Question 5 of 6: Rankine active pressure on a layered, partly submerged backfill

Question 5: Rankine active pressure on a layered, partly submerged backfill (Value: 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.

Find. The full Rankine active pressure distribution down the back of the wall, including the depths over which the soil goes into tension, and the total active thrust per metre run once those tension zones have cracked open.

Approach. Work down the wall in effective stress, computing the vertical effective stress at every change of layer or unit weight, converting it to a Rankine active pressure with each layer’s own $K_a$ and cohesion term, plotting the resulting diagram, and then integrating only the positive parts and adding the hydrostatic thrust below the water table.

1. Compute the two active earth-pressure coefficients and cohesion terms. For a vertical wall with a horizontal surface the Rankine coefficient is $K_a = \tan^{2}\left(45^{\circ} - \phi'/2\right)$, so $$K_a^{(1)} = \tan^{2}30^{\circ} = 0.333, \qquad K_a^{(2)} = \tan^{2}32.5^{\circ} = 0.406 .$$ The cohesion of the backfill reduces the active pressure by the constant amount $2c'\sqrt{K_a}$ throughout each layer: $$2c_1'\sqrt{K_a^{(1)}} = 2(20)(0.577) = 23.09\ \text{kPa}, \qquad 2c_2'\sqrt{K_a^{(2)}} = 2(25)(0.637) = 31.85\ \text{kPa} .$$ These are large relative to the 5 m of overburden available, which is the signal that this wall will spend much of its height in theoretical tension.
2. Build the vertical effective stress profile. The surcharge acts at the surface, layer 2 is moist down to the water table at 4 m and buoyant below it, where the effective unit weight is $\gamma_2' = \gamma_{2,sat} - \gamma_w = 20 - 9.81 = 10.19\ \text{kN/m}^{3}$. Working down, $$\sigma_v'(0) = q = 20.0\ \text{kPa},$$ $$\sigma_v'(3) = 20 + 18(3) = 74.0\ \text{kPa},$$ $$\sigma_v'(4) = 74.0 + 18.5(1) = 92.5\ \text{kPa},$$ $$\sigma_v'(5) = 92.5 + 10.19(1) = 102.7\ \text{kPa} .$$
3. Convert to active pressure at every breakpoint. Applying $\sigma_a' = \sigma_v' K_a - 2c'\sqrt{K_a}$ with each layer’s own constants, and noting that the layer boundary at 3 m carries two values because $K_a$ and $c'$ both change there while $\sigma_v'$ does not, $$\sigma_a'(0) = 20.0(0.333) - 23.09 = -16.43\ \text{kPa},$$ $$\sigma_a'(3^-) = 74.0(0.333) - 23.09 = +1.57\ \text{kPa},$$ $$\sigma_a'(3^+) = 74.0(0.406) - 31.85 = -1.82\ \text{kPa},$$ $$\sigma_a'(4) = 92.5(0.406) - 31.85 = +5.69\ \text{kPa},$$ $$\sigma_a'(5) = 102.7(0.406) - 31.85 = +9.82\ \text{kPa} .$$
4. Locate the two depths at which the pressure passes through zero. Setting $\sigma_a' = 0$ in layer 1 gives the classic tensile-crack depth, $$z_c = \frac{1}{\gamma_1}\left(\frac{2c_1'}{\sqrt{K_a^{(1)}}} - q\right) = \frac{1}{18}\left(\frac{23.09}{0.333} - 20\right) = \boxed{2.74\ \text{m}} ,$$ which falls inside layer 1 as required. Layer 2 then re-enters tension at its top because its larger cohesion term outweighs the small gain in $K_a$, and recovers at $$z = 3 + \frac{1}{\gamma_2}\left(\frac{2c_2'}{\sqrt{K_a^{(2)}}} - \sigma_v'(3)\right) = 3 + \frac{78.48 - 74.0}{18.5} = 3.24\ \text{m} .$$ The second tension zone is only 0.24 m deep, but it is real and must be shown on the diagram rather than smoothed over.
5. Add the water pressure below the water table. Below 4 m the wall carries hydrostatic pressure in addition to the effective earth pressure, $$u(5) = \gamma_w (5 - 4) = 9.81\ \text{kPa},$$ so the total lateral pressure at the base of the wall is $9.82 + 9.81 = 19.63\ \text{kPa}$ — the water contributes half of it, from only the bottom fifth of the wall.
6. Draw the distribution (part i). The diagram consists of a tension wedge from the surface to 2.74 m, a small compressive triangle from 2.74 m to the layer boundary at 3 m reaching 1.57 kPa, a sudden step down to $-1.82$ kPa at the top of layer 2, a second short tension wedge to 3.24 m, then a straight compressive line rising to 5.69 kPa at the water table and, on the flatter buoyant slope, to 9.82 kPa at the base, with the hydrostatic triangle superimposed over the bottom metre.
7. The completed pressure diagram.
   Rankine active pressure behind the wall. Dashed segments are theoretical tension and are discarded once the backfill cracks; the shaded red area is the retained effective earth pressure and the blue triangle is the hydrostatic thrust below the water table. Pressures in kPa.
8. Integrate the surviving areas (part ii). Once the backfill cracks, the tension zones transmit nothing and only the shaded areas contribute. Taking them in order down the wall, $$P_1 = \tfrac{1}{2}(3.00 - 2.74)(1.57) = 0.21\ \text{kN/m},$$ $$P_2 = \tfrac{1}{2}(4.00 - 3.24)(5.69) = 2.15\ \text{kN/m},$$ $$P_3 = \tfrac{1}{2}(5.69 + 9.82)(1.00) = 7.76\ \text{kN/m},$$ $$P_w = \tfrac{1}{2}(9.81)(1.00) = 4.91\ \text{kN/m} .$$
9. Sum to the design thrust. Adding the three soil areas and the water triangle, $$P_a = 0.21 + 2.15 + 7.76 + 4.91 = \boxed{15.0\ \text{kN/m}} .$$ Two features of that total are worth stating explicitly. The soil contributes only 10.1 kN/m and the water 4.9 kN/m, so nearly a third of the thrust on this wall comes from one metre of standing water — a direct argument for drainage behind the wall. And the resultant acts low, about 0.55 m above the base, because everything above 3.2 m has cracked away; the overturning moment is correspondingly small, but the wall must still be checked for the case in which the tension cracks fill with rainwater, which would add a further $\tfrac{1}{2}\gamma_w z_c^{2} = 36.8\ \text{kN/m}$ — more than twice the cracked thrust.