22-Agric-A2 Soil Physics and Mechanics · December 2019
Question 1 of 6: Bearing Capacity of a Circular Footing with a Rising Water Table
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 04-Agric-A2 Soil Physics & Mechanics,
National Exams December 2019 — a three-hour open-book examination;
any non-communicating calculator is permitted. The cover page states that five (5)
questions constitute a complete exam paper and that only the first five as they appear
in the answer book are marked, that each question is of equal value, and that some
questions require a written answer whose clarity and organization matter for marks. All
six printed questions are worked here, because the set is a study resource rather than a
timed attempt; on exam day a candidate submits only the first five, in order.
Find. FS with the water table at 5 m (a); the revised minimum FS with
the water table risen to 2 m (b); and the maximum allowable column load for FS = 3 (c).
Circular concrete footing, D = 1.5 m, buried Df = 1.0 m; the
figure models the footing as a solid concrete pier down to the base, matching the source
drawing.
Approach. Use Meyerhof's general bearing-capacity equation with circular
shape factors, correcting the surcharge and unit-weight terms for whichever position the
water table takes, and compare the resulting ultimate pressure against the gross applied
pressure from the column load plus the footing's own weight.
Bearing-capacity and shape factors (function of φ′ = 15° only,
unaffected by the water table).
$$\begin{aligned}
N_q&=e^{\pi\tan\phi'}\tan^2\!\left(45+\tfrac{\phi'}{2}\right)=3.941\\
N_c&=(N_q-1)\cot\phi'=10.98\\
N_\gamma&=2(N_q+1)\tan\phi'=2.648
\end{aligned}$$
For a circular footing (B/L = 1):
$$\begin{aligned}
S_c&=1+\frac{N_q}{N_c}=1.359\\
S_q&=1+\tan\phi'=1.268\\
S_\gamma&=0.6
\end{aligned}$$
Applied (gross) bearing pressure. Modelling the footing as a solid
concrete pier from the ground surface down to the base (matching the figure), area
$A=\frac{\pi}{4}D^2=1.767\ \text{m}^2$:
$$\begin{aligned}
W_{footing}&=\gamma_c A D_f=24(1.767)(1.0)=42.41\ \text{kN}\\
Q_{total}&=80+42.41=122.41\ \text{kN}
\end{aligned}$$
$$q_{applied}=\frac{Q_{total}}{A}=\boxed{69.27\ \text{kPa}}$$
a) Water table at 5 m — below the influence depth. Since
$D_w-D_f=5-1=4\ \text{m}$ exceeds B = 1.5 m, the water table lies below the failure zone and
plays no role; use the full unit weight throughout:
$$\begin{aligned}
q'&=\gamma D_f=23(1.0)=23\ \text{kPa}\\
\gamma_{avg}&=\gamma=23\ \text{kN/m}^3
\end{aligned}$$
$$\begin{aligned}
q_{ult}&=c'N_cS_c+q'N_qS_q+\tfrac12\gamma_{avg}BN_\gamma S_\gamma\\
&=20(10.98)(1.359)+23(3.941)(1.268)\\
&\quad+\tfrac12(23)(1.5)(2.648)(0.6)\\
&=440.7\ \text{kPa}
\end{aligned}$$
$$FS_a=\frac{q_{ult}}{q_{applied}}=\frac{440.7}{69.27}=\boxed{6.36}$$
b) Water table risen to 2 m — inside the influence depth. The
water table (2 m) is still below the footing base (1 m), so the surcharge term q′ is
unchanged, but it now sits within B = 1.5 m of the base (D1 = 2−1 = 1 m
< B), so the Nγ term uses the depth-weighted average unit weight with
γ′ = γ − γw = 23−9.81 = 13.19 kN/m³
below the water table:
$$\gamma_{avg}=\gamma'+\frac{D_1}{B}(\gamma-\gamma')=13.19+\frac{1.0}{1.5}(9.81)=19.73\ \text{kN/m}^3$$
$$\begin{aligned}
q_{ult}&=20(10.98)(1.359)+23(3.941)(1.268)\\
&\quad+\tfrac12(19.73)(1.5)(2.648)(0.6)\\
&=436.8\ \text{kPa}
\end{aligned}$$
$$FS_b=\frac{436.8}{69.27}=\boxed{6.31}$$
The revised minimum FS is only marginally lower than case (a), because the Nγ
term is a small fraction of qult for this shallow, small-diameter footing.
c) Maximum allowable load for FS = 3. Governed by the lower (case-b,
worst-case water table) capacity, since the table may rise at any point in the design life:
$$\begin{aligned}
q_{allow}&=\frac{q_{ult,b}}{3}=\frac{436.8}{3}=145.6\ \text{kPa}\\
Q_{allow}&=q_{allow}\,A=145.6(1.767)=257.3\ \text{kN}
\end{aligned}$$
$$P_{allow}=Q_{allow}-W_{footing}=257.3-42.4=\boxed{214.9\ \text{kN}}$$
