Long Term Consolidation in a Clay Layer

Updated to properly consider pressure at middle of clay layer.

A building is constructed above the soil profile shown below. Calculate the long-term consolidation settlement in the middle of the clay layer given the following: 

A) 4.3 inch


B) 5.6 inch


C) 6.0 inch


D) 8.3 inch


Solution

Answer: D) 6.8 inch

Answer: D) 8.3 inch

Long-term settlement is caused by the clay layer.

Calculate the original pressure in the middle of the clay layer:

\[p_0= \gamma _1 h_1+ \gamma _2 h_2+( \gamma '_3h_3)/2=80 \times 10+120 \times 15 + ((105-62.4) \times 20)/\sout{10}=\sout{2685\ PSF}\]

\[p_0= \gamma _1 h_1+ \gamma _2 h_2+( \gamma '_3h_3)/2=80 \times 10+120 \times 15 + ((105-62.4) \times 20)/2=3026\ PSF\]

Additional Pressure caused by the building is given in the middle of the clay layer:

\[\triangle p =1200\ PSF\]

Final Pressure at Middle of Clay Layer:

\[p_f=p_0+∆p=\sout{2685}+1200 = \sout{3885\ PSF}\]

\[p_f=p_0+∆p=3026+1200 =4226\ PSF\]Original void content in clay, where w is clay water content: 

\[e_0≈w(SG)=0.40 \times 2.6=1.04\]

Compression index:

\[C_c = 0.009(LL-10) = 0.009  \times  (50-10) =0.36\]

Long-term settlement in clay layer Long-term settlement in the clay layer:

\[S =  \frac{H  C_{c} Log_{10} (p_{f} /p_{0}) }{(1+ e_{0}) } =\frac{(20\times0.36 \times Log(\sout{3885}/2685)}{(1+1.04)} = \sout{\ 0.566\ ft=6.8\ inch}\]

\[S =  \frac{H  C_{c} Log_{10} (p_{f} /p_{0}) }{(1+ e_{0}) } =\frac{(20\times0.36 \times Log(4226/2685)}{(1+1.04)} =\ 0.70\ ft=8.3\ inch\]