Structural Concrete - Hassoun

482 Chapter 13 Footings
1. For the first method, assume a footing depth of 20 in. (d = 16.5 in.) and assume the weight of soil
is 100 pcf. Net upward pressure is
5000 − 20
(
12 × 150 (footing)− 5 − 20 )
× 100 = 4417 psf
12
Area of footing = 385 = 87.1 ft2
4.42
Assume a footing width of 9 ft; then the footing length is 87.1/9 = 9.7 ft, say, 10 ft. Choose a
footing 9 × 10 ft and place the column eccentrically, as shown in Fig. 13.27d. The center of the
footing is 10 in. away from the center of the column.
2. The design procedure now is similar to that for a single footing. Check the depth for two-way
and one-way shear action. Determine the bending moment at the face of the column for the
longitudinal and transverse directions. Due to the eccentricity of the footing, the critical section
will be on the left face of the column in Fig. 13.30d. The distance to the end of footing is
(5 × 12) − 2 = 58 in. = 4.833 ft.
P u = 1.2D + 1.6L = 1.2 × 200 + 1.6 × 165 = 504 K
q u = 504 = 5.6 ksf
9 × 10
Maximum M u =(5.6 × 9)× (4.833)2 = 588.6K⋅ ft (in 9 − ft width)
2
In the transverse direction,
M u =(5.6 × 10)× (4)2 = 448 K ⋅ ft
2
Revise the assumed depth if needed and choose the required reinforcement in both directions of
the footing, as was explained in the single-footing example.
3. For the second method, calculate the area of the footing in the same way as explained in the first
method; then calculate the maximum soil pressure and compare it with that allowable using actual
loads:
Total load P = 385 K
Size of footing = 10 × 9ft
Because the eccentricity is e = 10 in. 4.42 ksf
9(10) 2
The footing is not safe. Try a footing 9.25 × 13 ft (area = 120.25 ft 2 ).
q max = 385
120.25 + 6 × 320 = 4.22 ksf < 4.42 ksf
9.25(13) 2
q min = 3.2 − 1.22 = 1.98 ksf
4. Calculate the factored upward pressure using factored loads; then calculate moments and shears,
as explained in previous examples.