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
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in ETABS.
The numerical example is a flat slab that has three 24-foot-long spans in each
direction, as shown in Figure 1.

A

C

B

1'

24'

D

24'

24'

1'

2'

4

17

18

19

13

14

15

20

10" thick flat slab

24'

3

Columns are 12" x 36"
with long side parallel
to the Y-axis, typical

24'

9

2

10

11

12

Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi

24'

Y
5

1

X

6

7

Loading
DL = Self weight + 20 psf
LL = 80 psf

8

2'

Figure 1: Flat Slab For Numerical Example

ACI 318-08 RC-PN EXAMPLE 001 - 1

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The slab overhangs the face of the column by 6 inches along each side of the
structure. The columns are typically 12 inches wide by 36 inches long, with the
long side parallel to the Y-axis. The slab is typically 10 inches thick. Thick shell
properties are used for the slab.
The concrete has a unit weight of 150 pcf and an f 'c of 4000 psi. The dead load
consists of the self weight of the structure plus an additional 20 psf. The live load
is 80 psf.

TECHNICAL FEATURES OF ETABS TESTED
 Calculation of punching shear capacity, shear stress, and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the ETABS punching shear capacity, shear
stress ratio, and D/C ratio with the punching shear capacity, shear stress ratio and
D/C ratio obtained by the analytical method. They match exactly for this
example.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2

Method

Shear Stress Shear Capacity
(ksi)
(ksi)
D/C ratio

ETABS

0.1930

0.158

1.22

Calculated

0.1930

0.158

1.22

COMPUTER FILE: ACI 318-08 RC-PN EX001.EDB
CONCLUSION
The ETABS results show an exact comparison with the independent results.

ACI 318-08 RC-PN EXAMPLE 001 - 2

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HAND CALCULATION
Hand Calculation for Interior Column Using ETABS Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
20.5"

Y
4.25"

6" 6"

Critical section for
punching shear shown
dashed.

4.25"

A

B

Column

4.25"

18"

Side 3

Side 1

Side 2

X

Center of column is
point (x1, y1). Set
this equal to (0,0).

44.5"

18"

4.25"
Side 4
D

C

Figure 2: Interior Column, Grid B-2 in ETABS Model



V2

1



V3

1

1
 2  44.5
1  
 3  20.5
1
 2  20.5
1  
 3  44.5

 0.4955

 0.3115

The coordinates of the center of the column (x1, y1) are taken as (0, 0).

ACI 318-08 RC-PN EXAMPLE 001 - 3

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The following table is used for calculating the centroid of the critical section for
punching shear. Side 1, Side 2, Side 3, and Side 4 refer to the sides of the critical
section for punching shear, as identified in Figure 2.
Item
x2
y2
L
d
Ld
Ldx2
Ldy2

Side 1
10.25
0
44.5
8.5
378.25
3877.06
0

x3 

 Ldx

y3 

 Ldy

2

Ld
2

Ld

Side 2
0
22.25
20.5
8.5
174.25
0
3877.06



0
 0"
1105



0
 0"
1105

Side 3
10.25
0
44.5
8.5
378.25
3877.06
0

Side 4
0
22.25
20.5
8.5
174.25
0
3877.06

Sum
N.A.
N.A.
b0 = 130
N.A.
1105
0
0

The following table is used to calculate IXX, IYY and IXY. The values for IXX, IYY and IXY
are given in the “Sum” column.
Item
L
d
x2 - x3
y2 - y3
Parallel to
Equations
IXX
IYY
IXY

Side 1
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0

Side 2
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0

From the ETABS output at Grid B-2:
VU = 189.45 k

 V 2 M U 2 = 156.39 k-in
 V 3 M U 3 = 91.538 k-in

ACI 318-08 RC-PN EXAMPLE 001 - 4

Side 3
44.5
8.5
10.25
0
Y-Axis
5b, 6b, 7
64696.5
39739.9
0

Side 4
20.5
8.5
0
22.25
X-axis
5a, 6a, 7
86264.6
7151.5
0

Sum
N.A.
N.A.
N.A.
N.A.
N.A.
N.A.
301922.3
93782.8
0

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At the point labeled A in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU 

156.39 93782.8  22.25  0    0   10.25  0  
189.45


130  8.5
(301922.3)(93782.8)  (0)2
91.538 301922.3  10.25  0    0   22.25  0  

 301922.3 93782.8    0 2

vU = 0.1714  0.0115  0.0100 = 0.1499 ksi at point A
At the point labeled B in Figure 2, x4 = 10.25 and y4 = 22.25, thus:

vU 

156.39 93782.8  22.25  0    0  10.25  0  
189.45


130  8.5
 301922.3 93782.8    0 2
91.538 301922.3 10.25  0    0   22.25  0  

 301922.3 93782.8    0 2

vU = 0.1714  0.0115 + 0.0100 = 0.1699 ksi at point B
At the point labeled C in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU 

156.39 93782.8  22.25  0    0  10.25  0  
189.45


130  8.5
 301922.3 93782.8    0 2
91.538 301922.3 10.25  0    0   22.25  0  

 301922.3 93782.8    0 2

vU = 0.1714 + 0.0115 + 0.0100 = 0.1930 ksi at point C
At the point labeled D in Figure 2, x4 = 10.25 and y4 = 22.25, thus:
vU 

156.39 93782.8  22.25  0    0   10.25  0  
189.45


130  8.5
 301922.3 93782.8    0 2
91.538 301922.3  10.25  0    0   22.25  0  

 301922.3 93782.8    0 2

vU = 0.1714 + 0.0115 - 0.0100 = 0.1729 ksi at point D

ACI 318-08 RC-PN EXAMPLE 001 - 5

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Point C has the largest absolute value of vu, thus vmax = 0.1930 ksi

The shear capacity is calculated based on the smallest of ACI 318-08 equations 11-34,
11-35 and 11-36 with the b0 and d terms removed to convert force to stress.
4 

0.75  2 
 4000
36 /12 

 0.158 ksi in accordance with equation 11-34
 vC 
1000
 40  8.5

0.75 
 2  4000
 130

 0.219 ksi in accordance with equation 11-35
 vC 
1000

 vC 

0.75  4  4000
 0.190 ksi in accordance with equation 11-36
1000

Equation 11-34 yields the smallest value of vC = 0.158 ksi and thus this is the shear
capacity.

Shear Ratio 

ACI 318-08 RC-PN EXAMPLE 001 - 6

vU
0.193

 1.22
0.158
 vC
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