SME
Reinforced Concrete Design
Slabs and footings
Use this for “A 16 in column carries 200 kips service (272 kips factored) on soil allowing 4 ksf. Size the square footing and check punching shear with d = 15 in and f′c = 4000 psi.”
Diagram
b₀
=
4(c
+
d)
=
4
×
(40.64
cm
+
38.1
cm)
=
315
cm,
d
2
out
from
each
face
of
the
column.
The shaded soil pressure outside the perimeter is the shear Vᵤ the slab must carry.
This is an example. Type your own values to start a new problem.
P
Service load
example · kN
q
Net allowable soil pressure
example · kPa
B
Footing side
calculated · m
Pᵤ
Factored load
example · kN
c
Column side
example · cm
d
Effective depth
example · cm
f′c
Concrete strength
example
b₀
Critical perimeter
calculated · cm
Vᵤ
Punching shear
calculated · kN
φVc
Punching strength
calculated · kN
K
Punching check
worked out (not typed)
Formulas
B
=
⌈4
×
√(
P
q
)⌉
÷
4
7.25
=
⌈4
×
√(
200
4
)⌉
÷
4
b₀
=
4
×
(
c
+
d
)
124 = 4 × (16 + 15)
Vᵤ
=
(
Pᵤ
B
2
)
×
(
B
2
−
(
c
+
d
12
)²)
237.4655
=
(
272
7.25
2
)
×
(
7.25
2
−
(
16 + 15
12
)²)
φVc
=
0.75
×
4
×
√f′c
×
b₀
×
d
1,000
352.9102
=
0.75
×
4
×
√
4
,000
×
124
×
15
1,000
K
=
passes
if
Vᵤ
≤
φVc,
else
fails
237.4655 ≤ 352.9102, so passes: Vᵤ ≤ φVc
Formulas are worked in kip, ksf, ft, in. The step-by-step shows the conversions.
Assumptions
•
ACI 318-19: a square column centered on a square footing; normal-weight concrete, f′c in psi under the root.
•
An interior column, so the 4√f′c term governs two-way (punching) shear, with φ = 0.75.
•
Soil pressure is uniform under the factored load; b₀ runs d ÷ 2 out from each column face.
•
The side B is √(P ÷ q) rounded up to the next 3 in; 12 in to a foot.
Step-by-step
WE KNOW
P = 889.6443 kN, q = 191.521 kPa, Pᵤ = 1,209.9163 kN, c = 40.64 cm, d = 38.1 cm, f′c = 4,000 psi
FIND
footing side (B), critical perimeter (b₀), punching shear (Vᵤ), punching strength (φVc), punching check (K)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to kip, ksf, ft, in).
P = 889.6443 kN = 200 kip (1 kip = 4.44822 kN)
q = 191.521 kPa = 4 ksf (1 ksf = 47.8803 kPa)
Pᵤ = 1,209.9163 kN = 272 kip (1 kip = 4.44822 kN)
c = 40.64 cm = 16 in (1 in = 2.54 cm)
d = 38.1 cm = 15 in (1 in = 2.54 cm)
Step 2 · Find footing side (B)
Use B = ⌈4 × √(P ÷ q)⌉ ÷ 4
The
area
P
q
keeps
the
soil
at
its
allowable
pressure;
its
side,
rounded
up
to
a
quarter
foot.
B
=
⌈4
×
√(
200
4
)⌉
÷
4
B
=
7.25
ft
Step 3 · Find critical perimeter (b₀)
Use b₀ = 4 × (c + d)
Four
sides,
each
the
column
side
plus
d
2
on
both
faces.
b₀
=
4
×
(16
+
15)
b₀
=
4
×
31
b₀
=
124
in
Step 4 · Find punching shear (Vᵤ)
Use Vᵤ = (Pᵤ ÷ B²) × (B² − ((c + d) ÷ 12)²)
The
factored
soil
pressure
times
the
footing
area
outside
the
perimeter
(
inches
12
for
feet).
Vᵤ
=
(
272
7.25
2
)
×
(
7.25
2
−
(
16 + 15
12
)²)
Vᵤ
=
272
7.25
2
×
(
7.25
2
−
(
31
12
)
2
)
Vᵤ
=
272
7.25
2
×
(
7.25
2
−
2.583
2
)
Vᵤ
=
272
7.25
2
×
(52.5625
−
6.672)
Vᵤ
=
272
7.25
2
×
45.8905
Vᵤ
=
272
52.5625
×
45.8905
Vᵤ = 5.175 × 45.8905
Vᵤ = 237.4655 kip
Step 5 · Find punching strength (φVc)
Use φVc = 0.75 × 4 × √f′c × b₀ × d ÷ 1,000
The
concrete’s
shear
stress
4√f′c
(psi)
over
the
area
b₀
d
,
in
pounds
1000
for
kips.
φVc
=
0.75
×
4
×
√
4
,000
×
124
×
15
1,000
φVc
=
0.75
×
4
×
63.25
×
124
×
15
1,000
φVc
=
3
×
63.25
×
124
×
15
1,000
φVc
=
189.75
×
124
×
15
1,000
φVc
=
23,529
×
15
1,000
φVc
=
352,935
1,000
φVc = 352.9102 kip
Step 6 · Find punching check (K)
Use K = passes if Vᵤ ≤ φVc, else fails
The concrete carries the punching shear without shear reinforcement.
237.4655 ≤ 352.9102: passes: Vᵤ ≤ φVc
K
=
compare
Vᵤ
with
φVc
K
=
compare
237.4655
with
352.9102
K
=
passes:
Vᵤ
≤
φVc
Step 7 · Convert the answers
Change each answer back to the unit you chose.
B = 7.25 ft = 2.2098 m (1 m = 3.28084 ft)
b₀ = 124 in = 314.96 cm (1 in = 2.54 cm)
Vᵤ = 237.4655 kip = 1,056.299 kN (1 kip = 4.44822 kN)
φVc = 352.9102 kip = 1,569.823 kN (1 kip = 4.44822 kN)
Check
Put the numbers back in. Both sides should match.
7.25
=
⌈4
×
√(
200
4
)⌉
÷
4
✓
124 = 4 × (16 + 15) ✓
237.4655
=
(
272
7.25
2
)
×
(
7.25
2
−
(
16 + 15
12
)²)
✓
352.9102
=
0.75
×
4
×
√
4
,000
×
124
×
15
1,000
✓
237.4655 ≤ 352.9102, so passes: Vᵤ ≤ φVc ✓
Topic check
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