Question 3 of 6: Differential Settlement of Two Foundations on a Variable-Thickness Clay

Question 3: Differential Settlement of Two Foundations on a Variable-Thickness Clay (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two identical 1.5 m × 1.5 m footings, each carrying 780 kN, founded on sand over a clay wedge (1.2 m thick under Foundation A, 2.8 m thick under Foundation B) over gravel.

Find. (a) the ultimate (expected total) differential settlement between A and B; (b) how much longer until the differential settlement reaches 24 mm.

Approach. Each footing's own clay settlement is $S=m_v\,\Delta\sigma_z\,H$, where $\Delta\sigma_z$ is the Boussinesq stress increase at the clay's mid-depth under the CENTRE of that footing (found by superposing four quarter-rectangle corner solutions) and $H$ is the clay thickness beneath that footing. The difference $S_A-S_B$ is the ultimate differential settlement; scaling that ultimate value by Terzaghi's average degree-of-consolidation curve $U(T_v)$, calibrated against the one observed data point (10 mm at 2 years), gives the time to reach any other differential settlement, since $T_v\propto t$ for fixed $c_v$ and drainage path.

1. Footing pressure. $q=P/(B L)=780/(1.5\times1.5)=346.7$ kPa (identical for A and B).
2. Stress increase at each clay mid-depth (Boussinesq, under footing centre). Mid-clay depths are $z_A=3.0+1.2/2=3.6$ m and $z_B=1.4+2.8/2=2.8$ m below the footing base. Superposing four $0.75\times0.75$ m corner solutions, $$\Delta\sigma_A=4qI(m,n)\Big|_{z=3.6}=\boxed{26.8\ \text{kPa}},\qquad \Delta\sigma_B=4qI(m,n)\Big|_{z=2.8}=\boxed{42.4\ \text{kPa}}.$$ (The influence factor $I$ is evaluated from the standard Newmark corner-of-rectangle formula.)
3. Ultimate settlement of each footing. $$S_A=m_v\Delta\sigma_A H_A=0.7\times10^{-3}\times26.8\times1.2\times1000=\boxed{22.5\ \text{mm}},$$ $$S_B=m_v\Delta\sigma_B H_B=0.7\times10^{-3}\times42.4\times2.8\times1000=\boxed{83.1\ \text{mm}}.$$
4. Part (a) — ultimate differential settlement. $$\Delta S_{ult}=S_B-S_A=83.1-22.5=\boxed{60.6\ \text{mm}}.$$
5. Part (b) — calibrate the consolidation-rate curve. The observed 10 mm at $t=2$ yr corresponds to an average degree of consolidation $U_1=10/60.6=16.5\%$, giving (from $T_v=\tfrac{\pi}{4}U^2$ for $U\le60\%$) $T_{v,1}=0.0214$. The 24 mm jamming threshold needs $U_2=24/60.6=39.6\%$, giving $T_{v,2}=0.123$.
6. Solve for the required elapsed time. Since $T_v=c_v t/d^2\propto t$ for the same soil and drainage path, $t_2=t_1\,T_{v,2}/T_{v,1}=2\times(0.123/0.0214)=\boxed{11.5\ \text{years since construction}}$, i.e. a further $t_2-2=\boxed{9.5\ \text{years (≈114 months) from today}}$.