18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2015
Question 6 of 6: Stresses Under a Point Load (Boussinesq)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 04-Env-A3 / Geotechnical & Hydrogeological Engineering. 3 hours duration; open book exam, any non-communicating calculator permitted. The first five questions as they appear in the answer book are marked (20 marks each, 100 marks total); all six are solved below for completeness.
Reference texts. Braja M. Das, Principles of Geotechnical Engineering (9th ed.) — weight–volume relations, USCS classification, seepage/flow nets, slope-stability and stress-distribution chapters; Craig & Knappett, Craig's Soil Mechanics (8th ed.) — flow-net construction and Taylor's stability-number method; Freeze & Cherry, Groundwater (1979) — Darcy's law and the Dupuit–Thiem equation for radial flow to a well.
Question 6: Stresses Under a Point Load (Boussinesq) (20 marks)
Find. (a) total vertical, lateral and shear stress at A; (b) principal stresses and maximum shear stress at A.
Fig. Q6 — point A directly beneath the surface point load, with the total vertical and lateral stresses shown on a small soil element.
Approach. Superpose the pre-existing (geostatic, $K_0$) stress state with the Boussinesq point-load stress increment; along the load axis ($r=0$) both states share the same vertical/horizontal principal directions, so no rotation of axes is needed.
Geostatic (pre-load) total stresses at A. With $e=n/(1-n)=0.5/0.5=1.0$, $\gamma_{sat}=\dfrac{G_s+e}{1+e}\gamma_w=\dfrac{3.7}{2.0}\times9.81=18.15\ \text{kN/m}^3$ and $\gamma'=\gamma_{sat}-\gamma_w=8.34\ \text{kN/m}^3$. At $z=0.5$ m (water table at the surface):
$$u_0=\gamma_w z=4.91\ \text{kPa},\quad \sigma'_{v0}=\gamma'z=4.17\ \text{kPa},\quad \sigma_{v0}=\sigma'_{v0}+u_0=9.07\ \text{kPa},$$
$$\sigma'_{h0}=K_0\,\sigma'_{v0}=0.3\times4.17=1.25\ \text{kPa},\quad \sigma_{h0}=\sigma'_{h0}+u_0=\boxed{6.16\ \text{kPa}}.$$
Boussinesq stress increment directly under the load ($r=0$). At $r=0$, $R=z$, and the standard point-load solution reduces to
$$\Delta\sigma_z=\frac{3Q}{2\pi z^2}=\frac{3\times6.867}{2\pi\times0.5^2}=\boxed{13.12\ \text{kPa}},\qquad \Delta\sigma_r=\Delta\sigma_\theta=-\frac{Q(1-2\nu)}{4\pi z^2},\qquad \Delta\tau_{rz}=0.$$
With $\nu=0.5$, $(1-2\nu)=0$, so $\Delta\sigma_r=\Delta\sigma_\theta=0$ exactly — the point load adds no lateral stress on the axis when the soil is treated as incompressible ($\nu=0.5$).
Part (a) — total stresses at A. Adding the increment to the geostatic state:
$$\sigma_{z}=\sigma_{v0}+\Delta\sigma_z=9.07+13.12=\boxed{22.19\ \text{kPa}},$$
$$\sigma_{h}=\sigma_{h0}+\Delta\sigma_r=6.16+0=\boxed{6.16\ \text{kPa}},\qquad \tau=0+0=\boxed{0}.$$
Part (b) — principal stresses and maximum shear. Since $\tau=0$ on the vertical/horizontal planes at A, these ARE the principal planes: $\sigma_1=\sigma_z=22.19\ \text{kPa}$ (vertical), $\sigma_2=\sigma_3=\sigma_h=6.16\ \text{kPa}$ (horizontal, equal in every direction by axisymmetry). The greatest shear stress is
$$\tau_{max}=\frac{\sigma_1-\sigma_3}{2}=\frac{22.19-6.16}{2}=\boxed{8.02\ \text{kPa}}.$$
Check: $\Delta\sigma_r=0$ on the load axis is a special consequence of the assumed $\nu=0.5$ (undrained/incompressible) value — for any $\nu<0.5$ the point load would add a small (relieving) lateral increment here; the exam's choice of $\nu=0.5$ for a saturated soil under a surface load is deliberate and should not be treated as coincidence.
