24-Bld-A6 Geotechnical Materials and Analysis, December 2016, Q3

Question 3 (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. An L-shaped loaded area (a 4 m × 4 m footprint with a 2 m × 2 m corner omitted), uniformly loaded at q = 200 kPa; point A on the loaded edge at the mid-width of the base, point B at the re-entrant (inside) corner of the L.

Figure 3 — L-shaped loaded area (shaded), with points A (edge midpoint) and B (re-entrant corner).

Find. The increase in vertical stress $\Delta\sigma_z$ at 2.0 m depth below A and below B, and a qualitative comparison at 4 m depth.

Approach. Newmark's chart is a graphical tool for evaluating the same integral that the printed formula sheet gives algebraically for a rectangular loaded area, $$I(m,n)=\frac{1}{4\pi}\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot\frac{m^2+n^2+2}{m^2+n^2+1}+\arctan\!\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right],\quad m=\frac{B}{z},\ n=\frac{L}{z};$$ because neither A nor B sits at a corner of the L-shape itself, each is treated as the shared CORNER of two or three rectangles that together tile the loaded area (the method a Newmark chart overlay performs by counting influence blocks), and the influence factors are summed.

Decompose the loaded area at point A. A lies on the bottom edge at mid-width. Splitting the L-shape about the vertical line through A gives a fully loaded 2×4 m rectangle to one side (corner at A, $m=B/z=2/2=1$, $n=L/z=4/2=2$) and a fully loaded 2×2 m rectangle to the other side (corner at A, $m=n=1$); together these two rectangles reproduce the whole loaded area with no gaps or overlaps.
Decompose the loaded area at point B. B is the re-entrant corner. The full 4×4 m square, split into its four 2×2 m quadrants about B, has three quadrants loaded (the omitted corner is the fourth), so B is the shared corner of three loaded 2×2 m rectangles ($m=n=1$ each).
Evaluate the influence factors at z = 2 m. From the formula above, $I(1,1)=0.1752$ and $I(1,2)=0.1999$, so $$I_A = I(1,2)+I(1,1) = 0.1999+0.1752 = 0.3751,\qquad I_B = 3\times I(1,1) = 3(0.1752) = 0.5257.$$
Stress increase at 2 m depth. $$\Delta\sigma_{z,A} = qI_A = 200(0.3751) = 75.0\ \text{kPa},\qquad \Delta\sigma_{z,B} = qI_B = 200(0.5257) = 105.1\ \text{kPa}.$$ $$\boxed{\Delta\sigma_{z,A}=75.0\ \text{kPa}\ (2\ \text{m below A}),\qquad \Delta\sigma_{z,B}=105.1\ \text{kPa}\ (2\ \text{m below B})}$$
Comment on 4 m depth, without further calculation. Doubling the depth halves every $m=B/z$ and $n=L/z$ ratio, and the influence factor $I(m,n)$ decreases monotonically as m and n shrink (a loaded area subtends a smaller solid angle from a point twice as far below it), so BOTH $\Delta\sigma_{z,A}$ and $\Delta\sigma_{z,B}$ at 4 m will be noticeably smaller than at 2 m — roughly half to just over half their 2 m values, since $I$ falls off somewhat faster than linearly with depth for these plan proportions. The GAP between the two points should also narrow with depth: B currently carries more stress than A because three loaded quadrants surround it against A's two, but as z grows large compared with the 2–4 m plan dimensions, the loaded area increasingly resembles a point load viewed from directly below, and the stress distribution beneath the whole footprint tends toward uniformity — so $\Delta\sigma_{z,B}/\Delta\sigma_{z,A}$ should move from its 2 m-depth ratio of about 1.40 toward 1 as depth increases.

Question 3 — final results
Quantity	Value
Influence factor at A, z = 2 m	I_A = 0.375
Influence factor at B, z = 2 m	I_B = 0.526
$\Delta\sigma_z$ at A, 2 m depth	75.0 kPa
$\Delta\sigma_z$ at B, 2 m depth	105.1 kPa
Expected trend at 4 m depth	both smaller; B/A ratio moves toward 1

Concept & Theory

Topic. Vertical stress beneath an irregular loaded area by superposition of rectangular-corner Boussinesq/Newmark influence factors.

Key relations. Newmark's influence-factor equation for a uniformly loaded rectangle, corner-of-rectangle superposition for points not on a corner of the actual footprint (add rectangles where the point lies at a shared corner of loaded panels; subtract where a panel must be removed).

Why this works. The Boussinesq stress field is linear in the applied pressure, so the vertical stress from an arbitrary polygon can always be built from rectangles that share a corner at the point of interest — exactly what a Newmark chart overlay does by counting influence blocks under a scaled tracing of the loaded plan; the algebraic form used here is the same integral, evaluated exactly rather than by counting.

Common pitfall. Forgetting that $m$ and $n$ are interchangeable but must each use the correct HALF-width when the point is not centred, and double counting or omitting a panel when the point sits at a re-entrant rather than a convex corner.

Source. Das, Principles of Geotechnical Engineering, Ch. 8 (stress distribution); exam formula sheet, rectangular-loading and Newmark's-chart equations.

Question 3 of 5

Question 3 (20 marks)