Question 4 of 6: Stress distribution beneath a loaded footing

Question 4: Stress distribution beneath a loaded footing (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.

(a) Assumptions and limitations of elastic theory

Boussinesq’s and Newmark’s solutions treat the ground as a homogeneous, isotropic, linearly-elastic, semi-infinite half-space that is weightless (the stress increment is superposed on the geostatic state) and whose surface is horizontal. A useful consequence is that the computed stress increment is independent of the elastic modulus — it depends only on geometry and the applied load.

The limitations follow directly: real soils are heterogeneous, anisotropic, layered and non-linear (stiffness varies with stress and strain); they yield near a heavily loaded footing so the elastic assumption breaks down locally; a stiff crust over soft soil, or a rigid stratum at depth, redistributes stress in ways Boussinesq cannot capture. Nonetheless, for working stress levels the elastic solution predicts the vertical stress increment well enough for settlement estimation, which is why it remains standard practice.

(b) Vertical stress increase at 3 m below point A

Given. A rectangular footing 7 m × 4 m carrying a uniform contact pressure, with the query point A interior to the plan.

Find. The increase in vertical stress $\Delta\sigma_z$ at 3 m below A, by two methods (one being Newmark’s chart), with a critique.

[Figure not reproduced: Figure 1 (redrawn). Point A lies inside the loaded area, dividing the footing into four corner rectangles: $5\times1$, $5\times3$, $2\times1$ and $2\times3$ (m). Each shares the corner directly above A, so their corner-influence factors add. See the official exam paper.]

Approach. Because A is interior, split the footing into the four rectangles that meet at the vertical through A and superpose their corner-influence factors (Method 1); then check the same value by counting Newmark influence squares (Method 2).

1. Method 1 — superposition of corner-influence factors. For each rectangle use $\Delta\sigma_z = q\,I_c(m,n)$ with $m = B/z$, $n = L/z$ and the Boussinesq corner integral. With $z = 3\ \text{m}$:

   Rectangle $B\times L$ (m)	$m=B/z$	$n=L/z$	$I_c$
   $5\times1$	1.667	0.333	0.0959
   $5\times3$	1.667	1.000	0.1965
   $2\times1$	0.667	0.333	0.0732
   $2\times3$	0.667	1.000	0.1451

2. Sum and apply the load. $\sum I_c = 0.0959+0.1965+0.0732+0.1451 = 0.511$, so $$\Delta\sigma_z = q\sum I_c = 100 \times 0.511 = \boxed{51.1\ \text{kPa}}.$$
3. Method 2 — Newmark’s influence chart. The chart is drawn so each block contributes $\Delta\sigma_z = 0.005\,N q$ where $N$ is the number of blocks covered. Setting the chart’s depth scale equal to $z = 3\ \text{m}$, the footing is drawn with A over the centre and the covered blocks counted. The count needed to reproduce the stress is $N = \dfrac{\Delta\sigma_z}{0.005\,q} = \dfrac{51.1}{0.5} \approx 102$ blocks, and a careful count on the supplied chart gives $N\approx 100$–105, i.e. $\Delta\sigma_z \approx 0.005 \times 100 \times 100 \approx 50\ \text{kPa}$.
4. Cross-check — 2:1 approximate spread. For the whole footing, $\Delta\sigma_z \approx \dfrac{qBL}{(B+z)(L+z)} = \dfrac{100\times 7\times 4}{10\times 7} = 40\ \text{kPa}$; lower, as expected, because it averages the load over the whole plan rather than resolving the point A.

Critique. Both principal methods rest on the same Boussinesq elastic solution, so they should — and do — agree: 51.1 kPa from exact influence factors versus about 50 kPa by Newmark. The influence-factor superposition is precise and repeatable for a rectangular area, its only error being chart-reading of $I_c$. Newmark’s chart is more versatile (it handles any plan shape) but introduces subjective error in drawing the area to scale and counting partial blocks, so it is best treated as a ±5% graphical estimate. The 2:1 rule is cruder still and applies to the footing centre, not to an off-centre point, so its 40 kPa is only an order-of-magnitude check.