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:
- Additional pressure caused by the building is 1200 psf.
- The water level is at the top of the clay layer.
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:
\[S = Â \frac{H Â C_{c} Log_{10} (p_{f} /p_{0}) }{(1+ e_{0}) } =\frac{(20\times0.36 \times Log(\sout{3885}/2685)}{(1+1.04)} = \sout{\ 0.566\ ft=6.8\ inch}\]
\[S = Â \frac{H Â C_{c} Log_{10} (p_{f} /p_{0}) }{(1+ e_{0}) } =\frac{(20\times0.36 \times Log(4226/2685)}{(1+1.04)} =\ 0.70\ ft=8.3\ inch\]