Question 8 of 10: Stress increase beneath an embankment at points A, B and C

Question 8: Stress increase beneath an embankment at points A, B and C (24 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 symmetrical embankment 10 m high with a 5 m wide crest and side slopes of 1V:2H, built of fill of unit weight 17 kN/m³ on a silty clay. Point A lies on the centre line, point B beneath the left crest break, and point C beneath the left toe, all at 4 m depth.

Find. The vertical stress increase $\Delta\sigma_z$ at A, B and C.

Approach. The embankment is long compared with its cross-section, so the problem is one of plane strain and Osterberg's influence factor for an embankment loading applies. Each point is treated by superposing one or two Osterberg wedges, each consisting of a strip of full height $B_1$ adjacent to the vertical through the point followed by a ramp of horizontal extent $B_2$ falling to zero.

1. Establish the applied pressure and the geometry. The bearing pressure under the full height of the embankment is $$q_0 = \gamma H = 17(10) = 170\ \text{kPa}$$ With side slopes of 1V:2H and a height of 10 m, each slope runs $2 \times 10 = 20\ \text{m}$ horizontally, so the base width is $5 + 2(20) = 45\ \text{m}$. Measuring offsets from the centre line, the crest breaks are at $\pm 2.5\ \text{m}$ and the toes at $\pm 22.5\ \text{m}$; A, B and C are at offsets 0, 2.5 and 22.5 m respectively.
2. State the influence factor. Osterberg (1957), reproduced as Fig. 6.24 of Das, gives the vertical stress beneath the vertical edge of an embankment-shaped strip load as $\Delta\sigma_z = q_0 I'$, with $$I' = \frac{1}{\pi}\left[\frac{B_1 + B_2}{B_2}\left(\alpha_1 + \alpha_2\right) - \frac{B_1}{B_2}\,\alpha_2\right]$$ $$\alpha_2 = \tan^{-1}\!\frac{B_1}{z}, \qquad \alpha_1 = \tan^{-1}\!\frac{B_1 + B_2}{z} - \tan^{-1}\!\frac{B_1}{z}$$ with the angles in radians. Since $\alpha_1 + \alpha_2 = \tan^{-1}\left[(B_1+B_2)/z\right]$ this collapses to the compact form used below, $$I' = \frac{1}{\pi}\left[\frac{B_1 + B_2}{B_2}\tan^{-1}\!\frac{B_1 + B_2}{z} - \frac{B_1}{B_2}\tan^{-1}\!\frac{B_1}{z}\right]$$ Evaluating the closed form rather than reading the printed chart removes the chart-scaling error, which on the flat part of Osterberg's curves can reach several per cent.
3. Point A, on the centre line. By symmetry the embankment splits into two identical halves at A, each consisting of a full-height strip $B_1 = 2.5\ \text{m}$ (half the crest) followed by a ramp $B_2 = 20\ \text{m}$. With $z = 4\ \text{m}$, $$I'_A = \frac{1}{\pi}\left[\frac{22.5}{20}\tan^{-1}\!\frac{22.5}{4} - \frac{2.5}{20}\tan^{-1}\!\frac{2.5}{4}\right] = \frac{1}{\pi}\left[1.125(1.3934) - 0.125(0.5586)\right] = 0.4773$$ and, adding the two halves, $$\boxed{\Delta\sigma_A = 2 q_0 I'_A = 2(170)(0.4773) = 162.3\ \text{kPa}}$$ This is 95 per cent of the applied 170 kPa, as expected: at only 4 m below a 45 m wide load the point is effectively under a semi-infinite pressure.
4. Point B, beneath the crest break. B is not on an axis of symmetry, so the two sides are treated separately. Looking to the right of B the load is full height for the whole 5 m crest and then ramps down over 20 m, giving $B_1 = 5\ \text{m}$, $B_2 = 20\ \text{m}$; looking to the left the ramp begins immediately at B, giving $B_1 = 0$, $B_2 = 20\ \text{m}$. Hence $$I'_{B,right} = \frac{1}{\pi}\left[\frac{25}{20}\tan^{-1}\!\frac{25}{4} - \frac{5}{20}\tan^{-1}\!\frac{5}{4}\right] = 0.4906$$ $$I'_{B,left} = \frac{1}{\pi}\left[\frac{20}{20}\tan^{-1}\!\frac{20}{4} - 0\right] = \frac{1}{\pi}(1.3734) = 0.4372$$ Adding the two contributions, $$\boxed{\Delta\sigma_B = q_0\left(I'_{B,right} + I'_{B,left}\right) = 170(0.4906 + 0.4372) = 157.7\ \text{kPa}}$$
5. Point C, beneath the toe. The whole embankment lies to one side of C, and the load first rises away from C rather than falling, so the profile is obtained as the difference of two Osterberg wedges anchored at C. A wedge with $B_1 = 25\ \text{m}$, $B_2 = 20\ \text{m}$ places full pressure from C out to 25 m and then ramps to zero at 45 m; subtracting a wedge with $B_1 = 0$, $B_2 = 20\ \text{m}$ removes exactly the block that the real embankment does not have over the first 20 m and leaves the rising ramp. Therefore $$I'_{C} = \frac{1}{\pi}\left[\frac{45}{20}\tan^{-1}\!\frac{45}{4} - \frac{25}{20}\tan^{-1}\!\frac{25}{4}\right] - 0.4372 = 0.4996 - 0.4372 = 0.0625$$ $$\boxed{\Delta\sigma_C = q_0 I'_{C} = 170(0.0625) = 10.6\ \text{kPa}}$$
6. Check the results against physical expectation. Directly beneath the toe of a wide load the classical result is that the stress increase tends to half the applied pressure at shallow depth if the load were a full-height strip ending at that point; here the load does not end abruptly but tapers to zero over 20 m, so only about 6 per cent of $q_0$ is felt, which is consistent. The three values also decay in the right order and by the right amounts: A and B differ by less than 3 per cent because both sit under essentially the full crest at only 4 m depth, whereas C, 22.5 m away, feels almost nothing. An independent numerical integration of the Boussinesq plane-strain line-load solution $\Delta\sigma_z = \int \dfrac{2 p(x) z^3}{\pi\left[(x - x_0)^2 + z^2\right]^2}\,dx$ over the actual trapezoidal pressure distribution reproduces 162.3, 157.7 and 10.6 kPa exactly, confirming both the influence-factor algebra and the superposition used at C.

For a settlement calculation these values would be the stress increments at mid-depth of the first compressible sublayer; the near-uniformity of the increment between A and B is what makes an embankment of this width settle almost uniformly across its crest, while the sharp fall towards the toe is what produces the differential settlement and the longitudinal cracking so often seen at the shoulder of a high fill.