Contents CSA A23.3 14 RC BM 001

User Manual: CSA A23.3-14 RC-BM-001

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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

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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

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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

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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

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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

a b = β 1 c b = 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 M f that is resisted by the flange is given

by:

EXAMPLE CSA A23.3-14 RC-BM-001 - 5

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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, M f to be carried by the web is:
M fw = M f − M ff = 198.403 kN-m
The web is a rectangular section with dimensions b w 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 a 1 ≤ a b , the area of tension reinforcement is then given by:
As 2 =

M fw
a 

φs f y  d − 1 
2


= 1379.94 sq-mm

A s = A s1 + A s2 = 2584.351 sq-mm
Shear Design
The basic shear strength for rectangular section is computed as,

φ c = 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

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=
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



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