ACI 318 08 RC PN Example 001
User Manual: ACI 318-08 RC-PN Example 001
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ACI 318-08 RC-PN EXAMPLE 001 - 1
ACI 318-08 RC-PN EXAMPLE 001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.
4
A
3
2
1
BC D
X
Y
1'1' 24'24'24'
2'
2'
24'
24'
24'
10" thick flat slab
Loading
DL = Self weight + 20 psf
LL = 80 psf
Columns are 12" x 36"
with long side parallel
to the Y-axis, typical
Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
5678
910
11 12
13 14 15
17 18 19 20
4
A
3
2
1
BC D
X
Y
1'1' 24'24'24'
2'
2'
24'
24'
24'
10" thick flat slab
Loading
DL = Self weight + 20 psf
LL = 80 psf
Columns are 12" x 36"
with long side parallel
to the Y-axis, typical
Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
5678
910
11 12
13 14 15
17 18 19 20
Figure 1: Flat Slab For Numerical Example
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ACI 318-08 RC-PN EXAMPLE 001 - 2
The slab overhangs the face of the column by 6 inches along each side of the
structure. The columns are typically 12 inches wide by 36 inches long, with the
long side parallel to the Y-axis. The slab is typically 10 inches thick. Thick shell
properties are used for the slab.
The concrete has a unit weight of 150 pcf and an f 'c of 4000 psi. The dead load
consists of the self weight of the structure plus an additional 20 psf. The live load
is 80 psf.
TECHNICAL FEATURES OF ETABS TESTED
Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS punching shear capacity, shear
stress ratio, and D/C ratio with the punching shear capacity, shear stress ratio and
D/C ratio obtained by the analytical method. They match exactly for this
example.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method Shear Stress
(ksi) Shear Capacity
(ksi) D/C ratio
ETABS 0.1930 0.158 1.22
Calculated 0.1930 0.158 1.22
COMPUTER FILE: ACI 318-08 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.
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ACI 318-08 RC-PN EXAMPLE 001 - 3
HAND CALCULATION
Hand Calculation for Interior Column Using ETABS Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
4.25"
18"
18"
4.25"
6" 6" 4.25"4.25"
X
Y
44.5"
20.5"
Side 2
Side 3
AB
C
D
Column
Side 1
Side 4
Center of column is
point (x1, y1). Set
this equal to (0,0).
Critical section for
punching shear shown
dashed.
4.25"
18"
18"
4.25"
6" 6" 4.25"4.25"
X
Y
44.5"
20.5"
Side 2
Side 3
AB
C
D
Column
Side 1
Side 4
Center of column is
point (x1, y1). Set
this equal to (0,0).
Critical section for
punching shear shown
dashed.
Figure 2: Interior Column, Grid B-2 in ETABS Model
21
1 0.4955
2 44.5
13 20.5
V
31
1 0.3115
2 20.5
13 44.5
V
The coordinates of the center of the column (x1, y1) are taken as (0, 0).
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ACI 318-08 RC-PN EXAMPLE 001 - 4
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 Side 1 Side 2 Side 3 Side 4 Sum
x2 10.25 0 10.25 0 N.A.
y2 0 22.25 0
22.25 N.A.
L 44.5 20.5 44.5 20.5 b0 = 130
d 8.5 8.5 8.5 8.5 N.A.
Ld 378.25 174.25 378.25 174.25 1105
Ldx2 3877.06 0 3877.06 0 0
Ldy2 0 3877.06 0
3877.06 0
2
300"
1105
Ldx
xLd
2
300"
1105
Ldy
yLd
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 Side 1 Side 2 Side 3 Side 4 Sum
L 44.5 20.5 44.5 20.5 N.A.
d 8.5 8.5 8.5 8.5 N.A.
x2 - x3 10.25 0 10.25 0 N.A.
y2 - y3 0 22.25 0
22.25 N.A.
Parallel to Y-Axis X-axis Y-Axis X-axis N.A.
Equations 5b, 6b, 7 5a, 6a, 7 5b, 6b, 7 5a, 6a, 7 N.A.
IXX 64696.5 86264.6 64696.5 86264.6 301922.3
IYY 39739.9 7151.5 39739.9 7151.5 93782.8
IXY 0 0 0 0 0
From the ETABS output at Grid B-2:
VU = 189.45 k
22VU
M
= 156.39 k-in
33VU
M
= 91.538 k-in
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ACI 318-08 RC-PN EXAMPLE 001 - 5
At the point labeled A in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
2
2
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5 (301922.3)(93782.8) (0)
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0
U
v
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:
2
2
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5 301922.3 93782.8 0
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0
U
v
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:
2
2
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5 301922.3 93782.8 0
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0
Uv
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:
2
2
156.39 93782.8 22.25 0 0 10.25 0
189.45
130 8.5 301922.3 93782.8 0
91.538 301922.3 10.25 0 0 22.25 0
301922.3 93782.8 0
Uv
vU = 0.1714 + 0.0115 - 0.0100 = 0.1729 ksi at point D
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ACI 318-08 RC-PN EXAMPLE 001 - 6
Point C has the largest absolute value of vu, thus vmax = 0.1930 ksi
The shear capacity is calculated based on the smallest of ACI 318-08 equations 11-34,
11-35 and 11-36 with the b0 and d terms removed to convert force to stress.
4
0.75 2 4000
36/12 0.158
1000
Cv
ksi in accordance with equation 11-34
40 8.5
0.75 2 4000
130 0.219
1000
Cv
ksi in accordance with equation 11-35
0 75 4 4000 0190
1000
C.
v.
ksi in accordance with equation 11-36
Equation 11-34 yields the smallest value of
vC = 0.158 ksi and thus this is the shear
capacity.
0.193 1.22
0.158
U
C
v
Shear Ratio v
