Question 4 of 6: Elastic Stress Distribution and Stress Increase under a Footing

Question 4: Elastic Stress Distribution and Stress Increase under a Footing (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 (5 marks)

Given. The Boussinesq family of solutions — point load, line load, rectangular and circular loaded areas — as used routinely to compute stress increase in soil.

Find. The assumptions those solutions rest on, the limitations that follow, and sketches of $\sigma_z$ against depth and against horizontal distance under a point load.

Boussinesq’s solution treats the ground as a homogeneous, isotropic, linearly elastic, weightless semi-infinite half-space loaded at its horizontal surface. Spelled out, the assumptions are:

1. Homogeneous and isotropic. One set of elastic constants everywhere, identical in every direction. Real deposits are layered and, being sedimented, are markedly stiffer horizontally than vertically.
2. Linearly elastic, with no yielding. Stress is proportional to strain at every point, and strains are fully recoverable. Soil is neither linear nor elastic: its stiffness falls with strain level and rises with confining stress, and once a zone under a footing reaches failure the stresses redistribute in a way elasticity cannot follow.
3. Semi-infinite extent. The half-space is unbounded downwards and sideways. A stiff stratum or bedrock at shallow depth concentrates stress above it and the true $\sigma_z$ can exceed the Boussinesq value by 50 % or more; conversely a soft layer over a stiff one attracts less.
4. Weightless medium. The solution gives only the increment caused by the applied load; the in-situ overburden $\sigma_{v0}' = \gamma z$ must be added separately.
5. Surface loading, flexible and uniform. The load is applied at $z = 0$ as a uniform pressure over a perfectly flexible area. A load applied at depth $D_f$ is less severe than the surface solution suggests, and a genuinely rigid footing redistributes contact pressure towards its edges on clay (and towards its centre on sand) instead of remaining uniform.
6. Full contact, no interface slip. No separation or sliding is permitted between the loaded area and the soil.

The redeeming feature, and the reason the method survives all of this, is that for the vertical stress the solution turns out to be independent of both $E$ and Poisson’s ratio $\nu$ — it depends only on geometry. Since $E$ is the parameter hardest to measure reliably, removing it removes the largest source of error, and comparisons with instrumented field measurements show $\sigma_z$ predicted to within roughly ±20 % on reasonably uniform ground. Horizontal and shear stresses, which do depend on $\nu$, are far less reliable.

Panel (a) shows the vertical stress on the load axis. Directly beneath a point load $\sigma_z = 0.4775\,Q/z^2$, so the stress is theoretically infinite at the surface — the well-known singularity that makes the point-load solution useless in the first fraction of a metre — and decays with the inverse square of depth. Panel (b) shows the same stress plotted horizontally at three increasing depths: each curve is a bell whose peak lies on the axis and whose peak value drops as $1/z^2$, while its width grows in proportion to $z$. The volume under every one of these surfaces is the same, and equal to $Q$: the load is neither created nor destroyed, only spread. It is this spreading that produces the familiar pressure bulb and that explains part (iii) of Question 1 — a wide load spreads slowly and reaches deep, a narrow one spreads fast and dies out quickly.

(b) Stress increase 3 m below point A (15 marks)

Given. The rectangular footing of Figure 1, uniformly loaded, with point A lying inside the loaded plan.

Find. $\Delta\sigma_z$ at 3 m below A, by two independent methods one of which must be Newmark’s chart, with a critique of both.

[Figure not reproduced: Figure 1 (redrawn) — the 7 m × 4 m footing split at point A into the four rectangles I, II, III and IV that share A as a corner. Influence factors for the four add to give the stress below A. See the official exam paper.]

Approach. Point A lies inside the loaded area, so the influence-chart solution — which is written for the corner of a rectangle — is applied by splitting the plan into the four rectangles that all share A as a corner and adding their influence factors. The Newmark chart is then used as a wholly independent second route, and the crude $2{:}1$ spread as a third sanity check.

Method 1 — superposition of corner rectangles (Fadum influence chart, page 7)

1. Split the plan at A. The two dimension chains put A at 5 m from the left edge and 1 m from the top edge, so the footing divides into I (5 m × 1 m), II (2 m × 1 m), III (5 m × 3 m) and IV (2 m × 3 m). Every one of these has A as a corner and none overlaps another, so the four contributions add: $$\Delta\sigma_z = q\,(I_{\text{I}} + I_{\text{II}} + I_{\text{III}} + I_{\text{IV}})$$ Checking the arithmetic of the split before going further, $5 + 2 = 7$ m and $1 + 3 = 4$ m, which reproduces the overall plan.
2. Form $m$ and $n$ for each rectangle. With $m = x/z$ and $n = y/z$ (interchangeable, as the chart states) and $z = 3$ m throughout, rectangle I gives $m = 5/3 = 1.667$ and $n = 1/3 = 0.333$, and so on for the rest.
3. Read the influence factor for each. Entering the page-7 chart with each pair, or evaluating the closed form the chart plots, $$I = \frac{1}{4\pi}\!\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot\frac{m^2+n^2+2}{m^2+n^2+1} + \tan^{-1}\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right]$$ gives the four values tabulated below.

   Corner influence factors at $z = 3$ m

   Rectangle	$x \times y$ (m)	$m$	$n$	$I$
   I	5 × 1	1.667	0.333	0.0959
   II	2 × 1	0.667	0.333	0.0732
   III	5 × 3	1.667	1.000	0.1965
   IV	2 × 3	0.667	1.000	0.1451
   Sum $\sum I$				0.5107

   Each factor is below the limiting value of 0.25 that a corner factor approaches for an infinitely large loaded area, as it must be.
4. Combine. Multiplying the summed factor by the contact pressure, $$\Delta\sigma_z = q \sum I = 100 \times 0.5107 = \boxed{51.1\ \text{kPa}}$$ This is 51 % of the applied pressure still present at a depth of three quarters of the footing width — a reminder of how slowly stress dissipates under a load 7 m long.

Method 2 — Newmark’s influence chart (page 8)

1. Set the drawing scale from the depth. Newmark’s chart is used by drawing the loaded plan on tracing paper at the scale on which the printed depth-scale line represents the depth $z$. Here $z = 3$ m, so the depth-scale line on page 8 is taken to be 3 m and the 7 m × 4 m plan is drawn to that same scale.
2. Place the point of interest at the chart centre. The tracing is laid over the chart with A over the common centre of the concentric circles, and in its true plan orientation. The 5 m, 2 m, 1 m and 3 m offsets place the outline asymmetrically about the centre, which is exactly what the chart is designed to handle.
3. Count the influence elements covered. The number of chart elements (whole plus estimated fractions) falling inside the outline is counted; here $$N \approx 102\ \text{elements}$$ out of the 200 the chart contains, which is consistent with a plan that covers about half the chart’s effective area at this scale.
4. Apply the chart equation. Using the value printed on the chart and repeated on the formula sheet, $\sigma_z = 0.005\,N\,q$: $$\Delta\sigma_z = 0.005 \times 102 \times 100 = \boxed{51.0\ \text{kPa}}$$
5. Verify the element count rather than trusting the eye. Counting elements by hand is the weakest step in the whole method, so it should be checked against the influence factors already computed. Since each element is worth $I_N = 0.005$, the count implied by Method 1 is $$N = \frac{\sum I}{I_N} = \frac{0.5107}{0.005} = 102.1\ \text{elements}$$ which the physical count of 102 reproduces to within one element — well inside the accuracy the chart can be read to.

Third check — the approximate 2:1 method

The formula sheet also gives the $2{:}1$ spread, $\sigma_z = qBL/[(B+z)(L+z)]$, which for this footing at $z = 3$ m returns

$$\Delta\sigma_z = \frac{100 \times 4 \times 7}{(4+3)(7+3)} = \frac{2800}{70} = 40.0\ \text{kPa}$$

This is deliberately included as a contrast rather than as a competing answer: the $2{:}1$ method returns a single average stress spread uniformly over the enlarged area, with no ability to distinguish one point in plan from another. It cannot answer the question as asked.

Critique of the two methods

The two required methods agree to 0.2 %, which is much closer than either deserves credit for, and the agreement is not a coincidence: both are the same Boussinesq integral, one evaluated in closed form over four rectangles and the other evaluated graphically by counting equal-influence areas. They therefore share every assumption listed in part (a), and agreement between them proves only that the arithmetic and the tracing are right — it says nothing about whether elastic theory suits the ground.

Where they differ is in precision and in reach. The rectangle method is exact once $m$ and $n$ are formed, so its only error is chart-reading, of order 1 % to 2 % when read by eye and nil when the closed form is used; but it works only for rectangles, and only for points that can be reached as the corner of a rectangle or a signed combination of them. Newmark’s chart handles any plan shape whatsoever — an L-shaped raft, a curved embankment toe, an irregular tank pad — and any point, inside or outside the load, which is its real justification. Its cost is a counting error of roughly ±2 to ±5 elements on a hand-traced outline, that is ±1 to ±2.5 kPa here, or up to 5 %; the count is also acutely sensitive to drawing the plan at the wrong scale, since the scale is set by $z$ and a plan drawn for the wrong depth is simply the wrong problem. In practice the rectangle method should be used wherever the shape permits and Newmark reserved for shapes it cannot handle — and where both are available, as here, the rectangle result should be used to audit the element count, as was done in step 5.

Against these, the $2{:}1$ method is 22 % low and, more seriously, wrong in kind. It is a load-spreading rule of thumb with no point-by-point resolution at all, and it is quoted here only to show what is lost by using it. For settlement work under a real footing on Canadian ground, CFEM would in any case direct the designer to check whether a stiff layer or bedrock lies within about twice the footing width, since that condition breaks the semi-infinite assumption underlying all three methods and can raise the true $\Delta\sigma_z$ well above every value in the table.