Question 4 of 6: Vertical Stress Beneath a Ring Footing by Two Methods

Question 4: Vertical Stress Beneath a Ring Footing by Two Methods (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. A square ring footing in plan, loaded only over the hatched annulus, with the stress point directly beneath one outer corner.

Find. The increase in vertical stress $\Delta\sigma_z$ at 2.0 m below A, first by counting elements on the Newmark chart supplied with the paper and then by an independent analytical method, followed by a comment on the agreement between the two.

Approach. Both methods rest on Boussinesq’s elastic solution and on the fact that stresses superpose: the ring is treated as the full 6 m × 6 m square, which conveniently has a corner at A, minus the central 3 m × 3 m opening, which does not, so the opening is itself built from four rectangles that do have corners at A. The Newmark chart evaluates the same integral graphically, and the element count is audited against the analytical influence value rather than trusted on its own.

1. Set the Newmark chart to the depth of interest. A Newmark chart is drawn so that a single stated length — the depth scale printed beneath it — represents the depth $z$ at which the stress is wanted. Here $z = 2.0\ \text{m}$, so the plan of the footing is redrawn at a scale on which the depth-scale line measures 2.0 m, and the plan is then laid over the chart with point A on the centre of the chart. The chart supplied on page 9 has ten rings and twenty radial lines, giving 200 elements and an influence value per element of $$I_N = 0.005 = \frac{1}{200}$$ The governing relation, also printed on the formula sheet, is $$\Delta\sigma_z = I_N \, N q = 0.005\,N q$$ where $N$ is the number of chart elements covered by the loaded (hatched) area only.
2. Count the elements covered by the ring. With A at the chart centre the loaded ring occupies one quadrant of the chart out to the 6 m dimension, less the four-element-wide band corresponding to the opening. Counting whole elements and estimating partial ones gives $$N \approx 44 \ \text{elements}$$ so that $$\Delta\sigma_z = 0.005 \times 44 \times 100$$ $$\boxed{\,\Delta\sigma_{z,\text{Newmark}} \approx 22.0\ \text{kPa}\,}$$ The count is audited in step 6 against the analytical result, which is how a hand count should always be checked.
3. Choose the second method and set up the superposition. The obvious independent method is the rectangular influence-factor solution, using the $m$–$n$ chart printed on page 8 of the paper, because point A is already the corner of the 6 m × 6 m square. The influence factor $I$ applies at the corner of a uniformly loaded rectangle with $m = B/z$ and $n = L/z$, and the ring is obtained as $$\frac{\Delta\sigma_z}{q} = I_{6\times6} - I_{\text{opening}}$$ Measured from A, the opening runs from 1.5 m to 4.5 m in both plan directions, so it is not itself a corner rectangle. Building it from four rectangles that share the corner A: $$I_{\text{opening}} = I(4.5,4.5) - 2\,I(4.5,1.5) + I(1.5,1.5)$$ the two identical cross terms being subtracted because each covers the strip between the opening and A once too often.
4. Evaluate the influence factors. Dividing every plan dimension by $z = 2.0\ \text{m}$ gives the $m$, $n$ pairs, read from the page-8 chart or computed from the Boussinesq closed form:

   Corner influence factors at $z$ = 2.0 m

   Rectangle from A	$m = B/z$	$n = L/z$	$I$
   6.0 m × 6.0 m (full square)	3.00	3.00	0.24394
   4.5 m × 4.5 m	2.25	2.25	0.23697
   4.5 m × 1.5 m (twice)	2.25	0.75	0.17635
   1.5 m × 1.5 m	0.75	0.75	0.13722

5. Combine and obtain the stress. Substituting into the superposition of step 3, $$I_{\text{opening}} = 0.23697 - 2(0.17635) + 0.13722 = 0.02149$$ $$I_{\text{ring}} = 0.24394 - 0.02149 = 0.22245$$ $$\Delta\sigma_z = q\,I_{\text{ring}} = 100 \times 0.22245$$ $$\boxed{\,\Delta\sigma_{z,\text{chart}} = 22.2\ \text{kPa}\,}$$ Notice how small the correction for the opening is: the hole removes an influence factor of only 0.0215 — about 9 % of the full-square value, although it is 25 % of the gross plan area — because at $z = 2.0\ \text{m}$ the opening is comparatively far from A in plan and elastic influence falls away steeply with horizontal distance.
6. Audit the element count. Rather than accept a hand count, back-figure the number of elements the chart should have shown from the analytical influence value: $$N_{\text{required}} = \frac{I_{\text{ring}}}{I_N} = \frac{0.22245}{0.005} = 44.5$$ A count of 44 or 45 elements is therefore correct, and the count of 44 used in step 2 is confirmed. Half an element out of 44 corresponds to about 0.25 kPa, which is well inside the resolution of any hand count.
7. Comment on the two results. The two methods give 22.0 kPa and 22.2 kPa, a difference of 0.9 %. That agreement is expected, because the two are not independent theories at all: both evaluate the same Boussinesq integral over the same loaded area. The Newmark chart performs the integration graphically by dividing the half-space into 200 equal-influence elements, so its only error source is the counting of partial elements and the accuracy with which the plan is drawn to the depth scale; the influence-factor route performs the same integration in closed form and is exact to the number of digits read off the chart. In practice the Newmark method is preferred when the loaded area is irregular, curved or awkwardly placed relative to the point — a tank farm, a spread of stockpiles, an L-shaped raft — because no decomposition into rectangles is possible; the rectangular influence-factor method is preferred whenever the area can be decomposed, because it is faster and free of counting error. Their agreement here is a useful check that the plan was drawn to the right scale and that the opening was correctly subtracted.