Question 4 of 6: Vertical stress increase beneath a footing with a central opening

Question 4: Vertical stress increase beneath a footing with a central opening (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. Loaded (shaded) area $=$ a 6 m $\times$ 6 m square with a concentric 3 m $\times$ 3 m unloaded opening (1.5 m margin on every side); uniform contact pressure $q = 100\ \text{kPa}$. Point A is at the outer top-right corner of the loaded square. Depth of interest $z = 2.0$ m (part i).

Find. $\Delta\sigma_z$ below A at 2.0 m, by Newmark’s chart and by an independent method, with a comparison; then a qualitative discussion of the trend at 4.0 and 6.0 m.

Approach. The frame-shaped loaded area is treated by superposition about corner A: (full 6 m outer square) $-$ (central 3 m void), each expressed through the corner influence factor $I(m,n)$ for a uniformly loaded rectangle, with $m=B/z$, $n=L/z$. Newmark’s influence chart provides the independent check.

1. Outer 6 m × 6 m square, corner at A. With $m=n=6/2=3$, the Boussinesq corner factor is $$I(3,3) = 0.2439.$$
2. Central 3 m void, referred to corner A. The void spans, measured from A, from 1.5 m to 4.5 m in both directions; superposing the four corner rectangles about A, $$I_{\text{void}} = I(2.25,2.25) - 2\,I(2.25,0.75) + I(0.75,0.75) = 0.0215.$$
3. Net influence factor of the frame. $$I_{\text{net}} = 0.2439 - 0.0215 = 0.2225.$$
4. Stress increase (influence-factor method). $$\Delta\sigma_z = q\,I_{\text{net}} = (100)(0.2225) = \boxed{22.3\ \text{kPa}}.$$
5. Newmark’s chart check. With $\Delta\sigma_z = 0.005\,N\,q$, the number of influence squares implied is $$N = \dfrac{\Delta\sigma_z}{0.005\,q} = \dfrac{22.25}{(0.005)(100)} = 44.5.$$ Drawing the frame to the chart’s depth scale (length $AB$ set equal to $z = 2$ m) and counting the influence elements covered by the shaded area gives $N \approx 44$–45, i.e. $\Delta\sigma_z \approx 22$ kPa. The two methods agree to within the counting tolerance of the chart — that agreement is the required comment: the analytical superposition and Newmark’s graphical integration are two evaluations of the same Boussinesq integral.

(ii) Trend at 4.0 m and 6.0 m depth. The vertical stress increase decreases with depth. As $z$ grows, the fixed loaded footprint subtends an ever smaller solid angle at the point and the load effectively spreads over a larger horizontal area (the 2:1 spreading picture, or simply the $1/z^2$ decay of the Boussinesq kernel), so $\Delta\sigma_z$ below the same footing falls monotonically — from about 22 kPa at 2 m to roughly 17 kPa at 4 m and 13 kPa at 6 m. Hence at both 4.0 m and 6.0 m the vertical stress increase is smaller than at 2.0 m.