Bearing Capacity of a Square Footing: 18-Env-A3 Geotechnical and Hydrogeological Engineering, May 2016, Q6

Question 6: Bearing Capacity of a Square Footing (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data
Quantity	Symbol	Value
Footing plan size (design)	$B\times L$	2.0 m × 2.0 m
Embedment depth	$D_f$	1.0 m
Friction angle	$\phi'$	20°
Cohesion	$c'$	15 kPa
Soil unit weight	$\gamma$	19.0 kN/m³
Concrete unit weight	$\gamma_c$	24 kN/m³
Required factor of safety	FS	3.0

Figure 4 — stepped square footing: a 0.5 m × 0.5 m × 0.5 m pedestal on a 2.0 m × 2.0 m × 0.5 m base slab, base at $D_f=1.0$ m.

Find. (a) maximum allowable load $P$; (b) factor of safety if the as-built base shrinks to 1.9 m × 1.9 m.

Approach. Compute Meyerhof's general ultimate bearing capacity for the square footing (shape and depth factors included), divide by the required FS to get the allowable gross bearing pressure, then back out the allowable COLUMN load $P$ once the footing's own concrete weight and the backfill resting on its steps are subtracted from the total allowable load the soil can carry.

Bearing-capacity factors ($\phi'=20^\circ$, Meyerhof). $$\begin{aligned} N_q&=e^{\pi\tan\phi'}\tan^2\!\left(45+\frac{\phi'}{2}\right)=\boxed{6.40}, \\ N_c&=\frac{N_q-1}{\tan\phi'}=\boxed{14.83}, \\ N_\gamma&=(N_q-1)\tan(1.4\phi')=\boxed{2.87}. \end{aligned}$$
Part (a) — shape, depth factors and ultimate bearing capacity (B = L = 2.0 m). For a square footing ($B/L=1$), $$\begin{aligned} F_{cs}&=1+\frac{B}{L}\frac{N_q}{N_c}=1.431, \\ F_{qs}&=1+\frac{B}{L}\tan\phi'=1.364, \\ F_{\gamma s}&=1-0.4\frac{B}{L}=0.600. \end{aligned}$$ With $D_f/B=1/2=0.5$, $$\begin{aligned} F_{cd}&=1+0.4\frac{D_f}{B}=1.200, \\ F_{qd}&=1+2\tan\phi'(1-\sin\phi')^2\frac{D_f}{B}=1.158, \\ F_{\gamma d}&=1. \end{aligned}$$ The overburden at the base is $q=\gamma D_f=19.0\times1.0=19.0$ kPa, so $$\begin{aligned} q_u&=c'N_cF_{cs}F_{cd}+qN_qF_{qs}F_{qd}+\tfrac12\gamma BN_\gamma F_{\gamma s}F_{\gamma d} \\ &=382.3+192.0+32.7=\boxed{607.0\ \text{kPa}}. \end{aligned}$$
Allowable pressure and footing/backfill self-weight. $$q_{all}=\frac{q_u}{FS}=\frac{607.0}{3.0}=\boxed{202.3\ \text{kPa}}.$$ The concrete volume (pedestal $0.5\times0.5\times0.5$ + base slab $2.0\times2.0\times0.5$) is $V_c=0.125+2.0=2.125\ \text{m}^3$, so $W_c=24\times2.125=51.0$ kN. The backfill sitting on the base slab's exposed steps (area $2.0^2-0.5^2=3.75\ \text{m}^2$, 0.5 m deep) weighs $W_s=19.0\times3.75\times0.5=35.6$ kN. Together, $$W_{extra}=51.0+35.6=\boxed{86.6\ \text{kN}}.$$
Solve for the allowable column load. The total load the soil must carry (column load + footing + backfill) cannot exceed $q_{all}\times B^2$: $$P_{max}=q_{all}\,B^2-W_{extra}=202.3\times4.0-86.6=809.3-86.6=\boxed{723\ \text{kN}}.$$
Part (b) — as-built 1.9 m × 1.9 m footing, same design load $P=723$ kN. Recomputing with $B=1.9$ m ($D_f/B=0.526$; shape factors unchanged since the footing is still square): $$\begin{aligned} F_{cd}&=1.210,\quad F_{qd}=1.166, \\ q_u'&=385.7+193.4+31.1=\boxed{610.0\ \text{kPa}}. \end{aligned}$$ The smaller footing's concrete + backfill weight is $W_{extra}'=46.3+31.9=78.2$ kN (same pedestal, smaller 1.9×1.9×0.5 m slab), so the actual applied gross pressure with the SAME column load is $$q_{applied}=\frac{P+W_{extra}'}{B^2}=\frac{723+78.2}{1.9^2}=\frac{801.2}{3.61}=\boxed{221.9\ \text{kPa}}.$$ $$FS=\frac{q_u'}{q_{applied}}=\frac{610.0}{221.9}=\boxed{2.75}.$$

Check: engineering assumption — the "maximum allowable load P" is taken as the load applied AT the column (on top of the pedestal), so the footing's own concrete and the backfill soil resting on its exposed steps are additional load the soil must carry and are subtracted from $q_{all}B^2$ to isolate $P$; this is why the concrete unit weight is given in the problem. Building the base 0.1 m undersized on each side (2.0→1.9 m) drops the factor of safety from the required 3.0 down to about 2.75 — a real (if modest) erosion of margin from what looks like a small construction tolerance, worth flagging in a field QA context.

Quantity	Value
$N_c$, $N_q$, $N_\gamma$ ($\phi'=20^\circ$)	14.83, 6.40, 2.87
$q_u$ (2.0 m design footing)	607.0 kPa
Footing + backfill weight (2.0 m)	86.6 kN
(a) Max. allowable load, $P$	723 kN
$q_u$ (1.9 m as-built footing)	610.0 kPa
(b) FS, as-built 1.9 m × 1.9 m	2.75

Concept & Theory

Topic. Meyerhof's general bearing-capacity equation for a shallow square footing, with shape and depth correction factors.

Key relations. $q_u=c'N_cF_{cs}F_{cd}+qN_qF_{qs}F_{qd}+\tfrac12\gamma BN_\gamma F_{\gamma s}F_{\gamma d}$; $FS=q_u/q_{applied}$; Meyerhof factors $N_q=e^{\pi\tan\phi'}\tan^2(45+\phi'/2)$, $N_c=(N_q-1)\cot\phi'$, $N_\gamma=(N_q-1)\tan(1.4\phi')$.

Why this works. The three bearing-capacity terms superpose the independent contributions of cohesion, surcharge, and self-weight of the failure wedge; shape factors correct Terzaghi's original strip-footing solution for a finite (here square) plan, and depth factors credit the shear strength mobilized along the failure surface above footing-base level. Subtracting the footing's own weight and backfill from the allowable soil capacity is what converts a gross allowable PRESSURE into a net allowable COLUMN LOAD.

Common pitfall. Applying the design load $P$ from part (a) directly as $P/B^2$ against the NEW footing's $q_u$ in part (b) without re-adding the (now different) footing self-weight, or forgetting that a smaller base also changes the depth factors (through $D_f/B$) and not just the $\gamma BN_\gamma$ term.

Source. Das, Principles of Geotechnical Engineering, ch. 3–4 (bearing capacity of shallow foundations, Meyerhof's general equation and correction factors).

Question 6 of 6: Bearing Capacity of a Square Footing

Question 6: Bearing Capacity of a Square Footing (20 marks)