16-Civ-A4 Geotechnical Materials and Analysis · May 2018
Question 6 of 6: Vertical Stress Increase below a Rectangular Load
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examinations — 16-Civ-A4 Geotechnical Materials and Analysis, May 2018.
Closed book, 3 hours, 100 marks. Answer all six questions. All required charts (rectangular-area influence chart, Newmark chart) and a formula sheet are provided at the back of the exam.
Reference texts: R.F. Craig & J. Knappett, Craig’s Soil Mechanics (8th ed.); B.M. Das, Principles of Geotechnical Engineering; R.D. Holtz, W.D. Kovacs & T.C. Sheahan, An Introduction to Geotechnical Engineering (2nd ed.); M. Budhu, Soil Mechanics and Foundations. Take $\gamma_w = 9.81\ \text{kN/m}^3$ throughout.
Given. A $4\ \text{m}\times2\ \text{m}$ flexible rectangle carrying $q=90\ \text{kN/m}^2$. Point A is the bottom-right corner; point C lies 2 m to the right of A on the extended bottom edge (outside the area); point B is inside the area, 1.6 m from the right edge and 0.8 m from the bottom edge. Depth of interest $z=2\ \text{m}$.
Find. $\Delta\sigma_z$ at $z=2\ \text{m}$ below C (by the influence-coefficient chart) and below B (by Newmark’s chart).
Figure 3 plan: the $4\times2$ m loaded area with the exterior point C (2 m beyond corner A) and the interior point B.
Approach. Both methods use the corner influence factor $I(m,n)$ for a uniformly loaded rectangle. For C (an exterior point) superpose two rectangles that share a corner at C; for B (an interior point) split the area into the four rectangles meeting at B and add. Newmark’s chart gives the same interior result as a block count, $\Delta\sigma_z=0.005\,N\,q$.
Influence factor. For the corner of a uniform rectangle, with $m=B/z$ and $n=L/z$ (interchangeable),
$$I=\frac{1}{4\pi}\!\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1+m^2n^2}\cdot\frac{m^2+n^2+2}{m^2+n^2+1}+\tan^{-1}\!\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right].$$
(i) Below C by superposition. C sits on the extended bottom edge, 2 m beyond A. Take the large rectangle from C back across the full 6 m width (0 to 6 m) and subtract the 2 m strip (4 to 6 m) that carries no load. Both share the corner above C. At $z=2$:
large $6\times2$: $m=6/2=3,\ n=2/2=1\Rightarrow I(3,1)=0.2034$;
strip $2\times2$: $m=1,\ n=1\Rightarrow I(1,1)=0.1752$.
$$\Delta\sigma_{z,C}=q\,[I(3,1)-I(1,1)] = 90(0.2034-0.1752)=90(0.0282)=\boxed{2.54\ \text{kN/m}^2}.$$
(ii) Below B by four rectangles. B is interior, so the four rectangles meeting at B have plan sides (left 2.4, right 1.6) × (bottom 0.8, top 1.2). At $z=2$ the corner factors ($m=\text{side}/z$) are
$I(1.2,0.4)=0.1063$, $I(1.2,0.6)=0.1431$, $I(0.8,0.4)=0.0931$, $I(0.8,0.6)=0.1247$; sum $=0.4673$.
$$\Delta\sigma_{z,B}=q\sum I = 90\times0.4673=\boxed{42.1\ \text{kN/m}^2}.$$
Newmark’s chart (check on B). Drawing the loaded area to the chart’s depth scale ($z=2$ m sets the scale line), with B at the centre, the area covers about $N\approx93$ influence blocks. With $I_N=0.005$:
$$\Delta\sigma_{z,B}=0.005\,N\,q = 0.005(93)(90)=41.9\ \text{kN/m}^2,$$
matching the influence-factor value to within the block-counting tolerance.
