ACI 318-2025 PUNCHING SHEAR CHECK
1.0 INPUT DATA
Material Properties
f'c = psi Specified compressive strength of concrete
fy = psi Specified yield strength of reinforcement
λ = Lightweight concrete factor (1.0 for normal weight)
Geometric Properties
h =
in Slab overall thickness
c =
in Concrete cover to centroid of reinforcement
φb = in Slab main bar diameter
c1 = in Column dimension C1
c2 = in Column dimension C2
ρl = Flexural reinforcement ratio (decimal)
Vu = lb Factored shear force from analysis
Column Type Selection
Column Type =
Figure 1: Geometry
Punching Shear Perimeter
2.0 CALCULATIONS
2.1 Effective Depths
Effective depth in X-direction (assuming bars perpendicular to C1)
dv = h − c −
φb/2 Effective depth for shear
dv = 12 - 1.5 - 0.75/2 = 9.625 in
Minimum effective depth check
dv ≥ max(0.5×h, 16×φb)
dv = 9.625 ≥ 12 in
FAIL
2.2 Critical Section Properties
Critical section at dv/2 from column face
Interior Column (Section 22.6.4.2)
b0 = 2×(c1 + c2) +
4×π×(dv/2) Perimeter of critical section
b0 = 2×(12 + 12) + 4×π×(9.625/2) =
60.532 in
Edge Column (Section 22.6.4.3)
b0 = c1 + 2×c2 +
2×π×(dv/2) Perimeter of critical section
b0 = 12 + 2×12 + 2×π×(9.625/2) =
42.266 in
Corner Column (Section 22.6.4.4)
b0 = c1 + c2 +
π×(dv/2) Perimeter of critical section
b0 = 12 + 12 + π×(9.625/2) =
31.133 in
2.3 Shear Strength of Concrete
Nominal shear strength of concrete (Section 22.6.4.2)
vc = min(4×λ×√(f'c), (2 +
4×β)×λ×√(f'c))
Where β is the ratio of long side to short side of column
β = max(c1/c2,
c2/c1) = max(12/12, 12/12) =
1.000
4×λ×√(f'c) = 4×1.0×√30 = 21.909 psi
(2 + 4×β)×λ×√(f'c) = (2 +
4×1.0)×1.0×√30 = 32.863 psi
vc = min(21.909, 32.863) = 21.909 psi
2.4 Shear Reinforcement Requirements
Maximum shear that can be carried by concrete
φvVc = 0.75 × vc ×
b0 × dv
φvVc = 0.75 × 21.909 ×
60.532 × 9.625 = 94964 lb
Maximum shear that can be carried by slab
φvVmax = 0.75 ×
8×λ×√(f'c) × b0 × dv
φvVmax = 0.75 ×
8×1.0×√30 × 60.532 × 9.625 = 189928
lb
Shear Reinforcement Check
check_shear_reinforcement = NO SHEAR REINFORCEMENT
REQUIRED
(Vu = 45000 ≤
φvVc = 94964
lb)
2.5 Design of Shear Reinforcement (if required)
Shear reinforcement required
Av,required = (Vu/φ −
Vc) ×
s/(fy × dv)
Minimum requirements for shear reinforcement:
Av,min = max(0.75×√(f'c) ×
b0 × s/fy,
50×b0 × s/fy)
Spacing limitations:
s ≤ min(dv/2, 24×φdb)
Where φdb is stirrup diameter
No shear reinforcement required
*** OUTPUT SUMMARY ***
Design Summary
Column Type = Interior
Column aspect ratio β = 1.000
Effective depth dv = 9.625 in
Critical section perimeter b0 = 60.532 in
Concrete shear strength vc = 21.909 psi
Factored shear force Vu = 45000 lb
Design shear capacity φvVc = 94964 lb
Shear check: NO SHEAR REINFORCEMENT REQUIRED
RESULTS
ALL CHECKS PASSED - DESIGN IS ADEQUATE
