Shear strength of sand, WT rises

Assumptions: $\gamma_w = 9.81$ kN/m³; sand above the water table is dry ($\gamma = \gamma_d = 17$); $c = 0$.

Void ratio

$$\gamma_d = \frac{G_s\gamma_w}{1+e} \;\Rightarrow\; e = \frac{2.7\times9.81}{17} - 1 = 0.558$$ $$\gamma_{sat} = \frac{(G_s+e)\gamma_w}{1+e} = \frac{(2.7+0.558)(9.81)}{1.558} = 20.51\ \text{kN/m}^3$$

Case 1: Water table at 2.5 m

$$\begin{aligned} \sigma' &= 17(2.5) + (20.51 - 9.81)(1.5) \\ &= 42.5 + 16.05 = 58.6\ \text{kPa} \end{aligned}$$ $$s = \sigma'\tan\phi = 58.6\tan35^\circ = 41.0\ \text{kPa}$$

Case 2: Water table at the surface

$$\sigma' = (20.51-9.81)(4) = 42.8\ \text{kPa}, \qquad s = 42.8\tan35^\circ = 30.0\ \text{kPa}$$

Change in shear strength: $30.0 - 41.0 = -11.0$ kPa.

Answer: shear strength is 41.0 kPa with the water table at 2.5 m; it falls by about 11.0 kPa (27%) to 30.0 kPa when the water table rises to the surface.