16-Civ-A4 Geotechnical Materials and Analysis · December 2016
Question 3 of 5: Vertical stress increase under an L-shaped foundation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination 98-Civ-A4 Geotechnical Materials and Analysis — December 2016. Closed book; 3 hours; total 100 marks; answer ALL five questions. Newmark / m–n influence charts and a formula sheet are supplied with the paper.
Reference texts: B.M. Das & K. Sobhan, Principles of Geotechnical Engineering, 9th ed. (Cengage); R.F. Craig, Craig's Soil Mechanics, 8th ed. (Spon Press); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed. (Pearson).
Question 3: Vertical stress increase under an L-shaped foundation (20 marks)
Given. Uniformly loaded L-shaped area, $q = 200$ kPa (the shaded region = a 4 m × 4 m square with the top-right 2 m × 2 m corner removed). Point B lies at the re-entrant corner; point A lies on the bottom edge, $2$ m in from the left.
Find. The increase in vertical stress $\Delta\sigma_z$ at $z = 2.0$ m below A and below B; comment on the values at $z=4$ m without calculation.
[Figure not reproduced: Figure 3 (redrawn): L-shaped 200 kPa load; dashed lines show the corner-rectangle decomposition through A and B. See the official exam paper.]
Approach. Split the loaded area into rectangles whose corners meet over each point, read the m–n influence factor $I$ for each (equivalently $\Delta\sigma_z = 0.005\,N q$ from Newmark's chart), and superpose.
Stress below B (re-entrant corner), $z=2$ m. Three of the four quadrants around B are loaded, each a $2\times2$ rectangle with a corner at B: $m=n=2/2=1$, so $I(1,1)=0.175$ ($N \approx 0.175/0.005 = 35$ blocks per quadrant).
$$\Delta\sigma_{z,B} = q\big[3\,I(1,1)\big] = 200(3\times0.175) = \boxed{105\ \text{kPa}}.$$
Stress below A (bottom edge), $z=2$ m. A sits on the edge, so all load is on one side. Split at the vertical through A: the left column is a $2\times4$ rectangle with a corner at A, $I(1,2)=0.200$; the loaded part to the right is the $2\times2$ bottom-right square with a corner at A, $I(1,1)=0.175$.
$$\Delta\sigma_{z,A} = q\big[I(1,2)+I(1,1)\big] = 200(0.200+0.175) = \boxed{75\ \text{kPa}}.$$
Newmark-chart cross-check. With $\Delta\sigma_z = 0.005\,Nq$, the block counts are $N_B = 105/(0.005\times200) = 105$ and $N_A = 75/(0.005\times200) = 75$ influence squares — consistent with counting the shaded area on the chart placed with the depth scale $= z$ at each point.
Values at $z = 4$ m (by inspection). Stress dissipates with depth, so both values fall. Recomputing the same superposition at $z=4$ m gives $\Delta\sigma_{z,B}\approx50$ kPa and $\Delta\sigma_{z,A}\approx41$ kPa. B remains larger than A because B is enclosed by load on three sides whereas A only has load reaching from one side.
Final results — Question 3
Point
$\Delta\sigma_z$ at $z=2$ m
$\Delta\sigma_z$ at $z=4$ m
B (re-entrant corner)
105 kPa
$\approx$ 50 kPa
A (bottom edge)
75 kPa
$\approx$ 41 kPa
The result illustrates two ideas at once: stress concentrates where a point is surrounded by load (B > A), and vertical stress attenuates markedly with depth — both points lose roughly half their surface-level intensity by $z = 4$ m.
