SAFE Verification
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Verification
SAFE®
DESIGN OF SLABS, BEAMS AND FOUNDATIONIS
REINFORCED AND POST-TENSIONED CONCRETE
Verification Manual
ISO SAF112816M6 Rev.0
Proudly developed in the United States of America
.
November 2016
Copyright
Copyright Computers & Structures, Inc., 1978-2016
All rights reserved.
The CSI Logo® and SAFE® are registered trademarks of Computers & Structures, Inc.
Watch & LearnTM is a trademark of Computers & Structures, Inc. Adobe and Acrobat are
registered trademarks of Adobe Systems Incorported. AutoCAD is a registered trademark
of Autodesk, Inc.
The computer program SAFE® and all associated documentation are proprietary and
copyrighted products. Worldwide rights of ownership rest with Computers & Structures,
Inc. Unlicensed use of this program 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 TESTING 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.
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CONTENTS
Introduction
Methodology
Acceptance Criteria
Summary of Examples
Analysis Examples
1
Simply Supported Rectangular Plate
2
Rectangular Plate with Built-In Edges
3
Rectangular Plate with Mixed Boundary
4
Rectangular Plate on Elastic Beams
5
Infinite Flat Plate on Equidistant Columns
6
Infinite Flat Plate on Elastic Subgrade
7
Skew Plate with Mixed Boundary
8
ACI Handbook Flat Slab Example 1
9
ACI Handbook Two-Way Slab Example 2
10
PCA Flat Plate Test
11
University of Illinois Flat Plate Test F1
12
University of Illinois Flat Slab Tests F2 and F3
13
University of Illinois Two-way Slab Test T1
14
University of Illinois Two-Way Slab Test T2
15
Temperature Loading
16
Cracked Slab Analysis
17
Crack Width Analysis
CONTENTS - 1
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Design Examples
ACI 318-14
ACI 318-14 PT-SL 001
Post-Tensioned Slab Design
ACI 318-14 RC-BM-001
Flexural and Shear Beam Design
ACI 318-14 RC-PN-001
Slab Punching Shear Design
ACI 318-14 RC-SL-001
Slab Flexural Design
ACI 318-11
ACI 318-11 PT-SL 001
Post-Tensioned Slab Design
ACI 318-11 RC-BM-001
Flexural and Shear Beam Design
ACI 318-11 RC-PN-001
Slab Punching Shear Design
ACI 318-11 RC-SL-001
Slab Flexural Design
ACI 318-08
ACI 318-08 PT-SL 001
Post-Tensioned Slab Design
ACI 318-08 RC-BM-001
Flexural and Shear Beam Design
ACI 318-08 RC-PN-001
Slab Punching Shear Design
ACI 318-08 RC-SL-001
Slab Flexural Design
AS 3600-09
AS 3600-01 PT-SL-001
Post-Tensioned Slab Design
AS 3600-01 RC-BM-001
Flexural and Shear Beam Design
AS 3600-01 RC-PN-001
Slab Punching Shear Design
AS 3600-01 RC-SL-001
Slab Flexural Design
AS 3600-01
CONTENTS - 2
AS 3600-01 PT-SL-001
Post-Tensioned Slab Design
AS 3600-01 RC-BM-001
Flexural and Shear Beam Design
AS 3600-01 RC-PN-001
Slab Punching Shear Design
AS 3600-01 RC-SL-001
Slab Flexural Design
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BS 8110-97
BS 8110-97 PT-SL-001
Post-Tensioned Slab Design
BS 8110-97 RC-BM-001
Flexural and Shear Beam Design
BS 8110-97 RC-PN-001
Slab Punching Shear Design
BS 8110-97 RC-SL-001
Slab Flexural Design
CSA A23.3-14
CSA 23.3-14 PT-SL-001
Post-Tensioned Slab Design
CSA A23.3-14 RC-BM-001
Flexural and Shear Beam Design
CSA A23.3-14 RC-PN-001
Slab Punching Shear Design
CSA A23.3-14 RC-SL-001
Slab Flexural Design
CSA A23.3-04
CSA 23.3-04 PT-SL-001
Post-Tensioned Slab Design
CSA A23.3-04 RC-BM-001
Flexural and Shear Beam Design
CSA A23.3-04 RC-PN-001
Slab Punching Shear Design
CSA A23.3-04 RC-SL-001
Slab Flexural Design
Eurocode 2-04
Eurocode 2-04 PT-SL-001
Post-Tensioned Slab Design
Eurocode 2-04 RC-BM-001
Flexural and Shear Beam Design
Eurocode 2-04 RC-PN-001
Slab Punching Shear Design
Eurocode 2-04 RC-SL-001
Slab Flexural Design
Hong Kong CP-13
Hong Kong CP-13 PT-SL-001
Post-Tensioned Slab Design
Hong Kong CP-13 RC-BM-001
Flexural and Shear Beam Design
Hong Kong CP-13 RC-PN-001
Slab Punching Shear Design
Hong Kong CP-13 RC-SL-001
Slab Flexural Design
CONTENTS - 3
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Hong Kong CP-04
Hong Kong CP-04 PT-SL-001
Post-Tensioned Slab Design
Hong Kong CP-04 RC-BM-001
Flexural and Shear Beam Design
Hong Kong CP-04 RC-PN-001
Slab Punching Shear Design
Hong Kong CP-04 RC-SL-001
Slab Flexural Design
IS 456-00
IS 456-00 PT-SL-001
Post-Tensioned Slab Design
IS 456-00 RC-BM-001
Flexural and Shear Beam Design
IS 456-00 RC-PN-001
Slab Punching Shear Design
IS 456-00 RC-SL-001
Slab Flexural Design
Italian NTC 2008
Italian NTC-2008 PT-SL-001
Post-Tensioned Slab Design
Italian NTC-2008 RC-BM-001
Flexural and Shear Beam Design
Italian NTC-2008 PN-001
Slab Punching Shear Design
Italian NTC-2008 RC-SL-001
Slab Flexural Design
NZS 3101-06
NZS 3101-06 PT-SL-001
Post-Tensioned Slab Design
NZS 3101-06 RC-BM-001
Flexural and Shear Beam Design
NZS 3101-06 RC-PN-001
Slab Punching Shear Design
NZS 3101-06 RC-SL-001
Slab Flexural Design
Singapore CP 65-99
CONTENTS - 4
Singapore CP 65-99 PT-SL-001
Post-Tensioned Slab Design
Singapore CP 65-99 RC-BM-001
Flexural and Shear Beam Design
Singapore CP 65-99 RC-PN-001
Slab Punching Shear Design
Singapore CP 65-99 RC- SL-001
Slab Flexural Design
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Turkish TS 500-2000
Turkish TS 500-2000 PT-SL-001
Post-Tensioned Slab Design
Turkish TS 500-2000 RC-BM-001
Flexural and Shear Beam Design
Turkish TS 500-2000 RC-PN-001
Slab Punching Shear Design
Turkish TS 500-2000 RC- SL-001
Slab Flexural Design
Conclusions
References
CONTENTS - 5
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SAFE Software Verification Log
Revision
Number
0
1
2
3
4
Date
Description
December 02,
Initial release for SAFE v 12.0.0
2008
February 19, Initial release for SAFE v12.1.0. Example 15 and Example 16
2009
were added.
Revised to reflect results obtained from Version 12.2.0. All
examples, including 1 through 16 and all code-specific examples
December 26,
(ACI 318-00, AS 3600-01, BS 8110-97, CSA A23.3-04, Eurocode
2009
2-04, Hong Kong CoP-04, IS 456-00, NZS 3101, and Singapore
CP 65-99 – PS-SL, RC-BM, RC-PN, RC-SL)
Minor changes have been made to the Examples supplied with the
software: (1) The documented results for Analysis Examples 1, 4,
5, 7, and 8 have been updated to correct for truncation error in the
reported values. The values actually calculated by the software
have not changed for these examples. (2) The input data file for
Example 16 has been updated to correct the creep and shrinkage
parameters used so that they match those of the benchmark
July 12, 2010
example, and the documented results updated accordingly. The
behavior of the software has not changed for this example. (3) All
Slab Design examples have been updated to report the slab design
forces rather than the strip forces. The design forces account for
twisting moment in slab, so their values are more meaningful for
design. The behavior of the software has not changed for these
examples.
Minor changes have been made to the Examples supplied with the
software: (1) The documented results for Analysis Examples 1 to
7 have been updated to include the results from thin-plate and
December 8,
thick-plate formulation. (2) The input data files for Australian AS
2010
3600-2009 have been added. (3) The Eurocode 2-2004 design
verification examples now include the verification for all available
National Annexes.
SAFE 2014 Software Verification Log
Revision
Number
0
Date
February
2014
Description
Initial release for SAFE 2014 v14.0.0.
New design examples have been added for the following codes:
ACI 318-11, Hong Kong CoP-2013, Italian NTC 2008 and
LOG - 1
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SAFE Software Verification Log
Revision
Number
Date
Description
Turkish TS 500-2000 (Incident 63082).
Documentation for the punching-shear design examples of the
following codes have been corrected for an error in the
documented calculation of the punching perimeter: CSA A23.304, IS 456-00, and NZS 3101-06. No calculated results have
changed. (Incident 46359)
Documentation for the beam and slab design examples of the AS
3600-09 code have been updated to account for a change made to
the software under Incident 35218 for version 12.3.2 that updated
Equation 8.1.3(2). No calculated results have changed. (Incident
46359)
Results for the area of reinforcing steel have changed for
Eurocode P/T slab example “Eurocode 2-04 PT-SL-001”.
(Incident 62486)
Documentation for analysis Example 17 has been corrected for an
error in the documented cracked width computed by SAFE. No
calculated results have changed. (Incident 63153)
Initial release for SAFE 2014 v14.1.0.
1
June 2015
New design examples have been added for the following codes:
ACI 318-14 (Incident 79838), and CSA A23.3-14 (Incident
71674).
SAFE 2016 Software Verification Log
Revision
Number
0
Date
November
2016
Description
For Example 5, stiffening elements were updated in the model.
Results have changed slightly.
For Examples 8-14, stiffening elements were updated in the
column areas, and all beam cross-section properties were updated
to reflect values calculated from the actual geometry.
For Examples 10-14, the models were changed to reflect the fact
that the slab should extend to the outside faces of the columns.
LOG - 2
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Revision
Number
Date
Description
The slight increase in slab area has changed reported results.
For design examples ACI 318-08 RC-PN-001, ACI 318-11 RCPN-001 and ACI 318-14 RC-PN-001, incorrectly sized and
redundant stiff areas were removed from the models. Reported
results have changed slightly.
For design examples CSA A23.3-04 RC-SL-001 and CSA A23.314 RC-SL-001, mesh size has been changed to 0.25m to be the
same as in the rest of the international code examples. Results
have changed slightly.
For all changed models, reported results that have changed have
been updated in the corresponding documentation.
Documentation for Example 5 has been updated for incorrect
modeling information regarding the column stiff area dimensions.
Documentation for Example 14 has been updated for incomplete
modeling information in the images.
Documentation for Eurocode 2-04 RC-PN-001was updated to
reflect changes in how the program is determining K2*Med2 and
K3*Med3, which has changed the results.
Documentation for all design examples of type RC-SL has had
wording updated to reflect the current state of the models.
0
November
2016
Minor typo error for units for modulus of elasticity of concrete
(Ec) and modulus of elasticity of steel (Es) and poison ratio value
have been updated.
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INTRODUCTION
SAFE is a software application, based on the finite element method, for the
engineering analysis, design and detailing of reinforced-concrete and post-tensioned
slabs, beams and foundations.
This document provides example problems used to test various features and capabilities
of the SAFE program. Users should supplement these examples as necessary for
verifying their particular application of the software.
METHODOLOGY
A comprehensive series of test problems, or examples, designed to test the various
analysis and design features of the program have been created. The results produced by
SAFE were compared to independent sources, such as hand calculated results and
theoretical or published results. The comparison of the SAFE results with results
obtained from independent sources is provided in tabular form as part of each example.
To validate and verify SAFE results, the test problems were run on a PC platform that
was an Lenovo ThinkCentre machine with a Core i5, 2.67 GHz processor and 8.0 GB of
RAM operating on a Windows 7 operating system.
ACCEPTANCE CRITERIA
The comparison of the SAFE 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 SAFE results and the independent
results within the larger of the accuracy of the typical SAFE output and the
accuracy of the independent result.
Acceptable: For force, moment and displacement values, the difference between
the SAFE results and the independent results does not exceed five percent (5%).
For internal force and stress values, the difference between the SAFE results and
the independent results does not exceed ten percent (10%). For experimental
values, the difference between the SAFE results and the independent results does
not exceed twenty five percent (25%).
Unacceptable: For force, moment and displacement values, the difference
between the SAFE results and the independent results exceeds five percent (5%).
INTRODUCTION - 1
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For internal force and stress values, the difference between the SAFE results and
the independent results exceeds ten percent (10%). For experimental values, the
difference between the SAFE results and the independent results exceeds twenty
five percent (25%).
The percentage difference between results is typically calculated using the following
formula:
SAFE Result - Independent Result
Percent Difference = 100
Maximum of Independent Result
SUMMARY OF EXAMPLES
Examples 1 through 7 verify the accuracy of the elements and the solution algorithms
used in SAFE. These examples compare displacements and member internal forces
computed by SAFE with known theoretical solutions for various slab support and load
conditions.
Examples 8 through 14 verify the applicability of SAFE in calculating design moments
in slabs by comparing results for practical slab geometries with experimental results
and/or results using ACI 318-95 recommendations. Examples 15 and 16 verify the
applicability of SAFE for temperature loading and cracked deflection analysis for creep
and shrinkage by comparing the results from published examples.
Design examples verify the design algorithms used in SAFE 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, using ACI 318-14, ACI 318-11, ACI
318-08, AS 3600-09, AS 3600-01, BS 8110-97, CSA A23.3-14, CSA A23.3-04,
Eurocode 2-02, Hong Kong CP-13, Hong Kong CP-04, IS 456-00, Italian NTC 2008,
NZS 31-01-06, Singapore CP 65-99 and Turkish TS 500-2000 codes, by comparing
SAFE results with hand calculations.
Methodology - 2
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EXAMPLE 1
Simply Supported Rectangular Plate
PROBLEM DESCRIPTION
A simply supported, rectangular plate is analyzed for three load conditions:
uniformly distributed load over the slab (UL), a concentrated point load at the
center of the slab (PL), and a line load along a centerline of the slab (LL).
To test convergence, the problem is analyzed employing three mesh sizes, 4 × 4,
8 × 8, and 12 × 12, as shown in Figure 1-2. The slab is modeled using plate
elements in SAFE. The simply supported edges are modeled as line supports with
a large vertical stiffness. Three load cases are considered. Self weight is not
included in these analyses.
To obtain design moments, the plate is divided into three strips ― two edge
strips and one middle strip ― each way, based on the ACI 318-95 definition of
design strip widths for a two-way slab system as shown in Figure 1-3.
For comparison with the theoretical results, load factors of unity are used and
each load case is processed as a separate load combination.
Closed-form solutions to this problem are given in Timoshenko and Woinowsky
(1959) employing a double Fourier Series (Navier’s solution) or a single series
(Lévy’s solution). The numerically computed deflections, local moments,
average strip moments, and local shears obtained from SAFE are compared with
the corresponding closed form solutions.
SAFE results are shown for both thin plate and thick plate element formulations.
The thick plate formulation is recommended for use in SAFE, as it gives more
realistic shear forces for design, especially in corners and near supports and other
discontinuities. However, thin plate formulation is consistent with the closedform solutions.
GEOMETRY, PROPERTIES AND LOADING
Plate size,
a×b
Plate thickness
T
Modulus of elasticity
E
Poisson's ratio
v
=
=
=
=
360 in × 240 in
8 inches
3000 ksi
0.3
EXAMPLE 1 - 1
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Load Cases:
(UL) Uniform load
(PL) Point load
(LL) Line load
q
P
q1
(3)
= 100 psf
= 20 kips
= 1 kip/ft
q1
(2)
(1)
P
q
a = 30 '
q1
(2)
P
(1)
b = 20 '
(3)
y
x
Figure 1-1 Simply Supported Rectangular Plate
EXAMPLE 1 - 2
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5'
2 @ 10'
SAFE
0
5'
4 @ 5'
4x4 Mesh
2 @ 2.5'
4 @ 5'
2 @ 2.5'
8 @ 2.5'
8x8 Mesh
3 @ 20"
6 @ 40"
3 @ 20"
12 @ 20"
12x12 Mesh
Figure 1-2 SAFE Meshes for Rectangular Plate
EXAMPLE 1 - 3
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b = 20'
Y
X
a = 30'
Y
Edge Strip
b/4 = 5'
10'
Middle Strip
b/4 = 5'
X
X Strips
Y
5'
b/4
20'
Middle Strip
5'
b/4
Edge Strip
X
Y Strips
Figure 1-3 SAFE Definition of Design Strips
EXAMPLE 1 - 4
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TECHNICAL FEATURES OF SAFE TESTED
Deflection of slab at various mesh refinements.
Local moments, average strip moments, and local shears
RESULTS COMPARISON
Table 1-1 shows the deflections of four different points for three different mesh
refinements for the three load cases. The theoretical solutions based on Navier’s
formulations also are shown for comparison. It can be observed from Table 1-1
that the deflection obtained from SAFE converges monotonically to the
theoretical solution with mesh refinement. Moreover, the agreement is acceptable
even for the coarse mesh (4 × 4).
Table 1-2 shows the comparison of the numerically obtained local-moments at
critical points with that of the theoretical values. Only results from the 8x8 mesh
are reported. The comparison with the theoretical results is acceptable.
Table 1-3 shows the comparison of the numerically obtained local-shears at
critical points with that of the theoretical values. The comparison here needs an
explanation. The theoretical values were presented for both thin plate and thick
plate formulations. The theoretical values are for a thin plate solution where
shear strains across the thickness of the plate are ignored. The SAFE results for
thick plate are for an element that does not ignore the shear strains. The thin plate
theory results in concentrated corner uplift; consideration of the shear strains
spreads this uplift over some length of the supports near the corners. The shears
reported by SAFE for thick plate are more realistic.
The results of Table 1-3 are plotted in Figures 1-4 to 1-15. In general, it can be
seen that the thin plate formulation more closely matches the closed-form
solution than does the thick plate solution, as expected. The closed-form solution
cannot be used to validate the thick plate shears, since behavior is fundamentally
different in the corners. This can be seen clearly in Figures 6, 7, 10, 11, 14 and
15 which show the shear forces trajectories for thin plate and thick plate
solutions. The thin plate solution unrealistically carries loads to corners, whereas
the thick plate solution carries the load more toward the middle of the sites.
Table 1-4 shows the comparison of the average strip-moments for the load cases
with the theoretical average strip-moments. The comparison is excellent. This
checks both the accuracy of the finite element analysis and the integration
scheme over the elements.
EXAMPLE 1 - 5
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It should be noted that in calculating the theoretical solution, a sufficient number
of terms from the series is taken into account to achieve the accuracy of the
theoretical solutions.
Table 1-1 Comparison of Displacements
Thin-Plate Formulation
Location
Load
Case
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
60
60
0.0491
0.0492
0.0493
0.0492961
60
120
0.0685
0.0684
0.0684
0.0684443
180
60
0.0912
0.0908
0.0907
0.0906034
180
120
0.1279
0.1270
0.1267
0.1265195
60
60
0.0371
0.0331
0.0325
0.0320818
60
120
0.0510
0.0469
0.0463
0.0458716
180
60
0.0914
0.0829
0.0812
0.0800715
180
120
0.1412
0.1309
0.1283
0.1255747
60
60
0.0389
0.0375
0.0373
0.0370825
60
120
0.0593
0.0570
0.0566
0.0562849
180
60
0.0735
0.0702
0.0696
0.0691282
180
120
0.1089
0.1041
0.1032
0.1024610
UL
PL
LL
EXAMPLE 1 - 6
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Thick-Plate formulation
Location
Load
Case
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
60
60
0.0485
0.0501
0.0501
0.0492961
60
120
0.0679
0.0695
0.0694
0.0684443
180
60
0.0890
0.0919
0.0917
0.0906034
180
120
0.1250
0.1284
0.1281
0.1265195
60
60
0.0383
0.0339
0.0330
0.0320818
60
120
0.0556
0.0474
0.0469
0.0458716
180
60
0.0864
0.0834
0.0821
0.0800715
180
120
0.1287
0.1297
0.1293
0.1255747
60
60
0.0387
0.0381
0.0378
0.0370825
60
120
0.0583
0.0579
0.0574
0.0562849
180
60
0.0719
0.0710
0.0703
0.0691282
180
120
0.1060
0.1053
0.1044
0.1024610
UL
PL
LL
EXAMPLE 1 - 7
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Table 1-2 Comparison of Local Moments
Thin-Plate Formulation
Moment (kip-in/in)
M11
Location
Load
Case
M22
M12
X (in)
Y (in)
SAFE
8×8
Analytical
(Navier)
SAFE
8×8
Analytical
(Navier)
SAFE
8×8
Analytical
(Navier)
150
15
0.42
0.45
0.73
0.81
0.31
0.30
150
45
1.16
1.18
1.95
2.02
0.26
0.26
150
75
1.66
1.69
2.69
2.77
0.17
0.17
150
105
1.92
1.95
3.04
3.12
0.06
0.06
150
15
0.37
0.37
0.36
0.36
0.44
0.47
150
45
1.11
1.13
1.13
1.14
0.48
0.51
150
75
1.92
1.90
2.16
2.20
0.56
0.59
150
105
2.81
2.41
3.85
3.75
0.42
0.47
150
15
0.26
0.26
0.34
0.34
0.24
0.24
150
45
0.77
0.77
1.06
1.08
0.21
0.20
150
75
1.25
1.25
1.91
1.92
0.14
0.14
150
105
1.69
1.68
2.94
3.03
0.05
0.05
UL
PL
LL
EXAMPLE 1 - 8
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SAFE
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Thick-Plate Formulation
Moment (kip-in/in)
M11
Location
Load
Case
M22
M12
X (in)
Y (in)
SAFE
8×8
Analytical
(Navier)
SAFE
8×8
Analytical
(Navier)
SAFE
8×8
Analytical
(Navier)
150
15
0.43
0.45
0.74
0.81
0.31
0.30
150
45
1.16
1.18
1.95
2.02
0.26
0.26
150
75
1.66
1.69
2.69
2.77
0.17
0.17
150
105
1.92
1.95
3.04
3.12
0.06
0.06
150
15
0.29
0.37
0.34
0.36
0.43
0.47
150
45
1.07
1.13
1.14
1.14
0.41
0.51
150
75
1.91
1.90
2.15
2.20
0.42
0.59
150
105
2.83
2.41
3.82
3.75
0.22
0.47
150
15
0.27
0.26
0.34
0.34
0.24
0.24
150
45
0.78
0.77
1.07
1.08
0.21
0.20
150
75
1.25
1.25
1.91
1.92
0.14
0.14
150
105
1.68
1.68
2.94
3.03
0.05
0.05
UL
PL
LL
EXAMPLE 1 - 9
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Table 1-3 Comparison of Local Shears
Thin-Plate Formulation
Shears (×10−3 kip/in)
V13
Location
Load
Case
V23
X (in)
Y (in)
SAFE
(8×8)
Analytical
(Navier)
SAFE
(8×8)
Analytical
(Navier)
15
45
−27.54
−35.2
−5.76
−7.6
45
45
−16.07
−21.2
−17.19
−21.0
90
45
−7.31
−10.5
−28.39
−33.4
150
45
−1.71
−3.0
−36.23
−40.7
15
45
−4.84
−8.7
−2.43
−2.6
45
45
−6.75
−9.8
−8.57
−8.3
90
45
−12.45
−13.1
−20.53
−19.2
150
45
−11.19
−11.2
−34.82
−43.0
15
45
−13.2
−15.7
−4.57
−5.7
45
45
−10.91
−13.0
−13.47
−16.2
90
45
−5.76
−7.6
−22.59
−26.5
150
45
−1.45
−2.2
−29.04
−32.4
UL
PL
LL
EXAMPLE 1 - 10
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Thick-Plate formulation
Shears (×10−3 kip/in)
V13
Location
Load
Case
V23
X (in)
Y (in)
SAFE
(8×8)
Analytical
(Navier)
SAFE
(8×8)
Analytical
(Navier)
15
45
−21.27
−35.2
24.75
−7.6
45
45
−7.57
−21.2
−6.35
−21.0
90
45
−2.30
−10.5
−29.83
−33.4
150
45
−0.92
−3.0
−43.13
−40.7
15
45
−0.66
−8.7
18.01
−2.6
45
45
1.83
−9.8
2.33
−8.3
90
45
−8.01
−13.1
−14.89
−19.2
150
45
−18.02
−11.2
−48.18
−43.0
15
45
−7.69
−15.7
19.71
−5.7
45
45
−2.07
−13.0
−4.89
−16.2
90
45
−1.43
−7.6
−23.51
−26.5
150
45
−0.63
−2.2
−34.25
−32.4
UL
PL
LL
EXAMPLE 1 - 11
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Table 1-4 Comparison of Average Strip Moments
Thin-Plate Formulation
SAFE Average Strip Moments
(kip-in/in)
Load
Case
Strip
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Average Strip
Moments
(kip-in/in)
Column
0.758
0.800
0.805
0.810
x = 180"
Middle
1.843
1.819
1.819
1.820
MB
Column
0.974
0.989
0.992
0.994
y = 120"
Middle
2.701
2.769
2.781
2.792
Column
0.992
0.958
0.926
0.901
x = 180"
Middle
3.329
3.847
3.963
3.950
MB
Column
0.440
0.548
0.546
0.548
y = 120"
Middle
3.514
3.364
3.350
3.307
Column
0.547
0.527
0.522
0.519
x = 180"
Middle
1.560
1.491
1.482
1.475
MB
Column
1.205
1.375
1.418
1.432
y = 120"
Middle
3.077
3.193
3.213
3.200
Moment Direction
MA
UL
MA
PL
MA
LL
EXAMPLE 1 - 12
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Thick-Plate Formulation
SAFE Average Strip Moments
(kip-in/in)
Load
Case
Strip
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Average Strip
Moments
(kip-in/in)
Column
0.716
0.805
0.799
0.810
x = 180"
Middle
1.757
1.855
1.832
1.820
MB
Column
1.007
0.968
0.984
0.994
y = 120"
Middle
2.65
2.80
2.805
2.792
Column
0.969
1.128
1.043
0.901
x = 180"
Middle
2.481
3.346
3.781
3.950
MB
Column
0.763
0.543
0.533
0.548
y = 120"
Middle
3.149
3.381
3.372
3.307
Column
0.489
0.526
0.517
0.519
x = 180"
Middle
1.501
1.520
1.493
1.475
MB
Column
1.254
1.338
1.408
1.432
y = 120"
Middle
2.840
3.205
3.233
3.200
Moment Direction
MA
UL
MA
PL
MA
LL
EXAMPLE 1 - 13
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Uniform Load
5
V13 Shears (x10-3 kip/in)
0
-5
-10
-15
-20
-25
-30
-35
-40
0
20
40
60
80
100
120
140
160
140
160
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-4 V12 Shear Force for Uniform Loading
Uniform Load
30
V23 Shears (x10-3 kip/in)
20
10
0
-10
-20
-30
-40
-50
0
20
40
60
80
100
120
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-5 V13 Shear Force for Uniform Loading
EXAMPLE 1 - 14
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Figure 1-6 Vmax for Uniform Load for Thin-Plate Formulation
Figure 1-7 Vmax for Uniform Load for Thick-Plate Formulation
EXAMPLE 1 - 15
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Point Load
5
V13 Shears (x10-3 kip/in)
0
-5
-10
-15
-20
0
20
40
60
80
100
120
140
160
140
160
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-8 V12 Shear Force for Point Loading
Point Load
30
V23 Shears (x10-3 kip/in)
20
10
0
-10
-20
-30
-40
-50
-60
0
20
40
60
80
100
120
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-9 V13 Shear Force for Point Loading
EXAMPLE 1 - 16
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Figure 1-10 Vmax for Point Load for Thin-Plate Formulation
Figure 1-11 Vmax for Point Load for Thick-Plate Formulation
EXAMPLE 1 - 17
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Line Load
5
V13 Shears (x10-3 kip/in)
0
-5
-10
-15
-20
0
20
40
60
80
100
120
140
160
140
160
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-12 V12 Shear Force for Line Loading
Line Load
30
V23 Shears (x10-3 kip/in)
20
10
0
-10
-20
-30
-40
0
20
40
60
80
100
120
X Ordinates (Inches)
SAFE Thin Plate
Analytical Thin Plate
SAFE Thick Plate
Figure 1-13 V13 Shear Force for Point Loading
EXAMPLE 1 - 18
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Figure 1-14 Vmax for Line Load for Thin-Plate Formulation
Figure 1-15 Vmax for Line Load for Thick-Plate Formulation
EXAMPLE 1 - 19
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COMPUTER FILE:
S01a-Thin.FDB, S01b-Thin.FDB, S01c-Thin.FDB, S01a-Thick.FDB, S01bThick.FDB and S01c-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 1 - 20
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EXAMPLE 2
Rectangular Plate with Fixed Edges
PROBLEM DESCRIPTION
A fully fixed rectangular plate is analyzed for three load conditions. The
geometric descriptions and material properties and the load cases are the same as
those of Example 1. However, the boundary conditions are different. All edges
are fixed, as shown in Figure 2-1. To test convergence, the problem is analyzed
using three mesh sizes, as shown in Figure 1-2: 4 × 4, 8 × 8, and 12 × 12. The
plate is modeled using plate elements available in SAFE. The fixed edges are
modeled as line supports with large vertical and rotational stiffnesses. The self
weight of the plate is not included in any of the load cases. The numerical data
for this problem are given in the following section.
A theoretical solution to this problem, employing a single series (Lévy’s
solution), is given in Timoshenko and Woinowsky (1959). The numerically
computed deflections obtained from SAFE are compared with the theoretical
values.
GEOMETRY, PROPERTIES AND LOADING
Plate size
Plate thickness
Modulus of Elasticity
Poisson's ratio
a×b
T
E
v
=
=
=
=
360" × 240"
8 inches
3000 ksi
0.3
q
P
q1
=
=
=
100 psf
20 kips
1 kip/ft
Load Cases:
(UL)
(PL)
(LL)
Uniform load
Point load
Live load
EXAMPLE 2 - 1
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(3)
q1
(2)
(1)
P
q
a = 30 '
q1
(2)
P
(1)
b = 20 '
(3)
y
x
Figure 2-1 Rectangular Plate with All Edges Fixed
TECHNICAL FEATURES OF SAFE TESTED
Comparison of slab deflection with bench mark solution.
RESULTS COMPARISON
The numerical displacements obtained from SAFE are compared with those
obtained from the theoretical solution in Table 2-1. The theoretical results are
based on tabular values given in Timoshenko and Woinowsky (1959). A
comparison of deflections for the three load cases shows a fast convergence to
the theoretical values with successive mesh refinement.
EXAMPLE 2 - 2
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Table 2-1 Comparison of Displacements
Thin Plate Formulation
Location
Load
Case
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
60
60
0.0098
0.0090
0.0089
60
120
0.0168
0.0153
0.0150
180
60
0.0237
0.0215
0.0210
180
120
0.0413
0.0374
0.0366
60
60
0.0065
0.0053
0.0052
60
120
0.0111
0.0100
0.0100
180
60
0.0315
0.0281
0.0272
180
120
0.0659
0.0616
0.0598
60
60
0.0079
0.0072
0.0071
60
120
0.0177
0.0161
0.0158
180
60
0.0209
0.0188
0.0184
180
120
0.0413
0.0375
0.0367
Theoretical
Displacement
(in)
UL
0.036036
PL
0.057453
LL
EXAMPLE 2 - 3
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Thick Plate Formulation
Location
Load
Case
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
60
60
0.0085
0.0093
0.0091
60
120
0.0147
0.0156
0.0154
180
60
0.0214
0.0219
0.0215
180
120
0.0397
0.0381
0.0374
60
60
0.0083
0.0056
0.0053
60
120
0.0169
0.0101
0.0102
180
60
0.0270
0.0283
0.0278
180
120
0.0545
0.0600
0.0605
60
60
0.0072
0.0073
0.0073
60
120
0.0149
0.0165
0.0163
180
60
0.0198
0.0191
0.0188
180
120
0.0399
0.0382
0.0375
Theoretical
Displacement
(in)
UL
0.036036
PL
0.057453
LL
COMPUTER FILE:
S02a-Thin.FDB. S02b-Thin.FDB, S02c-Thin.FDB, S02a-Thick.FDB. S02bThick.FDB, and S02c-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 2 - 4
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EXAMPLE 3
Rectangular Plate with Mixed Boundary
PROBLEM DESCRIPTION
The plate, shown in Figure 3-1, is analyzed for uniform load only. The edges
along x = 0 and x = a are simply supported, the edge along y = b is free, and the
edge along y = 0 is fully fixed. The geometrical description and material
properties of this problem are the same as those of Example 1. To test
convergence, the problem is analyzed employing three mesh sizes, as shown in
Figure 1-2: 4 × 4, 8 × 8, and 12 × 12. The plate is modeled using plate elements
available in SAFE. The two simply supported edges are modeled as line supports
with large vertical stiffnesses. The fixed edge is modeled as a line support with
large vertical and rotational stiffnesses. The self weight of the plate is not
included in the analysis.
An explicit analytical expression for the deflected surface is given in
Timoshenko and Woinowsky (1959). The deflections obtained from SAFE are
compared with the theoretical values.
GEOMETRY, PROPERTIES AND LOADING
Plate size
a×b
Plate thickness
T
Modulus of elasticity
E
Poisson's ratio
v
Load Cases:
Uniform load
q
=
=
=
=
360" × 240"
8 inches
3000 ksi
0.3
=
100 psf
TECHNICAL FEATURES OF SAFE TESTED
Comparison of deflection with bench-mark solution.
RESULTS COMPARISON
The numerical solution obtained from SAFE is compared with the theoretical
solution that is given by Lévy (Timoshenko and Woinowsky 1959). Comparison
of deflections shows monotonic convergence to the theoretical values with
successive mesh refinement as depicted in Table 3-1. It is to be noted that even
with a coarse mesh (4 × 4) the agreement is very good.
EXAMPLE 3 - 1
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Y
X
Figure 3-1 Rectangular Plate with Two Edges Simply Supported,
One Edge Fixed and One Edge Free
EXAMPLE 3 - 2
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Table 3-1 Comparison of Displacements
Thin Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
180
0
0.0000
0.0000
0.0000
0.0000
180
60
0.0849
0.0831
0.0827
0.08237
180
120
0.2410
0.2379
0.2372
0.23641
180
180
0.3971
0.3947
0.3940
0.39309
180
240
0.5537
0.5511
0.5502
0.54908
Thick Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
180
0
0.0000
0.0000
0.0000
0.0000
180
60
0.0806
0.0841
0.0839
0.08237
180
120
0.2338
0.2398
0.2392
0.23641
180
180
0.3837
0.3973
0.3970
0.39309
180
240
0.5322
0.5544
0.5542
0.54908
COMPUTER FILE:
S03a-Thin.FDB, S03b-Thin.FDB, S03c-Thin.FDB, S03a-Thick.FDB, S03bThick.FDB, and S03c-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 3 - 3
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EXAMPLE 4
Rectangular Plate on Elastic Beams
PROBLEM DESCRIPTION
The plate, shown in Figure 4-1, is analyzed for a uniformly distributed surface
load. The edges along x = 0 and x = a are simply supported, and the other two
edges are supported on elastic beams. It is assumed that the beams resist bending
in vertical planes only and do not resist torsion. A theoretical solution to this
problem is given in Timoshenko and Woinowsky (1959). The deflections of the
plate and the moments and shears of the edge beams are compared with both the
theoretical solution and the results obtained using the Direct Design Method as
outlined in ACI 318-95 for a relative stiffness factor, λ, equal to 4. The relative
stiffness, λ, is the ratio of the bending stiffness of the beam and the bending
stiffness of the slab with a width equal to the length of the beam and is given by
the following equation.
λ=
EI b
, where,
aD
D=
Eh 3
,
12 1 − v 2
(
)
Ib is the moment of inertia of the beam about the horizontal axis,
a is the length of the beam, which also is equal to the one side of the
slab, and
h is the thickness of the slab.
To test convergence of results, the problem is analyzed employing three mesh
sizes, as shown in Figure 1-2: 4 × 4, 8 × 8, and 12 × 12. The slab is modeled
using plate elements. The simply supported edges are modeled as line supports
with a large vertical stiffness and zero rotational stiffness. Beam elements, with
no torsional rigidity, are defined on edges y = 0 and y = b. The flexural stiffness
of edge beams is expressed as a λ factor of the plate flexural stiffness.
EXAMPLE 4 - 1
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SAFE
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The subdivision of the plate into column and middle strips and also the definition
of tributary loaded areas for shear calculations comply with ACI 318-95
provisions and shown in Figure 4-2. A load factor of unity is used and the self
weight of the plate is not included in the analysis.
Y
X
Figure 4-1 Rectangular Plate on Elastic Beams
EXAMPLE 4 - 2
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Column Strip
120"
Edge Beam
120"
MIddle Strip
1 = 240"
Plate
120"
Column Strip
Definition of Strips
a = 360"
Edge Beam
b = 240"
Tributary Loaded Area for Shear on Edge Beams
Figure 4-2 Definition of Slab Strips and Tributary Areas for Shear on Edge Beams
EXAMPLE 4 - 3
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GEOMETRY, PROPERTIES AND LOADING
Plate size
a×b
Plate thickness
T
Modulus of elasticity
E
Poisson's ration
v
Beam moment of inertia
Ib
Relative stiffness parameter
λ
Load Case:
q
=
=
=
=
=
=
360" × 240"
8 inches
3000 ksi
0.3
4
4
= 100 psf (Uniform load)
TECHNICAL FEATURES OF SAFE TESTED
Comparisons of deflection with benchmark solution.
RESULTS COMPARISON
Table 4-1 shows monotonic convergence of SAFE deflections for λ = 4 to the
theoretical values with successive mesh refinement. Table 4-2 shows the
variation of bending moment in the edge beam along its length for λ = 4. The
theoretical solution and the ACI approximation using the Direct Design Method
also are shown.
The value of λ is analogous to the ACI ratio α1l2 / l1 (refer to Sections 13.6.4.4
and 13.6.5.1 of ACI 318-95). The correlation between the numerical results from
SAFE and the theoretical results is excellent. For design purposes, the ACI
approximation (Direct Design Method) compares well with the theory. For the
Direct Design Method, the moments are obtained at the grid points. In obtaining
SAFE moments, averaging was performed at the grid points.
In obtaining the ACI moments, the following quantities were computed:
α1 = EcbIb/EcsIs = 6.59375,
l2/l1 = 240/360 = 0.667,
α1l2/l1 = 4.3958,
βt = 0,
M0 = 2700 k-in.
From ACI section 13.6.4.4 for l2 / l1 = 0.667 and α1l2 / l1 = 4.3958, it is determined
that the column strip supports 85% of the total positive moment. The beam and
slab do not carry any negative moment about the Y-axis because of the simply
supported boundary conditions at x = 0 and x = a.
EXAMPLE 4 - 4
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From ACI section 13.6.5.1 for α1l2 / l1 = 4.3958, it is determined that the beam
carries 85% of the total column strip moment. Since one beam supports only onehalf of the column strip, the maximum beam positive moment is as follows
Mpositivebeam = 0.85 × 0.85 × 0.5 × M0
= 0.36125 × 2700
= 975.375 k-in
The beam moments at other locations are obtained assuming a parabolic variation
along the beam length.
Table 4-3 shows the variation of shear in edge beams for λ = 4. The agreement is
good considering that the SAFE element considers shear strains and the
theoretical solution does not. The ACI values are calculated based on the
definition of loaded tributary areas given in Section 13.6.8.1 of ACI 318-95. The
shear forces were obtained at the middle of the grid points. In obtaining SAFE
shear, no averaging was required for the shear forces.
Table 4-1 Comparison of Displacements
Thin Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
180
120
0.1812
0.1848
0.1854
0.18572
180
60
0.1481
0.1523
0.1530
0.15349
180
0
0.0675
0.0722
0.0730
0.07365
Thick Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
180
120
0.1792
0.1856
0.1862
0.18572
180
60
0.1467
0.1529
0.1536
0.15349
180
0
0.0677
0.0721
0.0730
0.07365
EXAMPLE 4 - 5
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Table 4-2 Variation of Average Bending Moment in an Edge Beam (λ = 4)
Thin Plate Formulation
Location
Y (in)
0, 240
Edge Beam Moment (k-in)
X (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
ACI
Theoretical
0
0.571
0.12
0.05
0
0
30
―
313.0
―
298.031
313.4984
60
590.8
591.4
591.5
541.875
591.6774
120
―
984.9
―
867.000
984.7026
180
1120.9
1120.8
1120.4
975.375
1120.1518
Thick Plate Formulation
Location
Y (in)
0, 240
EXAMPLE 4 - 6
Edge Beam Moment (k-in)
X (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
ACI
Theoretical
0
5.3
31.5
25.2
0
0
30
―
309.2
―
298.031
313.4984
60
591.0
586.8
592.1
541.875
591.6774
120
―
981.3
―
867.000
984.7026
180
1120.2
1116.4
1118.4
975.375
1120.1518
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Table 4-3 Variation of Shear in an Edge Beam (λ = 4)
Thin Plate Formulation
Location
Y (in)
Edge Beam Shear (k)
X (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
ACI
Theoretical
10
―
―
10.58
9.9653
10.6122
15
―
10.4
―
9.9219
10.4954
30
9.80
―
9.96
9.6875
9.9837
9.26
―
9.2969
9.2937
9.02
9.1319
9.0336
7.23
7.7778
7.2458
45
―
―
―
―
―
90
4.40
6.55
―
7.1875
6.5854
120
―
―
4.48
5.0000
4.4821
150
―
2.26
―
2.5000
2.2656
160
―
―
1.51
1.6667
1.5133
50
0, 240
80
EXAMPLE 4 - 7
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Thick Plate Formulation
Location
Y (in)
Edge Beam Shear (k)
X (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
ACI
Theoretical
10
―
―
8.04
9.9653
10.6122
15
―
8.31
―
9.9219
10.4954
30
9.59
―
7.91
9.6875
9.9837
7.57
―
9.2969
9.2937
7.43
9.1319
9.0336
6.39
7.7778
7.2458
45
―
―
―
―
―
90
4.32
6.03
―
7.1875
6.5854
120
―
―
4.06
5.0000
4.4821
150
―
2.08
―
2.5000
2.2656
160
―
―
1.38
1.6667
1.5133
50
0, 240
80
COMPUTER FILE:
S04a-Thin.FDB, S04b-Thin.FDB, S04c-Thin.FDB, S04a-Thick.FDB, S04bThick.FDB, and S04c-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 4 - 8
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EXAMPLE 5
Infinite Flat Plate on Equidistant Columns
PROBLEM DESCRIPTION
The plate, shown in Figure 5-1, is analyzed for uniform load. The overall
dimensions of the plate are significantly larger than the column spacing (a and b
in Figure 5-1). Analysis is limited to a single interior panel because it can be
assumed that deformation is identical for all panels in the plate. An analytical
solution, based on the foregoing assumption, is given in Timoshenko and
Woinowsky (1959).
Three mesh sizes, as shown in Figure 1-2, are used to test the convergence
property of the model: 4 × 4, 8 × 8, and 12 × 12. The model consists of a panel of
uniform thickness supported at four corners point. The effect of column support
within a finite area is not modeled. Due to symmetry, the slope of the deflection
surface in the direction normal to the boundaries is zero along the edges and the
shearing force is zero at all points along the edges of the panel, except at the
corners. To model this boundary condition, line supports with a large rotational
stiffness about the support line are defined on all four edges. Additional point
supports are provided at the corners. The panel is modeled using plate elements
in SAFE. In doing so, the effect of shear distortion is included.
To compare the effects of corner stiffness at the column/slab intersection, a
duplicate model of the 12 x 12 mesh was created where this region is
approximately modeled. This was done by using a special stiff area section in the
region concerned, shown as the 40" × 40" area in Figure 5-2, of which a 20” x
20” portion lies within the modeled region. To obtain design moments, the panel
is divided into three strips both ways, two column strips and one middle strip,
based on the ACI 318-95 definition of design strip widths, as shown in Figure 52 and in Figure 5-3. A load factor of unity is used. The self weight of the panel is
not included in the analysis.
Tables 5-1 through 5-3 show the comparison of the numerically computed
deflection, local moments, and local shears obtained from SAFE with their
theoretical counterparts.
Table 5-4 shows the comparison of the average design strip moments obtained
from SAFE with those obtained from the theoretical method and two ACI
alternative methods: the Direct Design Method (DDM) and the Equivalent Frame
Method (EFM).
EXAMPLE 5 - 1
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Floor Plan
a = 30'
y
b = 20'
Point Support at
Corner
x
Point Support at
Corner
A Typical Bay
Point Support at
Corner
Point Support at
Corner
Figure 5-1 Infinite Plate on Equidistant Columns
and Detail of Panel used in Analysis
Material Properties and Load
Modulus of Elasticity
= 3000 ksi
Poisson's Ratio
= 0.3
Uniform Load
= 100 psf
EXAMPLE 5 - 2
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360"
40" x 40"
Corner Stiffening
Column Strip
120"
120"
Middle Strip
240"
120"
Column Strip
Typical Interior Panel
−M
i
= 1800 k-in
M0 = 2700 k-in
+M
−M
i
= 1422 k-in
m
= 900 k-in
M0 = 2133 k-in
+M
m
Slab Corners Non-Rigid
Slab Corners Rigid
= 711 k-in
Figure 5-2 Definition of X-Strips
(Moment values obtained by EFM)
EXAMPLE 5 - 3
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40" x 40"
Corner Stiffening
Column Strip
120"
Middle Strip
360"
240"
120"
Column Strip
240"
−M
i
= 1200 k-in
M0 = 1800 k-in
+M
−M
i
= 833 k-in
m
= 600 k-in
M0 = 1250 k-in
+M
m
Slab Corners Non-Rigid
Slab Corners Rigid
= 417 k-in
Figure 5-3 Definition of Y-Strips
(Moment values obtained by EFM)
EXAMPLE 5 - 4
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GEOMETRY, PROPERTIES AND LOADING
Plate size
a×b
Plate thickness
T
Modulus of elasticity
E
Poisson's ration
v
Load Case:
q
=
=
=
=
SAFE
0
360" × 240"
8 inches
3000 ksi
0.3
= 100 psf (Uniform load)
TECHNICAL FEATURES OF SAFE TESTED
Comparisons of deflection with benchmark solution.
RESULTS COMPARISON
Table 5-1 shows the comparison of the numerical and the theoretical deflections.
The data indicates monotonic convergence of the numerical solution to the
theoretical values with successive mesh refinement.
The SAFE results for local moment and shear also compare closely with the
theoretical values, as shown in Table 5-2 and Table 5-3, respectively.
In Table 5-4 average strip moments obtained from SAFE are compared with both
the ACI and the theoretical values. EFM is used to calculate the interior span
moments as depicted in Figure 5-2 and Figure 5-3. The agreement between the
SAFE and the theoretical solution is excellent. ACI approximations, employing
either DDM or EFM, however, deviate from the theory. It should be noted that,
regardless of the method used, the absolute sum of positive and negative
moments in each direction equals the total static moment in that direction.
Table 5-5 shows the effect of corner rigidity. Comparisons with the EFM method
are shown.
EXAMPLE 5 - 5
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Table 5-1 Comparison of Displacements
Thin Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
0
0
0.263
0.278
0.280
0.280
0
60
0.264
0.274
0.275
0.275
0
120
0.266
0.271
0.271
0.270
120
0
0.150
0.153
0.153
0.152
120
120
0.101
0.101
0.100
0.098
180
0
0.114
0.108
0.106
0.104
180
60
0.072
0.069
0.067
0.065
180
120
0.000
0.000
0.000
0.000
Thick Plate Formulation
Location
EXAMPLE 5 - 6
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
0
0
0.249
0.279
0.284
0.280
0
60
0.252
0.276
0.280
0.275
0
120
0.252
0.273
0.275
0.270
120
0
0.139
0.155
0.157
0.152
120
120
0.082
0.101
0.103
0.098
180
0
0.094
0.109
0.110
0.104
180
60
0.052
0.069
0.070
0.065
180
120
0.000
0.000
0.000
0.000
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Table 5-2 Comparison of Local Moments
Thin Plate Formulation
Moments (k-in/in)
M11
Location
M22
X (in)
Y (in)
SAFE
(8×8)
Theoretical
SAFE
(8×8)
Theoretical
30
15
3.093
3.266
1.398
1.470
30
105
3.473
3.610
0.582
0.580
165
15
−2.948
−3.142
1.887
1.904
165
105
−9.758
−9.804
−7.961
−7.638
Thick Plate Formulation
Moments (k-in/in)
M11
Location
M22
X (in)
Y (in)
SAFE
(8×8)
Theoretical
SAFE
(8×8)
Theoretical
30
15
3.115
3.266
1.394
1.470
30
105
3.446
3.610
0.583
0.580
165
15
−2.977
−3.142
1.846
1.904
165
105
−9.686
−9.804
−7.894
−7.638
EXAMPLE 5 - 7
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Table 5-3 Comparison of Local Shears
Thin Plate Formulation
Shears (×10−3 k)
V13
Location
V23
X (in)
Y (in)
SAFE
(8×8)
Theoretical
SAFE
(8×8)
Theoretical
30
45
20.9
17.3
8.2
2.2
30
105
21.2
23.5
3.1
5.4
165
15
17.3
14.7
19.1
23.8
165
105
357.1
329.0
350.4
320.0
Thick Plate Formulation
Shears (×10−3 k)
V13
Location
EXAMPLE 5 - 8
V23
X (in)
Y (in)
SAFE
(8×8)
Theoretical
SAFE
(8×8)
Theoretical
30
45
20.2
17.3
8.7
2.2
30
105
24.3
23.5
8.1
5.4
165
15
26.7
14.7
24.7
23.8
165
105
287.5
329.0
277.6
320.0
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Table 5-4 Comparison of Average Strip Moments
Thin Plate Formulation
SAFE Moments
(k-in/in)
Average
Moment
Location
MA
x = 180"
MA
MB
MB
ACI 318-95
(k-in/in)
Strip
4×4
Mesh
8×8
Mesh
12 × 12
Mesh
Theoretical
(k-in/in)
DDM
EFM
Column
4.431
3.999
3.922
3.859
4.725
4.500
Middle
4.302
3.805
3.711
3.641
3.150
3.000
Column
−10.184
−10.865
−10.971
−11.091
−10.968
−11.250
Middle
−3.524
−3.777
−3.843
−3.891
−3.656
−3.750
Column
2.265
2.028
1.971
1.925
3.150
3.000
Middle
1.674
1.561
1.547
1.538
1.050
1.000
Column
−8.236
−8.902
−9.000
−9.139
−7.313
−7.500
Middle
−0.551
−0.449
−0.442
−0.430
−1.219
−1.250
x = 360"
y= 120"
y = 240"
Thick Plate Formulation
SAFE Moments
(k-in/in)
Average
Moment
MA
MA
Location
ACI 318-95
(k-in/in)
Strip
4×4
Mesh
8×8
Mesh
12 × 12
Mesh
Theoretical
(k-in/in)
DDM
EFM
Column
4.802
4.079
3.952
3.859
4.725
4.500
Middle
3.932
3.726
3.682
3.641
3.150
3.000
Column
−8.748
−10.691
−10.993
−11.091
−10.968
−11.250
Middle
−4.965
−3.954
−3.823
−3.891
−3.656
−3.750
x = 180"
x = 360"
EXAMPLE 5 - 9
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Thick Plate Formulation
SAFE Moments
(k-in/in)
Average
Moment
MB
MB
Location
y= 120"
y = 240"
ACI 318-95
(k-in/in)
Strip
4×4
Mesh
8×8
Mesh
12 × 12
Mesh
Theoretical
(k-in/in)
DDM
EFM
Column
2.361
2.078
2.000
1.925
3.150
3.000
Middle
1.628
1.537
1.533
1.538
1.050
1.000
Column
−6.321
−8.670
−9.025
−9.139
−7.313
−7.500
Middle
−1.514
−0.567
−0.431
−0.430
−1.219
−1.250
Table 5-5 Comparison of Average Strip Moments : Effect of Corner Rigidity
Thin Plate Formulation
SAFE Moments
(12×12 Mesh)
(k-in/in)
Average
Moment
Location
MA
x = 180"
MA
MB
MB
EXAMPLE 5 - 10
ACI 318-95
(EFM Method)
(k-in/in)
Strip
Slab Corner
Non-Rigid
Slab Corner
Rigid
Slab Corner
Non-Rigid
Slab Corner
Rigid
Column
3.922
3.472
4.500
3.555
Middle
3.711
3.285
3.000
2.370
Column
−10.971
−8.110
—
−8.887
Middle
−3.843
−2.863
—
−2.962
Column
1.971
1.470
3.000
2.085
Middle
1.547
1.361
1.000
0.695
Column
−4.807
−5.489
—
−5.206
Middle
−0.272
−0.347
—
−0.867
x = 360"
y= 120"
y = 240"
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Thick Plate Formulation
SAFE Moments
(12×12 Mesh)
(k-in/in)
Average
Moment
MA
MA
Location
Strip
Slab Corner
Non-Rigid
Slab Corner
Rigid
Slab Corner
Non-Rigid
Slab Corner
Rigid
Column
3.952
3.459
4.500
3.555
Middle
3.682
3.219
3.000
2.370
Column
−10.993
−8.249
—
−8.887
Middle
−3.823
−2.806
—
−2.962
Column
2.000
1.456
3.000
2.085
Middle
1.533
1.327
1.000
0.695
Column
−9.025
−5.742
—
−5.206
Middle
−0.431
−0.263
—
−0.867
x = 180"
x = 360"
MB
MB
ACI 318-95
(EFM Method)
(k-in/in)
y= 120"
y = 240"
COMPUTER FILE:
S05a-Thin.FDB, S05b-Thin.FDB, S05c-Thin.FDB, S05d.FDB, S05a-Thick.FDB,
S05b-Thick.FDB, S05c-Thick.FDB, and S05d-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 5 - 11
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EXAMPLE 6
Infinite Flat Plate on Elastic Subgrade
PROBLEM DESCRIPTION
An infinite plate resting on elastic subgrade and carrying equidistant and equal
loads, P, is shown in Figure 6-1. Each load is assumed to be distributed
uniformly over the area u × v of a rectangle. A theoretical double series solution
to this example is given in Timoshenko and Woinowsky (1959).
The numerically computed deflections and local moments obtained from SAFE
are compared to the theoretical values, as shown in Table 6-1 and Table 6-2.
Analysis is confined to a single interior panel. To model the panel, three mesh
sizes, as shown in Figure 1-2, are used: 4 × 4, 8 × 8, and 12 × 12. The slab is
modeled using plate elements and the elastic support is modeled as a surface
support with a spring constant of k, the modulus of subgrade reaction. The edges
are modeled as line supports with a large rotational stiffness about the support
line. Point loads P/4 are defined at the panel corners. In the theoretical
formulation (Timoshenko and Woinowsky 1959), each column load P is assumed
to be distributed over an area u × v of a rectangle, as shown in Figure 6-1. To
apply the theoretical formulation to this problem, concentrated corner loads are
modeled as a uniformly distributed load acting over a very small rectangular area
where u and v are very small.
GEOMETRY, PROPERTIES AND LOADING
Plate size
a×b
Plate thickness
T
Modulus of elasticity
E
Poisson's ratio
v
Modulus of subgrade reaction k
=
=
=
=
=
360" × 240"
15 inches
3000 ksi
0.2
1 ksi/in
Loading: Point Load
P = 400 kips
(assumed to be uniformly distributed over an area u × v)
EXAMPLE 6 - 1
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v
2
u
(Typ.)
2
u
2
(Typ.)
P
uv
v
2
P
uv
P
uv
FLOOR PLAN
b = 20'
a = 30'
v
2
v
2
Y
X
T = 15"
u
2
u
2
A Typical Panel
Figure 6-1 Rectangular Plate on Elastic Subgrade
EXAMPLE 6 - 2
k
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TECHNICAL FEATURES OF SAFE TESTED
Comparison of deflection on elastic foundation.
RESULTS COMPARISON
Good agreement has been found between the numerical and theoretical deflection
for k = 1 ksi/in, as shown in Table 6-1, except near the concentrated load. The
consideration of shear strains in the SAFE element makes it deflect more near the
concentrated load. As the modulus k is changed, the distribution of pressure
between the plate and the subgrade changes accordingly. The particular case, as k
approaches 0, corresponds to a uniformly distributed subgrade reaction, i.e., to
the case of a “reversed flat slab” uniformly loaded with q = P/ab. In fact the
problem changes to that of Example 5, with the direction of vertical axis
reversed. In Example 5, for a uniform load of 100 psf (P = 60 kips), the
maximum relative deflection is calculated as 0.280. Applying the formulation
used here with k = 1 × 10-6 yields a deflection value of 0.279". Table 6-2 shows
the comparison of the SAFE local moments using the 12 × 12 mesh with the
theoretical results. The results agree well.
Table 6-1 Comparison of Displacements
Thin Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
0
0
−0.0493
−0.05410
−0.05405
−0.05308
180
60
0.00091
0.00076
0.00080
0.00096
180
120
0.00040
0.00060
0.00064
0.00067
Thick Plate Formulation
Location
SAFE Displacement (in)
X (in)
Y (in)
4×4 Mesh
8×8 Mesh
12×12 Mesh
Theoretical
Displacement
(in)
0
0
−0.0436
−0.06011
−0.06328
−0.05308
180
60
0.00130
0.00074
0.00076
0.00096
180
120
−0.0019
0.00050
0.00059
0.00067
EXAMPLE 6 - 3
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Table 6-2 Comparison of Local Moments
Thin Plate Formulation
Moments (kip-in/in)
M11
Location
M22
X (in)
Y (in)
SAFE
(12×12)
Theoretical
SAFE
(12×12)
Theoretical
10
10
37.99
35.97
37.97
35.56
10
50
7.38
7.70
−6.74
−6.87
10
110
−0.30
−0.27
−5.48
−5.69
80
10
−6.52
−6.89
1.98
1.72
80
50
−3.58
−3.78
−0.93
−1.02
80
110
−0.88
−0.98
−1.86
−1.69
Thick Plate Formulation
Moments (kip-in/in)
M11
Location
EXAMPLE 6 - 4
M22
X (in)
Y (in)
SAFE
(12×12)
Theoretical
SAFE
(12×12)
Theoretical
10
10
36.77
35.97
36.73
35.56
10
50
7.13
7.70
−6.37
−6.87
10
110
−0.21
−0.27
−5.17
−5.69
80
10
−6.11
−6.89
2.05
1.72
80
50
−3.56
−3.78
−0.82
−1.02
80
110
−0.87
−0.98
−1.86
−1.69
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COMPUTER FILE:
S06a-Thin.FDB, S06b-Thin.FDB, S06c-Thin.FDB, S06a-Thick.FDB, S06bThick.FDB and S06c-Thick.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 6 - 5
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EXAMPLE 7
Skewed Plate with Mixed Boundary
PROBLEM DESCRIPTION
A skewed plate under uniform load, as shown in Figure 7-1, is analyzed for two
different support configurations. In the first case, all the edges are assumed to be
simply supported. In the second case, the edges y = 0 and y = b are released, i.e.,
the plate is assumed to be supported on its oblique edges only. A theoretical
solution to this problem is given in Timoshenko and Woinowsky (1959). In both
cases, the maximum deflection and the maximum moment are compared with the
corresponding theoretical values.
An 8 × 24 base mesh is used to model the plate, as shown in Figure 7-1. A large
vertical stiffness is defined for supports, and support lines are added on all four
edges for the first case and along the skewed edges only for the second case. A
load factor of unity is used. The self weight of the plate is not included in the
analysis.
GEOMETRY, PROPERTIES, AND LOADING
Plate size
Plate thickness
Modulus of elasticity
Poisson’s ratio
Load Cases:
a×b
T
E
v
Uniform load, q
=
=
=
=
480" × 240"
8 inches
3,000 ksi
0.2
= 100 psf
TECHNICAL FEATURES OF SAFE TESTED
Comparison of deflection and moments on skewed plate.
EXAMPLE 7 - 1
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40'
20'
20'
Geometry
16 @ 2.5'
8 @ 2.5'
8 @ 2.5'
Y
Mesh
Support Conditions:
(1) Simply supported on all edges
(2) Simply supported on oblique edges
Figure 7-1 Skew Plate
EXAMPLE 7 - 2
X
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RESULTS COMPARISON
Under the simply supported boundary condition, maximum deflection occurs at
the plate center and the maximum principal moment acts nearly in the direction
of the short span. Under the simply supported condition on the oblique edges and
free boundary conditions on the other two edges, maximum deflection occurs at
the free edge as expected.
Table 7-1 Comparison of Deflections and Bending Moments
Boundary
Condition
Simply supported
on all edges
Simply supported
on oblique edges
Simply supported
on oblique edges
Responses
SAFE
Theoretical
Thin
Plate
Thick
Plate
Maximum displacement
(inches)
0.156
0.160
0.162
Maximum bending moment
(k-in)
3.66
3.75
3.59
Maximum displacement at
the free edges (in)
1.51
1.52
1.50
Maximum bending moment
of the free edges (k-in)
12.03
12.28
11.84
Displacement at the center
(in)
1.21
1.23
1.22
Maximum bending moment
at the center (k-in)
11.78
11.81
11.64
COMPUTER FILES
S07a-Thin.FDB, S07b-Thin.FDB, S07a-Thick.FDB and S07b-Thick.FDB
CONCLUSION
The comparison of SAFE and the theoretical results is acceptable, as shown in
Table 7-1.
EXAMPLE 7 - 3
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EXAMPLE 8
ACI Handbook Flat Slab Example 1
PROBLEM DESCRIPTION
The flat slab system, arranged three-by-four, is shown in Figure 8-1. The slab
consists of twelve 7.5-inch-thick 18' × 22' panels. Edge beams on two sides
extend 16 inches below the slab soffit. Details are shown in Figure 8-2. There are
three sizes of columns and in some locations, column capitals. Floor to floor
heights below and above the slab are 16 feet and 14 feet respectively. A full
description of this problem is given in Example 1 of ACI 340.R-97 (ACI
Committee 340, 1997). The total factored moments in an interior E-W design
frame obtained from SAFE are compared with the corresponding results obtained
by the Direct Design Method, the Modified Stiffness Method, and the Equivalent
Frame Method.
The computational model uses a 10 × 10 mesh of elements per panel, as shown in
Figure 8-3. The mesh contains gridlines at column centerlines, column faces, and
the edges of column capitals. The grid lines extend to the slab edges. The regular
slab thickness is 7.5". A slab thickness of 21.5" is used to approximately model a
typical capital. The slab is modeled using plate elements. The columns are
modeled as point supports with vertical and rotational stiffnesses. Stiffness
coefficients used in the calculation of support flexural stiffness are all reproduced
from ACI Committee 340 (1997). Beams are defined on two slab edges, as
shown in Figure 8-1.
The model is analyzed for a uniform factored load of 0.365 ksf (wu = 1.4wd + 1.7
wt) in total, including self weight. To obtain factored moments in an E-W interior
design frame, the slab is divided into strips in the X-direction (E-W direction), as
shown in Figure 8-4. An interior design frame consists of one column strip and
two halves of adjacent middle strips.
EXAMPLE 8 - 1
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C
B
A
22'-0"
22'-0"
D
22'-0"
1
Edge beam
18'-0"
S1
S4
S7
S2
S5
S8
2
18'-0"
4'-0"
2'-6"
3
Design Frame
4'-0"
A
4
18'-0"
5
Edge beam
18'-0"
S2
S5
S8
S3
S6
S9
3'-6"
No edge beams on lines D and 5
Figure 8-1 Flat Slab from ACI Handbook
EXAMPLE 8 - 2
A
3'-6"
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A
B
D
C
22'-0"
22'-0"
19'-4"
22'-0"
18'-0"
18'-3"
Col. 16"x18"
Col. 18"x18"
Col. 18"x18"
Col. 16"x18"
Col. 20"x20"
Col. 18"x18"
Detail "a"
Detail "b"
Detail "c"
SECTION A—A
D
B
A
28"
30"
48"
16"
18"
18"
7.5"
14"
23.5"
20"
12"
Detail "a"
Detail "b"
Detail "c"
Figure 8-2 Sections and Details of ACI Handbook Flat Slab Example
EXAMPLE 8 - 3
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Figure 8-3 SAFE Mesh (10 × 10 per panel)
EXAMPLE 8 - 4
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A
B
C
5'-2"
SAFE
0
D
9'-8"
1
Exterior
Design Frame
2
18'-0"
Interior
Design Frame
7 @ 9'-0"
3
4
5
5'-3"
Interior Column Strip
Middle Strip
Exterior Column Strip
Figure 8-4 Definition of E-W Design Frames and Strips
EXAMPLE 8 - 5
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Figure 8-5 Comparison of Total Factored Moments (E-W Design Frame)
EXAMPLE 8 - 6
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GEOMETRY, PROPERTIES AND LOADING
Materials:
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Poisson's ratio
fc'
fy
γc
Ec
v
=
=
=
=
=
3
40
150
3320
0.2
SAFE
0
ksi
ksi
pcf
ksi
TECHNICAL FEATURES OF SAFE TESTED
Comparison of factored moments in slab.
RESULTS COMPARISON
The SAFE results for the total factored moments in an interior E-W design frame
are compared in Figure 8-5 with the results obtained by the Direct Design
Method (DDM), the Modified Stiffness Method (MSM), and the Equivalent
Frame Method (EFM). Only uniform loading with load factors of 1.4 and 1.7 has
been considered. The DDM, MSM, and EFM results are all reproduced from
Example 1 of ACI Committee 340 (1997), the Alternative Example 1 of ACI
Committee 340 (1991), and from Example 3 of ACI Committee 340 (1991),
respectively. Moments reported are calculated at the face of column capitals.
Overall, they compare well. A noticeable discrepancy is observed in the negative
column moment in the west side of the exterior bay (the edge beam side). In
contrast to the EFM, the DDM appears to underestimate this moment. The SAFE
result are between the two extreme values. The basic cause of this discrepancy is
the way in which each method accounts for the combined flexural stiffness of
columns framing into the joint. The DDM uses a stiffness coefficient kc of 4 in
the calculation of column and slab flexural stiffnesses. The EFM, on the other
hand, uses higher value of kc to allow for the added stiffness of the capital and
the slab-column joint. The use of MSM affects mainly the exterior bay moments,
which is not the case when the DDM is employed. In SAFE, member
contributions to joint stiffness are dealt with more systematically than any of the
preceding approaches. Hence, the possibility of over designing or under
designing a section is greatly reduced.
The factored strip moments are compared in Table 8-1. There is a discrepancy in
the end bays, particularly on the edge beam (west) side, where the SAFE and
EFM results for exterior negative column strip moment show the greatest
difference. This is expected because EFM simplifies a 3D structure to a 2D
structure, thereby neglecting the transverse interaction between adjacent strips.
Except for this localized difference, the comparison is good.
EXAMPLE 8 - 7
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Table 8-1 Comparison of Total Factored Strip Moments (k-ft) (Interior E-W Design Frame)
Factored Strip Moment (k-ft)
Span AB
Strip
Column
Strip
Middle
Strip
Method
−
M
+M
DDM
86
92
MSM
122
EFM
Span BC
−
−
Span CD
M
+M
−
130
143
85
71
56
130
140
72
117
144
44
145
161
62
125
159
128
58
121
138
72
88
62
54
43
37
43
48
57
0
10
55
52
43
37
43
46
48
0
EFM
10
55
53
48
29
48
54
41
0
SAFE
7
78
62
51
48
46
52
62
13
M
+M
161
130
56
83
157
130
140
83
157
SAFE
69
85
DDM
6
MSM
M
−
M
−
M
COMPUTER FILE:
S08.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 8 - 8
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EXAMPLE 9
ACI Handbook Two-Way Slab Example 2
PROBLEM DESCRIPTION
The two-way slab system arranged three-by-three is shown in Figure 9-1. The
slab consists of nine 6.5-inch-thick 20-foot × 24-foot panels. Beams extend 12
inches below the slab soffit. Details are shown in Figure 9-2. Sixteen inch × 16
inch columns are used throughout the system. Floor to floor height is 15 feet. A
full description of this problem is given in Example 2 of ACI 340.R-91 (ACI
Committee 340, 1991). The total factored moments in an interior design frame
obtained from SAFE are compared with the Direct Design Method, the Modified
Stiffness Method, and the Equivalent Frame Method.
The computational model uses a 10 × 10 mesh of elements per panel, as shown in
Figure 9-3. The mesh contains grid lines at both column centerlines and column
faces. The grid lines are extended to the slab edges. The slab is modeled using
plate elements. The columns are modeled as point supports with vertical and
rotational stiffnesses. A stiffness coefficient of 4 EI/L is used in the calculation of
support flexural stiffness. Torsional constants of 4790 in4 and 5478 in4 are
defined for the edge and interior beams respectively, in accordance with Section
13.7.5 of ACI 318-89 and Section 13.0 of ACI 318-95 code. The model is
analyzed for uniform factored total load of 0.347 ksf (wu = 1.4wd + 1.7w1),
including self weight. To obtain factored moments in an interior design frame,
the slab is divided into strips in the X-direction (E-W direction), as shown in
Figure 9-4. An interior design frame consists of one column strip and two halves
of adjacent middle strips.
GEOMETRY, PROPERTIES AND LOADING
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Poisson's ratio
Live load
Mechanical load
Exterior wall load
fc'
fy
wc
Ec
v
=
3
=
40
= 150
= 3120
=
0.2
w1
=
wd
=
wwall =
ksi
ksi
psf
ksi
125 psf
15 psf
400 plf
EXAMPLE 9 - 1
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A
B
20'-0"
C
20'-0"
D
20'-0"
1
24'-0"
2
A
24'-0"
Column
16"x16"
(Typ.)
3
24'-0"
4
Figure 9-1 ACI Handbook Two-Way Slab Example
EXAMPLE 9 - 2
A
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20'-0"
20'-0"
SAFE
0
20'-0"
Col. 16"x16"
(Typ.)
Detail 'a'
Detail 'b'
SECTION A-A
A
B
22"
16"
16"
6.5"
12"
18.5"
12"
10"
Detail 'a'
Detail 'b'
Figure 9-2 Details of Two-Way Slab Example from ACI Handbook
EXAMPLE 9 - 3
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Figure 9-3 SAFE Mesh (10 × 10 per panel)
EXAMPLE 9 - 4
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A
1
B
C
SAFE
0
D
5' - 8"
12' - 8"
Exterior
Design Frame
14' - 0"
24' - 0" Interior
2 10' - 0"
Design Frame
14' - 0"
3 10' - 0"
14' - 0"
4
5' - 8"
Interior Column Strip
Middle Strip
Exterior Column Strip
Figure 9-4 Definition of E-W Design Frames and Strips
EXAMPLE 9 - 5
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Figure 9-5 Comparison of Total Factored Moments (k-ft)
in an Interior E-W Design Frame
EXAMPLE 9 - 6
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TECHNICAL FEATURES OF SAFE TESTED
Calculation of factored moments in slab.
RESULTS COMPARISON
The SAFE results for the total factored moments in an interior E-W design frame
are compared with the results obtained by the Direct Design Method (DDM), the
Modified Stiffness Method (MSM), and the Equivalent Frame Method (EFM) as
shown in Figure 9-5. The results are for uniform loading with load factors. The
results are reproduced from ACI Committee 340 (1991). Moments reported are
calculated at the column face. For all practical purposes they compare well. At
the end bays, the MSM appears to overestimate the exterior column negative
moments with the consequent reduction in the mid-span moments.
The distribution of total factored moments to the beam, column strip, and middle
strip is shown in Table 9-1. The middle strip moments compare well. The total
column strip moments also compare well. The distribution of the column strip
moments between the slab and the beam has a larger scatter.
EXAMPLE 9 - 7
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Table 9-1 Comparison of Total Factored Moments (kip-ft)
Total Factored Moments in an E-W Design Frame(kip-ft)
Exterior Span
Strip
Method
Slab
Column
Strip
Slab
Middle
Strip
−
Interior Span
M
+M
−
M
+M
−
DDM
9
23
28
25
14
25
MSM
13
21
28
25
14
25
EFM
12
21
30
27
11
27
SAFE
22
27
62
58
14
58
DDM
3
69
84
76
41
76
MSM
5
63
84
76
41
76
EFM
4
63
89
82
34
82
SAFE
6
71
73
73
49
73
DDM
50
129
160
143
77
143
MSM
72
121
160
143
77
143
EFM
68
119
169
156
66
156
SAFE
62
102
141
122
60
122
M
−
M
Beam
COMPUTER FILE:
S09.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 9 - 8
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EXAMPLE 10
PCA Flat Plate Test
PROBLEM DESCRIPTION
This example models the flat plate structure tested by the Portland Cement
Association (Guralnick and LaFraugh 1963). The structure consists of nine 5.25inch-thick 15-foot × 15-foot panels arranged 3 × 3, as shown in Figure 10-1.
Deep and shallow beams are used on the exterior edges. The structure is
symmetric about the diagonal line through columns A1, B2, C3, and D4, except
the columns themselves are not symmetric about this line. The corner columns
are 12 inches × 12 inches and the interior columns are 18 inches × 18 inches.
Columns along the edges are 12 inches × 18 inches, with the longer dimension
parallel to the plate edge. A typical section of the plate and details of edge beams
are given in Figure 10-2. The total moments in an interior frame obtained
numerically from SAFE are compared with the test results and the numerical
values obtained by the Equivalent Frame Method (EFM).
A finite element model, shown in Figure 10-3, with 6 × 6 mesh per panel is
employed in the analysis. The slab is modeled using the plate elements in SAFE.
The columns are modeled as point supports with vertical and rotational
stiffnesses. The reduced-height columns in the test structure are fixed at the base.
Hence, rotational stiffnesses of point supports are calculated using a stiffness
coefficient of 4 and an effective height of 39.75 inches (Kc = 4EI / lc). In order to
account for rigidity of the slab-column joint, the portion of slab occupying the
column area is modeled as rigid by using a special stiff area element. A total
uniformly distributed design load of 156 psf (not factored) is applied to all the
panels.
To obtain design moment coefficients, the plate is divided into column and
middle strips. An interior design frame consists of one column strip and half of
each adjacent middle strip. Normalized values of design moments are used in the
comparison.
EXAMPLE 10 - 1
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B
A
15'− 0"
D
C
15'− 0"
15'− 0"
12"
Shallow Beam 12" x 8 ¼"
15'− 0"
2
A
15'− 0"
Shallow Beam 12" x 8 ¼"
12"
18"
18"
12"
Design Frame
Shallow Beam Side
Deep Beam 6" x 15 ¾"
1
18"
18"
3
Design Frame
Deep Beam Side
15'− 0"
4
Deep Beam 6" x 15 ¾"
Figure 10-1 PCA Flat Plate Example
EXAMPLE 10 - 2
12"
A
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B
15'−0"
C
15'−0"
D
15'−0"
Shallow Beam
Deep Beam
8 ¼"
12"
SAFE
0
15 ¾"
18"
5 ¼"
18"
3' −3 ¾"
12"
2'−9"
Anchor Bolts
SECTION A-A
Figure 10-2 Section and Details of PCA Flat Plate Example
Figure 10-3 SAFE Mesh (6 × 6 per panel)
EXAMPLE 10 - 3
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GEOMETRY, PROPERTIES AND LOADING
Concrete strength
fc'
Yield strength of steel fy
Concrete unit weight
wc
Modulus of elasticity
Ec
Poisson's ratio
v
Live load
Dead load
=
=
=
=
=
w1 =
wd =
4.1
40
150
3670
0.2
ksi
ksi
pcf
ksi
70 psf
86 psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of factored forces in slab.
RESULTS COMPARISON
The SAFE results for the total non-factored moments in an interior frame are
compared with test results and the Equivalent Frame Method (EFM). The test
and EFM results are all obtained from Corley and Jirsa (1970). The moments are
compared in Table 10-1. The negative design moments reported are at the faces
of the columns. Overall, the agreement between the SAFE and EFM results is
good. The experimental negative moments at exterior sections, however, are
comparatively lower. This may be partially the result of a general reduction of
stiffness due to cracking in the beam and column connection at the exterior
column, which is not accounted for in an elastic analysis. It is interesting to note
that even with an approximate representation of the column flexural stiffness, the
comparison of negative exterior moments between EFM and SAFE is excellent.
EXAMPLE 10 - 4
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Table 10-1 Comparison of Measured and Computer Moments
Moments in an Interior Design Frame (M / Wl1 *)
End Span
(Shallow Beam Side)
−
M
+M
PCA Test
0.037
0.047
EFM
0.044
SAFE
(Shallow Beam Slide)
SAFE
(Deep Beam Slide)
Method
−
−
M
+M
0.068
0.068
0.031
0.048
0.067
0.062
0.040
0.051
0.069
0.040
0.051
0.068
M
End Span
(Deep Beam Side)
Middle Span
−
−
−
M
+M
0.073
0.073
0.042
0.031
0.038
0.062
0.068
0.049
0.043
0.062
0.041
0.062
0.068
0.052
0.039
0.062
0.041
0.062
0.068
0.052
0.039
M
M
* Wl1 = 526.5 kip-ft
COMPUTER FILE:
S10.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 10 - 5
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EXAMPLE 11
University of Illinois Flat Plate Test F1
PROBLEM DESCRIPTION
This example models the flat plate structure tested at the University of Illinois by
Hatcher, Sozen, and Siess (1965). The structure consists of nine 1.75-inch-thick
5-foot × 5-foot panels arranged 3 × 3 as shown in Figure 11-1. Two adjacent
edges are supported by 2.00-inch-wide × 5.25-inch-deep beams and the other two
edges by shallow beams, 4 inches wide by 2.75 inches deep, producing a single
diagonal line of symmetry through columns A1, B2, C3, and D4. A typical
section and details of columns and edge beams are shown in Figure 11-2. The
moments computed numerically using SAFE are compared with the test results
and the EFM results.
The computational model uses a 6 × 6 mesh of elements per panel, as shown in
Figure 11-3. The mesh contains grid lines at column centerlines as well as
column faces. The slab is modeled using slab area elements and the columns are
modeled as point supports with vertical and rotational stiffnesses. The reducedheight columns in the test structure are pinned at the base. Hence, an approximate
value of 3(Kc = 3EI/lc) is used to calculate flexural stiffness of the supports,
taking the column height as 9.5 inches. In order to account for rigidity of the
slab-column joint, the portion of slab occupying the column area is modeled as
rigid by using a special stiff area element. Shallow and deep beams are defined
on the edges with properties derived from cross-section geometry. The model is
analyzed for uniform total load of 140 psf.
To obtain maximum factored moments in an interior design frame, the plate is
divided into columns and middle strips. An interior design frame consists of one
column strip and half of each adjacent middle strip.
GEOMETRY, PROPERTIES AND LOADING
Material:
Concrete strength
Yield strength of steel
Modulus of elasticity
Poisson’s ratio
f'c
fy
Ec
v
Loading:
Total uniform load
w =
=
2.5 ksi
= 36.7 ksi
= 2400 ksi
=
0.2
140 psf
EXAMPLE 11 - 1
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Figure 11-1 University of Illinois Flat Plate Test F1
EXAMPLE 11 - 2
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A
C
B
5' − 0"
Shallow Beam
D
5' − 0"
5' − 0"
Deep Beam
2¾"
11¼"
SAFE
0
5¼"
4"
Side Col. 4"x 6"
6"
1¾"
6"
Interior Col. 6" sq.
4"
Corner Col. 4" sq.
SECTION A-A
Figure 11-2 Sections and Details of University of Illinois Flat Plate Test F1
Figure 11-3 SAFE Mesh (6 × 6 per panel)
EXAMPLE 11 - 3
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TECHNICAL FEATURES OF SAFE TESTED
Calculation frame moments for uniform loading.
RESULTS COMPARISON
Table 11-1 shows the comparison of the SAFE results for uniform load moments
for an interior frame with the experimental and EFM results. The experimental
and EFM results are all obtained from Corley and Jirsa (1970).
Table 11-1 Comparison of Measured and Computed Moments
Moments in an Interior Design Frame (M/Wl1 *)
End Span
(Shallow Beam Side)
−
Method
M
+M
TEST F1
0.027
EFM
−
End Span
(Deep Beam Side)
Middle Span
M
-M
+M
0.049
0.065
0.064
0.040
0.047
0.044
0.072
0.066
SAFE
(Shallow Beam Side)
0.044
0.047
0.066
SAFE
(Deep Beam Side)
0.043
0.047
0.064
−
−
−
M
+M
0.058
0.058
0.047
0.034
0.034
0.067
0.073
0.044
0.046
0.060
0.039
0.059
0.065
0.048
0.043
0.059
0.039
0.058
0.064
0.047
0.042
M
M
* Wl1 = 17.5 kip-ft
The negative design moments reported are at the faces of the columns. From a
practical standpoint, even with a coarse mesh, the agreement between the SAFE
and EFM results is good. In general the experimentally obtained moments at
exterior sections are low, implying a loss of stiffness in the beam-column joint
area.
In comparing absolute moments at a section, the sum of positive and average
negative moments in the bay should add up to the total static moment. The SAFE
and EFM results comply with this requirement within an acceptable margin of
accuracy. The experimental results are expected to show greater discrepancy
because of the difficulty in taking accurate strain measurements.
COMPUTER FILE: S11.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 11 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE 12
University of Illinois Flat Slab Tests F2 and F3
PROBLEM DESCRIPTION
This example models F2 and F3, the flat slab structures tested at the University
of Illinois by Hatcher, Sozen, and Siess (1969) and Jirsa, Sozen, and Siess (1966)
respectively. A typical structure used in tests F2 and F3 is shown in Figure 12-1.
The fundamental difference between these two test structures is in the type of
reinforcement used. In test F2, the slab is reinforced with medium grade
reinforcement, whereas in test F3, welded wire fabrics are used. The structure
consists of nine 5-foot × 5-foot panels arranged 3 × 3. Two adjacent edges are
supported by deep beams, 2 inches wide by 6 inches deep, and the other two
edges by shallow beams, 4.5 inches wide by 2.5 inches deep, producing a single
diagonal line of symmetry through columns A1, B2, C3, and D4. A typical
section and details of columns, drop panels, and column capitals are shown in
Figure 12-2. For both structures, the numerical results obtained for an interior
frame by SAFE are compared with the experimental results and the EFM results
due to uniformly distributed load.
The computational model uses an 8 × 8 mesh of elements per panel, as shown in
Figure 12-3. The mesh contains grid lines at the column centerlines as well as the
edges of drop panels and interior column capitals. The slab thickness is increased
to 2.5 inches over the drop panels. A thickness of 4.5 inches is used to
approximately model the interior capitals. Short deep beams are used to model
the edge column capitals. In this model, the slab is modeled using plate elements
and the columns are modeled as point supports with vertical and rotational
stiffnesses. A stiffness coefficient of 4.91 (Kc = 4.91EIc / lc) is used in the
calculation of the support flexural stiffness based on a column height of 21.375
inches, measured from the mid-depth of the slab to the support center. Due to the
presence of capitals, columns are treated as non-prismatic. Shallow and deep
beams are defined on the edges with properties derived from their cross-section
geometry.
The test problems use two different concrete moduli of elasticity, Ec = 2100ksi
and Ec = 3700 ksi for the beams and slab. However, both test problems are
modeled in SAFE with concrete modulus of elasticity of 2100 ksi. This affects
the slab, beam, and column stiffness since the distribution of moment depends on
the relative stiffness.
EXAMPLE 12 - 1
Software Verification
SAFE
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PROGRAM NAME:
REVISION NO.:
A
B
5'− 0"
1
C
5'− 0"
Shallow Beam
D
5'− 0"
A
A
5'− 0"
3
5'− 0"
Deep Beam
2
Shallow Beam
5'− 0"
Design Frame
Deep Beam Side
Design Frame
Deep Beam Side
4
Deep Beam
Figure 12-1 University of Illinois Flat Slab Tests F2 and F3
The model is analyzed for uniform load. To obtain maximum factored moments
in an interior design frame, the slab is divided into two interior and two exterior
design frames spanning in the X direction (E-W direction). Because of
symmetry, results are shown for X strips only. An interior design frame consists
of one column strip and half of each adjacent middle strip.
EXAMPLE 12 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
A
B
5'− 0"
2½"
D
C
5'− 0"
5'− 0"
Deep Beam
Shallow Beam
1¾"
Side Column
3½" x 5"
SAFE
0
Interior Column
3¾" sq.
1' − 10¼"
Corner Column
3½" sq.
SECTION A−A
20"
12"
0.5"
1.75"
Drop panel
Capital
22.25"
1C = 17.375"
INTERIOR COLUMNS
Figure 12-2 Sections and Details of Flat Slabs F2 and F3
EXAMPLE 12 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Figure 12-3 SAFE Mesh (8 × 8 per mesh)
GEOMETRY, PROPERTIES AND LOADING
Concrete strength:
fc′ = 2.76 ksi (Test F2)
fc′ = 3.76 ksi (Test F3)
Yield strength of slab reinforcement:
fy = 49 ksi (Test F2)
fy = 54 ksi (Test F3)
EXAMPLE 12 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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Modulus of elasticity:
Ec = 2100 ksi (Test F2)
Ec = 3700 ksi (Test F3)
Poisson’s ratio:
ν = 0.2
Loading:
Total uniform design load, w = 280 psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation frame moments.
RESULTS COMPARISON
Table 12-1 shows the comparison of the SAFE results for moments in an interior
frame with the experimental and EFM results for both structures F2 and F3. The
experimental and EFM results are all obtained from Corley and Jirsa (1970).
Table 12-1 Comparison of Measured and Computer Moments
Moments in an Interior Design Frame (M / Wl1 *)
End Span
(Shallow Beam Side)
Method
−
M
+
M
−
M
End Span
(Deep Beam Side)
Middle Span
−
M
+
M
−
M
−
M
+
M
−
M
TEST F2
0.025
0.042
0.068
0.062
0.029
0.061
0.065
0.038
0.025
TEST F3
0.029
0.038
0.057
0.055
0.023
0.058
0.060
0.034
0.024
EFM
0.021
0.044
0.057
0.050
0.026
0.049
0.057
0.044
0.021
SAFE
(Shallow Beam Side)
0.026
0.042
0.067
0.058
0.025
0.057
0.066
0.042
0.024
SAFE
(Deep Beam Side)
0.026
0.041
0.066
0.057
0.024
0.057
0.066
0.042
0.024
* Wl1 = 35.0 k-ft
EXAMPLE 12 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Moments are compared at the edge of column capitals. Table 12-1 shows that the
SAFE and the EFM results are in excellent agreement. In general, the measured
positive moments appear to be lower than the SAFE and EFM values.
COMPUTER FILE: S12.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 12 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE 13
University of Illinois Two-Way Slab Test T1
PROBLEM DESCRIPTION
This example models the slab structure tested at the University of Illinois by
Gamble, Sozen, and Siess (1969). The structure is a two-way slab, 1.5 inches
thick, in which each panel is supported along all four edges by beams, as shown
in Figure 13-1. The structure consists of nine 5-foot × 5-foot panels arranged 3 ×
3. The edge beams extend 2.75 inches below the soffit of the slab and the interior
beams have an overall depth of 5 inches. The corner columns are 4 inches × 4
inches and the interior columns are 6 inches × 6 inches. Edge columns are 4
inches × 6 inches with the longer dimension parallel to the slab edge. A typical
section of the slab and details are shown in Figure 13-2. The moments in an
interior design frame due to uniform loads obtained from SAFE are compared
with the corresponding experimental results and the numerical values obtained
from the EFM.
The computational model uses a 6 × 6 mesh of elements per panel, as shown in
Figure 13-3. Grid lines are defined at column faces as well as the column
centerlines. The slab is modeled using the plate elements available in SAFE. The
columns are modeled as supports with both vertical and rotational stiffnesses. A
stiffness coefficient of 8.0 is used in the calculation of support flexural stiffnesses
based on a column height of 15.875 inches, measured from the mid-depth of the
slab to the support center. The column is assumed to be infinitely rigid over the
full depth of the beams framing into it. The value of 8.0 is 75% of the figure
obtained from Table 6.2 of ACI Committee 340 (1997) to approximately account
for the pinned end condition at the column base. In order to account for rigidity
of the slab-column joint, the portion of slab occupying the column area is
modeled as rigid by using a special stiff area element. Edge beam properties are
derived from their cross-section geometries.
To obtain maximum factored moments in an interior design frame, the slab is
divided into two interior and two exterior design frames spanning in the
X direction (E-W direction). Because of double symmetry, comparison is
confined to X strips only. An interior design frame consists of one column strip
and half of each adjacent middle strip.
EXAMPLE 13 - 1
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
A
B
5'− 0"
C
D
5'− 0"
5'− 0"
1
5'− 0"
2
Typical Design Frames
A 5'− 0"
A
3
5'− 0"
4
Figure 13-1 University of Illinois Two-Way Slab Example T1
GEOMETRY, PROPERTIES AND LOADING
EXAMPLE 13 - 2
Concrete strength
Yield strength of reinforcements
Modulus of elasticity
Poisson’s ratio
fc’
fy
Ec
ν
=
=
=
=
3 ksi
42 ksi
3000 ksi
0.2
Loading: Total uniform load
w
=
150 psf
Software Verification
PROGRAM NAME:
REVISION NO.:
A
B
5
16 "
8
Detail 'a'
D
C
5'− 0"
5'− 0"
SAFE
0
5'− 0"
Detail 'b'
Edge Col. 4" x 6"
Corner Col. 4" x 4"
Interior Col. 6" x 6"
SECTION A-A
A
B
3"
3"
1
4 "
4
1
1 "
2
1
3 "
2
5"
Detail 'a'
Detail 'b'
Figure 13-2 Sections and Details of Slab T1
EXAMPLE 13 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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Figure 13-3 SAFE Mesh of Slab T1 (6 × 6 per panel)
TECHNICAL FEATURES OF SAFE TESTED
Calculation frame moments and comparison with experimental and FEM
results.
RESULTS COMPARISON
Table 13.-1 shows the comparison of the moments in an interior design frame
obtained numerically from SAFE with the experimental results and the EFM
results. The experimental and EFM results are all obtained from Corley and Jirsa
(1970).
EXAMPLE 13 - 4
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SAFE
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PROGRAM NAME:
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Table 13-1 Comparison of Measured and Computer Moments
Moments in an Interior Design Frame ( M / Wl1 *)
Exterior Span
Method
−
M
+
M
Middle Span
−
M
−
M
+
M
−
M
Test T1
0.043
0.046
0.079
0.071
0.036
0.071
EFM
0.035
0.047
0.079
0.066
0.034
0.066
SAFE
0.044
0.049
0.071
0.061
0.041
0.061
* Wl1 = 18.75 k-ft
The negative design moments reported are at the face of columns. The
comparison is excellent. The minor discrepancy is attributed to the loss of
stiffness due to the development of cracks and the difficulty in measuring strains
accurately at desired locations.
COMPUTER FILE: S13.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 13 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE 14
University of Illinois Two-Way Slab Test T2
PROBLEM DESCRIPTION
This example models the slab structure tested at the University of Illinois by
Vanderbilt, Sozen, and Siess (1969). The structure is a two-way slab arranged in
3 × 3 panels in which each panel is supported along all four edges by beams, as
shown in Figure 14-1. The structure consists of nine 1.5-inch thick 5-foot × 5foot panels. The edge beams and the interior beams extend 1.5 inches below the
soffit of the slab. The corner columns are 4 inches × 4 inches and the interior
columns are 6 inches × 6 inches. Edge columns are 4 inches × 6 inches with the
longer dimension parallel to the slab edge. A typical section of the slab and
details is shown in Figure 14-2.
The computational model uses a 6 × 6 mesh of elements per panel, as shown in
Figure 14-3. Grid lines are defined at column faces as well as the column
centerlines. The slab is modeled using plate elements and the columns are
modeled as supports with both vertical and rotational stiffnesses. A stiffness
coefficient of 6.33 is used in the calculation of support flexural stiffnesses based
on a column height of 13.125 inches, measured from the mid-depth of the slab to
the support center. The column stiffness is assumed to be infinitely rigid over the
full depth of the beams framing into it. The value of 6.33 is 75% of the figure
obtained from Table A7 of Portland Cement Association (1990) to approximately
account for the pinned end condition at the column base. In order to account for
rigidity of the slab-column joint, the portion of slab occupying the column area is
modeled as rigid by using a special stiff area element. Edge beam properties are
derived from their cross-section geometries.
To obtain maximum factored moments in an interior design frame, the slab is
divided into two interior and two exterior design frames spanning in the X
direction (E-W direction). An interior design frame consists of one column strip
and half of each adjacent middle strip.
GEOMETRY, PROPERTIES AND LOADING
Concrete strength
Yield strength of reinforcement
Modulus of elasticity
Poisson’s ratio
Loading: Total uniform load
f c’
fy
Ec
ν
w
=
3 ksi
= 47.6 ksi
= 3000 ksi
= 0.2
= 139 psf
EXAMPLE 14 - 1
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PROGRAM NAME:
REVISION NO.:
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A
B
5'− 0"
C
5'− 0"
D
5'− 0"
1
5'− 0"
2
Typical Design Frame
A 5'− 0"
A
3
5'− 0"
4
Figure 14-1 University of Illinois Two-Way Floor Slab T2
EXAMPLE 14 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
A
B
5'− 0"
7
13 "
8
Detail 'a'
D
C
5'− 0"
SAFE
0
5'− 0"
Detail 'b'
Interior Col. 6" x 6"
Corner Col. 4" x 4"
Edge Col. 4" x 6"
SECTION A−A
A
B
3"
3"
3’’
1
1 "
2
1
1 "
2
3"
Detail 'a'
Detail 'b'
Figure 14-2 Sections and Details of Floor Slab T2
EXAMPLE 14 - 3
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PROGRAM NAME:
REVISION NO.:
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Figure 14-3 SAFE Mesh of Slab T2 (6 × 6 per panel)
TECHNICAL FEATURES OF SAFE TESTED
Calculation of frame forces and comparison with experimental and FEM
results.
RESULTS COMPARISON
Table 14-1 shows the comparison of the moments in an interior design frame
obtained numerically from SAFE with the experimental results and the EFM
results. The experimental and EFM results are all obtained from Corley and Jirsa
(1970).
EXAMPLE 14 - 4
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SAFE
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PROGRAM NAME:
REVISION NO.:
Table 14-1 Comparison of Measured and Computer Moments
Moments in an Interior Design Frame (M / Wl1 *)
Exterior Span
−
M
+M
TEST T1
0.036
0.056
EFM
0.046
SAFE
0.046
Method
Middle Span
−
−
−
M
+M
0.069
0.061
0.045
0.061
0.044
0.074
0.066
0.034
0.066
0.047
0.067
0.060
0.039
0.060
M
M
* Wl1 = 17.375 kip-ft
The negative design moments reported are at the face of columns. The
comparison is excellent except for the negative exterior moments where the
experimental results are lower than both the SAFE and the EFM results. The
discrepancy is attributed not only to the loss of stiffness due to the development
of cracks, but also to the difficulty in taking accurate strain measurements at
desired locations.
COMPUTER FILE: S14.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 14 - 5
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PROGRAM NAME:
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SAFE
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EXAMPLE 15
Temperature Loading
PROBLEM DESCRIPTION
In SAFE, two types of temperature loads can be applied to slab elements: an
overall change in temperature or a temperature gradient across the slab thickness.
This example tests each of these temperature loading methods using a 10-inchdeep x 12-inch-wide concrete slab. The slab is restrained in four different ways,
and different temperature loads are applied and analyzed using SAFE. The results
are compared to hand calculations and summarized in Table 15-1.
Temp, T1 = 100 degrees, F, Temp, T2 = 0 degrees, F, Span = 24 ft
EXAMPLE 15 - 1
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Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Slab thickness
Slab width
Clear span
Concrete strength
Modulus of elasticity
Poisson’s ratio
Temp, T1
Temp, T2
h
b
L
f c′
Ec
ν
T1
T2
=
10 in
=
12 in
=
288 in
= 4,000 psi
= 3,605 ksi
= 0.001
=
100 degrees, F
=
0 degrees, F
TECHNICAL FEATURES OF SAFE TESTED
Temperature and Temperature Gradient Loading
RESULTS COMPARISON
The force, reaction, or displacements are found using the SAFE program for the
cases described previously. The SAFE values were then compared to the
independent hand calculations and summarized in Table 15-1.
EXAMPLE 15 - 2
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PROGRAM NAME:
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Table 15-1 Comparison of Results
CASE AND FEATURE TESTED
Case 1, Force, F11 (k/ft)
INDEPENDEN
T RESULTS
237.93
SAFE
RESULTS
DIFFEREN
CE
237.95
0.01%
Case 2, Reaction, RB (k)
1.033
1.032
0.09%
Case 3, Mid-Span Deflection, (in)
0.570
0.570
0.00%
−2.281
−2.281
0.00%
Case 4, Free-End Displacement, (in)
COMPUTER FILES: S15a.FDB, S15b.FDB, S15c.FDB, S15d.FDB
CONCLUSION
The SAFE results show an exact or nearly exact comparison with the
independent hand-calculated results.
COMMENT
In Case 4, a stiffness modifier of 100 for V13 and V23 is used to avoid shear
deformation in plate.
The vertical offset of a slab can have a significant effect on the thermal loading
results. Therefore, it is recommended that users turn off the option to ignore the
vertical offsets when temperature loading is considered in a model (see the Run
menu > Ignore Vertical Offsets in Non P/T Models command).
EXAMPLE 15 - 3
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CALCULATIONS:
Design Parameters: T1 = 100, T2 = 0, h= 10, L = 24 ft (288 in), b = 36, ε = 5.5E-06
Case 1:
Slab Force, F11 =
εtAE =
0.0000055(100)(10 ×12)(3605) =
237.93 k/ft
Case 2:
3EI ε
3EI ε
2
From Roark and Young, p. 107
T 2 − T 1) L=
(T 2 − T1)
3 (
2hL
2hL
3(3605)(1000)(0.0000055)
= =
(100) 1.033 kips
2(10)(288)
Reaction, RB
=
Case 3:
Deflection, Z =
Case 4:
Deflection, Z =
EXAMPLE 15 - 4
−ε
−0.0000055
(T 2 − T1) L2 =
( −100 ) (288)2 = 0.5702 in
8h
8(10)
Roark…, p. 108
−ε
−0.0000055
(T 2 − T1) L2 =
( −100 ) (288)2 = 2.281 in
2h
2(10)
Roark…, p. 108
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EXAMPLE 16
Cracked Slab Analysis
CRACKED ANALYSIS METHOD
The moment curvature diagram shown in Figure 16-1 depicts a plot of the
uncracked and cracked conditions,Ψ 1 State 1, and,Ψ 2 State 2, for a reinforced
beam or slab. Plot A-B-C-D shows the theoretical moment versus curvature of a
slab or beam. The slope of the moment curvature between points A and B
remains linearly elastic until the cracking moment, Mr, is reached. The increase
in moment curvature between B-C at the cracking moment, Mr, accounts for the
introduction of cracks to the member cross-section. The slope of the moment
curvature between point C-D approaches that of the fully cracked condition, Ψ 2
State 2, as the moment increases.
Since the moments vary along the span of a slab or beam, it is generally not
accurate to assign the same cracked section effective moment of inertia along the
entire length of a span. A better approach and the one recently added to the SAFE
program is to account for the proper amount of curvature for each distinct finite
element of the slab or beam that corresponds to the amount of moment being
applied to that element. After the moment curvatures are known for each
element, the deflections can be calculated accordingly.
This verification example will compare the results from Example 8.4, Concrete
Structures, Stresses and Deformations, Third Edition, A Ghali, R Favre and M
Elbadry, pages 285-289, with the results obtained from SAFE. Both the
calculations and the SAFE analysis use the cracked analysis methodology
described in the preceding paragraphs.
PROBLEM DESCRIPTION
The slab used in this example has dimensions b = 0.3 m and h = 0.6 m. The slab
spans 8.0 m and has an applied load of 17.1 KN/m.
EXAMPLE 16 - 1
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Figure 16-1 Moment versus curvature for a reinforced slab member
Figure 16-2 One-Way Slab
EXAMPLE 16 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Slab thickness
h
Slab width
b
Clear span
L
Concrete Ultimate Strength h
fc′
Concrete cracking strength
f cr
Modulus of elasticity, Conc.
Es
Modulus of elasticity, Steel
Ec
Poisson’s ratio
ν
Uniform load
w
Creep coefficient
ϕ ( t,t0 )
Free shrinkage
ε CS ( t,t0 )
=
=
=
=
=
=
=
=
=
=
0.65
0.3
8.0
30
2.5
30
200
0.2
17.1
2.5
SAFE
0
m
m
m
MPa
MPa
GPa
GPa
KN/m
= −250E-6
Note: The concrete cracking strength of f cr = 2.5 MPa was used in this example
using the Run menu > Cracking Analysis Option command.
TECHNICAL FEATURES OF SAFE TESTED
Cracked Slab Analysis
RESULTS COMPARISON
SAFE calculated the displacements using a Nonlinear Cracked Load Case (see
Figure 16-1). The first nonlinear load case was calculated without creep and
shrinkage effects and the second nonlinear load case included creep and
shrinkage effects. Table 16-1 shows the results obtained from SAFE compared
with the referenced example.
Table 16-1 Comparison of Results
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERE
NCE
Mid-Span Displacement No Creep /
Shrinkage (m)
14.4 mm
13.55 mm
5.90%
Mid-Span Displacement with Creep /
Shrinkage (m)
23.9 mm
24.51 mm
2.51%
CASE AND FEATURE TESTED
EXAMPLE 16 - 3
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SAFE
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PROGRAM NAME:
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COMPUTER FILES: S16.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
CALCULATIONS:
Design Parameters:
Es = 200 GPa, Ec = 30 GPa, h = 0.65 m, b = 0.3m,
As = 1080 mm2, As′ = 270 mm2, Center of reinf. at 0.05 m
Span = 8.0 m, Uniform Load = 17.1 KN/m
Figure 16-3 Slab Cross-Section
Case 1 – Nonlinear cracked slab analysis without creep and shrinkage
1.1 Transformed Uncracked Section Properties:
Area, A = 0.2027m2
Y = 0.319m
I, transformed = 7.436E-03 m4
1.2 Transformed Cracked Section Properties:
Area, A = 0.2027 m2
C = 0.145 m
I, cracked = 1.809E-03 m4
EXAMPLE 16 - 4
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1.3 Cracked Bending Moment, Mr = 23.3E-03 × 2.5 × 10E6 = 58.3 KN-m
2
2
M
58.3
1.4 Interpolation coefficient, ζ =
1 − 1.0
0.82
1 − β1 β 2 r =
=
136
M
where β1 = 1.0 and β 2 = 1.0
1.5 Curvature:
State1: Uncracked
136E-06
=
610E-06 / m
Ψ1 =
30 ×109 × 7.436E-03
State2: Fully Cracked
136E-06
=
2506E-06 / m
Ψ2 =
30 ×109 ×1.809E-03
Interpolated curvature:
Ψm =
2157E-06 / m
(1 − ζ )Ψ 1 + ζ (Ψ 2 ) =−
(1 0.82 ) (610E-06 / m) + 0.82 ( 2506E-06 ) =
1.6 Slab Curvature:
Figure 16-4 Span-Curvature Diagram
1.7 Deflection:
By assuming a parabolic distribution of curvature across the entire span (see the
Mean Curvature over Entire Span plot in Figure 16-4), the deflection can be
calculated as,
EXAMPLE 16 - 5
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PROGRAM NAME:
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Deflection
= 0.002157
82
×10
=
×1000 14.4 mm (See Table 16-1)
96
Case 2 – Nonlinear cracked slab analysis with creep and shrinkage
2.1 Aged adjusted concrete modulus,
E C ( t , t0 )
=
EC ( t0 )
30e9
=
= 10GPa
1 + X ϕ ( t , t0 ) 1 + 0.8(2.5)
Where X ( t,t0 ) = 0.8 (SAFE Program Default), ϕ ( t,t0 ) = 2.5 (aging coefficient, see
Figure 16-5 below)
=
n
ES
200
= = 20
E C ( t,t0 ) 10
2.2 Age-adjusted transformed section in State1:
A1 = 0.2207 m 2
NA = 0.344 m from top of slab
=
I 1 8.724 ×10−3 m 4
yC = −0.020 m, distance from top of slab to the centroid of the concrete area
AC = 0.1937 m 2 , area of concrete
=
I C 6.937 ×10−3 m 4 , moment of inertia of AC about NA
IC
=
35.34 ×10−3 m 2
rC2 =
AC
κ=
1
I C 6.937 ×10−3
=
= 0.795, curvature reduction factor
I 8.724 ×10−3
2.3 Age adjusted transformed section in State2:
A2 = 0.0701m 2
NA = 0.233m from top of slab
=
I 2 4.277 ×10−3 m 4
yC = −0.161m, distance from top of slab to the centroid of the concrete area
EXAMPLE 16 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
AC = 0.0431m 2 , area of concrete
=
I C 1.190 ×10−3 m 4 , moment of inertia of AC about NA
IC
=
27.62 ×10−3 m 2
rC2 =
AC
κ=
2
I C 1.190 ×10−3
=
= 0.278
4.277 ×10−3
I
2.4 Changes in curvature due to creep and shrinkage:
State 1, Change in curvature between period t0 to t,
y
y
=
ψ κ ϕ ( t,t0 ) ψ ( t0 ) + ε 0 ( t0 ) C2 + ε CS ( t,t0 ) ε C2
Delta
rC
rC
−0.020
−0.020
−6
= 0.795 2.5 610 ×10−6 + 8 ×10−6
250
+
−
×
10
(
)
35.34 x10−3
35.34 ×10−3
= 1299 ×10−6 / m
The curvature at time t (State 1)
Ψ1 (t) =
1909x10−6 / m
( 610 + 1299 ) x10−6 / m =
State 2, Change in curvature between period t0 to t ,
y
y
=
ψ κ ϕ ( t,t0 ) ψ ( t0 ) + ε 0 ( t0 ) C2 + ε CS ( t,t0 ) ε C2
Delta
rC
rC
−0.161
−0.161
= 0.278 2.5 2506 ×10−6 + 222 ×10−6
+ −250 × 10−6 )
−3 (
27.62 x10
27.62 ×10−3
= 1248 ×10−6 / m
The curvature at time t (State 2)
Ψ 2 ( t ) = ( 2506 + 1248 ) ×10−6 / m =3754 ×10−6 / m
Interpolated curvature:
Ψ t =−
3584 ×10−6 / m
(1 ζ )Ψ 1 ( t ) + ζ (Ψ 2 ( t ) ) =(1 − 0.91) (1909 ×10−6 ) + 0.91( 3754 ×10−6 ) =
EXAMPLE 16 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
2.5 Deflection at center at time, t:
By assuming a parabolic distribution of curvature across the entire span, the
deflection can be calculated as,
82
Deflection
= 0.003584 ×10
=
×1000 23.90 mm (See Table 16-1)
96
2.6 The Load Case Data form for Nonlinear Long-Term Cracked Analysis:
The Creep Coefficient and Shrinkage Strain values must be user defined. For this
example, a shrinkage strain value of −250E-6 was used. Note that the value is input
as a positive value.
Figure 16-5 Load Case Data form for Nonlinear Long-Term Cracked Analysis
EXAMPLE 16 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE 17
Crack Width Analysis
The crack width, wk, is calculated using the methodology described in the
Eurocode EN 1992-1-1:2004, Section 7.3.4, which makes use of the following
expressions:
(1) =
wk sr ,max (ε sm − εcm )
(eq. 7.8)
where
sr,max is the maximum crack spacing
(2)
εsm
is the mean strain in the reinforcement under the relevant combination
of loads, including the effect of imposed deformations and taking into
account the effects of tension stiffening. Only the additional tensile
strain beyond the state of zero strain of the concrete at the same level
is considered.
εcm
is the mean strain in the concrete between cracks
εsm − εcm may be calculated from the expression
σs − kt
=
ε sm − εcm E
fct ,eff
(1 + αeρ p.eff )
ρ p,eff
Es
≥ 0.6
σs
Es
(eq. 7.9)
where
σs
is the stress in the tension reinforcement assuming a cracked section.
For pretensioned members, σs may be replaced by ∆σs, the stress
variation in prestressing tendons from the state of zero strain of the
concrete at the same level.
αe
is the ratio Ec / Ecm
ρp,eff is As / Ac,eff
Ap′ and Ac,eff ; Ap′ is the area of tendons within Ac,eff, and Ac,eff is the area of
tension concrete surrounding the reinforcing.
EXAMPLE 17 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
kt
(3)
is a factor dependent on the duration of the load
kt = 0.6 for short term loading
kt = 0.4 for long-term loading
In situations where bonded reinforcement is fixed at reasonably close
centers within the tension zone [spacing ≤ 5(c + φ / 2)], the maximum final
crack spacing may be calculated from
sr,max = k3c + k1k2k4φ / ρp,eff
(eq. 7.11)
where
φ is the bar diameter. Where a mixture of bar diameters is used in a
section, an equivalent diameter, φeq, should be used. For a section with
n1 bars of diameter φ1 and n2 bars of diameter φ2, the following equation
should be used:
n φ2 + n2 φ22
φeq =1 1
n1φ1 + n2 φ2
(eq. 7.12)
where
c
is the cover to the longitudinal reinforcement
k1 is a coefficient that takes into account the bond properties of the bonded
reinforcement:
= 0.8 for high bond bars
= 1.6 for bars with an effectively plain surface (e.g., prestressing
tendons)
k2 is a coefficient that takes into account the distribution of strain:
= 0.5 for bending
= 1.0 for pure tension
k3 and k4 are recommended as 3.4 and 0.425 respectively. See the
National Annex for more information.
For cases of eccentric tension or for local areas, intermediate values of k2
should be used that may be calculated from the relation:
k2 = (ε1 + ε2) / 2ε1
(eq. 7.13)
where ε1 is the greater and ε2 is the lesser tensile strain at the boundaries of
the section considered, assessed on the basis of a cracked section.
EXAMPLE 17 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
PROBLEM DESCRIPTION
The purpose of this example is to verify that the crack width calculation
performed by SAFE is consistent with the methodology described above. Hand
calculations using the Eurocode EN 1992-1-1:2004, Section 7.3.4 are shown
below as well as a comparison of the SAFE and hand calculated results.
A one-way, simply supported slab is modeled in SAFE. The modeled slab is 254
mm thick by 914 mm wide and spans 9,754 mm, as shown in Figure 17-1, and is
the same slab used to validate the Eurocode PT design (see design verification
example Eurocode 2-04 PT-SL-001). To test the crack width calculation, seven
#5 longitudinal bars have been added to the slab. The total area of mild steel
reinforcement is 1,400mm2. Currently, SAFE will account for some of the PT
effects. SAFE accounts for the PT effects on the moments and reinforcing
stresses but the tendon areas are not considered effective to resist cracking.
Prestressing tendon, Ap
Mild Steel, As
229 mm
254 mm
25 mm
Length, L = 9754 mm
Elevation
914 mm
Section
Figure 17-1 One-Way Slab
EXAMPLE 17 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 loads are as follows:
Loads:
Dead = self weight
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
ν
=
=
=
=
=
=
=
=
=
30
400
1862
1210
198
23.56
25000
200,000
0
MPa
MPa
MPa
MPa
mm2
KN/m3
N/mm3
N/mm3
Dead load
wd
=
self
KN/m2
TECHNICAL FEATURES OF SAFE TESTED
Calculation of the reported crack widths.
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE crack widths to those calculated by
hand.
EXAMPLE 17 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
FEATURE TESTED
Crack Widths (mm)
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
0.151mm
0.161mm
6.62%
COMPUTER FILE: S17.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE 17 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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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
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)
ω =5.984 kN/m2 x 0.914 m = 5.469 kN/m
Ultimate Moment, M U =
wl12
= 5.469 x (9.754)2/8 = 65.0 kN-m
8
Reinforcing steel stress, σ = 207N / mm 2 (calculated but not reported by SAFE)
EXAMPLE 17 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Check of Concrete Stresses at Midspan:
Figure 17-1 Settings used for this example
EXAMPLE 17 - 7
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
Calculation of Crack Width:
=
wk sr ,max (ε sm − ε cm )
where
f
σ s − kt ct ,eff (1 + α eρ p ,eff )
ρ p ,eff
σ
=
ε sm − εcm
≥ 0.6 s , where
Es
Es
As / A
ρ=
=
1.53mm 2 / mm / ( 60mm 2 / mm )
p ,eff
c .eff
ρ p ,eff = 0.026
1.744 N / mm 2
206 N / mm − 0.4
(1 + 8 ( 0.026 ) )
206
0.026
≥ 0.6
199948
199948
2
ε sm − ε cm
ε sm − ε cm
= 0.0009 ≥ 0.0006
sr ,max= k3c + k1=
k2 k4 φ / ρ p ,eff 3.4 (19.0mm ) + 0.8 ( 0.5 )( 0.425 )15.8mm / 0.026
= 168mm
Total crack width,
=
wk sr ,max ( ε sm − ε cm ) = 168mm ( 0.0009 ) = 0.151mm
EXAMPLE 17 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE ACI 318-14 PT-SL 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 SAFE. 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 SAFE results
and summarized for verification and validation of the SAFE results.
Loads: Dead = self weight, Live = 100psf
EXAMPLE ACI 318-14 PT-SL 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
T, h
d
L
f 'C
fy
=
=
=
=
=
10
9
384
4,000
60,000
in
in
in
psi
psi
Prestressing, ultimate
f pu =
270,000 psi
Prestressing, effective
=
175,500 psi
=
=
=
=
=
=
=
0.153
0.150
3,600
29,000
0
self
100
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
ν
Dead load,
wd
Live load,
wl
sq in
pcf
ksi
ksi
psf
psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
EXAMPLE ACI 318-14 PT-SL 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
RESULTS COMPARISON
The SAFE 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
SAFE
RESULTS
DIFFERENCE
1429.0
1428.3
0.05%
2.20
2.20
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-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-14 PT-SL 001 - 3
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
φ =0.9
Mild Steel Reinforcing
f′c = 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
ω =0.225 ksf × 3 ft = 0.675 klf,
Ultimate Moment, M U =
ω u = 0.310 ksf × 3ft = 0.930 klf
wl12
= 0.310 klf × 322/8 = 119.0 k-ft = 1429.0 k-in
8
EXAMPLE ACI 318-14 PT-SL 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
f 'c
(span-to-depth ratio > 35)
300 ρ 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 +
) 61.08 kips
Ultimate force in PT, =
Fult , PT A=
2 ( 0.153)(199.62
=
P ( f PS )
( 60.0 ) 120.0 kips
Ultimate force in RC, =
Fult , RC A=
2.00(assumed)=
s ( f )y
Total Ultimate force, Fult ,Total = 61.08 + 120.0 = 181.08 kips
Stress block =
depth, a
Fult ,Total
181.08
=
= 1.48 in
0.85 f ' cb 0.85 ( 4 ) ( 36 )
a
1.48 ( )
Ultimate moment due to PT, M ult , PT
= Fult , PT d − =
φ 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
φ 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
EXAMPLE ACI 318-14 PT-SL 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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,
=
57.83
=
M PT F=
( 4 in ) 231.3 k-in
PTI (sag)
FPTI M D − M PT −57.83 576.0 − 231.3
Stress in concrete, f =
, where S = 600 in3
±
= ±
10 ( 36 )
600
A
S
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,
=
53.70
=
M PT F=
( 4 in ) 214.8 k-in
PTI (sag)
FPTI M D + L − M PT −53.70 1037.0 − 214.8
Stress in concrete for (D + L+ PTF), f =
±
=±
10 ( 36 )
600
A
S
f =
−0.149 ± 1.727 ± 0.358
f = −1.518(Comp) max,1.220(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 = 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,
=
M PT F=
53.70
=
( 4 in ) 214.8 k-in
PTI (sag)
EXAMPLE ACI 318-14 PT-SL 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Stress in concrete for (D + 0.5L + PTF(L)),
FPTI M D + 0.5 L − M PT −53.70 806.0 − 214.8
±
=
±
f =
10 ( 36 )
600
A
S
f =
−0.149 ± 0.985
f = −1.134(Comp) max, 0.836(Tension) max
EXAMPLE ACI 318-14 PT-SL 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE ACI 318-14 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the beam flexural design in SAFE. 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 ACI 318-14.
The average shear stress in the beam falls below the maximum shear stress
allowed by ACI 318-14, requiring design shear reinforcement.
A simple-span, 20-foot-long, 12-inch-wide, and 18-inch-deep T beam with a
flange 4 inches thick and 24 inches wide is modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size is
specified as 6 inches. The beam is supported by columns without rotational
stiffnesses and with very large vertical stiffness (1 × 1020 kip/in).
The beam is loaded with symmetric third-point loading. One dead load (DL02)
case and one live load (LL30) case, with only symmetric third-point loads of
magnitudes 3, and 30 kips, respectively, are defined in the model. One load
combination (COMB30) is defined using the ACI 318-14 load combination
factors of 1.2 for dead load and 1.6 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. The total factored
moment and shear force are compared with the SAFE results and found to be
identical. After completing the analysis, the design is performed using the ACI
318-14 code in SAFE and also by hand computation. Table 1 shows the
comparison of the design longitudinal reinforcement. Table 2 shows the
comparison of the design shear reinforcement.
EXAMPLE ACI 318-14 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE ACI 318-14 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span
Overall depth
Flange thickness
Width of web
Width of flange,
Depth of tensile reinf.
Effective depth
Depth of comp. reinf.
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
240
18
4
12
24
3
15
3
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
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0.2
Dead load
Live load
Pd
Pl
=
=
2
30
SAFE
0
in
in
in
in
in
in
in
in
psi
psi
pcf
ksi
ksi
kips
kips
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the beam
with the moments obtained by the analytical method. They match exactly for this
problem. Table 1 also shows the comparison of the design reinforcement.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-in)
Method
Moment
(k-in)
As+
SAFE
4032
5.808
Calculated
4032
5.808
A +s ,min = 0.4752 sq-in
EXAMPLE ACI 318-14 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-in/ft)
Shear Force (kip)
SAFE
Calculated
50.40
0.592
0.592
COMPUTER FILE: ACI 318-14 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-14 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9
Ag = 264 sq-in
As,min = 0.0018Ag = 0.4752 sq-in
f c′ − 4000
0.85
=
1000
β1 =
0.85 − 0.05
=
cmax
0.003
=
d 5.625 in
0.003 + 0.005
amax = β1cmax = 4.78125 in
As = min[As,min, (4/3) As,required] = min[0.4752, (4/3)5.804] = 0.4752 sq-in
COMB30
Pu = (1.2Pd + 1.6Pt) = 50.4 k
Mu =
Pu l
= 4032 k-in
3
The depth of the compression block is given by:
a =−
d
d2 −
2 Mu
0.85 f c'ϕ b f
= 4.2671 in (a > ds)
Calculation for As is performed in two parts. The first part is for balancing the
compressive force from the flange, Cf, and the second part is for balancing the
compressive force from the web, Cw. Cf is given by:
Cf
= 0.85fc' (bf − bw) ds = 163.2 k
The portion of Mu that is resisted by the flange is given by:
Muf =
d
Cf d − s ϕ = 1909.44 k-in
2
EXAMPLE ACI 318-14 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Therefore, the area of tensile steel reinforcement to balance flange compression
is:
As1
=
M uf
f y (d − d s 2 ) ϕ
= 2.7200 sq-in
The balance of the moment to be carried by the web is given by:
Muw = Mu − Muf = 2122.56 k-in
The web is a rectangular section with dimensions bw and d, for which the design
depth of the compression block is recalculated as
= d−
a1
d2 −
2 M uw
= 4.5409 in (a1 ≤ amax)
0.85 f c′ ϕ bw
The area of tensile steel reinforcement to balance the web compression is then
given by:
As2
=
M uw
= 3.0878 sq-in
a1
ϕ fy d − ϕ
2
The area of total tensile steel reinforcement is then given by:
As
= As1 + As2 = 5.808 sq-in
Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
Check the limit of
0.75
f c′ :
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bwd
= 17.076 k
The maximum shear that can be carried by reinforcement is given by:
EXAMPLE ACI 318-14 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ϕ Vs =
ϕ8
SAFE
0
f c′ bwd = 68.305 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 8.538 k
(ϕ Vc + ϕ 50 bwd)
= 23.826 k
Vmax = ϕ Vc + ϕ Vs = 85.381 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ (Vc/2) ϕ,
Av
= 0,
s
else if (Vc/2) ϕ < Vu ≤ (ϕVc + ϕ 50 bwd),
50 bw
Av
=
,
fy
s
else if (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
(V − ϕVc )
Av
= u
ϕ f ys d
s
else if Vu > ϕ Vmax,
a failure condition is declared.
For each load combination, the Pu and Vu are calculated as follows:
Pu = 1.2Pd + 1.6P1
Vu = Pu
(COMB30)
Pd = 2 k
Pl
= 30 k
Pu = 50.4 k
Vu = 50.4 k, (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
Av
=
s
(Vu − ϕVc )
= 0.04937 sq-in/in or 0.592 sq-in/ft
ϕf ys d
EXAMPLE ACI 318-14 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-14 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
B
A
1'
24'
C
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
EXAMPLE ACI 318-14 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 plate
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 SAFE TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
SAFE
0.192
0.158
1.21
Calculated
0.193
0.158
1.22
COMPUTER FILE: ACI 318-14 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-14 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation for Interior Column Using SAFE 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 SAFE Model
γ
V2
γ
V3
1
1−
0.4955
=
=
2 44.5
1+
3 20.5
1
1−
0.3115
=
=
2 20.5
1+
3 44.5
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE ACI 318-14 RC-PN-001 - 3
Software Verification
SAFE
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
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
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
0
= 0"
1105
2
Ld
0
= 0"
1105
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 SAFE output at Grid B-2:
VU = 189.45 k
γ V 2 M U 2 = −156.39 k-in
γ V 3 M U 3 = 91.538 k-in
EXAMPLE ACI 318-14 RC-PN-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.:
SAFE
0
At the point labeled A in Figure 2, x4 = −10.25 and y4 = 22.25, thus:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
vU =
−
−
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:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
−
−
vU =
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
EXAMPLE ACI 318-14 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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.
ϕ vC
=
ϕ vC
=
=
ϕ vC
4
0.75 2 +
4000
36 /12
= 0.158 ksi in accordance with equation 11-34
1000
40 • 8.5
+ 2 4000
0.75
130
=
0.219 ksi in accordance with equation 11-35
1000
0.75 • 4 • 4000
= 0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of φvC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE ACI 318-14 RC-PN-001 - 6
vU
=
ϕ vC
0.193
= 1.22
0.158
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-14 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. 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 SAFE. The maximum element size is specified to be 36 inches. To
obtain factored moments and flexural reinforcement in a design strip, one onefoot-wide strip is defined in the X-direction on the slab, as shown in Figure 1.
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 SAFE results. After completing
the analysis, design is performed in accordance with ACI 318-14 using SAFE
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
GEOMETRY, PROPERTIES AND LOADING
Thickness
T, h =
6 in
EXAMPLE ACI 318-14 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Depth of tensile reinf.
Effective depth
Clear span
dc =
d
=
ln, l1 =
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
ν
=
=
=
=
=
=
Dead load
Live load
wd
wl
=
=
1 in
5 in
144 in
4,000
60,000
0
3,600
29,000
0
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
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-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-14 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
ϕ = 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f ′ − 4000
0.85 − 0.05 c
0.85
β1 =
=
1000
0.003
=
d 1.875 in
0.003 + 0.005
amax = β1cmax = 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'ϕ b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
= 0.213 sq-in > As,min
a
ϕ fy d −
2
As = 0.2114 sq-in
As =
EXAMPLE ACI 318-14 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-11 PT-SL 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 SAFE. 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 SAFE results
and summarized for verification and validation of the SAFE results.
Loads: Dead = self weight , Live = 100psf
EXAMPLE ACI 318-11 PT-SL 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
T, h
d
L
f 'C
fy
=
=
=
=
=
10
9
384
4,000
60,000
in
in
in
psi
psi
Prestressing, ultimate
f pu =
270,000 psi
Prestressing, effective
=
175,500 psi
=
=
=
=
=
=
=
0.153
0.150
3,600
29,000
0
self
100
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
ν
Dead load,
wd
Live load,
wl
sq in
pcf
ksi
ksi
psf
psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
EXAMPLE ACI 318-11 PT-SL 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
RESULTS COMPARISON
The SAFE 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
SAFE
RESULTS
DIFFERENCE
1429.0
1428.3
0.05%
2.20
2.20
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-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-11 PT-SL 001 - 3
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
φ =0.9
Mild Steel Reinforcing
f′c = 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
ω =0.225 ksf × 3 ft = 0.675 klf,
Ultimate Moment, M U =
ω u = 0.310 ksf × 3ft = 0.930 klf
wl12
= 0.310 klf × 322/8 = 119.0 k-ft = 1429.0 k-in
8
EXAMPLE ACI 318-11 PT-SL 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
f 'c
(span-to-depth ratio > 35)
300 ρ 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 +
) 61.08 kips
Ultimate force in PT, =
Fult , PT A=
2 ( 0.153)(199.62
=
P ( f PS )
( 60.0 ) 120.0 kips
Ultimate force in RC, =
Fult , RC A=
2.00(assumed)=
s ( f )y
Total Ultimate force, Fult ,Total = 61.08 + 120.0 = 181.08 kips
Stress block =
depth, a
Fult ,Total
181.08
=
= 1.48 in
0.85 f ' cb 0.85 ( 4 ) ( 36 )
a
1.48 ( )
Ultimate moment due to PT, M ult , PT
= Fult , PT d − =
φ 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
φ 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
EXAMPLE ACI 318-11 PT-SL 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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,
=
57.83
=
M PT F=
( 4 in ) 231.3 k-in
PTI (sag)
FPTI M D − M PT −57.83 576.0 − 231.3
Stress in concrete, f =
, where S = 600 in3
±
= ±
10 ( 36 )
600
A
S
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,
=
53.70
=
M PT F=
( 4 in ) 214.8 k-in
PTI (sag)
FPTI M D + L − M PT −53.70 1037.0 − 214.8
Stress in concrete for (D + L+ PTF), f =
±
=±
10 ( 36 )
600
A
S
f =
−0.149 ± 1.727 ± 0.358
f = −1.518(Comp) max,1.220(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 = 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,
=
M PT F=
53.70
=
( 4 in ) 214.8 k-in
PTI (sag)
EXAMPLE ACI 318-11 PT-SL 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Stress in concrete for (D + 0.5L + PTF(L)),
FPTI M D + 0.5 L − M PT −53.70 806.0 − 214.8
±
=
±
f =
10 ( 36 )
600
A
S
f =
−0.149 ± 0.985
f = −1.134(Comp) max, 0.836(Tension) max
EXAMPLE ACI 318-11 PT-SL 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-11 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the beam flexural design in SAFE. 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 ACI 318-11.
The average shear stress in the beam falls below the maximum shear stress
allowed by ACI 318-11, requiring design shear reinforcement.
A simple-span, 20-foot-long, 12-inch-wide, and 18-inch-deep T beam with a
flange 4 inches thick and 24 inches wide is modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size is
specified as 6 inches. The beam is supported by columns without rotational
stiffnesses and with very large vertical stiffness (1 × 1020 kip/in).
The beam is loaded with symmetric third-point loading. One dead load (DL02)
case and one live load (LL30) case, with only symmetric third-point loads of
magnitudes 3, and 30 kips, respectively, are defined in the model. One load
combination (COMB30) is defined using the ACI 318-11 load combination
factors of 1.2 for dead load and 1.6 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. The total factored
moment and shear force are compared with the SAFE results and found to be
identical. After completing the analysis, the design is performed using the ACI
318-11 code in SAFE and also by hand computation. Table 1 shows the
comparison of the design longitudinal reinforcement. Table 2 shows the
comparison of the design shear reinforcement.
EXAMPLE ACI 318-11 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE ACI 318-11 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span
Overall depth
Flange thickness
Width of web
Width of flange,
Depth of tensile reinf.
Effective depth
Depth of comp. reinf.
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
240
18
4
12
24
3
15
3
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
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0.2
Dead load
Live load
Pd
Pl
=
=
2
30
SAFE
0
in
in
in
in
in
in
in
in
psi
psi
pcf
ksi
ksi
kips
kips
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the beam
with the moments obtained by the analytical method. They match exactly for this
problem. Table 1 also shows the comparison of the design reinforcement.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-in)
Method
Moment
(k-in)
As+
SAFE
4032
5.808
Calculated
4032
5.808
A +s ,min = 0.4752 sq-in
EXAMPLE ACI 318-11 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-in/ft)
Shear Force (kip)
SAFE
Calculated
50.40
0.592
0.592
COMPUTER FILE: ACI 318-11 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-11 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9
Ag = 264 sq-in
As,min = 0.0018Ag = 0.4752 sq-in
f c′ − 4000
0.85
=
1000
β1 =
0.85 − 0.05
=
cmax
0.003
=
d 5.625 in
0.003 + 0.005
amax = β1cmax = 4.78125 in
As = min[As,min, (4/3) As,required] = min[0.4752, (4/3)5.804] = 0.4752 sq-in
COMB30
Pu = (1.2Pd + 1.6Pt) = 50.4 k
Mu =
Pu l
= 4032 k-in
3
The depth of the compression block is given by:
a =−
d
d2 −
2 Mu
0.85 f c'ϕ b f
= 4.2671 in (a > ds)
Calculation for As is performed in two parts. The first part is for balancing the
compressive force from the flange, Cf, and the second part is for balancing the
compressive force from the web, Cw. Cf is given by:
Cf
= 0.85fc' (bf − bw) ds = 163.2 k
The portion of Mu that is resisted by the flange is given by:
Muf =
d
Cf d − s ϕ = 1909.44 k-in
2
EXAMPLE ACI 318-11 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Therefore, the area of tensile steel reinforcement to balance flange compression
is:
As1
=
M uf
f y (d − d s 2 ) ϕ
= 2.7200 sq-in
The balance of the moment to be carried by the web is given by:
Muw = Mu − Muf = 2122.56 k-in
The web is a rectangular section with dimensions bw and d, for which the design
depth of the compression block is recalculated as
= d−
a1
d2 −
2 M uw
= 4.5409 in (a1 ≤ amax)
0.85 f c′ ϕ bw
The area of tensile steel reinforcement to balance the web compression is then
given by:
As2
=
M uw
= 3.0878 sq-in
a1
ϕ fy d − ϕ
2
The area of total tensile steel reinforcement is then given by:
As
= As1 + As2 = 5.808 sq-in
Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
Check the limit of
0.75
f c′ :
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bwd
= 17.076 k
The maximum shear that can be carried by reinforcement is given by:
EXAMPLE ACI 318-11 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ϕ Vs =
ϕ8
SAFE
0
f c′ bwd = 68.305 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 8.538 k
(ϕ Vc + ϕ 50 bwd)
= 23.826 k
Vmax = ϕ Vc + ϕ Vs = 85.381 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ (Vc/2) ϕ,
Av
= 0,
s
else if (Vc/2) ϕ < Vu ≤ (ϕVc + ϕ 50 bwd),
50 bw
Av
=
,
fy
s
else if (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
(V − ϕVc )
Av
= u
ϕ f ys d
s
else if Vu > ϕ Vmax,
a failure condition is declared.
For each load combination, the Pu and Vu are calculated as follows:
Pu = 1.2Pd + 1.6P1
Vu = Pu
(COMB30)
Pd = 2 k
Pl
= 30 k
Pu = 50.4 k
Vu = 50.4 k, (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
Av
=
s
(Vu − ϕVc )
= 0.04937 sq-in/in or 0.592 sq-in/ft
ϕf ys d
EXAMPLE ACI 318-11 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-11 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
B
A
1'
24'
C
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
EXAMPLE ACI 318-11 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 plate
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 SAFE TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
SAFE
0.192
0.158
1.21
Calculated
0.193
0.158
1.22
COMPUTER FILE: ACI 318-11 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-11 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation for Interior Column Using SAFE 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 SAFE Model
γ
V2
γ
V3
1
1−
0.4955
=
=
2 44.5
1+
3 20.5
1
1−
0.3115
=
=
2 20.5
1+
3 44.5
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE ACI 318-11 RC-PN-001 - 3
Software Verification
SAFE
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
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
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
0
= 0"
1105
2
Ld
0
= 0"
1105
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 SAFE output at Grid B-2:
VU = 189.45 k
γ V 2 M U 2 = −156.39 k-in
γ V 3 M U 3 = 91.538 k-in
EXAMPLE ACI 318-11 RC-PN-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.:
SAFE
0
At the point labeled A in Figure 2, x4 = −10.25 and y4 = 22.25, thus:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
vU =
−
−
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:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
−
−
vU =
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
EXAMPLE ACI 318-11 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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.
ϕ vC
=
ϕ vC
=
=
ϕ vC
4
0.75 2 +
4000
36 /12
= 0.158 ksi in accordance with equation 11-34
1000
40 • 8.5
+ 2 4000
0.75
130
=
0.219 ksi in accordance with equation 11-35
1000
0.75 • 4 • 4000
= 0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of φvC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE ACI 318-11 RC-PN-001 - 6
vU
=
ϕ vC
0.193
= 1.22
0.158
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-11 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. 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 SAFE. The maximum element size is specified to be 36 inches. To
obtain factored moments and flexural reinforcement in a design strip, one onefoot-wide strip is defined in the X-direction on the slab, as shown in Figure 1.
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 SAFE results. After completing
the analysis, design is performed in accordance with ACI 318-11 using SAFE
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
GEOMETRY, PROPERTIES AND LOADING
Thickness
T, h =
6 in
EXAMPLE ACI 318-11 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Depth of tensile reinf.
Effective depth
Clear span
dc =
d
=
ln, l1 =
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
ν
=
=
=
=
=
=
Dead load
Live load
wd
wl
=
=
1 in
5 in
144 in
4,000
60,000
0
3,600
29,000
0
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
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-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-11 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
ϕ = 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f ′ − 4000
0.85 − 0.05 c
0.85
β1 =
=
1000
0.003
=
d 1.875 in
0.003 + 0.005
amax = β1cmax = 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'ϕ b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
= 0.213 sq-in > As,min
a
ϕ fy d −
2
As = 0.2114 sq-in
As =
EXAMPLE ACI 318-11 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-08 PT-SL 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 SAFE. 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 SAFE results
and summarized for verification and validation of the SAFE results.
Loads: Dead = self weight , Live = 100psf
EXAMPLE ACI 318-08 PT-SL 001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Figure 1 One-Way Slab
GEOMETRY, PROPERTIES AND LOADING
Thickness,
Effective depth,
Clear span,
Concrete strength,
Yield strength of steel,
T, h
d
L
f 'C
fy
=
=
=
=
=
10
9
384
4,000
60,000
in
in
in
psi
psi
Prestressing, ultimate
f pu =
270,000 psi
Prestressing, effective
=
175,500 psi
=
=
=
=
=
=
=
0.153
0.150
3,600
29,000
0
self
100
fe
Area of Prestress (single strand), AP
Concrete unit weight,
wc
Modulus of elasticity,
Ec
Modulus of elasticity,
Es
Poisson’s ratio,
ν
Dead load,
wd
Live load,
wl
sq in
pcf
ksi
ksi
psf
psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live and post-tensioning
loads.
EXAMPLE ACI 318-08 PT-SL 001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
RESULTS COMPARISON
The SAFE 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
SAFE
RESULTS
DIFFERENCE
1429.0
1428.3
0.05%
2.20
2.20
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 318-05 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-08 PT-SL 001 - 3
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
CALCULATIONS:
Design Parameters:
φ =0.9
Mild Steel Reinforcing
f′c = 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
ω =0.225 ksf × 3 ft = 0.675 klf,
Ultimate Moment, M U =
ω u = 0.310 ksf × 3ft = 0.930 klf
wl12
= 0.310 klf × 322/8 = 119.0 k-ft = 1429.0 k-in
8
EXAMPLE ACI 318-08 PT-SL 001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
f 'c
(span-to-depth ratio > 35)
300 ρ 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 +
) 61.08 kips
Ultimate force in PT, =
Fult , PT A=
2 ( 0.153)(199.62
=
P ( f PS )
( 60.0 ) 120.0 kips
Ultimate force in RC, =
Fult , RC A=
2.00(assumed)=
s ( f )y
Total Ultimate force, Fult ,Total = 61.08 + 120.0 = 181.08 kips
Stress block =
depth, a
Fult ,Total
181.08
=
= 1.48 in
0.85 f ' cb 0.85 ( 4 ) ( 36 )
a
1.48 ( )
Ultimate moment due to PT, M ult , PT
= Fult , PT d − =
φ 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
φ 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
EXAMPLE ACI 318-08 PT-SL 001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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,
=
57.83
=
M PT F=
( 4 in ) 231.3 k-in
PTI (sag)
FPTI M D − M PT −57.83 576.0 − 231.3
Stress in concrete, f =
, where S = 600 in3
±
= ±
10 ( 36 )
600
A
S
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,
=
53.70
=
M PT F=
( 4 in ) 214.8 k-in
PTI (sag)
FPTI M D + L − M PT −53.70 1037.0 − 214.8
Stress in concrete for (D + L+ PTF), f =
±
=±
10 ( 36 )
600
A
S
f =
−0.149 ± 1.727 ± 0.358
f = −1.518(Comp) max,1.220(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 = 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,
=
M PT F=
53.70
=
( 4 in ) 214.8 k-in
PTI (sag)
EXAMPLE ACI 318-08 PT-SL 001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Stress in concrete for (D + 0.5L + PTF(L)),
FPTI M D + 0.5 L − M PT −53.70 806.0 − 214.8
±
=
±
f =
10 ( 36 )
600
A
S
f =
−0.149 ± 0.985
f = −1.134(Comp) max, 0.836(Tension) max
EXAMPLE ACI 318-08 PT-SL 001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-08 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify the beam flexural design in SAFE. 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 ACI 318-08.
The average shear stress in the beam falls below the maximum shear stress
allowed by ACI 318-08, requiring design shear reinforcement.
A simple-span, 20-foot-long, 12-inch-wide, and 18-inch-deep T beam with a
flange 4 inches thick and 24 inches wide is modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size is
specified as 6 inches. The beam is supported by columns without rotational
stiffnesses and with very large vertical stiffness (1 × 1020 kip/in).
The beam is loaded with symmetric third-point loading. One dead load (DL02)
case and one live load (LL30) case, with only symmetric third-point loads of
magnitudes 3, and 30 kips, respectively, are defined in the model. One load
combination (COMB30) is defined using the ACI 318-08 load combination
factors of 1.2 for dead load and 1.6 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. The total factored
moment and shear force are compared with the SAFE results and found to be
identical. After completing the analysis, the design is performed using the ACI
318-08 code in SAFE and also by hand computation. Table 1 shows the
comparison of the design longitudinal reinforcement. Table 2 shows the
comparison of the design shear reinforcement.
EXAMPLE ACI 318-08 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE ACI 318-08 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span
Overall depth
Flange thickness
Width of web
Width of flange,
Depth of tensile reinf.
Effective depth
Depth of comp. reinf.
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
240
18
4
12
24
3
15
3
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
=
=
=
=
=
=
4,000
60,000
0
3,600
29,000
0.2
Dead load
Live load
Pd
Pl
=
=
2
30
SAFE
0
in
in
in
in
in
in
in
in
psi
psi
pcf
ksi
ksi
kips
kips
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the beam
with the moments obtained by the analytical method. They match exactly for this
problem. Table 1 also shows the comparison of the design reinforcement.
Table 1 Comparison of Moments and Flexural Reinforcements
Reinforcement Area (sq-in)
Method
Moment
(k-in)
As+
SAFE
4032
5.808
Calculated
4032
5.808
A +s ,min = 0.4752 sq-in
EXAMPLE ACI 318-08 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-in/ft)
Shear Force (kip)
SAFE
Calculated
50.40
0.592
0.592
COMPUTER FILE: ACI 318-08 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-08 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.9
Ag = 264 sq-in
As,min = 0.0018Ag = 0.4752 sq-in
f c′ − 4000
0.85
=
1000
β1 =
0.85 − 0.05
=
cmax
0.003
=
d 5.625 in
0.003 + 0.005
amax = β1cmax = 4.78125 in
As = min[As,min, (4/3) As,required] = min[0.4752, (4/3)5.804] = 0.4752 sq-in
COMB30
Pu = (1.2Pd + 1.6Pt) = 50.4 k
Mu =
Pu l
= 4032 k-in
3
The depth of the compression block is given by:
a =−
d
d2 −
2 Mu
0.85 f c'ϕ b f
= 4.2671 in (a > ds)
Calculation for As is performed in two parts. The first part is for balancing the
compressive force from the flange, Cf, and the second part is for balancing the
compressive force from the web, Cw. Cf is given by:
Cf
= 0.85fc' (bf − bw) ds = 163.2 k
The portion of Mu that is resisted by the flange is given by:
Muf =
d
Cf d − s ϕ = 1909.44 k-in
2
EXAMPLE ACI 318-08 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Therefore, the area of tensile steel reinforcement to balance flange compression
is:
As1
=
M uf
f y (d − d s 2 ) ϕ
= 2.7200 sq-in
The balance of the moment to be carried by the web is given by:
Muw = Mu − Muf = 2122.56 k-in
The web is a rectangular section with dimensions bw and d, for which the design
depth of the compression block is recalculated as
= d−
a1
d2 −
2 M uw
= 4.5409 in (a1 ≤ amax)
0.85 f c′ ϕ bw
The area of tensile steel reinforcement to balance the web compression is then
given by:
As2
=
M uw
= 3.0878 sq-in
a1
ϕ fy d − ϕ
2
The area of total tensile steel reinforcement is then given by:
As
= As1 + As2 = 5.808 sq-in
Shear Design
The following quantities are computed for all of the load combinations:
ϕ
=
Check the limit of
0.75
f c′ :
f c′ = 63.246 psi < 100 psi
The concrete shear capacity is given by:
ϕ Vc =
ϕ2
f c′ bwd
= 17.076 k
The maximum shear that can be carried by reinforcement is given by:
EXAMPLE ACI 318-08 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
ϕ Vs =
ϕ8
SAFE
0
f c′ bwd = 68.305 k
The following limits are required in the determination of the reinforcement:
(ϕ Vc/2)
= 8.538 k
(ϕ Vc + ϕ 50 bwd)
= 23.826 k
Vmax = ϕ Vc + ϕ Vs = 85.381 k
Given Vu, Vc and Vmax, the required shear reinforcement in area/unit length for
any load combination is calculated as follows:
If Vu ≤ (Vc/2) ϕ,
Av
= 0,
s
else if (Vc/2) ϕ < Vu ≤ (ϕVc + ϕ 50 bwd),
50 bw
Av
=
,
fy
s
else if (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
(V − ϕVc )
Av
= u
ϕ f ys d
s
else if Vu > ϕ Vmax,
a failure condition is declared.
For each load combination, the Pu and Vu are calculated as follows:
Pu = 1.2Pd + 1.6P1
Vu = Pu
(COMB30)
Pd = 2 k
Pl
= 30 k
Pu = 50.4 k
Vu = 50.4 k, (ϕVc + ϕ 50 bwd) < Vu ≤ ϕ Vmax
Av
=
s
(Vu − ϕVc )
= 0.04937 sq-in/in or 0.592 sq-in/ft
ϕf ys d
EXAMPLE ACI 318-08 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-08 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
B
A
1'
24'
C
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
EXAMPLE ACI 318-08 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 plate
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 SAFE TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
SAFE
0.192
0.158
1.21
Calculated
0.193
0.158
1.22
COMPUTER FILE: ACI 318-08 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE ACI 318-08 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation for Interior Column Using SAFE 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 SAFE Model
γ
V2
γ
V3
1
1−
0.4955
=
=
2 44.5
1+
3 20.5
1
1−
0.3115
=
=
2 20.5
1+
3 44.5
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE ACI 318-08 RC-PN-001 - 3
Software Verification
SAFE
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
Side 2
0
22.25
20.5
8.5
174.25
0
3877.06
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
0
= 0"
1105
2
Ld
0
= 0"
1105
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 SAFE output at Grid B-2:
VU = 189.45 k
γ V 2 M U 2 = −156.39 k-in
γ V 3 M U 3 = 91.538 k-in
EXAMPLE ACI 318-08 RC-PN-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.:
SAFE
0
At the point labeled A in Figure 2, x4 = −10.25 and y4 = 22.25, thus:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
vU =
−
−
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:
156.39 93782.8 ( 22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) (10.25 − 0 )
189.45
−
−
vU =
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:
156.39 93782.8 ( −22.25 − 0 ) − ( 0 ) ( −10.25 − 0 )
189.45
−
−
vU =
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
EXAMPLE ACI 318-08 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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.
ϕ vC
=
ϕ vC
=
=
ϕ vC
4
0.75 2 +
4000
36 /12
= 0.158 ksi in accordance with equation 11-34
1000
40 • 8.5
+ 2 4000
0.75
130
=
0.219 ksi in accordance with equation 11-35
1000
0.75 • 4 • 4000
= 0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of φvC = 0.158 ksi and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE ACI 318-08 RC-PN-001 - 6
vU
=
ϕ vC
0.193
= 1.22
0.158
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE ACI 318-08 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE
A one-way, simple-span slab supported by walls on two opposite edges is
modeled using SAFE. 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 SAFE. The maximum element size is specified to be 36 inches. To
obtain factored moments and flexural reinforcement in a design strip, one onefoot-wide strip is defined in the X-direction on the slab, as shown in Figure 1.
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 SAFE results. After completing
the analysis, design is performed in accordance with ACI 318-08 using SAFE
and also by hand computation. Table 1 shows the comparison of the moments
and design reinforcements computed using the two methods.
GEOMETRY, PROPERTIES AND LOADING
Thickness
T, h =
6 in
EXAMPLE ACI 318-08 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Depth of tensile reinf.
Effective depth
Clear span
dc =
d
=
ln, l1 =
Concrete strength
Yield strength of steel
Concrete unit weight
Modulus of elasticity
Modulus of elasticity
Poisson’s ratio
fc
fy
wc
Ec
Es
ν
=
=
=
=
=
=
Dead load
Live load
wd
wl
=
=
1 in
5 in
144 in
4,000
60,000
0
3,600
29,000
0
psi
psi
pcf
ksi
ksi
80 psf
100 psf
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
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-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE ACI 318-08 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
ϕ = 0.9
b = 12 in
As,min = 0.0018bh = 0.1296 sq-in
f ′ − 4000
0.85 − 0.05 c
0.85
β1 =
=
1000
0.003
=
d 1.875 in
0.003 + 0.005
amax = β1cmax = 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'ϕ b
= 0.3128 in < amax
The area of tensile steel reinforcement is then given by:
Mu
= 0.213 sq-in > As,min
a
ϕ fy d −
2
As = 0.2114 sq-in
As =
EXAMPLE ACI 318-08 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE AS 3600-09 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE AS 3600-09 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification
and validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing and slab stresses with the independent hand calculations.
EXAMPLE AS 3600-09 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
156.12
156.14
0.01%
Area of Mild Steel req’d, As
(sq-cm)
16.55
16.59
0.24%
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
−10.465
0.10%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.407
0.05%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
−7.817
−7.817
0.00%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.759
0.00%
FEATURE TESTED
COMPUTER FILE: AS 3600-09 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE AS 3600-09 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
φ = 0.80
α=
1.0 − 0.003 f 'c = 0.91 > 0.85, Use α 2 = 0.85
2
=
γ 1.0 − 0.003 f 'c = 0.91 > 0.85, Use γ = 0.85
amax = γk 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
ω =10.772 kN/m2 x 0.914m = 9.846 kN/m, ω u = 14.363 kN/m2 x 0.914m = 13.128 kN/m
Ultimate Moment, M U =
EXAMPLE AS 3600-09 PT-SL-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 +
SAFE
0
f 'C bef d P
300 AP
30 ( 914 )( 229 )
= 1210 + 70 +
300 (198 )
= 1386 MPa ≤ f SE +=
200 1410 MPa
) 1000 273.60 kN
Ultimate force in PT,=
Fult , PT A=
197.4 (1386=
P ( f PS )
Total Ultimate force, Fult ,Total = 273.60 + 560.0 = 833.60 kN
2M*
Stress block depth, a =−
d
d −
0.85 f 'c φ 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 ,=
Fult , PT d − =
φ 273.60 229 −
= 45.65 kN-m
0.80 ) 1000
PT
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
φ 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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
=
257.4 (102 mm
=
M PT F=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT (1.15 )( −257.4 ) ( 0.80 ) 65.04 − (1.15 ) 26.23
±
=
±
Stress in concrete, f =
0.254 ( 0.914 )
0.00983
A
S
where S = 0.00983m3
f =
−1.275 ± 2.225 MPa
f = −3.500(Comp) max, 0.950(Tension) max
EXAMPLE AS 3600-09 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 )2 8 52.04 kN-m
Moment
due to live load, M L 4.788
=
=
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment
due to dead load, M L 4.788
=
=
2
Moment due to PT,
M PT F=
=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE AS 3600-09 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE AS 3600-09 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements automatically generated by SAFE. The maximum element size has been
specified to be 200 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. The total factored
moment and shear force are compared with the SAFE results and found to be
identical. After completing the analysis, the design is performed using the AS
3600-09 code in SAFE and also by hand computation. Table 1 shows the
comparison of the design longitudinal reinforcements. Table 2 shows the
comparison of the design shear reinforcements.
EXAMPLE AS 3600-09 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE AS 3600-09 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. 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+
SAFE
462
33.512
Calculated
462
33.512
A +s ,min = 3.00 sq-cm
EXAMPLE AS 3600-09 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
231
12.05
12.05
COMPUTER FILE: AS 3600-09 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-09 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.8
α=
1.0 − 0.003 f 'c = 0.91 > 0.85, Use α 2 = 0.85
2
=
γ 1.05 − 0.007 f 'c = 0.84 < 0.85, Use γ = 0.84
amax = γk u d = 0.84 • 0.36 • 425 = 128.52 mm
2
D f ′ct , f
Ast .min = α b
bw d , where
d f sy
for L- and T-Sections with the web in tension:
1/4
b
b
D
α b = 0.20 + f − 1 0.4 s − 0.18 ≥ 0.20 f
D
bw
bw
=0.2378
2
D f ′ct , f
Ast .min = 0.2378
bd
d f sy
= 0.2378 • (500/425)2 • 0.6 • SQRT(30)/460 • 300*425
= 299.8 sq-mm
COMB130
N* = (1.2Nd + 1.5Nt) = 231kN
M* =
N *l
= 462 kN-m
3
The depth of the compression block is given by:
a =−
d
d2 −
2M *
= 100.755 mm (a > Ds)
0.85 f 'c φ bef
The compressive force developed in the concrete alone is given by:
The first part is for balancing the compressive force from the flange, Cf, and the
second part is for balancing the compressive force from the web, Cw, 2. Cf is
given by:
EXAMPLE AS 3600-09 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
C f = 0.85 f 'c (bef − bw )× min (Ds , amax ) = 765 kN
Therefore, As1 =
Cf
f sy
and the portion of M* that is resisted by the flange is given by:
min (Ds , amax )
M uf = φC f d −
= 229.5 kN-m
2
As1 =
Cf
f sy
= 1663.043 sq-mm
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
0.85 f ′c φ bw
If a1 ≤ amax, the area of tension reinforcement is then given by:
As 2 =
M uw
= 1688.186 sq-mm
a1
φ f sy d −
2
Ast = As1 + As 2 = 3351.23 sq-mm = 33.512 sq-cm
Shear Design
The shear force carried by the concrete, Vuc, is calculated as:
13
A
Vuc = β1β 2 β3bw d o f 'cv st = 0 kN
bw d o
where,
f 'cv = ( f 'c ) = 3.107 N/mm2 ≤ 4MPa
1/3
EXAMPLE AS 3600-09 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
β1 = 1.11.6 −
SAFE
0
do
≥ 1.1 =1.2925, β2 = 1 and β3 = 1
1000
The shear force is limited to a maximum of:
Vu . max = 0.2 f 'c bd o = 765 kN
Given V*, Vuc, and Vu.max, the required shear reinforcement is calculated as
follows, where, φ, the strength reduction factor, is 0.7.
If V * ≤ φVuc / 2,
Asv
= 0 , if D ≤ 750 mm, otherwise Asv.min shall be provided.
s
If φVu.min < V * ≤ φVu.max ,
(
)
V * − φVuc
Asv
=
,
s
φ f sy. f d o cot θ v
and greater than Asv.min, defined as:
Asv. min
b
= 0.35 w
s
f sy . f
= 0.22826 sq-mm/mm = 228.26 sq-mm/m
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*=φ Vu,max = 35.52
degrees
If V * > φVmax , a failure condition is declared.
For load combination, the N* and V* are calculated as follows:
N* = 1.2Nd + 1.5N1
V* = N*
(COMB130)
Nd = 30 kips
Nl
= 130 kips
N* = 231 kN
EXAMPLE AS 3600-09 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
N* = 231 kN, ( φVu.min < V * ≤ φVu.max , )
(
)
V * − φVuc
Asv
=
, = 1.205 sq-mm/mm or 12.05 sq-cm/m
s
φ f sy. f do cot θ v
EXAMPLE AS 3600-09 RC-BM-001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE AS 3600-09 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate properties are used
for the slab.
EXAMPLE AS 3600-09 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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.
TECHNICAL FEATURES OF SAFE 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 SAFE 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)
SAFE
1.811
1.086
1.67
Calculated
1.811
1.086
1.67
COMPUTER FILE: AS 3600-09 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-09 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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
150
A
Column
Critical section for
punching shear shown
dashed.
109
B
109
450
Side 3
Side 1
Side 2
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 SAFE Model
From the SAFE output at grid line B-2:
V* = 1126.498 kN
Mv2 = −51.991 kN-m
Mv3 = 45.723 kN-m
EXAMPLE AS 3600-09 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The maximum design shear stress is computed along the major and minor axis of column
separately:
uM v
V*
=
vmax
1.0 + *
ud om
8V ad om
1126.498 • 103
3272 • 51.991 • 106
=
vmax, X
• 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 +
βh
ϕ fcv = min
0.34ϕ f ′c
ϕ
f ′c
= 1.803N/mm2 in accordance with AS 9.2.3(a)
AS 9.2.3(a) yields the smallest value of ϕ fcv = 1.086 N/mm2, and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE AS 3600-09 RC-PN-001 - 4
vU
=
ϕ f cv
1.811
= 1.67
1.086
Software Verification
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EXAMPLE AS 3600-2009 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE
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.
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 SAFE results. After completing
analysis, design is performed using the AS 3600-2009 code using SAFE and also
by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
EXAMPLE AS 3600-2009 RC-SL-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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
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-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-2009 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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HAND CALCULATION
The following quantities are computed for the load combination:
ϕ = 0.8
b = 1000 mm
α=
1.0 − 0.003 f 'c = 0.91 > 0.85, Use α 2 = 0.85
2
=
γ 1.05 − 0.007 f 'c = 0.84 < 0.85, Use γ = 0.84
amax = γk 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 φ b
EXAMPLE AS 3600-2009 RC-SL-001 - 3
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PROGRAM NAME:
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The area of tensile steel reinforcement is then given by:
Ast =
M*
= 557.966 sq-mm > As,min
a
φf sy d −
2
As = 5.57966 sq-cm
EXAMPLE AS 3600-2009 RC-SL-001 - 4
Software Verification
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EXAMPLE AS 3600-01 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE AS 3600-01 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
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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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing and slab stresses with the independent hand calculations.
EXAMPLE AS 3600-01 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
156.12
156.14
0.01%
16.55
16.59
0.24%
−3.500
−3.498
0.06%
0.950
0.948
0.21%
−10.460
−10.465
0.10%
8.402
8.407
0.05%
−7.817
−7.817
0.00%
5.759
5.759
0.00%
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
COMPUTER FILE: AS 3600-01 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE AS 3600-01 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
φ = 0.80
γ = [0.85 − 0.007( f 'c −28)]= 0.836
amax = γk 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
ω =10.772 kN/m2 x 0.914m = 9.846 kN/m, ω u = 14.363 kN/m2 x 0.914m = 13.128 kN/m
Ultimate Moment, M U =
EXAMPLE AS 3600-01 PT-SL-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 +
SAFE
0
f 'C bef d P
300 AP
30 ( 914 )( 229 )
= 1210 + 70 +
300 (198 )
200 1410 MPa
= 1386 MPa ≤ f SE +=
) 1000 273.60 kN
Ultimate force in PT,=
Fult , PT A=
197.4 (1386=
P ( f PS )
Total Ultimate force, Fult ,Total = 273.60 + 560.0 = 833.60 kN
2M*
Stress block depth, a =−
d
d −
0.85 f 'c φ 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 ,=
Fult , PT d − =
φ 273.60 229 −
= 45.65 kN-m
0.80 ) 1000
PT
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
φ 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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
=
257.4 (102 mm
=
M PT F=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT (1.15 )( −257.4 ) ( 0.80 ) 65.04 − (1.15 ) 26.23
±
=
±
Stress in concrete, f =
0.254 ( 0.914 )
0.00983
A
S
where S = 0.00983m3
f =
−1.275 ± 2.225 MPa
f = −3.500(Comp) max, 0.950(Tension) max
EXAMPLE AS 3600-01 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 )2 8 52.04 kN-m
Moment
due to live load, M L 4.788
=
=
Moment due to PT,
=
238.9 (102 mm
=
M PT F=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment
due to dead load, M L 4.788
=
=
2
Moment due to PT,
M PT F=
=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE AS 3600-01 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE AS 3600-01 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 3600-01.
The average shear stress in the beam is below the maximum shear stress
allowed by AS 3600-01, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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 3600-01 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. The total factored
moment and shear force are compared with the SAFE results and found to be
identical. After completing the analysis, the design is performed using the AS
3600-01 code in SAFE and also by hand computation. Table 1 shows the
comparison of the design longitudinal reinforcements. Table 2 shows the
comparison of the design shear reinforcements.
EXAMPLE AS 3600-01 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE AS 3600-01 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. 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+
SAFE
462
33.512
Calculated
462
33.512
A +s ,min = 3.92 sq-cm
EXAMPLE AS 3600-01 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
231
12.05
12.05
COMPUTER FILE: AS 3100-01 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-01 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
ϕ = 0.8
γ = [0.85 − 0.007( f 'c −28)]= 0.836
amax = γk u d = 0.836 • 0.4 • 425 = 142.12 mm
D f ′cf
Ast .min = 0.22
Ac
d f sy
2
= 0.22 • (500/425)2 • 0.6 • SQRT(30)/460 • 180,000
= 391.572 sq-mm
COMB130
N* = (1.2Nd + 1.5Nt) = 231kN
M* =
N *l
= 462 kN-m
3
The depth of the compression block is given by:
a =−
d
d2 −
2M *
= 100.755 mm (a > Ds)
0.85 f 'c φ bef
The compressive force developed in the concrete alone is given by:
The first part is for balancing the compressive force from the flange, Cf, and the
second part is for balancing the compressive force from the web, Cw, 2. Cf is
given by:
C f = 0.85 f 'c (bef − bw )× min (Ds , amax ) = 765 kN
Therefore, As1 =
Cf
f sy
and the portion of M* that is resisted by the flange is given by:
min (Ds , amax )
M uf = φC f d −
= 229.5 kN-m
2
EXAMPLE AS 3600-01 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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As1 =
Cf
f sy
= 1663.043 sq-mm
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
0.85 f ′c φ bw
If a1 ≤ amax, the area of tension reinforcement is then given by:
As 2 =
M uw
= 1688.186 sq-mm
a1
φ f sy d −
2
Ast = As1 + As 2 = 3351.23 sq-mm = 33.512 sq-cm
Shear Design
The shear force carried by the concrete, Vuc, is calculated as:
13
A f'
Vuc = β1 β 2 β 3bw d o st c = 0 kN
bw d o
d
where, β1 = 1.11.6 − o ≥ 1.1 =1.2925, β2 = 1 and β3 = 1
1000
The shear force is limited to a maximum of:
Vu . max = 0.2 f 'c bd o = 765 kN
Given V*, Vuc, and Vu.max, the required shear reinforcement is calculated as
follows, where, φ, the strength reduction factor, is 0.7.
If V * ≤ φVuc / 2,
Asv
= 0 , if D ≤ 750 mm, otherwise Asv.min shall be provided.
s
EXAMPLE AS 3600-01 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
If φVu.min < V * ≤ φVu.max ,
(
)
V * − φVuc
Asv
=
,
s
φ f sy. f d o cot θ v
and greater than Asv.min, defined as:
Asv. min
b
= 0.35 w
s
f sy . f
= 0.22826 sq-mm/mm = 228.26 sq-mm/m
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*=φ Vu,max = 35.52
degrees
If V * > φVmax , a failure condition is declared.
For load combination, the N* and V* are calculated as follows:
N* = 1.2Nd + 1.5N1
V* = N*
(COMB130)
Nd = 30 kips
Nl
= 130 kips
N* = 231 kN
N* = 231 kN, ( φVu.min < V * ≤ φVu.max , )
(
)
V * − φVuc
Asv
, = 1.205 sq-mm/mm or 12.05 sq-cm/m
=
s
φ f sy. f do cot θ v
EXAMPLE AS 3600-01 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE AS 3600-01 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate properties are used
for the slab.
EXAMPLE AS 3600-01 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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.
TECHNICAL FEATURES OF SAFE 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 SAFE 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)
SAFE
1.811
1.086
1.67
Calculated
1.811
1.086
1.67
COMPUTER FILE: AS 3600-01 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-01 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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
150
A
Column
Critical section for
punching shear shown
dashed.
109
B
109
450
Side 3
Side 1
Side 2
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 SAFE Model
From the SAFE output at grid line B-2:
V* = 1126.498 kN
Mv2 = −51.991 kN-m
Mv3 = 45.723 kN-m
EXAMPLE AS 3600-01 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The maximum design shear stress is computed along the major and minor axis of column
separately:
uM v
V*
=
vmax
1.0 + *
ud om
8V ad om
1126.498 • 103
3272 • 51.991 • 106
=
vmax, X
• 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 +
βh
ϕ fcv = min
0.34ϕ f ′c
ϕ
f ′c
= 1.803N/mm2 in accordance with AS 9.2.3(a)
AS 9.2.3(a) yields the smallest value of ϕ fcv = 1.086 N/mm2, and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE AS 3600-01 RC-PN-001 - 4
vU
=
ϕ f cv
1.811
= 1.67
1.086
Software Verification
PROGRAM NAME:
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EXAMPLE AS 3600-2001 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE
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.
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 SAFE results. After completing
analysis, design is performed using the AS 3600-2001 code using SAFE and also
by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
EXAMPLE AS 3600-2001 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
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-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE AS 3600-2001 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
ϕ = 0.8
b = 1000 mm
γ = [0.85 − 0.007( f 'c −28)]= 0.836
amax = γk 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 φ b
The area of tensile steel reinforcement is then given by:
Ast =
M*
= 557.966 sq-mm > As,min
a
φf sy d −
2
EXAMPLE AS 3600-2001 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
As = 5.57966 sq-cm
EXAMPLE AS 3600-2001 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE BS 8110-97 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE BS 8110-97 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with the independent hand calculations.
EXAMPLE BS 8110-97 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.79
0.71%
−5.058
−5.057
0.02%
2.839
2.839
0.00%
−10.460
−10.465
0. 50%
8.402
8.407
0.06%
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
COMPUTER FILE: BS 8110-97 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE BS 8110-97 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
γm, steel = 1.15
γm, 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 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
ω =10.772 kN/m2 x 0.914m = 9.846 kN/m, ωu = 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 =
7000
1862(198)
=
1210 +
1 − 1.7
9.754 / 0.229
30(914)(229)
= 1358 MPa ≤ 0.7 f pu
= 1303 MPa
EXAMPLE BS 8110-97 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
K factor used to determine the effective depth is given as:
174.4
M
=
= 0.1213 < 0.156
K=
2
2
f cu bd
30000 ( 0.914 )( 0.229 )
K
≤ 0.95d = 192.2 mm
z = d 0.5 + 0.25 −
0
.
9
) 1000 257.2 KN
Fult , PT A=
197.4 (1303=
Ultimate force in PT, =
P ( f PS )
) 1.15 43.00 kN-m
Ultimate moment due to PT,
=
M ult , PT F=
257.2 ( 0.192
=
ult , PT ( z ) / γ
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
) 1000 26.25 kN-m
Moment due to PT, =
257.4 (102mm
=
M PT F=
PTI (sag)
2
FPTI M D − M PT
−257.4
65.04 − 26.23
Stress in concrete, f =
±
=
±
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 )2 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
EXAMPLE BS 8110-97 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 9.431
f = −10.460(Comp) max, 8.402(Tension) max
EXAMPLE BS 8110-97 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE BS 8110-97 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 BS 8110-97.
The average shear stress in the beam is below the maximum shear stress
allowed by BS 8110-97, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20 and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined using the BS 8110-97 load combination
factors of 1.4 for dead loads and 1.6 for live loads. The model is analyzed for
both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the BS 8110-97 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE BS 8110-97 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE BS 8110-97 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span
Overall depth
Flange thickness
Width of web
Width of flange,
Depth of tensile reinf.
Effective depth
Depth of comp. reinf.
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. 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+
SAFE
312
20.90
Calculated
312
20.90
A +s ,min = 195.00 sq-mm
EXAMPLE BS 8110-97 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
6.50
6.50
COMPUTER FILE: BS 8110-97 RCBM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE BS 8110-97 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
As ,min = 0.0013bw h
= 195.00 sq-mm
COMB80
P = (1.4Pd + 1.6Pt) =156 kN
M* =
N *l
= 312 kN-m
3
The depth of the compression block is given by:
M
= 0.095963 < 0.156
f cu b f d 2
K=
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 373.4254 mm
0.9
The depth of the neutral axis is computed as:
x=
1
(d − z) = 114.6102 mm
0.45
And the depth of the compression block is given by:
a = 0.9x = 103.1492 mm > hf
The ultimate resistance moment of the flange is given by:
M=
f
0.67
γc
fcu ( b f − bw ) h f ( d − 0.5h f ) = 150.75 kN-m
The moment taken by the web is computed as:
EXAMPLE BS 8110-97 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
M w = M − M f = 161.25 kN-m
and the normalized moment resisted by the web is given by:
Mw
= 0.0991926 < 0.156
f cu bw d 2
Kw =
If Kw ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam. The reinforcement is calculated as the sum of two parts: one to balance
compression in the flange and one to balance compression in the web.
K
z = d 0.5 + 0.25 − w ≤ 0.95d = 371.3988 mm
0.9
Mf
=
As
fy
γs
( d − 0.5h )
f
+
Mw
= 2090.4 sq-mm
fy
z
γs
Shear Design
v=
V
≤ vmax = 1.2235 MPa
bw d
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
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
γ m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
f 3
40
k2 = cu = 1.06266, 1 ≤ k2 ≤
25
25
γm = 1.25
100 As
= 0.15
bd
EXAMPLE BS 8110-97 RC-BM-001 - 6
1
3
Software Verification
PROGRAM NAME:
REVISION NO.:
400
d
1
SAFE
0
4
=1
However, the following limitations also apply:
0.15 ≤
100 As
≤3
bd
400
d
1
4
≥1
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Given v, vc, and vmax, the required shear reinforcement is calculated as follows:
If v ≤ (vc + 0.4)
Asv
0.4bw
=
sv
0.87 f yv
If (vc + 0.4) < v ≤ vmax
Asv (v − vc )bw
=
sv
0.87 f yv
If v > vmax, a failure condition is declared.
(COMB80)
Pd = 20 kN
Pl
= 80 kN
V
= 156 kN
Asv (v − vc )bw
= 0.64967 sq-mm/mm = 649.67 sq-mm/m
=
sv
0.87 f yv
EXAMPLE BS 8110-97 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE BS 8110-97 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE BS 8110-97 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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)
SAFE
1.105
0.625
1.77
Calculated
1.105
0.625
1.77
COMPUTER FILE: BS 8110-97 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE BS 8110-97 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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 SAFE Model
From the SAFE output at Grid B-2:
V = 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
EXAMPLE BS 8110-97 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Maximum design shear stress in computed in along major and minor axis of column:
veff , x
=
1.5M
V
x
f +
ud
Vy
(BS 3.7.7.3)
1126.498 • 103
1.5 • 51.9908 • 106
veff , x =
+
1
0
1.1049 (Govern)
.
=
5016 • 218
1126.498 • 103 • 954
=
veff , y
1.5M
V
y
f+
ud
Vx
1126.498 • 103
1.5 • 45.7234 • 106
veff , y =
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
γ 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
γm = 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
EXAMPLE BS 8110-97 RC-PN-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
=
v
1.1049
= 1.77
0.6247
EXAMPLE BS 8110-97 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE BS 8110-97 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
the analysis, design was performed using the BS 8110-97 code by SAFE and also
by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
EXAMPLE BS 8110-97 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
27.197
5.853
Calculated
27.200
5.850
As+
Medium
A +s ,min = 162.5 sq-mm
EXAMPLE BS 8110-97 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: BS 8110-97 RC-SL-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE BS 8110-97 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
γm, steel
= 1.15
γm, 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
EXAMPLE BS 8110-97 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-14 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE CSA A23.3-14 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
EXAMPLE CSA A23.3-14 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
SAFE
RESULTS
DIFFERENCE
159.4
159.4
0.00%
16.25
16.32
0.43%
−5.058
−5.057
-0.02%
2.839
2.839
0.00%
−10.460
−10.465
0.05%
8.402
8.407
0.06%
−7.817
−7.817
0.00%
5.759
5.759
0.00%
COMPUTER FILE: CSA A23.3-14 PT-SL-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA A23.3-14 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
φc = 0.65 , φS = 0.85
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 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
ω =10.772 kN/m2 x 0.914m = 9.846 kN/m, ω u = 16.039 kN/m2 x 0.914m = 13.401 kN/m
Ultimate Moment, M U =
EXAMPLE CSA A23.3-14 PT-SL-001 - 4
wl12
= 13.401 x (9.754)2/8 = 159.42 kN-m
8
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Stress in strand, f pb =
f pe +
=
cy
SAFE
0
8000
( d p − cy )
lo
φ p Ap f pr + φs As f y 0.9 (197 )(1347 ) + 0.85 (1625 )( 400 )
=
= 61.66 mm
0.805 ( 0.65 )( 30.0 )( 0.895 )( 914 )
α1φc f 'c β1b
1210 +
f pb =
8000
( 229 − 61.66 ) =1347 MPa
9754
Depth of the compression block, a, is given as:
Stress block depth, a =−
d
d2 −
2M *
α 1 f 'c φc b
=
0.229 − 0.2292 −
2 (159.42 )
=
55.18
0.805 ( 30000 )( 0.65 )( 0.914 )
) 1000 265.9 kN
Fult , PT A=
197 (1347=
Ultimate force in PT, =
P ( f PS )
Ultimate moment due to PT,
a
55.18
) 45.52 kN-m
=
M ult ,=
Fult , PT d − =
φ 265.9 0.229 −
( 0.85
PT
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
EXAMPLE CSA A23.3-14 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
65.04 − 26.23
−257.4
Stress in concrete, f =
±
=
±
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
M PT F=
=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE CSA A23.3-14 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-14 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 CSA A23.3-14.
The average shear stress in the beam is below the maximum shear stress
allowed by CSA A23.3-14, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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 (LL100) with only symmetric third-point loads of
magnitudes 30, and 100 kN, respectively, are defined in the model. One load
combinations (COMB100) is defined using the CSA A23.3-14 load combination
factors of 1.25 for dead loads and 1.5 for live loads. The model is analyzed for
both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the CSA A23.3-14 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE CSA A23.3-14 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE CSA A23.3-14 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
100
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. 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+
SAFE
375
25.844
Calculated
375
25.844
A +s ,min = 535.82 sq-m
EXAMPLE CSA A23.3-14 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
187.5
12.573
12.573
COMPUTER FILE: CSA A23.3-14 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA A23.3-14 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
bw h = 357.2 sq-mm
fy
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.895
cb =
700
d = 256.46 mm
700 + f y
ab = β1cb = 229.5366 mm
As = min[As,min, (4/3) As,required] = min[357.2, (4/3)2445] = 357.2 sq-mm
COMB100
P = (1.25Pd + 1.5Pt) =187.5kN
M* =
Pl
= 375 kN-m
3
Mf = 375 kN-m
The depth of the compression block is given by:
C f α1 f ′c ( b f − bw ) min ( hs , ab ) = 724.5 kN
=
Therefore, As1 =
C f φc
f yφ s
and the portion of Mf that is resisted by the flange is given
by:
EXAMPLE CSA A23.3-14 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
C f φc
As1 =
f yφ s
= 1204.411 sq-mm
min (hs , ab )
M ff = C f d −
φc = 176.596 kN-m
2
Therefore, the balance of the moment, Mf to be carried by the web is:
Mfw = Mf − Mff = 198.403 kN-m
The web is a rectangular section with dimensions bw and d, for which the design
depth of the compression block is recalculated as:
a1 = d − d 2 −
2 M fw
α1 f 'c φc bw
= 114.5745 mm
If a1 ≤ ab, the area of tension reinforcement is then given by:
As 2 =
M fw
a
φs f y d − 1
2
= 1379.94 sq-mm
As = As1 + As2 = 2584.351 sq-mm
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 0.9d or 0.72h =
382.5 mm (governing) or 360 mm.
S ze = 300 if minimum transverse reinforcement
εx =
M f d v + V f + 0.5 N f
EXAMPLE CSA A23.3-14 RC-BM-001 - 6
2(E s As )
and ε x ≤ 0.003
Software Verification
PROGRAM NAME:
REVISION NO.:
=
β
SAFE
0
0.40
1300
= 0.07272
•
(1 + 1500ε x ) (1000 + S ze )
Vc = φc λβ f ′c bw dv = 29.708 kN
Vr ,max = 0.25φc f 'c bw d = 621.56 kN
θ = 50
Av (V f − Vc ) tan θ
= 1.2573 mm2/mm = 12.573 cm2/m.
=
s
φ s f yt d v
EXAMPLE CSA A23.3-14 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-14 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE CSA A23.3-14 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.792
1.127
1.59
Calculated
1.792
1.127
1.59
COMPUTER FILE: CSA A23.3-14 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA A23.3-14 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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
A
Column
Critical section for
punching shear shown
dashed.
109
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 SAFE Model
γ
V2
γ
V3
1
1−
0.495
=
=
2 1118
1+
3 518
1
1−
0.312
=
=
2 518
1+
3 1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE CSA A23.3-14 RC-PN-001 - 3
Software Verification
SAFE
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
2
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
= 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 SAFE output at Grid B-2:
Vf = 1126.498 kN
γ
γ
V2
Mf,2 = −25.725 kN-m
V3
Mf,3 = 14.272 kN-m
EXAMPLE CSA A23.3-14 RC-PN-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.:
SAFE
0
At the point labeled A 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 = 1.3666 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:
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 = 1.792 N/mm2 at point C
At the point labeled D 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 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
EXAMPLE CSA A23.3-14 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The shear capacity is calculated based on the minimum of the following three limits:
2
φc 1 + β 0.19λ f ′c
c
αd
vv min φc 0.19 + s λ f ′c
=
b0
φ 0.38λ 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
=
ϕ vv
1.792
= 1.59
1.127
EXAMPLE CSA A23.3-14 RC-PN-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-14 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
the analysis, design is performed using the CSA A23.3-14 code by SAFE and
also by hand computation. Table 1 show the comparison of the design
reinforcements computed using the two methods.
EXAMPLE CSA A23.3-14 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
25.00
5.414
Calculated
25.00
5.528
As+
Medium
A +s ,min = 357.2 sq-mm
EXAMPLE CSA A23.3-14 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: CSA A23.3-14 RC-SL-001.FDB
CONCLUSION
The SAFE results show a very close comparison with the independent results.
EXAMPLE CSA A23.3-14 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
φc = 0.65 for concrete
φs = 0.85 for reinforcement
As,min =
0.2 f ′c
bw h = 357.2 sq-mm
fy
b = 1000 mm
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.895
cb =
700
d = 75.43 mm
700 + f y
ab = β1cb = 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:
EXAMPLE CSA A23.3-14 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
a = d − d2 −
2M f
α 1 f 'c φc b
SAFE
0
= 13.769 mm < amax
The area of tensile steel reinforcement is then given by:
As =
Mf
a
φs f y d −
2
= 552.77 sq-mm > As,min
As = 5.528 sq-cm
EXAMPLE CSA A23.3-14 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-04 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE CSA A23.3-04 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.792
1.127
1.59
Calculated
1.792
1.127
1.59
COMPUTER FILE: CSA A23.3-04 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA A23.3-04 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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
A
Column
Critical section for
punching shear shown
dashed.
109
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 SAFE Model
γ
V2
γ
V3
1
1−
0.495
=
=
2 1118
1+
3 518
1
1−
0.312
=
=
2 518
1+
3 1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE CSA A23.3-04 RC-PN-001 - 3
Software Verification
SAFE
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
2
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
= 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 SAFE output at Grid B-2:
Vf = 1126.498 kN
γ
γ
V2
Mf,2 = −25.725 kN-m
V3
Mf,3 = 14.272 kN-m
EXAMPLE CSA A23.3-04 RC-PN-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.:
SAFE
0
At the point labeled A 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 = 1.3666 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:
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 = 1.792 N/mm2 at point C
At the point labeled D 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 = 1.6004 N/mm2 at point D
Point C has the largest absolute value of vu, thus vmax = 1.792 N/mm2
EXAMPLE CSA A23.3-04 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The shear capacity is calculated based on the minimum of the following three limits:
2
φc 1 + β 0.19λ f ′c
c
αd
vv min φc 0.19 + s λ f ′c
=
b0
φ 0.38λ 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
=
ϕ vv
1.792
= 1.59
1.127
EXAMPLE CSA A23.3-04 RC-PN-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA 23.3-04 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE CSA 23.3-04 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE TESTED
Calculation of the required flexural reinforcement
Check of slab stresses due to the application of dead, live, and post-tensioning
loads.
EXAMPLE CSA 23.3-04 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
SAFE
RESULTS
DIFFERENCE
159.4
159.4
0.00%
16.25
16.32
0.43%
−5.058
−5.057
0.02%
2.839
2.839
0.00%
−10.460
−10.465
0.05%
8.402
8.407
0.06%
−7.817
−7.817
0.00%
5.759
5.759
0.00%
COMPUTER FILE: CSA A23.3-04 PT-SL-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA 23.3-04 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
φc = 0.65 , φS = 0.85
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 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
ω =10.772 kN/m2 x 0.914m = 9.846 kN/m, ω u = 16.039 kN/m2 x 0.914m = 13.401 kN/m
Ultimate Moment, M U =
EXAMPLE CSA 23.3-04 PT-SL-001 - 4
wl12
= 13.401 x (9.754)2/8 = 159.42 kN-m
8
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Stress in strand, f pb =
f pe +
=
cy
SAFE
0
8000
( d p − cy )
lo
φ p Ap f pr + φs As f y 0.9 (197 )(1347 ) + 0.85 (1625 )( 400 )
=
= 61.66 mm
0.805 ( 0.65 )( 30.0 )( 0.895 )( 914 )
α1φc f 'c β1b
1210 +
f pb =
8000
( 229 − 61.66 ) =1347 MPa
9754
Depth of the compression block, a, is given as:
Stress block depth, a =−
d
d2 −
2M *
α 1 f 'c φc b
=
0.229 − 0.2292 −
2 (159.42 )
=
55.18
0.805 ( 30000 )( 0.65 )( 0.914 )
) 1000 265.9 kN
Fult , PT A=
197 (1347=
Ultimate force in PT, =
P ( f PS )
Ultimate moment due to PT,
a
55.18
) 45.52 kN-m
=
M ult ,=
Fult , PT d − =
φ 265.9 0.229 −
( 0.85
PT
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
EXAMPLE CSA 23.3-04 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
−257.4
65.04 − 26.23
Stress in concrete, f =
±
=
±
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
=
238.9 (102 mm
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE CSA 23.3-04 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-04 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 CSA A23.3-04.
The average shear stress in the beam is below the maximum shear stress
allowed by CSA A23.3-04, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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 (LL100) with only symmetric third-point loads of
magnitudes 30, and 100 kN, respectively, are defined in the model. One load
combinations (COMB100) is defined using the CSA A23.3-04 load combination
factors of 1.25 for dead loads and 1.5 for live loads. The model is analyzed for
both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the CSA A23.3-04 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE CSA A23.3-04 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE CSA A23.3-04 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
100
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. 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+
SAFE
375
25.844
Calculated
375
25.844
A +s ,min = 535.82 sq-m
EXAMPLE CSA A23.3-04 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
187.5
12.573
12.573
COMPUTER FILE: CSA A23.3-04 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE CSA A23.3-04 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
bw h = 357.2 sq-mm
fy
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.895
cb =
700
d = 256.46 mm
700 + f y
ab = β1cb = 229.5366 mm
As = min[As,min, (4/3) As,required] = min[357.2, (4/3)2445] = 357.2 sq-mm
COMB100
P = (1.25Pd + 1.5Pt) =187.5kN
M* =
Pl
= 375 kN-m
3
Mf = 375 kN-m
The depth of the compression block is given by:
=
C f α1 f ′c ( b f − bw ) min ( hs , ab ) = 724.5 kN
Therefore, As1 =
C f φc
f yφ s
and the portion of Mf that is resisted by the flange is given
by:
EXAMPLE CSA A23.3-04 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
C f φc
As1 =
f yφ s
= 1204.411 sq-mm
min (hs , ab )
M ff = C f d −
φc = 176.596 kN-m
2
Therefore, the balance of the moment, Mf to be carried by the web is:
Mfw = Mf − Mff = 198.403 kN-m
The web is a rectangular section with dimensions bw and d, for which the design
depth of the compression block is recalculated as:
a1 = d − d 2 −
2 M fw
α1 f 'c φc bw
= 114.5745 mm
If a1 ≤ ab, the area of tension reinforcement is then given by:
As 2 =
M fw
a
φs f y d − 1
2
= 1379.94 sq-mm
As = As1 + As2 = 2584.351 sq-mm
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 0.9d or 0.72h =
382.5 mm (governing) or 360 mm.
S ze = 300 if minimum transverse reinforcement
εx =
M f d v + V f + 0.5 N f
EXAMPLE CSA A23.3-04 RC-BM-001 - 6
2(E s As )
and ε x ≤ 0.003
Software Verification
PROGRAM NAME:
REVISION NO.:
=
β
SAFE
0
0.40
1300
= 0.07272
•
(1 + 1500ε x ) (1000 + S ze )
Vc = φc λβ f ′c bw dv = 29.708 kN
Vr ,max = 0.25φc f 'c bw d = 621.56 kN
θ = 50
Av (V f − Vc ) tan θ
= 1.2573 mm2/mm = 12.573 cm2/m.
=
s
φ s f yt d v
EXAMPLE CSA A23.3-04 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE CSA A23.3-04 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
the analysis, design is performed using the CSA A23.3-04 code by SAFE and
also by hand computation. Table 1 show the comparison of the design
reinforcements computed using the two methods.
EXAMPLE CSA A23.3-04 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
25.00
5.414
Calculated
25.00
5.528
As+
Medium
A +s ,min = 357.2 sq-mm
EXAMPLE CSA A23.3-04 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: CSA A23.3-04 RC-SL-001.FDB
CONCLUSION
The SAFE results show a very close comparison with the independent results.
EXAMPLE CSA A23.3-04 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
φc = 0.65 for concrete
φs = 0.85 for reinforcement
As,min =
0.2 f ′c
bw h = 357.2 sq-mm
fy
b = 1000 mm
α1 = 0.85 – 0.0015f'c ≥ 0.67 = 0.805
β1 = 0.97 – 0.0025f'c ≥ 0.67 = 0.895
cb =
700
d = 75.43 mm
700 + f y
ab = β1cb = 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:
EXAMPLE CSA A23.3-04 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
a = d − d2 −
2M f
α 1 f 'c φc b
SAFE
0
= 13.769 mm < amax
The area of tensile steel reinforcement is then given by:
As =
Mf
a
φs f y d −
2
= 552.77 sq-mm > As,min
As = 5.528 sq-cm
EXAMPLE CSA A23.3-04 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Eurocode 2-04 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE Eurocode 2-04 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification
and validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with independent hand calculations.
EXAMPLE Eurocode 2-04 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
166.41
166.41
0.00%
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
−10.465
0.05%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.407
0.06%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
−7.817
−7.817
0.00%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.759
0.00%
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+
SAFE
166.41
15.39
Calculated
166.41
15.36
SAFE
166.41
15.89
Calculated
166.41
15.87
SAFE
166.41
15.96
Calculated
166.41
15.94
Finland , Singapore and UK
Denmark
COMPUTER FILE: EUROCODE 2-04 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Eurocode 2-04 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
γm, steel = 1.15
γm, concrete = 1.50
η = 1.0 for fck ≤ 50 MPa
λ = 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)
Total = 10.772 kN/m2 (D+L)
= 15.260 kN/m2 (D+L)ult
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 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
EXAMPLE Eurocode 2-04 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
f A
f SE + 7000d 1 − 1.36 PU P l
Ultimate Stress in strand, f PS =
fCK bd
1862(198) (
1210 + 7000(229) 1 − 1.36
=
9754 )
30(914) ( 229 )
= 1361 MPa
) 1000 269.5 kN
Fult , PT A=
2 ( 99 )(1361=
Ultimate force in PT,=
P ( f PS )
CEN Default, Norway, Slovenia and Sweden:
Design moment M = 166.4122 kN-m
M
Compression block depth ratio: m = 2
bd ηf cd
166.4122
= =
0.1736
( 0.914 )( 0.229 )2 (1) ( 30000 1.50 )
Required area of mild steel reinforcing,
0.1920
ω = 1 − 1 − 2m = 1 − 1 − 2(0.1736) =
η f cd bd
1(30 /1.5)(914)(229)
2
=
=
AEquivTotal ω=
0.1920
2311 mm
400 /1.15
f yd
1361
2
AEquivTotal AP
=
=
+ AS 2311 mm
400
1.15
1361
2311 − 198
1536 mm 2
AS =
=
400 /1.15
Finland, Singapore and UK:
Design moment M = 166.4122 kN-m
Compression block depth ratio: m =
M
bd 2ηf cd
166.4122
=
0.2042
( 0.914 )( 0.229 )2 ( 0.85 ) ( 30000 1.50 )
Required area of mild steel reinforcing,
ω = 1 − 1 − 2m = 1 − 1 − 2(0.2042) =
0.23088
=
EXAMPLE Eurocode 2-04 PT-SL-001 - 5
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
η f cd bd
0.85(30 /1.5)(914)(229)
2
=
=
AEquivTotal ω=
0.23088
2362 mm
400 /1.15
f yd
1361
2
=
=
AEquivTotal AP
+ AS 2362 mm
400 1.15
1361
2362 − 198
1587 mm 2
AS =
=
400 1.15
Denmark:
Design moment M = 166.4122 kN-m
Compression block depth ratio: m =
M
bd 2ηf cd
166.4122
=
0.1678
( 0.914 )( 0.229 )2 (1.0 ) ( 30000 1.45 )
Required area of mild steel reinforcing,
0.1849
ω = 1 − 1 − 2m = 1 − 1 − 2(0.1678) =
=
η f bd
1.0(30 /1.45)(914)(229)
2
AEquivTotal ω=
=
=
cd 0.1849
2402 mm
400 /1.20
f yd
1361
2
=
=
AEquivTotal AP
+ AS 2402 mm
400
1.2
1361
AS =
2402 − 198
1594 mm 2
=
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
65.04 − 26.23
−257.4
Stress in concrete, f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
Moment due to PT,
EXAMPLE Eurocode 2-04 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment
due to dead load M D 5.984
=
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment
due to live load M L 4.788
=
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE Eurocode 2-04 PT-SL-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Eurocode 2-04 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 Eurocode 2-04.
The average shear stress in the beam is below the maximum shear stress
allowed by Eurocode 2-04, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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 Eurocode 2-04 load combination
factors of 1.35 for dead loads and 1.5 for live loads. The model is analyzed for
both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Eurocode 2-04 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE Eurocode 2-04 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE Eurocode 2-04 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
425
75
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
f'ck
fy
wc
Ec
Es
v
=
=
=
=
=
=
30
460
0
25x105
2x108
0.2
Dead load,
Live load,
Pd
Pl
=
=
30
130
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. They match exactly
for this problem. Table 1 also shows the comparison of design reinforcements.
EXAMPLE Eurocode 2-04 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Moments and Flexural Reinforcements
National Annex
CEN Default, Norway, Slovenia
and Sweden
Moment
Reinforcement Area
(sq-cm)
Method
(kN-m)
As+
SAFE
471
31.643
Calculated
471
31.643
SAFE
471
32.98
Calculated
471
32.98
SAFE
471
32.83
Calculated
471
32.83
Finland , Singapore and UK
Denmark
A +s ,min = 2.09 sq-cm
Table 2 Comparison of Shear Reinforcements
Reinforcement Area ,
National Annex
CEN Default, Norway,
Slovenia and Sweden
Shear
Force
(sq-cm/m)
Method
(kN)
As+
SAFE
235.5
6.16
Calculated
235.5
6.16
SAFE
235.5
6.16
Calculated
235.5
6.16
SAFE
235.5
6.42
Calculated
235.5
6.42
Finland , Singapore and UK
Denmark
COMPUTER FILE: Eurocode 2-04 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Eurocode 2-04 RC-BM-001 - 4
Av
s
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for both of the load combinations:
γs
= 1.15
γc
= 1.50
f cd =
α cc f ck / γ c
=
f yd f yk / γ s
η = 1.0 for fck ≤ 50 MPa
λ = 0.8 for fck ≤ 50 MPa
As ,min = 0.26
f ctm
bd = 208.73 sq-mm
f yk
As ,min = 0.0013bw h = 195.00 sq-mm
For CEN Default, Norway, Slovenia and Sweden—COMB130:
γm, steel = 1.15
γm, concrete = 1.50
αcc = 1.0
The depth of the compression block is given by:
M
471 • 106
= 0.217301
=
m =
bd 2η f cd 600 • 4252 • 1.0 • 1.0 • 30 /1.5
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
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.448
=
k2
d lim
EXAMPLE Eurocode 2-04 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
x λx
mlim = λ 1 − = 0.29417
d lim 2 d lim
x
= 1 − 1 − 2mlim = 0.3584
d lim
amax = ωlimd = 152.32 mm
ωlim = λ
ω = 1 − 1 − 2m = 0.24807
a = ωd = 105.4299 mm ≤ amax
As 2 =
(b
f
− bw ) h f η f cd
f yd
hf
M 2 = As 2 f yd d −
2
= 1500 sq-mm
= 225 kN-m
M1 = M − M2 = 246 kN-m
m1 =
M1
= 0.2269896 ≤ mlim
bw d 2η f cd
ω1 = 1 − 1 − 2m1 = 0.2610678
η f b d
As1 = ω1 cd w = 1664.304 sq-mm
f yd
As = As1 + As2 = 3164.307 sq-mm
For Singapore and UK—COMB130:
γm, steel = 1.15
γm, concrete = 1.50
αcc = 0.85
The depth of the compression block is given by:
M
471 • 106
= 0.255648
=
m =
bd 2η f cd 600 • 4252 • 1.0 • 0.85 • 30 /1.5
For reinforcement with fyk ≤ 500 MPa, the following values are used:
k1 = 0.40
EXAMPLE Eurocode 2-04 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
k2 = (0.6 + 0.0014/εcu2) = 1.00
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.60
=
k2
d lim
x λx
mlim = λ 1 − = 0.3648
d lim 2 d lim
x
= 1 − 1 − 2mlim = 0.48
d lim
amax = ωlimd = 204 mm
ωlim = λ
ω = 1 − 1 − 2m = 0.300923
a = ωd = 127.8939 mm ≤ amax
As 2 =
(b
f
− bw ) h f η f cd
f yd
hf
M 2 = As 2 f yd d −
2
= 1275 sq-mm
=191.25 kN-m
M1 = M − M2 = 279.75 kN-m
m1 =
M1
= 0.30368 ≤ mlim
bw d 2η f cd
ω1 = 1 − 1 − 2m1 = 0.37339
ηf b d
As1 = ω1 cd w
f yd
0.85 • 30
1.0 • 1.5 • 300 • 425
= 0.37339
= 2023.307 sq-mm
400
As = As1 + As2 = 3298.31 sq-mm
For Finland—COMB130:
γm, steel = 1.15
EXAMPLE Eurocode 2-04 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
γm, concrete = 1.50
αcc = 0.85
The depth of the compression block is given by:
=
m
M
471 • 106
= 0.255648
=
bd 2η f cd 600 • 4252 • 1.0 • 0.85 • 30 /1.5
For reinforcement with fyk ≤ 500 MPa, the following values are used:
k1 = 0.44
k2 = 1.10
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.5091
=
k2
d lim
x λx
mlim = λ 1 − = 0.3243
d lim 2 d lim
x
= 1 − 1 − 2mlim = 0.40728
d lim
amax = ωlimd = 173.094 mm
ωlim = λ
ω = 1 − 1 − 2m = 0.300923
a = ωd = 127.8939 mm ≤ amax
As 2 =
(b
f
− bw ) h f η f cd
f yd
hf
M 2 = As 2 f yd d −
2
= 1275 sq-mm
=191.25 kN-m
M1 = M − M2 = 279.75 kN-m
m1 =
M1
= 0.30368 ≤ mlim
bw d 2η f cd
ω1 = 1 − 1 − 2m1 = 0.37339
ηf b d
As1 = ω1 cd w
f yd
EXAMPLE Eurocode 2-04 RC-BM-001 - 8
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
0.85 • 30
1.0 • 1.5 • 300 • 425
= 0.37339
= 2023.307 sq-mm
400
As = As1 + As2 = 3298.31 sq-mm
For Denmark—COMB130:
γm, steel = 1.20
γm, concrete = 1.45
αcc = 1.0
The depth of the compression block is given by:
=
m
M
471 • 106
= 0.210058
=
bd 2η f cd 600 • 4252 • 1.0 • 1.0 • 30 /1.45
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
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.448
=
k2
d lim
x λx
mlim = λ 1 − = 0.29417
d lim 2 d lim
x
= 1 − 1 − 2mlim = 0.3584
d lim
amax = ωlimd = 152.32 mm
ωlim = λ
ω = 1 − 1 − 2m = 0.238499
a = ωd = 101.3620 mm ≤ amax
As 2 =
(b
f
− bw ) h f η f cd
f yd
= 1619.19 sq-mm
EXAMPLE Eurocode 2-04 RC-BM-001 - 9
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
hf
M 2 = As 2 f yd d −
2
= 232.76 kN-m
M1 = M − M2 = 238.24 kN-m
m1 =
M1
= 0.21250 ≤ mlim
bw d 2η f cd
ω1 = 1 − 1 − 2m1 = 0.241715
η f b d
As1 = ω1 cd w = 1663.37 sq-mm
f yd
As = As1 + As2 = 3282.56 sq-mm
Shear Design
For CEN Default, Finland, Singapore, Slovenia and UK
C Rd ,c = 0.18 γ c = 0.18/1.5 = 0.12
For Denmark
CRd ,c = 0.18 γ c = 0.18/1.45 = 0.124
For Sweden and Norway
CRd ,c = 0.15 γ c = 0.15/1.5 = 0.10
k=
1+
200
=
1.686 ≤ 2.0 with d in mm
d
ρ1 = 0.0
=
σcp N Ed Ac < 0.2 f cd = 0.0 MPa
For CEN Default, Denmark, Norway, Singapore, Slovenia, Sweden and
UK:
ν min = 0.035k 3 2 f ck 1 2 = 0.419677
For Finland:
ν min = 0.035k 2/3 f ck1 2 = 0.271561
13
VRd ,c C Rd ,c k (100 ρ1 fck ) + k1σ cp bw d = 34.62 kN for Finland
=
EXAMPLE Eurocode 2-04 RC-BM-001 - 10
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
13
VRd ,c C Rd ,c k (100 ρ1 fck ) + k1σ cp bw d = 53.5 kN for all other NA
=
αcw = 1
ν 1 = 0.61 −
f ck
= 0.528
250
z = 0.9d = 382.5 mm
θ is taken as 1.
VRd ,max =
α cwbw zν 1 f cd
= 1253.54 kN for Denmark
cot θ + tan θ
VRd ,max =
α cwbw zν 1 f cd
= 1211.76 kN for all other NA
cot θ + tan θ
VR,dc < VEd ≤ VRd,max (govern)
Computing the angle using vEd :
vEd =
235.5 • 103
= 2.0522
0.9 • 425 • 300
θ = 0.5sin −1
vEd
0.2 f ck (1 − f ck 250 )
θ = 0.5sin −1
2.0522
= 11.43°
0.2 • 30 (1 − 30 250 )
21.8° ≤ θ ≤ 45° , therefore use=
θ 21.8°
Asw
vEd bw
=
s
f ywd cot θ
Asw
2.0522 • 300
= 0.64243 sq-mm/m = 6.42 sq-cm/m for Denmark
=
460 1.20 • 2.5
s
Asw
2.0522 • 300
= 0.61566 sq-mm/m = 6.16 sq-cm/m for all other NA
=
460 1.15 • 2.5
s
EXAMPLE Eurocode 2-04 RC-BM-001 - 11
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE EUROCODE 2-04 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE EUROCODE 2-04 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.100
0.578
1.90
Calculated
1.099
0.578
1.90
SAFE
1.100
0.5796
1.90
Calculated
1.099
0.5796
1.90
SAFE
1.100
0.606
1.82
Calculated
1.099
0.606
1.81
COMPUTER FILE: EUROCODE 2-04 RC-PN-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE EUROCODE 2-04 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation for Interior Column using SAFE Method
d=
[( 250 − 26 ) + ( 250 − 38)]
2 = 218 mm
Refer to Figure 2.
u1 = u = 2•300 + 2•900 + 2•π•436 = 5139.468 mm
1172
Note: All dimensions in millimeters
Critical section for
punching shear
shown dashed.
Y
436
150 150 436
B
A
Column
Side 2
Side 1
Side 3
436
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 SAFE Model
From the SAFE output at Grid B-2:
VEd = 1112.197 kN
k2MEd2 = 41.593 kN-m
k3MEd3 = 20.576 kN-m
EXAMPLE EUROCODE 2-04 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Maximum design shear stress in computed in along major and minor axis of column:
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
+
vEd =
1 +
ud
VEdW1,2
VEdW1,3
(EC2 6.4.4(2))
c2
W1 = 1 + c1c2 + 4c2 d + 16d 2 + 2π dc1
2
W1,2
=
9002
+ 300 • 900 + 4 • 300 • 218 + 16 • 2182 + 2π • 218 • 900
2
W1,2 = 2,929, 744.957 mm2
W
3
=
1,3
9002
+ 900 • 300 + 4 • 900 • 218 + 16 • 2182 + 2π • 218 • 300
2
W1,2 = 2, 271,104.319 mm2
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
+
vEd =
1 +
u1d
VEdW1,2
VEdW1,3
1112.197 • 103
41.593 • 106 • 5139.468
20.576 • 106 • 5139.468
1
vEd =
+
+
5139.468 • 218 1112.197 • 103 • 2929744.957 1112.197 • 103 • 2271104.319
vEd = 1.099 N/mm2
Thus vmax = 1.099 N/mm2
For CEN Default, Finland, Norway, Singapore, Slovenia, Sweden and UK:
C Rd ,c = 0.18 γ c = 0.18/1.5 = 0.12
(EC2 6.4.4)
For Denmark:
CRd ,c = 0.18 γ 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 ρ1 fck ) + k1σ cp
=
(EC2 6.4.4)
with a minimum of:
v=
Rd ,c
(v
min
+ k1σ cp )
EXAMPLE EUROCODE 2-04 RC-PN-001 - 4
(EC2 6.4.4)
Software Verification
PROGRAM NAME:
REVISION NO.:
k=
1+
200
≤ 2.0 = 1.9578
d
k1
= 0.15.
ρ1 =
As1
≤ 0.02
bw d
SAFE
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
ρ1 = 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
ρ1 = 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
ρ1 = 9020.548 ( 8000 • 218 ) = 0.005172 ≤ 0.02
EXAMPLE EUROCODE 2-04 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
For CEN Default, Denmark, Norway, Singapore, Slovenia, Sweden and UK:
ν min = 0.035k 3 2 f ck 1 2 = 0.035 (1.9578 )
3/2
( 30 )
1/2
= 0.525 N/mm2
For Finland:
ν min = 0.035k 2/3 f ck1 2 = 0.035 (1.9578 )
2/3
( 30 )
1/2
= 0.3000 N/mm2
For CEN Default, Norway, Slovenia and Sweden:
vRd ,c = 0.12 • 1.9578 (100 • 0.004955 • 30 )
13
+ 0 = 0.5777 N/mm2
For Finland, Singapore, and UK:
13
vRd ,c = 0.12 • 1.9578 (100 • 0.005015 • 30 ) + 0 = 0.5796 N/mm2
For Denmark:
13
vRd ,c = 0.124 • 1.9578 (100 • 0.005015 • 30 ) + 0 = 0.606 N/mm2
For CEN Default, Norway, Slovenia and Sweden:
Shear Ratio
=
v max
=
vRd ,c
1.092
= 1.90
0.5777
For Finland, Singapore and UK:
Shear Ratio
=
v max
=
vRd ,c
1.092
= 1.90
0.5796
For Denmark:
=
Shear Ratio
v max
=
vRd ,c
EXAMPLE EUROCODE 2-04 RC-PN-001 - 6
1.092
= 1.81
0.606
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Eurocode 2-04 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. These moments are
identical. After completing the analysis, design is performed using the Eurocode
2-04 code by SAFE and also by hand computation. Table 1 shows the
comparison of the design reinforcements computed by the two methods.
EXAMPLE Eurocode 2-04 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
Denmark
EXAMPLE Eurocode 2-04 RC-SL-001 - 2
Method
Strip Moment
(kN-m)
As+
SAFE
25.797
5.400
Calculated
25.800
5.400
SAFE
25.797
5.446
Calculated
25.800
5.446
SAFE
25.797
5.626
Software Verification
PROGRAM NAME:
REVISION NO.:
Calculated
25.800
SAFE
0
5.626
A +s ,min = 204.642 sq-mm
COMPUTER FILE: Eurocode 2-04 RC-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Eurocode 2-04 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
η = 1.0 for fck ≤ 50 MPa
λ = 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:
γm, steel = 1.15
γm, concrete = 1.50
αcc = 1.0
The depth of the compression block is given by:
=
m
M
25.8347 • 106
= 0.08267
=
bd 2η f cd 1000 • 1252 • 1.0 • 1.0 • 30 /1.5
EXAMPLE Eurocode 2-04 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.448
=
k2
d lim
x λx
mlim = λ 1 − = 0.294
d lim 2 d lim
ω = 1 − 1 − 2m = 0.08640
ηf bd
As = ω cd = 540.024 sq-mm > As,min
f
yd
As = 5.400 sq-cm
For Singapore and UK:
γm, steel = 1.15
γm, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
M
25.8347 • 106
= 0.097260
=
m =
bd 2η f cd 1000 • 1252 • 1.0 • 0.85 • 30 /1.5
x λx
mlim = λ 1 − = 0.48
d lim 2 d lim
δ − 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
EXAMPLE Eurocode 2-04 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
δ is assumed to be 1
ω = 1 − 1 − 2m = 0.10251
ηf bd
As = ω cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
For Finland:
γm, steel = 1.15
γm, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
=
m
M
25.8347 • 106
= 0.097260
=
bd 2η f cd 1000 • 1252 • 1.0 • 0.85 • 30 /1.5
x λx
mlim = λ 1 − = 032433
d lim 2 d lim
δ − 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
δ is assumed to be 1
ω = 1 − 1 − 2m = 0.10251
ηf bd
As = ω cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
EXAMPLE Eurocode 2-04 RC-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
For Denmark:
γm, steel = 1.20
γm, concrete = 1.45
αcc = 1.0
The depth of the compression block is given by:
=
m
M
25.8347 • 106
= 0.0799153
=
bd 2η f cd 1000 • 1252 • 1.0 • 1.0 • 30 /1.5
x λx
mlim = λ 1 − = 0.294
d lim 2 d lim
δ − 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
δ is assumed to be 1
ω = 1 − 1 − 2m = 0.08339
ηf bd
As = ω cd = 562.62 sq-mm > As,min
f
yd
As = 5.626 sq-cm
EXAMPLE Eurocode 2-04 RC-SL-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-04 PT-SL-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 SAFE. 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
EXAMPLE Hong Kong CP-04 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with independent hand calculations.
EXAMPLE Hong Kong CP-04 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
SAFE
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.79
0.35%
−5.056
−5.056
0.00%
2.836
2.839
0.11%
−10.547
−10.465
0.77%
8.323
8.407
1.01%
COMPUTER FILE: HONG KONG CP-04 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Hong Kong CP-04 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
γm, steel = 1.15
γm, 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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 16.039 kN/m2 × 0.914 m = 14.659 kN/m
Ultimate Moment, M U =
wl12
2
= 14.659 × ( 9.754 ) 8 = 174.4 kN-m
8
EXAMPLE Hong Kong CP-04 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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)
f pe
Ultimate Stress in strand, f pb =+
= 1358 MPa ≤ 0.7 f pu
= 1303 MPa
K factor used to determine the effective depth is given as:
174.4
M
=
= 0.1213 < 0.156
K=
2
f cu bd
30000(0.914)(0.229)2
K
≤ 0.95d = 192.2 mm
z = d 0.5 + 0.25 −
0
.
9
) 1000 257.2 KN
Fult , PT A=
197.4 (1303=
Ultimate force in PT, =
P ( f PS )
) 1.15 43.00 KN-m
=
M ult , PT F=
=
257.2 ( 0.192
Ultimate moment due to PT,
ult , PT ( z ) / γ
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
131.40
(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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
Moment due to PT,
=
M PT F=
258.2 (101.6 mm
=
) 1000 26.23 kN-m
PTI (sag)
F
M
M
−258.2
65.04
26.23
Stress in concrete, f = PTI ± D ± PT =
±
±
(
)
A
S
S
0.254 0.914 0.00983 0.00983
where S = 0.00983 m3
f =
−1.112 ± 6.6166 ± 2.668 MPa
f = −5.060(Comp) max, 2.836(Tension) max
EXAMPLE Hong Kong CP-04 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 ( 2 ) ( 99 ) 1000 = 239.5 kN
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment
due to live load, M L 4.788
=
=
2
Moment due to PT,
=
M PT F=
239.5 (101.6 mm
=
) 1000 24.33 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
F
M
M
−258.2
117.08
24.33
±
±
f = PTI ± D ± PT =
0.254 ( 0.914 ) 0.00983 0.00983
A
S
S
f =
−1.112 ± 11.910 ± 2.475
f = −10.547(Comp) max, 8.323(Tension) max
EXAMPLE Hong Kong CP-04 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE HONG KONG CP-04 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 Hong Kong CP 2004.
The average shear stress in the beam is below the maximum shear stress
allowed by Hong Kong CP 2004, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20, and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined using the Hong Kong CP 2004 load
combination factors of 1.4 for dead loads and 1.6 for live loads. The model is
analyzed for both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Hong Kong CP 2004 code in SAFE and also by hand computation. The
design longitudinal reinforcements are compared in Table 1. The design shear
reinforcements are compared in Table 2.
EXAMPLE HONG KONG CP-04 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE HONG KONG CP-04 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
25×105
2×108
0.2
Dead load,
Live load,
Pd
Pl
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
312
20.904
Calculated
312
20.904
A +s ,min = 195.00 sq-mm
EXAMPLE HONG KONG CP-04 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
6.50
6.50
COMPUTER FILE: Hong Kong CP-04 RC-BM-001.FDB
CONCLUSION
The SAFE results show an approximate comparison with the independent results.
EXAMPLE HONG KONG CP-04 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
As ,min = 0.0013bw h
= 195.00 sq-mm
COMB80
P = (1.4Pd + 1.6Pt) =156 kN
M* =
N *l
= 312 kN-m
3
The depth of the compression block is given by:
M
= 0.095963 < 0.156
f cu b f d 2
K=
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 373.4254 mm
0.9
The depth of the neutral axis is computed as:
x=
1
(d − z) = 114.6102 mm
0.45
And the depth of the compression block is given by:
a = 0.9x = 103.1492 mm > hf
The ultimate resistance moment of the flange is given by:
M=
f
0.67
γc
fcu ( b f − bw ) h f ( d − 0.5h f ) = 150.75 kN-m
EXAMPLE HONG KONG CP-04 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The moment taken by the web is computed as:
M w = M − M f = 161.25 kN-m
and the normalized moment resisted by the web is given by:
Kw =
Mw
= 0.0991926 < 0.156
f cu bw d 2
If Kw ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam. The reinforcement is calculated as the sum of two parts: one to balance
compression in the flange and one to balance compression in the web.
K
z = d 0.5 + 0.25 − w ≤ 0.95d = 371.3988 mm
0.9
=
As
Mf
fy
γs
( d − 0.5h )
+
f
Mw
= 2090.4 sq-mm
fy
z
γs
Shear Design
v=
V
≤ vmax = 1.2235 MPa
bw d
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
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
γ m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
f 3
40
k2 = cu = 1.06266, 1 ≤ k2 ≤
25
25
γm = 1.25
100 As
bd
= 0.15
EXAMPLE HONG KONG CP-04 RC-BM-001 - 6
1
3
Software Verification
PROGRAM NAME:
REVISION NO.:
400
d
1
SAFE
0
4
=1
However, the following limitations also apply:
0.15 ≤
100 As
≤3
bd
400
d
1
4
≥1
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Given v, vc, and vmax, the required shear reinforcement is calculated as follows:
If v ≤ (vc + 0.4),
Asv
0.4bw
=
sv
0.87 f yv
If (vc + 0.4) < v ≤ vmax,
Asv (v − vc )bw
=
sv
0.87 f yv
If v > vmax, a failure condition is declared.
(COMB80)
Pd = 20 kN
Pl
= 80 kN
V
= 156 kN
*
V
ν =
= 2.0 MPa (φsνc < ν* ≤ φsνmax)
bw d
*
Asv (v − vc )bw
=
= 0.64967sq-mm/mm = 6.50 sq-cm/m
sv
0.87 f yv
EXAMPLE HONG KONG CP-04 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-04 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE Hong Kong CP-04 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.105
0.625
1.77
Calculated
1.105
0.625
1.77
COMPUTER FILE: HONG KONG CP-04 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Hong Kong CP-04 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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 SAFE Model
From the SAFE output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
EXAMPLE Hong Kong CP-04 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Maximum design shear stress in computed in along major and minor axis of column:
=
veff , x
1.5M
V
x
f +
ud
Vy
1126.498 • 103
1.5 • 51.9908 • 106
1.0
1.1049 (Govern)
+
veff , x =
=
5016 • 218
1126.498 • 103 • 954
=
veff , y
1.5M
V
y
f+
ud
Vx
1126.498 • 103
1.5 • 45.7234 • 106
1.0
+
1.0705
veff , y =
=
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
γ 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
γm = 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
EXAMPLE Hong Kong CP-04 RC-PN-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
=
v
1.1049
= 1.77
0.6247
EXAMPLE Hong Kong CP-04 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-04 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
analysis, design is performed using the Hong Kong CP-04 code by SAFE and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
EXAMPLE Hong Kong CP-04 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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
Reinforcement
Area (sq-cm)
Method
Strip
Moment
(kN-m)
SAFE
27.197
5.853
Calculated
27.200
5.842
Load
Level
As+
Medium
A +s ,min = 162.5 sq-mm
EXAMPLE Hong Kong CP-04 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: Hong Kong CP-04 RC-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Hong Kong CP-04 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
γm, steel
= 1.15
γm, 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
EXAMPLE Hong Kong CP-04 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-2013 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with independent hand calculations.
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
SAFE
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.79
0.35%
−5.056
−5.056
0.00%
2.836
2.839
0.11%
−10.547
−10.465
0.77%
8.323
8.407
1.01%
COMPUTER FILE: HONG KONG CP-13 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
γm, steel = 1.15
γm, 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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 16.039 kN/m2 × 0.914 m = 14.659 kN/m
Ultimate Moment, M U =
wl12
2
= 14.659 × ( 9.754 ) 8 = 174.4 kN-m
8
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
=
= 0.1213 < 0.156
K=
2
30000(0.914)(0.229)2
f cu bd
K
≤ 0.95d = 192.2 mm
z = d 0.5 + 0.25 −
0
.
9
) 1000 257.2 KN
Fult , PT A=
197.4 (1303=
Ultimate force in PT, =
P ( f PS )
) 1.15 43.00 KN-m
=
M ult , PT F=
=
257.2 ( 0.192
Ultimate moment due to PT,
ult , PT ( z ) / γ
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
131.40
(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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
Moment due to PT,
=
M PT F=
258.2 (101.6 mm
=
) 1000 26.23 kN-m
PTI (sag)
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
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 ( 2 ) ( 99 ) 1000 = 239.5 kN
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment
due to live load, M L 4.788
=
=
2
Moment due to PT,
=
M PT F=
239.5 (101.6 mm
=
) 1000 24.33 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
F
M
M
−258.2
117.08
24.33
±
±
f = PTI ± D ± PT =
0.254 ( 0.914 ) 0.00983 0.00983
A
S
S
f =
−1.112 ± 11.910 ± 2.475
f = −10.547(Comp) max, 8.323(Tension) max
EXAMPLE Hong Kong CP-2013 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE HONG KONG CP-2013 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 Hong Kong CP 2013.
The average shear stress in the beam is below the maximum shear stress
allowed by Hong Kong CP 2013, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20, and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined using the Hong Kong CP 2013 load
combination factors of 1.4 for dead loads and 1.6 for live loads. The model is
analyzed for both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Hong Kong CP 2013 code in SAFE and also by hand computation. The
design longitudinal reinforcements are compared in Table 1. The design shear
reinforcements are compared in Table 2.
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
25×105
2×108
0.2
Dead load,
Live load,
Pd
Pl
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
312
20.904
Calculated
312
20.904
A +s ,min = 195.00 sq-mm
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
6.50
6.50
COMPUTER FILE: Hong Kong CP-13 RC-BM-001.FDB
CONCLUSION
The SAFE results show an approximate comparison with the independent results.
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
As ,min = 0.0013bw h
= 195.00 sq-mm
COMB80
P = (1.4Pd + 1.6Pt) =156 kN
M* =
N *l
= 312 kN-m
3
The depth of the compression block is given by:
M
= 0.095963 < 0.156
f cu b f d 2
K=
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 373.4254 mm
0.9
The depth of the neutral axis is computed as:
x=
1
(d − z) = 114.6102 mm
0.45
And the depth of the compression block is given by:
a = 0.9x = 103.1492 mm > hf
The ultimate resistance moment of the flange is given by:
M=
f
0.67
γc
fcu ( b f − bw ) h f ( d − 0.5h f ) = 150.75 kN-m
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The moment taken by the web is computed as:
M w = M − M f = 161.25 kN-m
and the normalized moment resisted by the web is given by:
Kw =
Mw
= 0.0991926 < 0.156
f cu bw d 2
If Kw ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam. The reinforcement is calculated as the sum of two parts: one to balance
compression in the flange and one to balance compression in the web.
K
z = d 0.5 + 0.25 − w ≤ 0.95d = 371.3988 mm
0.9
=
As
Mf
fy
γs
( d − 0.5h )
+
f
Mw
= 2090.4 sq-mm
fy
z
γs
Shear Design
v=
V
≤ vmax = 1.2235 MPa
bw d
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
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
γ m bd d
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
1
f 3
40
k2 = cu = 1.06266, 1 ≤ k2 ≤
25
25
γm = 1.25
100 As
bd
= 0.15
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 6
1
3
Software Verification
PROGRAM NAME:
REVISION NO.:
400
d
1
SAFE
0
4
=1
However, the following limitations also apply:
0.15 ≤
100 As
≤3
bd
400
d
1
4
≥1
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Given v, vc, and vmax, the required shear reinforcement is calculated as follows:
If v ≤ (vc + 0.4),
Asv
0.4bw
=
sv
0.87 f yv
If (vc + 0.4) < v ≤ vmax,
Asv (v − vc )bw
=
sv
0.87 f yv
If v > vmax, a failure condition is declared.
(COMB80)
Pd = 20 kN
Pl
= 80 kN
V
= 156 kN
*
V
ν =
= 2.0 MPa (φsνc < ν* ≤ φsνmax)
bw d
*
Asv (v − vc )bw
=
= 0.64967sq-mm/mm = 6.50 sq-cm/m
sv
0.87 f yv
EXAMPLE HONG KONG CP-2013 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-2013 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE Hong Kong CP-2013 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.105
0.625
1.77
Calculated
1.105
0.625
1.77
COMPUTER FILE: HONG KONG CP-13 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Hong Kong CP-2013 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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 SAFE Model
From the SAFE output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
EXAMPLE Hong Kong CP-2013 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Maximum design shear stress in computed in along major and minor axis of column:
veff , x
=
1.5M
V
x
f +
ud
Vy
1126.498 • 103
1.5 • 51.9908 • 106
veff , x =
+
1.0
1.1049 (Govern)
=
5016 • 218
1126.498 • 103 • 954
=
veff , y
1.5M
V
y
f+
ud
Vx
1126.498 • 103
1.5 • 45.7234 • 106
veff , y =
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
γ 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
γm = 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
EXAMPLE Hong Kong CP-2013 RC-PN-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
=
v
1.1049
= 1.77
0.6247
EXAMPLE Hong Kong CP-2013 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Hong Kong CP-2013 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
analysis, design is performed using the Hong Kong CP-04 code by SAFE and
also by hand computation. Table 1 shows the comparison of the design
reinforcements computed using the two methods.
EXAMPLE Hong Kong CP-2013 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0
5.0
kPa
kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
27.197
5.853
Calculated
27.200
5.842
As+
Medium
A +s ,min = 162.5 sq-mm
EXAMPLE Hong Kong CP-2013 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: Hong Kong CP-13 RC-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Hong Kong CP-2013 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
γm, steel
= 1.15
γm, 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
EXAMPLE Hong Kong CP-2013 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE IS 456-00 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE IS 456-00 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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:
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 SAFE results and summarized for verification
and validation of the SAFE 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
ν
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with the independent hand calculations.
EXAMPLE IS 456-00 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
SAFE
RESULTS
DIFFERENCE
175.6
175.65
0.03%
19.53
19.768
1.22%
−5.058
−5.057
0.02%
2.839
2.839
0.00%
−10.460
−10.465
0.05%
8.402
8.407
0.06%
COMPUTER FILE: IS 456-00 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE IS 456-00 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fck = 30MPa
fy = 400MPa
γs = 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
γc = 1.50
α = 0.36
β = 0.42
f y − 250
xmax
= 0.53 − 0.05
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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 16.158 kN/m2 × 0.914 m = 14.768 kN/m
Ultimate Moment, M U =
EXAMPLE IS 456-00 PT-SL-001 - 4
wl12
2
= 14.768 × ( 9.754 ) 8 = 175.6 kN-m
8
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Ultimate Stress in strand, f PS = from Table 11: fp = 1435 MPa
) 1000 283.3 kN
Fult , PT A=
197.4 (1435=
Ultimate force in PT,=
P ( f PS )
Compression block depth ratio: m =
M
bd α 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 β m 1 − 1 − 4 ( 0.42 )( 0.3392 )
=
=
= 0.4094 > u ,max = 0.484
d
d
2β
2 ( 0.42 )
The area of tensile steel reinforcement is then given by:
x
z=
d 1 − β u =
229 (1 − 0.42 ( 0.4094 ) ) =
189.6 mm
d
ANET
=
Mu
=
( fy / γ s ) z
175.6
(1e6 ) 2663 mm 2
=
( 400 1.15)189.6
f
1435
As = ANET − AP P =
2663 − 198
1953 mm 2
=
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
Moment due to PT,
Stress in concrete,
257.4 (102 mm
=
M PT F=
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
−257.4
65.04 − 26.23
f =
±
=
±
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
EXAMPLE IS 456-00 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 )2 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
Moment due to PT,
238.9 (102 mm
=
=
M PT F=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 9.431
f = −10.460(Comp) max, 8.402(Tension) max
EXAMPLE IS 456-00 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE IS 456-00 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 IS 456-2000.
The average shear stress in the beam is below the maximum shear stress
allowed by IS 456-2000, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20, and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined using the IS 456-2000 load combination
factors of 1.5 for dead loads and 1.5 for live loads. The model is analyzed for
both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. The moment and
shear force are identical. After completing the analysis, design is performed
using the IS 456-2000 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE IS 456-00 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE IS 456-00 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
25×105
2×108
0.2
Dead load,
Live load,
Pd
Pl
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
312
21.13
Calculated
312
21.13
A +s ,min = 235.6 sq-mm
EXAMPLE IS 456-00 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
7.76
7.73
COMPUTER FILE: IS 456-00 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE IS 456-00 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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 = 235.6 sq-mm
fy
COMB80
P = (1.4Pd + 1.6Pt) =156 kN
N *l
= 312 kN-m
M =
3
*
xu ,max
d
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
xu ,max
= 0.4666
d
The normalized design moment, m, is given by
m=
Mu
b f d 2α f ck
M = 312x106/(600 • 4252 • 0.36 • 30) = 0.26656
Df
d
= 100/425 = 0.23529
EXAMPLE IS 456-00 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
xu 1 − 1 − 4 βm = 0.305848 > D f
=
d
2β
d
γf =
0.15 xu + 0.65 D f if
D f > 0.2 d = 84.49781
γf
M f = 0.45 f ck (b f − bw )γ f d − = 130.98359 kN-m
2
Mw = Mu − Mf. = 181.0164 kN-m
Mw,single = αfckbwd2
m=
x u,max
x u,max
1 − β
= 233.233 < Mw
d
d
Mw
= 0.309310
b f d 2α f ck
xu 1 − 1 − 4 βm = 0.36538
=
d
2β
Mf
Mw
= 2113 sq-mm
As =
+
( f y γ s )(d − 0.5 y f ) ( f y γ s ) z
Shear Design
τv =
Vu
= 1.2235
bd
τmax = 3.5 for M30 concrete
k = 1.0
=
δ 1
if Pu ≤ 0 , Under Tension
100 As
100 As
= 0.15 as 0.15 ≤
≤3
bd
bd
fck
25
1
4
= 1.0466
τ c = 0.29 From Table 19 of IS 456:2000 code
τcd = kδτc = 0.29
τcd +0.4 = 0.69
EXAMPLE IS 456-00 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The required shear reinforcement is calculated as follows:
If τcd + 0.4 < τv ≤ τc,max,
Asv (τ v − τ cd ) b
≥
=
7.73 sq-cm/m
sv
0.87 f y
EXAMPLE IS 456-00 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE IS 456-00 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE IS 456-00 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.792
1.141
1.57
Calculated
1.792
1.141
1.57
COMPUTER FILE: IS 456-00 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE IS 456-00 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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
Side 2
Side 1
Side 3
109
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 SAFE Model
γ
V2
γ
V3
1
1−
0.495
=
=
2 1118
1+
3 518
1
1−
0.312
=
=
2 518
1+
3 1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE IS 456-00 RC-PN-001 - 3
Software Verification
SAFE
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
2
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
= 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 SAFE output at Grid B-2:
VU = 1126.498 kN
γ
γ
V2
MU2 = −25.725 kN-m
V3
MU3 = 14.272 kN-m
EXAMPLE IS 456-00 RC-PN-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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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:
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 =1.5582 N/mm2 at point B
At the point labeled C 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 = 1.792 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
EXAMPLE IS 456-00 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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The shear capacity is calculated based on the minimum of the following three limits:
ks = 0.5 + βc ≤ 1.0 = 0.833
(IS 31.6.3.1)
τc = 0.25 = 1.127 N/mm2
(IS 31.6.3.1)
vc = ks τc= 1.141 N/mm2
(IS 31.6.3.1)
CSA 13.3.4.1 yields the smallest value of vc = 1.141 N/mm2, and thus this is the shear
capacity.
Shear Ratio
=
EXAMPLE IS 456-00 RC-PN-001 - 6
vU
=
vc
1.792
= 1.57
1.141
Software Verification
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EXAMPLE IS 456-00 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
analysis, design was performed using the IS 456-00 code by SAFE and also by
hand computation. Table 1 shows the comparison of the design reinforcements
computed using the two methods.
EXAMPLE IS 456-00 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
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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
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
Calculated
Reinforcement Area (sq-cm)
As+
As-
26.997
5.830
--
27.000
5.830
--
Medium
A +s ,min = 230.978 sq-mm
EXAMPLE IS 456-00 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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COMPUTER FILE: IS 456-00 RC-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE IS 456-00 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
γs = 1.15
γc = 1.50
α = 0.36
β = 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
EXAMPLE IS 456-00 RC-SL-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:
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SAFE
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xu ,max
= 0.466
d
The depth of the compression block is given by:
m=
Mu
bd 2αf ck
= 0.16
x
xu 1 − 1 − 4 β m
= 0.1727488 < u ,max
=
d
d
2β
The area of tensile steel reinforcement is given by:
x
=
z d 1 − β u . = 115.9307 mm
d
As =
Mu
, = 583.027 sq-mm > As,min
( fy /γ s ) z
As = 5.830 sq-cm
EXAMPLE IS 456-00 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE Italian NTC 2008 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE Italian NTC 2008 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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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, 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 SAFE results and summarized for verification
and validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with independent hand calculations.
EXAMPLE Italian NTC 2008 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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Table 1 Comparison of Results
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
Factored moment,
Mu (Ultimate) (kN-m)
165.9
165.9
0.00%
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
−10.465
0.05%
Normal Conc. Stress, bot
(D+L+PTF), MPa
8.402
8.407
0.06%
Long-Term Conc. Stress, top
(D+0.5L+PTF(L)), MPa
−7.817
−7.817
0.00%
Long-Term Conc. Stress, bot
(D+0.5L+PTF(L)), MPa
5.759
5.759
0.00%
FEATURE TESTED
Table 2 Comparison of Design Moments and Reinforcements
Reinforcement Area (sq-cm)
Method
Design Moment (kN-m)
As+
SAFE
165.9
16.39
Calculated
165.9
16.29
COMPUTER FILE: ITALIAN NTC 2008 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Italian NTC 2008 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
γm, steel = 1.15
γm, concrete = 1.50
η = 1.0 for fck ≤ 50 MPa
λ = 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)
Total = 10.772 kN/m2 (D+L)
= 15.260 kN/m2 (D+L)ult
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 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
EXAMPLE Italian NTC 2008 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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f A
f SE + 7000d 1 − 1.36 PU P l
Ultimate Stress in strand, f PS =
fCK bd
1862(198) (
=
1210 + 7000(229) 1 − 1.36
9754 )
30(914) ( 229 )
= 1361 MPa
) 1000 269.5 kN
Fult , PT A=
2 ( 99 )(1361=
Ultimate force in PT,=
P ( f PS )
Design moment M = 165.9 kN-m
Compression block depth ratio: m =
M
bd 2ηf cd
165.9
=
0.1731
( 0.914 )( 0.229 )2 (1) ( 30000 1.50 )
Required area of mild steel reinforcing,
ω = 1 − 1 − 2m = 1 − 1 − 2(0.1731) =
0.1914
=
η f bd
1(30 /1.5)(914)(229)
2
=
=
AEquivTotal ω=
cd 0.1914
2303 mm
400 /1.15
f yd
1366
2
AEquivTotal
= AP
=
+ AS 2311 mm
400
1361
2303 198
1629 mm 2
AS =−
=
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
=
257.4 (102 mm
=
M PT F=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
−257.4
65.04 − 26.23
Stress in concrete, f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
where S = 0.00983m3
Moment due to PT,
EXAMPLE Italian NTC 2008 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment
due to dead load M D 5.984
=
=
2
( 0.914 )( 9.754 )2 8 52.04 kN-m
Moment
due to live load M L 4.788
=
=
Moment due to PT,
M PT F=
=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE Italian NTC 2008 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE Italian NTC 2008 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 Italian NTC 2008.
The average shear stress in the beam is below the maximum shear stress
allowed by Italian NTC 2008, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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 Italian NTC 2008 load
combination factors of 1.35 for dead loads and 1.5 for live loads. The model is
analyzed for both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Italian NTC 2008 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE Italian NTC 2008 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE Italian NTC 2008 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
425
75
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
f'ck
fy
wc
Ec
Es
v
=
=
=
=
=
=
30
460
0
25x105
2x108
0.2
Dead load,
Live load,
Pd
Pl
=
=
30
130
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by the analytical method. They match exactly
for this problem. Table 1 also shows the comparison of design reinforcements.
EXAMPLE Italian NTC 2008 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Moments and Flexural Reinforcements
Moment
Reinforcement Area (sq-cm)
Method
(kN-m)
As+
SAFE
471
31.643
Calculated
471
31.643
A +s ,min = 2.09 sq-cm
Table 2 Comparison of Shear Reinforcements
Reinforcement Area ,
Shear Force
(sq-cm/m)
Method
(kN)
As+
SAFE
235.5
6.16
Calculated
235.5
6.16
COMPUTER FILE: Italian NTC 2008 RC-BM-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Italian NTC 2008 RC-BM-001 - 4
Av
s
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for both of the load combinations:
γs
= 1.15
γc
= 1.50
f cd =
α cc f ck / γ c
=
f yd f yk / γ s
η = 1.0 for fck ≤ 50 MPa
λ = 0.8 for fck ≤ 50 MPa
As ,min = 0.26
f ctm
bd = 208.73 sq-mm
f yk
As ,min = 0.0013bw h = 195.00 sq-mm
γm, steel = 1.15
γm, concrete = 1.50
αcc = 1.0
The depth of the compression block is given by:
M
471 • 106
= 0.217301
=
m =
bd 2η f cd 600 • 4252 • 1.0 • 1.0 • 30 /1.5
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
δ is assumed to be 1
δ − k1
x
for fck ≤ 50 MPa = 0.448
=
k2
d lim
EXAMPLE Italian NTC 2008 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
x λx
mlim = λ 1 − = 0.29417
d lim 2 d lim
x
= 1 − 1 − 2mlim = 0.3584
d lim
amax = ωlimd = 152.32 mm
ωlim = λ
ω = 1 − 1 − 2m = 0.24807
a = ωd = 105.4299 mm ≤ amax
As 2 =
(b
− bw ) h f η f cd
f
f yd
hf
M 2 = As 2 f yd d −
2
= 1500 sq-mm
= 225 kN-m
M1 = M − M2 = 246 kN-m
m1 =
M1
= 0.2269896 ≤ mlim
bw d 2η f cd
ω1 = 1 − 1 − 2m1 = 0.2610678
η f b d
As1 = ω1 cd w = 1664.304 sq-mm
f yd
As = As1 + As2 = 3164.307 sq-mm
Shear Design
C Rd ,c = 0.18 γ c = 0.18/1.5 = 0.12
k=
1+
200
=
1.686 ≤ 2.0 with d in mm
d
ρ1 = 0.0
=
σcp N Ed Ac < 0.2 f cd = 0.0 MPa
ν min = 0.035k 3 2 f ck 1 2 = 0.419677
EXAMPLE Italian NTC 2008 RC-BM-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
13
=
VRd ,c C Rd ,c k (100 ρ1 fck ) + k1σ cp bw d = 53.5 kN
αcw = 1
ν 1 = 0.61 −
f ck
= 0.528
250
z = 0.9d = 382.5 mm
θ is taken as 1.
VRd ,max =
α cwbw zν 1 f cd
= 1211.76 kN
cot θ + tan θ
VR,dc < VEd ≤ VRd,max (govern)
Computing the angle using vEd :
vEd =
235.5 • 103
= 2.0522
0.9 • 425 • 300
θ = 0.5sin −1
vEd
0.2 f ck (1 − f ck 250 )
θ = 0.5sin −1
2.0522
= 11.43°
0.2 • 30 (1 − 30 250 )
21.8° ≤ θ ≤ 45° , therefore use=
θ 21.8°
Asw
vEd bw
=
s
f ywd cot θ
Asw
2.0522 • 300
= 0.61566 sq-mm/m = 6.16 sq-cm/m
=
s
460 1.15 • 2.5
EXAMPLE Italian NTC 2008 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Italian NTC 2008 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE Italian NTC 2008 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.100
0.578
1.90
Calculated
1.099
0.578
1.90
COMPUTER FILE: ITALIAN NTC 2008 RC-PN-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Italian NTC 2008 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation for Interior Column using SAFE Method
d=
[( 250 − 26 ) + ( 250 − 38)]
2 = 218 mm
Refer to Figure 2.
u1 = u = 2•300 + 2•900 + 2•π•436 = 5139.468 mm
1172
Note: All dimensions in millimeters
Critical section for
punching shear
shown dashed.
Y
436
150 150 436
B
A
Column
Side 2
Side 1
Side 3
436
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 SAFE Model
From the SAFE output at Grid B-2:
VEd = 1112.197 kN
k2MEd2 = 41.593 kN-m
k3MEd3 = 20.576 kN-m
EXAMPLE Italian NTC 2008 RC-PN-001 - 3
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
Maximum design shear stress in computed in along major and minor axis of column:
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
+
vEd =
1 +
ud
VEdW1,2
VEdW1,3
(EC2 6.4.4(2))
c2
W1 = 1 + c1c2 + 4c2 d + 16d 2 + 2π dc1
2
=
W1,2
9002
+ 300 • 900 + 4 • 300 • 218 + 16 • 2182 + 2π • 218 • 900
2
W1,2 = 2,929, 744.957 mm2
W
=
3
1,3
9002
+ 900 • 300 + 4 • 900 • 218 + 16 • 2182 + 2π • 218 • 300
2
W1,2 = 2, 271,104.319 mm2
VEd k2 M Ed ,2u1 k3 M Ed ,3u1
vEd =
+
1 +
ud
VEdW1,2
VEdW1,3
1112.197 • 103
41.593 • 106 • 5139.468
20.576 • 106 • 5139.468
1
vEd =
+
+
5139.468 • 218 1112.197 • 103 • 2929744.957 1112.197 • 103 • 2271104.319
vEd = 1.099 N/mm2
Thus vmax = 1.099 N/mm2
C Rd ,c = 0.18 γ 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 ρ1 fck ) + k1σ cp
=
(EC2 6.4.4)
with a minimum of:
v=
Rd ,c
(v
min
k=
1+
k1
+ k1σ cp )
200
≤ 2.0 = 1.9578
d
= 0.15.
EXAMPLE Italian NTC 2008 RC-PN-001 - 4
(EC2 6.4.4)
(EC2 6.4.4(1))
(EC2 6.2.2(1))
Software Verification
PROGRAM NAME:
REVISION NO.:
ρ1 =
SAFE
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
ρ1 = 8641.661 ( 8000 • 218 ) = 0.004955 ≤ 0.02
ν 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
v max
=
vRd ,c
1/2
= 0.525 N/mm2
+ 0 = 0.5777 N/mm2
1.099
= 1.90
0.5777
EXAMPLE Italian NTC 2008 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Italian NTC 2008 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. These moments are
identical. After completing the analysis, design is performed using the Italian
NTC 2008 code by SAFE and also by hand computation. Table 1 shows the
comparison of the design reinforcements computed by the two methods.
EXAMPLE Italian NTC 2008 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
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 SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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+
SAFE
25.797
5.400
Calculated
25.800
5.400
A +s ,min = 204.642 sq-mm
COMPUTER FILE: Italian NTC 2008 RC-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Italian NTC 2008 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
η = 1.0 for fck ≤ 50 MPa
λ = 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
γm, steel = 1.15
γm, concrete = 1.50
αcc = 0.85:
The depth of the compression block is given by:
=
m
M
25.8347 • 106
= 0.097260
=
bd 2η f cd 1000 • 1252 • 1.0 • 0.85 • 30 /1.5
x λx
mlim = λ 1 − = 0.48
d lim 2 d lim
EXAMPLE Italian NTC 2008 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
δ − 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
δ is assumed to be 1
ω = 1 − 1 − 2m = 0.10251
ηf bd
As = ω cd = 544.61 sq-mm > As,min
f
yd
As = 5.446 sq-cm
EXAMPLE Italian NTC 2008 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE NZS 3101-06 PT-SL-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 SAFE. 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
Elevation
Section
Figure 1 One-Way Slab
EXAMPLE NZS 3101-06 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 SAFE results and summarized for verification and
validation of the SAFE 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
ν
=
=
=
=
=
=
=
=
=
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with the independent hand calculations.
EXAMPLE NZS 3101-06 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 1 Comparison of Results
FEATURE TESTED
INDEPENDENT
RESULTS
SAFE
RESULTS
DIFFERENCE
156.12
156.14
0.01%
14.96
15.08
0.74%
−5.058
−5.057
0.02%
2.839
2.839
0.00%
−10.460
−10.465
0.05%
8.402
8.407
0.06%
−7.817
−7.817
0.00%
5.759
5.759
0.00%
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
COMPUTER FILE: NZS 3101-06 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE NZS 3101-06 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
φb = 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
=
α1 0.85 for f ′c ≤ 55 MPa
=
β1 0.85 for f ′c ≤ 30,
cb =
εc
ε c + f y Es
d = 214.7
amax = 0.75β1cb = 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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 14.363 kN/m2 × 0.914 m = 13.128 kN/m
EXAMPLE NZS 3101-06 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
Ultimate Moment, M U =
SAFE
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 ρ P
30
300 ( 0.00095 )
200 1410 MPa
= 1385 MPa ≤ f SE +=
= 1210 + 70 +
) 1000 274.23 kN
Fult , PT A=
2 ( 99 )(1385=
Ultimate force in PT,=
P ( f PS )
d
d2 −
Stress block depth, a =−
2M *
α f 'c φ 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
= 49.01 kN-m
M ult ,=
Fult , PT d − =
φ 274.23 229 −
( 0.85 ) 1000
PT
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
AS =
=
=
(1e6) 1496 mm 2
a
0.03748
φ 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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
Stress in concrete,
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
65.04 − 26.23
−257.4
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
where S = 0.00983m3
EXAMPLE NZS 3101-06 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
=
238.9 (102 mm
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+0.5L+PTF(L)),
FPTI M D + 0.5 L − M PT
−238.9
91.06 − 24.33
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 6.788
f = −7.817(Comp) max, 5.759(Tension) max
EXAMPLE NZS 3101-06 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE NZS 3101-06 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 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 500-mm-deep T beam with a
flange 100 mm thick and 600 mm wide is modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(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
combinations (COMB130) is defined with 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 these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the NZS 3101-06 code in SAFE and also by hand computation. Table 1
shows the comparison of the design longitudinal reinforcements. Table 2 shows
the comparison of the design shear reinforcements.
EXAMPLE NZS 3101-06 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE NZS 3101-06 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
=
=
50
130
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
510
35.046
Calculated
510
35.046
A +s ,min = 535.82 sq-m
EXAMPLE NZS 3101-06 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
255
14.962
14.89
COMPUTER FILE: NZS 3101-06 RC-BM-001.FDB
CONCLUSION
The SAFE results show an acceptable close comparison with the independent
results.
EXAMPLE NZS 3101-06 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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 = 240.56 mm
amax = 0.75β1cb= 153.36 mm
As ,min
f ′c
Ac = 535.82
4 fy
sq-mm
= max
A
c
1.4 = 136.96
fy
= 535.82 sq-mm
COMB130
N* = (1.2Nd + 1.5Nt) = 255 kN
M* =
N *l
= 510 kN-m
3
The depth of the compression block is given by:
a =−
d
d −
2
2 M*
α1 f c'φbb f
= 105.322 mm (a > Ds)
The compressive force developed in the concrete alone is given by:
Cf is given by:
=
C f α1 f ′c ( b f − bw ) h f = 765 kN
EXAMPLE NZS 3101-06 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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Therefore, As1 =
Cf
fy
and the portion of M* that is resisted by the flange is given
by:
d
M *f = C f d − s φb = 243.84375 kN-m
2
Cf
As1 =
fy
= 1663.043 sq-mm
Therefore, the balance of the moment, M*, to be carried by the web is:
M*w = M* − M*f = 510 − 243.84375 = 266.15625 kN-m
The web is a rectangular section with dimensions bw and d, for which the depth of
the compression block is recalculated as:
a1 =−
d
d2 −
2 M w*
= 110.7354 mm ≤ amax
α1 f ′cφbbw
If a1 ≤ amax (NZS 9.3.8.1), the area of tension reinforcement is then given by:
M w*
As2 =
= 1841.577 sq-mm
a1
φb f y d −
2
As = As1 + As2 = 3504.62 sq-mm
Shear Design
The basic shear strength for rectangular section is computed as,
νb = 0.07 + 10
As
bw d
f ′c = 0.3834
f ′c ≤ 50 MPa, and
0.08
f ′c = 0.438 MPa ≤ νb ≤ 0.2
f ′c = 1.095 MPa
νc = kd ka νb = 0.438 where (kd =1.0, ka=1.0)
The average shear stress is limited to a maximum limit of,
vmax = min {0.2 f ′c , 8 MPa} = min{6, 8} = 6 MPa
EXAMPLE NZS 3101-06 RC-BM-001 - 6
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
The shear reinforcement is computed as follows:
If ν* ≤ φs (v c 2 ) or h ≤ max(300 mm, 0.5bw)
Av
=0
s
(NZS 9.3.9.4.13)
If φs (v c 2 ) < ν* ≤ φsνc,
Av
1
=
16
s
f ′c
bw
f yt
(NZS 9.3.9.4.15)
If φsνc < ν* ≤ φsνmax,
(NZS 9.3.9.4.2)
(
Av
v * − φ s vc
=
s
φ s f yt d
)
If ν* > νmax, a failure condition is declared.
For the load combination, the N* and V* are calculated as follows:
N* = 1.2Nd + 1.5N1
V* = N*
*
ν
*
V
=
bw d
(COMB130)
Nd = 50 kips
Nl
= 130 kips
V* = 255 kN
*
V
ν =
= 2.0 MPa (φsνc < ν* ≤ φsνmax)
bw d
*
(
)
v* − φs vc bw
Av
=
= 1.489 sq-mm/mm = 1489 sq-mm/m
φs f yt
s
EXAMPLE NZS 3101-06 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE NZS 3101-06 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE NZS 3101-06 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.792
1.141
1.57
Calculated
1.792
1.141
1.57
COMPUTER FILE: NZS 3101-06 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE NZS 3101-06 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE Method
d=
[( 259 − 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
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 SAFE Model
γ
V2
γ
V3
1
1−
0.495
=
=
2 1118
1+
3 518
1
1−
0.312
=
=
2 518
1+
3 1118
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
EXAMPLE NZS 3101-06 RC-PN-001 - 3
Software Verification
SAFE
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
2
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
= 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 SAFE output at Grid B-2:
VU = 1126.498 kN
γ
γ
V2
MU2 = −25.725 kN-m
V3
MU3 = 14.272 kN-m
EXAMPLE NZS 3101-06 RC-PN-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.:
SAFE
0
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:
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 =1.5582 N/mm2 at point B
At the point labeled C 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 = 1.792 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
EXAMPLE NZS 3101-06 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
6 1 + β ϕ f ′c
c
1 α d
ϕ vv min 1 + s ϕ f ′c = 1.141N/mm2 per
=
b0
6
1
ϕ f ′c
3
(NZS 12.7.3.2)
NZS 12.7.3.2 yields the smallest value of ϕ vv = 1.141 N/mm2, and thus this is the shear
capacity.
Shear Ratio
=
vU
=
ϕ vv
EXAMPLE NZS 3101-06 RC-PN-001 - 6
1.792
= 1.57
1.141
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE NZS 3101-06 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
analysis, design is performed using the NZS 3101-06 code by SAFE and also by
hand computation. Table 1 shows the comparison of the design reinforcements
computed using the two methods.
EXAMPLE NZS 3101-06 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
24.597
5.238
Calculated
24.6
5.238
As+
Medium
A +s ,min = 380.43 sq-mm
EXAMPLE NZS 3101-06 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: NZS 3101-06 RC-SL-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE NZS 3101-06 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for the load combination:
φb = 0.85
b = 1000 mm
=
α1 0.85 for f ′c ≤ 55MPa
=
β1 0.85 for f ′c ≤ 30,
cb =
εc
ε c + f y Es
d = 70.7547
amax = 0.75β1cb= 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 −
EXAMPLE NZS 3101-06 RC-SL-001 - 4
2 M*
= 9.449 mm < amax
α1 f ′ c φbb
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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The area of tensile steel reinforcement is then given by:
As =
M*
= 523.799 sq-mm > As,min
a
φb f y d −
2
As = 5.238 sq-cm
EXAMPLE NZS 3101-06 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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EXAMPLE Singapore CP 65-99 PT-SL-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 SAFE. 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
Elevation
914 mm
Section
Figure 1 One-Way Slab
EXAMPLE Singapore CP 65-99 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 mid-span of the slab. Independent hand
calculations are compared with the SAFE results and summarized for verification
and validation of the SAFE 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
ν
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with the independent hand calculations.
EXAMPLE Singapore CP 65-99 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
SAFE
RESULTS
DIFFERENCE
174.4
174.4
0.00%
19.65
19.79
0.71%
−5.058
−5.057
0.02%
2.839
2.839
0.00%
−10.460
−10.465
0. 50%
8.402
8.407
0.06%
COMPUTER FILE: SINGAPORE CP 65-99 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Singapore CP 65-99 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
f’c = 30MPa
fy = 400MPa
γm, 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
γm, 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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 16.039 kN/m2 × 0.914 m = 14.659 kN/m
wl12
= 14.659 × (9.754)2/8 = 174.4 kN-m
Ultimate Moment, M U =
8
EXAMPLE Singapore CP 65-99 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
=
= 0.1213 < 0.156
K=
2
30000(0.914)(0.229) 2
f cu bd
K
≤ 0.95d = 192.2 mm
z = d 0.5 + 0.25 −
0.9
) 1000 258.0 kN
Fult , PT A=
2 ( 99 )(1303=
Ultimate force in PT, =
P ( f PS )
Ultimate moment due to PT,
) 1.15 43.12 kN-m
=
M ult , PT F=
258.0 ( 0.192
=
ult , PT ( z ) / γ
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 ( 400 )(192 )
0.87 f y z X
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
Stress in concrete,
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
65.04 − 26.23
−257.4
f =
±
=
±
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
EXAMPLE Singapore CP 65-99 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
±
=
±
f =
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 9.431
f = −10.460(Comp) max, 8.402(Tension) max
EXAMPLE Singapore CP 65-99 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE SINGAPORE CP 65-99 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 Singapore CP 65-99.
The average shear stress in the beam is below the maximum shear stress
allowed by Singapore CP 65-99, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20, and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined with the Singapore CP 65-99 load
combination factors of 1.4 for dead loads and 1.6 for live loads. The model is
analyzed for both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Singapore CP 65-99 code in SAFE and also by hand computation.
Table 1 shows the comparison of the design longitudinal reinforcements. Table 2
shows the comparison of the design shear reinforcements.
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
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
25×105
2×108
0.2
Dead load,
Live load,
Pd
Pl
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
312
20.904
Calculated
312
20.904
A +s ,min = 195.00 sq-mm
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
6.50
6.50
COMPUTER FILE: SINGAPORE CP 65-99 RC-002.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
As ,min = 0.0013bw h
= 195.00 sq-mm
COMB80
P = (1.4Pd + 1.6Pt) = 156 kN
M* =
N *l
= 312 kN-m
3
The depth of the compression block is given by:
M
= 0.095963 < 0.156
f cu b f d 2
K=
Then the moment arm is computed as:
K
z = d 0.5 + 0.25 −
≤ 0.95d = 373.4254 mm
0
.
9
The depth of the neutral axis is computed as:
x=
1
(d − z) = 114.6102 mm
0.45
And the depth of the compression block is given by:
a = 0.9x = 103.1492 mm > hf
The ultimate resistance moment of the flange is given by:
M=
f
0.67
γc
f cu ( b f − bw ) h f ( d − 0.5h f
) = 150.75 kN-m
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The moment taken by the web is computed as:
M w = M − M f = 161.25 kN-m
And the normalized moment resisted by the web is given by:
Kw =
Mw
= 0.0991926 < 0.156
f cu bw d 2
If Kw ≤ 0.156 (BS 3.4.4.4), the beam is designed as a singly reinforced concrete
beam. The reinforcement is calculated as the sum of two parts: one to balance
compression in the flange and one to balance compression in the web.
K
z = d 0.5 + 0.25 − w ≤ 0.95d = 371.3988 mm
0.9
Mf
=
As
fy
γs
( d − 0.5h )
f
+
Mw
= 2090.4 sq-mm
fy
z
γs
Shear Design
v=
V
≤ vmax = 1.2235 MPa
bw d
vmax = min(0.8 fcu , 5 MPa) = 4.38178 MPa
The shear stress carried by the concrete, vc, is calculated as:
1
0.79k1k 2 100 As 3 400
vc =
γ m bd d
1
4
= 0.3568 MPa
k1 is the enhancement factor for support compression,
and is conservatively taken as 1 .
f
k2 = cu
25
1
3
40
= 1.06266, 1 ≤ k2 ≤
25
γm = 1.25
100 As
= 0.15
bd
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 6
1
3
Software Verification
PROGRAM NAME:
REVISION NO.:
400
d
1
SAFE
0
4
=1
However, the following limitations also apply:
0.15 ≤
100 As
≤3
bd
400
d
1
4
≥1
fcu ≤ 40 MPa (for calculation purposes only) and As is the area of tension
reinforcement.
Given v, vc, and vmax, the required shear reinforcement is calculated as follows:
If v ≤ (vc + 0.4),
Asv
0.4bw
=
sv
0.87 f yv
If (vc + 0.4) < v ≤ vmax ,
Asv (v − vc )bw
=
sv
0.87 f yv
If v > vmax, a failure condition is declared.
(COMB80)
Pd = 20 kN
Pl
= 80 kN
V
= 156 kN
*
V
ν =
= 2.0 MPa (φsνc < ν* ≤ φsνmax)
bw d
*
Asv (v − vc )bw
=
= 0.64967sq-mm/mm = 6.50 sq-cm/m
sv
0.87 f yv
EXAMPLE SINGAPORE CP 65-99 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Singapore CP 65-99 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE Singapore CP 65-99 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.105
0.625
1.77
Calculated
1.105
0.620
1.77
COMPUTER FILE: SINGAPORE CP 65-99 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Singapore CP 65-99 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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 1
Side 3
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 SAFE Model
From the SAFE output at Grid B-2:
V= 1126.498 kN
M2 = 51.9908 kN-m
M3 = 45.7234 kN-m
EXAMPLE Singapore CP 65-99 RC-PN-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Maximum design shear stress in computed in along major and minor axis of column:
=
veff , x
1.5M
V
x
f +
ud
Vy
(CP 3.7.7.3)
1126.498 • 103
1.5 • 51.9908 • 106
+
veff , x =
1.0
1.1049 (Govern)
=
5016 • 218
1126.498 • 103 • 954
=
veff , y
1.5M
V
y
f+
ud
Vx
1126.498 • 103
1.5 • 45.7234 • 106
veff , y =
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
γ 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
γm = 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
EXAMPLE Singapore CP 65-99 RC-PN-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
=
v
1.1049
= 1.77
0.6247
EXAMPLE Singapore CP 65-99 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Singapore CP 65-99 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
the analysis, design is performed using the Singapore CP 65-99 code by SAFE
and also by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
EXAMPLE Singapore CP 65-99 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
27.197
5.853
Calculated
27.200
5.850
As+
Medium
A +s ,min = 162.5 sq-mm
EXAMPLE Singapore CP 65-99 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: Singapore CP 65-99 RC-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Singapore CP 65-99 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, 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 ,min = 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
EXAMPLE Singapore CP 65-99 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Turkish TS 500-2000 PT-SL-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 SAFE. 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
Elevation
914 mm
Section
Figure 1 One-Way Slab
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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 mid-span of the slab. Independent hand
calculations are compared with the SAFE results and summarized for verification
and validation of the SAFE 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
ν
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 SAFE 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 SAFE total factored moments, required mild
steel reinforcing, and slab stresses with the independent hand calculations.
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
SAFE
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
−10.465
0. 50%
8.402
8.407
0.06%
COMPUTER FILE: TURKISH TS 500-2000 PT-SL-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATIONS:
Design Parameters:
Mild Steel Reinforcing
fck = 30MPa
fyk = 400MPa
γm, 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
γm, 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
ω =10.772 kN/m2 × 0.914 m = 9.846 kN/m, ωu = 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
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
f A
f pe + 7000d 1 − 1.36 PU P l
Ultimate Stress in strand, f Pd =
fCK bd
1862(198) (
1210 + 7000(229) 1 − 1.36
=
9754 )
30(914) ( 229 )
= 1361 MPa
) 1000 269.5 kN
Fult , PT A=
2 ( 99 )(1361=
Ultimate force in PT,=
P ( f PS )
d
d2 −
Stress block depth, a =−
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=
Fult , PT d −=
= 54.194 kN-m
269.5 229 −
1000
, PT
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
54.194
a
f yd d − ( 400 ) 229 −
2
2
K factor used to determine the effective depth is given as:
174.4
M
=
= 0.1819 < 0.156
K=
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
) 1000 258.0 kN
Ultimate force in PT, =
2 ( 99 )(1303=
Fult , PT A=
P ( f PS )
Ultimate moment due to PT,
) 1.15 43.12 kN-m
=
M ult , PT F=
258.0 ( 0.192
=
ult , PT ( z ) / γ
Net Moment to be resisted by As,
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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 =
)
(
)(
0.87 400 192
f yd z X
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
( 0.914 )( 9.754 ) 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
2
Moment due to PT,
Stress in concrete,
=
M PT F=
257.4 (102 mm
=
) 1000 26.25 kN-m
PTI (sag)
FPTI M D − M PT
−257.4
65.04 − 26.23
f =
±
=
±
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
( 0.914 )( 9.754 )2 8 65.04 kN-m
Moment=
due to dead load, M D 5.984
=
( 0.914 )( 9.754 ) 8 52.04 kN-m
Moment=
due to live load, M L 4.788
=
2
Moment due to PT,
=
M PT F=
238.9 (102 mm
=
) 1000 24.37 kN-m
PTI (sag)
Stress in concrete for (D+L+PTF),
FPTI M D + L − M PT
117.08 − 24.37
−238.8
f =
±
=
±
0.254 ( 0.914 )
0.00983
A
S
f =
−1.029 ± 9.431
f = −10.460(Comp) max, 8.402(Tension) max
EXAMPLE Turkish TS 500-2000 PT-SL-001 - 6
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Turkish TS 500-2000 RC-BM-001
Flexural and Shear Beam Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE. 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 TS 500-2000.
The average shear stress in the beam is below the maximum shear stress
allowed by TS 500-2000, 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 modeled using SAFE. The beam is
shown in Figure 1. The computational model uses a finite element mesh of frame
elements, automatically generated by SAFE. The maximum element size has
been specified to be 200 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
(DL20) and one live load case (LL80) with only symmetric third-point loads of
magnitudes 20, and 80 kN, respectively, are defined in the model. One load
combinations (COMB80) is defined with the Turkish TS 500-2000 load
combination factors of 1.4 for dead loads and 1.6 for live loads. The model is
analyzed for both of these load cases and the load combinations.
The beam moment and shear force are computed analytically. The total factored
moment and shear force are compared with the SAFE results. These moment and
shear force are identical. After completing the analysis, design is performed
using the Turkish TS 500-2000 code in SAFE and also by hand computation.
Table 1 shows the comparison of the design longitudinal reinforcements. Table 2
shows the comparison of the design shear reinforcements.
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
GEOMETRY, PROPERTIES AND LOADING
Clear span,
Overall depth,
Flange Thickness,
Width of web,
Width of flange,
Depth of tensile reinf.,
Effective depth,
Depth of comp. reinf.,
l
h
ds
bw
bf
dc
d
d'
=
=
=
=
=
=
=
=
6000
500
100
300
600
75
425
75
Concrete strength,
Yield strength of steel,
Concrete unit weight,
Modulus of elasticity,
Modulus of elasticity,
Poisson’s ratio,
f'ck
fyk
wc
Ec
Es
v
=
=
=
=
=
=
30
460
0
25×105
2×108
0.2
Dead load,
Live load,
Pd
Pl
=
=
20
80
SAFE
0
mm
mm
mm
mm
mm
mm
mm
mm
MPa
MPa
kN/m3
MPa
MPa
kN
kN
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural and shear reinforcement
Application of minimum flexural and shear reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE total factored moments in the design
strip with the moments obtained by 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-cm)
Method
Moment
(kN-m)
As+
SAFE
312
20.244
Calculated
312
20.244
A +s ,min = 325.9 sq-mm
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
Table 2 Comparison of Shear Reinforcements
Reinforcement Area,
Av
s
(sq-cm/m)
Shear Force (kN)
SAFE
Calculated
156
4.19
4.19
COMPUTER FILE: TURKISH TS 500-2000 RC-002.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Flexural Design
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
f=
cd
f ck 30
= = 20
γ mc 1.5
=
f yd
f yk 460
= = 400
γ ms 1.15
cb =
ε cu Es
ε cu Es + f yd
d = 255 mm
amax = 0.85k1cb = 177.7 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
γ mc
1.5
COMB80
Pd = (1.4PG + 1.6PQ) = 156 kN
Md =
Nd l
= 312 kN-m
3
The depth of the compression block is given by:
a =−
d
d2 −
2 Md
= 79.386 mm < 100 mm
0.85 fcd b
since a < amax,
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
a =−
d
d2 −
2M d
0.85 f cd b
a=
425 − 4252 −
(TS 7.1)
2 • 312 • 106
=
79.387 mm
0.85 • 20 • 600
If a ≤ amax (TS 7.1), the area of tensile steel reinforcement is then given by:
As
=
Md
312 • 106
=
= 2024.36 mm 2 , and
a
79.387
f yd d − 400 425 −
2
2
Shear Design
Pd = 20 kN
Pl
= 80 kN
V
= 156 kN
The shear force is limited to a maximum of,
=
Vmax 0.22
=
fcd Aw 561 kN
The nominal shear strength provided by concrete is computed as:
γN
=
Vcr 0.65 f ctd bw d 1 + d
Ag
= 105.9 kN, where N d = 0
=
Vc 0.8
=
Vcr 84.73 kN
The shear reinforcement is computed as follows:
If Vd ≤ Vcr
f ctd
mm 2
Asw
=
0.3
=
b
0.2876
f ywd
mm
s min
If Vcr ≤ Vd ≤ Vmax
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 6
(TS 8.1.5, Eqn 8.6)
Software Verification
PROGRAM NAME:
REVISION NO.:
Asw
=
s
Vd − Vc )
(=
f ywd d
0.419
mm 2
mm
SAFE
0
(TS 8.1.4, Eqn 8.5)
EXAMPLE Turkish TS 500-2000 RC-BM-001 - 7
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Turkish TS 500-2000 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
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 plate 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.
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
TECHNICAL FEATURES OF SAFE 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 SAFE 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
SAFE
1.690
1.278
1.32
Calculated
1.690
1.278
1.32
COMPUTER FILE: TURKISH TS 500-2000 RC-PN-001.FDB
CONCLUSION
The SAFE results show an exact comparison with the independent results.
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
Hand Calculation For Interior Column Using SAFE 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 SAFE Model
1
0.595
=
1118
1+
518
1−
η2 =
1
0.405
=
518
1+
1118
1−
η3 =
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.
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 3
Software Verification
SAFE
0
PROGRAM NAME:
REVISION NO.:
Item
x2
y2
L
d
Ld
Ldx2
Ldy2
Side 1
−259
0
1118
218
243724
−63124516
0
x3
=
Ldx
∑=
=
y3
Ldy
∑=
2
Ld
2
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
= 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
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 SAFE output at Grid B-2:
Vd= 1126.498 kN
0.4ηMd,2 = -8.4226 kN-m
0.4ηM d,3 = 10.8821 kN-m
Maximum design shear stress in computed in along major and minor axis of column:
V pd
0.4 M pd ,2u p d
0.4 M pd ,3u p d
+η
v pd =
1 + η
,
u p d
V pdWm ,2
V pdWm ,3
At the point labeled A in Figure 2, x4 = −259 and y4 = 559, thus:
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 4
(TS 8.3.1)
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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 = 1.4680 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
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
The concrete punching shear stress capacity of a section with punching shear
reinforcement is limited to:
v=
f=
0.35 f ck γ c
pr
ctd
v=
f=
0.35 30 1.5
= 1.278 N/mm2
pr
ctd
Shear Ratio
=
vpd
=
v pr
1.690
= 1.32
1.278
EXAMPLE Turkish TS 500-2000 RC-PN-001 - 6
(TS 8.3.1)
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
EXAMPLE Turkish TS 500-2000 RC-SL-001
Slab Flexural Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab flexural design in SAFE.
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.
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 SAFE results. After completing
the analysis, design is performed using the Turkish TS 500-2000 code by SAFE
and also by hand computation. Table 1 shows the comparison of the design
reinforcements computed by the two methods.
EXAMPLE Turkish TS 500-2000 RC-SL-001 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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
fyk
wc
Ec
Es
ν
=
=
=
=
=
=
30
460
0
25000
2×106
0
MPa
MPa
N/m3
MPa
MPa
Dead load
Live load
wd
wl
=
=
4.0 kPa
5.0 kPa
TECHNICAL FEATURES OF SAFE TESTED
Calculation of flexural reinforcement
Application of minimum flexural reinforcement
RESULTS COMPARISON
Table 1 shows the comparison of the SAFE 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)
SAFE
27.197
5.760
Calculated
27.200
5.760
As+
Medium
A +s ,min = 162.5 sq-mm
EXAMPLE Turkish TS 500-2000 RC-SL-001 - 2
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
COMPUTER FILE: Turkish TS 500-2000 RC-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.
EXAMPLE Turkish TS 500-2000 RC-SL-001 - 3
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
HAND CALCULATION
The following quantities are computed for all the load combinations:
γm, steel
= 1.15
γm, concrete = 1.50
f=
cd
f ck 30
= = 20
γ mc 1.5
=
f yd
f yk 460
= = 400
γ ms 1.15
cb =
ε cu Es
ε 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
γ 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:
EXAMPLE Turkish TS 500-2000 RC-SL-001 - 4
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
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
a
13.5518
f yd d − 400 125 −
2
2
EXAMPLE Turkish TS 500-2000 RC-SL-001 - 5
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
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CONCLUSIONS
The conclusions are presented separately for analysis, reinforced concrete beam and slab
design, and post-tensioned slab design in the following subsections.
ANALYSIS
The SAFE verification and validation example problems for analysis show Acceptable
comparison with the independent solutions. The accuracy of the SAFE results for certain
examples depends on the discretization of the area objects. For those examples, as the
discretization is refined, the solution becomes more accurate.
DESIGN
The design results for flexural and shear design for reinforced concrete beams; flexural
design for reinforced concrete and post-tensioned slab and stress checks for posttensioned slabs show exact comparison with hand calculations.
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 - 1
Software Verification
PROGRAM NAME:
REVISION NO.:
SAFE
0
References
ACI Committee 435, 1984. Deflection of Two-way Reinforced Concrete Floor Systems:
State-of-the-Art Report, (ACI 435-6R-74), (Reaffirmed 1984), American
Concrete Institute, Detroit, Michigan.
ACI Committee 336, 1988. Suggested Analysis and Design Procedures for Combined
Footings and Mats (ACI 336-2R-88), American Concrete Institute, Detroit,
Michigan.
ACI Committee 340, 1991. Design Handbook In Accordance with the Strength Design
Method of ACI 318-89, Volume 3, Two-way Slabs (ACI 340.4R-91), American
Concrete Institute, Detroit, Michigan.
ACI Committee 340, 1997. ACI Design Handbook, Design of Structural Reinforced
Concrete Elements in Accordance with the Strength Design Method of ACI 31895 (ACI 340R-97), American Concrete Institute, Detroit, Michigan.
ACI Committee 318, 1995. Building Code Requirements for Reinforced Concrete (ACI
318-95) and Commentary (ACI 318R-95), American Concrete Institute, Detroit,
Michigan.
Corley, W. G. and J. O. Jirsa, 1970. Equivalent Frame Analysis for Slab Design, ACI
Journal, Vol. 67, No. 11, November.
Gamble, W. L., M. A. Sozen, and C. P. Siess, 1969. Tests of a Two-way Reinforced
Concrete Floor Slab, Journal of the Structural Division, Proceedings of the
ASCE, Vol. 95, ST6, June.
Guralnick, S. A. and R. W. LaFraugh, 1963. Laboratory Study of a 45-Foot Square Flat
Plate Structure, ACI Journal, Vol. 60, No.9, September.
Hatcher, D. S., M. A. Sozen, and C. P. Siess, 1965. Test of a Reinforced Concrete Flat
Plate, Journal of the Structural Division, Proceedings of the ASCE, Vol. 91,
ST5, October.
Hatcher, D. S., M. A. Sozen, and C. P. Siess, 1969. Test of a Reinforced Concrete Flat
Slab, Journal of the Structural Division, Proceedings of the ASCE, Vol. 95, ST6,
June.
Jirsa, J. O., M. A. Sozen, and C. P. Siess, 1966. Test of a Flat Slab Reinforced with
Welded Wire Fabric, Journal of the Structural Division, Proceedings of the
ASCE, Vol. 92, ST3, June.
PCA, 1990. Notes on ACI 318-89 Building Code Requirements for Reinforced Concrete
with Design Applications, Portland Cement Association, Skokie, Illinois.
References - 1
Software Verification
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REVISION NO.:
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PCA, 1996. Notes on ACI 318-95 Building Code Requirements for Reinforced Concrete
with Design Applications, Portland Cement Association, Skokie, Illinois.
Roark, Raymond J., and Warren C. Young, 1975. Formulas for Stress and Strain, Fifth
Edition, Table 3, p. 107-108. McGraw-Hill, 2 Penn Plaza, New York, NY
10121-0101.
Timoshenko, S. and S. Woinowsky-Krieger, 1959, Theory of Plates and Shells,
McGraw-Hill, 2 Penn Plaza, New York, NY 10121-0101.
Ugural, A. C. 1981, Stresses in Plates and Shells, McGraw-Hill, 2 Penn Plaza, New
York, NY 10121-0101.
Vanderbilt, M. D., M. A. Sozen, and C. P. Siess, 1969. Tests of a Modified Reinforced
Concrete Two-Way Slab, Journal of the Structural Division, Proceedings of the
ASCE, Vol. 95, ST6, June.
References - 2
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