Increase in vertical stress below point A by superposition: 16-Civ-A4 Geotechnical Materials and Analysis, May 2013, Q4

Question 4: Increase in vertical stress below point A 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. Point A is the top-right corner of the plan (Figure 3), with z = 5 m. Two footings: M1 (q₁ = 50 kPa), an L-shape occupying the left column and the bottom strip, whose nearest edge is about 30 m to the left and 40 m below A; and M2 (q₂ = 60 kPa), a Z-shape whose right column runs 10 m wide directly up to A. Read from the figure (metres), with distances measured to the left (ξ) and below (η) of A:

Given data — loaded areas referenced to point A (ξ = left of A, η = below A)
Area	q (kPa)	Rectangles (ξ₁–ξ₂) × (η₁–η₂), m
M1 (m–n)	50	left column 30–50 × 10–50; bottom strip 0–50 × 40–50
M2 (Newmark)	60	right column 0–10 × 0–38 (corner at A); tongue 10–20 × 20–30

Find. Δσ_z at 5 m below A = Δσ_z1 (M1, influence factors) + Δσ_z2 (M2, Newmark chart).

Figure 3 — plan view. M1 (blue) is the L-shaped footing; M2 (orange) is the Z-shaped footing whose right column abuts point A. Overall envelope 50 m × 50 m; z is measured 5 m vertically below A.

Check: the plan dimensions are read from Figure 3 and close exactly: the M2 column is 20 + 10 + 8 = 38 m long, a 2 m unloaded gap separates it from the 10 m M1 strip (38 + 2 + 10 = 50 m), and the tongue sits under the second 10 m top dimension. Because the influence factor decays rapidly with horizontal offset, boundaries more than ~15 m from A (all of M1) contribute negligibly, so a ±1 m reading tolerance changes Δσ_z by well under 1 %. Assumptions are stated per exam Note 3.

Approach. Use the corner influence factor I(m, n) for a uniformly loaded rectangle (Boussinesq/Newmark rectangular chart), m = B/z and n = L/z interchangeable, and superpose rectangles that each share the corner at A: Δσ_z = q[I(ξ₂,η₂) − I(ξ₁,η₂) − I(ξ₂,η₁) + I(ξ₁,η₁)].

M1 by the m–n method (q₁ = 50 kPa). M1 = (left column 30–50 × 10–50) ∪ (bottom strip 0–50 × 40–50), less their overlap (30–50 × 40–50). Every rectangle's nearest corner to A is far off (≥ 30 m to the side or 40 m below, i.e. m or n ≥ 6 at z = 5 m), where the influence factor is only a few ten-thousandths: $$\Delta\sigma_{z1}=q_1\big[I_{\text{L-col}}+I_{\text{strip}}-I_{\text{overlap}}\big]=50\times(2.8\times10^{-4}).$$ $$\Delta\sigma_{z1}=\boxed{0.01\ \text{kPa (negligible)}}.$$ M1 is diagonally remote from A, so its stress at 5 m depth below A is essentially zero — the physically important lesson is that vertical stress increments dissipate quickly with horizontal distance.
M2 right column — a rectangle with its corner at A. The right column is 10 m wide (B = 10) by 38 m long (L = 38), with one corner exactly at A: $$m=\frac{B}{z}=\frac{10}{5}=2,\qquad n=\frac{L}{z}=\frac{38}{5}=7.6.$$ From the rectangular influence chart, I(m = 2, n = 7.6) = 0.240 (approaching the 0.250 limit of an infinite quarter-space). Then $$\Delta\sigma_{z,\text{col}}=q_2\,I=60\times0.240=14.4\ \text{kPa}.$$
M2 tongue — superposition of A-cornered rectangles. The tongue spans ξ = 10–20 m, η = 20–30 m: $$I_{\text{tongue}}=I(20,30)-I(10,30)-I(20,20)+I(10,20)=3\times10^{-4}.$$ Its contribution 60 × 3×10⁻⁴ ≈ 0.02 kPa is negligible (the tongue lies 20–30 m below A).
M2 total by the Newmark form. Adding the column and tongue, I_M2 = 0.240. Expressed through the Newmark chart, the equivalent number of influence squares covered by M2 with A at the chart centre is $$N=\frac{I_{\text{M2}}}{I_N}=\frac{0.2401}{0.005}\approx 48\ \text{squares},$$ $$\Delta\sigma_{z2}=I_N\,N\,q_2=0.005\times48\times60=\boxed{14.4\ \text{kPa}}.$$
Superpose the two footings. $$\Delta\sigma_z=\Delta\sigma_{z1}+\Delta\sigma_{z2}=0.01+14.4=\boxed{14.4\ \text{kPa}}.$$ The stress increase 5 m below A is controlled almost entirely by M2, whose loaded area sits directly against the point; the distant M1 adds nothing measurable.

Vertical stress increase 5 m below point A
Source	Method	Influence	Δσ_z (kPa)
M1 (q₁ = 50 kPa)	m–n coefficients	ΣI ≈ 2.8×10⁻⁴	≈ 0.01
M2 (q₂ = 60 kPa)	Newmark (N ≈ 48)	I ≈ 0.240	14.4
Total Δσ_z			≈ 14.4

Topic: Stresses in a soil mass — superposition with rectangular influence factors and the Newmark chart.

Key relations: corner factor I(m, n), m = B/z, n = L/z; four-rectangle superposition for a point at an arbitrary location; Newmark Δσ_z = I_N·N·q with I_N = 0.005 (200-square chart).

Why this works: Both charts integrate Boussinesq's elastic solution for a point load over an area. The m–n factor gives the stress under a rectangle's corner; any point is handled by adding and subtracting corner rectangles. The Newmark chart is the same integral drawn so that each influence square contributes an equal 0.005q, so counting covered squares evaluates the area integral graphically.

Common pitfall: Forgetting that the m–n factor is a corner value (using it as if the point were under the centre doubles or quadruples the stress), and mis-placing the Newmark centre — the point of interest, here A, must sit at the chart's centre before squares are counted.

Source: Das & Sobhan, Principles of Geotechnical Engineering, Ch. 10 (stresses in a soil mass; rectangular-area and Newmark influence charts).

Question 4 of 6: Increase in vertical stress below point A by superposition

Question 4: Increase in vertical stress below point A by superposition (20 marks)