The end column along a property line is connected to an interior column by a trapezoidal footing. The following data are given with reference to Fig. 14.3(b):
Column Loads: Q_{1}=2016 kN , Q_{2}=1560 kN.
Size of columns: 0.46 x 0.46 m.
L_{c}=5.48 m.
Determine the dimensions a and b of the trapezoidal footing. The net allowable bearing pressure q_{n a}=190 kPa.
Determine the center of bearing pressure x_{2} from the center of Column 1. Taking moments of all the loads about the center of Column 1, we have
(2016+1560) x_{2}=1560 \times 5.48
x_{2}=\frac{1560 \times 5.48}{3576}=2.39 m
Now x_{1}=2.39+\frac{0.46}{2}=2.62 m
Point O in Fig. 14.3(b) is the center of the area coinciding with the center of pressure. From the allowable pressure q_{a}=190 kPa, the area of the combined footing required is
A=\frac{3576}{190}=18.82 sq \cdot m
From geometry, the area of the trapezoidal footing (Fig. 14.3(b)) is
A=\frac{(a+b) L}{2}=\frac{(a+b)}{2}(5.94)=18.82
or (a+b)=6.34 m
where, L=L_{c}+b_{1}=5.48+0.46=5.94 m
From the geometry of the Fig. (14.3b), the distance of the center of area x_{1} can be written in terms of a, b and L as
x_{1}=\frac{L}{3} \frac{2 a+b}{a+b}
or \frac{2 a+b}{a+b}=\frac{3 x_{1}}{L}=\frac{3 \times 2.62}{5.94}=1.32 m
but a+b=6.32 m \text { or } b=6.32-a. Now substituting for b we have,
\frac{2 a+6.34-a}{6.34}=1.32
and solving, a = 2.03 m, from which, b = 6.34 – 2.03 = 4.31 m.
