Aci 318-14 RC-PN-001
Aci 318-14 RC-PN-001
Aci 318-14 RC-PN-001
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.
A B C D
1' 24' 24' 24' 1'
2'
4 17 18 19 20
10" thick flat slab
24'
13 14 15
3
Columns are 12" x 36"
24' with long side parallel
to the Y-axis, typical
9 10 11 12
2 Concrete Properties
Unit weight = 150 pcf
f'c = 4000 psi
24'
Y Loading
DL = Self weight + 20 psf
5 X 6 7 8 LL = 80 psf
1
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 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.
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.
CONCLUSION
The SAFE results show an exact comparison with the independent results.
HAND CALCULATION
Hand Calculation for Interior Column Using SAFE Method
d = [(10 - 1) + (10 - 2)] / 2 = 8.5"
Refer to Figure 2.
b 0 = 44.5 + 20.5 + 44.5 + 20.5 = 130"
20.5"
Y
Critical section for
4.25" 6" 6" 4.25"
punching shear shown
dashed.
A B
Column Side 2
4.25"
18"
Side 1
Side 3
X
44.5"
Center of column is
18"
point (x1, y1). Set
this equal to (0,0).
4.25"
Side 4
D
C
γ V2 =
1−
1
=
0.4955
2 44.5
1+
3 20.5
γ V3 =
1−
1
=
0.3115
2 20.5
1+
3 44.5
The coordinates of the center of the column (x 1 , 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.
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
=
x3
∑ Ldx
=
2 0
= 0"
Ld 1105
=
y3
∑ Ldy
=
2 0
= 0"
Ld 1105
The following table is used to calculate I XX , I YY and IXY . The values for I XX , I YY and I XY
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
Point C has the largest absolute value of v u , thus v max = 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 b 0 and d terms removed to convert force to stress.
4
0.75 2 + 4000
=ϕ vC 36 /12
= 0.158 ksi in accordance with equation 11-34
1000
40 • 8.5
0.75 + 2 4000
=ϕ vC = 130 0.219 ksi in accordance with equation 11-35
1000
0.75 • 4 • 4000
=ϕ vC = 0.190 ksi in accordance with equation 11-36
1000
Equation 11-34 yields the smallest value of φv C = 0.158 ksi and thus this is the shear
capacity.
vU 0.193
=
Shear Ratio = = 1.22
ϕ vC 0.158