Question 2 of 5: Stress Increase under an Embankment and Plate-Load-Test Scaling

Question 2: Stress Increase under an Embankment and Plate-Load-Test Scaling (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.

Part (a) — stress increase beneath the embankment

Given.

Find. The vertical stress increase $\Delta\sigma$ produced by the embankment fill at $A_1$ and at $A_2$.

[Figure not reproduced: Embankment cross-section as dimensioned on the exam paper. The lower chain 5 + 11.5 + 16.5 = 33 m closes against the upper chain 14 + 5 + 14 = 33 m, which places A₂ 5 m inside the left toe and A₁ on the centreline. See the official exam paper.]

Approach. Treat the fill as a flexible embankment load on an elastic half-space and use Osterberg's influence factor, which gives the vertical stress under the vertical face of a load consisting of a slope of horizontal projection $B_1$ rising to height $H$ followed by a level crest of width $B_2$. Any embankment can be built up from such units by superposition, adding for material present and subtracting for material absent.

1. State Osterberg's influence factor. For the standard unit $$\Delta\sigma = I\,q_0, \qquad q_0 = \gamma H$$ $$I = \frac{1}{\pi}\left[\frac{B_1 + B_2}{B_1}\left(\alpha_1 + \alpha_2\right) - \frac{B_2}{B_1}\,\alpha_2\right]$$ with $\alpha_2 = \tan^{-1}(B_2/z)$ and $\alpha_1 = \tan^{-1}\!\left[(B_1 + B_2)/z\right] - \alpha_2$ in radians. This is the equation plotted as Osterberg's chart in Das Fig. 5.x; reading the chart to two decimal places gives the same answers as evaluating it directly, and the direct evaluation is used here so the arithmetic is reproducible. For the full-height fill, $q_0 = 17.5 \times 7 = 122.5\ \text{kN/m}^2$.
2. Point $A_1$ — exploit symmetry. $A_1$ sits 16.5 m from each toe, exactly on the centreline of a 33 m base, so the embankment splits into two identical half-embankments each having its vertical face on the centreline. Each half has $B_1 = 14$ m (the slope) and $B_2 = 2.5$ m (half the crest), and $z = 5$ m, so $B_1/z = 2.8$ and $B_2/z = 0.5$.
3. Evaluate the influence factor for one half. $$\alpha_2 = \tan^{-1}(0.5) = 0.46365\ \text{rad}, \qquad \alpha_1 = \tan^{-1}(3.3) - 0.46365 = 1.27637 - 0.46365 = 0.81272\ \text{rad}$$ $$I = \frac{1}{\pi}\left[\frac{16.5}{14}(1.27637) - \frac{2.5}{14}(0.46365)\right] = \frac{1.42150}{\pi} = 0.4525$$
4. Double it for the two halves. Both halves load $A_1$ equally, so $$\Delta\sigma_{A_1} = 2 I q_0 = 2(0.4525)(122.5) = \boxed{110.9\ \text{kN/m}^2}$$ The point receives about 90 % of the full crest pressure, which is reasonable: at only 5 m depth beneath a 33 m-wide fill the loading is effectively one-dimensional.
5. Point $A_2$ — find the local fill height first. $A_2$ is 5 m inside the left toe, on the sloping face. The slope rises 7 m over 14 m, so directly above $A_2$ the fill is $$h_{A_2} = 7 \times \frac{5}{14} = 2.5\ \text{m}$$ Split the whole embankment on the vertical line through $A_2$ and treat the two sides separately.
6. Left of the split: a triangular wedge. The fill from the toe to $A_2$ is a triangle 5 m wide rising to 2.5 m, i.e. a standard unit with $B_1 = 5$ m, $B_2 = 0$ and $q_0 = 17.5(2.5) = 43.75\ \text{kN/m}^2$. With $B_2 = 0$ the influence factor collapses to $I = \tan^{-1}(B_1/z)/\pi$: $$I_L = \frac{\tan^{-1}(1.0)}{\pi} = 0.2500, \qquad \Delta\sigma_L = 43.75 \times 0.2500 = 10.94\ \text{kN/m}^2$$
7. Right of the split: add a full block, subtract the missing wedge. The fill to the right of $A_2$ is not a standard unit, so build it as a difference. First add the unit that has a vertical face 7 m high at $A_2$, a crest running 14 m to the far shoulder and a 14 m slope beyond it ($B_1 = 14$, $B_2 = 14$, $q_0 = 122.5$): $$\alpha_2 = \tan^{-1}(2.8) = 1.22777, \quad \alpha_1 = \tan^{-1}(5.6) - 1.22777 = 0.16618$$ $$I_R = \frac{1}{\pi}\left[\frac{28}{14}(1.39395) - (1.22777)\right] = 0.4966, \qquad \Delta\sigma_R = 122.5 \times 0.4966 = 60.84\ \text{kN/m}^2$$
8. Subtract the wedge that is not really there. The block just added assumes full height from $A_2$ to the shoulder, whereas the real slope only reaches 7 m at 14 m from the toe. The surplus is a triangle with a 4.5 m vertical face at $A_2$ falling to zero 9 m to the right, i.e. $B_1 = 9$, $B_2 = 0$, $q_0 = 17.5(4.5) = 78.75\ \text{kN/m}^2$: $$I_W = \frac{\tan^{-1}(1.8)}{\pi} = 0.3386, \qquad \Delta\sigma_W = 78.75 \times 0.3386 = 26.66\ \text{kN/m}^2$$
9. Superpose the three contributions. $$\Delta\sigma_{A_2} = \Delta\sigma_L + \Delta\sigma_R - \Delta\sigma_W = 10.94 + 60.84 - 26.66 = \boxed{45.1\ \text{kN/m}^2}$$ The result passes a sanity check: the fill standing directly over $A_2$ weighs only 43.75 kN/m², and the large mass of fill to the right adds a little more, so a value just above the local overburden is exactly what the geometry demands. It is also far below the 110.9 kN/m² at the centreline, which is why differential settlement between crest and toe drives embankment design.

Part (b) — sizing a footing from a plate load test

Given.

Find. The side length $B_F$ of the square column footing.

[Figure not reproduced: Plate load–settlement curve read from the exam figure, with the adopted trial marked. Readings used: 150→4.3, 250→7.3, 300→9.0, 400→13.2, 500→19.0 (kN/m² → mm). See the official exam paper.]

Approach. In sand, settlement scales with footing width through Terzaghi and Peck's empirical relation, so the problem is a one-unknown iteration: guess $B_F$, get the bearing pressure it implies, read the plate settlement at that pressure, scale it up, and compare with 25 mm.

1. State the scaling law. For a footing on sand carrying the same bearing pressure as the test plate, $$S_F = S_P\left[\frac{2B_F}{B_F + B_P}\right]^2$$ The bracket approaches 2 for a large footing, so a wide footing settles up to about four times as much as the plate at the same pressure — the scale effect that makes a raw plate result unusable without correction.
2. Note that the pressure depends on the answer. The footing must carry 2500 kN, so the bearing pressure it applies is $$q = \frac{Q}{B_F^2} = \frac{2500}{B_F^2}$$ A larger footing settles more per unit pressure but applies less pressure, so the two effects compete and the equation must be solved by trial.
3. Trial 1: $B_F = 3.0$ m. Then $q = 2500/3.0^2 = 278\ \text{kN/m}^2$, and the plate curve gives $S_P \approx 8.2$ mm at that pressure. Scaling, $$S_F = 8.2\left[\frac{2(3.0)}{3.0 + 0.305}\right]^2 = 8.2 \times 3.296 = 27.0\ \text{mm}$$ which exceeds the 25 mm limit, so the footing is too small.
4. Trial 2: $B_F = 3.1$ m. Now $q = 2500/3.1^2 = 260\ \text{kN/m}^2$ and $S_P \approx 7.6$ mm, giving $$S_F = 7.6 \times \left[\frac{6.2}{3.405}\right]^2 = 7.6 \times 3.316 = 25.3\ \text{mm}$$ essentially at the limit. Interpolating between the two trials, the required width is $$B_F \approx \boxed{3.1\ \text{m}}$$
5. Trial 3: confirm the practical size. Rounding up to a constructible 3.2 m gives $q = 2500/3.2^2 = 244\ \text{kN/m}^2$, $S_P \approx 7.1$ mm and $$S_F = 7.1 \times \left[\frac{6.4}{3.505}\right]^2 = 7.1 \times 3.334 = 23.7\ \text{mm} < 25\ \text{mm}$$ so a 3.2 m × 3.2 m footing satisfies the settlement limit with a small margin and is the size to specify.
6. Check that bearing capacity is not the binding constraint. At 244 kN/m² the plate is on the near-linear part of its curve and its ultimate pressure is roughly 700 kN/m², so the working pressure carries a margin of about three against plate failure — and the prototype footing, being 10 times wider, is stronger still in a frictional soil. Settlement governs, as it almost always does for footings on sand.