18-Env-A3 Geotechnical and Hydrogeological Engineering · May 2018
Question 5 of 6: Bearing Capacity of a Square Footing (Concentric and Eccentric Loading)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams May 2018 — 04-Env-A3, Geotechnical and Hydrogeological Engineering — 3 hours, open book. All SIX questions are answered below (the exam marks only the first five as submitted; all six are solved here as a complete study resource). Marking scheme: each question 20 marks, equal value.
Find. (a) the maximum allowable concentric column load at FS = 3.0; (b) the maximum allowable load at eccentricity $e=0.3$ m and FS = 2.0.
Figure 3: square column foundation, stepped base (0.5 m thick, 2.5×2.5 m) and pedestal (0.7×0.7 m) up to grade, 1.0 m total embedment.
Approach. Use Meyerhof's general bearing-capacity equation with his own shape and depth factors (a self-consistent single-author framework) to get the gross ultimate bearing pressure, divide by the required FS to get the total allowable load the base can carry, then subtract the footing's own concrete weight and the backfill soil weight above the base to isolate the allowable column load $P$. For the eccentric case, Meyerhof's effective-width method models the load as centred on a smaller effective footing.
Bearing-capacity and shape/depth factors at $\phi'=30^{\circ}$. $$N_q = e^{\pi\tan\phi'}\tan^2\!\left(45+\tfrac{\phi'}{2}\right) = 18.40, \qquad N_c=(N_q-1)\cot\phi' = 30.14, \qquad N_\gamma = (N_q-1)\tan(1.4\phi') = 15.67$$ With $K_p=\tan^2(45+\phi'/2)=3.0$, the square-footing ($B/L=1$) shape factors are $$s_c = 1+0.2K_p\left(\tfrac{B}{L}\right)=1.60, \qquad s_q=s_\gamma=1+0.1K_p\left(\tfrac{B}{L}\right)=1.30$$ and with $D_f/B=1.0/2.5=0.4$, the depth factors are $$d_c = 1+0.2\sqrt{K_p}\left(\tfrac{D_f}{B}\right)=1.139, \qquad d_q=d_\gamma=1+0.1\sqrt{K_p}\left(\tfrac{D_f}{B}\right)=1.069$$
Allowable total load and self-weight deduction. $$q_{all}=\dfrac{q_u}{FS}=\dfrac{1394}{3.0}=464.6\text{ kPa} \quad\Rightarrow\quad Q_{allow,total} = q_{all}A = 464.6(2.5^2)=2904\text{ kN}$$ Footing concrete volume $=(2.5)^2(0.5)+(0.7)^2(0.5)=3.37\text{ m}^3 \Rightarrow W_{footing}=24(3.37)=80.9$ kN. Backfill soil around the pedestal, above the base, up to grade: volume $=[(2.5)^2-(0.7)^2](0.5)=2.88\text{ m}^3 \Rightarrow W_{backfill}=16(2.88)=46.1$ kN. $$P_{max} = Q_{allow,total} - W_{footing} - W_{backfill} = 2904-80.9-46.1 = \boxed{2777\text{ kN}}$$
(b) Effective width for eccentric loading. Meyerhof's effective-area method centres the load on a reduced footing: $$B' = B-2e = 2.5-2(0.3)=1.9\text{ m}, \qquad L'=L=2.5\text{ m}, \qquad A'=B'L'=4.75\text{ m}^2$$ Re-computing the (now non-square, $B'/L'=0.76$) shape factors with the same $K_p$, and keeping the depth factors from Step 1 (the embedment geometry itself hasn't changed): $$s_c'=1+0.2K_p(0.76)=1.456, \qquad s_q'=s_\gamma'=1+0.1K_p(0.76)=1.228$$
Ultimate and allowable eccentric load. Use $B'$ in the self-weight (third) term only, per Meyerhof's method: $$q_u' = c'N_cs_c'd_c + qN_qs_q'd_q + 0.5\gamma B'N_\gamma s_\gamma'd_\gamma \approx \boxed{1199\text{ kPa}}$$ $$Q_{ult}=q_u'A' = 1199(4.75) = 5695\text{ kN} \quad\Rightarrow\quad Q_{allow}=\dfrac{Q_{ult}}{FS}=\dfrac{5695}{2.0}=2848\text{ kN}$$ $$P_{max,e} = Q_{allow} - W_{footing} - W_{backfill} = 2848-80.9-46.1 = \boxed{2721\text{ kN}}$$
Even though the required factor of safety is relaxed from 3.0 to 2.0, the eccentricity shrinks the effective bearing area to $4.75\text{ m}^2$ (76% of the full $6.25\text{ m}^2$) and lowers the self-weight term through the smaller $B'$ — the two effects roughly cancel the relaxed FS, so the maximum allowable column load barely changes (2721 kN vs. 2777 kN). This is the expected behaviour of the effective-width method: eccentricity is penalized primarily through lost bearing area, not just a nominally lower capacity.
