16-Civ-A4 Geotechnical Materials and Analysis · May 2018

Question 5 of 6: Effective Stress under Downward and Upward Seepage

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 5: Effective Stress under Downward and Upward Seepage (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. A 4 m saturated soil column ($\gamma_{sat}=20\ \text{kN/m}^3$) with plane X-X at its mid-height (2 m below the top face, 2 m above the base). Two constant-head reservoirs feed the column; taking the datum at the base of the soil, one reservoir surface is at elevation 7 m and the other at 5 m. In situation (1) the high (7 m) head is applied at the top face and the low (5 m) head at the base → downward seepage. In situation (2) the connections are reversed — high head at the base, low head at the top → upward seepage. The head loss across the 4 m of soil is $2\ \text{m}$ in both cases, so $i=2/4=0.5$.

Find. The effective stress $\sigma'$ on plane X-X for each situation.

Approach. At X-X the total stress is the weight of soil plus the free water standing above the top face; the pore pressure comes from the total head there (which, at the mid-point of the column, is the average of the two boundary heads). Effective stress is the difference. A seepage-force check ($\sigma'=\gamma'z\pm i\gamma_w z$) confirms each result.

1. Pore pressure at X-X (same for both). The total head varies linearly through the soil, so at the mid-height X-X it equals the mean boundary head, $h_{XX}=\tfrac{1}{2}(7+5)=6\ \text{m}$. With X-X at elevation 2 m, the pressure head is $6-2=4\ \text{m}$, giving $$u = \gamma_w(h_{XX}-z_{XX}) = 9.81\times 4 = \boxed{39.2\ \text{kN/m}^2}.$$ This is identical in (1) and (2) because X-X sits exactly midway between the two heads.
2. Situation (1) — downward seepage. The top face connects to the 7 m reservoir, so free water stands $7-4=3\ \text{m}$ above the soil. Total stress at X-X: $$\sigma_{(1)} = \gamma_w(3) + \gamma_{sat}(2) = 9.81(3)+20(2) = 29.4+40 = 69.4\ \text{kN/m}^2.$$ Effective stress: $\sigma_{(1)}' = 69.4 - 39.2 = \boxed{30.2\ \text{kN/m}^2}.$
3. Situation (2) — upward seepage. Now the top face connects to the 5 m reservoir, so only $5-4=1\ \text{m}$ of free water stands above the soil. Total stress at X-X: $$\sigma_{(2)} = \gamma_w(1) + \gamma_{sat}(2) = 9.81+40 = 49.8\ \text{kN/m}^2.$$ Effective stress: $\sigma_{(2)}' = 49.8 - 39.2 = \boxed{10.6\ \text{kN/m}^2}.$
4. Seepage-force cross-check. With $\gamma'=20-9.81=10.19\ \text{kN/m}^3$ over the 2 m of soil above X-X and $i=0.5$: downward seepage adds the seepage force, $\sigma'=\gamma'z+i\gamma_w z = 10.19(2)+0.5(9.81)(2)=20.4+9.8=30.2$; upward seepage subtracts it, $\sigma'=20.4-9.8=10.6$. Both agree exactly with the values above.