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Question 7.4: A column of a building carries a load of 1000 kips. The load...

A column of a building carries a load of 1000 kips. The load is transferred to sub soil through a square footing of size 16 x 16 ft founded at a depth of 6.5 ft below ground level. The soil below the footing is fine sand up to a depth of 16.5 ft and below this is a soft compressible clay of thickness 16 ft. The water table is found at a depth of 6.5 ft below the base of the footing. The specific gravities of the solid particles of sand and clay are 2.64 and 2.72 and their natural moisture contents are 25 and 40 percent respectively. The sand above the water table may be assumed to remain saturated. If the plastic limit and the plasticity index of the clay are 30 and 40 percent respectively, estimate the probable settlement of the footing

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1. Required  Δp\Delta p at the middle of the clay layer using the Boussinesq equation

 

zB=24.516=1.53<3.0\frac{z}{B}=\frac{24.5}{16}=1.53<3.0

 

Divide the footing into 4 equal parts so that ZIB > 3

 

The concentrated load at the center of each part = 250 kips

 

Radial distance, r = 5.66 ft

 

By the Boussinesq equation the excess pressure Δp\Delta p at depth 24.5 ft is (IB=0.41)\left(I_{B}=0.41\right)

 

Δp=4×Qz2IB=4×25024.52×0.41=0.683k/ft2\Delta p=4 \times \frac{Q}{z^{2}} I_{B}=\frac{4 \times 250}{24.5^{2}} \times 0.41=0.683 \mathrm{k} / \mathrm{ft}^{2}

 

2. Void ratio and unit weights

 

Per the procedure explained in Ex. 7.3

 

For sand  γt=124lb/ft3γb=61.6lb/ft3\gamma_{t}=124 \mathrm{lb} / \mathrm{ft}^{3} \quad \gamma_{b}=61.6 \mathrm{lb} / \mathrm{ft}^{3}

 

For clay  γb=51.4lb/ft3e0=1.09\gamma_{b}=51.4 \mathrm{lb} / \mathrm{ft}^{3} \quad e_{0}=1.09

 

3. Overburden pressure  p0p_{0}

 

p0=8×51.4+10×62+13×124=26391 b/ft2p_{0}=8 \times 51.4+10 \times 62+13 \times 124=26391 \mathrm{~b} / \mathrm{ft}^{2}

 

4. Compression index

 

wl=Ip+wp=40+30=70%,Cc=0.009(70−10)=0.54w_{l}=I_{p}+w_{p}=40+30=70 \%, \quad C_{c}=0.009(70-10)=0.54

 

Settlement  St=0.541+1.09×16×log⁡2639+6832639=0.413ft=4.96inS_{t}=\frac{0.54}{1+1.09} \times 16 \times \log \frac{2639+683}{2639}=0.413 \mathrm{ft}=4.96 \mathrm{in} .

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