Question

The results of two plate load tests for a settlement of 25 mm are given below:

Plate diameter	Load
0.3 m 0.6 m	30 kN 70 kN

What will be the size of square footing designed to carry a load of 900 kN as per Housel method ?

• A. 2.4 m
• B. 2.7 m
• C. 3 m
• D. 2.9 m

Answer 1

The correct option is D 2.9 m

$A_1=\frac{\pi }{4}\times {0.3}^2=0.07068m^2,Q_1=30kN,P_1=\pi \times 0.3=0.9425m$

$A_2=\frac{\pi }{4}\times {0.6}^2=0.28274m^2,Q_2=70kN,P_2=\pi \times 0.6=1.885m$

$Q_1=m{A}_1+n{P}_1\Rightarrow 30=m\times \left(0.07068\right)+n\left(0.9425\right)$ .........(1)

$Q_2=m{A}_2+n{P}_2\Rightarrow 70=m\times \left(0.28274\right)+n\left(1.885\right)$ .........(2)

Solving equation (1) and (2), we get

m = 70.73, n = 26.52

For Q = 900 kN

$900=70.75B^2+\left(4B\right)\times 26.52$

B = 2.895 m = 2.9 m