Question 4 of 6: Vertical Stress Increase by Superposition

Question 4: Vertical Stress Increase by Superposition (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. The plan of Figure 3 sits inside a 50 m × 50 m envelope with point A at its north-east corner; the two flexible foundations and the depth of interest are listed below.

Find. The total increase in vertical stress $\Delta\sigma_z = \Delta\sigma_{z1} + \Delta\sigma_{z2}$ at 5 m depth vertically below point A.

[Figure not reproduced: Figure 3 (redrawn to scale) — plan of the two multiplex footprints. Origin at the south-west corner, x east and y north; A is at (50, 50). See the official exam paper.]

Approach. Boussinesq’s solution is linear in the applied load, so any footprint can be built up as a signed sum of rectangles that each have one corner vertically above the point of interest. For M1 the four such rectangles are read off the m–n chart; for M2 the irregular L-shape is handled with the Newmark influence chart as instructed, and the count is checked against the equivalent corner-rectangle solution.

1. Fix a frame and express both footprints as corner rectangles. Take the origin at the south-west corner of the 50 m square, $x$ east and $y$ north, so $A = (50,\,50)$. Write each rectangle in terms of its distance west of A ($a = 50 - x$) and south of A ($b = 50 - y$), because a rectangle described that way automatically has a corner over A. The M1 footprint is the union of the south strip $a \in [0,50],\ b \in [40,50]$ and the west column $a \in [30,50],\ b \in [10,40]$; the M2 footprint is the union of the east column $a \in [0,10],\ b \in [0,38]$ and the arm $a \in [10,20],\ b \in [20,30]$.
2. Write the signed superposition for M1. Denoting by $R(a,b)$ the influence factor of a rectangle $a \times b$ with its corner over A, the strip is $R(50,50) - R(50,40)$ and the column is $R(50,40) - R(50,10) - R(30,40) + R(30,10)$. The $R(50,40)$ terms cancel, leaving $$I_{M1} = R(50,50) - R(50,10) - R(30,40) + R(30,10)$$
3. Convert each rectangle to m and n at z = 5 m. With $m = B/z$ and $n = L/z$ (interchangeable), the four rectangles and their chart values are tabulated below. The values are the standard Boussinesq corner factors and can be read directly off the m–n chart supplied with the paper.

4. Evaluate the M1 contribution. Multiplying the net influence factor by the contact stress, $$\Delta\sigma_{z1} = q_1\,I_{M1} = 50 \times 0.00028 = 0.014\ \text{kPa}$$ so that $$\boxed{\;\Delta\sigma_{z1} \approx 0.01\ \text{kPa} \approx 0\;}$$ The four chart factors are each close to the limiting value 0.25 because every one of these rectangles is large compared with $z = 5$ m; the physics is carried entirely by the differences between them, and those differences very nearly cancel.
5. Interpret that result before moving on. The nearest loaded element of M1 to point A is the north-east corner of the south strip, at $(50,\,10)$, which is 40 m away in plan. The Boussinesq point-load solution decays as $\sigma_z \propto z^3/R^5$, and the familiar 2:1 construction spreads load only about one horizontal unit for every two vertical, so at a depth of only 5 m the stress bulb of M1 has long since died out at A. The superposition is not wasted work — it is the calculation that demonstrates the answer — but the engineering conclusion is that M1 has no practical influence on the stress 5 m below A.
6. Set up the Newmark chart for M2. The chart supplied has an influence value $I_N = 0.005$, that is 200 influence elements. Draw the plan of M2 on tracing paper to the scale set by the chart’s depth line, $\overline{AB} = z = 5\ \text{m}$, lay the tracing over the chart with point A on the centre of the circles, and count the elements covered by the loaded area.
7. Count the elements. The 10 m × 38 m east column has one corner exactly at A and, being only two depth-units wide, covers a compact block of elements around the centre; the 10 m × 10 m arm lies between two and four depth-units away and covers a negligible fraction of an element. Counting gives $$N \approx 48\ \text{elements}$$
8. Evaluate the M2 contribution. Applying the chart formula, $$\Delta\sigma_{z2} = I_N\,N\,q_2 = 0.005 \times 48 \times 60$$ $$\boxed{\;\Delta\sigma_{z2} = 14.4\ \text{kPa}\;}$$
9. Check the count against the corner-rectangle solution. A Newmark count is only as good as the tracing, so verify it. The east column has its corner at A with $m = 10/5 = 2.0$ and $n = 38/5 = 7.6$, giving $I = 0.2398$; the arm contributes $R(20,30) - R(20,20) - R(10,30) + R(10,20) = 0.0003$. Hence $I_{M2} = 0.2401$ and $\Delta\sigma_{z2} = 60 \times 0.2401 = 14.4\ \text{kPa}$, which back-figures to $N = 14.4/(0.005 \times 60) = 48.0$ elements — the count is confirmed.
10. Superpose. Adding the two contributions, $$\Delta\sigma_z = \Delta\sigma_{z1} + \Delta\sigma_{z2} = 0.014 + 14.406$$ $$\boxed{\;\Delta\sigma_z \approx 14.4\ \text{kPa at 5 m below point A}\;}$$