Structural Concrete - Hassoun

Review Problems on Concrete Building Components 967
d. Design the Footing at C
1. Calculate the effective soil pressure.
Assume a total depth of footing 1.5 ft
The weight of the footing is 1.5 × 150 = 225 psf
(Assume concrete unit weight = 150 pcf)
The weight of soil above the footing is (4 − 1.5) ×130 = 325 psf
(Assume soil unit weight = 130 pcf)
Effective soil pressure = 4 − 225∕1000 − 325∕1000 = 3.45 ksf
2. Calculate the area of the footing:
Actual loads = DL + LL = 30 + 8 = 38 K
Area of footing = 38 = 11.01 ft2
3.45
Side of footing = 3.32 ft, use 4 ft
3. Calculate net upward pressure equals (factored load)/(area of footing):
P u = 1.2DL + 1.6LL
P u = 1.2(30)+ 1.6(8) = 48.80 K
Net upward pressure, q u = P u
A = 48.80 = 3.05 ksf
(4 × 4p)
4. Check depth due to two-way shear. Assume bar no. 8 bars both ways. Calculate d to the centroid
of the top steel layer:
d = 18 − 3(cover)−1.5(bar diameters) =13.5 in.
b 0 = 4(c + d) =4(12 + 13.5) =102 in.
c + d = 12 + 13.5 = 25.5in. = 2.125 ft
V u2 = P u − q u (c + d) 2 = 48.80 − 48.8(2.125) 2 = 35.03 K
V
Required d 1 = u2
4φλ √ 35.03 (1000)
=
f ′ c b 0 4(0.75)(1) √ = 1.81 in.
4000(102)
β = L W = 4 = 1 (Eq. 13.9)
4
since β ≤ 2
35.03 (1000)
Required d 2 = ( )
0.75 20×13.5
+ 2 ( √ = 1.56 in. (not critical)
4000)(102)
102
(α s = 20 for corner columns.) Thus, the assumed depth is adequate.
5. Check depth due to one-way shear. The critical section is at distance d from the face of the
column:
( L
Distance to critical section =
2 − c ) ( 4
2 − d =
2 − 1 2 − 13.5 )
= 0.375 ft
12
( L
V u1 = q u b
2 − c )
2 − d = 3.05 × 4 × 0.375 = 4.58 K
The depth required for one-way shear is
d =
V u1
2φλ √ f ′ c b = 4.58 (1000)
2(0.75)(1) √ = 1.00 in. < 13.5
4000(4 × 12)