3 • 44.25 50.45
12000
50.45
2
M n = Fcr S 33 ≤ M p
Mn =
86.5 • 1457.3 =
126.056kN-m
φb M=
0.9 • 126.056
n
φb M n = 113.45 kN-m
KBC 2009 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
KBC 2009 Example 002
BUILT UP WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
A demand capacity ratio is calculated for the built-up, ASTM A572 grade 50,
column shown below. An axial load of 300 kips (D) and 900 kips (L) is applied to
a simply supported column with a height of 5m.
GEOMETRY, PROPERTIES AND LOADING
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Warping constant calculation, Cw
Member compression capacity with slenderness reduction
KBC 2009 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared with the results from
ETABS.
ETABS
Independent
Percent
Difference
Compactness
Slender
Slender
0.00%
φcPn (kN)
2056.7
2056.7
0.00 %
Output Parameter
COMPUTER FILE: KBC 2009 EX002
CONCLUSION
The results show an exact comparison with the independent results.
KBC 2009 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material:
E = 205,000 MPa, Fy =345 MPa
Section: Built-Up Wide Flange
d = 432 mm, bf = 203 mm, tf = 25 mm, h = 382 mm, tw = 7 in.
Ignoring fillet welds:
A = 2(203)(25) + (382)(7) = 128.24 cm2
2(25)(203)3 (382)(7)3
+
= 34.867 E 06 mm3
12
12
Iy
34.867 E 06
=
ry =
= 52.1 mm.
A
12824
I y=
I x = ∑ Ad 2 + ∑ I x
Cw = 1443463.1 cm 6
=
J
bt 3
=
∑ 3 216.1 cm4
Member:
K = 1.0 for a pinned-pinned condition
L=5m
Loadings:
Pu = 1.2(300) + 1.6(700) = 1800 kN
Section Compactness:
Check for slender elements using Specification KBC 2009:
-3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Localized Buckling for Flange:
b 101.5
=
= 4.06
t
25
E
205, 000
=
λ p 0.38
= 0.38 = 9.263
Fy
345
λ=
λ < λ p , No localized flange buckling
Flange is Compact.
Localized Buckling for Web:
h 382
=
= 54.57 ,
7
t
E
205, 000
=
= 1.49 = 36.32
λr 1.49
Fy
345
λ=
λ > λr , Localized web buckling
Web is Slender.
Section is Slender
Member Compression Capacity:
Elastic Flexural Buckling Stress
Since the unbraced length is the same for both axes, the y-y axis will govern by
inspection.
KLy 1.0 • 5000
=
= 95.97
ry
52.1
=
Fe
π 2E
π 2 • 205, 000
= 219.68 MPa
=
2
2
( 95.97 )
KL
r
Elastic Critical Torsional Buckling Stress
Note: Torsional buckling will not govern if KLy > KLz, however, the check is included
here to illustrate the calculation.
-4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
π 2 EC w
1
+ GJ
Fe =
2
(K z L )
Ix + Iy
π 2 • 205, 000 • 1443463.1E 06
1
=
+ 78846.15 • 216.1E 04
Fe
2
( 45338 + 3486.7 ) E 04
( 5000 )
= 588 MPa > 288.84 MPa
Therefore, the flexural buckling limit state controls.
Fe = 220 MPa
Section Reduction Factors
Since the flange is not slender,
Qs = 1.0
Since the web is slender,
Take f as Fcr with Q = 1.0
4.71
KLy
E
205, 000
= 4.71
= 114.8 >
= 95.97
QFy
ry
1.0 ( 345 )
So
QFy
1.0( 345 )
Fe
f = Fcr =Q 0.658 Fy =1.0 0.658 220 • 345 =178.98 MPa
E 0.34 E
1 −
≤ b, where b = h
f (b t ) f
205, 000
0.34
205, 000
be = 1.92 ( 7 )
1 −
≤ 359.12 mm
178.98 ( 382 7 ) 178.98
=
be 359.12 mm ≤ 382 mm
be = 1.92t
therefore compute Aeff with reduced effective web width.
Aeff =
betw + 2b f t f =
( 359.12 )( 7 ) + 2 ( 203)( 25) =12663.84 mm 2
where Aeff is effective area based on the reduced effective width of the web, be.
-5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Aeff 12663.84
=
= 0.9875
A
12824
=
Q Q=
=
(1.00 )( 0.9875
) 0.9875
s Qa
=
Qa
Critical Buckling Stress
Determine whether Specification Equation E7-2 or E7-3 applies
4.71
KLy
E
205, 000
= 4.71
= 138.47 >
= 95.97
QFy
ry
0.9875 ( 345 )
Therefore, Specification Equation E7-2 applies.
When 4.71
E
KL
≥
QFy
r
QFy
0.9875( 345 )
Fe
Fy 0.9875 0.658 220=
=
=
Fcr Q 0.658
• 345 178.2 MPa
Nominal Compressive Strength
Pn = Fcr Ag = 12824 • 178.2 = 2285236.8 N
φc =0.90
φc P=
Fcr Ag= 0.90 ( 2285.24=) 2056.7 kN > 1800 kN
n
-6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NTC 2008 Example 001
WIDE FLANGE SECTION UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
In this example a continuous beam-column is subjected to factored axial load P =
1400 kN and major-axis bending moment M = 200 kN-m. The beam is
continuously braced to avoid any buckling effects. This example was tested using
the Italian NTC-2008 steel frame design code. The design capacities are
compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
A
M
P
A
L
Section A-A
L = 0.4 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80769 MPa
Loading
P = 1400 kN
M = 200 kN-m
Design Properties
fy = 235 MPa
Section: 457x191x98 UB
TECHNICAL FEATURES TESTED
Section compactness check (beam-column)
Section compression capacity
Section shear capacity
Section bending capacity with compression & shear reductions
Interaction capacity, D/C
NTC 2008 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-NTC-2008.pdf,” which is
available through the program “Help” menu. Examples were taken from Example
6.6 on pp. 57-59 from the book “Designers’ Guide to EN1993-1-1” by R.S.
Narayanan & A. Beeby.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 2
Class 2
0.00%
2797.6
2797.6
0.00%
667.5
667.5
0.00%
499.1
499.1
0.00%
310.8
310.8
0.00%
470.1
470.1
0.00%
0.644
0.644
0.00%
Section Compression Resistance,
Nc,Rd (kN)
Section Shear Resistance,
Vc,Rd,y (kN)
Section Plastic Bending Resistance,
Mc,y,Rd (kN-m)
Section Bending Resistance Axially
Reduced,
MN,y,Rd (kN-m)
Section Bending Resistance Shear
Reduced,
MV,y,Rd (kN-m)
Interaction Capacity, D/C
COMPUTER FILE: NTC 2008 EX001
CONCLUSION
The results show an exact comparison with the independent results.
NTC 2008 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Material: S275 Steel
E = 210000 MPa
fy = 235 MPa
Section: 457x191x98 UB
A = 12,500 mm2
b = 192.8 mm, tf = 19.6 mm, h = 467.2 mm, tw = 11.4 mm, r = 0 mm
hw = h − 2t f = 467.2 − 2 • 19.6 = 428 mm
d = h − 2 ( t f + r ) = 467.2 − 2 • (19.6 + 0 ) = 428 mm
=
c
b − tw − 2r 192.8 − 11.4 − 2 • 0
=
= 90.7 mm
2
2
Wpl,y = 2,230,000 mm3
Other:
γM 0 =
1.05
Loadings:
P = 1400 kN axial load
M y = 200 kN-m bending load at one end
Results in the following internal forces:
N Ed = 1400 kN
VEd = 500 kN
M y , Ed = 200 kN-m
NTC 2008 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Section Compactness:
235
=
fy
=
ε
−1 ≤=
α
=
α
235
= 1
235
N Ed
1
1 −
2 2htw f y
≤ 1
1
1, 400, 000
=
1 −
2.7818 > 1, so
2 2 • 467.2 • 11.4 • 235
α =1.0
Localized Buckling for Flange:
For the tip in compression under combined bending & compression
λ cl .1 =
λe =
9ε 9 • 1
=
= 9
α
1
c 90.7
=
= 4.63
t f 19.6
=
λ e 4.63 < λ=
9
cl .1
So Flange is Class 1 in combined bending and compression
Localized Buckling for Web:
α > 0.5, so
λ=
cl .1
λe =
396ε
396 • 1
=
= 33.00 for combined bending & compression
13α − 1 13 • 1 − 1
d 428
=
= 37.54
tw 11.4
=
λ e 37.54 > =
λ cl .1 33.00
λ=
cl .2
456ε
456 • 1
=
= 38.00
13α − 1 13 • 1 − 1
=
λ e 37.54 < λ
=
38.00
cl .2
NTC 2008 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
So Web is Class 2 in combined bending & compression.
Since Web is Class 2, Section is Class 2 in combined bending & compression.
Section Compression Capacity
N=
N pl=
c , Rd
, Rd
Af y 12,500 • 235
=
γM 0
1.05
N c , Rd = 2797.6 kN
Section Shear Capacity
AV , y = A − 2bt f + t f ( tw + 2r ) = 12,500 − 2 • 192.8 • 19.6 + 19.6 (11.4 + 2 • 0 )
AV , y = 5,165.7 mm 2
=
Vc , Rd , y
fy
235
=
Avy
• 5,165.7
γM 0 3
1.05 3
Vc , Rd , y = 667.5 kN
η =1.0
hw 428
72 235 72 235
37.5 <
72
==
=
=
1.0 235
tw 11.4
η
fy
So no shear buckling needs to be checked.
Section Bending Capacity
M
=
M=
c , y , Rd
pl , y , Rd
W pl , y f y 2, 230, 000 • 235
=
γM 0
1.05
M c , y , Rd = 499.1kN-m
NTC 2008 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
with Shear Reduction
V=
500 kN > 0.5 • Vc ,=
333.7 kN Shear Reduction is needed
Ed
Rd
Av =htw =467.2 • 11.4 =4,879.2 mm 2
2
2V
2 • 500 2
ρ
= Ed − 1=
− 1= 0.2482
V
667.5
,
c
Rd
ρAv 2
0.2482 • 4879.22
W
f
−
2,
230,
000
−
pl , y 4t yk
• 235
4 • 11.4
w
M y ,V , Rd =
=
≤ M y ,c , Rd
1.05
γM 0
M V ,r , Rd = 470.1kN-m
with Compression Reduction
=
n
N Ed
1400
=
= 0.50
N pl , Rd 2797.6
A − 2bt f 12,500 − 2 • 192.8 • 19.6
=
= 0.40 ≤ 0.5
12,500
A
1− n
1 − 0.5
= 499.1 •
M N=
M pl , y , Rd
, y , Rd
1 − 0.5a
1 − 0.5 • 0.4
=
a
M N , y , Rd = 310.8 kN-m
Interaction Capacity: Compression & Bending
Section Bending & Compression Capacity
Formula NTC 4.2.39
2
D
=
C
5n
2
M y , Ed M z , Ed
M y , Ed
200
0
0.414
+
=
+
=
≤
= 0.644
310.8
M N , y , Rd
M N , y , Rd M N , z , Rd
D
= 0.644 (Governs)
C
NTC 2008 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NTC 2008 Example 002
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
In this example a continuous beam-column is subjected to factored axial load P =
1400 kN, major-axis bending moment My = 200 kN-m, and a minor axis bending
moment of Mz = 100 kN-m. This example was tested using the Italian NTC-2008
steel frame design code. The design capacities are compared with independent
hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
Y-Axis
Z-Axis
Y-Y
My,top
Mz,top
Z-Z
P
L
A
A
Mz,bot
My,bot
Section A-A
L = 0.4 m
Material Properties
E = 210x103 MPa
v = 0.3
G = 80769 MPa
Design Properties
fy = 235 MPa
P = 1400 kN
Section: 457x191x98 UB
Mz,top = 100 kN-m
Mz,bot = -100 kN-m
My,top = 200 kN-m
My,bot = 0
Loading
NTC 2008 Example 002 - 1
Software Verification
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TECHNICAL FEATURES TESTED
Section compactness check (beam-column)
Section compression capacity
Section shear capacity for major & minor axes
Section bending capacity for major & minor axes
Member compression capacity for major & minor axes
Member bending capacity
Interaction capacity, D/C, for major & minor axes
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-NTC-2008.pdf,” which is
available through the program “Help” menu. Examples were taken from Example
6.6 on pp. 57-59 from the book “Designers’ Guide to EN1993-1-1” by R.S.
Narayanan & A. Beeby.
NTC 2008 Example 002 - 2
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PROGRAM NAME:
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Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Class 2
Class 2
0.00%
2,797.6
2,797.6
0.00%
2,797.6
2,797.6
0.00%
2,797.6
2,797.6
0.00%
499.1
499.1
0.00%
84.8
84.8
0.00%
470.1
470.1
0.00%
310.8
310.8
0.00%
82.26
82.26
0.00%
499.095
499.095
0.00%
667.5
667.5
0.00%
984.7
984.7
0.00%
2.044
2.044
0.00%
Section Compression Resistance,
Nc,Rd (kN)
Buckling Resistance in Compression,
Nbyy,Rd (kN)
Buckling Resistance in Compression,
Nbzz,Rd (kN)
Section Plastic Bending Resistance,
Mc,y,Rd (kN-m)
Section Plastic Bending Resistance,
Mc,z,Rd (kN-m)
Section Bending Resistance Shear Reduced,
MV,y,Rd (kN-m)
Section Bending Resistance Axially Reduced,
MN,y,Rd (kN-m)
Section Bending Resistance Axially Reduced,
MN,z,Rd (kN-m)
Member Bending Resistance,
Mb,Rd (kN-m)
Section Shear Resistance,
Vc,y,Rd (kN)
Section Shear Resistance,
Vc,z,Rd (kN)
Interaction Capacity, D/C
COMPUTER FILE: NTC 2008 EX002
CONCLUSION
The results show an exact comparison with the independent results.
NTC 2008 Example 002 - 3
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HAND CALCULATION
Properties:
Material: S275 Steel
E = 210,000 MPa
G = 80,769 MPa
fy = 235 MPa
Section: 457x191x98 UB
A = 12,500 mm2
b = 192.8 mm, tf = 19.6 mm, h = 467.2 mm, tw = 11.4 mm, r = 0 mm
hw = h − 2t f = 467.2 − 2 • 19.6 = 428 mm
d = h − 2 ( t f + r ) = 467.2 − 2 • (19.6 + 0 ) = 428 mm
=
c
b − tw − 2r 192.8 − 11.4 − 2 • 0
=
= 90.7 mm
2
2
Wpl,y = 2,230,000 mm3
Wpl,z = 379,000 mm3
ryy = 191.3 mm
rzz = 43.3331 mm
Izz = 23,469,998 mm4
=
I w 1.176 • 1012 mm 6
IT = 1,210,000 mm4
Member:
L = Lb = Lunbraced = 400 mm
Kyy = 1.0, Kzz = 1.0
Other:
γM 0 =
1.05
γM1 =
1.05
NTC 2008 Example 002 - 4
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Loadings:
P = 1400 kN axial load
M z −1 = 100 kN-m
M z − 2 = −100 kN-m
M y −1 = 200 kN-m
M y − 2 = 0 kN-m
Results in the following internal forces:
N Ed = 1400 kN
M y , Ed = 200 kN-m
M z , Ed = 100 kN-m
Vy , Ed = 500 kN-m
Vz , Ed = 0 kN-m
Section Compactness:
=
ε
235
=
fy
−1 ≤=
α
=
α
235
= 1
235
N Ed
1
1 −
2 2htw f y
≤ 1
1
1, 400, 000
=
1 −
2.7818 > 1, so
2 2 • 467.2 • 11.4 • 235
α =1.0
Localized Buckling for Flange:
For the tip in compression under combined bending & compression
λ cl .1 =
9ε 9 • 1
=
= 9
α
1
NTC 2008 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
λe =
ETABS
0
c 90.7
=
= 4.63
t f 19.6
=
λ e 4.63 < λ=
9
cl .1
So Flange is Class 1 in combined bending and compression
Localized Buckling for Web:
α > 0.5, so
λ=
cl .1
λe =
396ε
396 • 1
=
= 33.00 for combined bending & compression
13α − 1 13 • 1 − 1
d 428
=
= 37.54
tw 11.4
=
λ e 37.54 > =
λ cl .1 33.00
λ=
cl .2
456ε
456 • 1
=
= 38.00
13α − 1 13 • 1 − 1
=
λ e 37.54 < λ
=
38.00
cl .2
So Web is Class 2 in combined bending & compression.
Since Web is Class 2, Section is Class 2 in combined bending & compression.
Section Compression Capacity
N=
N pl=
c , Rd
, Rd
Af y 12,500 • 235
=
γM 0
1.05
N c , Rd = 2, 797.6 kN
Section Shear Capacity
For major y-y axis
AV , y = A − 2bt f + t f ( tw + 2r ) = 12,500 − 2 • 192.8 • 19.6 + 19.6 (11.4 + 2 • 0 )
NTC 2008 Example 002 - 6
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PROGRAM NAME:
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ETABS
0
AV , y = 5,165.7 mm 2
=
Vc , y , Rd
fy
235
=
Avy
• 5,165.7
γM 0 3
1.05 3
Vc , y , Rd = 667.5 kN
For minor z-z axis
AV , z =A − hwtw =
12,500 − 428 • 11.4 =
7, 620.8 mm 2
Vc , z , Rd
=
fy
235
Avy
=
• 7, 620.8
1.05 3
γM 0 3
Vc , z , Rd = 984.7 kN
η =1.0
hw 428
72 235 72 235
37.5 <
72
==
=
=
1.0 235
tw 11.4
η
fy
So no shear buckling needs to be checked.
Section Bending Capacity
For major y-y axis
M
=
M=
c , y , Rd
pl , y , Rd
W pl , y f y 2, 230, 000 • 235
=
γM 0
1.05
M c , y , Rd = 499.1kN-m
For minor z-z axis
M
=
M=
c , z , Rd
pl , z , Rd
W pl , z f y 379, 000 • 235
=
γM 0
1.05
M c , z , Rd = 84.8 kN-m
NTC 2008 Example 002 - 7
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With Shear Reduction
For major y-y axis
Vy ,=
500 kN > 0.5 • Vc , y ,=
333.7 kN Shear Reduction is needed
Ed
Rd
Av =htw =467.2 • 11.4 =4,879.2 mm 2
2
2VEd
2 • 500 2
ρ
=
− 1=
− 1= 0.2482
V
667.5
,
c
Rd
ρAv 2
0.1525 • 4879.22
W
f
−
2,
230,
000
−
pl , y 4t yk
• 235
4 • 11.4
w
M y ,V , Rd =
=
≤ M y ,c , Rd
1.05
γM 0
M V ,r , Rd = 470.1kN-m
For minor z-z axis
Vz , Ed = 0 kN < 0.5 • Vc , z , Rd No shear Reduction is needed
With Compression Reduction
=
n
=
a
N Ed
1400
=
= 0.50
N pl , Rd 2797.6
A − 2bt f 12,500 − 2 • 192.8 • 19.6
=
= 0.40 ≤ 0.5
A
12,500
For major y-y axis
1− n
1 − 0.5
= 499.1 •
1 − 0.5a
1 − 0.5 • 0.4
= 310.8 kN-m
M N=
M pl , y , Rd
, y , Rd
M N , y , Rd
NTC 2008 Example 002 - 8
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For minor z-z axis
n 1.2
b f 192.8
and
=
t f 19.6 mm < 40 mm
So we should use the Buckling Curve ‘a’ for the z-z axis and Buckling Curve ‘b’
for the y-y axis (NTC 2008, Table 4.2.VI).
Y-Y Axis Parameters:
For buckling curve a, α =0.21 (NTC 2008, Table 4.2 VI)
K y = 1.00
1 • 400 =
400 mm,
Lcr , y =
K y Ly =
=
N cr , y
=
λy
Lcr , y
ry
400
2.091
=
=
191.3
π2 E
π2 • 210, 000
=
=
5,925, 691kN
2
2
K y Ly 12,500 • ( 2.091)
A
ry
Af y
=
N cr , y
12,500 • 235
= 0.022
5,925, 691
NTC 2008 Example 002 - 9
Software Verification
PROGRAM NAME:
REVISION NO.:
(
ETABS
0
)
2
0.5 1 + 0.21( 0.022 − 0.2 ) + 0.0222
=
φ y 0.5 1 + α y λ y − 0.2 + λ=
y
φ y =0.482
Stress Reduction
Factor: χ y
=
1
1
=
= 1.0388
2
2
2
+
− 0.0222
0.482
0.482
φy + φy − λ y
=
χ y 1.0388 > 1.0, so=
χ y 1.0
N=
byy , Rd
χ y Af y 1.0 • 12,500 • 235
=
γM1
1.05
N byy , Rd = 2, 797.6 kN
Z-Z Axis Parameters:
For buckling curve b, α =0.34 (NTC 2008, Table 4.2 VI)
K z = 1.00
1 • 400 =
400 mm,
Lcr , z =
K z Lz =
=
N cr , z
=
λz
Lcr , z
rz
400
9.231
=
=
43.33
π2 E
π2 • 210, 000
=
=
304, 052 kN
2
2
K z Lz 12,500 • ( 9.231)
A
rz
12,500 • 235
= 0.098
304, 052
Af y
=
N cr , z
(
)
2
0.5 1 + 0.34 ( 0.098 − 0.2 ) + 0.0982
=
φ z 0.5 1 + α z λ z − 0.2 + λ=
z
φ z =0.488
Stress Reduction
Factor: χ z
=
1
1
=
= 1.0362
2
2
2
0.488
0.488
+
− 0.0982
φz + φz − λ z
NTC 2008 Example 002 - 10
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
=
χ z 1.0362 > 1.0, so=
χ z 1.0
N=
bzz , Rd
χ z Af y 1.0 • 12,500 • 235
=
γM1
1.05
N bzz , Rd = 2, 797.6 kN
Member Bending Capacity:
h 467.2
=
= 2.4 > 2
b f 192.8
So we should use the Buckling Curve ‘c’ for lateral-torsional buckling (NTC
2008, Table 4.2.VII).
α LT =
0.49
for rolled section)
λ LT ,0 =(default
0.4
β =0.75 (default for rolled section)
=
M B M=
0, =
MA M
=
200 kN-m
y −2
y −1
2
M
M
0
0
=
ψ 1.75 − 1.05 B + 0.3 B =
+ 0.3
=
1.75 − 1.05
1.75
MA
200
200
MA
2
Corrective Factor is determined from NTC 2008, Table 4.2 VIII
=
kc
1
1
=
= 1.329
1.33 − 0.33ψ 1.33 − 0.33 • 1.75
π2 EI z I w ( Lcr , z ) GIT
M cr =
ψ
+
2
I
π2 EI z
L
( cr , z ) z
2
NTC 2008 Example 002 - 11
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π2 • 210, 000 • 23, 469,998 1.176 • 1012 4002 • 80, 769 • 1, 210, 000
M cr =
1.75 •
+ 2
4002
23, 469,998 π • 210, 000 • 23, 469,998
M cr = 119, 477, 445,900 N-mm
=
λ LT
2, 230, 000 • 235
= 0.066
119, 477, 445,900
W pl , y f y
=
M cr
(
)
2
0.5 1 + 0.49 • ( 0.066 − 0.4 ) + 0.75 • 0.0662
=
φ LT 0.5 1 + α LT λ LT − λ LT ,0 + βλ=
LT
φ LT =
0.420
(
)
2
2
f =
1 − 0.5 (1 − kc ) 1 − 2 λ LT − 0.8 =
1 − 0.5 (1 − 1.329 ) 1 − 2 ( 0.066 − 0.8 ) =
0.987
=
χ LT
χ LT
1
1
1 1
≤ 1.0 or
2
f φ LT + φ LT 2 + βλ LT 2
λ LT f
1
1
1
1
≤ 1.0 or
2
0.066 0.987
0.987 0.420 + 0.4202 + 0.75 • 0.0662
1.2118 ≤ (1.0 or 230.9 )
χ=
LT
so
χ LT =
1.0
fy
235
M b , Rd =
χ LT W pl , y
=
1.0 • 2, 230, 000 •
γM1
1.05
M b , Rd = 499.095 kN-m
Interaction Capacity: Compression & Bending
Section Bending & Compression Capacity
NTC 2008 Example 002 - 12
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Formula NTC 4.2.39
2
D M y , Ed M z , Ed
=
+
C M N , y , Rd M N , z , Rd
5n
2
200 100
=
310.8 + 82.3
5•0.5
=
0.414 + 1.630
D
= 2.044 (Governs)
C
Member Bending & Compression Capacity: Method B
k factors used are taken from the software, and determined from Method 2 in
Annex B of Eurocode 3.
k yy = 0.547
k yz = 0.479
k zy = 0.698
k zz = 0.798
Formula NTC 4.2.37
M y , Ed
M z , Ed
N Ed
D
=
+ k yy
+ k yz
W f
W pl , z f yk
C χ y Af yk
χ LT pl , y yk
γM1
γM1
γM1
D
C
=
1, 400
200
100
+ 0.547 ×
+ 0.479 ×
1×12,500 × 235
2, 230, 000 × 235
379, 000 × 235
1×
1.05
1.05
1.05
D
=0.5 + 0.22 + 0.56
C
D
= 1.284
C
NTC 2008 Example 002 - 13
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Formula NTC 4.2.38
M y , Ed
M z , Ed
N Ed
D
=
+ k zy
+ k zz
W f
W pl , z f yk
C χ z Af yk
χ LT pl , y yk
γM1
γM1
γM1
1, 400
200
100
D
=
+ 0.698 ×
+ 0.798 ×
2, 230, 000 × 235
379, 000 × 235
C 1×12,500 × 235
1×
1.05
1.05
1.05
D
=0.5 + 0.28 + 0.941
C
D
= 1.721
C
NTC 2008 Example 002 - 14
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NZS 3404-1997 Example 001
WIDE FLANGE MEMBER UNDER COMPRESSION
EXAMPLE DESCRIPTION
The frame object axial strengths are tested in this example.
A continuous column is subjected to factored load N = 200 kN. This example
was tested using the NZS 3404-1997 steel frame design code. The design
capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
N
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
N
= 200 kN
Design Properties
fy = 250 MPa
Section: 350WC197
TECHNICAL FEATURES TESTED
Section compactness check (compression)
Section compression capacity
Member compression capacity
NZS 3404-1997 Example 001 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-NZS-3404-1997.pdf,” which is
available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0.00%
Section Axial Capacity, Ns (kN)
6275
6275
0.00%
Member Axial Capacity, Nc (kN)
4385
4385
0.00%
COMPUTER FILE: NZS 3404-1997 EX001
CONCLUSION
The results show an exact comparison with the independent results.
NZS 3404-1997 Example 001 - 2
Software Verification
PROGRAM NAME:
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HAND CALCULATION
Properties:
Material:
fy = 250 MPa
Section: 350WC197
Ag = An = 25100 mm2
bf = 350 mm, tf = 28 mm, h = 331 mm, tw = 20 mm
r33 = 139.15 mm, r22 = 89.264 mm
Member:
le33 = le22 = 6000 mm (unbraced length)
Considered to be a braced frame
Loadings:
N * = 200 kN
Section Compactness:
Localized Buckling for Flange:
=
λe
(b f − tw ) f y
350 − 20 250
=
= 5.89
2•tf
250
2 • 28
250
Flange is under uniform compression, so:
λ ep = 9, λ ey = 16, λ ew = 90
=
λ e 5.89 < λ=
9 , No localized flange buckling
ep
Flange is compact
NZS 3404-1997 Example 001 - 3
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Localized Buckling for Web:
=
λe
fy
h
331 250
=
= 16.55
tw 250 20 250
Web is under uniform compression, so:
λ ep = 30, λ ey = 45, λ ew = 180
=
λ e 16.55 <=
λ ep 30 , No localized web buckling
Web is compact.
Section is Compact.
Section Compression Capacity:
Section is not Slender, so Kf = 1.0
Ns =
K f An f y =
1 • 25,100 • 250 /103
N s = 6275kN
Member Weak-Axis Compression Capacity:
Frame is considered a braced frame in both directions, so k=
k=
1
e 22
e 33
le 22
le 33
6000
6000
=
= 67.216 and
=
= 43.119
r22 89.264
r33 139.15
Buckling will occur on the 22-axis.
λ=
n 22
=
α a 22
le 22
r22
K f fy
6000
=
•
250
89.264
(1 • 250=
)
250
67.216
2100(λ n 22 − 13.5)
= 20.363
λ n 22 2 − 15.3λ n 22 + 2050
α b 22 =
0.5 since cross-section is not a UB or UC section
NZS 3404-1997 Example 001 - 4
Software Verification
PROGRAM NAME:
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ETABS
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λ 22 = λ n 22 + α a 22 α b 22 = 67.216 + 20.363 • 0.5 = 77.398
=
η22 0.00326(λ 22 − 13.5)
= 0.2083 ≥ 0
λ 22
77.398
+ 1 + η22
+ 1 + 0.2083
90
90
=
ξ 22
=
=
1.317
2
2
λ 22
77.398
2
2
90
90
2
α c 22
=
ξ 22 1 −
2
90 2
1 −
ξ 22 λ 22
α=
1.317 1 −
c 22
2
90
0.6988
1 −
=
1.317 • 77.398
N c 22 =
α c 22 N s ≤ N s
N c 22 = 0.6988 • 6275= 4385 kN
NZS 3404-1997 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
1
NZS 3404-1997 Example 002
WIDE FLANGE MEMBER UNDER BENDING
EXAMPLE DESCRIPTION
The frame object bending strengths are tested in this example.
A continuous column is subjected to factored moment Mx = 1000 kN-m. This
example was tested using the NZS 3404-1997 steel frame design code. The
design capacities are compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
Mx
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
Mx = 1000 kN-m
Design Properties
fy = 250 MPa
Section: 350WC197
TECHNICAL FEATURES TESTED
Section compactness check (bending)
Section bending capacity
Member bending capacity
NZS 3404-1997 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
1
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-NZS-3404-1997.pdf, ” which is
available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness
Compact
Compact
0%
837.5
837.5
0%
837.5
837.5
0%
Section Bending Capacity,
Ms,major (kN-m)
Member Bending Capacity,
Mb (kN-m)
COMPUTER FILE: NZS 3404-1997 EX002
CONCLUSION
The results show an exact comparison with the independent results.
NZS 3404-1997 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
1
HAND CALCULATION
Properties:
Material:
fy = 250 MPa
Section: 350WC197
bf = 350 mm, tf = 28 mm, h = 331 mm, tw = 20 mm
I22 = 200,000,000 mm4
Z33 = 2,936,555.891 mm2
S33 = 3,350,000 mm2
J = 5,750,000 mm4
Iw = 4,590,000,000,000 mm6
Member:
le22 = 6000 mm (unbraced length)
Considered to be a braced frame
Loadings:
M m * = 1000 kN-m
This leads to:
M 2 * = 250 kN-m
M 3 * = 500 kN-m
M 4 * = 750 kN-m
Section Compactness:
Localized Buckling for Flange:
=
λe
(b f − tw ) f y
350 − 20 250
=
= 5.89
2•tf
250
2 • 28
250
NZS 3404-1997 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
1
Flange is under uniform compression, so:
λ ep = 9, λ ey = 16, λ ew = 90
=
λ e 5.89 < λ=
9 , No localized flange buckling
ep
Flange is compact
Localized Buckling for Web:
=
λe
fy
h
331 250
=
= 16.55
tw 250 20 250
Web is under bending, so:
λ ep = 82, λ ey = 115, λ ew = 180
=
λ e 16.55 <=
λ ep 30 , No localized web buckling
Web is compact.
Section is Compact.
Section Bending Capacity:
Z=
Z=
min( S ,1.5Z ) for compact sections
e
c
3,350, 000 mm 2
Z=
Z=
e 33
c 33
250 • 3,350, 000 /10002
M=
M s ,major
= f y Z=
s 33
e 33
=
M s 33 M
=
837.5 kN-m
s ,major
Member Bending Capacity:
kt = 1 (Program default)
kl = 1.4 (Program default)
kr = 1 (Program default)
lLTB = le22 = 6000 mm
NZS 3404-1997 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
1
le = kt kl kr lLTB = 1 • 1.4 • 1 • 6000 = 8400 mm 2
π2 EI 22
π2 EI w
GJ
+
2
le 2
le
M=
M
=
oa
o
π2 • 2 • 105 • 2 • 108
π2 • 2 • 105 • 4.59 • 1012
76,923.08
5,
750,
000
M oa =
Mo =
•
+
8, 4002
8, 4002
M=
M
=
1786.938 kN-m
oa
o
2
2
M
Ms
837.5
837.5
s
3
0.6
3
=
α s 0.6
+
=
−
+
−
1786.938
M oa
M oa
1786.938
α s =0.7954
=
αm
α=
m
1.7 M m *
( M 2 *) + ( M 3 *) + ( M 4 *)
2
2
1.7 • 1000
2
≤ 2.5
= 1.817 ≤ 2.5
( 250 ) + ( 500 ) + ( 750 )
2
2
2
M b =α m α s M s =0.7954 • 1.817 • 837.5 ≤ M s
M b 1210.64 kN-m ≤ 837.5 kN-m
=
NZS 3404-1997 Example 002 - 5
Software Verification
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0
NZS 3404-1997 Example 003
WIDE FLANGE MEMBER UNDER COMBINED COMPRESSION & BENDING
EXAMPLE DESCRIPTION
The frame object interacting axial and bending strengths are tested in this
example.
A continuous column is subjected to factored loads and moments N= 200 kN;
Mx= 1000 kN-m. This example was tested using the NZS 3404-1997 steel frame
design code. The design capacities are compared with independent hand
calculated results.
GEOMETRY, PROPERTIES AND LOADING
Mx
N
L
A
A
Section A-A
L=6m
Material Properties
E = 200x103 MPa
v = 0.3
G = 76923.08 MPa
Loading
Design Properties
fy = 250 MPa
N = 200 kN
Section: 350WC197
Mx = 1000 kN-m
NZS 3404-1997 Example 003 - 1
Software Verification
PROGRAM NAME:
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TECHNICAL FEATURES TESTED
Section compactness check (compression & bending)
Section bending capacity with compression reduction
Member in-plane bending capacity with compression reduction
Member out-of-plane bending capacity with compression reduction
RESULTS COMPARISON
Independent results are taken from hand calculations based on the CSI steel
design documentation contained in the file “SFD-NZS-3404-1997.pdf,” which is
available through the program “Help” menu.
Output Parameter
ETABS
Independent
Percent
Difference
Compactness`
Compact
Compact
0.00%
837.5
837.5
0.00%
823.1
823.1
0.00%
837.5
837.5
0.00%
Reduced Section Bending Capacity,
Mrx (kN-m)
Reduced In-Plane Member Bending
Capacity,
Mix (kN-m)
Reduced Out-of-Plane Member
Bending Capacity, Mo (kN-m)
COMPUTER FILE: NZS 3404-1997 EX003
CONCLUSION
The results show an exact comparison with the independent results.
NZS 3404-1997 Example 003 - 2
Software Verification
PROGRAM NAME:
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0
HAND CALCULATION
Properties:
Section: 350WC197
Ag = An = 25100 mm2
I22 = 200,000,000 mm4
I33 = 486,000,000 mm4
J = 5,750,000 mm4
Iw = 4,590,000,000,000 mm6
Member:
lz=le33 = le22 = 6000 mm (unbraced length)
Considered to be a braced frame
φ =0.9
Loadings:
N * = 200 kN
M m * = 1000 kN-m
Section Compactness:
From example SFD – IN-01-1, section is Compact in Compression
From example SFD – IN-01-2, section is Compact in Bending
Section Compression Capacity:
From example SFD – IN-01-1, N s = 6275kN
NZS 3404-1997 Example 003 - 3
Software Verification
PROGRAM NAME:
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Member Compression Capacity:
From example SFD – IN-01-1, N c 22 = 4385 kN
Section Bending Capacity:
From example SFD – IN-01-2,=
M s 33 M
=
837.5 kN-m
s ,major
Section Interaction: Bending & Compression Capacity:
N*
200
M r 33 =1.18M s 33 1 −
=1.18 • 837.5 1 −
≤ M s 33 =837.5
0.9 • 6275
φN s
=
M r 33 953.252 ≤ 837.5
M r 33 = 837.5kN-m
Member Strong-Axis Compression Capacity:
Strong-axis buckling strength needs to be calculated:
Frame is considered a braced frame in both directions, so ke 33 = 1
le 33
6000
=
= 43.119
r33 139.15
λ=
n 33
=
α a 33
le 33
r33
K f fy
6000
=
•
250 139.15
(1 • 250=
)
250
43.119
2100(λ n 33 − 13.5)
= 19.141
λ n 332 − 15.3λ n 33 + 2050
α b 33 =
0.5 since cross-section is not a UB or UC section
λ 33 = λ n 33 + α a 33α b 33 = 43.119 + 19.141 • 0.5 = 52.690
=
η33 0.00326(λ 33 − 13.5)
= 0.1278 ≥ 0
NZS 3404-1997 Example 003 - 4
Software Verification
PROGRAM NAME:
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0
λ 33
52.690
90 + 1 + η33
+ 1 + 0.1278
90
=
ξ33
=
=
2.145
2
2
λ 33
52.690
2
2
90
90
2
α c 33
=
ξ33 1 −
2
90 2
1 −
ξ33λ 33
1 −
α
=
2.145
c 33
2
90
0.8474
1 −
=
•
2.145
50.690
N c 33 =
α c 33 N s ≤ N s
=
N c 33 0.8474 • 6275
N c 33 = 5318 kN
Member Interaction: In-Plane Bending & Compression Capacity:
β=
m
M min
0
=
= 0
M max 1000
Since the section is compact,
3
1 + β 3
N*
N*
1 + βm
m
+ 1.18
M i = M s 33 1 −
1−
1 −
φN c 33
2
2 φN c 33
3
1 + 0 3
200
200
1+ 0
Mi =
837.5 1 −
1
−
+
1.18
1−
2 0.9 • 5318
2
0.9
•
5318
M i = 823.11 kN-m
NZS 3404-1997 Example 003 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS 2013
0
Member Interaction: Out-of-Plane Bending & Compression Capacity:
1
α bc =
3
1 − βm 1 − βm
N *
0.4
0.23
+
−
φN c 22
2
2
1
α bc =
3
1− 0 1− 0
200
0.4
0.23
+
−
0.9 • 4385
2
2
α bc =
4.120
π2 EI w
π2 • 2 • 106 • 4.59 • 1012
2
lz
60002
N oz = GJ +
= 76923.08 • 5.75 • 106 +
I 33 + I 22
( 4.86 + 2 ) •108
Ag
25100
=
N oz 4.423 • 1011 kN
M b 33o =α m α s M sx w/ an assumed uniform moment such that α m =1.0
M b 33o =
1.0 • 0.7954 • 837.5 =
666.145 kN-m
N *
N*
M o 33 =
α bc M b 33o 1 −
1 −
φN c 22 φN oz
≤ M r 33
200
200
4.12 • 666.145 1 −
2674 ≤ 837.5
M o 33 =
=
1 −
11
0.9 • 4385 0.9 • 4.423 • 10
M o 33 = 837.5 kN-m
NZS 3404-1997 Example 003 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-08 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear design of a rectangular concrete beam is calculated in this
example.
A simply supported beam is subjected to an ultimate uniform load of 9.736 k/ft.
This example is tested using the ACI 318-08 concrete design code. The flexural
and shear reinforcing computed is compared with independent hand calculated
results.
GEOMETRY, PROPERTIES AND LOADING
CL
10"
A
13.5"
2.5"
A
Section A-A
10' = 120"
Material Properties
E=
3600 k/in2
ν=
0.2
G=
1500 k/in2
Section Properties
d = 13.5 in
b = 10.0 in
I = 3,413 in4
Design Properties
f’c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
ACI 318-08 Example 001 - 1
Software Verification
PROGRAM NAME:
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RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 6.1 in Notes on ACI 318-08 Building Code.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (k-in)
1460.4
1460.4
0.00%
Tension Reinf, As (in2)
2.37
2.37
0.00%
Design Shear Force, Vu
37.73
37.73
0.00%
Shear Reinf, Av/s (in2/in)
0.041
0.041
0.00%
COMPUTER FILE: ACI 318-08 Ex001
CONCLUSION
The computed results show an exact match for the flexural and the shear
reinforcing.
ACI 318-08 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9, Ag = 160 sq-in
As,min =
=
200
bw d = 0.450 sq-in (Govern)
fy
3 f c'
fy
bw d = 0.427 sq-in
f c′ − 4000
0.85
=
1000
0.85 − 0.05
β1 =
0.003
=
d 5.0625 in
0.003 + 0.005
=
cmax
amax = β1cmax = 4.303 in
Combo1
wu = (1.2wd + 1.6wl) = 9.736 k/ft
Mu =
∙
wu l 2
= 9.736 102/8 = 121.7 k-ft = 1460.4 k-in
8
The depth of the compression block is given by:
a = d − d2 −
2Mu
0.85 f c'ϕb
= 4.183 in (a < amax)
The area of tensile steel reinforcement is given by:
Mu
1460.4
=
a
0.9 • 60 • (13.5 − 4.183 / 2 )
ϕ fy d −
2
As
=
As
= 2.37 sq-in
ACI 318-08 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
0.75
f c′ :
Check the limit of
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bd
= 12.807 k
The maximum shear that can be carried by reinforcement is given by:
ϕ Vs =
ϕ8
f c′ bd = 51.229 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 6.4035 k
ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ ϕ (Vc/2),
Av
= 0,
s
else if ϕ (Vc/2) < Vu ≤ ϕ Vmax
Av
(V − φVc ) Av
= u
≥
φ f ys d
s
s min
where:
b
Av
w
= max 50
s min
f yt
bw
,
f yt
3
4
f c′
else if Vu > ϕ Vmax,
a failure condition is declared.
ACI 318-08 Example 001 - 4
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Combo1
∙
Vu = 9.736 (5-13.5/12) = 37.727 k
φ (Vc =
/ 2 ) 6.4035 k ≤=
Vu 37.727 k ≤ φ V
=
64.036 k
max
10 10 3
Av
,
4, 000
= max 50
s min
60, 000 60, 000 4
in 2
Av
=
max
=
0.0083,
0.0079
0.0083
{
}
in
s min
Av
=
s
(Vu − φVc )
in 2
in 2
= 0.041
= 0.492
φ f ys d
in
ft
ACI 318-08 Example 001 - 5
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PROGRAM NAME:
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ACI 318-08 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected factored axial load Pu = 398.4 k and
moments Muy = 332 k-ft. This column is reinforced with 4 #9 bars. The total area
of reinforcement is 8.00 in2. The design capacity ratio is checked by hand
calculations and result is compared.
GEOMETRY, PROPERTIES AND LOADING
Pu=398.4 kips
22"
Muy=332k-ft
A
A
14"
2.5"
10’
Section A-A
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
Design Properties
b =
d =
f’c = 4 k/in2
fy = 60 k/in2
14 in
19.5 in
ACI 318-08 Example 002 - 1
Software Verification
PROGRAM NAME:
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TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Demand/Capacity Ratio
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.000
1.00
0.00%
COMPUTER FILE: ACI 318-08 Ex002
CONCLUSION
The computed results show an exact match with the independent results.
ACI 318-08 Example 002 - 2
Software Verification
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HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
f’c = 4 ksi
b = 14 inch
Pu = 398.4 kips
fy = 60 ksi
d = 19.5 inch
Mu = 332 k-ft
1) Because e = 10 inch < (2/3)d = 13 inch., assume compression failure. This assumption will be
checked later. Calculate the distance to the neutral axis for a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
87
87
dt =
(19.5) = 11.54 inch
87 + f y
87 + 60
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = 0.85 f c' ab = 0.85 • 4 • 14a = 47.6a
(
)
Cs = As' f y - 0.85 f c' = 4 ( 60 - 0.85 • 4 ) = 226.4 kips
Assume compression steels yields, (this assumption will be checked later).
T = As f s = 4 f s f s < f y
(
)
(Eqn. 1)
Pn = 47.6a + 226.4 - 4 f s
3) Taking moments about As:
a
'
Cc d - 2 + Cs d - d
The plastic centroid is at the center of the section and d " = 8.5 inch
e' = e + d " = 10 + 8.5 = 18.5 inch.
1
a
Pn =
47.6a 19.5 - + 226.4 (19.5 - 2.5 )
18.5
2
Pn =
1
e'
(
)
ACI 318-08 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
Pn = 50.17 a - 1.29a 2 + 208
ETABS
0
(Eqn. 2)
4) Assume c = 13.45 inch, which exceed cb (11.54 inch).
a = 0.85 • 13.45 = 11.43 inch
Substitute in Eqn. 2:
Pn = 50.17 • 11.43 - 1.29 • (11.43) + 208 = 612.9 kips
2
5) Calculate fs from the strain diagram when c = 13.45 inch.
19.5 -13.45
fs =
87 = 39.13 ksi
13.45
εs = εt = f s Es = 0.00135
6) Substitute a = 13.45 inch and fs = 39.13 ksi in Eqn. 1 to calculate Pn2:
Pn2 = 47.6 (11.43) + 226.4 - 4 ( 39.13) = 613.9 kips
Which is very close to the calculated Pn2 of 612.9 kips (less than 1% difference)
10
M n = Pn e = 612.9 = 510.8 kips-ft
12
7) Check if compression steels yield. From strain diagram,
13.45 - 2.5
ε s' =
( 0.003) = 0.00244 > ε y = 0.00207 ksi
13.45
Compression steels yields, as assumed.
8) Calculate φ ,
dt = d = 19.5 inch,
c = 13.45 inch
19.45 -13.45
= 0.00135
13.45
ε t (at the tension reinforcement level) = 0.003
Since ε t < 0.002 , then φ = 0.65
φ Pn = 0.65 ( 612.9 ) = 398.4 kips
φ M n = 0.65 ( 510.8 ) = 332 k-ft.
ACI 318-08 Example 002 - 4
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ACI 318-11 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear design of a rectangular concrete beam is calculated in this
example.
A simply supported beam is subjected to an ultimate uniform load of 9.736 k/ft.
This example is tested using the ACI 318-11 concrete design code. The flexural
and shear reinforcing computed is compared with independent hand calculated
results.
GEOMETRY, PROPERTIES AND LOADING
CL
10"
A
13.5"
2.5"
A
Section A-A
10' = 120"
Material Properties
E=
3600 k/in2
ν=
0.2
G=
1500 k/in2
Section Properties
d = 13.5 in
b = 10.0 in
I = 3,413 in4
Design Properties
f’c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
ACI 318-11 Example 001 - 1
Software Verification
PROGRAM NAME:
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ETABS
0
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 6.1 in Notes on ACI 318-11 Building Code.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (k-in)
1460.4
1460.4
0.00%
Tension Reinf, As (in2)
2.37
2.37
0.00%
Design Shear Force, Vu
37.73
37.73
0.00%
Shear Reinf, Av/s (in2/in)
0.041
0.041
0.00%
COMPUTER FILE: ACI 318-11 Ex001
CONCLUSION
The computed results show an exact match for the flexural and the shear
reinforcing.
ACI 318-11 Example 001 - 2
Software Verification
PROGRAM NAME:
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HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9, Ag = 160 sq-in
As,min =
=
200
bw d = 0.450 sq-in (Govern)
fy
3 f c'
fy
bw d = 0.427 sq-in
f c′ − 4000
0.85
=
1000
0.85 − 0.05
β1 =
0.003
=
d 5.0625 in
0.003 + 0.005
=
cmax
amax = β1cmax = 4.303 in
Combo1
wu = (1.2wd + 1.6wl) = 9.736 k/ft
Mu =
∙
wu l 2
= 9.736 102/8 = 121.7 k-ft = 1460.4 k-in
8
The depth of the compression block is given by:
a = d − d2 −
2Mu
0.85 f c'ϕb
= 4.183 in (a < amax)
The area of tensile steel reinforcement is given by:
Mu
1460.4
=
a
0.9 • 60 • (13.5 − 4.183 / 2 )
ϕ fy d −
2
As
=
As
= 2.37 sq-in
ACI 318-11 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
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Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
0.75
f c′ :
Check the limit of
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bd
= 12.807 k
The maximum shear that can be carried by reinforcement is given by:
ϕ Vs =
ϕ8
f c′ bd = 51.229 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 6.4035 k
(ϕ Vc + ϕ 50 bd)
= 11.466 k
ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ ϕ (Vc/2),
Av
= 0,
s
else if ϕ (Vc/2) < Vu ≤ ϕ Vmax
Av
(V − φVc ) Av
≥
= u
φ f ys d
s
s min
where:
b
Av
w
= max 50
s min
f yt
bw
,
f yt
3
4
f c′
else if Vu > ϕ Vmax,
a failure condition is declared.
ACI 318-11 Example 001 - 4
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Combo1
∙
Vu = 9.736 (5-13.5/12) = 37.727 k
φ (Vc =
/ 2 ) 6.4035 k ≤=
Vu 37.727 k ≤ φ V
=
64.036 k
max
10 10 3
Av
,
4, 000
= max 50
s min
60, 000 60, 000 4
in 2
Av
=
max
=
0.0083,
0.0079
0.0083
{
}
in
s min
Av
=
s
(Vu − φVc )
in 2
in 2
= 0.041
= 0.492
φ f ys d
in
ft
ACI 318-11 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-11 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected factored axial load Pu = 398.4 k and
moments Muy = 332 k-ft. This column is reinforced with 4 #9 bars. The total area
of reinforcement is 8.00 in2. The design capacity ratio is checked by hand
calculations and result is compared.
GEOMETRY, PROPERTIES AND LOADING
Pu=398.4 kips
22"
Muy=332k-ft
A
A
14"
2.5"
10’
Section A-A
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
Design Properties
b =
d =
f’c = 4 k/in2
fy = 60 k/in2
14 in
19.5 in
ACI 318-11 Example 002 - 1
Software Verification
PROGRAM NAME:
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ETABS
0
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Demand/Capacity Ratio
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.000
1.00
0.00%
COMPUTER FILE: ACI 318-11 Ex002
CONCLUSION
The computed results show an exact match with the independent results.
ACI 318-11 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
f’c = 4 ksi
b = 14 inch
Pu = 398.4 kips
fy = 60 ksi
d = 19.5 inch
Mu = 332 k-ft
1) Because e = 10 inch < (2/3)d = 13 inch., assume compression failure. This assumption will be
checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
87
87
dt =
(19.5) = 11.54 inch
87 + f y
87 + 60
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = 0.85 f c' ab = 0.85 • 4 • 14a = 47.6a
(
)
Cs = As' f y - 0.85 f c' = 4 ( 60 - 0.85 • 4 ) = 226.4 kips
Assume compression steels yields, (this assumption will be checked later).
T = As f s = 4 f s f s < f y
(
)
(Eqn. 1)
Pn = 47.6a + 226.4 - 4 f s
3) Taking moments about As:
a
'
Cc d - 2 + Cs d - d
The plastic centroid is at the center of the section and d " = 8.5 inch
e' = e + d " = 10 + 8.5 = 18.5 inch.
1
a
Pn =
47.6a 19.5 - + 226.4 (19.5 - 2.5 )
18.5
2
Pn =
1
e'
(
)
ACI 318-11 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
Pn = 50.17 a - 1.29a 2 + 208
ETABS
0
(Eqn. 2)
4) Assume c = 13.45 inch, which exceed cb (11.54 inch).
a = 0.85 • 13.45 = 11.43 inch
Substitute in Eqn. 2:
Pn = 50.17 • 11.43 - 1.29 • (11.43) + 208 = 612.9 kips
2
5) Calculate fs from the strain diagram when c = 13.45 inch.
19.5 -13.45
fs =
87 = 39.13 ksi
13.45
εs = εt = f s Es = 0.00135
6) Substitute a = 13.45 inch and fs = 39.13 ksi in Eqn. 1 to calculate Pn2:
Pn2 = 47.6 (11.43) + 226.4 - 4 ( 39.13) = 613.9 kips
Which is very close to the calculated Pn2 of 612.9 kips (less than 1% difference)
10
M n = Pn e = 612.9 = 510.8 kips-ft
12
7) Check if compression steels yield. From strain diagram,
13.45 - 2.5
ε s' =
( 0.003) = 0.00244 > ε y = 0.00207 ksi
13.45
Compression steels yields, as assumed.
8) Calculate φ ,
dt = d = 19.5 inch,
c = 13.45 inch
19.45 -13.45
= 0.00135
13.45
ε t (at the tension reinforcement level) = 0.003
Since ε t < 0.002 , then φ = 0.65
φ Pn = 0.65 ( 612.9 ) = 398.4 kips
φ M n = 0.65 ( 510.8 ) = 332 k-ft.
ACI 318-11 Example 002 - 4
Software Verification
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REVISION NO.:
ETABS
0
ACI 318-14 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear design of a rectangular concrete beam is calculated in this
example.
A simply supported beam is subjected to an ultimate uniform load of 9.736 k/ft.
This example is tested using the ACI 318-14 concrete design code. The flexural
and shear reinforcing computed is compared with independent hand calculated
results.
GEOMETRY, PROPERTIES AND LOADING
CL
10"
A
13.5"
2.5"
A
Section A-A
10' = 120"
Material Properties
E=
3600 k/in2
ν=
0.2
G=
1500 k/in2
Section Properties
d = 13.5 in
b = 10.0 in
I = 3,413 in4
Design Properties
f’c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
ACI 318-14 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 6.1 in Notes on ACI 318-14 Building Code.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (k-in)
1460.4
1460.4
0.00%
Tension Reinf, As (in2)
2.37
2.37
0.00%
Design Shear Force, Vu
37.73
37.73
0.00%
Shear Reinf, Av/s (in2/in)
0.041
0.041
0.00%
COMPUTER FILE: ACI 318-14 Ex001
CONCLUSION
The computed results show an exact match for the flexural and the shear
reinforcing.
ACI 318-14 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9, Ag = 160 sq-in
As,min =
=
200
bw d = 0.450 sq-in (Govern)
fy
3 f c'
fy
bw d = 0.427 sq-in
f c′ − 4000
0.85
=
1000
0.85 − 0.05
β1 =
0.003
=
d 5.0625 in
0.003 + 0.005
=
cmax
amax = β1cmax = 4.303 in
Combo1
wu = (1.2wd + 1.6wl) = 9.736 k/ft
Mu =
∙
wu l 2
= 9.736 102/8 = 121.7 k-ft = 1460.4 k-in
8
The depth of the compression block is given by:
a = d − d2 −
2Mu
0.85 f c'ϕb
= 4.183 in (a < amax)
The area of tensile steel reinforcement is given by:
Mu
1460.4
=
a
0.9 • 60 • (13.5 − 4.183 / 2 )
ϕ fy d −
2
As
=
As
= 2.37 sq-in
ACI 318-14 Example 001 - 3
Software Verification
PROGRAM NAME:
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0
Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
0.75
f c′ :
Check the limit of
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bd
= 12.807 k
The maximum shear that can be carried by reinforcement is given by:
ϕ Vs =
ϕ8
f c′ bd = 51.229 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 6.4035 k
(ϕ Vc + ϕ 50 bd)
= 11.466 k
ϕ Vmax = ϕ Vc + ϕ Vs = 64.036 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ ϕ (Vc/2),
Av
= 0,
s
else if ϕ (Vc/2) < Vu ≤ ϕ Vmax
Av
(V − φVc ) Av
≥
= u
φ f ys d
s
s min
where:
b
Av
w
= max 50
s min
f yt
bw
,
f yt
3
4
f c′
else if Vu > ϕ Vmax,
a failure condition is declared.
ACI 318-14 Example 001 - 4
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Combo1
∙
Vu = 9.736 (5-13.5/12) = 37.727 k
φ (Vc =
/ 2 ) 6.4035 k ≤=
Vu 37.727 k ≤ φ V
=
64.036 k
max
10 10 3
Av
,
4, 000
= max 50
s min
60, 000 60, 000 4
in 2
Av
=
max
=
0.0083,
0.0079
0.0083
{
}
in
s min
Av
=
s
(Vu − φVc )
in 2
in 2
= 0.041
= 0.492
φ f ys d
in
ft
ACI 318-14 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-14 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected factored axial load Pu = 398.4 k and
moments Muy = 332 k-ft. This column is reinforced with 4 #9 bars. The total area
of reinforcement is 8.00 in2. The design capacity ratio is checked by hand
calculations and result is compared.
GEOMETRY, PROPERTIES AND LOADING
Pu=398.4 kips
22"
Muy=332k-ft
A
A
14"
2.5"
10’
Section A-A
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
Design Properties
b =
d =
f’c = 4 k/in2
fy = 60 k/in2
14 in
19.5 in
ACI 318-14 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Demand/Capacity Ratio
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.000
1.00
0.00%
COMPUTER FILE: ACI 318-14 Ex002
CONCLUSION
The computed results show an exact match with the independent results.
ACI 318-14 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
f’c = 4 ksi
b = 14 inch
Pu = 398.4 kips
fy = 60 ksi
d = 19.5 inch
Mu = 332 k-ft
1) Because e = 10 inch < (2/3)d = 13 inch., assume compression failure. This assumption will be
checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
87
87
dt =
(19.5) = 11.54 inch
87 + f y
87 + 60
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = 0.85 f c' ab = 0.85 • 4 • 14a = 47.6a
(
)
Cs = As' f y - 0.85 f c' = 4 ( 60 - 0.85 • 4 ) = 226.4 kips
Assume compression steels yields, (this assumption will be checked later).
T = As f s = 4 f s f s < f y
(
)
(Eqn. 1)
Pn = 47.6a + 226.4 - 4 f s
3) Taking moments about As:
a
'
Cc d - 2 + Cs d - d
The plastic centroid is at the center of the section and d " = 8.5 inch
e' = e + d " = 10 + 8.5 = 18.5 inch.
1
a
Pn =
47.6a 19.5 - + 226.4 (19.5 - 2.5 )
18.5
2
Pn =
1
e'
(
)
ACI 318-14 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
Pn = 50.17 a - 1.29a 2 + 208
ETABS
0
(Eqn. 2)
4) Assume c = 13.45 inch, which exceed cb (11.54 inch).
a = 0.85 • 13.45 = 11.43 inch
Substitute in Eqn. 2:
Pn = 50.17 • 11.43 - 1.29 • (11.43) + 208 = 612.9 kips
2
5) Calculate fs from the strain diagram when c = 13.45 inch.
19.5 -13.45
fs =
87 = 39.13 ksi
13.45
εs = εt = f s Es = 0.00135
6) Substitute a = 13.45 inch and fs = 39.13 ksi in Eqn. 1 to calculate Pn2:
Pn2 = 47.6 (11.43) + 226.4 - 4 ( 39.13) = 613.9 kips
Which is very close to the calculated Pn2 of 612.9 kips (less than 1% difference)
10
M n = Pn e = 612.9 = 510.8 kips-ft
12
7) Check if compression steels yield. From strain diagram,
13.45 - 2.5
ε s' =
( 0.003) = 0.00244 > ε y = 0.00207 ksi
13.45
Compression steels yields, as assumed.
8) Calculate φ ,
dt = d = 19.5 inch,
c = 13.45 inch
19.45 -13.45
= 0.00135
13.45
ε t (at the tension reinforcement level) = 0.003
Since ε t < 0.002 , then φ = 0.65
φ Pn = 0.65 ( 612.9 ) = 398.4 kips
φ M n = 0.65 ( 510.8 ) = 332 k-ft.
ACI 318-14 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AS 3600-2009 Example 001
Shear and Flexural Reinforcement Design of a Singly Reinforced T-Beam
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design. The load level is
adjusted for the case corresponding to the following conditions:
The stress-block extends below the flange but remains within the balanced
condition permitted by AS 3600-09.
The average shear stress in the beam is below the maximum shear stress
allowed by AS 3600-09, requiring design shear reinforcement.
A simple-span, 6-m-long, 300-mm-wide, and 500-mm-deep T-beam with a
flange 100 mm thick and 600 mm wide is considered. The beam is shown in
Figure 1. The computational model uses a finite element mesh of frame elements
automatically generated. The maximum element size has been specified to be
500 mm. The beam is supported by columns without rotational stiffnesses and
with very large vertical stiffness (1 × 1020 kN/m).
The beam is loaded with symmetric third-point loading. One dead load case
(DL30) and one live load case (LL130), with only symmetric third-point loads of
magnitudes 30, and 130 kN, respectively, are defined in the model. One load
combinations (COMB130) is defined using the AS 3600-09 load combination
factors of 1.2 for dead load and 1.5 for live load. The model is analyzed for both
of these load cases and the load combination.
The beam moment and shear force are computed analytically. Table 1 shows the
comparison of the design longitudinal reinforcements. Table 2 shows the
comparison of the design shear reinforcements.
AS 3600-2009 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
600 mm
75 mm
100 mm
500 mm
75 mm
300 mm
Beam Section
2000 mm
2000 mm
2000 mm
Shear Force
Bending Moment
Figure 1 The Model Beam for Flexural and Shear Design
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
AS 3600-2009 Example 001 - 2
L
h
Ds
bw
bf
dsc
=
=
=
=
=
=
6000
500
100
300
600
75
mm
mm
mm
mm
mm
mm
Software Verification
PROGRAM NAME:
REVISION NO.:
Effective depth,
Depth of comp. reinf.,
d
d'
=
=
425
75
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
f’c
fy
wc
Ec
Es
v
=
=
=
=
=
=
30
460
0
25x105
2x108
0.2
Dead load,
Live load,
Pd
Pl
=
=
30
130
ETABS
0
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the total factored moments in the design strip.
They match exactly for this problem. Table 1 also shows the design
reinforcement comparison.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-cm)
Method
Moment (kN-m)
As+
ETABS
462
33.512
Calculated
462
33.512
A +s ,min = 3.00 sq-cm
AS 3600-2009 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
ETABS
Calculated
231
12.05
12.05
COMPUTER FILE: AS 3600-2009 EX001
CONCLUSION
The computed results show an exact match for the flexural and the shear
reinforcing.
AS 3600-2009 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
φ
= 0.8 for bending
0.67 ≤ α 2 ≤ 0.85, where α 2 = 1.0 − 0.003 f c ' = 0.91 , use α 2 = 0.85
0.67 ≤ γ ≤ 0.85, where α 2= 1.05 − 0.007 f c =' 0.84 , use γ = 0.84
ku ≤ 0.36
=
amax γ=
ku d 0.840.36=
425 128.52 mm
2
D f ′ct , f
Ast .min = α b
bw d
d f sy
where for L- and T-Sections with the web in tension:
D= h= 500 mm
1/4
bf
bf
D
− 1 0.4 s − 0.18 ≥ 0.20
D
bw
bw
α b = 0.20 +
= 0.2378
30 3.3 MPa
f 'ct , f 0.5
f 'c 0.5
=
=
=
f sy
= f=
460 MPa ≤ 500 MPa
y
2
D f ′ct , f
Ast .min = 0.2378
bd
d f sy
= 0.2378 • (500/425)2 •3.3/460 • 300•425
= 299.9 mm2
COMB130
V* = (1.2Pd + 1.5Pl) = 231kN
M* =
V *L
= 462 kN-m
3
The depth of the compression block is given by:
AS 3600-2009 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
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a =−
d
d2 −
2M *
= 100.755 mm (a > Ds), so design as a T-beam.
α 2 f 'c φ b f
The compressive force developed in the concrete alone is given by the following
methodology:
The first part of the calculation is for balancing the compressive force from the
flange, Cf, and the second part of the calculation is for balancing the compressive
force from the web, Cw. Cf is given by:
Cf =
α 2 f c′ ( b f − bw )min ( Ds , amax ) =
765 kN
Therefore,
A
=
s1
C f 765
= = 1663.043 mm 2
f sy 460
and the portion of M* that is resisted by the flange is given by:
min ( Ds , amax )
Muf =
φC f d −
= 229.5 kN-m
2
Again, the value for φ is 0.80 by default. Therefore, the balance of the
moment, M* to be carried by the web is:
M=
M * − Muf = 462 – 229.5 = 232.5
uw
The web is a rectangular section of dimensions bw and d, for which the design
depth of the compression block is recalculated as:
a1 =−
d
d2 −
2 M uw
= 101.5118 mm
α 2 f ′c φ bw
a1 ≤ amax , so no compression reinforcement is needed, and the area of tension
reinforcement is then given by:
As 2 =
M uw
= 1688.186 mm2
a
φ f sy d − 1
2
Ast = As1 + As 2 = 3351.23 sq-mm = 33.512 sq-cm
AS 3600-2009 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Shear Design
φ = 0.7 for shear
Calculated at the end of the beam, so M=0 and Ast = 0.
The shear force carried by the concrete, Vuc, is calculated as:
13
A
Vuc = β1β2β3bv d o f cv′ st
bv d o
= 0 kN
where,
f cv′ = ( f c′) = 3.107 N/mm2 ≤ 4MPa
1/3
d
=
β1 1.11.6 − o ≥ 1.1 =1.2925,
1000
β2 = 1 since no significant axial load is present
β3 = 1
bv = bw = 300mm as there are no grouted ducts
do = d = 425 mm
The shear force is limited to a maximum of:
Vu .max = 0.2 f c′ bd o = 765 kN
And the beam must have a minimum shear force capacity of:
Vu .min =
Vuc + 0.6bw d o =
0 + 0.6300425 =
77 kN
=
V * 231 kN > φV=
0 , so reinforcement is needed.
uc / 2
=
V * 231 kN ≤ φV=
535.5 kN , so concrete crushing does not occur.
u .max
f 'c bv
bw
mm 2
Asv
=
=
max
0.35 , 0.06
max {228.26, 214.33}
f sy
f sy
m
s min
mm
Asv
= 228.26
m
s min
2
AS 3600-2009 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
COMB130
Since φV
=
53.55 kN=
< V * 231 kN ≤ φV=
535.5 kN
u .min
u .max
(
)
V * − φVuc
Asv
A
=
≥ sv
φf sy d o cot θv s min
s
θv = the angle between the axis of the concrete compression strut and the
longitudinal axis of the member, which varies linearly from 30
degrees when V*=φVu,min to 45 degrees when V*=φVu,max = 35.52
degrees
θv = 35.52 degrees
( 213 − 0 )
Asv
mm 2 Asv
=
=
≥
1205.04
s
m
s
0.7460425cot 35.52o
(
Asv
cm 2
= 12.05
s
m
AS 3600-2009 Example 001 - 8
)
mm 2
=
228.26
m
min
Software Verification
PROGRAM NAME:
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ETABS
2
AS 3600-2009 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected to factored axial load
N = 1733 kN and moment My = 433 kN-m. This column is reinforced with five
25M bars. The design capacity ratio is checked by hand calculations and the
result is compared with computed results. The column is designed as a short,
non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1733 kN
My= 433 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fcu = 30 MPa
fy = 460 MPa
350 m
490 mm
AS 3600-2009 Example 002 - 1
Software Verification
PROGRAM NAME:
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ETABS
2
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.089
1.00
8.9%
COMPUTER FILE: AS 3600-2009 EX002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
AS 3600-2009 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
2
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fy = 460 MPa
fcu = 30 MPa
b = 350 mm
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
cb =
600
600
dt =
( 490 ) = 277.4 mm
600 + f y
600 + 460
2) From the equation of equilibrium:
N = Cc + C s − T
where
′
=
Cc α=
0.85 • 30 •=
350a 8925a
2 fc ab
Cs = As ( f y − α 2 fc′) = 2500 ( 460 − 0.85 • 30 ) = 1, 086, 250 N
Assume compression steel yields, (this assumption will be checked later).
T = As f = 2500 f s = 2500 f s ( f s < f y )
N1 =
8925a + 1.086, 250 − 2500 fs
(Eqn. 1)
3) Taking moments about As:
=
N2
1
a
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
=
N
1
a
8925a 490 − + 1, 086, 250 ( 490 − 60 )
465
2
N = 9404.8a − 9.597 a 2 + 1, 004, 489
(Eqn. 2)
AS 3600-2009 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
2
4) Assume c = 333.9 mm, which exceeds cb (296 mm).
=
a 0.84
=
• 333.9 280.5 mm
Substitute in Eqn. 2:
N 2 = 8925 • 280.5 − 9.597 ( 280.5 ) + 1, 004, 489= 2,888, 240 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 − 333.9
=
fs =
600 280.5 MPa
333.9
ε s =εt =fs Es = 0.0014
6) Substitute a = 280.5 mm and fs = 280.5 MPa in Eqn. 1 to calculate N1:
N1= 8925 ( 280.5 ) + 1, 086, 250 − 2500 ( 280.5 )= 2,887,373 N
which is very close to the calculated N2 of 2,888,240 (less than 1% difference)
250
M
= Ne
= 2888
=
722 kN-m
1000
7) Check if compression steel yields. From strain diagram,
333.9 − 60
=
ε′s
=
> ε y 0.0023
) 0.0025=
( 0.003
333.9
Compression steel yields, as assumed.
8) Therefore, section capacity is
N = φ • 2888 =
1733 kN
e
250
M = φ • 2888 • = 0.60 • 2888 • = 433 kN-m
1000
1000
AS 3600-2009 Example 002 - 4
Software Verification
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0
PROGRAM NAME:
REVISION NO.:
BS 8110-1997 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
Example Description
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simple supported beam is subjected to a uniform factored load
of 36.67 kN/m. This example was tested using the BS 8110-97 concrete design
code. The flexural and shear reinforcing computed is compared with independent
results.
GEOMETRY, PROPERTIES AND LOADING
CL
230mm
A
550 mm
60 mm
A
Section A-A
6m
Material Properties
E=
25x106 kN/m2
ν=
0.2
G=
10416666.7kN/m2
Section Properties
d = 490 mm
W = 36.67 kN/m
Design Properties
fcu = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
BS 8110-1997 Example 001 - 1
Software Verification
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0
PROGRAM NAME:
REVISION NO.:
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 7.2 on page 149 of Reinforced Concrete
Design by W. H. Mosley, J. H. Bungey & R. Hulse.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (kN-m)
165.02
165.02
0.00%
Tension Reinf, As (mm2)
964.1
964.1
0.00%
Design Shear, Vu (kN)
92.04
92.04
0.00%
Shear Reinf, Asv/sv (mm2/mm)
0.231
0.231
0.00%
COMPUTER FILE: BS 8110-1997 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
BS 8110-1997 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
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0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
=
As ,min 0.0013
=
bw h 0.0013=
230550 164.45 mm 2
Design Combo COMB1
wu = =36.67 kN/m
Mu =
wu l 2
= 165 kN-m
8
The depth of the compression block is given by:
K=
M
= 0.0996 < 0.156
f cu b d 2
If K ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam.
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 427.90 mm
0
.
9
The ultimate resistance moment is given by:
As =
M
= 964.1 sq-mm
( f y 1.15) z
BS 8110-1997 Example 001 - 3
Software Verification
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PROGRAM NAME:
REVISION NO.:
Shear Design
L2
− d=
ω
V=
92.04 kN at distance, d, from support
U
U
2
v=
VU
= 0.8167 MPa
bd
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
v ≤ vmax , so no concrete crushing
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.79k1k 2 100 As 3 400 4
vc =
= 0.415 MPa
γ m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
f 3
k2 = cu = 1.06266, 1 ≤ k2 ≤
25
40
25
1
3
γm, concrete = 1.25
0.15 ≤
100 As
≤3
bd
100 As 100266
=
= 0.2359
bd
230490
1
400 4
400
=
0.95 ≥ 1, so
d
d
1
4
is taken as 1.
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
If (vc + 0.4) < v ≤ vmax
Asv (v − vc )bw
=
sv
0.87 f yv
BS 8110-1997 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
Asv
=
sv
v − vc ) bw ( 0.8167 − 0.4150 )
(=
0.87 f yv
0.87 • 460
ETABS
0
= 0.231 sq-mm/mm
BS 8110-1997 Example 001 - 5
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PROGRAM NAME:
REVISION NO.:
ETABS
0
BS 8110-1997 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected to factored axial load
N = 1971 kN and moment My = 493 kN-m. This column is reinforced with five
25M bars. The design capacity ratio is checked by hand calculations and the
result is compared with computed results. The column is designed as a short,
non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1971 kN
My= 493 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fcu = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
BS 8110-1997 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.994
1.00
0.40%
COMPUTER FILE: BS 8110-1997 EX002
CONCLUSION
The computed result shows an acceptable comparison with the independent
result.
BS 8110-1997 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Column Strength under compression control
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
=
cb
700
700
=
dt
=
( 490 ) 312 mm
700 + f y / γ s
700 + 460 /1.15
2) From the equation of equilibrium:
N = Cc + C s − T
where
0.67
=
fcu ab 0.67 1.5 •=
30 • 350a 4667 a
γΜ
As′
2500
971, 014 N
Cs =
f y − 0.4467 fcu ) = ( 460 − 0.4467 • 30 ) =
(
1.15
γs
Assume compression steel yields (this assumption will be checked later).
As fs 2500 fs
=
T =
= 2174 fs ( fs < f y )
γs
1.15
=
Cc
N1 =4, 667 a + 971, 014 − 2174 fs
(Eqn. 1)
3) Taking moments about As:
=
N
1
a
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
=
N
1
a
4, 667 a 490 − + 971, 014 ( 490 − 60 )
465
2
N = 4917.9a − 5.018a 2 + 897,926
(Eqn. 2)
BS 8110-1997 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Assume c = 364 mm, which exceed cb (296 mm).
=
a 0.9
=
• 364 327.6 mm
Substitute in Eqn. 2:
N 2 4917.9 • 327.6 − 5.018 ( 327.6 ) + 897,926
=
= 1,970,500 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 − 364
=
fs =
700 242.3 MPa
364
ε=
ε=
fs Es = 0.0012
s
t
6) Substitute a = 327.6 mm and fs = 242.3 MPa in Eqn. 1 to calculate N1:
=
N1 4, 667 ( 327.6 ) + 971, 014 − 2174 ( 242.3
=
) 1,973,163 N
which is very close to the calculated N2 of 1,970,500 (less than 1% difference)
250
M = Ne = 1971
= 493 kN-m
1000
7) Check if compression steels yield. From strain diagram,
365 − 60
=
ε′s
=
=
> ε y 0.0023
) 0.00292
( 0.0035
365
Compression steel yields, as assumed.
8) Therefore, the section capacity is
N = 1971 kN
M = 493 kN-m
BS 8110-1997 Example 002 - 4
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CSA A23.3-04 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simply supported beam is subjected to a uniform factored load
of 92.222 kN/m. This example is tested using the CSA A23.3-04 concrete design
code. The flexural and shear reinforcing computed is compared with independent
results.
GEOMETRY, PROPERTIES AND LOADING
CL
400mm
A
600 mm
54 mm
A
Section A-A
6m
Material Properties
E=
25x106 kN/m2
ν=
0.2
G=
10416666.7kN/m2
Section Properties
d = 546 mm
W = 92.222 kN/m
Design Properties
f’c = 40 MPa
fy = 400 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
CSA A23.3-04 Example 001 - 1
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 2.2 on page 2-12 in Part II on Concrete Design
Handbook of Cement Association of Canada.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mf (kN-m)
415.00
415.00
0.00%
Tension Reinf, As (mm2)
2466
2466
0.00%
Design Shear, Vf (kN)
226.31
226.31
0.00%
Shear Reinf, Av/s (mm2/mm)
0.379
0.379
0.00%
COMPUTER FILE: CSA A23.3-04 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
CSA A23.3-04 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
φc = 0.65 for concrete
φs = 0.85 for reinforcement
As,min =
0.2 f ′c
b h = 758.95 mm2
fy
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.79
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.87
cb =
700
d = 347.45 mm
700 + f y
ab = β1cb = 302.285 mm
COMB1
wu l 2
= 415 kN-m
Mf =
8
The depth of the compression block is given by:
a = d − d2 −
2M f
α 1 f 'c φc b
= 102.048 mm
If a ≤ ab, the area of tension reinforcement is then given by:
As =
Mf
a
φs f y d −
2
= 2466 mm2
CSA A23.3-04 Example 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
4
4
2
=
=
=
As ,min min
As ,min , As ,required min
758.95, 2466 758.95 mm
3
3
Shear Design
The basic shear strength for rectangular section is computed as,
φc = 0.65 for shear
λ = {1.00, for normal density concrete
d v is the effective shear depth. It is taken as the greater of
0.72h = 432 mm or 0.9d = 491.4 mm (governing).
β = 0.18 since minimum transverse reinforcement is provided
∙
V f = 92.222 (3 - 0.546) = 226.31 kN
Vc = φc λβ f ′c bw dv = 145.45 kN
Vr ,max = 0.25φc f 'c bw d = 1419.60 kN
θ = 35º since f y ≤ 400 MPa and f 'c ≤ 60 MPa
Av (V f − Vc ) tan θ
= 0.339 mm2/mm
=
s
φ s f yt d v
fc
Av
b = 0.379 mm2/mm (Govern)
= 0.06
fy
s min
'
CSA A23.3-04 Example 001 - 4
Software Verification
ETABS
4
PROGRAM NAME:
REVISION NO.:
CSA A23.3-04 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load N
= 2098 kN and moment My = 525 kN-m. This column is reinforced with 5 T25
bars. The design capacity ratio is checked by hand calculations and results are
compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
2098 kN
My= 525 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
f’c = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
CSA A23.3-04 Example 002 - 1
Software Verification
ETABS
4
PROGRAM NAME:
REVISION NO.:
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.9869
1.00
-1.31%
COMPUTER FILE: CSA A23.3-04 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
CSA A23.3-04 Example 002 - 2
Software Verification
ETABS
4
PROGRAM NAME:
REVISION NO.:
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
700
700
dt =
( 490) = 296 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
Pr = Cc + Cs − T
where
Cc = φc • α1 f c' ab = 0.65 • 0.805 • 30 • 350a = 5494.1a
(
)
Cs = φs As' f y - 0.805 f c' = 0.85 • 2500 ( 460 - 0.805 • 30 ) = 926,181 N
Assume compression steels yields, (this assumption will be checked later).
T = φs As f s = 0.85 • 2500 f s =
2125 f s ( f s < f y )
Pr = 5, 494.1a + 926,181 - 2125 f s
(Eqn. 1)
3) Taking moments about As:
a
'
Cc d - 2 + C s d - d
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
Pr =
5, 494.1a 490 - + 926,181 ( 490 - 60 )
465
2
2
Pr = 5789.5a - 5.91a + 856, 468.5
Pr =
1
e'
(
)
(Eqn. 2)
CSA A23.3-04 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
4) Assume c = 355 mm, which exceed cb (296 mm).
a = 0.895 • 355 = 317.7 mm
Substitute in Eqn. 2:
Pr = 5789.5 • 317.7 - 5.91 ( 317.7 ) + 856, 468.5 = 2,099,327.8 N
2
5) Calculate fs from the strain diagram when c = 350 mm.
490 - 355
fs =
700 = 266.2 MPa
355
εs = εt = f s Es = 0.0013
6) Substitute a = 317.7 mm and fs = 266.2 MPa in Eqn. 1 to calculate Pr2:
Pr2 = 5, 494.1 ( 317.7 ) + 926,181 - 2125 ( 266.2 ) = 2,106,124.9 N
Which is very close to the calculated Pr1 of 2,012,589.8 (less than 1% difference)
250
M r = Pr e = 2100
= 525 kN-m
1000
7) Check if compression steels yield. From strain diagram,
355 - 60
ε s' =
( 0.0035) = 0.00291 > ε y = 0.0023
355
Compression steels yields, as assumed.
8) Therefore, section capacity is
Pr = 2098 kN
M r = 525 kN-m
CSA A23.3-04 Example 002 - 4
Software Verification
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CSA A23.3-14 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simply supported beam is subjected to a uniform factored load
of 92.222 kN/m. This example is tested using the CSA A23.3-14 concrete design
code. The flexural and shear reinforcing computed is compared with independent
results.
GEOMETRY, PROPERTIES AND LOADING
CL
400mm
A
600 mm
54 mm
A
Section A-A
6m
Material Properties
E=
25x106 kN/m2
ν=
0.2
G=
10416666.7kN/m2
Section Properties
d = 546 mm
W = 92.222 kN/m
Design Properties
f’c = 40 MPa
fy = 400 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
CSA A23.3-14 Example 001 - 1
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 2.2 on page 2-12 in Part II on Concrete Design
Handbook of Cement Association of Canada.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mf (kN-m)
415.00
415.00
0.00%
Tension Reinf, As (mm2)
2466
2466
0.00%
Design Shear, Vf (kN)
226.31
226.31
0.00%
Shear Reinf, Av/s (mm2/mm)
0.379
0.379
0.00%
COMPUTER FILE: CSA A23.3-14 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
CSA A23.3-14 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
φc = 0.65 for concrete
φs = 0.85 for reinforcement
As,min =
0.2 f ′c
b h = 758.95 mm2
fy
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.79
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.87
cb =
700
d = 347.45 mm
700 + f y
ab = β1cb = 302.285 mm
COMB1
wu l 2
= 415 kN-m
Mf =
8
The depth of the compression block is given by:
a = d − d2 −
2M f
α 1 f 'c φc b
= 102.048 mm
If a ≤ ab, the area of tension reinforcement is then given by:
As =
Mf
a
φs f y d −
2
= 2466 mm2
CSA A23.3-14 Example 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
4
4
2
=
=
=
As ,min min
As ,min , As ,required min
758.95, 2466 758.95 mm
3
3
Shear Design
The basic shear strength for rectangular section is computed as,
φc = 0.65 for shear
λ = {1.00, for normal density concrete
d v is the effective shear depth. It is taken as the greater of
0.72h = 432 mm or 0.9d = 491.4 mm (governing).
β = 0.18 since minimum transverse reinforcement is provided
∙
V f = 92.222 (3 - 0.546) = 226.31 kN
Vc = φc λβ f ′c bw dv = 145.45 kN
Vr ,max = 0.25φc f 'c bw d = 1419.60 kN
θ = 35º since f y ≤ 400 MPa and f 'c ≤ 60 MPa
Av (V f − Vc ) tan θ
= 0.339 mm2/mm
=
s
φ s f yt d v
fc
Av
b = 0.379 mm2/mm (Govern)
= 0.06
fy
s min
'
CSA A23.3-14 Example 001 - 4
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
CSA A23.3-14 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load N
= 2098 kN and moment My = 525 kN-m. This column is reinforced with 5 T25
bars. The design capacity ratio is checked by hand calculations and results are
compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
2098 kN
My= 525 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
f’c = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
CSA A23.3-14 Example 002 - 1
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.9869
1.00
-1.31%
COMPUTER FILE: CSA A23.3-14 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
CSA A23.3-14 Example 002 - 2
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
700
700
dt =
( 490) = 296 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
Pr = Cc + Cs − T
where
Cc = φc • α1 f c' ab = 0.65 • 0.805 • 30 • 350a = 5494.1a
(
)
Cs = φs As' f y - 0.805 f c' = 0.85 • 2500 ( 460 - 0.805 • 30 ) = 926,181 N
Assume compression steels yields, (this assumption will be checked later).
T = φs As f s = 0.85 • 2500 f s =
2125 f s ( f s < f y )
Pr = 5, 494.1a + 926,181 - 2125 f s
(Eqn. 1)
3) Taking moments about As:
a
'
Cc d - 2 + C s d - d
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
Pr =
5, 494.1a 490 - + 926,181 ( 490 - 60 )
465
2
2
Pr = 5789.5a - 5.91a + 856, 468.5
Pr =
1
e'
(
)
(Eqn. 2)
CSA A23.3-14 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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4) Assume c = 355 mm, which exceed cb (296 mm).
a = 0.895 • 355 = 317.7 mm
Substitute in Eqn. 2:
Pr = 5789.5 • 317.7 - 5.91 ( 317.7 ) + 856, 468.5 = 2,099,327.8 N
2
5) Calculate fs from the strain diagram when c = 350 mm.
490 - 355
fs =
700 = 266.2 MPa
355
εs = εt = f s Es = 0.0013
6) Substitute a = 317.7 mm and fs = 266.2 MPa in Eqn. 1 to calculate Pr2:
Pr2 = 5, 494.1 ( 317.7 ) + 926,181 - 2125 ( 266.2 ) = 2,106,124.9 N
Which is very close to the calculated Pr1 of 2,012,589.8 (less than 1% difference)
250
M r = Pr e = 2100
= 525 kN-m
1000
7) Check if compression steels yield. From strain diagram,
355 - 60
ε s' =
( 0.0035) = 0.00291 > ε y = 0.0023
355
Compression steels yields, as assumed.
8) Therefore, section capacity is
Pr = 2098 kN
M r = 525 kN-m
CSA A23.3-14 Example 002 - 4
Software Verification
ETABS
2
PROGRAM NAME:
REVISION NO.:
EN 2-2004 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simple supported beam is subjected to a uniform factored load
of 36.67 kN/m. This example is tested using the Eurocode concrete design code.
The flexural and shear reinforcing computed is compared with independent
results.
GEOMETRY, PROPERTIES AND LOADING
CL
230mm
A
550 mm
60 mm
A
Section A-A
6m
Material Properties
E=
25x106 kN/m2
ν=
0.2
G=
10416666.7kN/m2
Section Properties
d = 490 mm
b = 230 mm
Design Properties
fck = 30 MPa
fyk = 460 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
EN 2-2004 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
2
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution.
Country
γc
γs
α cc
k1
k2
k3
k4
CEN Default, Slovenia, Sweden, Portugal
1.5
1.15
1.0
0.44
1.25
0.54
1.25
UK
1.5
1.15
0.85
0.40
1.25
0.40
1.25
Norway
1.5
1.15
0.85
0.44
1.25
0.54
1.25
Singapore
1.5
1.15
0.85
0.40
1.25
0.54
1.25
Finland
1.5
1.15
0.85
0.44
1.10
0.54
1.25
Denmark
1.45
1.2
1.0
0.44
1.25
0.54
1.25
Germany
1.5
1.15
0.85
0.64
0.80
0.72
0.80
Poland
1.4
1.15
1.0
0.44
1.25
0.54
1.25
EN 2-2004 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
Country
Design
Moment,
MEd (kN-m)
Tension
Reinforcing,
As+ (sq-mm)
Design Shear,
VEd
(kN)
ETABS
2
Shear
Reinforcing,
Asw/s (sqmm/m)
% diff.
Method
ETABS
Hand
ETABS
Hand
ETABS
Hand
ETABS
Hand
0.00%
CEN Default,
Slovenia,
Sweden, Portugal
165
165
916
916
110
110
249.5
249.5
0.00%
UK
165
165
933
933
110
110
249.5
249.5
0.00%
Norway
165
165
933
933
110
110
249.5
249.5
0.00%
Singapore
165
165
933
933
110
110
249.5
249.5
0.00%
Finland
165
165
933
933
110
110
249.5
249.5
0.00%
Denmark
165
165
950
950
110
110
249.5
249.5
0.00%
Germany
165
165
933
933
110
110
249.5
249.5
0.00%
Poland
165
165
925
925
110
110
249.5
249.5
0.00%
COMPUTER FILE: EN 2-2004 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
EN 2-2004 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
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HAND CALCULATION
Flexural Design
The following quantities are computed for both of the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
α cc = 1.0
k1 = 0.44
k2 =
k4 =
1.25 ( 0.6 + 0.0014 / ε cu 2 ) =
1.25 k3 = 0.54
f cd = α cc f ck / γ c = 1.0(30)/1.5 = 20 MPa
=
f yd f yk / γ s = 460/1.15 = 400 Mpa
f=
f yk / γ s = 460/1.15 = 400 Mpa
ywd
η = 1.0 for fck ≤ 50 MPa
λ = 0.8 for fck ≤ 50 MPa
As ,min = 0.26
f ctm
bd = 184.5 sq-mm,
f yk
2/3
= 0.3(30)2/3 = 2.896 N/sq-mm
where f ctm = 0.3 f cwk
As,min = 0.0013bh = 164.5 sq-mm
COMB1
The factored design load and moment are given as,
wu = 36.67 kN/m
M =
∙
wu l 2
= 36.67 62/8 = 165.0 kN-m
8
EN 2-2004 Example 001 - 4
Software Verification
PROGRAM NAME:
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ETABS
2
The limiting value of the ratio of the neutral axis depth at the ultimate limit state
to the effective depth, ( x / d )lim , is given as,
δ − k1
x
for fck ≤ 50 MPa ,
=
k2
d lim
where δ = 1 , assuming no moment redistribution
δ − k1
x
=
=
k2
d lim
(1 − 0.44
)
=
1.25
0.448
The normalized section capacity as a singly reinforced beam is given as,
x λx
mlim = λ 1 − = 0.29417
d lim 2 d lim
The limiting normalized steel ratio is given as,
x
= 1 − 1 − 2mlim = 0.3584
d lim
ωlim = λ
The normalized moment, m, is given as,
165 • 106
=
=0.1494 < mlim so a singly reinforced
m= 2
bd η f cd
230 • 4902 • 1.0 • 20
beam will be adequate.
M
ω = 1 − 1 − 2m = 0.16263 < ωlim
η f bd
1.0 • 20 • 230 • 490
As = ω cd = 0.1626
=916 sq-mm
400
f yd
Shear Design
The shear force demand is given as,
=
VEd ω=
L / 2 110.01 kN
The shear force that can be carried without requiring design shear reinforcement,
EN 2-2004 Example 001 - 5
Software Verification
PROGRAM NAME:
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ETABS
2
1/3
=
VRd ,c CRd ,c k (100 ρ1 f ck ) + k1σ cp bw d
1/3
VRd ,c= 0.12 • 1.6389 (100 • 0.0 • 30 ) + 0.0 230 • 490= 0 kN
with a minimum of:
vmin + k1σ cp bd = [ 0.4022 + 0.0] 230 • 490 =
45.3 kN
V=
Rd , c
where,
k=
1+
=
ρ1
200
≤ 2.0 = 1.6389
d
AS
0
=
= 0.0 ≤ 0.02
bd 230490
As = 0 for ρl at the end of a simply-supported beam as it taken as the tensile
reinforcement at the location offset by d+ldb beyond the point considered.
(EN 1992-1-1:2004 6.2.2(1) Figure 6.3)
=
σ cp
N Ed
= 0.0
Ac
CRd ,c = 0.18 / γ c =0.12
vmin 0.035
k 3/2 fck 1/2 0.4022
=
=
The maximum design shear force that can be carried without crushing of the
notional concrete compressive struts,
=
VRd ,max α cwbzv1 f cd / ( cot θ + tan θ )
where,
α cw = 1.0
=
z 0.9
=
d 441.0 mm
f
v1 = 0.6 1 − ck = 0.528
250
EN 2-2004 Example 001 - 6
Software Verification
PROGRAM NAME:
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ETABS
2
vEd
−1
=
θ 0.5sin
=
5.33
0.2 f ck (1 − f ck / 250 )
where,
=
vEd
VEd
= 0.9761
bw d
21.8 ≤ θ ≤ 45 , therefore use θ = 21.8
=
VRd ,max α cwbzv1 f cd / ( cot θ
=
+ tan θ ) 369 kN
VRd ,max > VEd , so there is no concrete crushing.
The required shear reinforcing is,
Asw
VEd
110.01 • 1e6
=
=
= 249.5 sq-mm/m
s
zf ywd cot θ 441 • 460 • 2.5
1.15
EN 2-2004 Example 001 - 7
Software Verification
PROGRAM NAME:
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ETABS
0
EN 2-2004 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected to factored axial load
N = 2374 kN and moment My = 593 kN-m. This column is reinforced with five
25 bars. The design capacity ratio is checked by hand calculations and the result
is compared with computed results. The column is designed as a short, non-sway
member.
GEOMETRY, PROPERTIES AND LOADING
2374 kN
My= 593 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fck = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
EN 2-2004 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.009
1.00
0.90%
COMPUTER FILE: EN 2-2004 EX002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
EN 2-2004 Example 002 - 2
Software Verification
PROGRAM NAME:
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ETABS
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HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fy = 460 MPa
fck = 30 MPa
b = 350 mm
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
cb =
700
700
dt =
( 490 ) = 312 mm
700 + f y /γ s
700 + 460 / 1.15
2) From the equation of equilibrium:
N = Cc + C s − T
where
f
30
Cc =
α cc ck ab =
1.0
• 350a =
7000a
γc
1.5
=
Cs
As′
fck 2500
30
f y − α cc=
460 − 1.0 •=
956,521.7 N
1.5
γs
γ c 1.15
Assume compression steel yields (this assumption will be checked later).
As fs 2500 fs
=
= 2174 fs ( fs < f y )
γs
1.15
N1 =
7, 000a + 956,521.7 − 2174 fs
=
T
(Eqn. 1)
3) Taking moments about As:
1
a
=
N2
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
=
N2
1
a
7, 000a 490 − + 956,521.7 ( 490 − 60 )
465
2
N 2 = 7376.3a - 7.527a 2 + 884,525.5
(Eqn. 2)
EN 2-2004 Example 002 - 3
Software Verification
PROGRAM NAME:
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ETABS
0
4) Assume c = 356 mm, which exceed cb (312 mm).
=
a 0.8
=
• 356 284.8 mm
Substitute in Eqn. 2:
N=
7376.3• 284.8 − 7.527 ( 284.8 ) + 884,525.5
= 2,374,173 N
2
2
5) Calculate fs from the strain diagram when c = 356 mm.
490 − 356
=
fs =
700 263.4 MPa
356
ε s =εt =fs Es = 0.00114
6) Substitute a = 284.8 mm and fs = 263.4 MPa in Eqn. 1 to calculate N1:
=
=
N1 7, 000 ( 284.8 ) + 956,522 − 2174 ( 263.5
) 2,377, 273 N
which is very close to the calculated N2 of 2,374,173 (less than 1% difference)
250
M
= Ne
= 2374
=
593.5 kN-m
1000
7) Check if compression steels yield. From strain diagram,
356 − 60
=
ε′s
=
> ε y 0.0023
) 0.0029=
( 0.0035
356
Compression steel yields, as assumed.
8) Therefore, section capacity is
N = 2,374 kN
M = 593 kN-m
EN 2-2004 Example 002 - 4
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PROGRAM NAME:
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HK CP-2004 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected to factored axial load
N = 1971 kN and moment My = 493 kN-m. This column is reinforced with five
25M bars. The design capacity ratio is checked by hand calculations and the
result is compared with the computed results. The column is designed as a short,
non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1971 kN
My= 493 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fcu = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
HK CP-2004 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.994
1.00
0.60%
COMPUTER FILE: HK CP-2004 EX002
CONCLUSION
The computed result shows an acceptable comparison with the independent
result.
HK CP-2004 Example 002 - 2
Software Verification
PROGRAM NAME:
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HAND CALCULATION
Column Strength under compression control
fy = 460 MPa
fcu = 30 MPa
b = 350 mm
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
=
cb
700
700
=
dt
=
( 490 ) 312 mm
700 + f y / γ s
700 + 460 /1.15
2) From the equation of equilibrium:
N = Cc + C s − T
where
0.67
fcu ab 0.67 1.5 •=
30 • 350a 4667 a
=
γΜ
As′
2500
971, 014 N
Cs =
f y − 0.4467 fcu ) = ( 460 − 0.4467 • 30 ) =
(
1.15
γs
Assume compression steels yields, (this assumption will be checked later).
As fs 2500 fs
=
T =
= 2174 fs ( fs < f y )
γs
1.15
N1 =4, 667 a + 971, 014 − 2174 fs
Cc
=
(Eqn. 1)
3) Taking moments about As:
=
N
1
a
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
=
N
1
a
4, 667 a 490 − + 971, 014 ( 490 − 60 )
465
2
N = 4917.9a − 5.018a 2 + 897,926
(Eqn. 2)
HK CP-2004 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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4) Assume c = 364 mm, which exceed cb (312 mm).
=
a 0.9
=
• 364 327.6 mm
Substitute in Eqn. 2:
N 2 4917.9 • 327.6 − 5.018 ( 327.6 ) + 897,926
=
= 1,970,500 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 − 364
=
fs =
700 242.3 MPa
364
ε s =εt =fs Es = 0.0012
6) Substitute a = 327.6 mm and fs = 242.3 MPa in Eqn. 1 to calculate N1:
=
N1 4, 667 ( 327.6 ) + 971, 014 − 2174 ( 242.3
=
) 1,973,163 N
which is very close to the calculated N2 of 1,970,500 (less than 1% difference)
250
M = Ne = 1971
= 493 kN-m
1000
7) Check if compression steel yields. From strain diagram,
365 − 60
=
ε′s
=
=
> ε y 0.0023
) 0.00292
( 0.0035
365
Compression steel yields, as assumed.
8) Therefore, the section capacity is
N = 1971 kN
M = 493 kN-m
HK CP-2004 Example 002 - 4
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
IS 456-2000 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simply supported beam is subjected to a uniform factored load
of 37.778 kN/m. This example is tested using the IS 456-2000 concrete design
code. The flexural and shear reinforcing computed is compared with independent
results.
GEOMETRY, PROPERTIES AND LOADING
CL
300mm
A
600 mm
37.5 mm
A
Section A-A
6m
Material Properties
E=
19.365x106 kN/m2
ν=
0.2
G=
8068715.3kN/m2
Section Properties
d = 562.5 mm
w = 37.778 kN/m
Design Properties
fck = 15 MPa
fy = 415 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
IS 456-2000 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
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RESULTS COMPARISON
The example problem is same as Example-1 given in SP-16 Design Aids for
Reinforced Concrete published by Bureau of Indian Standards. For this example
a direct comparison for flexural steel only is possible as corresponding data for
shear steel reinforcement is not available in the reference for this problem.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (kN-m)
170.00
170.00
0.00%
Tension Reinf, As (mm2)
1006
1006
0.00%
Design Shear, Vu (kN)
113.33
113.33
0.00%
Shear Reinf, Asv/s (mm2/mm)
0.333
0.333
0.00%
COMPUTER FILE: IS 456-2000 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
IS 456-2000 Example 001 - 2
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
α = 0.36
β = 0.42
As ,min ≥
0.85
bd = 345.63 sq-mm
fy
COMB1
Mu = 170 kN-m
Vu = 113.33 kN-m
xu ,max
d
Xu , max
d
0.53
0.53 − 0.05 f y − 250
165
=
f
0.48 − 0.02 y − 415
85
0.46
if
f y ≤ 250 MPa
if 250 < f y ≤ 415 MPa
if 415 < f y ≤ 500 MPa
if
f y ≥ 500 MPa
= 0.48
The normalized design moment, m, is given by
m=
Mu
= 0.33166
bw d 2α f ck
Mw,single = αfckbwd2
x u,max
x u,max
1 − β
= 196.436 kN-m > Mu
d
d
IS 456-2000 Example 001 - 3
Software Verification
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PROGRAM NAME:
REVISION NO.:
So no compression reinforcement is needed
= 0.3983
xu 1 − 1 − 4 β m
=
d
2β
x
z = d 1 − β u = 562.5{1 − 0.42 • 0.3983}= 468.406
d
M
u
= 1006 sq-mm
As =
( fy γ s ) z
Shear Design
τv =
Vu
= 0.67161
bd
τmax = 2.5 for M15 concrete
k = 1.0
=
δ 1
0.15 ≤
if Pu ≤ 0 , Under Tension
100 As
≤3
bd
100 As
= 0.596
bd
τ c = 0.49 From Table 19 of IS 456:2000 code, interpolating between rows.
τcd = kδτc = 0.49
The required shear reinforcement is calculated as follows:
Since τv > τcd
Asv
0.4b (τ v − τ cd ) b
0.4 • 300 ( 0.67161 − 0.49 ) • 300
max
,
,
=
=
max
s
415
415 1.15 )
1.15
f
f
γ
γ
(
)
(
(
)
(
)
y
y
y
Asv
mm 2
=
max
=
{0.333,0.150} 0.333
s
mm
IS 456-2000 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
IS 456-2000 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load N
= 1913 kN and moment My= 478 kN-m. This column is reinforced with 5 25M
bars. The design capacity ratio is checked by hand calculations and computed
result is compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1913 kN
My= 478 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
f’c = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
IS 456-2000 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.997
1.00
0.30%
COMPUTER FILE: IS 456-2000 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results. The larger variation is due to equivalent rectangular compression block
assumption.
IS 456-2000 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
700
700
dt =
( 490) = 296 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
N = Cc + C s − T
where
0.36
Cc =
f ck ab = 0.4286 • 30 • 350a = 4500a
0.84
A'
2500
Cs = s f y - 0.4286 f ck =
( 460 - 0.4286 • 30 ) = 972, 048 N
γs
1.15
Assume compression steels yields, (this assumption will be checked later).
Af
2500 f s
= 2174 f s ( f s < f y )
T= s s=
γs
1.15
(Eqn. 1)
N1 = 4500a + 972, 048 - 2174 f s
(
)
3) Taking moments about As:
1
a
N 2 = ' Cc d - + Cs d - d '
e
2
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
N2 =
4500a 490 - + 972, 048 ( 490 - 60 )
465
2
2
N 2 = 4742a - 4.839a + 898,883
(
)
(Eqn. 2)
IS 456-2000 Example 002 - 3
Software Verification
PROGRAM NAME:
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ETABS
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4) Assume c = 374 mm, which exceed cb (296 mm).
a = 0.84 • 374 = 314.2 mm
Substitute in Eqn. 2:
N 2 = 4742 • 314.2 - 4.039 ( 314.2 ) + 898,883 = 1,911, 037 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 - 374
fs =
700 = 217.11 MPa
374
εs = εt = f s Es = 0.0011
6) Substitute a = 314.2 mm and fs = 217.11 MPa in Eqn. 1 to calculate N1:
N1 = 4500 ( 314.2 ) + 972, 048 - 2174 ( 217.4 ) = 1,913, 765 N
Which is very close to the calculated N2 of 1,911,037 (less than 1% difference)
250
M = Ne = 1913
= 478 kN-m
1000
7) Check if compression steels yield. From strain diagram,
374 - 60
ε s' =
( 0.0035 ) = 0.0029 > ε y = 0.0023
374
Compression steels yields, as assumed.
8) Therefore, section capacity is
N = 1913 kN
M = 478 kN-m
IS 456-2000 Example 002 - 4
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NTC 2008 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
In the example a simple supported beam is subjected to a uniform factored load
of 36.67 kN/m. This example is tested using the Italian NTC 2008 concrete
design code. The flexural and shear reinforcing computed is compared with
independent results.
GEOMETRY, PROPERTIES AND LOADING
CL
230mm
A
550 mm
60 mm
A
Section A-A
6m
Material Properties
E=
25x106 kN/m2
ν=
0.2
G=
10416666.7kN/m2
Section Properties
d = 490 mm
b = 230 mm
Design Properties
fck = 30 MPa
fyk = 460 MPa
TECHNICAL FEATURES TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
NTC 2008 Example 001 - 1
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RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, MEd (kNm)
165.00
165.00
0.00%
Tension Reinf, As (mm2)
933
933
0.00%
Design Shear, VEd (kN)
110.0
110.0
0.00%
Shear Reinf, Asw/s (mm2/m)
345.0
345.0
0.00%
COMPUTER FILE: NTC 2008 Ex001
CONCLUSION
The computed results show an exact match with the independent results.
NTC 2008 Example 001 - 2
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HAND CALCULATION
Flexural Design
The following quantities are computed for both of the load combinations:
γc, concrete = 1.50
α cc = 0.85
k1 = 0.44
k2 =
k4 =
1.25 ( 0.6 + 0.0014 / ε cu 2 ) =
1.25 k3 = 0.54
f cd = α cc f ck / γ c = 0.85(30)/1.5 = 17 MPa
f yd =
f y 460
= 400 Mpa
γ s 1.15
η = 1.0 for fck ≤ 50 MPa
λ = 0.8 for fck ≤ 50 MPa
As ,min = 0.26
f ctm
bd = 184.5 sq-mm,
f yk
2/3
= 0.3(30)2/3 = 2.896 N/sq-mm
where f ctm = 0.3 f cwk
As,min = 0.0013bh = 164.5 sq-mm
COMB1
The factored design load and moment are given as,
wu = 36.67 kN/m
M =
∙
wu l 2
= 36.67 62/8 = 165.0 kN-m
8
The limiting value of the ratio of the neutral axis depth at the ultimate limit state
to the effective depth, ( x / d )lim , is given as,
NTC 2008 Example 001 - 3
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δ − k1
x
for fck ≤ 50 MPa ,
=
k2
d lim
where δ = 1 , assuming no moment redistribution
δ − k1
x
=
=
k2
d lim
(1 − 0.44
)
=
1.25
0.448
The normalized section capacity as a singly reinforced beam is given as,
x λx
mlim = λ 1 − = 0.29417
d lim 2 d lim
The limiting normalized steel ratio is given as,
x
= 1 − 1 − 2mlim = 0.3584
d lim
ωlim = λ
The normalized moment, m, is given as,
165 • 106
M
=
=0.1758 < mlim so a singly reinforced beam
230 • 4902 • 17
bd 2 f cd
will be adequate.
m=
ω = 1 − 1 − 2m = 0.1947 < ωlim
f bd
17 • 230 • 490
As = ω cd = 0.1947
=933 sq-mm
400
f yd
Shear Design
The shear force demand is given as,
=
VEd ω=
L / 2 110.0 kN
The shear force that can be carried without requiring design shear reinforcement,
1/3
=
VRd ,c CRd ,c k (100 ρ1 f ck ) + k1σ cp bw d
1/3
VRd ,c= 0.12 • 1.6389 (100 • 0.0 • 30 ) + 0.0 230 • 490= 0 kN
NTC 2008 Example 001 - 4
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with a minimum of:
vmin + k1σ cp bd = [ 0.4022 + 0.0] 230 x 490 =
45.3 kN
V=
Rd , c
where,
k=
1+
=
ρ1
200
≤ 2.0 = 1.6389
d
AS
0
=
= 0.0 ≤ 0.02
bd 230490
As = 0 for ρl at the end of a simply-supported beam as it taken as the tensile
reinforcement at the location offset by d+ldb beyond the point considered.
(EN 1992-1-1:2004 6.2.2(1) Figure 6.3)
=
σ cp
N Ed
= 0.0
Ac
CRd ,c = 0.18 / γ c =0.12
=
vmin 0.035
=
k 3/2 fck 1/2 0.4022
The maximum design shear force that can be carried without crushing of the
notional concrete compressive struts,
cot α + cot ϑ
=
=
α c f 'cd
VRd ,max zb
297 kN
2
1 + cot ϑ
where,
=
z 0.9
=
d 441.0 mm
α c = 1.0 since there is no axial compression
f 'cd = 0.5 f cd
α = 900 for vertical stirrups
vEd
−1
=
ϑ 0.5sin
=
5.33
0.2 f ck (1 − f ck / 250 )
where,
NTC 2008 Example 001 - 5
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=
vEd
ETABS
0
VEd
= 0.9761
bw d
21.8 ≤ ϑ ≤ 45 , therefore use ϑ = 21.8
The required shear reinforcing is,
Asw VEd
mm 2
1
110.0106
=
= = 249.4
s
zf ywd ( cot α + cot ϑ ) sin α 441 460 2.5
m
1.15
The minimum required shear reinforcing is,
mm 2
Asw
(controls)
=
=
=
b
1.5
1.5
230
345.0
m
s min
NTC 2008 Example 001 - 6
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NTC 2008 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load N
= 2174 kN and moment My = 544 kN-m. This column is reinforced with 5-25
bars. The design capacity ratio is checked by hand calculations and computed
result is compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
2174 kN
My= 544 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fck = 25 MPa
fy = 460 MPa
350 m
490 mm
NTC 2008 Example 002 - 1
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TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.092
1.00
9.20%
COMPUTER FILE: EN 2-2004 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
NTC 2008 Example 002 - 2
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HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fcu = 25 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3) d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis for a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
700
700
dt =
( 490) = 296 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
N = Cc + C s − T
where
α f ck
0.85 • 30
• 350a = 5950a
Cc =
ab =
γc
1.5
As'
α f ck 2500
0.85 • 30
460 = 963, 043 N
Cs =
fy =
1.5
γs
γ c 1.15
Assume compression steels yields, (this assumption will be checked later).
Af
2500 f s
T= s s=
= 2174 f s ( f s < f y )
γs
1.15
(Eqn. 1)
N1 = 5,950a + 963, 043 - 2174 f s
3) Taking moments about As:
1
a
N 2 = ' Cc d - + Cs d - d '
e
2
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
N2 =
5950a 490 - + 963, 043 ( 490 - 60 )
465
2
(
)
NTC 2008 Example 002 - 3
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N 2 = 6270a - 6.3978a 2 + 890,556
ETABS
2
(Eqn. 2)
4) Assume c = 365 mm, which exceed cb (296 mm).
a = 0.8 • 365 = 292 mm
Substitute in Eqn. 2:
N 2 = 6270 • 292 - 6.3978 ( 292 ) + 890,556 = 2,175,893 N
2
5) Calculate fs from the strain diagram when c = 356 mm.
490 - 365
fs =
700 = 240.0 MPa
365
εs = εt = f s Es = 0.0012
6) Substitute a = 284.8 mm and fs = 263.4 MPa in Eqn. 1 to calculate N1:
N1 = 5950 ( 292 ) + 963, 043 - 2174 ( 240.0 ) = 2,178, 683 N
Which is very close to the calculated N2 of 2,175,893 (less than 1% difference)
250
M = Ne = 2175
= 544 kN-m
1000
7) Check if compression steels yield. From strain diagram,
365 - 60
ε s' =
( 0.0035 ) = 0.0029 > ε y = 0.0023
365
Compression steels yields, as assumed.
8) Therefore, section capacity is
N = 2,174 kN
M = 544 kN-m
NTC 2008 Example 002 - 4
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KBC 2009 Example 001
Shear and Flexural Reinforcement Design of a Singly Reinforced Rectangle
PROBLEM DESCRIPTION
The purpose of this example is to verify the flexural and shear design. A simplespan, 6-m-long, 300-mm-wide, and 560-mm-deep beam is modeled. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated. The maximum element size has been
specified to be 200 mm. The beam is supported by joint restraints that have no
rotational stiffness. One end of the beam has no longitudinal stiffness.
The beam is loaded with symmetric third-point loading. One dead load case
(DL50) and one live load case (LL130) with only symmetric third-point loads of
magnitudes 50, and 130 kN, respectively, are defined in the model. One load
combination (COMB130) is defined using the KBC 2009 load combination
factors of 1.2 for dead loads and 1.5 for live loads. The model is analyzed for
both of those load cases and the load combinations.
Table 1 shows the comparison of the design longitudinal reinforcements. Table 2
shows the comparison of the design shear reinforcements.
GEOMETRY, PROPERTIES AND LOADING
Clear span,
L
=
Overall depth,
h
=
Width of beam,
b
=
Effective depth,
d
=
Depth of comp. reinf.,
d'
=
Concrete strength,
fck =
Yield strength of steel,
fy
=
Concrete unit weight,
wc =
Modulus of elasticity,
Ec =
Modulus of elasticity,
Es =
Poisson’s ratio,
v
=
Dead load,
Live load,
Pd
Pl
=
=
6000
560
300
500
60
30
460
0
25x105
2x105
0.2
50
130
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
KBC 2009 Example 001 - 1
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Figure 1 The Model Beam for Flexural and Shear Design
KBC 2009 Example 001 - 2
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TECHNICAL FEATURES TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the total factored moments in the design strip
with the moments obtained using the analytical method. They match exactly for
this problem. Table 1 also shows the comparison of design reinforcements.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-mm)
Method
Moment
(kN-m)
As+
As-
ETABS
360
2109
0
Calculated
360
2109
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-mm/m)
Shear Force (kN)
ETABS
Calculated
180
515.3
515.4
COMPUTER FILE: KBC 2009 EX001
CONCLUSION
The computed results show an exact comparison with the independent results.
KBC 2009 Example 001 - 3
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HAND CALCULATION
Flexural Design
The following quantities are computed for the load combination:
φb = 0.85
=
β1 0.85 −.007(30 − 28)
= 0.836 for =
f ck 30MPa,
cmax =
εc
ε c + f y Es
d = 187.5 mm
amax = β1cmax= 156.75 mm
=
Ac b=
d 150, 000 mm 2
As ,min
0.25 f ck
Ac = 446.5
fy
mm2
= max
1.4 Ac = 456.5
fy
= 456.5 mm2
COMB130
Vu = (1.0Pd + 1.0Pl) = 180 kN – Loads were Ultimate
Mu =
Vu L
= 360 kN-m
3
The depth of the compression block is given by:
a =−
d
d2 −
2 Mu
0.85 f ck φbb
= 26.81 mm ; a < amax
Since a < amax , compression reinforcing is NOT required.
KBC 2009 Example 001 - 4
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The required tension reinforcing is:
Mu
2108.9 mm 2
=
a
f y d − φb
2
=
As
Shear Design
The following quantities are computed for all of the load combinations:
φ
=
0.75
The concrete limit is:
f ck = 5.48 MPa < 8.4 MPa
The concrete shear capacity is given by:
φVc =
1/6φ f ck bd
= 102.69 kN
The maximum shear that can be carried by reinforcement is given by:
φVs= 0.25φ f ck bd = 154.05 kN
The following limits are required in the determination of the reinforcement:
φVc/2
= 51.35 kN
φVmax = φVc + φVs
= 256.75 kN
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ φ(Vc/2),
Av
= 0,
s
else if φ(Vc/2) < Vu ≤ φVmax
Av
(V − φVc ) Av
= u
≥
φ f ys d
s
s min
where:
b
Av
w
= max 3.5
s min
f y
bw
,
fy
0.2 f ck
KBC 2009 Example 001 - 5
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else if Vu > φVmax,
a failure condition is declared.
Combo1
Vu = 180 kN
φ (Vc /=
2 ) 51.35 kN ≤ =
Vu 180 kN ≤ φ Vmax
= 256.75 kN
300 0.2 30
Av
=
max
3.5
,
300
s min
420 420
mm 2
Av
=
max
=
2.5,
0.78
0.0083
{
}
mm
s min
Av
(Vu − φVc )
mm 2
mm 2
= = 0.5154
= 515.4
φ f yd
mm
m
s
KBC 2009 Example 001 - 6
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KBC 2009 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load Pu
= 1879 kN and moment Mu = 470 kN-m. This column is reinforced with 5 T25
bars. The design capacity ratio is checked by hand calculations and computed
result is compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1879 kN
My= 470 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fck = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
KBC 2009 Example 002 - 1
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RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
1.003
1.00
0.30%
COMPUTER FILE: KBC 2009 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
KBC 2009 Example 002 - 2
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HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fck = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cmax =
0.003
0.003
d=
( 490 ) = 183.75 mm
0.003 + 0.005
0.003 + 0.005
2) From the equation of equilibrium:
Pu = Cc + Cs − T
where
Cc = 0.85 f ck ab = 0.85 • 30 • 350a = 8925a
(
)
Cs = As' f y - 0.85 f ck = 2500 ( 460 - 0.85 • 30 ) = 1, 086, 250 N
Assume compression steels yields, (this assumption will be checked later).
T = As f s = 2500 f s ( f s < f y )
Pu = 8,925a +1, 086, 250 - 2500 f s
(Eqn. 1)
3) Taking moments about As:
1
a
C d - + Cs d - d '
' c
e
2
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
Pu =
8,925a 490 - + 1, 086, 250 ( 490 - 60 )
465
2
2
Pu = 9, 404.8a - 9.6a +1, 004, 489.2
Pu =
(
)
(Eqn. 2)
KBC 2009 Example 002 - 3
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4) Assume c = 335 mm, which exceed cmax (183.75 mm).
=
β1 0.85 −.007(30 − 28)
= 0.836 for =
f ck 30MPa,
a = 0.836 • 335 = 280 mm
Substitute in Eqn. 2:
Pu = 9, 404.8 • 280 - 9.6 ( 280 ) +1, 004, 489.2 = 2,885,193.2 N
2
5) Calculate fs from the strain diagram when c = 335 mm.
490 - 335
fs =
600 = 277.8 MPa
335
εs = εt = f s Es = 0.00138
6) Substitute a = 280 mm and fs = 277.7 MPa in Eqn. 1 to calculate Pu2 :
Pu2 = 8,925 ( 280 ) +1, 086, 250 - 2500 ( 277.8 ) = 2,890, 750 N
Which is very close to the calculated Pu1 of 2,885,193.2 (less than 1% difference)
250
M u = Pu e = 2890
= 722.5 kN-m
1000
7) Check if compression steels yield. From strain diagram,
335 - 60
ε s' =
( 0.003) = 0.00263 > ε y = 0.0023
335
Compression steels yields, as assumed.
8) Therefore, section capacity is
Pu = 0.65 • 2890 = 21879 kN
M u = 0.65 • 722.5 = 470 kN-m
KBC 2009 Example 002 - 4
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RCDF 2004 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
In the example a simple supported beam is subjected to a uniform factored load
of 6.58 Ton/m (64.528 kN/m). This example was tested using the Mexican
RCDF 2004 concrete design code. The computed moment and shear strengths are
compared with independent hand calculated results.
GEOMETRY, PROPERTIES AND LOADING
CL
W Ton/m
A
b
h
r
A
L
L=6m
Material Properties
E=
ν=
G=
1979899 kg/cm2
0.2
824958 kg/cm2
Section Properties
h
r
b
W
= 0.65 m
= 0.05 m
= 0.30 m
= 6.58 Ton/m
(64.528 kN/m)
Design Properties
f’c = 200 kg/cm2 (19.6133 MPa)
fy = 4200 kg/cm2 (411.88 MPa)
RCDF 2004 Example 001 - 1
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TECHNICAL FEATURES TESTED
Design moment calculation, M and factored moment resistance, Mu.
Minimum reinforcement calculation, As
Design Shear Strength, V, and factored shear strength, Vu
RESULTS COMPARISON
Independent results are hand calculated based on the equivalent rectangular stress
distribution described in Example 5.2 on page 92 of “Aspectos Fundamentales
del Concreto Reforzado” Fourth Edition by Óscar M. González Cuevas and
Francisco Robles Fernández-Villegas.
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment (kN-m)
290.38
290.38
0%
As (mm2)
1498
1498
0%
Design Shear (kN)
154.9
154.9
0%
Av/s (mm2/m)
563
563
0%
COMPUTER FILE: RCDF 2004 EX001
CONCLUSION
The computed results show an acceptable comparison with the independent
results for bending and an acceptable-conservative comparison for shear.
RCDF 2004 Example 001 - 2
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GEOMETRY AND PROPERTIES
Clear span,
Overall depth,
Width of beam,
Effective depth,
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
L
h
b
d
f’c
fy
wc
Ec
Es
v
=
6
=
650
=
300
=
600
=
19.61
= 411.88
=
0
= 20.6x103
= 20.0x104
=
0.2
ETABS
0
m
mm
mm
mm
N/ mm2
N/ mm2
kN/m3
N/ mm2
N/ mm2
HAND CALCULATION
Flexural Design
The following quantities are computed for the load combination:
=
f c*
cb =
f 'c 19.61
=
= 15.69 MPa
1.25 1.25
ε c Es
ε c Es + f yd
d = 355.8 mm
amax = β1cb = 302.4 mm
where,
=
As ,min
f c*
=
β1 1.05 −
, 0.65 ≤ β1 ≤ 0.85
140
0.22 f 'c
=
bd 425.8 mm 2
fy
COMB1
ωu = 6.58 ton/cm (64.528kN/m)
RCDF 2004 Example 001 - 3
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Mu =
ωu l 2
8
ETABS
0
∙
= 64.528 6.02/8 = 290.376 kN-m
The depth of the compression block is given by:
a =−
d
d2 −
2 Mu
0.85 f c* FR b
(RCDF-NTC 2.1, 1.5.1.2)
= 154.2 mm
where FR = 0.9
Compression steel not required since a < amax.
The area of tensile steel reinforcement is given by:
As
=
Mu
290376000
=
= 1498 mm 2
a
0.9(411.88)
600
−
154.2
/
2
(
)
FR f y d −
2
Shear Design
The shear demand is computed as:
=
Vu ω ( L / 2 − d ) =15.79 ton (154.9 kN) at distance, d, from support for
this example
The shear force is limited to a maximum of,
(
)
Vmax
= VcR + 0.8 f c* Acv
The nominal shear strength provided by concrete is computed as:
=
VcR 0.3FRv ( 0.2 + 20 ρ ) f c* Acv = 0.3 • 0.8 ( 0.3665 ) 15.69 • 300 • 600
=43.553 kN
where FRv = 0.8
The shear reinforcement is computed as follows:
0.1 f c '
mm 2
Av
=
=
b 289
fy
m
s min
(RCDF-NTC 2.5.2.3, Eqn 2.22)
RCDF 2004 Example 001 - 4
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Av
=
s
Vu − FRvVcR )
(=
FRv f ys d
ETABS
0
154.9 − 0.8 • 43.553
mm 2
= 563
0.8 • 411.88 • 600
m
(RCDF-NTC 2.5.2.3, Eqn 2.23)
RCDF 2004 Example 001 - 5
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RCDF 2004 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected to factored axial load N = 1794 kN
and moment My = 448 kN-m. This column is reinforced with five 25M bars. The
design capacity ratio is checked by hand calculations and the result is compared
with a computed result. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1794 kN
My= 448 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fcu = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
RCDF 2004 Example 002 - 1
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PROGRAM NAME:
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0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: RCDF 2004 EX002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
RCDF 2004 Example 002 - 2
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HAND CALCULATION
Column Strength under compression control
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
=
cb
600
600
=
dt
=
( 490 ) 277 mm
600 + f y
600 + 460
2) From the equation of equilibrium:
N = Cc + C s − T
where
=
Cc 0.85
=
f *c ab 0.85 • 0.8 • =
30 • 350a 7140a
Cs =
As′ ( f y − 0.85 f *c ) =
2500 ( 460 − 0.85 • 0.8 • 30 ) =
1, 099, 000 N
Assume compression steels yields, (this assumption will be checked later).
=
T A=
2500 fs ( fs < f y )
s fs
N1 =
7140a + 1, 099, 000 − 2500 fs
(Eqn. 1)
3) Taking moments about As:
1
a
=
N
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
1
a
=
N
7140a 490 − + 1, 099, 000 ( 490 − 60 )
465
2
2
N 2 =7542a − 7.677 a + 1, 016, 280
(Eqn. 2)
RCDF 2004 Example 002 - 3
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4) Assume c = 347 mm, which exceeds cb (277 mm).
a = β1a = 0.836 • 347 = 290 mm
Substitute in Eqn. 2:
N 2 = 7542 • 290 − 7.677 ( 290 ) + 1, 016, 280 = 2,557,824 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 − 347
=
fs =
600 247.3 MPa
347
ε s =εt =fs Es = 0.0012
6) Substitute a = 290 mm and fs = 247.3 MPa in Eqn. 1 to calculate N1:
N1 = 7140 ( 290 ) + 1, 099, 000 − 2500 ( 247.3) = 2,551,350 N
which is very close to the calculated N2 of 2,557,824 (less than 1% difference)
250
M
= Ne
= 2552
=
638 kN-m
1000
7) Check if compression steel yields. From strain diagram,
347 - 60
ε s' =
( 0.003) = 0.0025 > ε y = 0.0023
347
Compression steel yields, as assumed.
8) Therefore, section capacity is
N = FR ( 2551) = 1794 kN
M = FR ( 638 ) = 448 kN-m
RCDF 2004 Example 002 - 4
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NZS 3101-2006 Example 001
Shear and Flexural Reinforcement Design of a Singly Reinforced Rectangle
PROBLEM DESCRIPTION
The purpose of this example is to verify the flexural and shear design. The load
level is adjusted for the case corresponding to the following conditions:
The stress-block dimension, a, extends below, amax , which requires that
compression reinforcement be provided as permitted by NZS 3101-06.
The average shear stress in the beam is below the maximum shear stress
allowed by NZS 3101-06, requiring design shear reinforcement.
A simple-span, 6-m-long, 300-mm-wide, and 560-mm-deep beam is modeled.
The beam is shown in Figure 1. The computational model uses a finite
element mesh of frame elements, automatically generated. The maximum
element size has been specified to be 200 mm. The beam is supported by joint
restraints that have no rotational stiffness. One end of the beam has no
longitudinal stiffness.
The beam is loaded with symmetric third-point loading. One dead load case
(DL50) and one live load case (LL130) with only symmetric third-point loads of
magnitudes 50, and 130 kN, respectively, are defined in the model. One load
combination (COMB130) is defined using the NZS 3101-06 load combination
factors of 1.2 for dead loads and 1.5 for live loads. The model is analyzed for
both of those load cases and the load combinations.
Table 1 shows the comparison of the design longitudinal reinforcements. Table 2
shows the comparison of the design shear reinforcements.
NZS 3101-2006 Example 001 - 1
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Figure 1 The Model Beam for Flexural and Shear Design
NZS 3101-2006 Example 001 - 2
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GEOMETRY, PROPERTIES AND LOADING
Clear span,
L
=
Overall depth,
h
=
Width of beam,
b
=
Effective depth,
d
=
Depth of comp. reinf.,
d'
=
Concrete strength,
f’c =
Yield strength of steel,
fy
=
Concrete unit weight,
wc =
Modulus of elasticity,
Ec =
Modulus of elasticity,
Es =
Poisson’s ratio,
v
=
6000
560
300
500
60
30
460
0
25x105
2x105
0.2
=
=
50
130
Dead load,
Live load,
Pd
Pl
ETABS
0
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the total factored moments in the design strip
with the moments obtained using the analytical method. They match exactly for
this problem. Table 1 also shows the comparison of design reinforcements.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-mm)
Method
Moment
(kN-m)
As+
As-
ETABS
510
3170
193
Calculated
510
3170
193
NZS 3101-2006 Example 001 - 3
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Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-mm/m)
Shear Force (kN)
ETABS
Calculated
255
1192.5
1192.5
COMPUTER FILE: NZS 3101-2006 EX001
CONCLUSION
The computed results show an exact comparison with the independent results.
NZS 3101-2006 Example 001 - 4
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HAND CALCULATION
Flexural Design
The following quantities are computed for the load combination:
φb = 0.85
=
α1 0.85 for f ′c ≤ 55MPa
=
β1 0.85 for f ′c ≤ 30,
cb =
εc
ε c + f y Es
d = 283.02 mm
amax = 0.75β1cb= 180.42 mm
=
Ac b=
d 150, 000 mm 2
As ,min
f ′c
Ac = 446.5
4 fy
mm2
= max
1.4 Ac = 456.5
fy
= 456.5 mm2
COMB130
V* = (1.2Pd + 1.5Pl) = 255 kN
M* =
V *L
= 510 kN-m
3
The depth of the compression block is given by:
a =−
d
d −
2
2 M*
α1 fc'φb b f
= 194.82 mm ; a > amax
Since a ≥ amax , compression reinforcing is required.
NZS 3101-2006 Example 001 - 5
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The compressive force, C, developed in the concrete alone is given by:
=
C α=
1,380.2 kN
1 f ′c bamax
The resisting moment by the concrete compression and tension reinforcement is:
a
M c* = C d − max φb = 480.8 kN-m
2
Therefore the moment required by concrete compression and tension
reinforcement is:
M s* = M * − M c* = 29.2 kN − m
The required compression reinforcing is given by:
=
As′
M s*
193 mm 2 , where
=
f s′ − α1 f 'c ( d − d ′ ) φb
)
(
cb=
,max
amax
= 0.75=
cb 0.75283.02
= 212.26 mm
β1
c
− d '
f s′ ε c ,max Es b ,max
=
≤ fy ;
cb ,max
212.26 − 60
=
f s′ 0.003200, 000
= 430 MPa =
≤ f y 460 MPa
212.26
f s′ = 430 MPa
The required tension reinforcing for balancing the compression in the concrete is:
=
As1
M c*
= 3, 001 mm 2
a
f y d − max φb
2
And the tension required for balancing the compression reinforcement is given
by:
=
As 2
M s*
=
169.9 mm 2
f y ( d − d ') φb
Therefore, the total tension reinforcement, A
=
As1 + As 2 is given by:
s
NZS 3101-2006 Example 001 - 6
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As = As1 + As 2 = 3001 + 169.9 = 3170.5 mm 2
Shear Design
The nominal shear strength provided by concrete is computed as:
VC = vC ACV , where
vC = kd ka vb , and
kd = 1.0 since shear reinforcement provided will be equal
to or greater than the nominal amount required.
ka = 1.0 (Program default)
A
=
vb 0.07 + 10 s f 'C , except vb is neither less than
bd
0.08 f 'C nor greater than 0.2 f 'C and f 'C ≤ 50 MPa
vC = 0.4382
The average shear stress is limited to a maximum limit of,
vmax = min {0.2 f ′c , 8 MPa} = min{6, 8} = 6 MPa
For this example, the nominal shear strength provided by concrete is:
VC= vC ACV= 0.4382 • 300 • 500= 65.727 kN
*
V
=
= 1.7 MPa < vmax , so there is no concrete crushing.
v
bw d
*
If ν* > νmax, a failure condition is declared.
For this example the required shear reinforcing strength is:
φs = 0.75
V=
S
V*
255
− VC =
− 65.727 = 274.3 kN
φS
0.75
The shear reinforcement is computed as follows:
Since h =
560 mm > max {300 mm, 0.5bw =
0.5300 =
150 mm}
φsνc = 0.328 MPa
NZS 3101-2006 Example 001 - 7
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φsνmax = 4.5 MPa
So φsνc < ν* ≤ φsνmax, and shear reinforcement is required and calculate as:
Av
=
s
VS
274.27 • 1E6
=
= 1192.5 mm 2
f yt d
460 • 500
NZS 3101-2006 Example 001 - 8
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NZS 3101-2006 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
In this example, a reinforced concrete column is subjected factored axial load N*
= 2445 kN and moment My = 611 kN-m. This column is reinforced with 5 T25
bars. The design capacity ratio is checked by hand calculations and computed
result is compared. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
2445 kN
My= 611 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
f’c = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied Reinforced Concrete Column Design
NZS 3101-2006 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.994
1.00
0.60%
COMPUTER FILE: NZS 3101-2006 Ex002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
NZS 3101-2006 Example 002 - 2
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HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This assumption will
be checked later. Calculate the distance to the neutral axis fro a balanced condition, cb:
Position of neutral axis at balance condition:
cb =
600
600
dt =
( 490) = 277 mm
600 + f y
600 + 460
2) From the equation of equilibrium:
N * = Cc + C s − T
where
Cc = 0.85 f c'ab = 0.85 • 30 • 350a = 8925a
(
)
Cs = As' f y - 0.85 f c' = 2500 ( 460 - 0.85 • 30 ) = 1,086, 250 N
Assume compression steels yields, (this assumption will be checked later).
T = As f s = 2500 f s ( f s < f y )
N * = 8,925a + 1,086, 250 - 2500 f s
(Eqn. 1)
3) Taking moments about As:
a
'
Cc d - 2 + C s d - d
The plastic centroid is at the center of the section and d " = 215 mm
e' = e + d " = 250 + 215 = 465 mm
1
a
N* =
8,925a 490 - + 1,086, 250 ( 490 - 60 )
465
2
*
2
N = 9, 404.8a - 9.6a + 1,004, 489.2
N* =
1
e'
(
)
(Eqn. 2)
NZS 3101-2006 Example 002 - 3
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4) Assume c = 330 mm, which exceed cb (296 mm).
a = 0.85 • 330 = 280.5 mm
Substitute in Eqn. 2:
N * = 9, 404.8 • 280.5 - 9.6 ( 280.5) + 1,004, 489.2 = 2,887, 205.2 N
2
5) Calculate fs from the strain diagram when c = 330 mm.
490 - 330
fs =
600 = 290.9 MPa
330
εs = εt = f s Es = 0.00145
6) Substitute a = 280.5 mm and fs = 290.9 MPa in Eqn. 1 to calculate N*2:
N 2* = 8,925 ( 280.5) + 1,086, 250 - 2500 ( 290.9 ) = 2,862, 462.5 N
Which is very close to the calculated Pr1 of 2,887,205.2 (less than 1% difference)
250
M * = N * e = 2877
= 719 kN-m
1000
7) Check if compression steels yield. From strain diagram,
330 - 60
ε s' =
( 0.003) = 0.00245 > ε y = 0.0023
330
Compression steels yields, as assumed.
8) Therefore, section capacity is
N * = 0.85 • 2877 = 2445 kN
M * = 0.85 • 719 = 611 kN-m
NZS 3101-2006 Example 002 - 4
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SS CP 65-1999 Example 001
SHEAR AND FLEXURAL REINFORCEMENT DESIGN OF A SINGLY REINFORCED RECTANGLE
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
A simply supported beam is subjected to a uniform unfactored dead load and
imposed load of 25 and 19 kN/m respectively spanning 6m. This example is
tested using the Singapore CP65-99 concrete design code. The flexural and shear
reinforcing computed is compared with independent results.
GEOMETRY, PROPERTIES AND LOADING
Dead Load=25kN/m
Live Load=19kN/m
CL
b=300mm
A
d=490 mm
A
300mm
300mm
6m
h=600 mm
Section A-A
Design Properties
fcu = 30 MPa
fy = 460 MPa
fyv = 250 MPa
TECHNICAL FEATURES OF TESTED
Calculation of Flexural reinforcement, As
Enforcement of Minimum tension reinforcement, As,min
Calculation of Shear reinforcement, Av
Enforcement of Minimum shear reinforcing, Av,min
SS CP 65-1999 Example 001 - 1
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RESULTS COMPARISON
The detailed work-out of the example above can be obtained from Example 3.4
of Chanakya Arya (1994). “Design of Structural Elements.” E & FN Spon, 54-55
Output Parameter
ETABS
Independent
Percent
Difference
Design Moment, Mu (kN-m)
294.30
294.30
0.00%
Tension Reinf, As (mm2)
1555
1555
0.00%
Design Shear, Vu (kN)
160.23
160.23
0.00%
Shear Reinf, Asv/sv (mm2/mm)
0.730
0.730
0.00 %
COMPUTER FILE: SS CP 65-1999 EX001
CONCLUSION
The computed flexural results show an exact match with the independent results.
SS CP 65-1999 Example 001 - 2
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HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
As ,min = 0.0013bh , where b=300mm, h=600mm
= 234.00 sq-mm
Design Combo COMB1
wu = =65.4 kN/m
Mu =
Vu =
wu l 2
= 294.3 kN-m
8
wu l
− wu d = 160.23 kN
2
The depth of the compression block is given by:
K=
M
= 0.108 < 0.156
f cu b d 2
If K ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam.
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 473.221 mm, where d=550 mm
0.9
The ultimate resistance moment is given by:
As =
M
= 1555 sq-mm
( f y 1.15) z
SS CP 65-1999 Example 001 - 3
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Shear Design
Vu = =160.23 kN at distance, d, from support
v=
V
= 0.9711 MPa
bw d
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
v ≤ vmax , so no concrete crushing
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.84k1 k 2 100 As 3 400 4
vc =
= 0.4418 MPa
γ m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
f 3
80
k2 = cu = 1.0, 1 ≤ k2 ≤
30
30
1
3
γm = 1.25
0.15 ≤
100 As
≤3
bd
100 As 100 • 469
= 0.2842
=
300 • 550
bd
1
400 4
400
=
0.95 ≥ 1, so
d
d
1
4
is taken as 1.
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
If (vc + 0.4) < v ≤ vmax
Asv
=
sv
v − vc ) bw ( 0.9711 − 0.4418 )
(=
0.87 f yv
0.87250
= 0.730 sq-mm/mm
SS CP 65-1999 Example 001 - 4
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SS CP 65-1999 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected to factored axial load N = 1971 kN
and moment My = 493 kN-m. This column is reinforced with five 25M bars. The
design capacity ratio is checked by hand calculations and the result is compared
with the calculated result. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1971 kN
My = 493 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fcu = 30 MPa
fy = 460 MPa
350 m
490 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
SS CP 65-1999 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.994
1.00
0.60%
COMPUTER FILE: SS CP 65-1999 EX002
CONCLUSION
The computed results show an acceptable comparison with the independent
results.
SS CP 65-1999 Example 002 - 2
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HAND CALCULATION
Column Strength under compression control
fcu = 30 MPa
b = 350 mm
fy = 460 MPa
d = 490 mm
1) Because e = 250 mm < (2/3)d = 327 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balanced condition:
=
cb
700
700
=
dt
=
( 490 ) 296 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
N = Cc + C s − T
where
0.67
Cc =
fcu ab 0.67 1.5 •=
30 • 350a 4667 a
=
γΜ
As′
2500
971, 014 N
Cs =
f y − 0.4467 fcu ) = ( 460 − 0.4467 • 30 ) =
(
1.15
γs
Assume compression steel yields (this assumption will be checked later).
As fs 2500 fs
=
T =
= 2174 fs ( fs < f y )
γs
1.15
N1 =4, 667 a + 971, 014 − 2174 fs
(Eqn. 1)
3) Taking moments about As:
=
N
1
a
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d " = 215 mm
e′ =e + d ′′ =250 + 215 =465 mm
=
N
1
a
4, 667 a 490 − + 971, 014 ( 490 − 60 )
465
2
N = 4917.9a − 5.018a 2 + 897,926
(Eqn. 2)
SS CP 65-1999 Example 002 - 3
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4) Assume c = 364 mm, which exceeds cb (296 mm).
a = 0.9 • 364 = 327.6 mm
Substitute in Eqn. 2:
N 2 4917.9 • 327.6 − 5.018 ( 327.6 ) + 897,926
=
= 1,970,500 N
2
5) Calculate fs from the strain diagram when c = 365 mm.
490 − 364
=
fs =
700 242.3 MPa
364
ε s =εt =fs Es = 0.0012
6) Substitute a = 327.6 mm and fs = 242.3 MPa in Eqn. 1 to calculate N1:
=
N1 4, 667 ( 327.6 ) + 971, 014 − 2174 ( 242.3
=
) 1,973,163 N
which is very close to the calculated N2 of 1,970,500 (less than 1% difference)
250
M = Ne = 1971
= 493 kN-m
1000
7) Check if compression steel yields. From strain diagram,
364 − 60
=
ε′s
=
> ε y 0.0023
) 0.0029=
( 0.0035
364
Compression steel yields, as assumed.
8) Therefore, section capacity is
N = 1971 kN
M = 493 kN-m
SS CP 65-1999 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TS 500-2000 Example 001
Shear and Flexural Reinforcement Design of a Singly Reinforced Rectangle
EXAMPLE DESCRIPTION
The flexural and shear strength of a rectangular concrete beam is tested in this
example.
A simply supported beam is subjected to a uniform factored load of 36.67 kN/m.
This example is tested using the Turkish TS 500-2000 concrete design code. The
flexural and shear reinforcing computed is compared with independent results.
GEOMETRY, PROPERTIES AND LOADING
230mm
A
CL
550 mm
60 mm
A
Section A-A
6m
Material Properties
E=
25.000x106 kN/m2
ν=
0.2
Clear span,
Overall depth,
Width of beam,
Effective depth,
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
Section Properties
d = 543.75 mm
L
h
b
d
fck
fyk
wc
Ec
Es
v
=
=
=
=
=
=
=
=
=
=
6000
550
230
490
30
420
0
25x103
2x105
0.2
Design Properties
fck = 30 MPa
fy = 420 MPa
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
-1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
ETABS
Independent
Percent
Difference
165.02
165.02
0.00%
Tension Reinf, As (mm2)
1022
1022
0.00%
Design Shear, Vd (kN)
110.0
110.0
0.00%
Shear Reinf, Asw/s (mm2/mm)
0.2415
0.2415
0.00%
Output Parameter
Design Moment, Md (kN-m)
COMPUTER FILE: TS 500-2000 EX001
CONCLUSION
The computed results show an exact match with the independent results.
TS 500-2000 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Flexural Design
The following quantities are computed for the load combination:
f=
cd
f ck 30
= = 20
γ mc 1.5
=
f yd
f yk 420
= = 365
γ ms 1.15
cb =
ε cu Es
ε cu Es + f yd
d = 304.6 mm
amax = 0.85k1cb = 212.3 mm
k1 = 0.85 − 0.006 ( f ck − 25 ) = 0.82 , 0.70 ≤ k1 ≤ 0.85
where,
=
As ,min
0.8 f ctd
=
bd 315.5 mm 2
f yd
Where
=
f ctd
0.35 f cu 0.35 30
= = 1.278
γ mc
1.5
COMB1
ωd = 36.67 kN/m
Md =
∙
ωd L2
= 36.67 62/8 = 165.02 kN-m
8
The depth of the compression block is given by:
a =−
d
d2 −
2 Md
= 95.42 mm
0.85 fcd b
Compression steel not required since a < amax.
The area of tensile steel reinforcement is given by:
As =
Md
165E6
=
a
365 • ( 490 − 95.41/ 2 )
f yd d −
2
As = 1022 mm2
TS 500-2000 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Shear Design
The shear demand is computed as:
Vd =
ωL
=110.0 kN at face of support for this example
2
The shear force is limited to a maximum of,
=
Vmax 0.22
=
fcd Aw 496 kN
The nominal shear strength provided by concrete is computed as:
γN
=
Vcr 0.65 fctd bd 1 + d
Ag
=93.6 kN, where N d = 0
=
Vc 0.8
=
Vcr 74.9 kN
The shear reinforcement is computed as follows:
If Vd ≤ Vcr
f ctd
mm 2
Asw
=
=
b
0.3
0.2415
(min. controls)
f ywd
mm
s min
(TS 8.1.5, Eqn 8.6)
If Vcr ≤ Vd ≤ Vmax
Asw
=
s
Vd − Vc )
(=
f ywd d
0.1962
mm 2
mm
(TS 8.1.4, Eqn 8.5)
TS 500-2000 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TS 500-2000 Example 002
P-M INTERACTION CHECK FOR COMPRESSION-CONTROLLED RECTANGULAR COLUMN
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete column is subjected to factored axial load N = 1908 kN
and moment My = 477 kN-m. This column is reinforced with five 25M bars. The
design capacity ratio is checked by hand calculations and the result is compared
with the computed result. The column is designed as a short, non-sway member.
GEOMETRY, PROPERTIES AND LOADING
1908 kN
My = 477 kN-m
550mm
3m
A
A
350mm
60 mm
Section A-A
Material Properties
Ec = 25x106 kN/m2
ν = 0.2
G = 10416666.7kN/m2
Section Properties
Design Properties
b =
d =
fck = 25 MPa
fyk = 420 MPa
350 mm
550 mm
TECHNICAL FEATURES TESTED
Tied reinforced concrete column design
TS 500-2000 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are hand calculated and compared.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.992
1.00
0.80%
COMPUTER FILE: TS 500-2000 EX002
CONCLUSION
The computed result shows an acceptable comparison with the independent
result.
TS 500-2000 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Column Strength under compression control
fck = 25 MPa
b = 350 mm
fyk = 420 MPa
d = 490 mm
1) Because e = 167.46 mm < (2/3)d = 326.67 mm, assume compression failure. This
assumption will be checked later. Calculate the distance to the neutral axis for a
balanced condition, cb:
Position of neutral axis at balance condition:
cb =
0.003 • 2x105
600
dt =
( 490 ) = 305 mm
5
0.003 • 2x10 + f yk
600 + 420 / 1.15
2) From the equation of equilibrium:
N = Cc + C s − T
where
Cc = 0.85f ck ab = 0.85 • 25 / 1.5 • 350a = 4,958a
A′
f 2500
Cs = s f yk − 0.85 ck =
( 420 − 0.85 • 25 /1.5) =882, 246 N
γs
γ c 1.15
Assume compression steels yields, (this assumption will be checked later).
A f
2500 f s
T= s s=
= 2174f s f s < f y
1.15
γs
N1 =4,958a + 882, 246 − 2,174 fs
(
)
(Eqn. 1)
3) Taking moments about As:
1
a
=
N
Cc d − + C s ( d − d ′ )
e′
2
The plastic centroid is at the center of the section and d ′′ = 215 mm
e′ = e + d ′′ = 250 + 215 = 465 mm
1
a
=
N
4,958a 490 − + 882, 246 ( 490 − 60 )
465
2
N2 =
5525a − 5.3312a 2 + 815,840
(Eqn. 2)
TS 500-2000 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Assume c = 358.3 mm, which exceed cb (305 mm).
a = 0.85 • 358 = 304.6 mm
Substitute in Eqn. 2:
N=
5525 • 304.6 − 5.3312 ( 304.6 ) + 815,840
= 1,907, 643 N
2
2
5) Calculate fs from the strain diagram when c = 359 mm.
490 − 358.3
=
f s =
600 220.2 > 420 MPa
358.3
εs = εt = f s Es = 0.0011
6) Substitute a = 304.6m and fs 221.2 MPa in Eqn. 1 to calculate N1:
=
N1 4,958 ( 304.6 ) + 882, 246 − 2174 ( 220.2
=
) 1,907, 601 N
which is very close to the calculated N2 of 2,002,751 (less than 1% difference)
250
M = Ne = 1908
= 477 kN-m
1000
7) Check if compression steel yields. From strain diagram,
358 − 60
=
ε′s
=
> ε y 0.0021
) 0.0025=
( 0.003
358
Compression steel yields, as assumed.
8) Therefore, section capacity is
N = 1908 kN
M = 477 kN-m
TS 500-2000 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-08 Wall-001
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 735 k and
moments Muy = 1504 k-ft. This wall is reinforced with two #9 bars at each end
and #4 bars at 14.00 inches on center of each face. The total area of
reinforcement is 5.20 in2. The design capacity ratio is checked by hand
calculations and results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
tb = 12 in
h = 60 in
As1= As5
= 2-#9 (2.00 in^2)
As2, As3, As4 = 2-#4 (0.40 in^2)
Design Properties
f ′c = 4 k/in2
fy = 60 k/in2
EXAMPLE ACI 318-08 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.007
1.00
0.70%
COMPUTER FILE: ACI 318-08 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-08 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
1) A value of e = 24.58 inch was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model. The values of M u and Pu were large enough to
produce a flexural D/C ratio very close to or equal to one. The depth to the neutral
axis, c, was determined by iteration using an excel spreadsheet so that equations 1 and
2 below were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85=
• 4 •12a 40.8a
Cs =A1′ ( fs1 − 0.85 fc' ) + A2′ ( fs 2 − 0.85 fc' ) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5
Pn1 = 40.8a + A1′ ( fs1 − 0.85 fC′ ) + A2′ ( fs 2 − 0.85 fc′) A1′ ( fs1 − 0.85 fc′) +
A3′ ( fs 3 − 0.85 fc′) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) − Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
Cs1 A1′ ( f s1 − 0.85 f c′)=
where=
; Csn An′ ( f sn − 0.85 f c′) ; Tsn = f sn Asn ; and the bar strains
are determined below. The plastic centroid is at the center of the section and d " = 28
inch
e′ =e + d ′′ =24.54 + 28 =52.55 inch.
4) Using c = 30.1 inch (from iteration),
a = 0.85 • 30.1 = 25.58 inch
EXAMPLE ACI 318-08 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
5) Assuming the extreme fiber strain equals 0.003 and c= 30.1 inch, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.0028; f s =
ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.0014
f s 2 = 40.75 ksi
= 0.0000
f s 3 = 00.29 ksi
= 0.0014
f s 4 = 40.20 ksi
= 0.0028
f s 5 = 60.00 ksi
Substitute in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal gives,
Pn1 = 1035 k
Pn2 = 1035 k
M=
P=
1035(24.54) /12 = 2116 k-ft
n
ne
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00244 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.712
0.005 − ε y
( φt − φc )
7) Calculate φ ,
φPn = 0.711(1035 ) =
735 kips
φM n = 0.711( 2115 ) =
1504 k-ft.
EXAMPLE ACI 318-08 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-08 Wall-002
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. The wall is reinforced as shown below. The concrete core wall is
loaded with a factored axial load Pu = 2384 k and moments Mu3 = 9293k-ft. The
design capacity ratio is checked by hand calculations and results are compared
with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-08 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
ETABS
0
Design Properties
Section Properties
tb = 8 in
h = 98 in
As1= As6 = 2-#10,2#6 (5.96 in^2)
As2, As3, As4 and As5 = 2-#6 (0.88 in^2)
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: ACI 318-08 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-08 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength under compression and bending
1) A value of e = 46.78 inches was determined using e = M u / Pu where M u and Pu
were taken from the ETABS test model interaction diagram. The values of M u and
Pu were large enough to produce a flexural D/C ratio very close to or equal to one.
The depth to the neutral axis, c, was determined by iteration using an excel
spreadsheet so that equations 1 and 2 below were equal.
2) From the equation of equilibrium:
Pn1 = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
=
Ccw 0.85 fc′ • 8 • ( a − 8 )
=
Ccf 0.85 fc′ ( 8 • ( 98 − 40 ) )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5 + As6 f s6
=
Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • ( 98 − 40 ) ) + A1′ ( fs1 − 0.85 fc′) +
A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6
(Eqn. 1)
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) +
1 Ccf ( d - d ′ ) + Ccw d −
2
Pn 2 =
e′
Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 2)
EXAMPLE ACI 318-08 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where=
, Csn An′ ( fsn − 0.85 fc′) , Tsn = f sn Asn , and the bar strains
Cs1 A1′ ( fs1 − 0.85 fc′)=
are determined below. The plastic centroid is at the center of the section, d ′′ =
98 − 8
2
= 45 inches
e′= e + d ′′ = 46.78 + 45 = 91.78 inches
4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 44.58
inches.
=
a 0.85
=
• c 0.85 •=
44.58 37.89 inches
5) Assuming the extreme fiber strain equals 0.003 and c = 44.58 inches, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain,
then f s = f y :
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
c − 2s − d '
εs3 =
0.003
c
d − c − 2s
=
εs 4
εs6
d −c
d −c−s
=
εs5
εs6
d −c
d −c
εs6 =
0.003
c
= 0.00273;=
f s ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.00152
f s 2 = 44.07 ksi
= 0.00310
fs3
= 0.00090
f s 4 = 26.2 ksi
= 0.00211
fs5
= 60.00 ksi
= 0.00333
fs6
= 60.00 ksi
= 8.94 ksi
Substituting the above values of the compression block depth, a, and the rebar
stresses into equations Eqn. 1 and Eqn. 2 give
Pn1 = 3148 k
Pn2 = 3148 k
M=
P=
3148(46.78) /12 = 12,273 k-ft
n
ne
EXAMPLE ACI 318-08 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00332 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.757
−
ε
0.005
y
( φt − φc )
7) Calculate the capacity,
φPn = 0.757 ( 3148 ) =
2384 kips
φM n = 0.757 (12, 273) =
9293 k-ft.
EXAMPLE ACI 318-08 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-11 Wall-001
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 735 k and
moments Muy = 1,504 k-ft. This wall is reinforced with two #9 bars at each end
and #4 bars at 14.00 inches on center of each face. The total area of
reinforcement is 5.20 in2. The design capacity ratio is checked by hand
calculations and results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-11 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
tb = 12 in
h = 60 in
As1= As5
= 2-#9 (2.00 in^2)
As2, As3, As4 = 2-#4 (0.40 in^2)
ETABS
0
Design Properties
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.007
1.00
0.70%
COMPUTER FILE: ACI 318-11 WALL-001
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE ACI 318-11 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
1) A value of e = 24.58 inch was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model. The values of M u and Pu were large enough to
produce a flexural D/C ratio very close to or equal to one. The depth to the neutral
axis, c, was determined by iteration using an excel spreadsheet so that equations 1 and
2 below were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85=
• 4 •12a 40.8a
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5
Pn1 =40.8a + A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) A1′ ( fs1 − 0.85 fc′) +
A3′ ( fs 3 − 0.85 fc′) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) − Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
where
; Csn An′ ( fsn − 0.85 fc′) ; Tsn = f sn Asn ; and the bar strains
=
Cs1 A′ ( f s1 − 0.85 f c′)=
are determined below. The plastic centroid is at the center of the section and d ′′ = 28
inch
e′ =e + d ′′ =24.54 + 28 =52.55 inch.
4) Using c = 30.1 inch (from iteration),
a = 0.85 • 30.1 = 25.58 inches
5) Assuming the extreme fiber strain equals 0.003 and c= 30.1 inch, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
EXAMPLE ACI 318-11 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
87
c
ETABS
0
= 0.0028; f s =
ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.0014
f s 2 = 40.75 ksi
= 0.0000
f s 3 = 00.29 ksi
= 0.0014
f s 4 = 40.20 ksi
= 0.0028
f s 5 = 60.00 ksi
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 1035 k
Pn2 = 1035 k
M=
P=
1035(24.54) /12 = 2116 k-ft
n
ne
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00244 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.712
0.005 − ε y
( φt − φc )
7) Calculate φ ,
φPn = 0.711(1035 ) =
735 kips
φM n = 0.711( 2115 ) =
1504 k-ft.
EXAMPLE ACI 318-11 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-11 Wall-002
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. The wall is reinforced as shown below. The concrete core wall is
loaded with a factored axial load Pu = 2384 k and moments Mu3 = 9293k-ft. The
design capacity ratio is checked by hand calculations and results are compared
with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-11 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
ETABS
0
Design Properties
Section Properties
tb = 8 in
h = 98 in
As1= As6 = 2-#10,2#6 (5.96 in^2)
As2, As3, As4 and As5 = 2-#6 (0.88 in^2)
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: ACI 318-11 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-11 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength under compression and bending
1) A value of e = 46.78 inches was determined using e = M u / Pu where M u and Pu
were taken from the ETABS test model interaction diagram. The values of M u and
Pu were large enough to produce a flexural D/C ratio very close to or equal to one.
The depth to the neutral axis, c, was determined by iteration using an excel
spreadsheet so that equations 1 and 2 below were equal.
2) From the equation of equilibrium:
Pn1 = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
=
Ccw 0.85 fc′ • 8 • ( a − 8 )
=
Ccf 0.85 fc′ ( 8 • ( 98 − 40 ) )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5 + As6 f s6
=
Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • ( 98 − 40 ) ) + A1′ ( fs1 − 0.85 fc′) +
A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
e′
Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 1)
(Eqn. 2)
Cs1 A1′ ( fs1 − 0.85 fc′)=
where=
, Csn An′ ( f sn − 0.85 f c′) , Tsn = f sn Asn , and the bar
strains are determined below. The plastic centroid is at the center of the section, d ′′
98 − 8
= 45 inches
=
2
e′ =e + d ′′ =46.78 + 45 =91.78 inches
EXAMPLE ACI 318-11 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 44.58
inches.
a = 0.85 • c=0.85 • 44.58 = 37.89 inches
5) Assuming the extreme fiber strain equals 0.003 and c = 44.58 inches, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain
then, f s = f y :
c−d '
0.003
c
c−s−d '
ε s2 =
0.003
c
c − 2s − d '
ε s3 =
0.003
c
d − c − 2s
ε s4 =
ε s6
d −c
d −c−s
ε s5 =
ε s6
d −c
d −c
ε s6 =
0.003
c
ε s1 =
= 0.00273;=
f s ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.00152
f s 2 = 44.07 ksi
= 0.00310
f s 3 = 8.94 ksi
= 0.00090
f s 4 = 26.2 ksi
= 0.00211
f s 5 = 60.00 ksi
= 0.00333
f s 6 = 60.00 ksi
Substituting the above values of the compression block depth, a, and the rebar
stresses into equations Eqn. 1 and Eqn. 2 give
Pn1 = 3148 k
Pn2 = 3148 k
M=
P=
3148(46.78) /12 = 12,273 k-ft
n
ne
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00332 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.757
0.005
−
ε
y
( φt − φc )
EXAMPLE ACI 318-11 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
7) Calculate the capacity,
φPn = 0.757 ( 3148 ) =
2384 kips
φM n = 0.757 (12, 273) =
9, 293 k-ft.
EXAMPLE ACI 318-11 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-14 Wall-001
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 735 k and
moments Muy = 1,504 k-ft. This wall is reinforced with two #9 bars at each end
and #4 bars at 14.00 inches on center of each face. The total area of
reinforcement is 5.20 in2. The design capacity ratio is checked by hand
calculations and results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-14 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
Section Properties
tb = 12 in
h = 60 in
As1= As5
= 2-#9 (2.00 in^2)
As2, As3, As4 = 2-#4 (0.40 in^2)
ETABS
0
Design Properties
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.007
1.00
0.70%
COMPUTER FILE: ACI 318-14 WALL-001
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE ACI 318-14 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
COLUMN STRENGTH UNDER COMPRESSION CONTROL
1) A value of e = 24.58 inch was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model. The values of M u and Pu were large enough to
produce a flexural D/C ratio very close to or equal to one. The depth to the neutral
axis, c, was determined by iteration using an excel spreadsheet so that equations 1 and
2 below were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85=
• 4 •12a 40.8a
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5
Pn1 =40.8a + A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) A1′ ( fs1 − 0.85 fc′) +
A3′ ( fs 3 − 0.85 fc′) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) − Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
where
; Csn An′ ( fsn − 0.85 fc′) ; Tsn = f sn Asn ; and the bar strains
=
Cs1 A′ ( f s1 − 0.85 f c′)=
are determined below. The plastic centroid is at the center of the section and d ′′ = 28
inch
e′ =e + d ′′ =24.54 + 28 =52.55 inch.
4) Using c = 30.1 inch (from iteration),
a = 0.85 • 30.1 = 25.58 inches
5) Assuming the extreme fiber strain equals 0.003 and c= 30.1 inch, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
EXAMPLE ACI 318-14 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
87
c
ETABS
0
= 0.0028; f s =
ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.0014
f s 2 = 40.75 ksi
= 0.0000
f s 3 = 00.29 ksi
= 0.0014
f s 4 = 40.20 ksi
= 0.0028
f s 5 = 60.00 ksi
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 1035 k
Pn2 = 1035 k
M=
P=
1035(24.54) /12 = 2116 k-ft
n
ne
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00244 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.712
0.005 − ε y
( φt − φc )
7) Calculate φ ,
φPn = 0.711(1035 ) =
735 kips
φM n = 0.711( 2115 ) =
1504 k-ft.
EXAMPLE ACI 318-14 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 318-14 Wall-002
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. The wall is reinforced as shown below. The concrete core wall is
loaded with a factored axial load Pu = 2384 k and moments Mu3 = 9293k-ft. The
design capacity ratio is checked by hand calculations and results are compared
with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 318-14 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
ETABS
0
Design Properties
Section Properties
tb = 8 in
h = 98 in
As1= As6 = 2-#10,2#6 (5.96 in^2)
As2, As3, As4 and As5 = 2-#6 (0.88 in^2)
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.999
1.00
0.10%
COMPUTER FILE: ACI 318-14 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-14 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength under compression and bending
1) A value of e = 46.78 inches was determined using e = M u / Pu where M u and Pu
were taken from the ETABS test model interaction diagram. The values of M u and
Pu were large enough to produce a flexural D/C ratio very close to or equal to one.
The depth to the neutral axis, c, was determined by iteration using an excel
spreadsheet so that equations 1 and 2 below were equal.
2) From the equation of equilibrium:
Pn1 = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
=
Ccw 0.85 fc′ • 8 • ( a − 8 )
=
Ccf 0.85 fc′ ( 8 • ( 98 − 40 ) )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As4 f s4 + As5 f s5 + As6 f s6
=
Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • ( 98 − 40 ) ) + A1′ ( fs1 − 0.85 fc′) +
A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
e′
Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 1)
(Eqn. 2)
Cs1 A1′ ( fs1 − 0.85 fc′)=
where=
, Csn An′ ( f sn − 0.85 f c′) , Tsn = f sn Asn , and the bar
strains are determined below. The plastic centroid is at the center of the section, d ′′
98 − 8
= 45 inches
=
2
e′ =e + d ′′ =46.78 + 45 =91.78 inches
EXAMPLE ACI 318-14 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 44.58
inches.
a = 0.85 • c=0.85 • 44.58 = 37.89 inches
5) Assuming the extreme fiber strain equals 0.003 and c = 44.58 inches, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain
then, f s = f y :
c−d '
0.003
c
c−s−d '
ε s2 =
0.003
c
c − 2s − d '
ε s3 =
0.003
c
d − c − 2s
ε s4 =
ε s6
d −c
d −c−s
ε s5 =
ε s6
d −c
d −c
ε s6 =
0.003
c
ε s1 =
= 0.00273;=
f s ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.00152
f s 2 = 44.07 ksi
= 0.00310
f s 3 = 8.94 ksi
= 0.00090
f s 4 = 26.2 ksi
= 0.00211
f s 5 = 60.00 ksi
= 0.00333
f s 6 = 60.00 ksi
Substituting the above values of the compression block depth, a, and the rebar
stresses into equations Eqn. 1 and Eqn. 2 give
Pn1 = 3148 k
Pn2 = 3148 k
M=
P=
3148(46.78) /12 = 12,273 k-ft
n
ne
6) Determine if φ is tension controlled or compression controlled.
εt = 0.00332 , ε y = 0.0021
for ε y < εt < 0.005 ; φ =
0.005 − εt
= 0.757
0.005
−
ε
y
( φt − φc )
EXAMPLE ACI 318-14 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
7) Calculate the capacity,
φPn = 0.757 ( 3148 ) =
2384 kips
φM n = 0.757 (12, 273) =
9, 293 k-ft.
EXAMPLE ACI 318-14 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 530-11 Masonry Wall-001
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. A reinforced masonry wall is subjected to factored axial load Pu = 556
k and moments Muy = 1331 k-ft. This wall is reinforced with two #9 bars at each
end and #4 bars at 14.00 inches on center each of face module (The reinforcing is
not aligned with the conventional masonry block spacing for calculation
convenience. The same excel spreadsheet used in other concrete examples was
used here). The total area of reinforcement is 5.20 in2. The design capacity ratio
is checked by hand calculations and the results are compared with ETABS
program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 530-11 Masonry Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
2250 k/in2
0.2
750 k/in2
ETABS
0
Design Properties
Section Properties
tb = 12 in
h = 60 in
As1= As5
= 2-#9 (2.00 in^2)
As2, As3, As4 = 2-#4 (0.40 in^2)
f ′m = 2.5 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.939
1.00
-6.1%
COMPUTER FILE: ACI 530-11 MASONRY WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 530-11 Masonry Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Column Strength under compression control
1) A value of e = 28.722 inches was determined using e = M u / Pu where M u and Pu
were taken from the ETABS test model. The values of M u and Pu were large enough
to produce a flexural D/C ratio very close to or equal to one. The depth to the neutral
axis, c, was determined by iteration using an excel spreadsheet so that equations 1 and
2 below were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =
β1 fm′ ab =
0.8 • 2.5 •12a =
24.0a
Cs = A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) + A3′ ( fs 3 − 0.8 fm′ )
T = As4 f s4 + As5 f s5
Pn1 =
24a + A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) + A3′ ( fs 3 − 0.8 fm′ ) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) − Ts 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
Cs1 A1′ ( f s1 − 0.8 f m′ )=
where=
; Csn An′ ( f sn − 0.8 f m′ ) ; Tsn = f sn Asn ; and the bar strains
are determined below. The plastic centroid is at the center of the section and d ′′ = 28
inch
e′ =e + d ′′ =28.722 + 28 =56.72 inch.
4) Using c = 32.04 inch (from iteration),
a = 0.80 • 332.04 = 25.64 inch
EXAMPLE ACI 530-11 Masonry Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
5) Assuming the extreme fiber strain equals 0.0025 and c= 32.04 inch, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c−d '
ε s1 =
0.0025
c
c−s−d '
εs 2 =
0.0025
c
c − 2s − d '
εs3 =
0.0025
c
d −c−s
εs 4 =
0.0025
c
d −c
εs5 =
0.0025
c
= 0.00207; f s =
ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.00125
f s 2 = 36.30 ksi
= 0.00016
f s 3 = 4.62 ksi
= 0.00093
f s 4 = 27.10 ksi
= 0.00203
f s 5 = 58.70 ksi
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 618 k;
Pn2 = 618 k
M=
P=
618(28.72) /12 = 1479 k-ft
n
ne
6) Calculate φ ,
φPn = 0.9 ( 618 ) =
556 kips
φM n = 0.9 (1479 ) =
1331 k-ft.
EXAMPLE ACI 530-11 Masonry Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE ACI 530-11 Masonry Wall-002
P-M INTERACTION CHECK FOR WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. The wall is reinforced as shown below. The concrete core wall is
loaded with a factored axial load Pu = 1496 k and moments Mu3 = 7387 k-ft. The
design capacity ratio is checked by hand calculations and the results are
compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE ACI 530-11 Masonry Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
G=
3600 k/in2
0.2
1500 k/in2
ETABS
0
Design Properties
Section Properties
tb = 8 in
h = 98 in
As1= As6 = 2-#10,2#6 (5.96 in^2)
As2, As3, As4 and As5 = 2-#6 (0.88 in^2)
f ′c = 4 k/in2
fy = 60 k/in2
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Column Demand/Capacity Ratio
0.998
1.00
-0.20%
COMPUTER FILE: ACI 530-11 MASONRY WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE ACI 530-11 Masonry Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength under compression and bending
1) A value of e = 59.24 inches was determined using e = M u / Pu where M u and Pu
were taken from the ETABS test model interaction diagram. The values of M u and
Pu were large enough to produce a flexural D/C ratio very close to or equal to one.
The depth to the neutral axis, c, was determined by iteration using an excel
spreadsheet so that equations 1 and 2 below were equal.
2) From the equation of equilibrium:
Pn1 = Cc + Cs − T
where
Cc =
β1 fm′ ab =
0.8 • 2.5 •12a =
24.0a
Cs = A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) + A3′ ( fs 3 − 0.8 fm′ )
T = As4 f s4 + As5 f s5 + As6 f s6
Pn1 =
24a + A1′ ( fs1 − 0.8 fm′ ) + A2′ ( fs 2 − 0.8 fm′ ) +
(Eqn. 1)
A3′ ( fs 3 − 0.8 fm′ ) − As 4 fs 4 − As 5 fs 5 − As 6 fs 6
3) Taking moments about As6:
a −tf
1 Ccf ( d − d ') + Ccw d −
2
Pn 2 =
e′
Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
+ Cs1 ( d − d ') + Cs 2 ( 4s ) +
(Eqn. 2)
Cs1 A1′ ( f s1 − 0.8 f m′ )=
where=
; Csn An′ ( f sn − 0.8 f m′ ) ; Tsn = f sn Asn ; and the bar strains
are determined below. The plastic centroid is at the center of the section and d ′′ = 45
inch
e′ =e + d ′′ =59.24 + 45 =104.24 inch.
EXAMPLE ACI 530-11 Masonry Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Iterating on a value of c until equations 1 and 2 are equal c is found to be c = 41.15
inches.
• c 0.8 • 41.15
a 0.8
=
=
= 32.92 inches
5) Assuming the extreme fiber strain equals 0.0025 and c = 41.15 inches, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain,
then f s = f y :
c−d '
0.0025
c
c−s−d '
ε s2 =
0.0025
c
c − 2s − d '
ε s3 =
0.0025
c
d − c − 2s
ε s4 =
ε s6
d −c
d −c−s
ε s5 =
ε s6
d −c
d −c
ε s6 =
0.0025
c
ε s1 =
= 0.00226;=
f s ε s E ≤ Fy ; f s1 = 60.00 ksi
= 0.00116
f s 2 = 33.74 ksi
= 0.00007
f s 3 = 2.03 ksi
= 0.00102
f s 4 = 29.7 ksi
= 0.00212
f s 5 = 60.00 ksi
= 0.00321
f s 6 = 60.00 ksi
Substituting the above values of the compression block depth, a, and the rebar
stresses into equations Eqn. 1 and Eqn. 2 give
Pn1 = 1662 k
Pn2 = 1662 k
M=
P=
1662(41.15) /12 = 8208 k-ft
n
ne
6) Calculate the capacity,
φPn = 0.9 (1622 ) =
1496 kips
φM n = 0.9 ( 8208 ) =
7387 k-ft.
EXAMPLE ACI 530-11 Masonry Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
EXAMPLE AS 3600-09 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 3438 kN and
moments Muy = 2003 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE AS 3600-09 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
4
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Wall flexural Demand/Capacity ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.083
1.00
8.30%
COMPUTER FILE: AS 3600-09 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE AS 3600-09 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 582.6 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85 • 30
=
• 300a 7650a
Cs =
A1 ( fs1 − 0.85 • fc′) + A2 ( fs 2 − 0.85 • fc′) + A3 ( fs 3 − 0.85 • fc′)
T = As4 f s4 + As5 f s5
Pn1 =
7650a + A1 ( fs1 − 0.85 • fc′) + A2 ( fs 2 − 0.85 • fc′) +
(Eqn. 1)
A3 ( fs 3 − 0.85 • fc′) − As 4 fs 4 − As 5 fs 5
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ') + Cs 2 ( 3s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
where=
; Cs 2 A2 ( f s 2 − 0.85 • f c′) ; Cs 3 ( f s 3 − 0.85 • f c′) ;
Cs1 A1 ( fs1 − 0.85 • fc′)=
Ts 4 = f s 4 As 4 and the bar strains are determined below. The plastic centroid is at the
center of the section and d ′′ = 700mm
e′ =e + d ′′ =582.6 + 700 =1282.61 mm.
EXAMPLE AS 3600-09 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
4) Using c = 821.7 mm (from iteration),
′ 0.84
γ 1.05 − 0.007( f=
a = γ c = 0.84 • 821.7=690.2 mm, where=
c)
5) Assuming the extreme fiber strain equals 0.003 and c = 30 inch, the steel stresses and
strains can be calculated. When the bar strain exceeds the yield strain, then f s = f y :
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
2
d
c
s
−
−
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.0028; f s =
ε s E ≤ Fy ; f s1 = 460.00 ksi
= 0.0015
f s 2 = 307.9 ksi
= 0.0003
f s 3 = 52.3 ksi
= 0.0010
f s 4 = 203.2 ksi
= 0.0023
f s 5 = 458.8 ksi
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 5289 kN
Pn2 = 5289 kN
M=
P=
5289(582.6) /1000000 = 3081 k-ft
n
ne
6) Calculate φ ,
φPn = 0.65 ( 5289 ) =
3438 kN
φM n = 0.65 ( 3081) =
2003 kN-m
EXAMPLE AS 3600-09 Wall-001 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
2
EXAMPLE AS 3600-09 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
11175 kN and moments Muy = 12564 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE AS 3600-09 Wall-002 - 1
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
2
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.082
1.00
8.20%
COMPUTER FILE: AS 3600-09 WALL-002
CONCLUSION
The ETABS result shows an acceptable comparison with the independent result.
EXAMPLE AS 3600-09 Wall-002 - 2
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
2
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1124.3 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = 0.85 f c' ab = 0.85 • 30 • 300a = 7650a
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
Ccw 0.85 fc′ • 200 • ( a − 200 )
=
Ccf = 0.85 fc′ ( 200 • 2500 )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As 4 fs 4 + As 5 fs 5 + As 6 fs 6
=
Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • 98 ) + A1′ ( fs1 − 0.85 fc′) +
A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6
(Eqn. 1)
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) +
1 Ccf ( d − d ') + Ccw d −
2
Pn 2 =
e′
Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 2)
EXAMPLE AS 3600-09 Wall-002 - 3
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
2
where=
, Csn An′ ( fsn − 0.85 fc′) , Tsn = fsn Asn , and the bar strains
Cs1 A1′ ( fs1 − 0.85 fc′)=
are determined below. The plastic centroid is at the center of the section, d ′′
2500 − 200
= 1150 mm
=
2
e′ =e + d ′′ =1124.3 + 1150 =2274.3 mm
(4) Using c = 1341.6 mm (from iteration)
a=
β1c =
0.85 •1341.6 =
1140.4 mm
5) Assuming the extreme fiber strain equals 0.003 and c = 1341.6 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
c − 2s − d ′
εs3 =
0.003
c
d − c − 2s
=
εs 4
εs 6
d −c
d −c−s
=
εs5
εs 6
d −c
d −c
εs 6 =
0.003
c
= 0.00278; fs =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00199
f s 2 = 398.7 MPa
= 0.00121
f s 3 = 242.2 MPa
= 0.00080
f s 4 = 160.3 MPa
= 0.00158
f s 5 = 16.8 MPa
= 0.00237
f s 6 = 460.0 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give,
Pn1 = 17192 kN
Pn2 = 17192 kN
M=
P=
17192(1124.3) /1000000 = 19329 kN-m
n
ne
EXAMPLE AS 3600-09 Wall-002 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
2
6) Calculate φ ,
=
φPn 0.65=
(17192 ) 11175 kN
=
φM n 0.65 =
(19329 ) 12564 kN-m
EXAMPLE AS 3600-09 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE BS 8110-97 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 3246 kN and
moments Muy = 1969 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE BS 8110-97 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural Demand/Capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.997
1.00
0.30%
COMPUTER FILE: BS 8110-97 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE BS 8110-97 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
f ′c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 606.5 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
The distance to the neutral axis for a balanced condition, cb:
=
cb
700
700
=
dt
=
(1450 ) 922.7 mm
700 + f y / γ s
700 + 460 /1.15
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc
0.67
0.67
=
fcu ab
•=
30 • 300a 4020a
γm
1.5
As′1
0.67 As′2
0.67 As′3
0.67
fc′ +
fc′ +
fc′
fs1 −
fs 2 −
fs 3 −
γs
γm
γm
γm
γs
γs
As 4
A
T
fs 4 + s 5 fs 5
=
γs
γs
Cs =
As′1
0.67 As′2
0.67
fc′ +
fc′ +
fs1 −
fs 2 −
γs
γm
γm
γs
As′3
A
0.67 As 4
fc′ −
fs 4 + s 5 fs 5
fs 3 −
γs
γm
γs
γs
4709a +
Pn1 =
(Eqn. 1)
EXAMPLE BS 8110-97 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As5:
Pn 2
=
a
1
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
As1
As 2
As 3
0.67
0.67
0.67
′ ; Cs 2
f c=
f c′=
f c′ ;
f s1 −
fs2 −
; Cs 3
fs3 −
γs
γm
γs
γm
γs
γm
As 4
0.67
=
Ts 4
f c′ and the bar strains and stresses are determined below.
fs4 −
γs
γm
=
Cs1
where
The plastic centroid is at the center of the section and d ′′ = 700 mm
e′ =e + d ′′ =606.5 + 700 =1306.5 mm.
4) Using c = 875.2 mm (from iteration), which is more than cb (722.7mm).
a=
β1c =
0.9 • 875.2 =
787.7 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 643.6 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.00330; f s =
ε s E ≤ Fy ; f s1 = 460 MPa
= 0.00190
f s 2 = 380.1 MPa
= 0.00050
f s 3 = 100.1 MPa
= 0.00090
f s 4 = 179.8 MPa
= 0.00230
f s 5 = 459.7 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3246 kN
Pn2 = 3246 kN
M=
P=
3246(606.5) /1000 = 1969 kN-m
n
ne
EXAMPLE BS 8110-97 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE BS 8110-97 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
8368 kN and moments Muy = 11967 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE BS 8110-97 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.001
1.00
0.10%
COMPUTER FILE: BS 8110-97 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE BS 8110-97 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1430 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
0.67
=
Ccw
fcu • 200 • ( a − 200 )
γm
0.67
Ccf =
fcu ( 200 • 2500 )
γm
As′1
0.67 As′2
0.67 As′3
0.67
fc′ +
fc′ +
fc′
fs1 −
fs 2 −
fs 3 −
γs
γm
γm
γm
γs
γs
A
A
A
T = s4 f s4 + s5 f s5 + s6 f s6
γs
γs
γs
Cs =
=
Pn1
A′
0.67
0.67
0.67
fcu • 200 • ( a − 200 ) +
fcu ( 200 • 2500 ) + s1 fs1 −
fc′
γm
γm
γs
γm
(Eqn. 1)
As′2
As 5
As 6
0.67 As′3
0.67 As 4
+
fc′ +
fc′ −
fs 4 +
fs 5 +
fs 6
fs 2 −
fs 3 −
γs
γm
γm
γs
γs
γs
γs
EXAMPLE BS 8110-97 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As6:
a −tf
−tf
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
e′
+Cs 3 ( 3s ) − Ts 4 ( 2 s ) − Ts 5 ( s )
+ Cs1 ( d − d ′ ) + Cs 2 ( 4 s )
(Eqn. 2)
As1
Asn
Asn
0.67
0.67
0.67
f c′=
f c′=
f c′
f s1 −
; Csn
f sn −
; Tsn
f sn −
γs
γm
γs
γm
γs
γm
and the bar strains and stresses are determined below.
=
Cs1
where
The plastic centroid is at the center of the section and d ′′ = 1150 mm
e′ =e + d ′′ =1430 + 1150 =2580 mm.
4) Using c = 1160 mm (from iteration),
a=
β1c =
0.9 •1160 =
1044 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 1160 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c−d '
ε s1 =
0.0035
c
c−s−d '
εs 2 =
0.0035
c
c − 2s − d '
εs3 =
0.0035
c
d − c − 2s
=
εs 4
εs6
d −c
d −c−s
=
εs5
εs6
d −c
d −c
εs6 =
0.0035
c
= 0.00320; f s =
ε s E ≤ Fy ; f s1 = 460 MPa
= 0.00181
f s 2 = 362.0 MPa
= 0.00042
f s 3 = 84.4 MPa
= 0.00097
f s 4 = 193.2 MPa
= 0.00235
f s 5 = 460.00 MPa
= 0.00374
f s 6 = 460.00 MPa
EXAMPLE BS 8110-97 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 8368 kN
Pn2 = 8368 kN
M=
P=
8368(1430) /1000 = 11,967 kN-m
n
ne
EXAMPLE BS 8110-97 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
EXAMPLE CSA A23.3-04 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete column is subjected to factored axial load Pu = 3870 kN
and moments Muy = 2109 kN-m. This wall is reinforced with two 30M bars at
each end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE CSA A23.3-04 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
4
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Flexural Demand/Capacity ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.986
1.00
-1.40%
COMPUTER FILE: CSA A23.3-04 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE CSA A23.3-04 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
HAND CALCULATION
Wall Strength Determined as follows:
f ′ c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 545 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
The distance to the neutral axis for a balanced condition, cb:
cb
=
700
700
dt
=
=
(1450 ) 875 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = φc α1 fc′ab = 0.65 • 0.805 • 30 • 300a = 4709a
Cs = φs As′1 ( fs1 − α1 fc′) + φs As′2 ( fs 2 − α1 fc' ) + φs As′3 ( fs 3 − α1 fc′)
T = φs As 4 fs 4 + φAs 5 fs 5
=
Pn1 4709a + A1′ ( fs1 − 0.805 fc′) + A2′ ( fs 2 − 0.805 fc′) − φAs 3 fs 3 − φAs 4 fs 4 − φAs 5 fs 5 (Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
EXAMPLE CSA A23.3-04 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
where Cs1 = φs As′1 ( fs1 − α1 fc′) ; Cs 2 = φs As′2 ( f s 2 − α1 f c′) ; Cs 3 = φs As′3 ( f s 3 − α1 f c′) ;
Ts 4 = φs f s 4 As 4 and the bar strains are determined below. The plastic centroid is at the
center of the section and d ′′ = 700 mm
e′ =e + d ′′ =545 + 700 =1245 inch.
4) Using c = 894.5 mm (from iteration), which is more than cb (875mm).
a = β1c = 0.895 • 894.5 = 800.6 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 643.6 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.0035
c
= 0.00330; f s =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00193
f s 2 = 387.0 MPa
= 0.00057
f s 3 = 113.1 MPa
= 0.00080
f s 4 = 160.8 MPa
= 0.00217
f s 5 = 434.7 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3870 kN
Pn2 = 3870 kN
M=
P=
3870(545) / 1000 = 2109 kN-m
n
ne
EXAMPLE CSA A23.3-04 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE CSA A23.3-04 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
10687 kN and moments Muy = 13159 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE CSA A23.3-04 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.994
1.00
0.40%
COMPUTER FILE: CSA A23.3-04 WALL-002
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE CSA A23.3-04 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
WALL STRENGTH DETERMINED AS FOLLOWS:
f ʹc = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 1231.3 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = φc α1 f c′ab = 0.65 • 0.805 • 30 • 300a = 4709a
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
Ccw = φc α1 f c′•200• ( a - 200 )
Ccf = φc α1 f c′ ( 200•2500 )
Cs = φs As′1 ( fs1 − α1φc fc′) + φs As′2 ( fs 2 − α1φc fc′) + φs As′3 ( fs 3 − α1φc fc′)
T = φs As 4 fs 4 + φs As 5 fs 5 + φs As 6 fs 6
Pn1 = φc α1 fc′ • 200 • ( a − 200 ) + φc α1 fc′ ( 200 • 2500 ) + φs As′1 ( fs1 − α1φc fc′) +
φs As′2 ( fs 2 − α1φc fc′) + φs As′3 ( fs 3 − α1φc fc′) − φs As 4 fs 4 − φAs 5 fs 5 − φAs 6 fs 6
(Eqn. 1)
EXAMPLE CSA A23.3-04 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) + Cs 2 ( 4 s ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
(Eqn. 2)
Pn 2 =
e′
Cs 3 ( 3s ) − Ts 4 ( 2 s ) − Ts 5 ( s )
where Cs1 = φs As′1 ( fs1 − α1φc fc′) ; Csn = φs Asn′ ( fsn − α1φc fc′) ; Ts 4 = φs fsn Asn and the bar
strains are determined below. The plastic centroid is at the center of the section and
d ′′ = 700 mm
e′ =e + d ′′ =1231.3 + 1050 =2381.3 inch.
4) Using c = 1293.6 mm (from iteration),
a=
β1c =
0.895 •1293.6 =
1157.8 mm
5) Assuming the extreme fiber strain equals 0.0030 and c = 1293.6 mm, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain,
then f s = f y :
c − d′
ε s1 =
0.0035
c
c − s − d′
εs 2 =
0.0035
c
c − 2s − d ′
εs3 =
0.0035
c
d − c − 2s
=
εs 4
εs 6
d −c
d −c−s
=
εs5
εs 6
d −c
d −c
εs5 =
0.0035
c
= 0.00323; fs =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00198
f s 2 = 397.0 MPa
= 0.00074
f s 3 = 148.1 MPa
= 0.00175
f s 4 = 100.9 MPa
= 0.00299
f s 5 = 349.8 MPa
= 0.00230
f s 6 = 460.0 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 10687 kN
Pn2 = 10687 kN
M=
P=
10687(1231.3) /1000000 = 13159 kN-m
n
ne
EXAMPLE CSA A23.3-04 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE CSA A23.3-14 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete column is subjected to factored axial load Pu = 3870 kN
and moments Muy = 2109 kN-m. This wall is reinforced with two 30M bars at
each end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE CSA A23.3-14 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Flexural Demand/Capacity ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.986
1.00
-1.40%
COMPUTER FILE: CSA A23.3-14 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE CSA A23.3-14 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
f ′ c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 545 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
The distance to the neutral axis for a balanced condition, cb:
cb
=
700
700
dt
=
=
(1450 ) 875 mm
700 + f y
700 + 460
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = φc α1 fc′ab = 0.65 • 0.805 • 30 • 300a = 4709a
Cs = φs As′1 ( fs1 − α1 fc′) + φs As′2 ( fs 2 − α1 fc' ) + φs As′3 ( fs 3 − α1 fc′)
T = φs As 4 fs 4 + φAs 5 fs 5
=
Pn1 4709a + A1′ ( fs1 − 0.805 fc′) + A2′ ( fs 2 − 0.805 fc′) − φAs 3 fs 3 − φAs 4 fs 4 − φAs 5 fs 5 (Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
EXAMPLE CSA A23.3-14 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where Cs1 = φs As′1 ( fs1 − α1 fc′) ; Cs 2 = φs As′2 ( f s 2 − α1 f c′) ; Cs 3 = φs As′3 ( f s 3 − α1 f c′) ;
Ts 4 = φs f s 4 As 4 and the bar strains are determined below. The plastic centroid is at the
center of the section and d ′′ = 700 mm
e′ =e + d ′′ =545 + 700 =1245 inch.
4) Using c = 894.5 mm (from iteration), which is more than cb (875mm).
a = β1c = 0.895 • 894.5 = 800.6 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 643.6 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c−d '
ε s1 =
0.003
c
c−s−d '
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.0035
c
= 0.00330; f s =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00193
f s 2 = 387.0 MPa
= 0.00057
f s 3 = 113.1 MPa
= 0.00080
f s 4 = 160.8 MPa
= 0.00217
f s 5 = 434.7 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3870 kN
Pn2 = 3870 kN
M=
P=
3870(545) / 1000 = 2109 kN-m
n
ne
EXAMPLE CSA A23.3-14 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE CSA A23.3-14 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
10687 kN and moments Muy = 13159 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE CSA A23.3-14 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.994
1.00
0.40%
COMPUTER FILE: CSA A23.3-14 WALL-002
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE CSA A23.3-14 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
WALL STRENGTH DETERMINED AS FOLLOWS:
f ʹc = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 1231.3 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc = φc α1 f c′ab = 0.65 • 0.805 • 30 • 300a = 4709a
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
Ccw = φc α1 f c′•200• ( a - 200 )
Ccf = φc α1 f c′ ( 200•2500 )
Cs = φs As′1 ( fs1 − α1φc fc′) + φs As′2 ( fs 2 − α1φc fc′) + φs As′3 ( fs 3 − α1φc fc′)
T = φs As 4 fs 4 + φs As 5 fs 5 + φs As 6 fs 6
Pn1 = φc α1 fc′ • 200 • ( a − 200 ) + φc α1 fc′ ( 200 • 2500 ) + φs As′1 ( fs1 − α1φc fc′) +
φs As′2 ( fs 2 − α1φc fc′) + φs As′3 ( fs 3 − α1φc fc′) − φs As 4 fs 4 − φAs 5 fs 5 − φAs 6 fs 6
(Eqn. 1)
EXAMPLE CSA A23.3-14 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) + Cs 2 ( 4 s ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
(Eqn. 2)
Pn 2 =
e′
Cs 3 ( 3s ) − Ts 4 ( 2 s ) − Ts 5 ( s )
where Cs1 = φs As′1 ( fs1 − α1φc fc′) ; Csn = φs Asn′ ( fsn − α1φc fc′) ; Ts 4 = φs fsn Asn and the bar
strains are determined below. The plastic centroid is at the center of the section and
d ′′ = 700 mm
e′ =e + d ′′ =1231.3 + 1050 =2381.3 inch.
4) Using c = 1293.6 mm (from iteration),
a=
β1c =
0.895 •1293.6 =
1157.8 mm
5) Assuming the extreme fiber strain equals 0.0030 and c = 1293.6 mm, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain,
then f s = f y :
c − d′
ε s1 =
0.0035
c
c − s − d′
εs 2 =
0.0035
c
c − 2s − d ′
εs3 =
0.0035
c
d − c − 2s
=
εs 4
εs 6
d −c
d −c−s
=
εs5
εs 6
d −c
d −c
εs5 =
0.0035
c
= 0.00323; fs =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00198
f s 2 = 397.0 MPa
= 0.00074
f s 3 = 148.1 MPa
= 0.00175
f s 4 = 100.9 MPa
= 0.00299
f s 5 = 349.8 MPa
= 0.00230
f s 6 = 460.0 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 10687 kN
Pn2 = 10687 kN
M=
P=
10687(1231.3) /1000000 = 13159 kN-m
n
ne
EXAMPLE CSA A23.3-14 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Eurocode 2-2004 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 4340 kN and
moments Muy = 2503 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Eurocode 2-2004 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d=
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.993
1.00
0.70%
COMPUTER FILE: EUROCODE 2-2004 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Eurocode 2-2004 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
f ′c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 576.3 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
The distance to the neutral axis for a balanced condition, cb:
=
cb
700
700
=
dt
=
(1450 ) 922.7 mm
700 + f y / γ s
700 + 460 /1.15
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
α cc fck
1.0 • 30
=
Cc =
ab
=
• 300a 6000a
γm
1.5
As1
α cc fck
fs1 −
γs
γm
As 4
A
T
fs 4 + s 5 fs 5
=
γs
γs
Cs =
Pn1 =6000a +
As 2
α cc fck
+
fs 2 −
γm
γs
As1
α cc fck
fs1 −
γs
γm
As 3
α cc fck
fs 3 −
γs
γm
As 3
α cc fck
+
fs 3 −
γm
γs
As 2
α cc fck
+
fs 2 −
γm
γs
As 4
A
fs 4 − s 5 fs 5
−
γs
γs
+
(Eqn. 1)
3) Taking moments about As5:
EXAMPLE Eurocode 2-2004 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
Pn 2
=
a
1
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
=
Cs1
where
Ts 4 =
ETABS
0
(Eqn. 2)
As1
α cc fck
As 2
As 3
α cc f ck
α cc f ck
=
=
fs1 −
; Cs 2
fs3 −
;
fs2 −
; Cs 3
γs
γm
γs
γm
γs
γm
As 4
( f s 4 ) and the bar strains and stresses are determined below.
γs
The plastic centroid is at the center of the section and d ′′ = 700 mm
e′ =e + d ′′ =576.73 + 700 =1276.73 mm.
4) Using c = 885.33 mm (from iteration), which is more than cb (922.7mm).
a = λ1c = 0.80 • 885.33=708.3 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 885.33 mm, the steel
stresses and strains can be calculated. When the bar strain exceeds the yield strain,
then f s = f y :
c − d′
ε s E ≤ Fy ; f s1 = 460 MPa
ε s1 =
0.0035 = 0.00330; f s =
c
c − s − d′
f s 2 = 383.7 MPa
εs 2 =
0.0035 = 0.00192
c
d − c − 2s
f s 3 = 107.0 MPa
=
εs3
ε s 5 = 0.00054
d −c
d −c−s
= 0.00085
f s 4 = 169.7 MPa
=
εs 4
εs5
d −c
d −c
= 0.00223
f s 5 = 446.5 MPa
εs5 =
0.0035
c
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 4340 kN
Pn2 = 4340 kN
M=
P=
4340(708.3) /1000 = 2503 kN-m
n
ne
EXAMPLE Eurocode 2-2004 Wall-001 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
4
EXAMPLE Eurocode 2-2004 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to factored axial load Pu =
11605 kN and moments Muy = 15342 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Eurocode 2-2004 Wall-002 - 1
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
4
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.011
1.00
1.10%
COMPUTER FILE: EUROCODE 2-2004 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Eurocode 2-2004 Wall-002 - 2
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
4
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1322 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
Where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
α cc fck
0.85 • 30
Ccw=
• 200 • ( a − 200 )=
• 200 • ( a − 200 )= 3400(a − 200)
γm
1.5
α cc fck
0.85(30)
=
200 • ( 2500
=
− 1000 ) )
− 1000 ) ) 5,100, 000
Ccf
(
( 200 • ( 2500=
γm
1.5
α f
α f
Cs =A1′ fs1 − cc ck + A2′ fs 2 − cc ck
γm
γm
f
f
f
T = As 4 s 4 + As 5 s 4 + As 6 s 4
γs
γs
γs
Pn1= 3400(a − 200) + 5,100, 000 +
+
As 3
α cc fck
fs 3 −
γs
γm
α cc fck
+ A3′ fs 3 −
γm
As1
α cc fck
fs1 −
γs
γm
As 2
α cc fck
+
fs 2 −
γm
γs
As 4
A
A
fs 4 − s 5 fs 5 − s 6 fs 6
−
γs
γs
γs
(Eqn. 1)
EXAMPLE Eurocode 2-2004 Wall-002 - 3
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
4
3) Taking moments about As6:
a − tf
− tf + Cs1 ( d - d' ) +
1 Ccf ( d - d' ) + Ccw d 2
Pn2 =
e′
Cs2 ( 4s ) + Cs3 ( 3s ) − Ts4 ( 2s ) − Ts5 ( s )
=
Cs1
where
Ts 4 =
(Eqn. 2)
As1
α cc f ck
As 2
As 3
α cc f ck
α cc f ck
=
=
f s1 −
; Cs 2
fs3 −
fs2 −
; Cs 3
γs
γm
γs
γm
γs
γm
;
As 4
( f s 4 ) and the bar strains and stresses are determined below.
γs
The plastic centroid is at the center of the section and d ′′ = 700 mm
e′ =e + d ′′ =1322 + 700 =2472 mm.
4) Using c = 1299 mm (from iteration),
a=
β1c =
0.895 •1299 =
1163 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 1299 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c − d′
ε s1 =
0.0035
c
c − s − d′
εs 2 =
0.0035
c
c − 2s − d ′
εs3 =
0.0035
c
d − c − 2s
=
εs 4
εs6
d −c
d −c−s
=
εs5
εs6
d −c
d −c
εs6 =
0.0035
c
= 0.00323; f s =
ε s E ≤ Fy ; f s1 = 460 MPa
= 0.00199
f s 2 = 398.2 MPa
= 0.00075
f s 3 = 150.3 MPa
= 0.00049
f s 4 = 97.5 MPa
= 0.00173
f s 5 = 345.4 MPa
= 0.00297
f s 6 = 460.00 MPa
EXAMPLE Eurocode 2-2004 Wall-002 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
4
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 11605 kN
Pn2 = 11605 kN
M=
P=
11605(1322) /1000 = 15342 kN-m
n
ne
EXAMPLE Eurocode 2-2004 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Indian IS 456-2000 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected factored axial load Pu = 3146 kN and
moments Muy = 1875 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Indian IS 456-2000 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.035
1.00
3.50%
COMPUTER FILE: INDIAN IS 456-2000 WALL-001
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE Indian IS 456-2000 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
F ′c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 596 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
0.36
f ck ab = 0.4286 • 30 • 300a = 3857 a , where a = 0.84 xu
0.84
A′
A′
A′
Cs =s1 ( fs1 − 0.4286 fc′) + s 2 ( fs 2 − 0.4286 fc′) + s 3 ( fs 3 − 0.4286 fc′)
γs
γs
γs
A
A
T = s4 f s4 + s5 f s5
Cc =
γs
Pn1 =
3857 a +
γs
As′1
A′
( fs1 − 0.4286 fc′) + s 2 ( fs 2 − 0.4286 fc′) +
γs
γs
As′3
A
A
fs 3 − 0.4286 fc' ) − s 4 fs 4 + s 5 fs 5
(
γs
γs
γs
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
EXAMPLE Indian IS 456-2000 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
As1
As 2
′) ; C s 2
( f s1 − 0.4286 f c=
( f s 2 − 0.4286 f c′) ;
γs
γs
As 3
A
=
Cs 3
( f s 3 − 0.4286 f c′) ; Ts 4 = s 4 ( f s 4 ) and the bar strains and stresses are
γs
γs
determined below.
where
=
Cs1
The plastic centroid is at the center of the section and d " = 700 mm
e′ =e + d ′′ =596 + 700 =1296 mm.
4) Using c = 917.3 mm (from iteration)
a=
β1c =
0.84 • 917.3 =
770.5 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 917.3 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c−d '
= 0.00331;=
f s ε s E ≤ Fy ;
0.0035
c
c−s−d '
ε s2 =
0.0035 = 0.00197
c
c − 2s − d '
ε s3 =
0.0035 = 0.00064
c
d −c−s
= 0.00070
ε s4 =
ε s5
d −c
d −c
= 0.00203
ε s5 =
0.0035
c
ε s1 =
f s1 = 460 MPa
f s 2 = 394.8 MPa
f s 3 = 127.7 MPa
f s 4 = 139.4 MPa
f s 5 = 406.5 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3146 kN
Pn2 = 3146 kN
M=
P=
3146(596) /1000 = 1875 kN-m
n
ne
EXAMPLE Indian IS 456-2000 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Indian IS 456-2000 Wall-002
FRAME – P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to factored axial load Pu= 8426
kN and moments Muy= 11670 kN-m. This wall is reinforced as noted below.
The design capacity ratio is checked by hand calculations and results are
compared with ETABS program.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Indian IS 456-2000 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete Wall Demand/Capacity Ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.003
1.00
0.30%
COMPUTER FILE: INDIAN IS 456-2000 WALL-002
CONCLUSION
The ETABS results show a very close match with the independent results.
EXAMPLE Indian IS 456-2000 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
WALL STRENGTH DETERMINED AS FOLLOWS:
1) A value of e = 1385 mm was determined using e = M u / Pu where M u and Pu were taken
from the ETABS test model PMM interaction diagram for the pier, P1. Values for M u and
Pu were taken near the balanced condition and large enough to produce a flexural D/C ratio
very close to or equal to one. The depth to the neutral axis, c, was determined by iteration
using an excel spreadsheet so that equations 1 and 2 below were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
0.36
Ccw =
f ck •200• ( a - 200 ) , where a = 0.84x u
0.84
0.36
Ccf =
f ck 200 ( 2500-1000 )
0.84
A'
0.36 As2'
0.36 As3'
0.36
Cs = s1 f s1 f ck +
f
f ck +
f s3 f ck
s2
γs
0.84 γ s
0.84 γ s
0.84
A
A
A
T = s4 f s4 + s5 f s5 + s6 f s6
γs
γs
γs
As1'
0.36
0.36
0.36 As2'
0.36
Pn1 =
f ck •200• ( a - 200 ) +
f ck 200 ( 2500-1000 ) +
f s1 f ck +
f s2 f ck
0.84
0.84
0.84 γ s
0.84
γs
+
As3'
A
A
0.36 As4
f s3 f ck f s4 − s5 f s5 − s6 f s6
0.84 γ s
γs
γs
γs
(Eqn. 1)
EXAMPLE Indian IS 456-2000 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As6:
Pn2 =
a − tf
Ccf ( d - d' ) + Ccw d - 2 − tf + Cs1 ( d - d' ) + Cs2 ( 4s ) + Cs3 ( 3s ) − Ts4 ( 2s ) − Ts5 ( s )
(Eqn. 2)
A'
A'
A
0.36
0.36
Where Cs1 = s1 f s1 f ck ; Cs 2 = sn f sn f ck ; Ts 4 = sn ( f sn ) and the bar
γs
0.84
γs
0.84
γs
strains and stresses are determined below.
1
e'
The plastic centroid is at the center of the section and d " = 1150 mm
e' = e + d " = 1138 +1150 = 2535 mm.
4) Using c = 1298.1 mm (from iteration)
a = β1c = 0.84 • 1298.1=1090.4 mm
5) Assuming the extreme fiber strain equals 0.0035 and c= 1298.1 mm, the steel stresses and
strains can be calculated. When the bar strain exceeds the yield strain then, f s = f y :
c−d '
0.003
c
c−s−d '
ε s2 =
0.0035
c
c − 2s − d '
ε s3 =
0.0035
c
d − c − 2s
ε s4 =
ε s5
d −c
d −c−s
ε s5 =
ε s5
d −c
d −c
ε s6 =
0.0035
c
ε s1 =
=0.00323;=
f s ε s E ≤ Fy ;
f s1 = 460 MPa
=0.00199
f s 2 = 398.0 MPa
=0.00075
f s 3 = 150.0 MPa
=0.00049
f s 4 = 98.1 MPa
=0.00173
f s 5 = 346.1 MPa
=0.00297
f s 6 = 460.0 MPa
EXAMPLE Indian IS 456-2000 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Substitute in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the two equations
are equal gives,
Pn1 = 8426 kN
Pn2 = 8426 kN
M=
P=
8426(1385) /1000 = 11670 kN-m
n
ne
EXAMPLE Indian IS 456-2000 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE KBC 2009 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 4549 kN and
moments Muy = 2622 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE KBC 2009 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
t = 300 mm
h = 1500 mm
d=
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm2)
As2, As3, As4 = 2-15M (400 mm2)
f ck = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Flexural Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.002
1.00
0.2%
COMPUTER FILE: KBC 2009 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE KBC 2009 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 576.2 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
f ck ab 0.85 • 30
=
• 300a 7650a
Cs =
A1 ( f s1 − 0.85 • f ck ) + A2 ( f s 2 − 0.85 • f ck ) + A3 ( f s 3 − 0.85 • f ck )
T = As4 f s4 + As5 f s5
Pn1 =
7650a + A1 ( f s1 − 0.85 • f ck ) + A2 ( f s 2 − 0.85 • f ck ) +
(Eqn. 1)
A3 ( f s 3 − 0.85 • f ck ) − As 4 f s 4 − As 5 f s 5
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
where=
; Cs 2 A2 ( f s 2 − 0.85 • f ck ) ; Cs 3 ( f s 3 − 0.85 • f ck ) ;
Cs1 A1 ( f s1 − 0.85 • f ck )=
Ts 4 = f s 4 As 4 and the bar strains are determined below. The plastic centroid is at the
center of the section and d ′′ = 700mm
e′ =e + d ′′ =576.2 + 700 =1276.4 mm.
EXAMPLE KBC 2009 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Using c = 833.27 mm (from iteration),
a = β1c = 0.836 • 833.27=696.61 mm,
5) Assuming the extreme fiber strain equals 0.003 and c= 833.27 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
c − 2s − d '
εs3 =
0.003
c
d −c−s
εs 4 =
0.003
d −c
d −c
εs5 =
0.003
c
= 0.0028; f s =
ε s E ≤ Fy ;
f s1 = 460.00 MPa
= 0.0016
f s 2 = 312.0 MPa
= 0.0003
f s 3 = 60.0 MPa
= 0.00103
f s 4 = 259.5 MPa
= 0.0022
f s 5 = 444.1 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 5340 kN
Pn2 = 5340 kN
M=
P=
5340(576.4) /1000 = 3078 kN-m
n
ne
6) Calculate φ ,
φPn = 0.65 ( 5340 ) =
3471 kN
φM n = 0.65 ( 3078 ) =
2000.7 kN-m
EXAMPLE KBC 2009 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE KBC 2009 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
11256 kN and moments Muy = 1498 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE KBC 2009 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
t = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm2)
As2, As3, As4, As5 = 2-20M (600 mm2)
f ck = 30 MPa
fy = 420 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.007
1.00
0.7%
COMPUTER FILE: KBC 2009 WALL-002
CONCLUSION
The ETABS result shows a very close match with the independent result.
EXAMPLE KBC 2009 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1199.2 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
=
Ccw 0.85 f ck • 200 • ( a − 200 )
Ccf = 0.85 f ck ( 200 •1500 )
Cs =A1′ ( f s1 − 0.85 f ck ) + A2′ ( f s 2 − 0.85 f ck ) + A3′ ( f s 3 − 0.85 f ck )
T = As 4 fs 4 + As 5 fs 5 + As 6 fs 6
=
Pn1 0.85 f ck • 200 • ( a − 200 ) + 0.85 f ck ( 200 •1500 ) + A1′ ( f s1 − 0.85 f ck ) +
A2′ ( f s 2 − 0.85 f ck ) + A3′ ( f s 3 − 0.85 f ck ) + As 4 f s 4 + As 5 f s 5 + As 6 f s 6
(Eqn. 1)
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) + Cs 2 ( 4s ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
(Eqn. 2)
e′
Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
EXAMPLE KBC 2009 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where=
, Csn An′ ( f sn − 0.85 f ck ) , Tsn = fsn Asn , and the bar
Cs1 A1′ ( f s1 − 0.85 f ck )=
strains are determined below. The plastic centroid is at the center of the section, d ′′
2500 − 200
= 1150 mm
=
2
e′ =e + d ′′ =1199.2 + 1150 =2349.2 mm
4) Using c = 1480 mm (from iteration),
a=
β1c =
0.836 •1480 =
1237.28 mm
5) Assuming the extreme fiber strain equals 0.003 and c = 1480 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
c − 2s − d ′
εs3 =
0.003
c
d − c − 2s
=
εs 4
εs 6
d −c
d −c−s
=
εs5
εs 6
d −c
d −c
εs 6 =
0.003
c
= 0.0028; fs =
ε s E ≤ Fy ;
f s1 = 420.0 MPa
= 0.00186
f s 2 = 373.0 MPa
= 0.00093
f s 3 = 186.5 MPa
= 0.0000
f s 4 = 0.0 MPa
= 0.00093
f s 5 = 186.5 MPa
= 0.00272
f s 6 = 373.0 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
axial force from two equations are less than 1%
Pn1 = 13232 kN
Pn2 = 13250 kN , use average Pn = 13242 kN
M=
P=
13242(1199.2) /1000 = 15879.8 kN-m
n
ne
6) Calculate φ ,
=
φPn 0.85=
(13242 ) 11256 kN
=
φM n 0.85 (=
15879.8 ) 13498 kN-m
EXAMPLE KBC 2009 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Mexican RCDF-04 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete column is subjected factored axial load Pu = 3545 kN and
moments Muy = 1817 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Mexican RCDF-04 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d=
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Flexural Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.016
1.00
1.60%
COMPUTER FILE: MEXICAN RCDF-04 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Mexican RCDF-04 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
f ′c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 512.5 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc*ab 0.85 • 0.8 • 30
=
• 300a 6120a
Cs =−
A1 ( fs1 0.85 • 0.8 • fc* ) + A2 ( fs 2 − 0.85 • 0.8 • fc* ) + A3 ( fs 3 − 0.85 • 0.8 • fc* )
T = As4 f s4 + As5 f s5
Pn1 =+
6120a A1 ( fs1 − 0.85 • 0.8 • fc* ) + A2 ( fs 2 − 0.85 • 0.8 • fc* ) +
A3 ( fs 3 − 0.85 • 0.8 • fc* ) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
EXAMPLE Mexican RCDF-04 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where=
; Cs 2 A2 ( f s 2 − 0.85 • 0.8 • f c* ) ;
Cs1 A1 ( fs1 − 0.85 • 0.8 • fc* )=
Cs 3 ( f s 3 − 0.85 • 0.8 • f c* ) ; Ts 4 = f s 4 As 4 and the bar strains are determined below. The
plastic centroid is at the center of the section and d ′′ = 700mm
e′ =e + d ′′ =512.5 + 700 =1212.5 mm.
4) Using c = 936.2 mm (from iteration)
a =βc =0.85 • 916.2 =805 mm,
5) Assuming the extreme fiber strain equals 0.003 and c = 936.2 inch, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.0028; f s =
ε s E ≤ Fy ;
f s1 = 460.00 MPa
= 0.0017
f s 2 = 343.6 MPa
= 0.0005
f s 3 = 119.3 MPa
= 0.0060
f s 4 = 105.4 MPa
= 0.0175
f s 5 = 329.3 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 5064 kN
Pn2 = 5064 kN
M=
P=
5064(512.5) /1000000 = 2595 kN-m
n
ne
EXAMPLE Mexican RCDF-04 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
7) Calculate φPn and, φM n ,
φPn = 0.70 ( 5064 ) =
3545 kips
φM n = 0.70 ( 2595 ) =
1817 k-ft.
EXAMPLE Mexican RCDF-04 Wall-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Mexican RCDF-2004 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial loading and moment is tested in this
example. The wall is reinforced as shown below. The concrete core wall is
loaded with a factored axial load Pu = 10165 kN and moments Mu3 = 11430 kNm. The design capacity ratio is checked by hand calculations and results are
compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Mexican RCDF-2004 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.000
1.00
0.000%
COMPUTER FILE: MEXICAN RCDF-04 WALL-002
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE Mexican RCDF-2004 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1124.4 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
Ccw 0.85 • 0.8 fc′ • 200 • ( a − 200 )
=
Ccf = 0.85 • 0.8 fc′ ( 200 •1500 )
Cs =
A1 ( fs1 − 0.85 • 0.8 • fc′) + A2 ( fs 2 − 0.85 • 0.8 • fc′) + A3 ( fs 3 − 0.85 • 0.8 • fc′)
T = As4 f s4 + As5 f s5 + As6 f s6
Pn1 0.85 • 0.8 fc′ • 200 • ( a − 200 ) + 0.85 • 0.8 fc′ ( 200 •1500 ) + A1 ( fs1 − 0.85 • 0.8 • fc′)
=
+ A2 ( fs 2 − 0.85 • 0.8 • fc′) + A3 ( fs 3 − 0.85 • 0.8 • fc′) − As 4 fs 4 − As 5 fs 5 − As 6 fs 6
(Eqn. 1)
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
e′
Cs 2 ( 4s ) + Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
(Eqn. 2)
EXAMPLE Mexican RCDF-2004 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where
, Csn An ( f sn − 0.85 • 0.8 f c′) , Tsn = f sn Asn , and the
=
Cs1 A1 ( f s1 − 0.85 • 0.8 f c′)=
bar strains are determined below. The plastic centroid is at the center of the section,
2500 − 200
= 1150 mm
d ′′ =
2
e' = e + d " = 1124.4 +1150 = 2274.4 mm
4) Using c = 1413 mm (from iteration)
a = 0.85c = 0.85 • 1413=1201 mm
5) Assuming the extreme fiber strain equals 0.003 and c = 1413 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c−d '
0.003
c
c−s−d '
ε s2 =
0.003
c
c − 2s − d '
ε s3 =
0.003
c
d − c − 2s
ε s4 =
ε s6
d −c
d −c−s
ε s5 =
ε s6
d −c
d −c
ε s6 =
0.003
c
ε s1 =
= 0.00279;=
f s ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00181
f s 2 = 362.2 MPa
= 0.00083
f s 2 = 166.8 MPa
= 0.00014
f s 3 = 28.6 MPa
= 0.00112
f s 4 = 223.9 MPa
= 0.00210
f s 5 = 419.3 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 14522 kN
Pn2 = 14522 kN
M=
P=
14522(1124.4) /1000000 = 16328 kN-m
n
ne
6) Calculate φPn and φM n ,
=
φPn 0.70=
(14522 ) 10165 kN
=
φM n 0.70 (=
16382 ) 11430 kN-m
EXAMPLE Mexican RCDF-2004 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE NZS-3101-2006 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected to factored axial load Pu = 4549 kN and
moments Muy = 2622 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE NZS-3101-2006 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d=
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Flexural Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.000
1.00
0.00%
COMPUTER FILE: NZS 3101-06 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE NZS-3101-2006 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 576.2 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85 • 30
=
• 300a 7650a
Cs =
A1 ( fs1 − 0.85 • fc′) + A2 ( fs 2 − 0.85 • fc′) + A3 ( fs 3 − 0.85 • fc′)
T = As4 f s4 + As5 f s5
Pn1 =
7650a + A1 ( fs1 − 0.85 • fc′) + A2 ( fs 2 − 0.85 • fc′) +
A3 ( fs 3 − 0.85 • fc′) − As 4 fs 4 − As 5 fs 5
(Eqn. 1)
3) Taking moments about As5:
=
Pn 2
1
a
Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( 3s ) + Cs 3 ( 2s ) − Ts 4 ( s )
e′
2
(Eqn. 2)
where=
; Cs 2 A2 ( f s 2 − 0.85 • f c′) ; Cs 3 ( f s 3 − 0.85 • f c′) ;
Cs1 A1 ( fs1 − 0.85 • fc′)=
Ts 4 = f s 4 As 4 and the bar strains are determined below. The plastic centroid is at the
center of the section and d ′′ = 700mm
e′ =e + d ′′ =576.2 + 700 =1276.4 mm.
EXAMPLE NZS-3101-2006 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
4) Using c = 821.7 mm (from iteration),
a = γc = 0.85 • 821.7=698.46 mm,
5) Assuming the extreme fiber strain equals 0.003 and c= 821.7 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
−
−
2
d
c
s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.0028; f s =
ε s E ≤ Fy ;
f s1 = 460.00 MPa
= 0.0015
f s 2 = 307.9 MPa
= 0.0003
f s 3 = 52.3 MPa
= 0.0010
f s 4 = 203.2 MPa
= 0.0023
f s 5 = 458.8 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 5352 kN
Pn2 = 5352 kN
M=
P=
5352(576.4) /1000000 = 3085 kN-m
n
ne
6) Calculate φ ,
φPn = 0.85 ( 5352 ) =
4549 kN
φM n = 0.85 ( 3085 ) =
2622 kN-m
EXAMPLE NZS-3101-2006 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE NZS 3101-06 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
13625 kN and moments Muy = 16339 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE NZS 3101-06 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Demand/Capacity Ratio for a General Reinforcing pier section.
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.000
1.00
0.00%
COMPUTER FILE: NZS 3101-06 WALL-002
CONCLUSION
The ETABS result shows a very close match with the independent result.
EXAMPLE NZS 3101-06 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1199.2 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc 0.85
=
fc′ab 0.85 • 30
=
• 300a 7650a
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
Ccw 0.85 fc′ • 200 • ( a − 200 )
=
Ccf = 0.85 fc′ ( 200 • 2500 )
Cs =A1′ ( fs1 − 0.85 fc′) + A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′)
T = As 4 fs 4 + As 5 fs 5 + As 6 fs 6
=
Pn1 0.85 fc′ • 8 • ( a − 8 ) + 0.85 fc′ ( 8 • 98 ) + A1′ ( fs1 − 0.85 fc′) +
A2′ ( fs 2 − 0.85 fc′) + A3′ ( fs 3 − 0.85 fc′) + As 4 fs 4 + As 5 fs 5 + As 6 fs 6
(Eqn. 1)
3) Taking moments about As6:
a −tf
′
′
C
d
−
d
+
C
d
−
−
t
+
C
d
−
d
+
C
s
+
4
(
)
(
)
(
)
1 cf
cw
f
s1
s2
2
Pn 2 =
(Eqn. 2)
e′
Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
EXAMPLE NZS 3101-06 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where=
, Csn An′ ( fsn − 0.85 fc′) , Tsn = fsn Asn , and the bar strains
Cs1 A1′ ( fs1 − 0.85 fc′)=
are determined below. The plastic centroid is at the center of the section, d ′′
2500 − 200
= 1150 mm
=
2
e′ =e + d ′′ =1199.2 + 1150 =2349.2 mm
4) Using c = 1259.8 mm (from iteration),
a=
β1c =
0.85 •1259.8 =
1070.83 mm
5) Assuming the extreme fiber strain equals 0.003 and c = 1259.8 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
c − 2s − d ′
εs3 =
0.003
c
d − c − 2s
=
εs 4
εs 6
d −c
d −c−s
=
εs5
εs 6
d −c
d −c
εs 6 =
0.003
c
= 0.00276; fs =
ε s E ≤ Fy ; f s1 = 460.0 MPa
= 0.00167
f s 2 = 333.3 MPa
= 0.00057
f s 3 = 114.2 MPa
= 0.00052
f s 4 = 104.9 MPa
= 0.00167
f s 5 = 324.0 MPa
= 0.00272
f s 6 = 460.0 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 16029 kN
Pn2 = 16029 kN
M=
P=
16029(1199.2) /1000000 = 19222 kN-m
n
ne
6) Calculate φ ,
=
φPn 0.85=
(16029 ) 13625 kN
=
φM n 0.85 (=
19222 ) 16339 kN-m
EXAMPLE NZS 3101-06 Wall-002 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
EXAMPLE Singapore CP65-99 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected factored axial load Pu = 3246 kN and
moments Muy = 1969 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Singapore CP65-99 Wall-001 - 1
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d’ =
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.997
1.00
0.30%
COMPUTER FILE: SINGAPORE CP65-99 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Singapore CP65-99 Wall-001 - 2
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
HAND CALCULATION
Wall Strength Determined as follows:
f ’c = 30MPa
b = 300mm
fy = 460 MPa
h = 1500 mm
1) A value of e = 606.5 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
The distance to the neutral axis for a balanced condition, cb:
cb =
700
700
dt =
(1450 ) = 922.7 mm
700 + f y / γ s
700 + 460 /1.15
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
=
Cc
0.67
0.67
=
fcu ab
•=
30 • 300a 4020a
γm
1.5
As′1
0.67 As′2
0.67 As′3
0.67
fc′ +
fc′ +
fc′
fs1 −
fs 2 −
fs 3 −
γs
γm
γm
γm
γs
γs
As 4
A
=
T
fs 4 + s 5 fs 5
γs
γs
Cs =
As′1
0.67 As′2
0.67
fc′ +
fc′ +
fs1 −
fs 2 −
γs
γm
γ
γ
s
m
As′3
A
0.67 As 4
fc′ −
fs 4 + s 5 fs 5
fs 3 −
γs
γm
γs
γs
4709a +
Pn1 =
3) Taking moments about As5:
a
1 Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) +
2
Pn 2 =
e′
Cs 3 ( 2s ) − Ts 4 ( s )
(Eqn. 1)
(Eqn. 2)
EXAMPLE Singapore CP65-99 Wall-001 - 3
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
As1
As 2
0.67
0.67
′ ; Cs 2
f c=
f c′ ;
f s1 −
fs2 −
γs
γm
γs
γm
As 3
As 4
0.67
0.67
′
=
Cs 3
f=
f c′ and the bar strains and
fs3 −
fs4 −
c ; Ts 4
γs
γm
γs
γm
stresses are determined below.
=
Cs1
where
The plastic centroid is at the center of the section and d ′′ = 700 mm
e′ =e + d ′′ =606.5 + 700 =1306.5 mm.
4) Using c = 887.5 mm (from iteration), which is slightly more than cb (922.7mm).
a=
β1c =
0.90 • 875.2 =
787.6 mm
5) Assuming the extreme fiber strain equals 0.0035 and c = 875.2 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c − d′
ε s1 =
0.0035
c
c − s − d′
εs 2 =
0.0035
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.0035
c
= 0.00330; f s =
ε s E ≤ Fy ; f s1 = 460 MPa
= 0.00190
f s 2 = 380.1 MPa
= 0.00050
f s 3 = 100.1 MPa
= 0.00090
f s 4 = 179.8 MPa
= 0.00230
f s 5 = 459.7 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3246 kN
Pn2 = 3246 kN
M=
P=
3246(606.5) /1000 = 1969 kN-m
n
ne
EXAMPLE Singapore CP65-99 Wall-001 - 4
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
EXAMPLE Singapore CP65-99 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to a factored axial load Pu =
8368 kN and moments Muy = 11967 kN-m. This wall is reinforced as noted
below. The design capacity ratio is checked by hand calculations and the results
are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Singapore CP65-99 Wall-002 - 1
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 30 MPa
fy = 460 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
1.001
1.00
0.10%
COMPUTER FILE: SINGAPORE CP65-99 WALL-002
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Singapore CP65-99 Wall-002 - 2
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1430 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
0.67
=
Ccw
fcu • 200 • ( a − 200 )
γm
0.67
Ccf =
f cu ( 200•2500 )
γm
As′1
0.67 As′2
0.67 As′3
0.67
fc′ +
fc′ +
fc′
fs1 −
fs 2 −
fs 3 −
γs
γm
γm
γm
γs
γs
A
A
A
T = s 4 fs 4 + s 5 fs 5 + s 6 fs 6
γs
γs
γs
Cs =
=
Pn1
A′
0.67
0.67
0.67
fcu • 200 • ( a - 200 ) +
fcu ( 200 • 2500 ) + s1 fs1 −
fc′ +
γm
γm
γs
γm
(Eqn. 1)
As′2
As 5
As 6
0.67 As′3
0.67 As 4
fc′ +
fc′ −
fs 4 +
fs 5 +
fs 6
fs 2 −
fs 3 −
γs
γm
γm
γs
γs
γs
γs
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) + Cs 2 ( 4 s ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
e′
Cs 3 ( 3s ) − Ts 4 ( 2 s ) − Ts 5 ( s )
(Eqn. 2)
EXAMPLE Singapore CP65-99 Wall-002 - 3
Software Verification
PROGRAM NAME: ETABS
REVISION NO.:
0
As1
Asn
Asn
0.67
0.67
0.67
f c′=
f c′=
f c′
f s1 −
; Csn
f sn −
; Tsn
f sn −
γs
γm
γs
γm
γs
γm
and the bar strains and stresses are determined below.
=
Cs1
where
The plastic centroid is at the center of the section and d ′′ = 1150 mm
e′ =e + d ′′ =1430 + 1150 =2580 mm.
4) Using c = 1160 mm (from iteration),
a=
β1c =
0.9 •1160 =
1044 mm
5) Assuming the extreme fiber strain equals 0.0035 and c= 1160 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
= 0.00320; f s =
ε s E ≤ Fy ;
ε s1 =
0.0035
c
c−s−d
= 0.00181
εs 2 =
0.0035
c
c − 2s − d ′
εs3 =
0.0035 = 0.00042
c
d − c − 2s
= 0.00097
=
εs 4
εs6
d −c
d −c−s
= 0.00235
=
εs5
εs6
d −c
d −c
= 0.00374
εs6 =
0.0035
c
f s1 = 460 MPa
f s 2 = 362.0 MPa
f s 3 = 84.4 MPa
f s 4 = 193.2 MPa
f s 5 = 460.00 MPa
f s 6 = 460.00 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 8368 kN
Pn2 = 8368 kN
M=
P=
8368(1430) /1000 = 11,967 kN-m
n
ne
EXAMPLE Singapore CP65-99 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Turkish TS 500-2000 Wall-001
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example.
A reinforced concrete wall is subjected factored axial load Pu = 3132 kN and
moments Muy = 1956 kN-m. This wall is reinforced with two 30M bars at each
end and 15M bars at 350 mm on center of each face. The total area of
reinforcement is 4000 mm2. The design capacity ratio is checked by hand
calculations and the results are compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Turkish TS 500-2000 Wall-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 300 mm
h = 1500 mm
d=
50 mm
s=
350 mm
As1= As5 = 2-30M (1400 mm^2)
As2, As3, As4 = 2-15M (400 mm^2)
f ′c = 25 MPa
fy = 420 MPa
TECHNICAL FEATURES OF ETABS TESTED
Wall flexural demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.997
1.00
0.30%
COMPUTER FILE: TURKISH TS 500-2000 WALL-001
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EXAMPLE Turkish TS 500-2000 Wall-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
f’c = 25 MPa
b = 300mm
fy = 420 MPa
h = 1500 mm
1) A value of e = 715 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for pier P1. Values for
M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
fck
0.67
=
Cc 0.85
=
ab
•=
25 • 300a 3350a
γc
1.5
A′
0.85 As′2
0.85 As′3
0.85
Cs = s1 fs1 −
fck +
fck +
fck
fs 2 −
fs 3 −
γs
γc
γc
γc
γs
γs
As 4
A
=
T
fs 4 + s 5 fs 5
γs
γs
As′1
0.85 As′2
0.85
fck +
fck +
fs1 −
fs 2 −
γs
γc
γc
γs
As′3
A
0.85 As 4
fck −
fs 4 + s 5 fs 5
fs 3 −
γs
γc
γs
γs
Pn1 =
3350a +
(Eqn. 1)
3) Taking moments about As5:
a
1 Cc d − + Cs1 ( d − d ′ ) + Cs 2 ( d − d ′ − s ) +
2
Pn 2 =
e′
Cs 3 ( 2s ) − Ts 4 ( s )
(Eqn. 2)
EXAMPLE Turkish TS 500-2000 Wall-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
As1
As 2
0.85
0.85
f ck=
f ck ;
f s1 −
; Cs 2
fs2 −
γs
γc
γs
γc
As 3
As 4
0.85
0.85
=
f ck and the bar strains and stresses
Cs 3
f ck=
fs4 −
fs3 −
; Ts 4
γs
γc
γs
γc
are determined below.
=
Cs1
where
The plastic centroid is at the center of the section and d ′′ = 700 mm
e′ =e + d ′′ =715 + 700 =1415 mm.
4) Using c = 853.4 mm (from iteration),
=
a k=
0.85 • 853.4
= 725.4 mm
1c
5) Assuming the extreme fiber strain equals 0.0030 and c = 853.4 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain then,
fs = f y :
c − d′
ε s1 =
0.003
c
c − s − d′
εs 2 =
0.003
c
d − c − 2s
=
εs3
εs5
d −c
d −c−s
=
εs 4
εs5
d −c
d −c
εs5 =
0.003
c
= 0.00282; f s =
ε s E ≤ Fy ;
f s1 = 420.0 MPa
= 0.00159
f s 2 = 318.8 MPa
= 0.00036
f s 3 = 72.7 MPa
= 0.00087
f s 4 = 173.4 MPa
= 0.00210
f s 5 = 419.5 MPa
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 3132 kN
Pn2 = 3132 kN
M=
P=
3132(624.4) /1000 = 1956 kN-m
n
ne
EXAMPLE Turkish TS 500-2000 Wall-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EXAMPLE Turkish TS 500-2000 Wall-002
P-M INTERACTION CHECK FOR A WALL
EXAMPLE DESCRIPTION
The Demand/Capacity ratio for a given axial load and moment are tested in this
example. A reinforced concrete wall is subjected to factored axial load Pu = 9134
kN and moments Muy = 11952 kN-m. This wall is reinforced as noted below. The
design capacity ratio is checked by hand calculations and the results are
compared with ETABS program results.
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE Turkish TS 500-2000 Wall-002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Material Properties
E=
ν=
25000 MPa
0.2
ETABS
0
Design Properties
Section Properties
tb = 200 mm
H = 2500 mm
d=
2400 mm
s=
460 mm
As1= As5 = 4-35M+2-20M (4600 mm^2)
As2, As3, As4, As5 = 2-20M (600 mm^2)
f ′c = 25 MPa
fy = 420 MPa
TECHNICAL FEATURES OF ETABS TESTED
Concrete wall demand/capacity ratio
RESULTS COMPARISON
Independent results are hand calculated and compared with ETABS design
check.
Output Parameter
ETABS
Independent
Percent
Difference
Wall Demand/Capacity Ratio
0.996
1.00
0.40%
COMPUTER FILE: TURKISH TS 500-2000 WALL-002
CONCLUSION
The ETABS results show an acceptable match with the independent results.
EXAMPLE Turkish TS 500-2000 Wall-002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Wall Strength Determined as follows:
1) A value of e = 1308.6 mm was determined using e = M u / Pu where M u and Pu were
taken from the ETABS test model PMM interaction diagram for the pier, P1. Values
for M u and Pu were taken near the balanced condition and large enough to produce a
flexural D/C ratio very close to or equal to one. The depth to the neutral axis, c, was
determined by iteration using an excel spreadsheet so that equations 1 and 2 below
were equal.
2) From the equation of equilibrium:
Pn = Cc + Cs − T
where
Cc =Ccw +Ccf , where Ccw and Ccf are the area of the concrete web and flange in
compression
f
0.85 • 30
• 200 • ( a − 200 )= 3400(a − 200)
Ccw= ck • 200 • ( a − 200 )=
1.5
γc
f
0.85(30)
=
Ccf 0.85 • ck ( 200 • ( 2500
=
− 1000 ) )
=
− 1000 ) ) 5,100, 000
( 200 • ( 2500
1.5
γc
f
f
f
Cs =
A1′ fs1 − 0.85 • ck + A2′ fs 2 − 0.85 • ck + A3′ fs 3 − 0.85 • ck
γc
γc
γc
f
f
f
T = As 4 s 4 + As 5 s 4 + As 6 s 4
γs
γs
γs
Pn1= 3400(a − 200) + 5,100, 000 +
+
As1
fck As 2
fck
fs1 − 0.85 •
+
fs 2 − 0.85 •
γs
γc γs
γc
As 3
fck As 4
A
A
fs 4 − s 5 fs 5 − s 6 fs 6
fs 3 − 0.85 •
−
γs
γc γs
γs
γs
(Eqn. 1)
EXAMPLE Turkish TS 500-2000 Wall-002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
3) Taking moments about As6:
a −tf
− t f + Cs1 ( d − d ′ ) + Cs 2 ( 4s ) +
1 Ccf ( d − d ′ ) + Ccw d −
2
Pn 2 =
(Eqn. 2)
e′
Cs 3 ( 3s ) − Ts 4 ( 2s ) − Ts 5 ( s )
where
=
Cs1
As1
As 2
As 3
0.85
0.85
0.85
fck ;
fck=
fck=
fs 3 −
fs1 −
; Cs 2
fs 2 −
; Cs 3
γ
γ
γs
γc
γ
γ
s
c
s
c
As 4
and the bar strains and stresses are determined below.
γs
The plastic centroid is at the center of the section and d ′′ = 1150 mm
e′ =e + d ′′ =1308.6 + 1150 =2458.6 mm.
Ts 4 =
4) Using c = 1327 mm (from iteration)
=
a k=
0.85 •1327
= 1061.1 mm
1c
5) Assuming the extreme fiber strain equals 0.003 and c = 1327 mm, the steel stresses
and strains can be calculated. When the bar strain exceeds the yield strain, then
fs = f y :
c − d′
ε s1 =
0.003
c
c−s−d '
ε s2 =
0.003
c
c − 2s − d '
ε s3 =
0.003
c
d − c − 2s
ε s4 =
ε s6
d −c
d −c−s
ε s5 =
ε s6
d −c
d −c
ε s6 =
0.003
c
= 0.00277;=
f s ε s E ≤ Fy ; f s1 = 420.0 MPa
= 0.00173
f s 2 = 346.8 MPa
= 0.00069
f s 3 = 138.8 MPa
= 0.00035
f s 4 = 69.2 MPa
= 0.00139
f s 5 = 277.2 MPa
= 0.00243
f s 6 = 420.0 MPa
EXAMPLE Turkish TS 500-2000 Wall-002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Substituting in Eqn. 1 and 2 and iterating the value of the neutral axis depth until the
two equations are equal give
Pn1 = 9134 kN
Pn2 = 9134 kN
M=
P=
9134(1308.6) /1000 = 11952 kN-m
n
ne
EXAMPLE Turkish TS 500-2000 Wall-002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
AISC-360-05 Example 001
COMPOSITE GIRDER DESIGN
EXAMPLE DESCRIPTION
A series of 45-ft. span composite beams at 10 ft. o/c carry the loads shown
below. The beams are ASTM A992 and are unshored during construction. The
concrete has a specified compressive strength, fc′ = 4 ksi. Design a typical floor
beam with 3-in., 18-gage composite deck and 4 ½ in. normal weight concrete
above the deck, for fire protection and mass. Select an appropriate beam and
determine the required number of ¾ in.-diameter shear studs.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W21x55
E = 29000 ksi
Fy = 50 ksi
Loading
w = 830 plf (Dead Load)
w = 200 plf (Construction)
w = 100 plf (SDL)
w = 1000 plf (Live Load)
Geometry
Span, L = 45 ft
AISC-360-05 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
RESULTS COMPARISON
Independent results are referenced from Example I.1 from the AISC Design
Examples, Version 13.0.
ETABS
Independent
Percent
Difference
Pre-composite Mu (k-ft)
333.15
333.15
0.00%
Pre-composite ΦbMn (k-ft)
472.5
472.5
0.00%
2.3
2.3
0.00%
Required Strength Mu (k-ft)
687.5
687.5
0.00%
Full Composite ΦbMn (k-ft)
1027.1
1027.1
0.00%
Partial Composite ΦbMn (k-ft)
770.3
770.3
0.00 %
Shear Stud Capacity Qn
17.2
17.2
0.00 %
Shear Stud Distribution
35
34
2.9%
Live Load Deflection (in.)
1.35
1.30
3.70%
Required Strength Vu (kip)
61.1
61.1
0.00%
ΦVn (k)
234
234
0.00%
Output Parameter
Pre-composite Deflection (in.)
AISC-360-05 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
COMPUTER FILE: AISC-360-05 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
The live load deflection differs due to a difference in methodology. In the AISC
example, the live load deflection is computed based on a lower bound value of
the beam moment of inertia, whereas in ETABS, it is computed based on the
approximate value of the beam moment of inertia derived from Equation (C-I3-6)
from the Commentary on the AISC Load and Resistance Factor Design
Specification – Second Edition.
AISC-360-05 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
HAND CALCULATION
Properties:
Materials:
ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi, wsteel = 490 pcf
4000 psi normal weight concrete
Ec = 3,644 ksi, fc′ = 4 ksi, wconcrete = 145 pcf
Section:
W21x55
d = 20.8 in, bf = 8.22 in, tf = 0.522 in, tw = 0.38 in, h = 18.75 in., rfillet = 0.5 in.
Asteel = 16.2 in2, Ssteel = 109.6 in3, Zsteel = 126 in3, Isteel = 1140 in4
Deck:
tc =4 ½ in., hr = 3 in., sr =12 in., wr = 6 in.
Shear Connectors:
d = ¾ in, h =4 ½ in, Fu = 65 ksi
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
wD =(10 • 77.5 + 55.125) • 10−3 =0.830125 kip/ft
wL = 10 • 20 • 10−3 = 0.200 kip/ft
wu = 1.2 • 0.830125 + 1.6 • 0.200 = 1.31615 kip/ft
wu • L2 1.31615 • 452
=
Mu =
= 333.15 kip-ft
8
8
Moment Capacity:
Φ b M n =Φ b • Z s • Fy =( 0.9 • 126 • 50 ) 12 =472.5 kip-ft
AISC-360-05 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Pre-Composite Deflection:
0.830
4
• ( 45 • 12 )
5wD L
12
=
∆ nc
=
= 2.31 in.
384 EI 384 • 29, 000 • 1,140
4
5•
Design for Composite Flexural Strength:
Required Flexural Strength:
wu = 1.2 • 0.830 + 1.2 • 0.100 + 1.6 • 1 = 2.71 kip/ft
wu • L2 2.68 • 452
Mu =
=
= 687.5 kip-ft
8
8
Full Composite Action Available Flexural Strength:
Effective width of slab:
10.0
45.0 ft
beff =
• 2 sides =10.0 ft ≤
=11.25 ft
2
8
Resistance of steel in tension:
C = Py = As • Fy = 16.2 • 50 = 810 kips controls
Resistance of slab in compression:
Ac = beff • tc =
(10 • 12 ) • 4.5 =
540 in
2
C= 0.85 • f 'c A=
0.85 • 4 • 540
= 1836 kips
c
Depth of compression block within slab:
=
a
C
810
=
= 1.99 in.
0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4
Moment resistance of composite beam for full composite action:
d1 = (tc + hr ) −
2.00
a
= (4.5 + 3) −
= 6.51 in.
2
2
d
20.8 / 12
ΦM n =
Φ Py • d1 + Py • =
0.9 810 • 6.51 / 12 + 810 •
1027.1 kip-ft
=
2
2
AISC-360-05 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Partial Composite Action Available Flexural Strength:
Assume 36.1% composite action:
C= 0.361 • P=
0.361 • 810= 292.4 kips
y
Depth of compression block within concrete slab:
C
292.4
=
= 0.72 in.
0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4
=
a
( tc + hr ) −
d1 =
a
=
2
( 4.5 + 3) −
0.72
= 7.14 in.
2
Compression force within steel section:
(P
y
− C ) 2 =( 810 − 292.4 ) 2 =258.8 kips
Tensile resistance of one flange:
Fflange = b f • t f • Fy = 8.22 • 0.522 • 50 = 214.5 kip
Tensile resistance of web:
Fweb = T • tw • Fy = 18.75 • 0.375 • 50 = 351.75 kips
Tensile resistance of one fillet area:
Ffillet=
(P − 2• F
y
flange
− Fweb ) 2=
(810 − 2 • 214.5 − 351.2 )
2= 14.6 kips
Compression force in web:
Cweb =( Py − C ) / 2 − Fflange − Ffillet =258.8 − 214.5 − 14.6 =29.7 kips
Depth of compression block in web:
x=
Cweb
29.7
• T=
• 18.76= 1.584 in.
Fweb
351.75
Location of centroid of steel compression force measured from top of steel section:
d2
0.5 • t f • Fflange + ( t f + 0.5 • rfillet ) • Ffillet + ( t f + rfillet + 0.5 • x ) • C web
=
( Py − C ) / 2
0.5 • 0.522 • 214.5 + ( 0.522 + 0.5 • 0.5) • 14.6 + ( 0.522 + 0.5 + 0.5 • 1.58) • 29.7
= 0.467 in.
258.8
AISC-360-05 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Moment resistance of composite beam for partial composite action:
ΦM n =
Φ C • ( d1 + d 2 ) + Py • ( d 3 − d 2 )
20.8
12 770.3 kip-ft
= 0.9 292.4 • ( 7.14 + 0.467 ) + 810 •
− 0.467 =
2
Shear Stud Strength:
From AISC Manual Table 3.21 assuming one shear stud per rib placed in the
weak position, the strength of ¾ in.-diameter shear studs in normal weight
concrete with f c′ = 4 ksi and deck oriented perpendicular to the beam is:
Qn = 17.2 kips
Shear Stud Distribution:
=
n
ΣQn 292.4
=
= 17 from each end to mid-span, rounded up to 35 total
17.2
Qn
Live Load Deflection:
Modulus of elasticity ratio:
n E=
Ec 29, 000 3,=
=
644 8.0
Transformed elastic moment of inertia assuming full composite action:
Transformed
Area
A (in2)
Moment Arm
from
Centroid
y (in.)
Ay
(in.3)
Ay2
(in,4)
I0
(in.4)
Slab
67.9
15.65
1,062
16,620
115
W21x50
16.2
0
0
0
1,140
1,062
16,620
1,255
Element
84.1
I x =I 0 + Ay 2 =
1, 255 + 16,620 =
17,874 in.4
=
y
1, 062
= 12.6 in.
84.1
2
I tr = I x − A • y = 17,874 − 82.6 • 12.62 = 4, 458 in 4
AISC-360-05 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Effective moment inertia assuming partial composite action:
I equiv = I s + ΣQn Py ( I tr − I s ) = 1,140 + 0.361 ( 4, 458 − 1,140 ) = 3,133 in 4
I eff = 0.75 • I equiv = 0.75 • 3,133 = 2,350 in 4
5 • (1 12 ) • ( 30 • 12 )
5wL L4
=
∆ LL
=
= 1.35 in.
384 EI eff 384 • 29, 000 • 2,350
4
Design for Shear Strength:
Required Shear Strength:
wu = 1.2 • 0.830 + 1.2 • 0.100 + 1.6 • 1 = 2.71 kip/ft
wu • L 2.71 • 45
=
= 61.1 kip-ft
Vu =
2
2
Available Shear Strength:
1.0 • 0.6 • 20.8 • 0.375 • 50 =
234 kips
ΦVn =
Φ • 0.6 • d • t w • Fy =
AISC-360-05 Example 001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
AISC-360-10 Example 001
COMPOSITE GIRDER DESIGN
EXAMPLE DESCRIPTION
A typical bay of a composite floor system is illustrated below. Select an
appropriate ASTM A992 W-shaped beam and determine the required number of
¾ in.-diameter steel headed stud anchors. The beam will not be shored during
construction. To achieve a two-hour fire rating without the application of spray
applied fire protection material to the composite deck, 4 ½ in. of normal weight
(145 lb/ft3) concrete will be placed above the top of the deck. The concrete has a
specified compressive strength, fc′ = 4 ksi.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W21x50
E = 29000 ksi
Fy = 50 ksi
Loading
w = 800 plf (Dead Load)
w = 250 plf (Construction)
w = 100 plf (SDL)
w = 1000 plf (Live Load)
Geometry
Span, L = 45 ft
AISC-360-10 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
RESULTS COMPARISON
Independent results are referenced from Example I.1 from the AISC Design
Examples, Version 14.0.
ETABS
Independent
Percent
Difference
Pre-composite Mu (k-ft)
344.2
344.2
0.00%
Pre-composite ΦbMn (k-ft)
412.5
412.5
0.00%
2.6
2.6
0.00%
Required Strength Mu (k-ft)
678.3
678.4
0.01%
Full Composite ΦbMn (k-ft)
937.1
937.1
0.00%
Partial Composite ΦbMn (k-ft)
763.2
763.2
0.00%
Shear Stud Capacity Qn
17.2; 14.6
17.2; 14.6
0.00%
Shear Stud Distribution
46
46
0.00%
Live Load Deflection (in.)
1.34
1.26
6.0%
Required Strength Vu (kip)
60.3
60.3
0.00%
ΦVn (k)
237.1
237.1
0.00%
Output Parameter
Pre-composite Deflection (in.)
AISC-360-10 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
COMPUTER FILE: AISC-360-10 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
The live load deflection differs due to a difference in methodology. In the AISC
example, the live load deflection is computed based on a lower bound value of
the beam moment of inertia, whereas in ETABS, it is computed based on the
approximate value of the beam moment of inertia derived from Equation (C-I3-6)
from the Commentary on the AISC Load and Resistance Factor Design
Specification – Second Edition.
AISC-360-10 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
HAND CALCULATION
Properties:
Materials:
ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi, wsteel = 490 pcf
4000 psi normal weight concrete
Ec = 3,644 ksi, f c′ = 4 ksi, wconcrete = 145 pcf
Section:
W21x50
d = 20.8 in, bf = 6.53 in, tf = 0.535 in, tw = 0.38 in, k = 1.04 in
Asteel = 14.7 in2, Ssteel = 94.6 in3, Zsteel = 110 in3, Isteel = 984 in4
Deck:
tc =4 ½ in., hr = 3 in., sr =12 in., wr = 6 in.
Shear Connectors:
d = ¾ in, h =4 ½ in, Fu = 65 ksi
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
wD = (10 • 75 + 50) • 10−3 = 0.800 kip/ft
wL = 10 • 25 • 10−3 = 0.250 kip/ft
wu = 1.2 • 0.800 + 1.6 • 0.250 = 1.36 kip/ft
=
Mu
wu • L2 1.36 • 452
=
= 344.25 kip-ft
8
8
Moment Capacity:
Φ b M n =Φ b • Z s • Fy =(0.9 • 110 • 50) 12 =412.5 kip-ft
AISC-360-10 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Pre-Composite Deflection:
5wD L4
=
∆ nc
=
384 EI
0.800
4
• ( 45 • 12 )
12
= 2.59 in.
384 • 29, 000 • 984
5•
Camber
= 0.8 • ∆ nc
= 0.8 • 2.59
= 2.07 in., which is rounded down to 2 in.
Design for Composite Flexural Strength:
Required Flexural Strength:
wu = 1.2 • 0.800 + 1.2 • 0.100 + 1.6 • 1 = 2.68 kip/ft
wu • L2 2.68 • 452
=
Mu =
= 678.38 kip-ft
8
8
Full Composite Action Available Flexural Strength:
Effective width of slab:
10.0
45.0 ft
beff =
• 2 sides =10.0 ft ≤
=11.25 ft
2
8
Resistance of steel in tension:
C = Py = As • Fy = 14.7 • 50 = 735 kips controls
Resistance of slab in compression:
Ac = beff • tc =
(10 • 12 ) • 4.5 =
540 in
2
C= 0.85 • f 'c A=
0.85 • 4 • 540
= 1836 kips
c
Depth of compression block within slab:
a
=
C
735
=
= 1.80 in.
0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4
Moment resistance of composite beam for full composite action:
a
1.80
d1 = ( tc + hr ) − = ( 4.5 + 3) −
= 6.60 in.
2
2
d
20.8 /12
ΦM n =
Φ Py • d1 + Py • =
0.9 735 • 6.60 /12 + 735 •
937.1 kip-ft
=
2
2
AISC-360-10 Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Partial Composite Action Available Flexural Strength:
Assume 50.9% composite action:
C= 0.509 • P=
373.9 kips
y
Depth of compression block within concrete slab:
373.9
C
=
= 0.92 in.
0.85 • beff • f 'c 0.85 • (10 • 12 ) • 4
a
=
d =
1
(tc + hr ) − a2 =
( 4.5 + 3) −
0.92
= 7.04 in.
2
Compressive force in steel section:
Py − C 735 − 373.9
=
= 180.6 kips
2
2
Steel section flange ultimate compressive force:
C flange = b f • t f • Fy = 6.53 • 0.535 • 50 = 174.7 kips
Steel section web (excluding fillet areas) ultimate compressive force:
Cweb = ( d − 2 • k ) • tw • Fy = (20.8 − 2 • 1.04) • 0.38 • 50 = 355.7 kips
Steel section fillet ultimate compressive force:
=
C fillet
Py − (2 • C flange + Cweb ) 735 − (2 • 174.7 + 355.7)
=
= 14.5 kips
2
2
Assuming a rectangular fillet area, the distance from the bottom of the top flange to
the neutral axis of the composite section is:
( P − C ) / 2 − C flange
x =(k − t f ) • y
C fillet
180.6 − 174.7
=(1.04 − 0.535) •
=0.20 in.
14.98
AISC-360-10 Example 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Distance from the centroid of the compressive force in the steel section to the top of
the steel section:
d2 =
C flange • t f / 2 + ((Py − C ) / 2 − C flange ) • (t f + x / 2)
( Py − C ) / 2
174.7 • 0.535 / 2 + (180.6 − 174.7) • (0.535 + 0.2 / 2)
= 0.279 in.
180.6
Moment resistance of composite beam for partial composite action:
ΦM n =
Φ C • ( d1 + d 2 ) + Py • ( d 3 − d 2 )
20.8
12 763.2 kip-ft
= 0.9 373.9 • ( 7.04 + 0.279 ) + 735 •
− 0.279 =
2
Shear Stud Strength:
From AISC Manual Table 3.21, assuming the shear studs are placed in the weak
position, the strength of ¾ in.-diameter shear studs in normal weight concrete with
f c′ = 4 ksi and deck oriented perpendicular to the beam is:
Qn = 17.2 kips for one shear stud per deck flute
Qn = 14.6 kips for two shear studs per deck flute
Shear Stud Distribution:
There are at most 22 deck flutes along each half of the clear span of the beam.
ETABS only counts the studs in the first 21 deck flutes as the 22nd flute is potentially
too close to the point of zero moment for any stud located in it to be effective. With
two shear studs in the first flute, 20 in the next in the next twenty flutes, and one
shear stud in the 22nd flute, in each half of the beam, there is a total of 46 shear studs
on the beam, and the total force provided by the shear studs in each half span is:
ΣQn =2 • 14.6 + 20 • 17.2 =373.9 kip
AISC-360-10 Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Live Load Deflection:
Modulus of elasticity ratio:
644 8.0
n E=
Ec 29, 000 3,=
=
Transformed elastic moment of inertia assuming full composite action:
Transformed
Area
A (in2)
Moment Arm
from Centroid
y (in.)
Ay
(in.3)
Ay2
(in,4)
I0
(in.4)
Slab
67.9
15.65
1,062
16,620
115
W21x50
14.7
0
0
0
984
1,062
16,620
1,099
Element
82.6
Ix =
I 0 + Ay 2 =
1, 099 + 16, 620 =
17, 719 in.4
=
y
1, 062
= 12.9 in.
82.6
2
I tr = I x − A • y = 17, 719 − 82.6 • 12.92 = 4, 058 in 4
Effective moment inertia assuming partial composite action:
I equiv = I s + ΣQn / Py ( I tr − I s ) = 984 + 0.51(4,058 − 984) = 3,176 in 4
I eff = 0.75 • I equiv = 0.75 • 3,176 = 2,382 in 4
=
∆ LL
5wL L4
5 • (1 / 12) • (30 • 12) 4
=
= 1.34 in.
384 EI eff
384 • 29, 000 • 2, 382
Design for Shear Strength:
Required Shear Strength:
wu = 1.2 • 0.800 + 1.2 • 0.100 + 1.6 • 1 = 2.68 kip/ft
=
Vu
wu • L 2.68 • 45
=
= 60.3 kip-ft
2
2
AISC-360-10 Example 001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Available Shear Strength:
ΦVn =
Φ • 0.6 • d • tw • Fy =
1.0 • 0.6 • 20.8 • 0.38 • 50 =
237.1 kips
AISC-360-10 Example 001 - 9
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
AISC-360-10 Example 002
COMPOSITE GIRDER DESIGN
EXAMPLE DESCRIPTION
The design is checked for the composite girder shown below. The deck is 3 in.
deep with 4 ½″ normal weight (145 pcf) concrete cover with a compressive
strength of 4 ksi. The girder will not be shored during construction. The applied
loads are the weight of the structure, a 25 psf construction live load, a 10 psf
superimposed dead load and a 100 psf non-reducible service line load.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W24x76
E = 29000 ksi
Fy = 50 ksi
Loading
P = 36K (Dead Load)
P = 4.5K (SDL)
P = 45K (Live Load)
Geometry
Span, L = 45 ft
AISC-360-10 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
RESULTS COMPARISON
Independent results are referenced from Example I.2 from the AISC Design
Examples, Version 14.0.
ETABS
Independent
Percent
Difference
Pre-composite Mu (k-ft)
622.3
622.3
0.00%
Pre-composite ΦbMn (k-ft)
677.2
677.2
0.00%
1.0
1.0
0.00%
Required Strength Mu (k-ft)
1216.3
1216.3
0.00%
Full Composite ΦbMn (k-ft)
1480.1
1480.1
0.00%
Partial Composite ΦbMn (k-ft)
1267.3
1267.3
0.00%
Shear Stud Capacity Qn
21.54
21.54
0.00%
Shear Stud Distribution
26, 3, 26
26, 3, 26
0.00%
Live Load Deflection (in.)
0.63
0.55
12.7%
Required Strength Vu (kip)
122.0
122.0
0.00%
ΦVn (k)
315.5
315.5
0.00%
Output Parameter
Pre-composite Deflection (in.)
AISC-360-10 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
COMPUTER FILE: AISC-360-10 EXAMPLE 002.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
The live load deflection differs more markedly because of a difference in
methodology. In the AISC example, the live load deflection is computed based
on a lower bound value of the beam moment of inertia, whereas in ETABS, it is
computed based on the approximate value of the beam moment of inertia derived
from Equation (C-I3-6) from the Commentary on the AISC Load and Resistance
Factor Design Specification – Second Edition.
AISC-360-10 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
HAND CALCULATION
Properties:
Materials:
ASTM A572 Grade 50 Steel
E = 29,000 ksi, Fy = 50 ksi, wsteel = 490 pcf
4000 psi normal weight concrete
Ec = 3,644 ksi, f c′ = 4 ksi, wconcrete = 145 pcf
Section:
W24x76
d = 23.9 in, bf = 8.99 in, tf = 0.68 in, tw = 0.44 in
Asteel = 22.4 in2, Isteel = 2100 in4
Deck:
tc =4 ½ in., hr = 3 in., sr =12 in., wr = 6 in.
Shear Connectors:
d = ¾ in, h =4 ½ in, Fu = 65 ksi
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
22.4
w A
=
= 76.2 plf
•w =
sq .ft . • 490 pcf
steel steel 144
PD =
36 kips
[(45 ft)(10 ft)(75 psf ) + (50 plf )(45 ft)] (0.001 kip / lb) =
PL
=
ft)(10 ft)(25 psf )] (0.001 kip/lb)
[(45
11.25 kips
1.2 wL2
L
+ (1.2 PD + 1.6 PL )
8
3
2
76.2 • 30
30
= 1.2
+ (1.2 • 36 +1.6 • 11.25
=
622.3 kip-ft
)
8
3
Mu =
AISC-360-10 Example 002 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Moment Capacity:
Lb = 10 ft
Lp = 6.78 ft
Lr = 19.5 ft
ΦbBF = 22.6 kips
ΦbMpx = 750 kip-ft
Cb = 1.0
Φ b M=
Cb Φ b M px − Φ b BF ( Lb − L p )
n
=
1.0 750 − 22.6 • (10 − 6.78 )
=
677.2 kip-ft
Pre-Composite Deflection:
0.0762
• 3604
PD L
5wD L
36.0 • 360
12
∆
=
+
=
+
= 1.0
nc
28 EI
384 EI 28 • 29, 000 • 2,100 384 • 29, 000 • 2,100
3
4
3
5•
= 0.8 • ∆ nc
= 0.8 in. which is rounded down to ¾ in.
Camber
Design for Composite Flexural Strength:
Required Flexural Strength:
40.5 kips
P =
[ (45 ft)(10 ft)(75 +10psf ) + (50 plf)(45 ft)] (0.001 kip/lb) =
D
P
L
=
ft)(10 ft)(100 psf ) ] (0.001 kip/lb)
[ (45
45 kips
1.2 wL2
L
+ (1.2 PD + 1.6 PL )
8
3
2
1.2 • 76.22 • 30
30
1216.3 kip-ft
=
+ (1.2 • 40.5 +1.6
=
• 45)
8
3
Mu =
Full Composite Action Available Flexural Strength:
Effective width of slab:
=
b
eff
30.0 ft
= 7.5
=
ft 90 in.
8
AISC-360-10 Example 002 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Resistance of steel in tension:
C = Py = As • Fy = 22.4 • 50 = 1,120 kips controls
Resistance of slab in compression
Ac = beff • tc + (beff 2) • hr = (7.5 • 12) • 4.5 +
2
7.5 • 12
• 3= 540 in
2
C= 0.85 • f 'c A=
0.85 • 4 • 540
= 1836 kips
c
Depth of compression block within slab:
=
a
C
1,120
=
= 3.66 in.
0.85 • beff • f 'c 0.85 • (7.5 • 12) • 4
Moment resistance of composite beam for full composite action:
d1 = (tc + hr ) −
3.66
a
= (4.5 + 3) −
= 5.67 in.
2
2
d
ΦM =
Φ C • d + P •
n
1
y
2
23.9 12
=
0.9 • 1,120 • 5.67 / 12 + 1,120 •
1480.1 kip-ft
=
2
Partial Composite Action Available Flexural Strength:
Assume 50% composite action:
C = 0.5 • Py = 560 kips
Depth of compression block within slab
=
a
C
560
=
= 1.83 in.
0.85 • beff • f 'c 0.85 • (7.5 • 12) • 4
d1 = (tc + hr ) −
a
1.83
= (4.5 + 3) −
= 6.58 in.
2
2
Depth of compression block within steel section flange
=
x
Py − C
1,120 − 560
=
= 0.623 in.
2 • b f • Fy 2 • 8.99 • 50
=
d 2 x=
/ 2 0.311 in.
AISC-360-10 Example 002 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
M n =C • (d1 + d 2 ) + Py • (d3 − d 2 )
23.9
= 560 • (6.58 + 0.312) + 1,120 •
− 0.312 12 =1, 408 kip-ft
2
ΦM n = 0.9 M n = 0.9 • 1, 408 = 1, 267.3 kip-ft
Shear Stud Strength:
=
Qn 0.5 Asa
f 'c Ec ≤ Rg R p Asa Fu
0.442 in 2
Asa =
πd sa 2 4 =
π(0.75) 2 4 =
f c ' = 4 ksi
1.5
1.5
=
E w=
f c ' 145
=
4 3, 490 ksi
c
Rg = 1.0 Studs welded directly to the steel shape with the slab haunch
Rp = 0.75 Studs welded directly to the steel shape
Fu = 65 ksi
Qn = 0.5 • 0.4422 4 • 3, 490 ≤ 1.0 • 0.75 • 0.4422 • 65
= 26.1 kips ≥ 21.54 kips controls
Shear Stud Distribution:
n=
=
ΣQn
Qn
560
= 26 studs from each end to nearest concentrated load point
21.54
Add 3 studs between load points to satisfy maximum stud spacing requirement.
Live Load Deflection:
Modulus of elasticity ratio:
=
n E=
/ Ec 29, 000 / 3,
=
644 8.0
AISC-360-10 Example 002 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Transformed elastic moment of inertia assuming full composite action:
Transformed
Area
A (in2)
Moment Arm
from
Centroid
y (in.)
Ay
(in.3)
Ay2
(in,4)
I0
(in.4)
Slab
50.9
17.2
875
15,055
86
Deck ribs
17.0
13.45
228
3,069
13
W21x50
22.4
0
0
0
2,100
1,103
18,124
2,199
Element
89.5
I x =I 0 + Ay 2 =2,199 + 18,124 =20,323 in.4
=
y
1, 092
= 12.2 in.
89.5
2
I tr = I x − A • y = 20,323 − 90.3 • 12.22 = 6,831 in 4
Effective moment of inertia assuming partial composite action:
I equiv = I s + ΣQn Py ( I tr − I s ) = 2,100 + 0.5 ( 6,831 − 2,100 ) = 5, 446 in 4
I eff = 0.75 • I equiv = 0.75 • 5, 446 = 4, 084 in 4
=
∆ LL
PL L3
45.0 • (30 • 12)3
=
= 0.633 in.
28EI eff 28 • 29,000 • 4,084
Design for Shear Strength:
Required Shear Strength:
Pu = 1.2 • PD + 1.6 • PL = 1.2 • 40.5 + 1.6 • 45 = 120.6 kip
=
Vu
1.2 • w • L
1.2 • 0.076 • 30
120.6 121.2 kip-ft
=
+ Pu
+=
2
2
Available Shear Strength:
ΦVn =
Φ • 0.6 • d • tw • Fy =
1.0 • 0.6 • 23.9 • 0.44 • 50 =
315.5 kips
AISC-360-10 Example 002 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
BS-5950-90 Example-001
STEEL DESIGNERS MANUAL SIXTH EDITION - DESIGN OF SIMPLY SUPPORTED COMPOSITE
BEAM
EXAMPLE DESCRIPTION
Design a composite floor with beams at 3-m centers spanning 12 m. The
composite slab is 130 mm deep. The floor is to resist an imposed load of 5.0
kN/m2, partition loading of 1.0 kN/m2 and a ceiling load of 0.5 kN/m2. The floor
is to be un-propped during construction.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
UKB457x191x67
E = 205,000 MPa
Fy = 355 MPa
Loading
w = 6.67kN/m (Dead Load)
w = 1.5kN/m (Construction)
w = 1.5kN/m (Superimposed Load)
w = 18.00kN/m (Live Load)
Geometry
Span, L = 12 m
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
BS-5950-90 Example-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
RESULTS COMPARISON
Independent results are referenced from the first example, Design of Simply
Supported Composite Beam, in Chapter 21 of the Steel Construction Institute
Steel Designer’s Manual, Sixth Edition.
ETABS
Independent
Percent
Difference
211.2
211.3
0.05%
Construction Ms (kN-m)
522.2
522.2
0.00%
Construction Deflection (mm)
29.9
29.9
0.00%
Design Moment (kN-m)
724.2
724.3
0.01%
Full Composite Mpc (kN-m)
968.9
968.9
0.00%
Partial Composite Mc (kN-m)
910.8
910.9
0.01%
Shear Stud Capacity Qn (kN)
57.6
57.6
0.00%
Live Load Deflection (mm)
33.2
33.2
0.00%
Applied Shear Force Fv (kN)
241.4
241.4
0.00%
Shear Resistance Pv (kN)
820.9
821.2
0.00%
Output Parameter
Construction Design
Moment (kN-m)
COMPUTER FILE: BS-5950-90 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an excellent comparison with the independent results.
BS-5950-90 Example-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
HAND CALCULATION
Properties:
Materials:
S355 Steel:
E = 205,000 MPa, py = 355 MPa, γs = 7850 kg/m3
Light-weight concrete:
E = 24,855 MPa, fcu = 30 MPa, γc = 1800 kg/m3
Section:
UKB457x191x67
D = 453.6 mm, bf = 189.9 mm, tf = 12.7 mm, tw = 8.5 mm
Asteel = 8,550 mm2, Isteel = 29,380 cm4
Deck:
Ds =130 mm, Dp = 50 mm, sr = 300 mm, br = 150 mm
Shear Connectors:
d = 19 mm, h = 95 mm, Fu = 450 MPa
Loadings:
Self weight slab
= 2.0 kN/m2
Self weight beam
= 0.67 kN/m
Construction load
= 0.5 kN/m2
Ceiling
= 0.5 kN/m2
Partitions (live load)
= 1.0 kN/m2
Occupancy (live load)
= 5.0 kN/m2
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
wult construction = 1.4 • 0.67 + (1.4 • 2.0 + 1.6 • 0.5 ) • 3.0 = 11.74 kN/m
BS-5950-90 Example-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
wult construction • L2 11.74 • 122
=
=
= 211.3 kN-m
M ult construction
8
8
M s = S z • Py = 1, 471 • 103 • 355 • 10−6 = 522.2 kN-m
Pre-Composite Deflection:
wconstruction = 2.0 • 3.0 + 0.67 = 6.67 kN/m
5 • wconstruction • L4
5 • 6.67 • 12, 0004
=
δ
=
= 29.9 mm
384 • E • I
384 • 205, 000 • 29,380 • 104
Camber
= 0.8 •=
δ 24 mm, which is rounded down to 20 mm
Design for Composite Flexural Strength:
Required Flexural Strength:
wult = 1.4 • 0.67 + (1.4 • 2.0 + 1.6 • 1 + 1.6 • 5 ) • 3.0 = 40.24 kN/m
wult • L2 40.24 • 122
M
=
=
= 724.3 kN-m
ult
8
8
Full Composite Action Available Flexural Strength:
Effective width of slab:
B=
e
L 12,000
=
= 3, 000 mm ≤ 3,000 mmm
4
4
Resistance of slab in compression:
Rc = 0.45 • f cu • Be • ( Ds − D p )= 0.45 • 30 • 3, 000 • (130 − 50 ) • 10−3= 3, 240 kN
Resistance of steel in tension:
Rs = Py = As • p y = 8,550 • 355 • 10−3 = 3, 035 kN controls
Moment resistance of composite beam for full composite action:
D
R ( Ds − D p )
M=
Rs + Ds − s
for Rs ≤ Rc
pc
Rc
2
2
3,035 80
453.6
= 3,035
+ 130 −
• =
• 10−3 968.9 kN-m
3, 240 2
2
BS-5950-90 Example-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Partial Composite Action Available Flexural Strength:
Assume 72% composite action – the 75% assumed in the example requires more
shear studs than can fit on the beam given its actual clear length.
Rq = 0.72 • Rs = 2,189 kN
Tensile Resistance of web:
Rw = tw • ( D − 2 • t f ) • p y = 8.5 • ( 453.6 − 2 • 12.7 ) • 355 • 10−3 = 1, 292 kN
As Rq > Rw, the plastic axis is in the steel flange, and
Rq ( Ds − D p ) ( Rs − Rq ) 2 t f
D
M c =Rs + Rq Ds −
−
2
2
4
Rc
Rf
( 3,035 − 2,189 ) 12.7 • 10−3
453.6
2,189 80
= 3,035
• 10−3 + 2,1899 130 −
• • 10−3 −
2
3, 240 2
( 3,035 − 1, 292 ) 4
2
= 910.9 kN-m
Shear Stud Strength:
Characteristic resistance of 19 mm-diameter studs in normal weight 30 MPa
concrete:
Qk = 100 kN from BS 5950: Part 3 Table 5
Adjusting for light-weight concrete:
Qk = 90 kN
Reduction factor for profile shape with ribs perpendicular to the beam and two studs
per rib:
k = 0.6 •
br ( h − D p )
150 ( 95 − 50 )
•
= 0.6 •
•
= 1.62 but k ≤ 0.8
50
50
Dp
Dp
Design strength:
Q p =k • 0.8 • Qk =0.8 • 0.8 • 90 =57.6 kN
Shear Stud Distribution:
The example places two rows of shear studs and computes the numbers of deck ribs
available for placing shear studs based on the beam center to center span and the
deck rib spacing: 12 m / 300 mm = 40
BS-5950-90 Example-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
However, the number of deck ribs available for placing shear studs must be based on
the beam clear span, and since the clear beam span is somewhat less than the 12 m
center to center span there are only 39 deck ribs available.
ETABS selects 72% composite action, which is the highest achievable and sufficient
to meet the live load deflection criteria. ETABS satisfies this 72% composite action
by placing one stud per deck rib along the entire length of the beam, plus a second
stud per rib in all the deck ribs except the mid-span rib since this is the location of
the beam zero moment and a stud in that rib would not contribute anything to the
total resistance of the shear connectors. The total resistance of the shear connectors
is:
Rq =2 • 19 • Q p =38 • 57.6 =2,189 kN
Live Load Deflection:
The second moment of area of the composite section, based on elastic properties, Ic
is given by:
=
Ic
Asteel • ( D+ Ds + D p )
2
4 • (1 + α e • r )
+
beff • ( Ds − D p )
12 • α e
3
+ I steel
Asteel
8,550
=
= 0.0356
beff • ( Ds − D p ) 3,000 • (130 − 50 )
=
r
For light-weight concrete:
αs =
10
α l =25
Proportion of total loading which is long term:
=
ρl
wdl + wsdl + 0.33 • wlive 6.67 + 1.5 + 0.33 • 18
=
= 0.541
wdl + wsdl + wlive
6.67 + 1.5 + 18
α e = α s + ρl • ( α l − α s ) = 10 + 0.541 • ( 25 − 10 ) = 18.1
8,550 • ( 453.4 + 130 + 50 ) 3,000 • 803
=
Ic
+
+ 294 • 106
4 • (1 + 18.1 • 0.0356 )
12 • 18.1
2
=
( 521 + 7 + 294 ) • 106=
822 • 106 mm 4
Live load deflection assuming full composite action:
BS-5950-90 Example-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
5 • 18 • (12,000 )
5 • wlive • L4
=
δc
=
= 28.8 mm
384 • E • I c 384 • 205,000 • 822 • 106
4
Adjust for partial composite action:
5 • 18 • (12, 000 )
5 • wlive • L4
=
δs
=
384 • E • I c 384 • 205, 000 • 294 • 106
4
= 80.7 mm non-composite reference deflection
δ partial = δc + 0.3 • (1 − K ) • ( δ s − δc )
= 28.9 + 0.3 • (1 − 0.72 ) • ( 80.7 − 28.9 )= 33.2 mm
Design for Shear Strength:
Required Shear Strength:
=
Fv
wult • L 40.24 • 12
=
= 241.4 kN
2
2
Shear Resistance of Steel Section:
P = 0.6 • p y • Ds • tw = 0.6 • 355 • 453.4 • 8.5 • 10−3 = 820.9 kN
V
BS-5950-90 Example-001 - 7
Software Verification
ETABS
4
PROGRAM NAME:
REVISION NO.:
CSA-S16-09 Example-001
HANDBOOK OF STEEL CONSTRUCTION, TENTH EDITION - COMPOSITE BEAM
EXAMPLE DESCRIPTION
Design a simply supported composite beam to span 12 m and carry a uniformly
distributed specified load of 18 kN/m live load and 12 kN/m dead load. Beams
are spaced at 3 m on center and support a 75 mm steel deck (ribs perpendicular to
the beam) with a 65 mm cover slab of 25 MPa normal density concrete.
Calculations are based on Fy = 345 MPa. Live load deflections are limited to
L/300.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
W460x74
E = 205,000 MPa
Fy = 345 MPa
Loading
w = 8.0kN/m (Dead Load)
w = 2.5kN/m (Construction)
w = 4.0kN/m (Superimposed Load)
w = 18.00kN/m (Live Load)
Geometry
Span, L = 12 m
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
CSA-S16-09 Example-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
RESULTS COMPARISON
Independent results are referenced from the design example on page 5-25 of the
Handbook of Steel Construction, Tenth Edition.
ETABS
Independent
Percent
Difference
247.4
247.5
0.04%
Construction Ms (kN-m)
512.3
512.3
0.00%
Construction Deflection (mm)
32.4
32.4
0.00%
Design Moment (kN-m)
755.8
756
0.02%
Full Composite Mrc (kN-m)
946.7
946.7
0.00%
Partial Composite Mrc (kN-m)
783.6
783.6
0.00%
Shear Stud Capacity Qn (kN)
68.7
68.7
0.00%
30
30
0.00%
Live Load Deflection (mm)
32.9
32.9
0.00%
Bottom Flange Tension (MPa)
267.2
267.1
0.04%
Design Shear Force Vf (kN)
251.9
251.9
0.00%
Shear Resistance Vr (kN)
842.9
842.9
0.00%
Output Parameter
Construction Design
Moment (kN-m)
Shear Stud Distribution
COMPUTER FILE: CSA-S16-09 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
CSA-S16-09 Example-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
HAND CALCULATION
Properties:
Materials:
ASTM A992 Grade 50 Steel
E = 200,000 MPa, Fy = 345 MPa, γs = 7850 kg/m3
Normal weight concrete
E = 23,400 MPa, fcu = 20 MPa, γc = 2300 kg/m3
Section:
W460x74
d = 457 mm, bf = 190 mm, tf = 14.5 mm, tw = 9 mm, T = 395 mm, rfillet=16.5 mm
, Z s 1, 650 • 103 mm3 , =
As = 9,450 mm2=
I s 333 • 106 mm 4
Deck:
tc =65 mm, hr = 75 mm, sr = 300 mm, wr = 150 mm
Shear Connectors:
d = 19 mm, h = 115 mm, Fu = 450 MPa
Loadings:
Self weight slab
= 2.42 kN/m2
Self weight beam
= 0.73 kN/m
Construction load
= 0.83 kN/m2
Superimposed dead load
= 1.33 kN/m2
Live load
= 6.0 kN/m2
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
w f construction = 1.25 • 0.73 + (1.25 • 2.42 + 1.5 • 0.83) • 3.0 = 13.75 kN/m
w f construction • L2 13.75 • 122
M f construction
=
=
= 247.5 kN-m
8
8
CSA-S16-09 Example-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
Moment Capacity:
M s = Z s • 0.9 • Fy =1, 650 • 103 • 0.9 • 345 • 10−6 = 512.3 kN-m
Pre-Composite Deflection:
wconstruction = 2.42 • 3.0 + 0.73 = 8.0 kN/m
=
δ
5 • wconstruction • L4
5 • 8.0 •12, 0004
=
= 32.4 mm
384 • E • I
384 • 200, 000 • 33, 300 •104
= 0.8 •=
δ 25.9 mm, which is rounded down to 25 mm
Camber
Design for Composite Flexural Strength:
Required Flexural Strength:
w f = 1.25 • 0.73 + (1.25 • 2.42 + 1.25 • 1.33 + 1.5 • 6 ) • 3.0= 42 kN/m
w f • L2 42 • 122
Mf
=
=
= 756.0 kN-m
8
8
Full Composite Action Available Flexural Strength:
Effective width of slab:
=
b
l
L
12,000
=
= 3, 000 mm ≤ 3,000 mmm
4
4
Resistance of slab in compression:
f c′ 0.8125
α=1 0.85 − 0.0015 • =
C 'r = α1 • Φ c • t • b f • f c′ = 0.8125 • 0.65 • 65 • 3,000 • 25 • 10−3 = 2,574 kN controls
Resistance of steel in tension:
3
Φ • As • F=
0.9 • 9,450 • 345 • 10−=
2,934 kN
y
Depth of compression block within steel section top flange:
=
x
(Φ • As • Fy − C ' r ) 2
=
Φ • Fy • b f
− 2,547 ) • 103 2
( 2,934
=
0.9 • 345 • 190
3.05 mm
CSA-S16-09 Example-001 - 4
Software Verification
ETABS
4
PROGRAM NAME:
REVISION NO.:
Moment resistance of composite beam for full composite action:
t x
d x
M rc= C 'r • h r + c + + Φ • As • Fy • −
2 2
2 2
65 3
457 3
3
= 2,574 • 75 + + • 10−3 + 2,934 •
− • 10−=
946.7 kN-m
2 2
2
2
Partial Composite Action Available Flexural Strength:
Assume 40.0% composite action:
Qr = 0.4 • Rc = 0.4 • 2,574 = 1,031 kN
Depth of compression block within concrete slab:
Qr
1,031 • 103
=
a =
= 26 mm
α1 • Φ c • beff • f c′ 0.8125 • 0.65 • 3,000 • 25
Compression force within steel section:
Cr =
( Py − Qr ) 2 =
951.6 kN
( 2,934 − 1,031) 2 =
Tensile resistance of one flange:
F
= Φ • b • t • F = 0.9 • 190 • 14.5 • 345 • 10−3 = 855.4 kN
flange
y
f
f
Tensile resistance of web:
F
=
Φ • T • t • F =•
0.9 395 • 9 • 345 • 10−3 =
1,103.8 kN
web
w y
Tensile resistance of one fillet area:
Ffillet=
(P − 2 • F
y
flange
− Fweb ) 2=
( 2,934 − 2 • 855.4 − 1,103.8)
2= 59.8 kN
Compression force in web:
951.6 − 855.4 − 59.7 =
36.4 kN
Cweb =
Cr − Fflange − Ffillet =
Depth of compression block in web:
x=
Cweb
36.4
• T=
• 395= 13 mm
Fweb
1,103.8
CSA-S16-09 Example-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
Location of centroid of compressive force within steel section measured from top of
steel section:
d2
0.5 • t f • Fflange + ( t f + 0.5 • rfillet ) • Ffillet + ( t f + rfillet + 0.5 • x ) • C web
=
Cr
0.5 • 14 • 855 + (14 + 0.5 • 16.5) • 60 + (14 + 16.5 + 0.5 • 44 ) • 36.4
= 9.4 mm
951.6
Moment resistance of composite beam for partial composite action:
a
d
M rc = Qr • h r + tc − + d 2 + Py • − d 2
2
2
26
457
3
783.6 kN-m
= 1,031 • 75 + 65 − + 9.4 • 10−3 + 2,934 •
− 9.4 • 10−=
2
2
Shear Stud Strength:
From CISC Handbook of Steel Construction Tenth Edition for 19-mm-diameter
studs,
hd = 75 mm, wd/hd = 2.0, 25 MPa, 2,3000 kg/m3 concrete:
qrr = 68.7 kN
2 • Qr 2 • 1, 031
Total number of studs required ==
= 30
qrr
68.7
Live Load Deflection:
Modulus of elasticity ratio:
n E=
Ec 200, 000 23,=
=
400 8.55
Transformed elastic moment of inertia assuming full composite action:
Element
Transformed
Area
A (mn2)
Moment Arm
from Centroid
Ay
Ay2
I0
y (mm)
(103 mm3) (106 mm4) (106 mm4)
Slab
22,815
336
7,666
2,576
8
W460x74
9,450
0
0
0
333
7,666
2,576
341
32,265
CSA-S16-09 Example-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
I x =I 0 + Ay 2 =341 • 106 + 2,576 • 106 =2,917 • 106 mm 4
=
y
7, 666 • 106
= 238 mm
32, 265
2
I tr = I x − A • y = 2,917 • 106 − 32, 265 • 2382 = 1, 095 • 106 mm 4
Effective moment of inertia assuming partial composite action:
I eff =
I s + 0.85 p 0.25 ( I tr − I s )
= 333 + 0.85 • 0.400.25 • (1,095 − 333)
= 848 • 106 mm 4
5 • 18 • (12,000 )
5wL L4
∆ LL= 1.15 •
= 1.15 •
= 32.9 mm
384 EI eff
384 • 200,000 • 848 • 106
4
Bottom Flange Tension:
Stress in tension flange due to specified load acting on steel beam alone:
=
f1
M1
8 • 120002
=
= 98.6 MPa
S x 8 • 1460 • 103
Bottom section modulus based on transformed elastic moment of inertia assuming,
per the original example, full composite action:
=
St
I tr
1,095 • 106
=
= 1350 mm
d
(228.5
+
237.6)
( + y)
2
Stress in tension flange due to specified live and superimposed dead loads acting on
composite section:
M 2 (18 + 4) • 120002
=
f2 =
= 168.5 MPa
St
8 • 2350 • 103
f1 + f 2 =
98.6 +168.5=267.1 MPa
CSA-S16-09 Example-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
4
Design for Shear Strength:
Required Shear Strength:
=
Vf
wfactored • L 42 • 12
= = 252 kN
2
2
Shear Resistance of Steel Section:
Vr =Φ • Aw • Fs =0.9 • d • tw • 0.66 • Fy =0.9 • 457 • 9 • 0.66 • 345 =842.9 kN
CSA-S16-09 Example-001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
EC-4-2004 Example-001
STEEL DESIGNERS MANUAL SEVENTH EDITION - DESIGN OF SIMPLY SUPPORTED COMPOSITE
BEAM
EXAMPLE DESCRIPTION
Consider an internal secondary composite beam of 12-m span between columns
and subject to uniform loading. Choose a UKB457x191x74 in S 355 steel.
GEOMETRY, PROPERTIES AND LOADING
EC-4-2004 Example-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
Member Properties
UKB457x191x74
E = 205,000 MPa
fy = 355 MPa
ETABS
3
Geometry
Loading
Span, L = 12 m
w = 8.43kN/m (Dead Load)
w = 2.25kN/m (Construction)
Beam spacing, b =3 m
w = 1.5kN/m (Superimposed Load)
w = 15.00kN/m (Live Load)
TECHNICAL FEATURES OF ETABS TESTED
Composite beam design, including:
Selection of steel section, camber and shear stud distribution
Member bending capacities, at construction and in service
Member deflections, at construction and in service
RESULTS COMPARISON
Independent results are referenced from the first example, Design of Simply
Supported Composite Beam, in Chapter 22 of the Steel Construction Institute
Steel Designer’s Manual, Seventh Edition.
ETABS
Independent
Percent
Difference
250.4
250.4
0.00%
Construction Ma,pl,Rd (kN-m)
587
587
0.00%
Construction Deflection (mm)
32.5
32.5
0.00%
Design Moment (kN-m)
628.4
628.4
0.01%
Full Composite Mpc (kN-m)
1020
1020
0.00%
Partial Composite Mc (kN-m)
971.2
971.2
0.00%
Shear Stud Capacity PRd
Input
52.0
NA
Shear Stud Distribution
77
76
1.3%
19.3
19.1
1.03%
Output Parameter
Construction MEd (kN-m)
Live Load Deflection (mm)
EC-4-2004 Example-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
Output Parameter
Required Strength VEd (kN)
Vpl,Rd (kN)
ETABS
3
ETABS
Independent
Percent
Difference
209.5
209.5
0.00%
843
843
0.00%
COMPUTER FILE: EC-4-2004 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
The shear stud capacity Pr was entered as an overwrite, since it is controlled by
the deck profile geometry and the exact geometry of the example, which assumes
a deck profile with a rib depth of 60 mm, a depth above profile of 60 mm and a
total depth of 130 mm, cannot be modeled in ETABS, since in ETABS, only the
rib depth and depth above profile can be specified.
EC-4-2004 Example-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
HAND CALCULATION
Properties:
Materials:
S 355 Steel:
E = 210,000 MPa, fy = 355 MPa, partial safety factor γa = 1.0
Normal weight concrete class C25/30:
Ecm = 30,500 MPa, fcu = 30 MPa, density wc = 24 kN/m3
Section:
UKB457x191x74
ha = 457 mm, bf = 190.4 mm, tf = 14.5 mm, tw = 9 mm,
Aa = 9,460 mm2, Iay = 33,319 cm4, Wpl = 1,653 cm3
Deck:
Slab depth hs =130 mm, depth above profile hc = 60 mm,
Deck profile height hp = 60 mm, hd = hp + 10 mm for re-entrant stiffener,
sr = 300 mm, b0 = 150 mm
Shear Connectors:
d = 19 mm, h = 95 mm, Fu = 450 MPa
Loadings:
Self weight slab, decking, reinforcement
= 2.567 kN/m2
Self weight beam
= 0.73 kN/m
Construction load
= 0.75 kN/m2
Ceiling
= 0.5 kN/m2
Partitions (live load)
= 1.0 kN/m2
Occupancy (live load)
= 4.0 kN/m2
EC-4-2004 Example-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Design for Pre-Composite Condition:
Construction Required Flexural Strength:
wfactored construction = 1.25 • (2.567 • 3.0 + 0.73) + 1.5 • 0.75 • 3.0 = 13.91 kN/m
M
=
Ed
wfactored construction • L2 13.91 • 122
=
= 250.4 kN-m
8
8
Moment Capacity:
M a , pl ,Rd = W pl • f d = 1,653 • 103 • 355 • 10−6 = 587 kN-m
Pre-Composite Deflection:
wconstruction= 2.567 • 3.0 + 0.73= 8.43 kN/m
=
δ
5 • wconstruction • L4
5 • 8.43 • 12,0004
=
= 32.5 mm
384 • E • I ay
384 • 210,000 • 33,319 • 104
Camber
= 0.8 •=
δ 26 mm, which is rounded down to 25 mm
Design for Composite Flexural Strength:
Required Flexural Strength:
wfactored = 1.25 • 0.73 + (1.25 • 2.567 + 1.25 • 0.5 + 1.5 • 1 + 1.5 • 4.0) • 3.0 = 34.91 kN/m
M
=
Ed
wfactored • L2 34.91 • 122
=
= 628.4 kN-m
8
8
Full Composite Action Available Flexural Strength:
Effective width of slab:
=
beff
2 • L 2 • 12
= = 3m
8
8
Resistance of slab in compression:
Rc=
0.85 • f ck
• beff • hc= 0.85 • (25 /1.5) • 3, 000 • 60 • 10−3= 2,550 kN controls
γc
Resistance of steel section in tension:
Rs = f yd • Aa = 355 • 9, 460 • 10−3 = 3,358 kN
EC-4-2004 Example-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Depth of compression block within steel section flange:
=
x
Rs − Rc
3, 358 − 2, 250
=
= 6 mm
2 • b f • f yd 2 • 190.4 • 355
/ 2 0.273 in.
d 2 x=
=
The plastic axis is in the steel flange and the moment resistance for full composite
action is:
h ( R − Rc ) 2 t f
h
h
M a , pl ,RD =
Rs − d 2 +R c hs − c - s
Rf
2
4
2
2
453.6
60
(3,358 − 2,550) 2 14.5
• 10−3 + 2,550 130 − • 10−3 −
• 10−3
2
2
980
4
= 1020.0 kN-m
= 3,358
Partial Composite Action Available Flexural Strength:
Assume 77.5% composite action:
R=
0.775 • R=
0.775 • 3,358= 1,976 kN
q
s
Tensile Resistance of web:
Rw =
tw • ( D − 2 • t f ) • p y =
8.5 • (453.6 − 2 •12.7) • 355 •10−3 =
1, 292 kN
As Rq > Rw, the plastic axis is in the steel flange, and
M c =Rs
R h ( R − Rq ) 2 t f
h
+ Rq hs − q c − s
Rc 2
Rf
2
4
= 3,358
453.6
1,976 60
(3,358 − 1,976) 2 14.5
• 10−3 + 1,976 130 −
• • 10−3 −
• 10−3
2
2,
250
2
980
4
= 971.2 kN-m
Resistance of Shear Connector:
Resistance of shear connector in solid slab:
PRd= 0.29 • α • d 2 •
0.29 • α • d 2 •
d2
h 95
f ck • Ecm γ v ≤ 0.8 • fu • π γ v with α =1.0 for =
>4
d 19
4
f ck Ecm =
γ v 0.29 • 1.0 • 192 • 25 • 30,500 • 10−3 1.25
= 73 kN controls
EC-4-2004 Example-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
d2
19
0.8 • fu • π γ v =0.8 • 450 • π 1.25 =81.7 kN
4
4
Reduction factor for decking perpendicular to beam – assuming two studs per rib:
0.7
( b0 hp ) ( hsc hp ) − 1 ≤ 0.75 per EN 1994-1-1 Table 6.2
nr
=
kt
0.7 150
( 95 60 ) − 1= 0.72 ≤ 0.75
2 60
=
PRd = 0.72 • 73= 52 kN
Total resistance with two studs per rib and 19 ribs from the support to the mid-span:
Rq =2 • 19 • 52 =1,976 kN
Live Load Deflection:
The second moment of area of the composite section, based on elastic properties, Ic
is given by:
Aa • (h + 2 • hp + hc ) 2
=
Ic
4 • (1 + n • r )
beff • hc3
+
+ I ay
12 • n
Aa
9, 460
=
= 0.052
beff • hc 3, 000 • 60
=
r
n = modular ratio = 10 for normal weight concrete subject to variable loads
=
Ic
9, 460 • (457 + 2 • 70 + 60) 2 3, 000 • 603
+
+ 33,320 • 104
4 • (1 + 10 • 0.052)
12 • 10
= (6.69 + 0.05 + 3.33) • 108 = 10.08 • 108 mm 4
=
δlive
5 • wlive • L4
5 • 15 • (12, 000) 4
=
= 19.1 mm
384 • E • I c 384 • 210, 000 • 10.08 • 108
EC-4-2004 Example-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
3
Design for Shear Strength:
Required Shear Strength:
=
VEd
wfactored • L 34.91 • 12
= = 209.5 kN
2
2
Shear Resistance of Steel Section:
=
V pl , Rd
457 • 9.0 • 355
=
843 kN
3 • 10−3
EC-4-2004 Example-001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC-360-10 Example 001
COMPOSITE COLUMN DESIGN
EXAMPLE DESCRIPTION
Determine if the 14-ft.-long filled composite member illustrated below is
adequate for the indicated dead and live loads. The composite member consists
of an ASTM A500 Grade B HSS with normal weight (145 lb/ft3) concrete fill
having a specified compressive strength, fc′ = 5 ksi.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
HSS10x6 x⅜
E = 29,000 ksi
Fy = 46 ksi
Loading
PD = 32.0 kips
PL = 84.0 kips
Geometry
Height, L = 14 ft
AISC-360-10 Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURE OF ETABS TESTED
Compression capacity of composite column design.
RESULTS COMPARISON
Independent results are referenced from Example I.4 from the AISC Design
Examples, Version 14.0.
ETABS
Independent
Percent
Difference
Required Strength Pu (kip)
172.8
172.8
0.00%
Available Strength ΦPn (kip)
342.93
354.78
3.34%
Output Parameter
COMPUTER FILE: AISC-360-10 EXAMPLE 001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
AISC-360-10 Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Materials:
ASTM A500 Grade B Steel
E = 29,000 ksi, Fy = 46 ksi, Fu = 58 ksi
5000 psi normal weight concrete
Ec = 3,900 ksi, f c′ = 5 ksi, wconcrete = 145 pcf
Section dimensions and properties:
HSS10x6x⅜
H = 10.0 in, B= 6.00 in, t = 0.349 in
As = 10.4 in2, Isx = 137 in4, Isy = 61.8 in4
Concrete area
hi = H − 2 • t = 10 − 2 • 0.349 = 9.30 in.
bi = B − 2 • t = 6 − 2 • 0.349 = 5.30 in.
Ac= bi • hi − t 2 • (4 − π)= 5.30 • 9.30 − (0.349) 2 • (4 − π)= 49.2 in.2
Moment of inertia for bending about the y-y axis:
( H − 4 • t ) • bt3 t • ( B − 4 • t )3 (9π2 − 64) • t 4
B − 4•t 4•t
=
+
+
+ π • t2
−
I cy
12
6
36 • π
2
3• π
2
(10 − 4 • 0.349) • 5.303 0.349 • (6 − 4 • 0.349)3 (9π2 − 64) • 0.3494
+
+
+
12
6
36 • π
6 − 4 • 0.349 4 • 0.349 2
)
π • 0.3492 (
−
2
3• π
= 114.3 in.4
=
Design for Compression:
Required Compressive Strength:
Pu = 1.2 • PD + 1.6 • PL = 1.2 • 32.0 + 1.6 • 84.0 = 172.8 kips
AISC-360-10 Example 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Nominal Compressive Strength:
E
Pno = Pp = Fy • As + C2 • f c′ Ac + Asr s
Ec
where
C2 = 0.85 for rectangular sections
Asr = 0 when no reinforcing is present within the HSS
Pno = 46 • 10.4 + 0.85 • 5 • (49.2 + 0.0) = 687.5 kips
Weak-axis Elastic Buckling Force:
As
0.6 + 2
C3 =
≤ 0.9
Ac + As
10.4
=
0.6 + 2
≤ 0.9
49.2 + 10.4
0.9 controls
= 0.949 > 0.9
EI eff = Es • I sy + Es • I sr + C3 • Ec • I cy
= 29, 000 • 62.1 + 0 + 0.9 • 3,900 • 114.3
= 2, 201, 000 kip-in 2
Pe = π2 ( EI eff ) ( KL) 2 where K = 1.0 for a pin-ended member
=
Pe
π2 • 2, 201, 000
= 769.7 kips
1.0 • (14.0 • 12) 2
Available Compressive Strength:
Pno
688
=
= 0.893 < 2.25
Pe 769.7
Therefore, use AISC Specification Equation I2-2:
Pno
ΦPn =
ΦPno 0.658 Pe =
0.75 • 687.5 • (0.658)0.893 =
354.8 kips
AISC-360-10 Example 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC-360-10 Example 002
COMPOSITE COLUMN DESIGN
EXAMPLE DESCRIPTION
Determine if the 14-ft.-long filled composite member illustrated below is
adequate for the indicated dead load compression and wind load tension. The
entire load is applied to the steel section.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
HSS10x6 x⅜
E = 29,000 ksi
Fy = 46 ksi
Loading
PD = -32.0 kips
PW = 100.0 kips
Geometry
Height, L = 14 ft
TECHNICAL FEATURE OF ETABS TESTED
Tension capacity of composite column design.
AISC-360-10 Example 002 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Independent results are referenced from Example I.5 from the AISC Design
Examples, Version 14.0.
ETABS
Independent
Percent
Difference
Required Strength, Pu (kip)
71.2
71.2
0.00%
Available Strength, ΦPn (kip)
430.5
430.0
0.12%
Output Parameter
COMPUTER FILE: AISC-360-10 EXAMPLE 002.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
AISC-360-10 Example 002 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Materials:
ASTM A500 Grade B Steel
E = 29,000 ksi, Fy = 46 ksi, Fu = 58 ksi
5000 psi normal weight concrete
Ec = 3,900 ksi, f c′ = 5 ksi, wconcrete = 145 pcf
Steel section dimensions:
HSS10x6x⅜
H = 10.0 in, B = 6.00 in, t = 0.349 in, As = 10.4 in2
Design for Tension:
Required Compressive Strength:
The required compressive strength is (taking compression as negative and tension as
positive):
Pu = 0.9 • PD + 1.0 • PW = 0.9 • (−32.0) + 1.0 • 100.0 = 71.2 kips
Available Tensile Strength:
ΦPn =Φ ( As • Fy + Asr • Fysr ) =0.9(10.4 • 46 + 0 • 60) =430 kips
AISC-360-10 Example 002 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AISC-360-10 Example 003
COMPOSITE COLUMN DESIGN
EXAMPLE DESCRIPTION
Determine if the 14-ft.-long filled composite member illustrated below is
adequate for the indicated axial forces, shears, and moments. The composite
member consists of an ASTM A500 Grade B HSS with normal weight (145
lb/ft3) concrete fill having a specified compressive strength, f c′ = 5 ksi.
GEOMETRY, PROPERTIES AND LOADING
Member Properties
HSS10x6 x⅜
E = 29,000 ksi
Fy = 46 ksi
Loading
Pr = 129.0 kips
Mr = 120.0 kip-ft
Vr = 17.1 kips
Geometry
Height, L = 14 ft
AISC-360-10 Example 003 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURE OF ETABS TESTED
Tension capacity of composite column design.
RESULTS COMPARISON
Independent results are referenced from Example I.1 from the AISC Design
Examples, Version 14.0.
ETABS
Independent
Percent
Difference
129
129
0.00%
Available Strength, ΦPn (kip)
342.9
354.78
-3.35%
Required Strength, Mu (k-ft)
120
120
0.00%
130.58
130.5
0.06%
1.19
1.18
0.85%
Output Parameter
Required Strength, Fu (k)
Available Strength, ΦbMn (k-ft)
Interaction Equation H1-1a
COMPUTER FILE: AISC-360-10 EXAMPLE 003.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
AISC-360-10 Example 003 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Properties:
Materials:
ASTM A500 Grade B Steel
E = 29,000 ksi, Fy = 46 ksi, Fu = 58 ksi
5000 psi normal weight concrete
Ec = 3,900 ksi, f c′ = 5 ksi, wconcrete = 145 pcf
Section dimensions and properties:
HSS10x6x⅜
H = 10.0 in, B= 6.00 in, t = 0.349 in
As = 10.4 in2, Isx = 137 in4, Zsx=33.8 in3, Isy = 61.8 in4
Concrete area
ht = 9.30 in., bt = 5.30 in., Ac = 49.2 in.2, Icx = 353 in4, Icy = 115 in4
Compression capacity:
Nominal Compressive Strength:
ΦcPn= 354.78 kips as computed in Example I.4
Bending capacity:
Maximum Nominal Bending Strength:
Zsx = 33.8 in3
bi • hi 2
− 0.192 • ri 3 where ri = t
4
5.30 • (9.30) 2
3
114.7 in.3
=
− 0.192 • (0.349)
=
4
Zc =
0.85 • f c′ • Zc
2
0.85 • 5 • 115 1, 798.5 kip-in.
=46 • 33.8 +
=
=149.9 kip-ft
2
12 in./ft
M D = Fy • Z sx +
AISC-360-10 Example 003 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Available Bending Strength:
=
hn
0.85 • f c′ • Ac
h
≤ i
2(0.85 • f c′ • bi + 4 • t • Fy ) 2
0.85 • 5 • 49.2
9.30
≤
2(0.85 • 5 • 5.30 + 4 • 0.349 • 50)
2
= 1.205 ≤ 4.65
=1.205 in.
=
Z sn = 2 • t • hn2 = 2 • 0.349 • (1.205) 2 = 1.01 in.3
Z cn =bi • hn2 =5.30 • (1.205) 2 =7.70 in.3
0.85 • f c′ • Z cn
2
0.85 • 5 • 7.76 1, 740 kip-in.
= 1,800 − 46 • 1.02 −
=
= 144.63 kip-ft
2
12 in./ft
M nx = M D − Fy • Z sn −
Φ b M nx =0.9 • 144.63 =
130.16 kip-ft
Interaction Equation H1-1a:
Pu
8 Mu
+
Φ c • Pn 9 Φ b • M n
≤ 1.0
129
8 120
+
≤ 1.0
354.78 9 130.16
1.18 > 1.0 n.g.
AISC-360-10 Example 003 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-08 PT-SL Example 001
Design Verification of Post-Tensioned Slab using the ACI 318-08 code
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 10
inches thick by 36 inches wide and spans 32 feet, as shown in shown in Figure 1.
A 36-inch-wide design strip was centered along the length of the slab and was
defined as an A-Strip. B-strips were placed at each end of the span perpendicular
to the Strip-A (the B-Strips are necessary to define the tendon profile). A tendon,
with two strands having an area of 0.153 square inches each, was added to the AStrip. The self weight and live loads were added to the slab. The loads and posttensioning forces are shown below. The total factored strip moments, required
area of mild steel reinforcement, and slab stresses are reported at the mid-span of
the slab. Independent hand calculations were compared with the ETABS results
and summarized for verification and validation of the ETABS results.
Loads: Dead = self weight , Live = 100psf
ACI 318-08 PT-SL Example 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate
T, h =
d
=
L =
f 'C =
fy
=
f pu
Prestressing, effective
fe
Area of Prestress (single strand), A P
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
Dead load,
wd
Live load,
wl
10
9
384
4,000
60,000
in
in
in
psi
psi
=
270,000 psi
=
=
=
=
=
=
=
=
175,500 psi
0.153 sq in
0.150 pcf
3,600 ksi
29,000 ksi
0
self
psf
100
psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
ACI 318-08 PT-SL Example 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
The ETABS total factored moments, required mild steel reinforcing and slab
stresses are compared to the independent hand calculations in Table 1.
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
1429.0
1428.3
-0.05%
2.21
2.21
0.00%
0.734
0.735
0.14%
0.414
0.414
0.00%
1.518
1.519
0.07%
1.220
1.221
0.08%
1.134
1.135
0.09%
0.836
0.837
0.12%
Factored moment,
Mu (Ultimate) (k-in)
Area of Mild Steel req’d,
As (sq-in)
Transfer Conc. Stress, top
(D+PTI), ksi
Transfer Conc. Stress, bot
(D+PTI), ksi
Normal Conc. Stress, top
(D+L+PTF), ksi
Normal Conc. Stress, bot
(D+L+PTF), ksi
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), ksi
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), ksi
COMPUTER FILE: ACI 318-08 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-08 PT-SL Example 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
=0.9
Mild Steel Reinforcing
fc = 4000 psi
fy = 60,000 psi
Post-Tensioning
fj
Stressing Loss
Long-Term Loss
fi
fe
= 216.0 ksi
= 27.0 ksi
= 13.5 ksi
= 189.0 ksi
= 175.5 ksi
Loads:
Dead, self-wt =
Live,
10 / 12
ft 0.150 kcf = 0.125 ksf (D) 1.2 = 0.150 ksf (Du)
0.100 ksf (L) 1.6 = 0.160 ksf (Lu)
Total =0.225 ksf (D+L)
0.310 ksf (D+L)ult
=0.225 ksf 3 ft = 0.675 klf,
2
Ultimate Moment,
M
U
ACI 318-08 PT-SL Example 001 - 4
w l1
8
u
= 0.310 ksf 3ft = 0.930 klf
= 0.310 klf 322/8 = 119.0 k-ft = 1429.0 k-in
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Stress in strand,
f PS f SE 1 0 0 0 0
f 'c
300 P
1 7 5, 5 0 0 1 0, 0 0 0
ETABS
0
(span-to-depth ratio > 35)
4, 000
3 0 0 0 .0 0 0 9 4 4
1 9 9, 6 2 4 p si 2 0 5, 5 0 0 p si
Ultimate force in PT,
F u lt , P T A P
Ultimate force in RC,
F u lt , R C A s
f y
Total Ultimate force,
F u lt , T o ta l 6 1 .0 8 1 2 0 .0 1 8 1 .0 8 k ip s
Stress block depth,
F u lt , T o ta l
a
0 .8 5 f ' c b
Ultimate moment due to PT,
Net ultimate moment,
M
net
M
M
u lt , P T
U
f PS
2 0 .1 5 3 1 9 9 .6 2 6 1 .0 8 k ip s
2 .0 0 ( a s s u m e d ) 6 0 .0 1 2 0 .0 k ip s
1 8 1 .0 8
0 .8 5 4 3 6
1 .4 8 in
a
1 .4 8
F u lt , P T d 6 1 .0 8 9
0 .9 4 5 4 .1 k -in
2
2
M
u lt , P T
Required area of mild steel reinforcing,
1 4 2 9 .0 4 5 4 .1 9 7 4 .9 k -in
AS
M
net
a
fy d
2
9 7 4 .9
1 .4 8
0 .9 6 0 9
2
2 .1 8 in
2
Note: The required area of mild steel reinforcing was calculated from an assumed amount of
steel. Since the assumed value and the calculated value are not the same a second iteration can be
performed. The second iteration changes the depth of the stress block and the calculated area of
steel value. Upon completion of the second iteration the area of steel was found to be 2.21in2
ACI 318-08 PT-SL Example 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Check of Concrete Stresses at Mid-Span:
Initial Condition (Transfer), load combination (D + L + PTi) = 1.0D + 1.0PTI
The stress in the tendon at transfer = jacking stress stressing losses = 216.0 27.0
= 189.0 ksi
The force in the tendon at transfer, = 1 8 9 .0 2 0 .1 5 3 5 7 .8 3 kips
Moment due to dead load,
Moment due to PT,
Stress in concrete,
M
f
PT
M
0 .1 2 5 3 3 2
D
2
8 4 8 .0 k -f t 5 7 6 k -in
F P T I (s a g ) 5 7 .8 3 4 in
FPTI
M
A
D
M
PT
S
5 7 .8 3
10 36
2 3 1 .3 k -in
5 7 6 .0 2 3 1 .3
, where S = 600 in3
600
f 0 .1 6 1 0 .5 7 4 5
f = - 0 .7 3 5 ( C o m p ) m a x , 0 .4 1 4 ( T e n s io n ) m a x
Normal Condition, load combinations: (D + L + PTF) = 1.0D + 1.0L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 1 7 5 .5 2 0 .1 5 3 5 3 .7 0 kips
Moment due to dead load,
M
Moment due to dead load,
M
Moment due to PT,
M
0 .1 2 5 3 3 2
2
D
8 4 8 .0 k -f t 5 7 6 k -in
L
0 .1 0 0 3 3 2
2
8 3 8 .4 k -ft 4 6 1 k -in
PT
F P T I (s a g ) 5 3 .7 0 4 in
Stress in concrete for (D + L+ PTF),
f
FPTI
M
A
DL
M
2 1 4 .8 k -in
PT
S
5 3 .7 0
10 36
1 0 3 7 .0 2 1 4 .8
600
f 0 .1 4 9 1 .7 2 7 0 .3 5 8
f 1 .5 1 8 ( C o m p ) m a x , 1 .2 2 0 ( T e n s io n ) m a x
Long-Term Condition, load combinations: (D + 0.5L + PTF(L)) = 1.0D + 0.5L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 1 7 5 .5 2 0 .1 5 3 5 3 .7 0 kips
Moment due to dead load,
M
Moment due to dead load,
M
Moment due to PT,
M
ACI 318-08 PT-SL Example 001 - 6
0 .1 2 5 3 3 2
2
D
8 4 8 .0 k -f t 5 7 6 k -in
L
0 .1 0 0 3 3 2
2
8 3 8 .4 k -ft 4 6 0 k -in
PT
F P T I (s a g ) 5 3 .7 0 4 in
2 1 4 .8 k -in
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Stress in concrete for (D + 0.5L + PTF(L)),
f
FPTI
A
M
D 0 .5 L
S
M
PT
5 3 .7 0
10 36
8 0 6 .0 2 1 4 .8
600
f 0 .1 4 9 0 .9 8 5
f 1 .1 3 4 ( C o m p ) m a x , 0 .8 3 6 ( T e n s io n ) m a x
ACI 318-08 PT-SL Example 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-08 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
A
C
B
1'
24'
D
24'
24'
1'
2'
4
17
18
19
13
14
15
20
10" thick flat slab
24'
3
Columns are 12" x 36"
with long side parallel
to the Y-axis, typical
24'
9
2
10
11
12
Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
24'
Y
5
1
X
6
7
Loading
DL = Self weight + 20 psf
LL = 80 psf
8
2'
Figure 1: Flat Slab For Numerical Example
ACI 318-08 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The slab overhangs the face of the column by 6 inches along each side of the
structure. The columns are typically 12 inches wide by 36 inches long, with the
long side parallel to the Y-axis. The slab is typically 10 inches thick. Thick shell
properties are used for the slab.
The concrete has a unit weight of 150 pcf and an f 'c of 4000 psi. The dead load
consists of the self weight of the structure plus an additional 20 psf. The live load
is 80 psf.
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS punching shear capacity, shear
stress ratio, and D/C ratio with the punching shear capacity, shear stress ratio and
D/C ratio obtained by the analytical method. They match exactly for this
example.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(ksi)
(ksi)
D/C ratio
ETABS
0.1930
0.158
1.22
Calculated
0.1930
0.158
1.22
COMPUTER FILE: ACI 318-08 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
ACI 318-08 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Hand Calculation for Interior Column Using ETABS Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
20.5"
Y
4.25"
6" 6"
Critical section for
punching shear shown
dashed.
4.25"
A
B
Column
4.25"
18"
Side 3
Side 1
Side 2
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
44.5"
18"
4.25"
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 44.5
1
3 20.5
1
2 20.5
1
3 44.5
0.4955
0.3115
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
ACI 318-08 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for
punching shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical
section for punching shear, as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
10.25
0
44.5
8.5
378.25
3877.06
0
x3
Ldx
y3
Ldy
2
Ld
2
Ld
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
0
0"
1105
0
0"
1105
Side 3
10.25
0
44.5
8.5
378.25
3877.06
0
Side 4
0
22.25
20.5
8.5
174.25
0
3877.06
Sum
N.A.
N.A.
b0 = 130
N.A.
1105
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the “Sum” column.
Item
L
d
x2 - x3
y2 - y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 2
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
From the ETABS output at Grid B-2:
VU = 189.45 k
V 2 M U 2 = 156.39 k-in
V 3 M U 3 = 91.538 k-in
ACI 318-08 RC-PN EXAMPLE 001 - 4
Side 3
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 4
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
301922.3
93782.8
0
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
At the point labeled A in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
(301922.3)(93782.8) (0)2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 0.0100 = 0.1499 ksi at point A
At the point labeled B in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 + 0.0100 = 0.1699 ksi at point B
At the point labeled C in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 + 0.0100 = 0.1930 ksi at point C
At the point labeled D in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 - 0.0100 = 0.1729 ksi at point D
ACI 318-08 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Point C has the largest absolute value of vu, thus vmax = 0.1930 ksi
The shear capacity is calculated based on the smallest of ACI 318-08 equations 11-34,
11-35 and 11-36 with the b0 and d terms removed to convert force to stress.
4
0.75 2
4000
36 /12
0.158 ksi in accordance with equation 11-34
vC
1000
40 8.5
0.75
2 4000
130
0.219 ksi in accordance with equation 11-35
vC
1000
vC
0.75 4 4000
0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of vC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
ACI 318-08 RC-PN EXAMPLE 001 - 6
vU
0.193
1.22
0.158
vC
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-08 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using ETABS. The slab is 6 inches thick and spans 12 feet between
walls. The walls are modeled as line supports. The computational model uses a
finite element mesh, automatically generated by ETABS. The maximum element
size is specified to be 36 inches. To obtain factored moments and flexural
reinforcement in a design strip, one one-foot-wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
12 ft span
Simply
supported
edge at wall
Free edge
1ft design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL80) and one live load case (LL100) with uniformly
distributed surface loads of magnitudes 80 and 100 psf, respectively, are defined
in the model. A load combination (COMB100) is defined using the ACI 318-08
load combination factors, 1.2 for dead loads and 1.6 for live loads. The model is
analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed in accordance with ACI 318-08 using ETABS
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
ACI 318-08 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
6
1
5
144
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0
Dead load
Live load
wd
wl
=
=
in
in
in
in
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area (sq-in)
Method
Strip
Moment
(k-in)
ETABS
55.22
0.213
Calculated
55.22
0.213
As+
Medium
A s ,min = 0.1296 sq-in
COMPUTER FILE: ACI 318-08 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-08 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
= 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f 4000
1 0.85 0.05 c
0.85
1000
0.003
d 1.875 in
0.003 0.005
amax = 1cmax = 1.59375 in
For the load combination, w and Mu are calculated as follows:
w = (1.2wd + 1.6wt) b / 144
cmax
wl12
Mu
8
As = min[As,min, (4/3) As,required] = min[0.1296, (4/3)2.11] = 0.1296 sq-in
COMB100
wd = 80 psf
wt = 100 psf
w = 21.33 lb/in
Mu-strip = 55.22 k-in
Mu-design = 55.629 k-in
The depth of the compression block is given by:
a d d2
2 Mu
0.85 f c' b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
As
= 0.213 sq-in > As,min
a
fy d
2
As = 0.2114 sq-in
ACI 318-08 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-11 PT-SL EXAMPLE 001
Design Verification of Post-Tensioned Slab using the ACI 318-11 code
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 10
inches thick by 36 inches wide and spans 32 feet, as shown in shown in Figure 1.
A 36-inch-wide design strip was centered along the length of the slab and was
defined as an A-Strip. B-strips were placed at each end of the span perpendicular
to the Strip-A (the B-Strips are necessary to define the tendon profile). A tendon,
with two strands having an area of 0.153 square inches each, was added to the AStrip. The self weight and live loads were added to the slab. The loads and posttensioning forces are shown below. The total factored strip moments, required
area of mild steel reinforcement, and slab stresses are reported at the mid-span of
the slab. Independent hand calculations were compared with the ETABS results
and summarized for verification and validation of the ETABS results.
Loads: Dead = self weight , Live = 100psf
ACI 318-11 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate
T, h
d
L
f 'C
fy
=
=
=
=
=
f pu =
Prestressing, effective
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
Dead load,
wd
Live load,
wl
10
9
384
4,000
60,000
in
in
in
psi
psi
270,000 psi
=
175,500 psi
=
=
=
=
=
=
=
0.153
0.150
3,600
29,000
0
self
100
sq in
pcf
ksi
ksi
psf
psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
ACI 318-11 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
The ETABS total factored moments, required mild steel reinforcing and slab
stresses are compared to the independent hand calculations in Table 1.
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
1429.0
1428.3
-0.05%
2.21
2.21
0.00%
0.734
0.735
0.14%
0.414
0.414
0.00%
1.518
1.519
0.07%
1.220
1.221
0.08%
1.134
1.135
0.09%
0.836
0.837
0.12%
Factored moment,
Mu (Ultimate) (k-in)
Area of Mild Steel req’d,
As (sq-in)
Transfer Conc. Stress, top
(D+PTI), ksi
Transfer Conc. Stress, bot
(D+PTI), ksi
Normal Conc. Stress, top
(D+L+PTF), ksi
Normal Conc. Stress, bot
(D+L+PTF), ksi
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), ksi
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), ksi
COMPUTER FILE: ACI 318-11 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-11 PT-SL EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
=0.9
Mild Steel Reinforcing
fc = 4000 psi
fy = 60,000 psi
Post-Tensioning
fj
Stressing Loss
Long-Term Loss
fi
fe
= 216.0 ksi
= 27.0 ksi
= 13.5 ksi
= 189.0 ksi
= 175.5 ksi
Loads:
Dead, self-wt = 10 / 12 ft 0.150 kcf = 0.125 ksf (D) 1.2 = 0.150 ksf (Du)
Live,
0.100 ksf (L) 1.6 = 0.160 ksf (Lu)
Total =0.225 ksf (D+L)
0.310 ksf (D+L)ult
=0.225 ksf 3 ft = 0.675 klf,
Ultimate Moment, M U
u = 0.310 ksf 3ft = 0.930 klf
wl12
= 0.310 klf 322/8 = 119.0 k-ft = 1429.0 k-in
8
ACI 318-11 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
f'c
(span-to-depth ratio > 35)
300 P
4, 000
175,500 10, 000
300 0.000944
199, 624 psi 205,500 psi
Ultimate Stress in strand, f PS f SE 10000
Ultimate force in PT, Fult , PT AP f PS 2 0.153199.62 61.08 kips
Ultimate force in RC, Fult , RC As f y 2.00(assumed) 60.0 120.0 kips
Total Ultimate force, Fult ,Total 61.08 120.0 181.08 kips
Stress block depth, a
Fult ,Total
0.85 f ' cb
181.08
1.48 in
0.85 4 36
a
1.48
Ultimate moment due to PT, M ult , PT Fult , PT d 61.08 9
0.9 454.1 k-in
2
2
Net ultimate moment, M net M U M ult , PT 1429.0 454.1 974.9 k-in
Required area of mild steel reinforcing, AS
M net
974.9
2.18 in 2
a
1.48
f y d 0.9 60 9
2
2
Note: The required area of mild steel reinforcing was calculated from an assumed amount of
steel. Since the assumed value and the calculated value are not the same a second iteration can be
performed. The second iteration changes the depth of the stress block and the calculated area of
steel value. Upon completion of the second iteration the area of steel was found to be 2.21in2
ACI 318-11 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Check of Concrete Stresses at Mid-Span:
Initial Condition (Transfer), load combination (D + L + PTi) = 1.0D + 1.0PTI
The stress in the tendon at transfer = jacking stress stressing losses = 216.0 27.0
= 189.0 ksi
The force in the tendon at transfer, = 189.0 2 0.153 57.83 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to PT, M PT FPTI (sag) 57.83 4 in 231.3 k-in
F
M M PT 57.83 576.0 231.3
Stress in concrete, f PTI D
, where S = 600 in3
A
S
10 36
600
f 0.161 0.5745
f = -0.735(Comp)max, 0.414(Tension)max
Normal Condition, load combinations: (D + L + PTF) = 1.0D + 1.0L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5 2 0.153 53.70 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to dead load, M L 0.100 3 32 8 38.4 k-ft 461 k-in
2
Moment due to PT,
M PT FPTI (sag) 53.70 4 in 214.8 k-in
FPTI M D L M PT 53.70 1037.0 214.8
10 36
600
A
S
f 0.149 1.727 0.358
f 1.518(Comp) max,1.220(Tension) max
Stress in concrete for (D + L+ PTF), f
Long-Term Condition, load combinations: (D + 0.5L + PTF(L)) = 1.0D + 0.5L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5 2 0.153 53.70 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to dead load, M L 0.100 3 32 8 38.4 k-ft 460 k-in
2
Moment due to PT,
ACI 318-11 PT-SL EXAMPLE 001 - 6
M PT FPTI (sag) 53.70 4 in 214.8 k-in
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Stress in concrete for (D + 0.5L + PTF(L)),
M
M PT 53.70 806.0 214.8
F
f PTI D 0.5 L
A
S
10 36
600
f 0.149 0.985
f 1.134(Comp) max, 0.836(Tension) max
ACI 318-11 PT-SL EXAMPLE 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-11 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
A
C
B
1'
24'
D
24'
24'
1'
2'
4
17
18
19
13
14
15
20
10" thick flat slab
24'
3
Columns are 12" x 36"
with long side parallel
to the Y-axis, typical
24'
9
2
10
11
12
Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
24'
Y
5
1
X
6
7
Loading
DL = Self weight + 20 psf
LL = 80 psf
8
2'
Figure 1: Flat Slab For Numerical Example
ACI 318-11 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The slab overhangs the face of the column by 6 inches along each side of the
structure. The columns are typically 12 inches wide by 36 inches long, with the
long side parallel to the Y-axis. The slab is typically 10 inches thick. Thick shell
properties are used for the slab.
The concrete has a unit weight of 150 pcf and an f 'c of 4000 psi. The dead load
consists of the self weight of the structure plus an additional 20 psf. The live load
is 80 psf.
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS punching shear capacity, shear
stress ratio, and D/C ratio with the punching shear capacity, shear stress ratio and
D/C ratio obtained by the analytical method. They match exactly for this
example.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(ksi)
(ksi)
D/C ratio
ETABS
0.1930
0.158
1.22
Calculated
0.1930
0.158
1.22
COMPUTER FILE: ACI 318-11 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
ACI 318-11 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Hand Calculation for Interior Column Using ETABS Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
20.5"
Y
4.25"
6" 6"
Critical section for
punching shear shown
dashed.
4.25"
A
B
Column
4.25"
18"
Side 3
Side 1
Side 2
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
44.5"
18"
4.25"
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 44.5
1
3 20.5
1
2 20.5
1
3 44.5
0.4955
0.3115
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
ACI 318-11 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for
punching shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical
section for punching shear, as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
10.25
0
44.5
8.5
378.25
3877.06
0
x3
Ldx
y3
Ldy
2
Ld
2
Ld
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
0
0"
1105
0
0"
1105
Side 3
10.25
0
44.5
8.5
378.25
3877.06
0
Side 4
0
22.25
20.5
8.5
174.25
0
3877.06
Sum
N.A.
N.A.
b0 = 130
N.A.
1105
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the “Sum” column.
Item
L
d
x2 - x3
y2 - y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 2
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
From the ETABS output at Grid B-2:
VU = 189.45 k
V 2 M U 2 = 156.39 k-in
V 3 M U 3 = 91.538 k-in
ACI 318-11 RC-PN EXAMPLE 001 - 4
Side 3
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 4
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
301922.3
93782.8
0
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
At the point labeled A in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
(301922.3)(93782.8) (0)2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 0.0100 = 0.1499 ksi at point A
At the point labeled B in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 + 0.0100 = 0.1699 ksi at point B
At the point labeled C in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 + 0.0100 = 0.1930 ksi at point C
At the point labeled D in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 - 0.0100 = 0.1729 ksi at point D
ACI 318-11 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Point C has the largest absolute value of vu, thus vmax = 0.1930 ksi
The shear capacity is calculated based on the smallest of ACI 318-11 equations 11-34,
11-35 and 11-36 with the b0 and d terms removed to convert force to stress.
4
0.75 2
4000
36 /12
0.158 ksi in accordance with equation 11-34
vC
1000
40 8.5
0.75
2 4000
130
0.219 ksi in accordance with equation 11-35
vC
1000
vC
0.75 4 4000
0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of vC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
ACI 318-11 RC-PN EXAMPLE 001 - 6
vU
0.193
1.22
0.158
vC
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-11 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using ETABS. The slab is 6 inches thick and spans 12 feet between
walls. The slab is modeled using thin plate elements. The walls are modeled as
line supports. The computational model uses a finite element mesh, automatically
generated by ETABS. The maximum element size is specified to be 36 inches.
To obtain factored moments and flexural reinforcement in a design strip, one
one-foot-wide strip is defined in the X-direction on the slab, as shown in Figure
1.
Simply
supported
edge at wall
12 ft span
Simply
supported
edge at wall
Free edge
1ft design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL80) and one live load case (LL100) with uniformly
distributed surface loads of magnitudes 80 and 100 psf, respectively, are defined
in the model. A load combination (COMB100) is defined using the ACI 318-11
load combination factors, 1.2 for dead loads and 1.6 for live loads. The model is
analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed in accordance with ACI 318-11 using ETABS
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
ACI 318-11 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
6
1
5
144
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0
Dead load
Live load
wd
wl
=
=
in
in
in
in
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area (sq-in)
Method
Strip
Moment
(k-in)
ETABS
55.22
0.213
Calculated
55.22
0.213
As+
Medium
A s ,min = 0.1296 sq-in
COMPUTER FILE: ACI 318-11 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-11 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
= 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f 4000
1 0.85 0.05 c
0.85
1000
0.003
d 1.875 in
0.003 0.005
amax = 1cmax = 1.59375 in
For the load combination, w and Mu are calculated as follows:
w = (1.2wd + 1.6wt) b / 144
cmax
wl12
Mu
8
As = min[As,min, (4/3) As,required] = min[0.1296, (4/3)2.11] = 0.1296 sq-in
COMB100
wd = 80 psf
wt = 100 psf
w = 21.33 lb/in
Mu-strip = 55.22 k-in
Mu-design = 55.629 k-in
The depth of the compression block is given by:
a d d2
2 Mu
0.85 f c' b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
As
= 0.213 sq-in > As,min
a
fy d
2
As = 0.2114 sq-in
ACI 318-11 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-14 PT-SL EXAMPLE 001
Design Verification of Post-Tensioned Slab using the ACI 318-14 code
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 10
inches thick by 36 inches wide and spans 32 feet, as shown in shown in Figure 1.
A 36-inch-wide design strip was centered along the length of the slab and was
defined as an A-Strip. B-strips were placed at each end of the span perpendicular
to the Strip-A (the B-Strips are necessary to define the tendon profile). A tendon,
with two strands having an area of 0.153 square inches each, was added to the AStrip. The self weight and live loads were added to the slab. The loads and posttensioning forces are shown below. The total factored strip moments, required
area of mild steel reinforcement, and slab stresses are reported at the mid-span of
the slab. Independent hand calculations were compared with the ETABS results
and summarized for verification and validation of the ETABS results.
Loads: Dead = self weight, Live = 100psf
ACI 318-14 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate
T, h
d
L
f 'C
fy
=
=
=
=
=
f pu =
Prestressing, effective
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
Dead load,
wd
Live load,
wl
10
9
384
4,000
60,000
in
in
in
psi
psi
270,000 psi
=
175,500 psi
=
=
=
=
=
=
=
0.153
0.150
3,600
29,000
0
self
100
sq in
pcf
ksi
ksi
psf
psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
ACI 318-14 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
The ETABS total factored moments, required mild steel reinforcing and slab
stresses are compared to the independent hand calculations in Table 1.
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
1429.0
1428.3
-0.05%
2.21
2.21
0.00%
0.734
0.735
0.14%
0.414
0.414
0.00%
1.518
1.519
0.07%
1.220
1.221
0.08%
1.134
1.135
0.09%
0.836
0.837
0.12%
Factored moment,
Mu (Ultimate) (k-in)
Area of Mild Steel req’d,
As (sq-in)
Transfer Conc. Stress, top
(D+PTI), ksi
Transfer Conc. Stress, bot
(D+PTI), ksi
Normal Conc. Stress, top
(D+L+PTF), ksi
Normal Conc. Stress, bot
(D+L+PTF), ksi
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), ksi
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), ksi
COMPUTER FILE: ACI 318-14 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-14 PT-SL EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
=0.9
Mild Steel Reinforcing
fc = 4000 psi
fy = 60,000 psi
Post-Tensioning
fj
Stressing Loss
Long-Term Loss
fi
fe
= 216.0 ksi
= 27.0 ksi
= 13.5 ksi
= 189.0 ksi
= 175.5 ksi
Loads:
Dead, self-wt = 10 / 12 ft 0.150 kcf = 0.125 ksf (D) 1.2 = 0.150 ksf (Du)
Live,
0.100 ksf (L) 1.6 = 0.160 ksf (Lu)
Total =0.225 ksf (D+L)
0.310 ksf (D+L)ult
=0.225 ksf 3 ft = 0.675 klf,
Ultimate Moment, M U
u = 0.310 ksf 3ft = 0.930 klf
wl12
= 0.310 klf 322/8 = 119.0 k-ft = 1429.0 k-in
8
ACI 318-14 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
f'c
(span-to-depth ratio > 35)
300 P
4, 000
175,500 10, 000
300 0.000944
199, 624 psi 205,500 psi
Ultimate Stress in strand, f PS f SE 10000
Ultimate force in PT, Fult , PT AP f PS 2 0.153199.62 61.08 kips
Ultimate force in RC, Fult , RC As f y 2.00(assumed) 60.0 120.0 kips
Total Ultimate force, Fult ,Total 61.08 120.0 181.08 kips
Stress block depth, a
Fult ,Total
0.85 f ' cb
181.08
1.48 in
0.85 4 36
a
1.48
Ultimate moment due to PT, M ult , PT Fult , PT d 61.08 9
0.9 454.1 k-in
2
2
Net ultimate moment, M net M U M ult , PT 1429.0 454.1 974.9 k-in
Required area of mild steel reinforcing, AS
M net
974.9
2.18 in 2
a
1.48
f y d 0.9 60 9
2
2
Note: The required area of mild steel reinforcing was calculated from an assumed amount of
steel. Since the assumed value and the calculated value are not the same a second iteration can be
performed. The second iteration changes the depth of the stress block and the calculated area of
steel value. Upon completion of the second iteration the area of steel was found to be 2.21in2
ACI 318-14 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Check of Concrete Stresses at Mid-Span:
Initial Condition (Transfer), load combination (D + L + PTi) = 1.0D + 1.0PTI
The stress in the tendon at transfer = jacking stress stressing losses = 216.0 27.0
= 189.0 ksi
The force in the tendon at transfer, = 189.0 2 0.153 57.83 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to PT, M PT FPTI (sag) 57.83 4 in 231.3 k-in
F
M M PT 57.83 576.0 231.3
Stress in concrete, f PTI D
, where S = 600 in3
A
S
10 36
600
f 0.161 0.5745
f = -0.735(Comp)max, 0.414(Tension)max
Normal Condition, load combinations: (D + L + PTF) = 1.0D + 1.0L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5 2 0.153 53.70 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to dead load, M L 0.100 3 32 8 38.4 k-ft 461 k-in
2
Moment due to PT,
M PT FPTI (sag) 53.70 4 in 214.8 k-in
FPTI M D L M PT 53.70 1037.0 214.8
10 36
600
A
S
f 0.149 1.727 0.358
f 1.518(Comp) max,1.220(Tension) max
Stress in concrete for (D + L+ PTF), f
Long-Term Condition, load combinations: (D + 0.5L + PTF(L)) = 1.0D + 0.5L + 1.0PTF
Tendon stress at normal = jacking stressing long-term = 216.0 27.0 13.5 = 175.5 ksi
The force in tendon at Normal, = 175.5 2 0.153 53.70 kips
2
Moment due to dead load, M D 0.125 3 32 8 48.0 k-ft 576 k-in
Moment due to dead load, M L 0.100 3 32 8 38.4 k-ft 460 k-in
2
Moment due to PT,
ACI 318-14 PT-SL EXAMPLE 001 - 6
M PT FPTI (sag) 53.70 4 in 214.8 k-in
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Stress in concrete for (D + 0.5L + PTF(L)),
M
M PT 53.70 806.0 214.8
F
f PTI D 0.5 L
A
S
10 36
600
f 0.149 0.985
f 1.134(Comp) max, 0.836(Tension) max
ACI 318-14 PT-SL EXAMPLE 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-14 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
A
C
B
1'
24'
D
24'
24'
1'
2'
4
17
18
19
13
14
15
20
10" thick flat slab
24'
3
Columns are 12" x 36"
with long side parallel
to the Y-axis, typical
24'
9
2
10
11
12
Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
24'
Y
5
1
X
6
7
Loading
DL = Self weight + 20 psf
LL = 80 psf
8
2'
Figure 1: Flat Slab For Numerical Example
ACI 318-14 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The slab overhangs the face of the column by 6 inches along each side of the
structure. The columns are typically 12 inches wide by 36 inches long, with the
long side parallel to the Y-axis. The slab is typically 10 inches thick. Thick shell
properties are used for the slab.
The concrete has a unit weight of 150 pcf and an f 'c of 4000 psi. The dead load
consists of the self weight of the structure plus an additional 20 psf. The live load
is 80 psf.
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS punching shear capacity, shear
stress ratio, and D/C ratio with the punching shear capacity, shear stress ratio and
D/C ratio obtained by the analytical method. They match exactly for this
example.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(ksi)
(ksi)
D/C ratio
ETABS
0.1930
0.158
1.22
Calculated
0.1930
0.158
1.22
COMPUTER FILE: ACI 318-14 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
ACI 318-14 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Hand Calculation for Interior Column Using ETABS Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
20.5"
Y
4.25"
6" 6"
Critical section for
punching shear shown
dashed.
4.25"
A
B
Column
4.25"
18"
Side 3
Side 1
Side 2
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
44.5"
18"
4.25"
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 44.5
1
3 20.5
1
2 20.5
1
3 44.5
0.4955
0.3115
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
ACI 318-14 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for
punching shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical
section for punching shear, as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
10.25
0
44.5
8.5
378.25
3877.06
0
x3
Ldx
y3
Ldy
2
Ld
2
Ld
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
0
0"
1105
0
0"
1105
Side 3
10.25
0
44.5
8.5
378.25
3877.06
0
Side 4
0
22.25
20.5
8.5
174.25
0
3877.06
Sum
N.A.
N.A.
b0 = 130
N.A.
1105
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the “Sum” column.
Item
L
d
x2 - x3
y2 - y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 2
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
From the ETABS output at Grid B-2:
VU = 189.45 k
V 2 M U 2 = 156.39 k-in
V 3 M U 3 = 91.538 k-in
ACI 318-14 RC-PN EXAMPLE 001 - 4
Side 3
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0
Side 4
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
301922.3
93782.8
0
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
At the point labeled A in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
(301922.3)(93782.8) (0)2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 0.0100 = 0.1499 ksi at point A
At the point labeled B in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 0.0115 + 0.0100 = 0.1699 ksi at point B
At the point labeled C in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 + 0.0100 = 0.1930 ksi at point C
At the point labeled D in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5
301922.3 93782.8 0 2
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0 2
vU = 0.1714 + 0.0115 - 0.0100 = 0.1729 ksi at point D
ACI 318-14 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Point C has the largest absolute value of vu, thus vmax = 0.1930 ksi
The shear capacity is calculated based on the smallest of ACI 318-14 equations 11-34,
11-35 and 11-36 with the b0 and d terms removed to convert force to stress.
4
0.75 2
4000
36 /12
0.158 ksi in accordance with equation 11-34
vC
1000
40 8.5
0.75
2 4000
130
0.219 ksi in accordance with equation 11-35
vC
1000
vC
0.75 4 4000
0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of vC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
ACI 318-14 RC-PN EXAMPLE 001 - 6
vU
0.193
1.22
0.158
vC
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
ACI 318-14 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using ETABS. The slab is 6 inches thick and spans 12 feet between
walls. The walls are modeled as line supports. The computational model uses a
finite element mesh, automatically generated by ETABS. The maximum element
size is specified to be 36 inches. To obtain factored moments and flexural
reinforcement in a design strip, one one-foot-wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
12 ft span
Simply
supported
edge at wall
Free edge
1ft design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL80) and one live load case (LL100) with uniformly
distributed surface loads of magnitudes 80 and 100 psf, respectively, are defined
in the model. A load combination (COMB100) is defined using the ACI 318-14
load combination factors, 1.2 for dead loads and 1.6 for live loads. The model is
analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed in accordance with ACI 318-14 using ETABS
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
ACI 318-14 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
6
1
5
144
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0
Dead load
Live load
wd
wl
=
=
in
in
in
in
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area (sq-in)
Method
Strip
Moment
(k-in)
ETABS
55.22
0.213
Calculated
55.22
0.213
As+
Medium
A s ,min = 0.1296 sq-in
COMPUTER FILE: ACI 318-14 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
ACI 318-14 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
= 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f 4000
1 0.85 0.05 c
0.85
1000
0.003
d 1.875 in
0.003 0.005
amax = 1cmax = 1.59375 in
For the load combination, w and Mu are calculated as follows:
w = (1.2wd + 1.6wt) b / 144
cmax
wl12
Mu
8
As = min[As,min, (4/3) As,required] = min[0.1296, (4/3)2.11] = 0.1296 sq-in
COMB100
wd = 80 psf
wt = 100 psf
w = 21.33 lb/in
Mu-strip = 55.22 k-in
Mu-design = 55.629 k-in
The depth of the compression block is given by:
a d d2
2 Mu
0.85 f c' b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
As
= 0.213 sq-in > As,min
a
fy d
2
As = 0.2114 sq-in
ACI 318-14 RC-SL EXAMPLE 001 - 3
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0
AS 3600-2001 PT-SL Example 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel reinforcing strength for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
AS 3600-2001 PT-SL Example 001 - 1
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0
A 914-mm-wide design strip is centered along the length of the slab and is defined
as an A-Strip. B-Strips have been placed at each end of the span, perpendicular to
Strip-A (the B-Strips are necessary to define the tendon profile). A tendon with
two strands, each having an area of 99 mm2, has been added to the A-Strip. The
self-weight and live loads were added to the slab. The loads and post-tensioning
forces are as follows:
Loads:
Dead = self weight, Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the midspan of the slab. Independent hand calculations
were compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
T, h =
d
=
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate
Prestressing, effective
Area of prestress (single tendon),
Concrete unit weight,
Concrete modulus of elasticity,
Rebar modulus of elasticity,
Poisson’s ratio,
L
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
=
Dead load,
Live load,
wd
wl
=
=
254 mm
229 mm
9754
30
400
1862
1210
198
23.56
25000
200,000
0
mm
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
self KN/m2
4.788 KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing and slab stresses with the independent hand calculations.
AS 3600-2001 PT-SL Example 001 - 2
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ETABS
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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(0.8D+1.15PTI), MPa
Transfer Conc. Stress, bot
(0.8D+1.15PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), MPa
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
156.12
156.17
0.03%
16.55
16.60
0.30%
3.500
3.498
-0.06%
0.950
0.948
-0.21%
10.460
10.467
0.07%
8.402
8.409
0.08%
7.817
7.818
0.01%
5.759
5.760
0.02%
COMPUTER FILE: AS 3600-2001 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
AS 3600-2001 PT-SL Example 001 - 3
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ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPai
fy = 400MPa
Post-Tensioning
fpu =
fpy =
Stressing Loss =
Long-Term Loss =
fi =
fe =
1862 MPa
1675 MPa
186 MPa
94 MPa
1490 MPa
1210 MPa
0.80
0.85 0.007 f 'c 28= 0.836
amax k u d = 0.836*0.4*229 = 76.5 mm
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.2 = 7.181 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) x 1.5 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 14.363 kN/m2 (D+L)ult
=10.772 kN/m2 x 0.914m = 9.846 kN/m, u = 14.363 kN/m2 x 0.914m = 13.128 kN/m
Ultimate Moment, M U
AS 3600-2001 PT-SL Example 001 - 4
wl12
= 13.128 x (9.754)2/8 = 156.12 kN-m
8
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Ultimate Stress in strand, f PS f SE 70
ETABS
0
f 'C bef d P
300 AP
30 914 229
1210 70
300 198
1386 MPa f SE 200 1410 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1386 1000 273.60 kN
Total Ultimate force, Fult ,Total 273.60 560.0 833.60 kN
2M*
Stress block depth, a d d
0.85 f 'c b
2
0.229 0.2292
2 159.12
40.90
0.85 30000 0.80 0.914
Ultimate moment due to PT,
a
40.90
M ult , PT Fult , PT d 273.60 229
0.80 1000 45.65 kN-m
2
2
Net ultimate moment, M net M U M ult , PT 156.1 45.65 110.45 kN-m
Required area of mild steel reinforcing,
M net
110.45
1e6 1655 mm 2
AS
0.04090
a
f y d 0.80 400000 0.229
2
2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (0.8D+1.15PTi) = 0.80D+0.0L+1.15PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT 1.15 257.4 0.80 65.04 1.15 26.23
Stress in concrete, f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
f 1.275 2.225 MPa
f 3.500(Comp) max, 0.950(Tension) max
AS 3600-2001 PT-SL Example 001 - 5
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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at Normal = jacking stressing long-term = 1490 186 94= 1210 MPa
The force in tendon at Normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at Normal = jacking stressing long-term =1490 186 94 = 1210 MPa
The force in tendon at Normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to dead load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
0.254 0.914
0.00983
A
S
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
AS 3600-2001 PT-SL Example 001 - 6
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AS 3600-2001 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m, with the long side parallel
to the Y-axis. The slab is typically 0.25-m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead
load consists of the self weight of the structure plus an additional 1 kN/m2. The
live load is 4 kN/m2.
AS 3600-2001 RC-PN EXAMPLE 001 - 1
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio,
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid Point B-2
Shear Stress Shear Capacity D/C ratio
Method
(N/mm2)
(N/mm2)
ETABS
1.799
1.086
1.66
Calculated
1.811
1.086
1.67
COMPUTER FILE: AS 3600-2001 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
AS 3600-2001 RC-PN EXAMPLE 001 - 2
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ETABS
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
dom = [(250 26) + (250 38)] / 2 = 218 mm
Refer to Figure 2.
U = 518+ 1118 + 1118 + 518 = 3272 mm
ax = 518 mm
ay = 1118 mm
Note: All dimensions in millimeters
518
Y
109
150
A
Critical section for
punching shear shown
dashed.
109
B
Side 2
109
Side 3
Side 1
Column
150
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at grid line B-2:
V* = 1126.498 kN
Mv2 = 51.991 kN-m
Mv3 = 45.723 kN-m
AS 3600-2001 RC-PN EXAMPLE 001 - 3
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The maximum design shear stress is computed along the major and minor axis of column
separately:
vmax
uM v
V*
1.0 *
ud om
8V ad om
vmax, X
1126.498 103
3272 51.991 106
1
3
3272 218 8 1126.498 10 1118 218
vmax, X = 1.579 1.0774 = 1.7013 N/mm2
vmax,Y
1126.498 103
3272 45.723 106
1
3
3272 218 8 1126.498 10 518 218
vmax,Y = 1.579 1.1470 = 1.811 N/mm2 (Govern)
The largest absolute value of vmax= 1.811 N/mm2
The shear capacity is calculated based on the smallest of AS 3600-01 equation 11-35,
with the dom and u terms removed to convert force to stress.
2
0.17 1
h
fcv min
0.34 f c
f c
= 1.803N/mm2 in accordance with AS 9.2.3(a)
AS 9.2.3(a) yields the smallest value of fcv = 1.086 N/mm2, and thus this is the shear
capacity.
Shear Ratio
vU
1.811
1.67
f cv 1.086
AS 3600-2001 RC-PN EXAMPLE 001 - 4
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ETABS
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AS 3600-2001 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa), with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the AS 36002001 load combination factors, 1.2 for dead loads and 1.5 for live loads. The
model is analyzed for both load cases and the load combinations.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design is performed using the AS 3600-2001 code using ETABS and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
AS 3600-2001 RC-SL EXAMPLE 001 - 1
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
24.597
5.58
Calculated
24.600
5.58
As+
Medium
A s ,min = 282.9 sq-mm
COMPUTER FILE: AS 3600-2001 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
AS 3600-2001 RC-SL EXAMPLE 001 - 2
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ETABS
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HAND CALCULATION
The following quantities are computed for the load combination:
= 0.8
b = 1000 mm
0.85 0.007 f 'c 28= 0.836
a max k u d = 0.836•0.4•125 = 41.8 mm
For the load combination, w and M* are calculated as follows:
w = (1.2wd + 1.5wt) b
wl12
Mu
8
D f cf
0.22
bd
d fsy
2
Ast .min
= 0.22•(150/125)2•0.6•SQRT(30)/460•100•125
= 282.9 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.3 kN/m
M-strip* = 24.6 kN-m
M-design* = 24.633 kN-m
The depth of the compression block is given by:
a d d2
2M *
= 10.065 mm < amax
0.85 f 'c b
The area of tensile steel reinforcement is then given by:
Ast
M*
= 557.966 sq-mm > As,min
a
f sy d
2
AS 3600-2001 RC-SL EXAMPLE 001 - 3
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ETABS
0
As = 5.57966 sq-cm
AS 3600-2001 RC-SL EXAMPLE 001 - 4
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ETABS
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AS 3600-2009 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel reinforcing strength for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
AS 3600-2009 PT-SL EXAMPLE 001 - 1
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ETABS
0
A 914-mm-wide design strip is centered along the length of the slab and is defined
as an A-Strip. B-Strips have been placed at each end of the span, perpendicular to
Strip-A (the B-Strips are necessary to define the tendon profile). A tendon with
two strands, each having an area of 99 mm2, has been added to the A-Strip. The
self weight and live loads were added to the slab. The loads and post-tensioning
forces are as follows:
Loads:
Dead = self weight, Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations were compared with the ETABS results and summarized for
verification and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
T, h =
d
=
Clear span,
Concrete strength,
Yield strength of steel,
Prestressing, ultimate
Prestressing, effective
Area of prestress (single tendon),
Concrete unit weight,
Concrete modulus of elasticity,
Rebar modulus of elasticity,
Poisson’s ratio,
L
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
=
Dead load,
Live load,
wd
wl
=
=
254 mm
229 mm
9754
30
400
1862
1210
198
23.56
25000
200,000
0
mm
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
self KN/m2
4.788 KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing and slab stresses with the independent hand calculations.
AS 3600-2009 PT-SL EXAMPLE 001 - 2
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REVISION NO.:
ETABS
0
Table 1 Comparison of Results
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
156.12
156.17
0.03%
Area of Mild Steel req’d, As
(sq-cm)
16.55
16.60
0.30%
Transfer Conc. Stress, top
(0.8D+1.15PTI), MPa
3.500
3.498
-0.06%
Transfer Conc. Stress, bot
(0.8D+1.15PTI), MPa
0.950
0.948
-0.21%
Normal Conc. Stress, top
(D+L+PTF), MPa
10.460
10.467
0.07%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.409
0.08%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
7.817
7.818
0.01%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.760
0.02%
FEATURE TESTED
COMPUTER FILE: AS 3600-2009 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
AS 3600-2009 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPai
fy = 400MPa
Post-Tensioning
fpu =
fpy =
Stressing Loss =
Long-Term Loss =
fi =
fe =
1862 MPa
1675 MPa
186 MPa
94 MPa
1490 MPa
1210 MPa
0.80
2 1.0 0.003 f 'c = 0.91 > 0.85, Use 2 0.85
1.0 0.003 f 'c = 0.91 > 0.85, Use 0.85
amax k u d = 0.85*0.36*229 = 70.07 mm
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.2 = 7.181 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) x 1.5 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 14.363 kN/m2 (D+L)ult
=10.772 kN/m2 x 0.914m = 9.846 kN/m, u = 14.363 kN/m2 x 0.914m = 13.128 kN/m
Ultimate Moment, M U
AS 3600-2009 PT-SL EXAMPLE 001 - 4
wl12
= 13.128 x (9.754)2/8 = 156.12 kN-m
8
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Stress in strand, f PS f SE 70
ETABS
0
f 'C bef d P
300 AP
30 914 229
1210 70
300 198
1386 MPa f SE 200 1410 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1386 1000 273.60 kN
Total Ultimate force, Fult ,Total 273.60 560.0 833.60 kN
2M*
Stress block depth, a d d
0.85 f 'c b
2
0.229 0.2292
2 159.12
40.90
0.85 30000 0.80 0.914
Ultimate moment due to PT,
a
40.90
M ult , PT Fult , PT d 273.60 229
0.80 1000 45.65 kN-m
2
2
Net ultimate moment, M net M U M ult , PT 156.1 45.65 110.45 kN-m
Required area of mild steel reinforcing,
M net
110.45
1e6 1655 mm 2
AS
0.04090
a
f y d 0.80 400000 0.229
2
2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (0.8D+1.15PTi) = 0.80D+0.0L+1.15PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT 1.15 257.4 0.80 65.04 1.15 26.23
Stress in concrete, f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
f 1.275 2.225 MPa
f 3.500(Comp) max, 0.950(Tension) max
AS 3600-2009 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at Normal = jacking stressing long-term = 1490 186 94= 1210 MPa
The force in tendon at Normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at Normal = jacking stressing long-term =1490 186 94 = 1210 MPa
The force in tendon at Normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to dead load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
0.254 0.914
0.00983
A
S
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
AS 3600-2009 PT-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
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ETABS
0
AS 3600-2009 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m, with the long side parallel
to the Y-axis. The slab is typically 0.25-m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
AS 3600-2009 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio,
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid Point B-2
Shear Stress Shear Capacity D/C ratio
Method
(N/mm2)
(N/mm2)
ETABS
1.793
1.127
1.60
Calculated
1.811
1.086
1.67
COMPUTER FILE: AS 3600-2009 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
AS 3600-2009 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
dom = [(250 26) + (250 38)] / 2 = 218 mm
Refer to Figure 2.
U = 518+ 1118 + 1118 + 518 = 3272 mm
ax = 518 mm
ay = 1118 mm
Note: All dimensions in millimeters
518
Y
109
150
A
Critical section for
punching shear shown
dashed.
109
B
Side 2
109
Side 3
Side 1
Column
150
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at grid line B-2:
V* = 1126.498 kN
Mv2 = 51.991 kN-m
Mv3 = 45.723 kN-m
AS 3600-2009 RC-PN EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The maximum design shear stress is computed along the major and minor axis of column
separately:
vmax
uM v
V*
1.0 *
ud om
8V ad om
vmax, X
1126.498 103
3272 51.991 106
1
3
3272 218 8 1126.498 10 1118 218
vmax, X = 1.579 1.0774 = 1.7013 N/mm2
vmax,Y
1126.498 103
3272 45.723 106
1
3
3272 218 8 1126.498 10 518 218
vmax,Y = 1.579 1.1470 = 1.811 N/mm2 (Govern)
The largest absolute value of vmax= 1.811 N/mm2
The shear capacity is calculated based on the smallest of AS 3600-09 equation 11-35,
with the dom and u terms removed to convert force to stress.
2
0.17 1
h
fcv min
0.34 f c
f c
= 1.803N/mm2 in accordance with AS 9.2.3(a)
AS 9.2.3(a) yields the smallest value of fcv = 1.086 N/mm2, and thus this is the shear
capacity.
Shear Ratio
vU
1.811
1.67
f cv 1.086
AS 3600-2009 RC-PN EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
AS 3600-2009 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa), with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the AS 36002009 load combination factors, 1.2 for dead loads and 1.5 for live loads. The
model is analyzed for both load cases and the load combinations.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design is performed using the AS 3600-2009 code using ETABS and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
AS 3600-2009 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
24.597
5.58
Calculated
24.600
5.58
As+
Medium
A s ,min = 370.356 sq-mm
COMPUTER FILE: AS 3600-2009 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
AS 3600-2009 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
= 0.8
b = 1000 mm
2 1.0 0.003 f 'c = 0.91 > 0.85, Use 2 0.85
1.05 0.007 f 'c = 0.84 < 0.85, Use 0.84
a max k u d = 0.84•0.36•125 = 37.80 mm
For the load combination, w and M* are calculated as follows:
w = (1.2wd + 1.5wt) b
wl12
Mu
8
2
'
h f ct , f
As 0.24
bh for flat slabs
d f sy , f
Ast .min
2
h f ct , f
0.24
bd
d f sy , f
= 0.24•(150/125)2•0.6•SQRT(30)/460•1000•150
= 370.356 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.3 kN/m
M-strip* = 24.6 kN-m
M-design* = 24.633 kN-m
The depth of the compression block is given by:
a d d2
2M *
= 10.065 mm < amax
0.85 f 'c b
AS 3600-2009 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The area of tensile steel reinforcement is then given by:
Ast
M*
= 557.966 sq-mm > As,min
a
f sy d
2
As = 5.57966 sq-cm
AS 3600-2009 RC-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
BS 8110-1997 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
BS 8110-1997 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, was added to the AStrip. The self-weight and live loads were added to the slab. The loads and posttensioning forces are as follows.
Loads:
Dead = self weight,
Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the midspan of the slab. Independent hand calculations
have been compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
Dead load
Live load
wd
wl
=
=
254 mm
229 mm
9754 mm
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
self kN/m2
4.788 kN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
BS 8110-1997 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.80
0.76%
5.058
5.057
0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
COMPUTER FILE: BS 8110-1997 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
BS 8110-1997 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fcu = 30 MPa
fy = 400 MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, steel = 1.15
m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Loads:
Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) x 1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult
=10.772 kN/m2 x 0.914m = 9.846 kN/m, u = 16.039 kN/m2 x 0.914m = 14.659 kN/m
wl12
= 14.659 x (9.754)2/8 = 174.4 kN-m
8
f pu Ap
7000
Ultimate Stress in strand, f pb f pe
1 1.7
l/d
f cu bd
Ultimate Moment, M U
1210
7000
1862(198)
1 1.7
9.754 / 0.229
30(914)(229)
1358 MPa 0.7 f pu 1303 MPa
BS 8110-1997 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
K factor used to determine the effective depth is given as:
174.4
M
K
0.1213 < 0.156
=
2
2
f cu bd
30000 0.914 0.229
K
0.95d = 192.2 mm
z d 0.5 0.25
0
.
9
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1303 1000 257.2 KN
Ultimate moment due to PT, M ult , PT Fult , PT ( z ) / 257.2 0.192 1.15 43.00 kN-m
Net Moment to be resisted by As, M NET MU M PT
174.4 43.00 131.40 kN-m
The area of tensile steel reinforcement is then given by:
As
M NET
131.4
1e6 1965 mm 2
=
0.87 f y z
0.87 400 192
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
Moment due to PT,
M PT FPTI (sag) 257.4 102mm 1000 26.25 kN-m
2
FPTI M D M PT
257.4
65.04 26.23
A
S
0.254 0.914
0.00983
where S=0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Stress in concrete, f
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at Normal = jacking stressing long-term = 1490 186 94= 1210 MPa
The force in tendon at Normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
BS 8110-1997 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
BS 8110-1997 PT-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
BS 8110-1997 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25-m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a fcu of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
BS 8110-1997 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Shear Stress Shear Capacity D/C ratio
Method
(N/mm2)
(N/mm2)
ETABS
1.119
0.660
1.70
Calculated
1.105
0.625
1.77
COMPUTER FILE: BS 8110-1997 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
BS 8110-1997 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
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ETABS
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 218 mm
Refer to Figure 2.
u = 954+ 1554 + 954 + 1554 = 5016 mm
Note: All dimensions in millimeters
954
Y
327
150 150
A
Column
Critical section for
punching shear shown
dashed.
327
B
Side 2
Side 3
Side 1
327
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1554
450
327
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
V = 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
BS 8110-1997 RC-PN EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Maximum design shear stress in computed in along major and minor axis of column:
veff , x
1.5M
V
x
f
ud
Vy
veff , x
1126.498 103
1.5 51.9908 106
1
.
0
1.1049 (Govern)
5016 218
1126.498 103 954
veff , y
1.5M
V
y
f
ud
Vx
veff , y
1126.498 103
1.5 45.7234 106
1.0
1.0705
5016 218
1126.498 103 1554
(BS 3.7.7.3)
The largest absolute value of v = 1.1049 N/mm2
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.79k1k 2 100 As 3 400 4
vc
= 0.3568 MPa
m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
1
f 3 30 3
k2 = cu = = 1.0627 > 1.0 OK
25
25
m = 1.25
400
d
1
4
= 1.16386 > 1 OK.
fcu 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Areas of reinforcement at the face of column for the design strips are as
follows:
As in Strip Layer A = 9494.296 mm2
As in Strip Layer B = 8314.486 mm2
BS 8110-1997 RC-PN EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Average As = (9494.296+8314.486)/2 = 8904.391 mm2
100 As
= 100 8904.391/(8000 218) = 0.51057
bd
vc
0.79 1.0 1.0627
1/ 3
0.51057 1.16386 = 0.6247 MPa
1.25
BS 3.7.7.3 yields the value of v = 0.625 N/mm2 , and thus this is the shear capacity.
Shear Ratio
vU
1.1049
1.77
v
0.6247
BS 8110-1997 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
BS 8110-1997 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the BS 8110-97
load combination factors, 1.4 for dead loads and 1.6 for live loads. The model is
analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design was performed using the BS 8110-97 code by ETABS and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
BS 8110-1997 RC-SL EXAMPLE 001 - 1
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ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
27.197
5.853
Calculated
27.200
5.850
As+
Medium
A s ,min = 162.5 sq-mm
COMPUTER FILE: BS 8110-1997 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
BS 8110-1997 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
m, steel
= 1.15
m, concrete = 1.50
b
= 1000 mm
For the load combination, w and M are calculated as follows:
w = (1.4wd + 1.6wt) b
M
wl12
8
As,min = 0.0013bwd
= 162.5 sq-mm
COMB100
wd =
4.0 kPa
wt
5.0 kPa
=
w = 13.6 kN/m
M-strip = 27.2 kN-m
M-design = 27.2366 kN-m
The depth of the compression block is given by:
K
M
= 0.05810 < 0.156
f cu bd 2
The area of tensile steel reinforcement is then given by:
K
0.95d =116.3283
z d 0.5 0.25
0
.
9
As
M
= 585.046 sq-mm > As,min
0.87 f y z
As = 5.850 sq-cm
BS 8110-1997 RC-SL EXAMPLE 001 - 3
Software Verification
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ETABS
0
CSA 23.3-04 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span
CSA 23.3-04 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads were added to the slab. The loads and posttensioning forces are as follows:
Loads:
Dead = self weight,
Live = 4.788 KN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the midspan of the slab. Independent hand calculations
have been compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L =
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
CSA 23.3-04 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), MPa
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
159.4
159.4
0.00%
16.25
16.33
0.49%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
7.817
7.818
0.01%
5.759
5.760
0.02%
COMPUTER FILE: CSA A23.3-04 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
CSA 23.3-04 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fcu = 30MPa
fy = 400MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
c 0.65 , S 0.85
1 = 0.85 – 0.0015f'c 0.67 = 0.805
1 = 0.97 – 0.0025f'c 0.67 = 0.895
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.25 = 7.480 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) x 1.50 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L)
= 14.662 kN/m2 (D+L)ult
=10.772 kN/m2 x 0.914m = 9.846 kN/m, u = 16.039 kN/m2 x 0.914m = 13.401 kN/m
Ultimate Moment, M U
CSA 23.3-04 PT-SL EXAMPLE 001 - 4
wl12
= 13.401 x (9.754)2/8 = 159.42 kN-m
8
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Ultimate Stress in strand, f pb f pe
cy
ETABS
0
8000
d p cy
lo
p Ap f pr s As f y 0.9 197 1347 0.85 1625 400
61.66 mm
0.805 0.65 30.0 0.895 914
1c f 'c 1b
f pb 1210
8000
229 61.66 1347 MPa
9754
Depth of the compression block, a, is given as:
Stress block depth, a d d 2
2M *
1 f 'c c b
0.229 0.2292
2 159.42
55.18
0.805 30000 0.65 0.914
Ultimate force in PT, Fult , PT AP ( f PS ) 197 1347 1000 265.9 kN
Ultimate moment due to PT,
a
55.18
M ult , PT Fult , PT d 265.9 0.229
0.85 45.52 kN-m
2
2
Net Moment to be resisted by As, M NET MU M PT
159.42 45.52 113.90 kN-m
The area of tensile steel reinforcement is then given by:
As
M NET
=
0.87 f y z
113.90
1e6 1625 mm 2
55.18
0.87 400 229
2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
CSA 23.3-04 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
Stress in concrete, f PTI D
0.254(0.914)
0.00983
A
S
where S = 0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
0.254 0.914
0.00983
A
S
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
CSA 23.3-04 PT-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA A23.3-04 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8 m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
CSA A23.3-04 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
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ETABS
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.793
1.127
1.59
Calculated
1.792
1.127
1.59
COMPUTER FILE: CSA A23.3-04 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
CSA A23.3-04 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
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ETABS
0
HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 218 mm
Refer to Figure 2.
b0 = 518+ 1118 + 1118 + 518 = 3272 mm
Note: All dimensions in millimeters
518
Y
109
150
150
109
A
Column
Critical section for
punching shear shown
dashed.
B
Side 2
109
Side 1
Side 3
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 1118
1
3 518
1
2 518
1
3 1118
0.495
0.312
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
CSA A23.3-04 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for punching
shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical section for
punching shear as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
259
0
1118
218
243724
63124516
0
x3
Ldx
y3
Ldy
2
Ld
Ld
2
Side 2
0
559
518
218
112924
0
63124516
0
0 mm
713296
0
0 mm
713296
Side 3
259
0
1118
218
243724
63124516
0
Side 4
0
559
518
218
112924
0
63124516
Sum
N.A.
N.A.
b0 = 3272
N.A.
713296
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the "Sum" column.
Item
L
d
x2 x3
y2 y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 2
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
From the ETABS output at Grid B-2:
Vf = 1126.498 kN
V2
Mf,2 = 25.725 kN-m
V3
Mf,3 = 14.272 kN-m
CSA A23.3-04 RC-PN EXAMPLE 001 - 4
Side 3
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 4
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
1.23E+11
3.86E+10
0
Software Verification
PROGRAM NAME:
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ETABS
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At the point labeled A in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 ( 259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 0.1169 0.0958 = 1.3666 N/mm2 at point A
At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus:
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
vf
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 (259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 0.1169 + 0.0958 =1.5582 N/mm2 at point B
At the point labeled C in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0)2
14.272 106 [1.23 1011 (259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 + 0.1169 + 0.0958 = 1.792 N/mm2 at point C
At the point labeled D in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 ( 259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 + 0.1169 0.0958 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
CSA A23.3-04 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The shear capacity is calculated based on the minimum of the following three limits:
2
c 1 0.19 f c
c
d
vv min c 0.19 s f c
b0
0.38 f
c
c
1.127 N/mm2 in accordance with CSA 13.3.4.1
CSA 13.3.4.1 yields the smallest value of vv = 1.127 N/mm2 , and thus this is the shear
capacity.
Shear Ratio
vU
1.792
1.59
1.127
vv
CSA A23.3-04 RC-PN EXAMPLE 001 - 6
Software Verification
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REVISION NO.:
ETABS
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CSA A23.3-04 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the CSA A23.304 load combination factors, 1.25 for dead loads and 1.5 for live loads. The
model is analyzed for these load cases and load combinations.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed using the CSA A23.3-04 code by ETABS and
also by hand computation. Table 1 show the comparison of the design
reinforcements computed using the two methods.
CSA A23.3-04 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
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ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip the moments obtained by the hand computation method. Table 1 also
shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
25.00
5.414
Calculated
25.00
5.528
As+
Medium
A s ,min = 357.2 sq-mm
CSA A23.3-04 RC-SL EXAMPLE 001 - 2
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ETABS
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COMPUTER FILE: CSA A23.3-04 RC-SL EX001.EDB
CONCLUSION
The ETABS results show a very close comparison with the independent results.
CSA A23.3-04 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
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ETABS
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HAND CALCULATION
The following quantities are computed for the load combination:
c = 0.65 for concrete
s = 0.85 for reinforcement
As,min =
0.2 f c
bw h = 357.2 sq-mm
fy
b = 1000 mm
1 = 0.85 – 0.0015f'c 0.67 = 0.805
1 = 0.97 – 0.0025f'c 0.67 = 0.895
cb =
700
d = 75.43 mm
700 f y
ab = 1cb = 67.5 mm
For the load combination, w and M* are calculated as follows:
w = (1.25wd + 1.5wt) b
wl12
Mu
8
As = min[As,min, (4/3) As,required] = min[357.2, (4/3)540.63] = 357.2 sq-mm
= 0.22•(150/125)2•0.6•SQRT(30)/460•100•125
= 282.9 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.5 kN/m
Mf-strip = 25.0 kN-m
Mf-design = 25.529 kN-m
The depth of the compression block is given by:
CSA A23.3-04 RC-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
a d d2
2M f
1 f 'c c b
ETABS
0
= 13.769 mm < amax
The area of tensile steel reinforcement is then given by:
As
Mf
a
s f y d
2
= 552.77 sq-mm > As,min
As = 5.528 sq-cm
CSA A23.3-04 RC-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
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ETABS
0
CSA A23.3-14 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
CSA A23.3-14 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads were added to the slab. The loads and posttensioning forces are as follows:
Loads:
Dead = self weight,
Live = 4.788 KN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the midspan of the slab. Independent hand calculations
have been compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L =
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
CSA A23.3-14 PT-SL EXAMPLE 001 - 2
Software Verification
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REVISION NO.:
ETABS
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RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), MPa
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
159.4
159.4
0.00%
16.25
16.33
0.49%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
7.817
7.818
0.01%
5.759
5.760
0.02%
COMPUTER FILE: CSA A23.3-14 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
CSA A23.3-14 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fcu = 30MPa
fy = 400MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
c 0.65 , S 0.85
1 = 0.85 – 0.0015f'c 0.67 = 0.805
1 = 0.97 – 0.0025f'c 0.67 = 0.895
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m x 23.56 kN/m3 = 5.984 kN/m2 (D) x 1.25 = 7.480 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) x 1.50 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L)
= 14.662 kN/m2 (D+L)ult
=10.772 kN/m2 x 0.914m = 9.846 kN/m, u = 16.039 kN/m2 x 0.914m = 13.401 kN/m
Ultimate Moment, M U
CSA A23.3-14 PT-SL EXAMPLE 001 - 4
wl12
= 13.401 x (9.754)2/8 = 159.42 kN-m
8
Software Verification
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Ultimate Stress in strand, f pb f pe
cy
ETABS
0
8000
d p cy
lo
p Ap f pr s As f y 0.9 197 1347 0.85 1625 400
61.66 mm
0.805 0.65 30.0 0.895 914
1c f 'c 1b
f pb 1210
8000
229 61.66 1347 MPa
9754
Depth of the compression block, a, is given as:
Stress block depth, a d d 2
2M *
1 f 'c c b
0.229 0.2292
2 159.42
55.18
0.805 30000 0.65 0.914
Ultimate force in PT, Fult , PT AP ( f PS ) 197 1347 1000 265.9 kN
Ultimate moment due to PT,
a
55.18
M ult , PT Fult , PT d 265.9 0.229
0.85 45.52 kN-m
2
2
Net Moment to be resisted by As, M NET MU M PT
159.42 45.52 113.90 kN-m
The area of tensile steel reinforcement is then given by:
As
M NET
=
0.87 f y z
113.90
1e6 1625 mm 2
55.18
0.87 400 229
2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
CSA A23.3-14 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
Stress in concrete, f PTI D
0.254(0.914)
0.00983
A
S
where S = 0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
0.254 0.914
0.00983
A
S
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
CSA A23.3-14 PT-SL EXAMPLE 001 - 6
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ETABS
0
CSA A23.3-14 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8 m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
CSA A23.3-14 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.793
1.127
1.59
Calculated
1.792
1.127
1.59
COMPUTER FILE: CSA A23.3-14 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
CSA A23.3-14 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
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ETABS
0
HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 218 mm
Refer to Figure 2.
b0 = 518+ 1118 + 1118 + 518 = 3272 mm
Note: All dimensions in millimeters
518
Y
109
150
150
109
A
Column
Critical section for
punching shear shown
dashed.
B
Side 2
109
Side 1
Side 3
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 1118
1
3 518
1
2 518
1
3 1118
0.495
0.312
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
CSA A23.3-14 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for punching
shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical section for
punching shear as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
259
0
1118
218
243724
63124516
0
x3
Ldx
y3
Ldy
2
Ld
Ld
2
Side 2
0
559
518
218
112924
0
63124516
0
0 mm
713296
0
0 mm
713296
Side 3
259
0
1118
218
243724
63124516
0
Side 4
0
559
518
218
112924
0
63124516
Sum
N.A.
N.A.
b0 = 3272
N.A.
713296
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the "Sum" column.
Item
L
d
x2 x3
y2 y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 2
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
From the ETABS output at Grid B-2:
Vf = 1126.498 kN
V2
Mf,2 = 25.725 kN-m
V3
Mf,3 = 14.272 kN-m
CSA A23.3-14 RC-PN EXAMPLE 001 - 4
Side 3
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 4
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
1.23E+11
3.86E+10
0
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
At the point labeled A in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 ( 259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 0.1169 0.0958 = 1.3666 N/mm2 at point A
At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus:
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
vf
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 (259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 0.1169 + 0.0958 =1.5582 N/mm2 at point B
At the point labeled C in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0)2
14.272 106 [1.23 1011 (259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 + 0.1169 + 0.0958 = 1.792 N/mm2 at point C
At the point labeled D in Figure 2, x4 = 259 and y4 = 559, thus:
vf
1126.498 103 25.725 106 [3.86 1010 (559 0) (0)(259 0)]
3272 218
(1.23 1011 )(3.86 1010 ) (0) 2
14.272 106 [1.23 1011 ( 259 0) (0)(559 0)]
(1.23 1011 )(3.86 1010 ) (0) 2
vf = 1.5793 + 0.1169 0.0958 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
CSA A23.3-14 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
The shear capacity is calculated based on the minimum of the following three limits:
2
c 1 0.19 f c
c
d
vv min c 0.19 s f c
b0
0.38 f
c
c
1.127 N/mm2 in accordance with CSA 13.3.4.1
CSA 13.3.4.1 yields the smallest value of vv = 1.127 N/mm2 , and thus this is the shear
capacity.
Shear Ratio
vU
1.792
1.59
1.127
vv
CSA A23.3-14 RC-PN EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
CSA A23.3-14 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the CSA A23.314 load combination factors, 1.25 for dead loads and 1.5 for live loads. The
model is analyzed for these load cases and load combinations.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed using the CSA A23.3-14 code by ETABS and
also by hand computation. Table 1 show the comparison of the design
reinforcements computed using the two methods.
CSA A23.3-14 RC-SL EXAMPLE 001 - 1
Software Verification
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REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip the moments obtained by the hand computation method. Table 1 also
shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
25.00
5.414
Calculated
25.00
5.528
As+
Medium
A s ,min = 357.2 sq-mm
CSA A23.3-14 RC-SL EXAMPLE 001 - 2
Software Verification
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ETABS
0
COMPUTER FILE: CSA A23.3-14 RC-SL EX001.EDB
CONCLUSION
The ETABS results show a very close comparison with the independent results.
CSA A23.3-14 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
c = 0.65 for concrete
s = 0.85 for reinforcement
As,min =
0.2 f c
bw h = 357.2 sq-mm
fy
b = 1000 mm
1 = 0.85 – 0.0015f'c 0.67 = 0.805
1 = 0.97 – 0.0025f'c 0.67 = 0.895
cb =
700
d = 75.43 mm
700 f y
ab = 1cb = 67.5 mm
For the load combination, w and M* are calculated as follows:
w = (1.25wd + 1.5wt) b
wl12
Mu
8
As = min[As,min, (4/3) As,required] = min[357.2, (4/3)540.63] = 357.2 sq-mm
= 0.22•(150/125)2•0.6•SQRT(30)/460•100•125
= 282.9 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.5 kN/m
Mf-strip = 25.0 kN-m
Mf-design = 25.529 kN-m
The depth of the compression block is given by:
CSA A23.3-14 RC-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
a d d2
2M f
1 f 'c c b
ETABS
0
= 13.769 mm < amax
The area of tensile steel reinforcement is then given by:
As
Mf
a
s f y d
2
= 552.77 sq-mm > As,min
As = 5.528 sq-cm
CSA A23.3-14 RC-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EN 2-2004 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
EN 2-2004 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, was added to the AStrip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows:
Loads: Dead = self weight,
Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations were compared with the ETABS results and summarized for
verification and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with independent hand calculations.
EN 2-2004 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 1 Comparison of Results
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
166.41
166.44
0.02%
Transfer Conc. Stress, top
(D+PTI), MPa
5.057
5.057
0.00%
Transfer Conc. Stress, bot
(D+PTI), MPa
2.839
2.839
0.00%
Normal Conc. Stress, top
(D+L+PTF), MPa
10.460
10.467
0.07%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.409
0.08%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
7.817
7.818
0.01%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.760
0.02%
FEATURE TESTED
Table 2 Comparison of Design Moments and Reinforcements
Reinforcement Area
(sq-cm)
National Annex
CEN Default, Norway,
Slovenia and Sweden
Method
Design Moment
(kN-m)
As+
ETABS
166.44
15.39
Calculated
166.41
15.36
ETABS
166.44
15.90
Calculated
166.41
15.87
ETABS
166.44
15.96
Calculated
166.41
15.94
Finland , Singapore and UK
Denmark
COMPUTER FILE: EN 2-2004 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EN 2-2004 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
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ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, steel = 1.15
m, concrete = 1.50
1.0 for fck ≤ 50 MPa
0.8 for fck ≤ 50 MPa
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.35 = 8.078 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.50 = 7.182 kN/m2 (Lu)
= 15.260 kN/m2 (D+L)ult
Total = 10.772 kN/m2 (D+L)
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 15.260 kN/m2 0.914 m = 13.948 kN/m
wl12
2
Ultimate Moment, M U
= 13.948 9.754 8 = 165.9 kN-m
8
EN 2-2004 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
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ETABS
0
f A
Ultimate Stress in strand, f PS f SE 7000d 1 1.36 PU P l
fCK bd
1862(198)
1210 7000(229) 1 1.36
9754
30(914) 229
1361 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 2 99 1361 1000 269.5 kN
CEN Default, Norway, Slovenia and Sweden:
Design moment M = 166.4122 kN-m
M
Compression block depth ratio: m 2
bd f cd
166.4122
0.1736
0.914 0.229 2 1 30000 1.50
Required area of mild steel reinforcing,
1 1 2m = 1 1 2(0.1736) 0.1920
f bd
1(30 /1.5)(914)(229)
2
AEquivTotal cd 0.1920
2311 mm
f yd
400 /1.15
1361
2
AEquivTotal AP
AS 2311 mm
400
1.15
1361
2
AS 2311 198
1536 mm
400 /1.15
Finland, Singapore and UK:
Design moment M = 166.4122 kN-m
Compression block depth ratio: m
M
bd 2f cd
166.4122
0.2042
0.914 0.229 2 0.85 30000 1.50
Required area of mild steel reinforcing,
1 1 2m = 1 1 2(0.2042) 0.23088
EN 2-2004 PT-SL EXAMPLE 001 - 5
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
f bd
0.85(30 /1.5)(914)(229)
2
AEquivTotal cd 0.23088
2362 mm
f yd
400 /1.15
1361
2
AEquivTotal AP
AS 2362 mm
400 1.15
1361
2
AS 2362 198
1587 mm
400 1.15
Denmark:
Design moment M = 166.4122 kN-m
Compression block depth ratio: m
M
bd 2f cd
166.4122
0.1678
0.914 0.229 2 1.0 30000 1.45
Required area of mild steel reinforcing,
1 1 2m = 1 1 2(0.1678) 0.1849
f bd
1.0(30 /1.45)(914)(229)
2
AEquivTotal cd 0.1849
2402 mm
f yd
400 /1.20
1361
2
AEquivTotal AP
AS 2402 mm
400
1.2
1361
2
AS 2402 198
1594 mm
400
1.2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
Stress in concrete, f PTI D
0.254 0.914
0.00983
A
S
Moment due to PT,
EN 2-2004 PT-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
where S = 0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term=1490 186 94 = 1210 MPa
The force in tendon at normal = 1210 197.4 1000 238.9 kN
Moment due to dead load M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
A
S
0.254 0.914
0.00983
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
EN 2-2004 PT-SL EXAMPLE 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
EN 2-2004 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. Thick shell properties are used for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
EN 2-2004 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
National Annex
CEN Default, Norway,
Slovenia and Sweden
Finland, Singapore and UK
Denmark
Method
Shear
Stress
(N/mm2)
Shear
Capacity
(N/mm2)
D/C
ratio
ETABS
1.107
0.610
1.82
Calculated
1.089
0.578
1.89
ETABS
1.107
0.612
1.81
Calculated
1.089
0.5796
1.88
ETABS
1.107
0.639
1.73
Calculated
1.089
0.606
1.80
COMPUTER FILE: EN 2-2004 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EN 2-2004 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
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ETABS
0
HAND CALCULATION
Hand Calculation for Interior Column using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 2.
u1 = u = 2300 + 2900 + 2436 = 5139.468 mm
1172
Note: All dimensions in millimeters
Critical section for
punching shear
shown dashed.
Y
436
150 150 436
B
A
Column
436
Side 3
Side 1
Side 2
450
X
1772
450
Center of column is
point (x1, y1). Set
this equal to (0,0).
436
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
VEd = 1112.197 kN
k2MEd2 = 38.933 kN-m
k3MEd3 = 17.633 kN-m
EN 2-2004 RC-PN EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Maximum design shear stress in computed in along major and minor axis of column:
vEd
W1
VEd
ud
k2 M Ed ,2u1 k3 M Ed ,3u1
1
VEdW1,2
VEdW1,3
(EC2 6.4.4(2))
c12
c1c2 4c2 d 16d 2 2 dc1
2
W1,2
9002
300 900 4 300 218 16 2182 2 218 900
2
W1,2 2,929, 744.957 mm2
W1,3 3
9002
900 300 4 900 218 16 2182 2 218 300
2
W1,2 2, 271,104.319 mm2
vEd
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
1
ud
VEdW1,2
VEdW1,3
vEd
1112.197 103
38.933 106 5139.468
17.633 106 5139.468
1
5139.468 218 1112.197 103 2929744.957 1112.197 103 2271104.319
vEd 1.089 N/mm2
Thus vmax = 1.089 N/mm2
For CEN Default, Finland, Norway, Singapore, Slovenia, Sweden and UK:
C Rd ,c 0.18 c = 0.18/1.5 = 0.12
(EC2 6.4.4)
For Denmark:
CRd ,c 0.18 c = 0.18/1.45 = 0.124
(EC2 6.4.4)
The shear stress carried by the concrete, VRd,c, is calculated as:
13
VRd ,c C Rd ,c k 100 1 fck k1 cp
(EC2 6.4.4)
with a minimum of:
vRd ,c vmin k1 cp
EN 2-2004 RC-PN EXAMPLE 001 - 4
(EC2 6.4.4)
Software Verification
PROGRAM NAME:
REVISION NO.:
k 1
200
2.0 = 1.9578
d
k1
= 0.15.
1 =
As1
0.02
bw d
ETABS
0
(EC2 6.4.4(1))
(EC2 6.2.2(1))
Area of reinforcement at the face of column for design strip are as follows:
For CEN Default, Norway, Slovenia and Sweden:
As in Strip Layer A = 9204.985 mm2
As in Strip Layer B = 8078.337 mm2
Average As = 9204.985 8078.337 2 = 8641.661 mm2
1 = 8641.661 8000 218 = 0.004955 0.02
For Finland, Singapore and UK:
As in Strip Layer A = 9319.248 mm2
As in Strip Layer B = 8174.104 mm2
Average As = 9319.248 8174.104 2 = 8746.676 mm2
1 = 8746.676 8000 218 = 0.005015 0.02
For Denmark:
As in Strip Layer A = 9606.651 mm2
As in Strip Layer B = 8434.444 mm2
Average As = 9606.651 8434.444 2 = 9020.548 mm2
1 = 9020.548 8000 218 = 0.005172 0.02
EN 2-2004 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
For CEN Default, Denmark, Norway, Singapore, Slovenia, Sweden and UK:
min 0.035k 3 2 f ck 1 2 = 0.035 1.9578
30
3/2
1/2
= 0.525 N/mm2
For Finland:
min 0.035k 2/3 f ck 1 2 = 0.035 1.9578
30
2/3
1/2
= 0.3000 N/mm2
For CEN Default, Norway, Slovenia and Sweden:
vRd ,c 0.12 1.9578 100 0.004955 30
13
2
0 = 0.5777 N/mm
For Finland, Singapore, and UK:
13
2
vRd ,c 0.12 1.9578 100 0.005015 30 0 = 0.5796 N/mm
For Denmark:
vRd ,c 0.124 1.9578 100 0.005015 30
13
2
0 = 0.606 N/mm
For CEN Default, Norway, Slovenia and Sweden:
Shear Ratio
v max
1.089
1.89
vRd ,c
0.5777
For Finland, Singapore and UK:
Shear Ratio
v max
1.089
1.88
vRd ,c
0.5796
For Denmark:
Shear Ratio
EN 2-2004 RC-PN EXAMPLE 001 - 6
v max
1.089
1.80
vRd ,c
0.606
Software Verification
PROGRAM NAME:
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ETABS
0
EN 2-2004 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
4 m span
Simply
supported
edge at wall
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Eurocode 204 load combination factors, 1.35 for dead loads and 1.5 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. These moments
are identical. After completing the analysis, design is performed using the
Eurocode 2-04 code by ETABS and also by hand computation. Table 1 shows
the comparison of the design reinforcements computed by the two methods.
EN 2-2004 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fck
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
EN 2-2004 RC-SL EXAMPLE 001 - 2
kPa
kPa
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip the moments obtained by the hand computation method. Table 1 also
shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Reinforcement Area
(sq-cm)
National Annex
CEN Default, Norway,
Slovenia and Sweden
Finland , Singapore and
UK
Method
Strip Moment
(kN-m)
As+
ETABS
25.797
5.400
Calculated
25.800
5.400
ETABS
25.797
5.446
Calculated
25.800
5.446
ETABS
25.797
5.626
Calculated
25.800
5.626
Denmark
A s ,min = 204.642 sq-mm
COMPUTER FILE: EN 2-2004 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
EN 2-2004 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
1.0 for fck ≤ 50 MPa
0.8 for fck ≤ 50 MPa
b = 1000 mm
For the load combination, w and M are calculated as follows:
w = (1.35wd + 1.5wt) b
wl12
M
8
As ,min
0.0013bw d
max
fctm
0
.
26
bd
f
yk
= 204.642 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.9 kN/m
M-strip = 25.8 kN-m
M-design= 25.8347 kN-m
For CEN Default, Norway, Slovenia and Sweden:
m, steel = 1.15
m, concrete = 1.50
αcc = 1.0
The depth of the compression block is given by:
m
M
bd 2 f cd
EN 2-2004 RC-SL EXAMPLE 001 - 4
25.8347 106
= 0.08267
1000 1252 1.0 1.0 30 /1.5
Software Verification
PROGRAM NAME:
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ETABS
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For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.44
k2 = k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1
k1
x
for fck 50 MPa = 0.448
k2
d lim
x x
mlim 1 = 0.294
d lim 2 d lim
1 1 2m = 0.08640
f bd
As cd = 540.024 sq-mm > As,min
f
yd
As = 5.400 sq-cm
For Singapore and UK:
m, steel = 1.15
m, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
m
M
bd 2 f cd
25.8347 106
= 0.097260
1000 1252 1.0 0.85 30 /1.5
x x
mlim 1 = 0.48
d lim 2 d lim
k1
x
for fck 50 MPa = 0.60
k2
d lim
For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.40
k2 = (0.6 + 0.0014/εcu2) = 1.00
EN 2-2004 RC-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
is assumed to be 1
1 1 2m = 0.10251
f bd
As cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
For Finland:
m, steel = 1.15
m, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
m
M
bd 2 f cd
25.8347 106
= 0.097260
1000 1252 1.0 0.85 30 /1.5
x x
mlim 1 = 032433
d lim 2 d lim
k1
x
for fck 50 MPa = 0.5091
k2
d lim
For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.44
k2 = 1.1
k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1
1 1 2m = 0.10251
f bd
As cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
EN 2-2004 RC-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
For Denmark:
m, steel = 1.20
m, concrete = 1.45
αcc = 1.0
The depth of the compression block is given by:
m
M
bd 2 f cd
25.8347 106
= 0.0799153
1000 1252 1.0 1.0 30 /1.5
x x
mlim 1 = 0.294
d lim 2 d lim
k1
x
for fck 50 MPa = 0.448
k2
d lim
For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.44
k2 = k4 = 1.25(0.6 + 0.0014/εcu2) = 1.25
is assumed to be 1
1 1 2m = 0.08339
f bd
As cd = 562.62 sq-mm > As,min
f
yd
As = 5.626 sq-cm
EN 2-2004 RC-SL EXAMPLE 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HK CP-2004 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
HK CP-2004 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
To ensure one-way action Poisson’s ratio is taken to be zero. A 254-mm-wide
design strip is centered along the length of the slab and has been defined as an
A-Strip. B-strips have been placed at each end of the span, perpendicular to StripA (the B-Strips are necessary to define the tendon profile). A tendon with two
strands, each having an area of 99 mm2, was added to the A-Strip. The self weight
and live loads were added to the slab. The loads and post-tensioning forces are as
follows:
Loads:
Live = 4.788 kN/m2
Dead = self weight,
The total factored strip moments, required area of mild steel reinforcement and
slab stresses are reported at the midspan of the slab. Independent hand calculations
were compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with independent hand calculations.
HK CP-2004 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (KN-m)
Area of Mild Steel req’d,
As (cm2)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.80
0.41%
5.056
5.057
0.02%
2.836
2.839
0.11%
10.547
10.467
-0.76%
8.323
8.409
1.03%
COMPUTER FILE: HK CP-2004 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
HK CP-2004 PT-SL EXAMPLE 001 - 3
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ETABS
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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fc = 30 MPa
fy = 400 MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, steel = 1.15
m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 16.039 kN/m2 0.914 m = 14.659 kN/m
Ultimate Moment, M U
HK CP-2004 PT-SL EXAMPLE 001 - 4
wl12
2
= 14.659 9.754 8 = 174.4 kN-m
8
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ETABS
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f pu Ap
7000
1 1.7
l/d
f cu bd
7000
1862(198)
1210
1 1.7
9.754 / 0.229
30(914)(229)
Ultimate Stress in strand, f pb f pe
1358 MPa 0.7 f pu 1303 MPa
K factor used to determine the effective depth is given as:
174.4
M
K
0.1213 < 0.156
=
2
30000(0.914)(0.229)2
f cu bd
K
0.95d = 192.2 mm
z d 0.5 0.25
0
.
9
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1303 1000 257.2 KN
Ultimate moment due to PT, M ult , PT Fult , PT ( z ) / 257.2 0.192 1.15 43.00 KN-m
Net Moment to be resisted by As, M NET MU M PT
174.4 43.00 131.40 kN-m
The area of tensile steel reinforcement is then given by:
As
131.40
M
1e6 1965mm 2
=
0.87 f y z
0.87 400 192
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 2 99 1000 258.2 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 258.2 101.6 mm 1000 26.23 kN-m
F
M
M
258.2
65.04
26.23
Stress in concrete, f PTI D PT
0.254 0.914 0.00983 0.00983
A
S
S
where S = 0.00983 m3
f 1.112 6.6166 2.668 MPa
f 5.060(Comp) max, 2.836(Tension) max
HK CP-2004 PT-SL EXAMPLE 001 - 5
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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 2 99 1000 239.5 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 239.5 101.6 mm 1000 24.33 kN-m
Stress in concrete for (D+L+PTF),
F
M
M
258.2
117.08
24.33
f PTI D PT
A
S
S
0.254 0.914 0.00983 0.00983
f 1.112 11.910 2.475
f 10.547(Comp) max, 8.323(Tension) max
HK CP-2004 PT-SL EXAMPLE 001 - 6
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ETABS
0
HK CP-2004 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a fcu of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
HK CP-2004 RC-PN EXAMPLE 001 - 1
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ETABS
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.116
0.662
1.69
Calculated
1.105
0.625
1.77
COMPUTER FILE: HK CP-2004 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
HK CP-2004 RC-PN EXAMPLE 001 - 2
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ETABS
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 1.
u = 954+ 1554 + 954 + 1554 = 5016 mm
Note: All dimensions in millimeters
954
Y
Critical section for
punching shear shown
dashed.
327 150 150 327
A
Column
B
Side 2
Side 3
Side 1
327
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1554
450
327
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
HK CP-2004 RC-PN EXAMPLE 001 - 3
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Maximum design shear stress in computed in along major and minor axis of column:
veff , x
1.5M
V
x
f
ud
Vy
veff , x
1126.498 103
1.5 51.9908 106
1.0
1.1049 (Govern)
5016 218
1126.498 103 954
veff , y
1.5M
V
y
f
ud
Vx
veff , y
1126.498 103
1.5 45.7234 106
1.0
1.0705
5016 218
1126.498 103 1554
The largest absolute value of v = 1.1049 N/mm2
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.79k1k 2 100 As 3 400 4
vc
= 0.3568 MPa
m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
1
f 3 30 3
k2 = cu = = 1.0627 > 1.0 OK
25
25
m = 1.25
400
d
1
4
= 1.16386 > 1 OK.
fcu 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Area of reinforcement at the face of column for design strip are as follows:
As in Strip Layer A = 9494.296 mm2
As in Strip Layer B = 8314.486 mm2
HK CP-2004 RC-PN EXAMPLE 001 - 4
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Average As = 9494.296 8314.486 2 = 8904.391 mm2
100 As
= 100 8904.391 8000 218 = 0.51057
bd
vc
0.79 1.0 1.0627
1/ 3
0.51057 1.16386 = 0.6247 MPa
1.25
BS 3.7.7.3 yields the value of v = 0.625 N/mm2 , and thus this is the shear capacity.
Shear Ratio
vU
1.1049
1.77
v
0.6247
HK CP-2004 RC-PN EXAMPLE 001 - 5
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ETABS
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HK CP-2004 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Hong Kong
CP-04 load combination factors, 1.4 for dead loads and 1.6 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design is performed using the Hong Kong CP-04 code by ETABS and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
HK CP-2004 RC-SL EXAMPLE 001 - 1
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ETABS
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement
Area (sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
27.197
5.853
Calculated
27.200
5.842
As+
Medium
A s ,min = 162.5 sq-mm
HK CP-2004 RC-SL EXAMPLE 001 - 2
Software Verification
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ETABS
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COMPUTER FILE: HK CP-2004 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
HK CP-2004 RC-SL EXAMPLE 001 - 3
Software Verification
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ETABS
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HAND CALCULATION
The following quantities are computed for the load combination:
m, steel
= 1.15
m, concrete = 1.50
b
= 1000 mm
For the load combination, the w and M are calculated as follows:
w = (1.4wd + 1.6wt) b
M
wl12
8
As,min = 0.0013bwd
= 162.5 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 13.6 kN/m
M-strip = 27.2 kN-m
M-design = 27.2366 kN-m
The depth of the compression block is given by:
K
M
= 0.05810 < 0.156
f cu bd 2
The area of tensile steel reinforcement is then given by:
K
0.95d =116.3283
z d 0.5 0.25
0
.
9
As
M
= 585.046 sq-mm > As,min
0.87 f y z
As = 5.850 sq-cm
HK CP-2004 RC-SL EXAMPLE 001 - 4
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ETABS
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HK CP-2013 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
HK CP-2013 PT-SL EXAMPLE 001 - 1
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ETABS
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To ensure one-way action Poisson’s ratio is taken to be zero. A 254-mm-wide
design strip is centered along the length of the slab and has been defined as an
A-Strip. B-strips have been placed at each end of the span, perpendicular to StripA (the B-Strips are necessary to define the tendon profile). A tendon with two
strands, each having an area of 99 mm2, was added to the A-Strip. The self weight
and live loads were added to the slab. The loads and post-tensioning forces are as
follows:
Loads:
Live = 4.788 kN/m2
Dead = self weight,
The total factored strip moments, required area of mild steel reinforcement and
slab stresses are reported at the midspan of the slab. Independent hand calculations
were compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with independent hand calculations.
HK CP-2013 PT-SL EXAMPLE 001 - 2
Software Verification
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ETABS
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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (KN-m)
Area of Mild Steel req’d,
As (cm2)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.80
0.41%
5.056
5.057
0.02%
2.836
2.839
0.11%
10.547
10.467
-0.76%
8.323
8.409
1.03%
COMPUTER FILE: HK CP-2013 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
HK CP-2013 PT-SL EXAMPLE 001 - 3
Software Verification
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ETABS
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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fc = 30 MPa
fy = 400 MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, steel = 1.15
m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 16.039 kN/m2 0.914 m = 14.659 kN/m
Ultimate Moment, M U
HK CP-2013 PT-SL EXAMPLE 001 - 4
wl12
2
= 14.659 9.754 8 = 174.4 kN-m
8
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ETABS
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f pu Ap
7000
1 1.7
l/d
f cu bd
7000
1862(198)
1210
1 1.7
9.754 / 0.229
30(914)(229)
Ultimate Stress in strand, f pb f pe
1358 MPa 0.7 f pu 1303 MPa
K factor used to determine the effective depth is given as:
174.4
M
K
0.1213 < 0.156
=
2
30000(0.914)(0.229)2
f cu bd
K
0.95d = 192.2 mm
z d 0.5 0.25
0
.
9
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1303 1000 257.2 KN
Ultimate moment due to PT, M ult , PT Fult , PT ( z ) / 257.2 0.192 1.15 43.00 KN-m
Net Moment to be resisted by As, M NET MU M PT
174.4 43.00 131.40 kN-m
The area of tensile steel reinforcement is then given by:
As
131.40
M
1e6 1965mm 2
=
0.87 f y z
0.87 400 192
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 2 99 1000 258.2 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 258.2 101.6 mm 1000 26.23 kN-m
F
M
M
258.2
65.04
26.23
Stress in concrete, f PTI D PT
0.254 0.914 0.00983 0.00983
A
S
S
where S = 0.00983 m3
f 1.112 6.6166 2.668 MPa
f 5.060(Comp) max, 2.836(Tension) max
HK CP-2013 PT-SL EXAMPLE 001 - 5
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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 2 99 1000 239.5 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT, M PT FPTI (sag) 239.5 101.6 mm 1000 24.33 kN-m
Stress in concrete for (D+L+PTF),
F
M
M
258.2
117.08
24.33
f PTI D PT
A
S
S
0.254 0.914 0.00983 0.00983
f 1.112 11.910 2.475
f 10.547(Comp) max, 8.323(Tension) max
HK CP-2013 PT-SL EXAMPLE 001 - 6
Software Verification
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ETABS
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HK CP-2013 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a fcu of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
HK CP-2013 RC-PN EXAMPLE 001 - 1
Software Verification
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REVISION NO.:
ETABS
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.116
0.662
1.69
Calculated
1.105
0.625
1.77
COMPUTER FILE: HK CP-2013 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
HK CP-2013 RC-PN EXAMPLE 001 - 2
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ETABS
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 1.
u = 954+ 1554 + 954 + 1554 = 5016 mm
Note: All dimensions in millimeters
954
Y
Critical section for
punching shear shown
dashed.
327 150 150 327
A
Column
B
Side 2
Side 3
Side 1
327
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1554
450
327
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
HK CP-2013 RC-PN EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Maximum design shear stress in computed in along major and minor axis of column:
veff , x
1.5M
V
x
f
ud
Vy
veff , x
1126.498 103
1.5 51.9908 106
1.0
1.1049 (Govern)
5016 218
1126.498 103 954
veff , y
1.5M
V
y
f
ud
Vx
veff , y
1126.498 103
1.5 45.7234 106
1.0
1.0705
5016 218
1126.498 103 1554
The largest absolute value of v = 1.1049 N/mm2
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.79k1k 2 100 As 3 400 4
vc
= 0.3568 MPa
m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
1
f 3 30 3
k2 = cu = = 1.0627 > 1.0 OK
25
25
m = 1.25
400
d
1
4
= 1.16386 > 1 OK.
fcu 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Area of reinforcement at the face of column for design strip are as follows:
As in Strip Layer A = 9494.296 mm2
As in Strip Layer B = 8314.486 mm2
HK CP-2013 RC-PN EXAMPLE 001 - 4
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Average As = 9494.296 8314.486 2 = 8904.391 mm2
100 As
= 100 8904.391 8000 218 = 0.51057
bd
vc
0.79 1.0 1.0627
1/ 3
0.51057 1.16386 = 0.6247 MPa
1.25
BS 3.7.7.3 yields the value of v = 0.625 N/mm2 , and thus this is the shear capacity.
Shear Ratio
vU
1.1049
1.77
v
0.6247
HK CP-2013 RC-PN EXAMPLE 001 - 5
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ETABS
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HK CP-2013 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Hong Kong
CP-04 load combination factors, 1.4 for dead loads and 1.6 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design is performed using the Hong Kong CP-04 code by ETABS and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
HK CP-2013 RC-SL EXAMPLE 001 - 1
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement
Area (sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
27.197
5.853
Calculated
27.200
5.842
As+
Medium
A s ,min = 162.5 sq-mm
HK CP-2013 RC-SL EXAMPLE 001 - 2
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COMPUTER FILE: HK CP-2013 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
HK CP-2013 RC-SL EXAMPLE 001 - 3
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HAND CALCULATION
The following quantities are computed for the load combination:
m, steel
= 1.15
m, concrete = 1.50
b
= 1000 mm
For the load combination, the w and M are calculated as follows:
w = (1.4wd + 1.6wt) b
M
wl12
8
As,min = 0.0013bwd
= 162.5 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 13.6 kN/m
M-strip = 27.2 kN-m
M-design = 27.2366 kN-m
The depth of the compression block is given by:
K
M
= 0.05810 < 0.156
f cu bd 2
The area of tensile steel reinforcement is then given by:
K
0.95d =116.3283
z d 0.5 0.25
0
.
9
As
M
= 585.046 sq-mm > As,min
0.87 f y z
As = 5.850 sq-cm
HK CP-2013 RC-SL EXAMPLE 001 - 4
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IS 456-2000 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
IS 456-2000 PT-SL EXAMPLE 001 - 1
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A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows:
Loads:
Dead = self weight,
Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations were compared with the ETABS results and summarized for
verification and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L =
254 mm
229 mm
9754 mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
30
=
400
=
1862
=
1210
=
198
=
23.56
= 25000
= 200,000
=
0
Dead load
Live load
wd
wl
=
=
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
self kN/m2
4.788 kN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
IS 456-2000 PT-SL EXAMPLE 001 - 2
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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
175.60
175.69
0.05%
19.53
19.775
1.25%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
COMPUTER FILE: IS 456-2000 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
IS 456-2000 PT-SL EXAMPLE 001 - 3
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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fck = 30MPa
fy = 400MPa
s = 1.15
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe =1210 MPa
c = 1.50
= 0.36
= 0.42
f 250
xmax
0.53 0.05 y
d
165
250 f y 415 MPa
if
xu ,max
0.484
d
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.50 = 8.976 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.50 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L)
= 16.158 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 16.158 kN/m2 0.914 m = 14.768 kN/m
Ultimate Moment, M U
IS 456-2000 PT-SL EXAMPLE 001 - 4
wl12
2
= 14.768 9.754 8 = 175.6 kN-m
8
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Ultimate Stress in strand, f PS from Table 11: fp = 1435 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 197.4 1435 1000 283.3 kN
Compression block depth ratio: m
M
bd f ck
2
175.6
0.3392
0.914 0.229 2 0.36 30000
Required area of mild steel reinforcing,
x
xu 1 1 4 m 1 1 4 0.42 0.3392
0.4094 > u ,max 0.484
d
d
2
2 0.42
The area of tensile steel reinforcement is then given by:
x
z d 1 u 229 1 0.42 0.4094 189.6 mm
d
ANET
Mu
175.6
1e6 2663 mm 2
f y / s z 400 1.15189.6
f
1435
2
As = ANET AP P 2663 198
1953 mm
fy
400
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT,
Stress in concrete,
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
f PTI D
A
S
0.254 0.914
0.00983
where S=0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
IS 456-2000 PT-SL EXAMPLE 001 - 5
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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term=1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
IS 456-2000 PT-SL EXAMPLE 001 - 6
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IS 456-2000 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
IS 456-2000 RC-PN EXAMPLE 001 - 1
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained in ETABS with the punching shear capacity, shear stress
ratio and D/C ratio obtained by the analytical method. They match exactly for
this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.793
1.141
1.57
Calculated
1.792
1.141
1.57
COMPUTER FILE: IS 456-2000 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
IS 456-2000 RC-PN EXAMPLE 001 - 2
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 1.
b0 = 518+ 1118 + 1118 + 518 = 3272 mm
Note: All dimensions in millimeters
518
Y
109 150
150
A
Column
Critical section for
punching shear shown
dashed.
109
B
109
Side 3
Side 1
Side 2
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V2
1
V3
1
1
2 1118
1
3 518
1
2 518
1
3 1118
0.495
0.312
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
IS 456-2000 RC-PN EXAMPLE 001 - 3
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ETABS
0
PROGRAM NAME:
REVISION NO.:
The following table is used for calculating the centroid of the critical section for punching
shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical section for
punching shear as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
259
0
1118
218
243724
63124516
0
x3
Ldx
y3
Ldy
2
Ld
Side 2
0
559
518
218
112924
0
63124516
0
0 mm
713296
0
0 mm
713296
2
Ld
Side 3
259
0
1118
218
243724
63124516
0
Side 4
0
559
518
218
112924
0
63124516
Sum
N.A.
N.A.
b0 = 3272
N.A.
713296
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the "Sum" column.
Item
L
d
x2 x3
y2 y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 2
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
From the ETABS output at Grid B-2:
VU = 1126.498 kN
V2
MU2 = 25.725 kN-m
V3
MU3 = 14.272 kN-m
IS 456-2000 RC-PN EXAMPLE 001 - 4
Side 3
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 4
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
1.23E+11
3.86E+10
0
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At the point labeled A in Figure 2, x4 = 259 and y4 = 559, thus:
6
10
1126.498 103 25.725 10 3.86 10 559 0 0 259 0
vU
1.23 1011 3.86 1010 0 2
3272 218
14.272 106 1.23 1011 259 0 0 559 0
1.23 1011 3.86 1010 0 2
vU = 1.5793 0.1169 0.0958 = 1.3666 N/mm2 at point A
At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus:
vU
6
10
1126.498 103 25.725 10 3.86 10 559 0 0 259 0
1.23 1011 3.86 1010 0 2
3272 218
14.272 106 1.23 1011 259 0 0 559 0
1.23 1011 3.86 1010 0 2
vU = 1.5793 0.1169 + 0.0958 =1.5582 N/mm2 at point B
At the point labeled C in Figure 2, x4 = 259 and y4 = 559, thus:
vU
6
10
1126.498 103 25.725 10 3.86 10 559 0 0 259 0
1.23 1011 3.86 1010 0 2
3272 218
14.272 106 1.23 1011 259 0 0 559 0
1.23 1011 3.86 1010 0 2
vU = 1.5793 + 0.1169 + 0.0958 = 1.792 N/mm2 at point C
At the point labeled D in Figure 2, x4 = 259 and y4 = 559, thus:
6
10
1126.498 103 25.725 10 3.86 10 559 0 0 259 0
vU
1.23 1011 3.86 1010 0 2
3272 218
14.272 106 1.23 1011 259 0 0 559 0
1.23 1011 3.86 1010 0 2
vU = 1.5793 + 0.1169 0.0958 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
IS 456-2000 RC-PN EXAMPLE 001 - 5
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The shear capacity is calculated based on the minimum of the following three limits:
ks = 0.5 + c 1.0 = 0.833
(IS 31.6.3.1)
c = 0.25 = 1.127 N/mm
(IS 31.6.3.1)
vc = ks c= 1.141 N/mm2
(IS 31.6.3.1)
2
CSA 13.3.4.1 yields the smallest value of vc = 1.141 N/mm2, and thus this is the shear
capacity.
Shear Ratio
IS 456-2000 RC-PN EXAMPLE 001 - 6
vU
1.792
1.57
vc
1.141
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IS 456-2000 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
Simply
supported
edge at wall
4 m span
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the IS 456-00
load combination factors, 1.5 for dead loads and 1.5 for live loads. The model is
analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design was performed using the IS 456-00 code by ETABS and also by
hand computation. Table 1 shows the comparison of the design reinforcements
computed using the two methods.
IS 456-2000 RC-SL EXAMPLE 001 - 1
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ETABS
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Method
Strip
Moment
(kN-m)
ETABS
Calculated
Reinforcement Area (sq-cm)
As+
As-
26.997
5.830
--
27.000
5.830
--
Medium
A s ,min = 230.978 sq-mm
IS 456-2000 RC-SL EXAMPLE 001 - 2
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COMPUTER FILE: IS 456-2000 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
IS 456-2000 RC-SL EXAMPLE 001 - 3
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HAND CALCULATION
The following quantities are computed for the load combination:
s = 1.15
c = 1.50
= 0.36
= 0.42
b = 1000 mm
For the load combination, w and M are calculated as follows:
w = (1.5wd + 1.5wt) b
wl12
M
8
As ,min
0.85
bd
fy
= 230.978 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 13.5 kN/m
M-strip = 27.0 kN-m
M-design = 27.0363 kN-m
xu ,max
d
IS 456-2000 RC-SL EXAMPLE 001 - 4
0.53
0.53 0.05 f y 250
165
0.48 0.02 f y 415
85
0.46
if
f y 250 MPa
if 250 f y 415 MPa
if 415 f y 500 MPa
if
f y 500 MPa
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
xu ,max
0.466
d
The depth of the compression block is given by:
m
Mu
bd 2f ck
= 0.16
x
xu 1 1 4 m
= 0.1727488 < u ,max
d
2
d
The area of tensile steel reinforcement is given by:
x
z d 1 u . = 115.9307 mm
d
As
Mu
, = 583.027 sq-mm > As,min
fy / s z
As = 5.830 sq-cm
IS 456-2000 RC-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NTC 2008 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
NTC 2008 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, was added to the AStrip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows:
Loads: Dead = self weight,
Live = 4.788 kN/m2
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations were compared with the ETABS results and summarized for
verification and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
254
229
9754
mm
mm
mm
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
Live load
wd
wl
=
=
self
4.788
KN/m2
KN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with independent hand calculations.
NTC 2008 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 1 Comparison of Results
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
165.90
165.93
0.02%
Transfer Conc. Stress, top
(D+PTI), MPa
5.057
5.057
0.00%
Transfer Conc. Stress, bot
(D+PTI), MPa
2.839
2.839
0.00%
Normal Conc. Stress, top
(D+L+PTF), MPa
10.460
10.467
0.07%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.409
0.08%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
7.817
7.818
0.01%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.760
0.02%
FEATURE TESTED
Table 2 Comparison of Design Moments and Reinforcements
Reinforcement Area (sq-cm)
Method
Design Moment (kN-m)
As+
ETABS
165.9
16.40
Calculated
165.9
16.29
COMPUTER FILE: NTC 2008 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
NTC 2008 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, steel = 1.15
m, concrete = 1.50
1.0 for fck ≤ 50 MPa
0.8 for fck ≤ 50 MPa
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.35 = 8.078 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.50 = 7.182 kN/m2 (Lu)
= 15.260 kN/m2 (D+L)ult
Total = 10.772 kN/m2 (D+L)
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 15.260 kN/m2 0.914 m = 13.948 kN/m
wl12
2
Ultimate Moment, M U
= 13.948 9.754 8 = 165.9 kN-m
8
NTC 2008 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
f A
Ultimate Stress in strand, f PS f SE 7000d 1 1.36 PU P l
fCK bd
1862(198)
1210 7000(229) 1 1.36
9754
30(914) 229
1361 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 2 99 1361 1000 269.5 kN
Design moment M = 165.9 kN-m
Compression block depth ratio: m
M
bd 2f cd
165.9
0.1731
0.914 0.229 2 1 30000 1.50
Required area of mild steel reinforcing,
1 1 2m = 1 1 2(0.1731) 0.1914
f bd
1(30 /1.5)(914)(229)
2
AEquivTotal cd 0.1914
2303 mm
f yd
400 /1.15
1366
2
AEquivTotal AP
AS 2311 mm
400
1361
2
AS 2303 198
1629 mm
400
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
Stress in concrete, f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
Moment due to PT,
NTC 2008 PT-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term=1490 186 94 = 1210 MPa
The force in tendon at normal = 1210 197.4 1000 238.9 kN
Moment due to dead load M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
A
S
0.254 0.914
0.00983
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
NTC 2008 PT-SL EXAMPLE 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NTC 2008 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. Thick shell properties are used for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
NTC 2008 RC-PN EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress
(N/mm2)
Shear Capacity
(N/mm2)
D/C ratio
ETABS
1.117
0.611
1.83
Calculated
1.092
0.578
1.89
COMPUTER FILE: NTC 2008 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
NTC 2008 RC-PN EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
Hand Calculation for Interior Column using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 2.
u1 = u = 2300 + 2900 + 2436 = 5139.468 mm
1172
Note: All dimensions in millimeters
Critical section for
punching shear
shown dashed.
Y
436
150 150 436
B
A
Column
436
Side 3
Side 1
Side 2
450
X
1772
450
Center of column is
point (x1, y1). Set
this equal to (0,0).
436
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
VEd = 1112.197 kN
k2MEd2 = 38.933 kN-m
k3MEd3 = 17.633 kN-m
NTC 2008 RC-PN EXAMPLE 001 - 3
Software Verification
ETABS
0
PROGRAM NAME:
REVISION NO.:
Maximum design shear stress in computed in along major and minor axis of column:
vEd
W1
VEd
ud
k2 M Ed ,2u1 k3 M Ed ,3u1
1
VEdW1,2
VEdW1,3
(EC2 6.4.4(2))
c12
c1c2 4c2 d 16d 2 2 dc1
2
W1,2
9002
300 900 4 300 218 16 2182 2 218 900
2
W1,2 2,929, 744.957 mm2
W1,3 3
9002
900 300 4 900 218 16 2182 2 218 300
2
W1,2 2, 271,104.319 mm2
vEd
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
1
ud
VEdW1,2
VEdW1,3
vEd
1112.197 103
38.933 106 5139.468
17.633 106 5139.468
1
5139.468 218 1112.197 103 2929744.957 1112.197 103 2271104.319
vEd 1.089 N/mm2
Thus vmax = 1.089 N/mm2
C Rd ,c 0.18 c = 0.18/1.5 = 0.12
(EC2 6.4.4)
The shear stress carried by the concrete, VRd,c, is calculated as:
13
VRd ,c C Rd ,c k 100 1 fck k1 cp
(EC2 6.4.4)
with a minimum of:
vRd ,c vmin k1 cp
k 1
k1
200
2.0 = 1.9578
d
= 0.15.
NTC 2008 RC-PN EXAMPLE 001 - 4
(EC2 6.4.4)
(EC2 6.4.4(1))
(EC2 6.2.2(1))
Software Verification
PROGRAM NAME:
REVISION NO.:
1 =
ETABS
0
As1
0.02
bw d
Area of reinforcement at the face of column for design strip are as follows:
As in Strip Layer A = 9204.985 mm2
As in Strip Layer B = 8078.337 mm2
Average As = 9204.985 8078.337 2 = 8641.661 mm2
1 = 8641.661 8000 218 = 0.004955 0.02
min 0.035k 3 2 f ck 1 2 = 0.035 1.9578
3/2
vRd ,c 0.12 1.9578 100 0.004955 30
30
13
Shear Ratio
1/2
= 0.525 N/mm2
2
0 = 0.5777 N/mm
v max
1.089
1.89
vRd ,c
0.5777
NTC 2008 RC-PN EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NTC 2008 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
4 m span
Simply
supported
edge at wall
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Italian NTC
2008 load combination factors, 1.35 for dead loads and 1.5 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. These moments
are identical. After completing the analysis, design is performed using the Italian
NTC 2008 code by ETABS and also by hand computation. Table 1 shows the
comparison of the design reinforcements computed by the two methods.
NTC 2008 RC-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fck
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2x106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip the moments obtained by the hand computation method. Table 1 also
shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Reinforcement Area (sq-cm)
Method
Strip Moment
(kN-m)
As+
ETABS
25.797
5.400
Calculated
25.800
5.400
A s ,min = 204.642 sq-mm
COMPUTER FILE: NTC 2008 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
NTC 2008 RC-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATION
The following quantities are computed for the load combination:
1.0 for fck ≤ 50 MPa
0.8 for fck ≤ 50 MPa
b = 1000 mm
For the load combination, w and M are calculated as follows:
w = (1.35wd + 1.5wt) b
wl12
M
8
As ,min
0.0013bw d
max
fctm
0
.
26
bd
f
yk
= 204.642 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.9 kN/m
M-strip = 25.8 kN-m
M-design= 25.8347 kN-m
m, steel = 1.15
m, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
m
M
bd 2 f cd
25.8347 106
= 0.097260
1000 1252 1.0 0.85 30 /1.5
x x
mlim 1 = 0.48
d lim 2 d lim
NTC 2008 RC-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
k1
x
for fck 50 MPa = 0.60
k2
d lim
For reinforcement with fyk 500 MPa, the following values are used:
k1 = 0.40
k2 = (0.6 + 0.0014/εcu2) = 1.00
is assumed to be 1
1 1 2m = 0.10251
f bd
As cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
NTC 2008 RC-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
NZS 3101-2006 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 915 mm wide and spans 9754 mm as, shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
NZS 3101-2006 PT-SL EXAMPLE 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows:
Loads:
Live = 4.788 kN/m2
Dead = self weight,
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the midspan of the slab. Independent hand calculations
were compared with the ETABS results and summarized for verification and
validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d
=
L
=
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
=
=
=
=
=
=
=
=
=
Dead load
Live load
wd
wl
=
=
254 mm
229 mm
9754 mm
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
self kN/m2
4.788 kN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
NZS 3101-2006 PT-SL EXAMPLE 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
Long-Term Conc. Stress,
top (D+0.5L+PTF(L)), MPa
Long-Term Conc. Stress,
bot (D+0.5L+PTF(L)), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
156.12
156.17
0.02%
14.96
15.08
0.74%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
7.817
7.818
0.01%
5.759
5.760
0.02%
COMPUTER FILE: NZS 3101-2006 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
NZS 3101-2006 PT-SL EXAMPLE 001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
b = 0.85
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
1 0.85 for f c 55 MPa
1 0.85 for f c 30,
cb
c
c f y Es
d = 214.7
amax = 0.751cb = 136.8 mm
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.2 = 7.181kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.5 = 7.182 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 14.363 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 14.363 kN/m2 0.914 m = 13.128 kN/m
NZS 3101-2006 PT-SL EXAMPLE 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Moment, M U
ETABS
0
wl12
= 13.128 (9.754)2/8 = 156.12 kN-m
8
Ultimate Stress in strand, f PS f SE 70
f'c
300 P
30
300 0.00095
1385 MPa f SE 200 1410 MPa
1210 70
Ultimate force in PT, Fult , PT AP f PS 2 99 1385 1000 274.23 kN
Stress block depth, a d d 2
2M *
f 'c b
0.229 0.2292
2 156.12
1e3 37.48 mm
0.85 30000 0.85 0.914
Ultimate moment due to PT,
a
37.48
M ult , PT Fult , PT d 274.23 229
0.85 1000 49.01 kN-m
2
2
Net ultimate moment, M net M U M ult , PT 156.1 49.10 107.0 kN-m
Required area of mild steel reinforcing,
M net
107.0
(1e6) 1496 mm 2
AS
a
0.03748
f y (d ) 0.85(400000) 0.229
2
2
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses =1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT,
Stress in concrete,
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
NZS 3101-2006 PT-SL EXAMPLE 001 - 5
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ETABS
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f 1.109 3.948MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
Long-Term Condition, load combinations: (D+0.5L+PTF(L)) = 1.0D+0.5L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+0.5L+PTF(L)),
M
M PT
F
238.9
91.06 24.33
f PTI D 0.5 L
A
S
0.254 0.914
0.00983
f 1.029 6.788
f 7.817(Comp) max, 5.759(Tension) max
NZS 3101-2006 PT-SL EXAMPLE 001 - 6
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NZS 3101-2006 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8 m spans in each direction, as
shown in Figure 1.
A
0 .3 m
B
8 m
C
8 m
D
0 .3 m
8 m
0 .6 m
4
0 .2 5 m t h ic k f la t s la b
8 m
3
C o lu m n s a r e 0 .3 m x 0 .9 m
w it h lo n g s id e p a r a lle l
8 m
t o t h e Y - a x is , t y p ic a l
C o n c r e t e P r o p e r t ie s
2
U n it w e ig h t = 2 4 k N /m
f 'c = 3 0 N /m m
3
2
8 m
Y
L o a d in g
D L = S e lf w e ig h t + 1 .0 k N /m
X
1
L L = 4 .0 k N /m
2
2
0 .6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
NZS 3101-2006 RC-PN EXAMPLE 001 - 1
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
(N/mm2)
D/C ratio
ETABS
1.793
1.141
1.57
Calculated
1.792
1.141
1.57
COMPUTER FILE: NZS 3101-2006 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
NZS 3101-2006 RC-PN EXAMPLE 001 - 2
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 259 26 250 38 2
= 218 mm
Refer to Figure 2.
b0 = 518+ 1118 + 1118 + 518 = 3272 mm
N o t e : A ll d im e n s io n s in m illim e t e r s
518
Y
C r it ic a l s e c t io n f o r
109
150
150
p u n c h in g s h e a r s h o w n
109
dashed.
A
C o lu m n
B
109
S id e 3
S id e 1
S id e 2
450
X
1118
C e n t e r o f c o lu m n is
450
p o in t ( x 1 , y 1 ) . S e t
t h is e q u a l t o ( 0 ,0 ) .
109
S id e 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
V 2
1
1
2
1
3
V 3
1
0 .4 9 5
1118
518
1
2
1
3
0 .3 1 2
518
1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
NZS 3101-2006 RC-PN EXAMPLE 001 - 3
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PROGRAM NAME:
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The following table is used for calculating the centroid of the critical section for punching
shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical section for
punching shear as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
259
0
1118
218
243724
63124516
0
x3
Ldx2
y3
Ldy 2
Ld
0
Ld
Side 2
0
559
518
218
112924
0
63124516
Side 3
259
0
1118
218
243724
63124516
0
Side 4
0
559
518
218
112924
0
63124516
Sum
N.A.
N.A.
b0 = 3272
N.A.
713296
0
0
0 mm
713296
0
0 mm
713296
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the "Sum" column.
Item
L
d
x2 x3
y2 y3
Parallel to
Equations
IXX
IYY
IXY
Side 1
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 2
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
From the ETABS output at Grid B-2:
VU = 1126.498 kN
V 2
MU2 = 25.725 kN-m
V 3
MU3 = 14.272 kN-m
NZS 3101-2006 RC-PN EXAMPLE 001 - 4
Side 3
1118
218
259
0
Y-Axis
5b, 6b, 7
2.64E+10
1.63E+10
0
Side 4
518
218
0
559
X-axis
5a, 6a, 7
3.53E+10
2.97E+09
0
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
1.23E+11
3.86E+10
0
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At the point labeled A in Figure 2, x4 = 259 and y4 = 559, thus:
vU
1 1 2 6 .4 9 8 1 0
3272 218
2 5 .7 2 5 1 0 3 .8 6 1 0
6
3
10
559
0 0 2 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
1 4 .2 7 2 1 0 1 .2 3 1 0
6
11
259
2
0 0 5 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
2
vU = 1.5793 0.1169 0.0958 = 1.3666 N/mm2 at point A
At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus:
vU
1 1 2 6 .4 9 8 1 0
3272 218
2 5 .7 2 5 1 0 3 .8 6 1 0
6
3
10
559
0 0 2 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
1 4 .2 7 2 1 0 1 .2 3 1 0
6
11
259
2
0 0 5 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
2
vU = 1.5793 0.1169 + 0.0958 =1.5582 N/mm2 at point B
At the point labeled C in Figure 2, x4 = 259 and y4 = 559, thus:
vU
1 1 2 6 .4 9 8 1 0
3272 218
2 5 .7 2 5 1 0 3 .8 6 1 0
6
3
10
559
0 0 2 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
1 4 .2 7 2 1 0 1 .2 3 1 0
6
11
259
2
0 0 5 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
2
vU = 1.5793 + 0.1169 + 0.0958 = 1.792 N/mm2 at point C
At the point labeled D in Figure 2, x4 = 259 and y4 = 559, thus:
vU
1 1 2 6 .4 9 8 1 0
3272 218
2 5 .7 2 5 1 0 3 .8 6 1 0
6
3
10
559
0 0 2 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
1 4 .2 7 2 1 0 1 .2 3 1 0
6
11
259
2
0 0 5 5 9 0
1 .2 3 1 0 1 1 3 .8 6 1 0 1 0 0
2
vU = 1.5793 + 0.1169 0.0958 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
NZS 3101-2006 RC-PN EXAMPLE 001 - 5
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PROGRAM NAME:
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ETABS
0
The shear capacity is calculated based on the smallest of NZS 3101-06, with the bo and u
terms removed to convert force to stress.
1
2
1
f c
6
c
1
sd
v v m in 1
f c
6
b
0
1
f c
3
= 1.141N/mm2 per
NZS 12.7.3.2 yields the smallest value of
capacity.
S h e a r R a tio
vU
vv
NZS 3101-2006 RC-PN EXAMPLE 001 - 6
1 .7 9 2
1 .1 4 1
1 .5 7
vv
(NZS 12.7.3.2)
= 1.141 N/mm2, and thus this is the shear
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ETABS
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NZS 3101-2006 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
4 m span
Simply
supported
edge at wall
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 kN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the NZS 310106 load combination factors, 1.2 for dead loads and 1.5 for live loads. The model
is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
analysis, design is performed using the NZS 3101-06 code by ETABS and also
by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
NZS 3101-2006 RC-SL EXAMPLE 001 - 1
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
24.597
5.238
Calculated
24.6
5.238
As+
Medium
A s ,min = 380.43 sq-mm
NZS 3101-2006 RC-SL EXAMPLE 001 - 2
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COMPUTER FILE: NZS 3101-2006 RC-SL EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
NZS 3101-2006 RC-SL EXAMPLE 001 - 3
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HAND CALCULATION
The following quantities are computed for the load combination:
b = 0.85
b = 1000 mm
1 0.85 for f c 55MPa
1 0.85 for f c 30,
cb
c
c f y Es
d = 70.7547
amax = 0.751cb= 45.106 mm
For the load combination, w and M* are calculated as follows:
w = (1.2wd + 1.5wt) b
Mu
wl12
8
As ,min
f c
bw d 372.09 sq-mm
4 fy
max
1.4 bw d 380.43 sq-mm
fy
= 380.43 sq-mm
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 12.3 kN/m
M*-strip = 24.6 kN-m
M*-design = 24.6331 kN-m
The depth of the compression block is given by:
a d d2
NZS 3101-2006 RC-SL EXAMPLE 001 - 4
2 M*
= 9.449 mm < amax
1 f c bb
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The area of tensile steel reinforcement is then given by:
As
M*
= 523.799 sq-mm > As,min
a
b f y d
2
As = 5.238 sq-cm
NZS 3101-2006 RC-SL EXAMPLE 001 - 5
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ETABS
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SS CP 65-99 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
SS CP 65-99 PT-SL EXAMPLE 001 - 1
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A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows.
Loads:
Live = 4.788 kN/m2
Dead = self weight,
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations are compared with the ETABS results and summarized for verification
and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d =
L =
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f 'c
fy
fpu
fe
Ap
wc
Ec
Es
Dead load
Live load
wd =
wl =
254
229
9754
=
30
=
400
=
1862
=
1210
=
198
=
23.56
= 25000
= 200,000
=
0
self
4.788
mm
mm
mm
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
kN/m2
kN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
SS CP 65-99 PT-SL EXAMPLE 001 - 2
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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.80
0.76%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
COMPUTER FILE: SS CP 65-1999 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
SS CP 65-99 PT-SL EXAMPLE 001 - 3
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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
m, steel = 1.15
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 16.039 kN/m2 0.914 m = 14.659 kN/m
wl12
Ultimate Moment, M U
= 14.659 (9.754)2/8 = 174.4 kN-m
8
SS CP 65-99 PT-SL EXAMPLE 001 - 4
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ETABS
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f pu Ap
7000
1 1.7
l/d
f cu bd
7000
1862(198)
1210
1 1.7
9754 / 229
30(914)(229)
Ultimate Stress in strand, f pb f pe
1358 MPa 0.7 f pu 1303 MPa
K factor used to determine the effective depth is given as:
174.4
M
K
0.1213 < 0.156
=
2
f cu bd
30000(0.914)(0.229) 2
K
0.95d = 192.2 mm
z d 0.5 0.25
0.9
Ultimate force in PT, Fult , PT AP ( f PS ) 2 99 1303 1000 258.0 kN
Ultimate moment due to PT,
M ult , PT Fult , PT ( z ) / 258.0 0.192 1.15 43.12 kN-m
Net Moment to be resisted by As,
M NET MU M PT
174.4 43.12 131.28 kN-m
The area of tensile steel reinforcement is then given by:
M NET
131.28
1e6 1965 mm 2
As
=
0.87 f y z X
0.87 400 192
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT,
Stress in concrete,
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
SS CP 65-99 PT-SL EXAMPLE 001 - 5
Software Verification
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ETABS
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Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
SS CP 65-99 PT-SL EXAMPLE 001 - 6
Software Verification
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ETABS
0
SS CP 65-1999 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a fcu of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
SS CP 65-1999 RC-PN EXAMPLE 001 - 1
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ETABS
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
D/C ratio
(N/mm2)
ETABS
1.116
0.662
1.69
Calculated
1.105
0.620
1.77
COMPUTER FILE: SS CP 65-1999 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
SS CP 65-1999 RC-PN EXAMPLE 001 - 2
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 = 218 mm
Refer to Figure 1.
u = 954+ 1554 + 954 + 1554 = 5016 mm
Note: All dimensions in millimeters
954
Y
327
150 150
A
Column
Critical section for
punching shear shown
dashed.
327
B
Side 2
Side 3
Side 1
327
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1554
450
327
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
From the ETABS output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
SS CP 65-1999 RC-PN EXAMPLE 001 - 3
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Maximum design shear stress in computed in along major and minor axis of column:
veff , x
1.5M
V
x
f
ud
Vy
veff , x
1126.498 103
1.5 51.9908 106
1.0
1.1049 (Govern)
5016 218
1126.498 103 954
veff , y
1.5M
V
y
f
ud
Vx
veff , y
1126.498 103
1.5 45.7234 106
1.0
1.0705
5016 218
1126.498 103 1554
(CP 3.7.7.3)
The largest absolute value of v = 1.1049 N/mm2
The shear stress carried by the concrete, vc, is calculated as:
1
1
0.79k1k 2 100 As 3 400 4
vc
= 0.3568 MPa
m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
1
f 3 30 3
k2 = cu = = 1.0627 > 1.0 OK
25
25
m = 1.25
400
d
1
4
= 1.16386 > 1 OK.
fcu 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Area of reinforcement at the face of column for design strip are as follows:
As in Strip Layer A = 9494.296 mm2
As in Strip Layer B = 8314.486 mm2
Average As = (9494.296+8314.486)/2 = 8904.391 mm2
SS CP 65-1999 RC-PN EXAMPLE 001 - 4
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100 As
= 100 8904.391/(8000 218) = 0.51057
bd
vc
0.79 1.0 1.0627
1/ 3
0.51057 1.16386 = 0.6247 MPa
1.25
BS 3.7.7.3 yields the value of v = 0.625 N/mm2, and thus this is the shear capacity.
Shear Ratio
vU
1.1049
1.77
v
0.6247
SS CP 65-1999 RC-PN EXAMPLE 001 - 5
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SS CP 65-1999 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
4 m span
Simply
supported
edge at wall
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 KN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Singapore
CP 65-99 load combination factors, 1.4 for dead loads and 1.6 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed using the Singapore CP 65-99 code by ETABS
and also by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
SS CP 65-1999 RC-SL EXAMPLE 001 - 1
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fsy
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
27.197
5.853
Calculated
27.200
5.850
As+
Medium
A s ,min = 162.5 sq-mm
SS CP 65-1999 RC-SL EXAMPLE 001 - 2
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COMPUTER FILE: SS CP 65-1999 RC EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
SS CP 65-1999 RC-SL EXAMPLE 001 - 3
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HAND CALCULATION
The following quantities are computed for all the load combinations:
m, steel
= 1.15
m, concrete = 1.50
b
= 1000 mm
For each load combination, the w and M are calculated as follows:
w = (1.4wd + 1.6wt) b
M
wl12
8
As ,m in 0 . 0013bw d
= 162.5 sq-mm
COMB100
wd =
4.0 kPa
wt
5.0 kPa
=
w = 13.6 kN/m
M-strip = 27.2 kN-m
M-design = 27.2366 kN-m
The depth of the compression block is given by:
K
M
= 0.05810 < 0.156
f cu bd 2
The area of tensile steel reinforcement is then given by:
K
0.95d =116.3283
z d 0.5 0.25
0.9
As
M
= 585.046 sq-mm > As,min
0.87 f y z
As = 5.850 sq-cm
SS CP 65-1999 RC-SL EXAMPLE 001 - 4
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TS 500-2000 PT-SL EXAMPLE 001
Post-Tensioned Slab Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the slab stresses and the required area of
mild steel strength reinforcing for a post-tensioned slab.
A one-way, simply supported slab is modeled in ETABS. The modeled slab is 254
mm thick by 914 mm wide and spans 9754 mm, as shown in shown in Figure 1.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
914 mm
Section
Elevation
Figure 1 One-Way Slab
TS 500-2000 PT-SL EXAMPLE 001 - 1
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A 254-mm-wide design strip is centered along the length of the slab and has been
defined as an A-Strip. B-strips have been placed at each end of the span,
perpendicular to Strip-A (the B-Strips are necessary to define the tendon profile).
A tendon with two strands, each having an area of 99 mm2, has been added to the
A-Strip. The self weight and live loads have been added to the slab. The loads and
post-tensioning forces are as follows.
Loads:
Live = 4.788 kN/m2
Dead = self weight,
The total factored strip moments, required area of mild steel reinforcement, and
slab stresses are reported at the mid-span of the slab. Independent hand
calculations are compared with the ETABS results and summarized for verification
and validation of the ETABS results.
GEOMETRY, PROPERTIES AND LOADING
Thickness
Effective depth
Clear span
T, h =
d =
L =
Concrete strength
Yield strength of steel
Prestressing, ultimate
Prestressing, effective
Area of Prestress (single strand)
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
f ck
fyk
fpu
fe
Ap
wc
Ec
Es
Dead load
Live load
wd =
wl =
254
229
9754
=
30
=
400
=
1862
=
1210
=
198
=
23.56
= 25000
= 200,000
=
0
self
4.788
mm
mm
mm
MPa
MPa
MPa
MPa
mm2
kN/m3
N/mm3
N/mm3
kN/m2
kN/m2
TECHNICAL FEATURES OF ETABS TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments, required
mild steel reinforcing, and slab stresses with the independent hand calculations.
TS 500-2000 PT-SL EXAMPLE 001 - 2
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Table 1 Comparison of Results
FEATURE TESTED
Factored moment,
Mu (Ultimate) (kN-m)
Area of Mild Steel req’d,
As (sq-cm)
Transfer Conc. Stress, top
(D+PTI), MPa
Transfer Conc. Stress, bot
(D+PTI), MPa
Normal Conc. Stress, top
(D+L+PTF), MPa
Normal Conc. Stress, bot
(D+L+PTF), MPa
INDEPENDENT
RESULTS
ETABS
RESULTS
DIFFERENCE
174.4
174.4
0.00%
14.88
14.90
0.13%
5.058
5.057
-0.02%
2.839
2.839
0.00%
10.460
10.467
0.07%
8.402
8.409
0.08%
COMPUTER FILE: TS 500-2000 PT-SL EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
TS 500-2000 PT-SL EXAMPLE 001 - 3
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HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fck = 30MPa
fyk = 400MPa
m, steel = 1.15
Post-Tensioning
fpu = 1862 MPa
fpy = 1675 MPa
Stressing Loss = 186 MPa
Long-Term Loss = 94 MPa
fi = 1490 MPa
fe = 1210 MPa
m, concrete = 1.50
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Loads:
Dead, self-wt = 0.254 m 23.56 kN/m3 = 5.984 kN/m2 (D) 1.4 = 8.378 kN/m2 (Du)
Live,
= 4.788 kN/m2 (L) 1.6 = 7.661 kN/m2 (Lu)
Total = 10.772 kN/m2 (D+L) = 16.039 kN/m2 (D+L)ult
=10.772 kN/m2 0.914 m = 9.846 kN/m, u = 16.039 kN/m2 0.914 m = 14.659 kN/m
wl12
Ultimate Moment, M U
= 14.659 (9.754)2/8 = 174.4 kN-m
8
TS 500-2000 PT-SL EXAMPLE 001 - 4
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f A
Ultimate Stress in strand, f Pd f pe 7000d 1 1.36 PU P l
fCK bd
1862(198)
1210 7000(229) 1 1.36
9754
30(914) 229
1361 MPa
Ultimate force in PT, Fult , PT AP ( f PS ) 2 99 1361 1000 269.5 kN
Stress block depth, a d d 2
2M d
0.85 f cd b
0.229 0.2292
2 174.4
1e3 55.816 mm
0.85 20000 0.914
Ultimate moment due to PT,
a
55.816
M ult , PT Fult , PT d 269.5 229
1000 54.194 kN-m
2
2
Net ultimate moment, M net M U M ult , PT 174.4 54.194 120.206 kN-m
Required area of mild steel reinforcing,
AS
M net
120.206 106
1488.4 mm 2
a
54.194
f yd d 400 229
2
2
K factor used to determine the effective depth is given as:
174.4
M
K
0.1819 < 0.156
=
2
f cu bd
30000 1.5(0.914)(0.229) 2
K
0.95d = 192.2 mm
z d 0.5 0.25
0
.
9
Ultimate force in PT, Fult , PT AP ( f PS ) 2 99 1303 1000 258.0 kN
Ultimate moment due to PT,
M ult , PT Fult , PT ( z ) / 258.0 0.192 1.15 43.12 kN-m
Net Moment to be resisted by As,
TS 500-2000 PT-SL EXAMPLE 001 - 5
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M NET MU M PT
174.4 43.12 131.28 kN-m
The area of tensile steel reinforcement is then given by:
M
131.28
1e6 1965 mm 2
As NET =
f yd z X
0.87 400 192
Check of Concrete Stresses at Midspan:
Initial Condition (Transfer), load combination (D+PTi) = 1.0D+0.0L+1.0PTI
Tendon stress at transfer = jacking stress stressing losses = 1490 186 = 1304 MPa
The force in the tendon at transfer, = 1304 197.4 1000 257.4 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to PT,
Stress in concrete,
M PT FPTI (sag) 257.4 102 mm 1000 26.25 kN-m
F
M M PT
257.4
65.04 26.23
f PTI D
A
S
0.254 0.914
0.00983
where S = 0.00983m3
f 1.109 3.948 MPa
f 5.058(Comp) max, 2.839(Tension) max
Normal Condition, load combinations: (D+L+PTF) = 1.0D+1.0L+1.0PTF
Tendon stress at normal = jacking stressing long-term = 1490 186 94 = 1210 MPa
The force in tendon at normal, = 1210 197.4 1000 238.9 kN
Moment due to dead load, M D 5.984 0.914 9.754 8 65.04 kN-m
2
Moment due to live load, M L 4.788 0.914 9.754 8 52.04 kN-m
2
Moment due to PT,
M PT FPTI (sag) 238.9 102 mm 1000 24.37 kN-m
Stress in concrete for (D+L+PTF),
F
M
M PT
238.8
117.08 24.37
f PTI D L
A
S
0.254 0.914
0.00983
f 1.029 9.431
f 10.460(Comp) max, 8.402(Tension) max
TS 500-2000 PT-SL EXAMPLE 001 - 6
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TS 500-2000 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m
A
B
8m
C
8m
D 0.3 m
8m
0.6 m
4
0.25 m thick flat slab
8m
3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical
8m
Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2
2
8m
Y
X
1
Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2
0.6 m
Figure 1: Flat Slab for Numerical Example
The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. The slab is typically 0.25 m thick. Thick shell properties are used
for the slab.
The concrete has a unit weight of 24 kN/m3 and a fck of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.
TS 500-2000 RC-PN EXAMPLE 001 - 1
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TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from ETABS with the punching shear capacity, shear
stress ratio and D/C ratio obtained by the analytical method. They match exactly
for this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method
Shear Stress Shear Capacity
(N/mm2)
D/C ratio
(N/mm2)
ETABS
1.695
1.278
1.33
Calculated
1.690
1.278
1.32
COMPUTER FILE: TS 500-2000 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
TS 500-2000 RC-PN EXAMPLE 001 - 2
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HAND CALCULATION
Hand Calculation For Interior Column Using ETABS Method
d 250 26 250 38 2 = 218 mm
b0 = 518+ 1118 + 1118 + 518 = 3272 mm
Note: All dimensions in millimeters
518
Y
109 150
150
A
Column
Critical section for
punching shear shown
dashed.
109
B
109
Side 3
Side 1
Side 2
450
X
Center of column is
point (x1, y1). Set
this equal to (0,0).
1118
450
109
Side 4
D
C
Figure 2: Interior Column, Grid B-2 in ETABS Model
2 1
1
0.595
1118
1
518
3 1
1
0.405
518
1
1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
The following table is used for calculating the centroid of the critical section for punching
shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical section for
punching shear as identified in Figure 2.
TS 500-2000 RC-PN EXAMPLE 001 - 3
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Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
259
0
1118
218
243724
63124516
0
x3
Ldx
y3
Ldy
2
Ld
Ld
2
Side 2
0
559
518
218
112924
0
63124516
0
0 mm
713296
0
0 mm
713296
Side 3
259
0
1118
218
243724
63124516
0
Side 4
0
559
518
218
112924
0
63124516
Sum
N.A.
N.A.
b0 = 3272
N.A.
713296
0
0
The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the "Sum" column.
Item
L
d
x2 x3
y2 y3
Parallel to
Equations
IXX
IYY
Side 1
1118
218
259
0
Y-Axis
5b, 6b
5.43E+07
6.31E+07
Side 2
518
218
0
559
X-axis
5a, 6a
6.31E+07
1.39E07
Side 3
1118
218
259
0
Y-Axis
5b, 6b
2.64E+10
1.63E+10
Side 4
518
218
0
559
X-axis
5a, 6a
3.53E+10
2.97E+09
Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
1.23E+11
3.86E+10
From the ETABS output at Grid B-2:
Vd= 1126.498 kN
0.4Md,2 = -8.4226 kN-m
0.4M d,3 = 10.8821 kN-m
Maximum design shear stress in computed in along major and minor axis of column:
v pd
V pd
0.4M pd ,2u p d
0.4M pd ,3u p d
1
,
u p d
V pdWm,2
V pdWm,3
At the point labeled A in Figure 2, x4 = 259 and y4 = 559, thus:
TS 500-2000 RC-PN EXAMPLE 001 - 4
(TS 8.3.1)
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vU
ETABS
0
6
10
1126.498 103 8.423 10 3.86 10 559 0
1.23 1011 3.86 1010
3272 218
10.882 106 1.23 1011 259 0
1.23 1011 3.86 1010
vU = 1.5793 0.0383 0.0730 = 1.4680 N/mm2 at point A
At the point labeled B in Figure 2, x4 = 259 and y4 = 559, thus:
6
10
1126.498 103 8.423 10 3.86 10 559 0
vU
1.23 1011 3.86 1010
3272 218
10.8821 106 1.23 1011 259 0
1.23 1011 3.86 1010
vU = 1.5793 0.0383 + 0.0730 =1.614 N/mm2 at point B
At the point labeled C in Figure 2, x4 = 259 and y4 = 559, thus:
6
10
1126.498 103 8.423 10 3.86 10 559 0
vU
1.23 1011 3.86 1010
3272 218
10.882 106 1.23 1011 259 0
1.23 1011 3.86 1010
vU = 1.5793 + 0.0383 + 0.0730 = 1.690 N/mm2 at point C
At the point labeled D in Figure 2, x4 = 259 and y4 = 559, thus:
6
10
1126.498 103 8.423 10 3.86 10 559 0
vU
1.23 1011 3.86 1010
3272 218
10.8821 106 1.23 1011 259 0
1.23 1011 3.86 1010
vU = 1.5793 + 0.383 0.0730 = 1.5446 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.690 N/mm2
TS 500-2000 RC-PN EXAMPLE 001 - 5
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The concrete punching shear stress capacity of a section with punching shear
reinforcement is limited to:
v pr f ctd 0.35 f ck c
(TS 8.3.1)
v pr f ctd 0.35 30 1.5 1.278 N/mm
2
Shear Ratio
vpd
1.690
1.32
v pr
1.278
TS 500-2000 RC-PN EXAMPLE 001 - 6
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0
TS 500-2000 RC-SL EXAMPLE 001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in ETABS.
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. The slab is 150 mm thick and spans 4 meters between
walls. The walls are modeled as pin supports. The computational model uses a
finite element mesh, automatically generated by SAFE. The maximum element
size is specified as 0.25 meters. To obtain factored moments and flexural
reinforcement in a design strip, one one-meter wide strip is defined in the Xdirection on the slab, as shown in Figure 1.
Simply
supported
edge at wall
4 m span
Simply
supported
edge at wall
Free edge
1 m design strip
Y
X
Free edge
Figure 1 Plan View of One-Way Slab
One dead load case (DL4KPa) and one live load case (LL5KPa) with uniformly
distributed surface loads of magnitudes 4 and 5 KN/m2, respectively, are defined
in the model. A load combination (COMB5kPa) is defined using the Turkish TS
500-2000 load combination factors, 1.4 for dead loads and 1.6 for live loads. The
model is analyzed for both load cases and the load combination.
The slab moment on a strip of unit width is computed analytically. The total
factored strip moments are compared with the ETABS results. After completing
the analysis, design is performed using the Turkish TS 500-2000 code by ETABS
and also by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
TS 500-2000 RC-SL EXAMPLE 001 - 1
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GEOMETRY, PROPERTIES AND LOADING
Thickness
Depth of tensile reinf.
Effective depth
Clear span
T, h
dc
d
ln, l1
=
=
=
=
150
25
125
4000
mm
mm
mm
mm
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fck
fyk
wc
Ec
Es
=
=
=
=
=
=
30
460
0
25000
2106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF ETABS TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS total factored moments in the
design strip with the moments obtained by the hand computation method. Table 1
also shows the comparison of the design reinforcements.
Table 1 Comparison of Design Moments and Reinforcements
Load
Level
Reinforcement Area
(sq-cm)
Method
Strip
Moment
(kN-m)
ETABS
27.197
5.760
Calculated
27.200
5.760
As+
Medium
A s ,min = 162.5 sq-mm
TS 500-2000 RC-SL EXAMPLE 001 - 2
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COMPUTER FILE: TS 500-2000 RC EX001.EDB
CONCLUSION
The ETABS results show an acceptable comparison with the independent results.
TS 500-2000 RC-SL EXAMPLE 001 - 3
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HAND CALCULATION
The following quantities are computed for all the load combinations:
m, steel
= 1.15
m, concrete = 1.50
f cd
f ck 30
20
mc 1.5
f yd
f yk
cb
ms
460
400
1.15
cu Es
cu Es f yd
d = 75 mm
amax = 0.85k1cb = 52.275 mm
where,
As ,min
k1 0.85 0.006 f ck 25 0.82 , 0.70 k1 0.85
0.8 f ctd
bd 325.9 mm 2
f yd
Where f ctd
0.35 f cu 0.35 30
1.278
mc
1.5
For each load combination, the w and M are calculated as follows:
w = (1.4wd + 1.6wt) b
wl12
M
8
COMB100
wd = 4.0 kPa
wt
= 5.0 kPa
w = 13.6 kN/m
M-strip = 27.2 kN-m
M-design = 27.2366 kN-m
The depth of the compression block is given by:
TS 500-2000 RC-SL EXAMPLE 001 - 4
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The depth of the compression block is given by:
a d d2
2 Md
0.85 fcd b
a 125 1252
(TS 7.1)
2 27.2366 106
13.5518 mm
0.85 20 1000
If a amax (TS 7.1), the area of tensile steel reinforcement is then given by:
As
Md
27.2366 106
576 mm 2
13.5518
a
f yd d 400 125
2
2
TS 500-2000 RC-SL EXAMPLE 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
ETABS
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REFERENCES
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REFERENCES
1
Software Verification
PROGRAM NAME:
REVISION NO.:
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REFERENCES
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Software Verification
PROGRAM NAME:
REVISION NO.:
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REFERENCES
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