Structural Concrete - Hassoun

13.6 Combined Footings 473
Figure 13.25
Analysis of combined footing in the transverse direction.
does not provide a detailed approach for the design of combined footings. The design, in general,
is based on structural analysis.
A simple method of analysis is to treat the footing as a beam in the longitudinal direction,
loaded with uniform upward pressure, q u . For the transverse direction, it is assumed that the column
load is spread over a width under the column equal to the column width plus d on each side,
whenever that is available. In other words, the column load acts on a beam under the column within
the footing, which has a maximum width of c + 2d and a length equal to the short side of the footing
(Fig. 13.25). A smaller width, down to c + d may be used. The next example explains the design
method in detail.
Example 13.7
Design a rectangular combined footing to support two columns, as shown in Fig. 13.26. The edge column,
I, has a section 16 × 16 in. and carries a DL of 180 K and an LL of 120 K. The interior column,
II, has a section 20 × 20 in. and carries a DL of 250 K and an LL of 140 K. The allowable soil pressure
is 5 ksf and the bottom of the footing is 5 ft below final grade. Design the footing using f c ′ = 4ksi,
f y = 60 ksi.
Solution
1. Determine the location of the resultant of the column loads. Take moments about the center of the
exterior column I:
(250 + 140)×16
x =
= 9 ft from column I
(250 + 140)+(180 + 120)
The distance of the resultant from the property line is 9 + 2 = 11.0 ft. The length of the footing is
2 × 11 = 22.0 ft. In this case the resultant of column loads will coincide with the resultant of the
upward pressure on the footing.
2. Determine the area of the footing. Assume the footing total depth is 36 in. (d = 36 − 4.5 = 31.5 in.).
Total actual (working) loads = 300 + 390 = 690 K
( ) 36
New upward pressure = 5000 −
12 × 150 −(2 × 100) =4350 psf