Rankine Active Force on a Retaining Wall — Two-Layer Sand with Groundwater and Surcharge: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2013, Q5

Question 5: Rankine Active Force on a Retaining Wall — Two-Layer Sand with Groundwater and Surcharge (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.

[Figure not reproduced: Figure 4 — retaining wall with surcharge, two-layer sand backfill and groundwater table (as printed). See the official exam paper or the cited reference text.]

Figure 4 — retaining wall with surcharge, two-layer sand backfill and groundwater table (as printed).

Given.

Given data
Quantity	Symbol	Value
Total wall height	$H$	6 m
Depth to groundwater table	$H_1$	2 m
Unit weight, sand above water table	$\gamma_1$	16 kN/m³
Saturated unit weight, sand below water table	$\gamma_2$	19 kN/m³
Friction angle, upper sand	$\phi_1$	32° ($c_1=0$)
Friction angle, lower (saturated) sand	$\phi_2$	36° ($c_2=0$)
Surface surcharge	$q$	15 kN/m²

Find. (a) total Rankine active force per unit length of wall, $P_a$; (b) height of the resultant above the wall base.

Approach. Compute Rankine active-pressure coefficients for each layer, then build the EFFECTIVE lateral-pressure diagram in three pieces — the upper (moist) layer, the lower (submerged) layer using the layer's BUOYANT unit weight, and a separate hydrostatic water-pressure triangle below the water table — noting that the coefficient jumps at the layer interface because $\phi_1\ne\phi_2$, even though effective vertical stress is continuous there. Sum forces and take moments about the wall base to locate the resultant.

Check: the source prints the surcharge $q$ with units "kN/m³" — a surface surcharge is a pressure (force per unit AREA), so this is treated as $q=15$ kN/m² (kPa), the physically consistent reading, throughout.

Compute the active earth-pressure coefficients. $$\begin{aligned} K_{a1}&=\tan^2\!\left(45-\frac{\phi_1}{2}\right)=\tan^2(29^\circ)=\boxed{0.307} \\ K_{a2}&=\tan^2\!\left(45-\frac{\phi_2}{2}\right)=\tan^2(27^\circ)=\boxed{0.260} \end{aligned}$$
Build the effective vertical stress profile. Using the lower layer's buoyant unit weight $\gamma_2'=\gamma_2-\gamma_w=19-9.81=9.19$ kN/m³ below the water table, $$\sigma_v'(0)=q=15,\quad \sigma_v'(H_1)=q+\gamma_1H_1=15+32=47,\quad \sigma_v'(H)=47+\gamma_2'(H-H_1)=47+9.19(4)=83.8\ \text{kPa}.$$
Convert to lateral effective pressure, layer by layer. In the upper layer ($K_{a1}$): $\sigma_h'$ runs from $K_{a1}q=4.61$ kPa at the surface to $K_{a1}(47)=14.44$ kPa at the water table. Immediately below the interface, the SAME vertical stress (47 kPa) is now multiplied by $K_{a2}$: $\sigma_h'=K_{a2}(47)=12.20$ kPa, rising to $K_{a2}(83.8)=21.75$ kPa at the base — the visible "step" in the pressure diagram at $z=H_1$ is exactly $(K_{a1}-K_{a2})\times47=2.24$ kPa, caused entirely by the change in friction angle.
Add the hydrostatic water-pressure triangle below the water table. $$u(H)=\gamma_w(H-H_1)=9.81(4)=39.24\ \text{kPa (triangular, zero at the water table)}.$$
Sum each piece's area (force per unit length) and part (a)'s total. Splitting each trapezoid/triangle in the usual way (area $=\tfrac12(\text{top}+\text{bottom})\times\text{height}$): $$P_1=19.05\ \text{kN/m (upper sand)},\quad P_2=67.89\ \text{kN/m (lower sand, effective)},\quad P_w=78.48\ \text{kN/m (water)},$$ $$P_a=P_1+P_2+P_w=\boxed{165.4\ \text{kN/m}}.$$ Water contributes nearly half the total thrust — a direct consequence of the shallow (2 m) water table.
Part (b) — locate the resultant by taking moments about the wall base. Each piece's centroid height above the base follows from the standard trapezoid/triangle centroid formulas ($\bar y=5.17$ m for $P_1$, $2.19$ m for $P_2$, $1.33$ m for $P_w$, all measured from the base). Then $$\bar y=\frac{P_1(5.17)+P_2(2.19)+P_w(1.33)}{P_a}=\boxed{2.13\ \text{m above the base}}.$$

Quantity	Value
$K_{a1}$ / $K_{a2}$	0.307 / 0.260
Upper-sand force, $P_1$	19.05 kN/m
Lower-sand (effective) force, $P_2$	67.89 kN/m
Water-pressure force, $P_w$	78.48 kN/m
Total active force, $P_a$	165.4 kN/m
Height of resultant above base	2.13 m

Concept & Theory

Topic. Rankine active earth pressure in a layered, submerged backfill with a surface surcharge.

Key relations. $K_a=\tan^2(45-\phi/2)$; $\sigma_h'=K_a\sigma_v'$ (effective stress); total lateral pressure below the water table is effective lateral pressure PLUS hydrostatic pore pressure, added separately because water pressure is isotropic (not scaled by $K_a$).

Why this works. Rankine theory assumes a frictionless, vertical wall and a horizontal backfill surface in a state of plastic equilibrium; each layer's own $\phi$ sets its own $K_a$, and effective (not total) vertical stress is what drives the horizontal effective stress — water pressure must be found independently and added afterward, never multiplied by $K_a$.

Common pitfall. Applying a single $K_a$ (usually the upper layer's) through the full depth, which misses the pressure discontinuity at the layer interface; or multiplying the water pressure by $K_a$ as though it were soil, which understates the true thrust because water offers no shear resistance at all.

Source. Das, Principles of Geotechnical Engineering, ch. 13 (lateral earth pressure); Craig's Soil Mechanics, ch. 6.

Question 5 of 6: Rankine Active Force on a Retaining Wall — Two-Layer Sand with Groundwater and Surcharge

Question 5: Rankine Active Force on a Retaining Wall — Two-Layer Sand with Groundwater and Surcharge (20 marks)