Question 4 of 6: Consolidation

Question 4: Consolidation (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. Sand 0–5 m, $\gamma_t=21\text{ kN/m}^3$ (above WT). Clay 5–8 m ($H=3$ m), $\gamma_t=19\text{ kN/m}^3$, $OCR=1.5$, $e_0=0.993$, $c_v=0.81\text{ m}^2/\text{yr}$, $C_c=0.15$, $C_s=0.02$. WT at 5 m. Part (a): 5 m×5 m footing, $q=500\text{ kPa}$. Part (b): $\Delta\sigma_{avg}=150$ kPa, $t=2$ yr.

Find. (a) Average $\Delta\sigma$ within the clay below the footing centre. (b) Settlement at $t=2$ yr.

Approach. (a) Use the 4-corner Newmark method (quarter the 5×5 footing into four 2.5×2.5 squares meeting at the centre) at the top, mid-depth and bottom of the clay, then Simpson's rule for the layer average. (b) Compare $\sigma'_{v0}$, $\sigma'_p=OCR\cdot\sigma'_{v0}$ and $\sigma'_{vf}=\sigma'_{v0}+\Delta\sigma$ to select the recompression/virgin-compression split, then scale the ultimate settlement by the degree of consolidation at $t=2$ yr.

1. Part (a) — Boussinesq stress increase at 3 depths. Below the centre of a 5×5 m footing, quartering into four 2.5×2.5 m squares gives $m=n=B/z$ per quadrant, with $\Delta\sigma=4qI_c(m,n)$ (Newmark's closed-form corner-influence factor). At the top ($z=5$ m), mid-depth ($z=6.5$ m) and base ($z=8$ m) of the clay: $$\Delta\sigma_{top}=168.05\text{ kPa},\qquad \Delta\sigma_{mid}=113.20\text{ kPa},\qquad \Delta\sigma_{bot}=80.16\text{ kPa}$$ Simpson's rule (0.25-point method) gives the layer-average increase: $$\Delta\sigma_{avg}=\frac{\Delta\sigma_{top}+4\Delta\sigma_{mid}+\Delta\sigma_{bot}}{6}=\frac{168.05+4(113.20)+80.16}{6}=\boxed{116.8\text{ kPa}}$$
2. Part (b) — effective stress history at clay mid-depth. Above the WT the sand is fully effective; below it, use the buoyant unit weight $\gamma'=19-9.81=9.19\text{ kN/m}^3$. $$\sigma_{v0}'=21(5)+9.19(1.5)=105+13.79=\boxed{118.8\text{ kPa}}$$ $$\sigma_p'=OCR\times\sigma_{v0}'=1.5(118.8)=\boxed{178.2\text{ kPa}}$$ $$\sigma_{vf}'=\sigma_{v0}'+150=118.8+150=\boxed{268.8\text{ kPa}}$$ Since $\sigma_{vf}'>\sigma_p'>\sigma_{v0}'$, the clay crosses its preconsolidation pressure — settlement is the sum of a recompression term ($\sigma_{v0}'\to\sigma_p'$) and a virgin-compression term ($\sigma_p'\to\sigma_{vf}'$).
3. Part (b) — ultimate consolidation settlement. $$S_{c,ult}=\frac{C_s}{1+e_0}H\log_{10}\!\frac{\sigma_p'}{\sigma_{v0}'}+\frac{C_c}{1+e_0}H\log_{10}\!\frac{\sigma_{vf}'}{\sigma_p'}$$ $$=\frac{0.02}{1.993}(3)\log_{10}\!\frac{178.2}{118.8}+\frac{0.15}{1.993}(3)\log_{10}\!\frac{268.8}{178.2}=0.00527+0.0403=\boxed{0.0456\text{ m (45.6 mm)}}$$
4. Part (b) — degree of consolidation and settlement at $t=2$ yr. The clay drains upward only (permeable sand above, impervious rock below), so $H_{dr}=H=3$ m. $$T_v=\frac{c_vt}{H_{dr}^2}=\frac{0.81(2)}{3^2}=\boxed{0.180}$$ With $T_v<0.287$ (i.e.\ $U<60\%$), use $T=(\pi/4)(U/100)^2$: $$U=100\sqrt{\frac{4T_v}{\pi}}=100\sqrt{\frac{4(0.180)}{\pi}}=\boxed{47.9\%}$$ $$S(t=2\text{ yr})=U\times S_{c,ult}=0.479(0.0456)=\boxed{0.0218\text{ m (21.8 mm)}}$$