16-Civ-A4 Geotechnical Materials and Analysis · May 2017
Question 5 of 5: Vertical-Stress Increase between Two Footings (Two Methods)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: PEO/EGBC National Examination 16-Civ-A4 — Geotechnical Materials and Analysis, May 2017. Closed book, 3 hours, 100 marks. Answer ALL questions (5 × 20 marks). All required charts and equations were supplied at the back of the paper.
Reference texts: Das, B.M. & Sobhan, K., Principles of Geotechnical Engineering, 9th ed. (Cengage); Holtz, Kovacs & Sheahan, An Introduction to Geotechnical Engineering, 2nd ed.; Knappett & Craig, Craig’s Soil Mechanics, 8th ed.; Budhu, Soil Mechanics and Foundations, 3rd ed.
Check (source figures): Q2’s flow net is a hand-drawn sketch; the counts used (Nd = 10, one drop to B and nine to A) were taken from the printed net and carry the usual ±1-field tolerance. Void ratio (e0), Cc and the preconsolidation break in Q4’s Figure 4(a), and the OCR–Af read in Q3’s Figure 3, are read off hand-plotted curves. Conclusions are robust to these reading tolerances; each is flagged where it bites.
Question 5: Vertical-Stress Increase between Two Footings (Two Methods) (20 marks)
Given. Two $2\ \text{m}\times2\ \text{m}$ footings on the ground surface, each applying $q=30\ \text{kPa}$, with a $2\ \text{m}$ clear gap between them. The gap is not dimensioned on Figure 5, but it scales equal to the 2 m footing width, which puts each footing centre $r=2$ m from A. Point A is on the ground-surface centreline midway between the footings, at depth $z=2\ \text{m}$.
Find. The increase in vertical stress $\Delta\sigma_v$ at A, by two independent methods.
Figure 5 (schematic): plan and section — two 2×2 m footings (q = 30 kPa) with a 2 m gap; A is 2 m below the midpoint.
Approach. A lies outside both loaded areas, so superpose their effects. Method 1: exact Boussinesq rectangular-corner influence factors, splitting each footing into rectangles that share a corner over A. Method 2: idealize each footing as an equivalent point load at its centre and use Boussinesq’s point-load equation. Both results are then summed over the two (symmetric) footings.
Method 1 — influence-factor superposition. Place A at the origin; one footing spans $1\!-\!3\ \text{m}$ horizontally from A and $\pm1\ \text{m}$ transversely. Each transverse half is the difference of two corner rectangles with sides $(3,1)$ and $(1,1)$ at $z=2$ ($m,n$ values $1.5,0.5$ and $0.5,0.5$): $$I(1.5,0.5)=0.131,\qquad I(0.5,0.5)=0.084.$$
Assemble both footings. Two transverse halves per footing and two footings: $$\Delta\sigma_v = q\cdot 4\,[\,I(1.5,0.5)-I(0.5,0.5)\,]=30(4)(0.131-0.084).$$ $$\boxed{\Delta\sigma_v \approx 5.7\ \text{kPa}\ \ (\text{influence-factor method})}$$
Method 2 — equivalent point loads. Each footing is replaced by $Q=q\,A=30(2\times2)=120\ \text{kN}$ at its centre; the horizontal distance from each centre to A is $r=2\ \text{m}$, depth $z=2\ \text{m}$ ($r/z=1$): $$\sigma_z=\frac{3Q}{2\pi z^{2}}\left[\frac{1}{1+(r/z)^{2}}\right]^{5/2}=\frac{3(120)}{2\pi(2)^{2}}(0.5)^{5/2}=2.53\ \text{kPa per footing}.$$
Sum the two point loads. $$\boxed{\Delta\sigma_v = 2(2.53)\approx 5.1\ \text{kPa}\ \ (\text{point-load method})}$$
The two methods agree to within about 10 %. The exact superposition ($\approx5.7\ \text{kPa}$) is slightly larger than the point-load idealization ($\approx5.1\ \text{kPa}$): concentrating each distributed load at a single point underestimates the near-surface spreading toward A. A reasonable design value is $\Delta\sigma_v(\text{A})\approx 5\text{-}6\ \text{kPa}$.