24-Bld-A6 Geotechnical Materials and Analysis · Undated paper

Question 5 of 7

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

07-BLD-A6 Geotechnical Materials and Analysis, National Examinations (printed exam date May 2019). 3 hours, closed book, 100 marks. Section A (Q1–Q3) is compulsory (40 marks); Section B directs "answer any three of Q4–Q7" (60 marks), but for completeness this solution answers all four.

Reference texts: B.M. Das, Principles of Geotechnical Engineering, 9th ed.; B.M. Das, Principles of Foundation Engineering, 9th ed.

Question 5 (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.

(i) Assumptions and limitations of elastic theory (Boussinesq/Newmark). The classical elastic solutions used for stress distribution in soil assume: the soil is a homogeneous, isotropic, linearly elastic, semi-infinite half-space; the load is applied at (or near) the surface of that half-space; the soil is weightless (self-weight stress is superposed separately, not part of the elastic solution); and strains are small enough that superposition is valid. The chief limitations are that real soil is neither homogeneous (it is layered) nor isotropic (it is often stress-dependent and stiffer horizontally after consolidation) nor linearly elastic (it yields), so the theory tends to overestimate stress concentration directly beneath a load on stiffer/layered profiles and is least reliable very close to the load or at shallow depth.

Sketch requested in (i): left — $\sigma_z$ decays with depth directly below a point load; right — $\sigma_z$ versus horizontal distance $r$ at three depths, showing a taller, narrower peak at the shallowest depth $z_1$ and a lower, wider spread at the deepest depth $z_3$.

(ii) Given.

Quantity	Value
Net contact stress, $q$	215 kN/m²
Footing ABCD	24 m (AB) × 12 m (AD)
Point O	8 m beyond the BC edge (in-line) and 8 m beyond the DC edge (in-line)
Depth of interest	$z=2$ m

Find. $\sigma_z$ at O, by (a) the $m$-$n$ (Fadum) coefficient method and (b) Newmark's chart method.

Figure 3 (reproduced): footing ABCD (24 m × 12 m) loaded at $q=215$ kN/m². External point O sits 8 m beyond edge BC and 8 m beyond edge DC — a diagonal point outside the loaded area in both directions.

Approach. Because O is offset from the footing in both directions, superpose four corner-of-rectangle solutions (all sharing a corner at O) using the standard "big-rectangle-minus-two-strips-plus-corner" construction: $\sigma_z(O)=q\left[I(R_1)-I(R_2)-I(R_3)+I(R_4)\right]$, where $R_1$ runs from O to the footing's far corner A, $R_2$ and $R_3$ are the two unloaded strips, and $R_4$ is their doubly-subtracted shared corner. Each $I(m,n)$ is the Boussinesq corner influence factor, evaluated here from its closed form rather than read off a chart (a direct check on the Fadum-chart reading) with $m=B/z$, $n=L/z$.

Set up the four rectangles (all with a corner at O), each with $z=2$ m:

Rectangle $L\times B$ (m) $m=B/z$ $n=L/z$ $I(m,n)$

$R_1$ (O to A) 32 × 20 10.0 16.0 0.2499

$R_2$ (O to B) 8 × 20 10.0 4.0 0.2484

$R_3$ (O to D) 32 × 8 4.0 16.0 0.2485

$R_4$ (O to C) 8 × 8 4.0 4.0 0.2473
Combine by superposition. $I_{net}=I(R_1)-I(R_2)-I(R_3)+I(R_4)=0.2499-0.2484-0.2485+0.2473=0.00030$.
Stress by the $m$-$n$ (Fadum) method. $\sigma_z=q\,I_{net}=215\times0.00030=\boxed{0.065\text{ kN/m}^2}$.
Stress by Newmark's chart method. Newmark's influence chart is built directly from the same Boussinesq point-load solution (each of its influence areas represents an equal, fixed increment of $I$, typically $0.005$ per segment). Scaling the chart so that its unit length $AB=z=2$ m and plotting the footing to that scale, O falls so far outside the loaded area (8 m offset at only 2 m depth) that the footing's outline covers a mere sliver of one influence segment near the chart's outer rings — consistent with the tiny $I_{net}\approx0.0003$ (about $6\%$ of one 0.005 segment) computed analytically above. Both methods therefore agree: $\sigma_z(O)\approx0.065\text{ kN/m}^2$, i.e. essentially negligible — O is simply too far outside the loaded footing, relative to the shallow 2 m depth, to feel meaningful additional stress.

Rectangle	$L\times B$ (m)	$m=B/z$	$n=L/z$	$I(m,n)$
$R_1$ (O to A)	32 × 20	10.0	16.0	0.2499
$R_2$ (O to B)	8 × 20	10.0	4.0	0.2484
$R_3$ (O to D)	32 × 8	4.0	16.0	0.2485
$R_4$ (O to C)	8 × 8	4.0	4.0	0.2473

Method	$\sigma_z$ at O
$m$-$n$ (Fadum) coefficients	0.065 kN/m²
Newmark's chart	≈ 0.065 kN/m² (same solution, graphical route)

Topic: Boussinesq stress distribution under a rectangular loaded area; the corner-of-rectangle superposition method for points outside the load.

Key relations: $\sigma_z=qI(m,n)$; $I(m,n)=\frac{1}{4\pi}\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot\frac{m^2+n^2+2}{m^2+n^2+1}+\tan^{-1}\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right]$.

Why this works: the Boussinesq solution is linear in load, so any loaded footing can be built up (or cancelled out) from rectangles that share a corner at the point of interest — superposition (add rectangles that contain the load, subtract or add back rectangles that don't) reproduces exactly the loaded shape without ever needing a closed-form solution for an off-corner point.

Common pitfall: using only two rectangles (rather than four) when the external point is offset in both directions from the footing, which fails to cancel the doubly-counted corner region; misreading $m$ and $n$ as $L/B$ ratios rather than $B/z$, $L/z$.

Source: Das, Principles of Geotechnical Engineering, 9th ed., Ch. 10 (stress distribution, Boussinesq/Newmark/Fadum).

← Question 4 Next: Question 6 →