16-Civ-B3 Geotechnical Design · May 2017

Question 7 of 9: Terzaghi and general bearing-capacity equations with a moving water table

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7: Terzaghi and general bearing-capacity equations with a moving water table (24 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. A foundation at D_f = 1.5 m in a c'–phi' soil whose unit weights must first be derived from the phase relationships printed on Figure 3.

Find. The allowable bearing capacity for (a) a strip footing with the water table 1.0 m down, (b) a 1.5 m square footing after the water table falls to 2.0 m, and (c) the same square footing carrying a load inclined at 15 degrees to the vertical, using the general bearing-capacity equation.

[Figure not reproduced: Figure 3 redrawn, showing the founding depth, the width and the two water-table positions used in parts (a) and (b)–(c). See the official exam paper.]

Approach. Convert the phase data into moist, saturated and submerged unit weights; establish the surcharge at founding level and the unit weight to be used in the width term for each water-table position; then apply the Terzaghi equation in its strip and square forms for parts (a) and (b), and the general (Meyerhof/Vesic) equation with shape, depth and inclination factors for part (c).

1. Derive the three unit weights from the phase relationships. The dry unit weight follows from the specific gravity and the void ratio, $$\gamma_d=\frac{G_s\gamma_w}{1+e}=\frac{(2.72)(9.81)}{1+0.72}=15.513\ \text{kN/m}^{3}$$ Above the water table the soil carries w = 12 per cent of water, so $$\gamma=\gamma_d(1+w)=(15.513)(1.12)=17.375\ \text{kN/m}^{3}$$ and below the water table it is saturated, $$\gamma_{sat}=\frac{(G_s+e)\gamma_w}{1+e}=\frac{(2.72+0.72)(9.81)}{1.72}=19.620\ \text{kN/m}^{3}$$ $$\boxed{\gamma=17.38,\ \gamma_{sat}=19.62,\ \gamma'=\gamma_{sat}-\gamma_w=9.81\ \text{kN/m}^{3}}$$ As a check on internal consistency, the degree of saturation implied above the water table is S = wG_s/e = (0.12)(2.72)/0.72 = 0.45, a sensible partially saturated value.
2. Record the Terzaghi bearing-capacity factors for phi' = 25 degrees. From Das, Principles of Foundation Engineering, Table 3.1 (Kumbhojkar, 1993): N_c = 25.13, N_q = 12.72, N_gamma = 8.34. The first two can be confirmed in closed form, $$N_q=\frac{e^{2\left(\tfrac{3\pi}{4}-\tfrac{\phi'}{2}\right)\tan\phi'}}{2\cos^{2}\!\left(45^\circ+\tfrac{\phi'}{2}\right)}=12.72,\qquad N_c=(N_q-1)\cot\phi'=25.13$$
3. Part (a): establish the surcharge and the width-term unit weight. The water table at 1.0 m lies between the ground surface and the base at 1.5 m, so the surcharge is built from moist soil above the water table and submerged soil below it, $$q=\gamma D_1+\gamma' D_2=(17.375)(1.0)+(9.81)(0.5)=22.280\ \text{kPa}$$ Because the water table is at or above founding level, the failure wedge beneath the footing is entirely submerged and the unit weight in the width term is gamma' = 9.81 kN/m³.
4. Part (a): apply the Terzaghi strip equation. For a continuous footing, $$q_u=c'N_c+qN_q+\tfrac{1}{2}\gamma' B N_\gamma$$ $$q_u=(25)(25.13)+(22.280)(12.72)+\tfrac{1}{2}(9.81)(1.5)(8.34)$$ $$q_u=628.3+283.4+61.4=973.0\ \text{kPa}$$ Dividing by the required factor of safety, $$\boxed{q_{all}=\frac{973.0}{3}=324.3\ \text{kPa}}$$ The cohesion term supplies 65 per cent of the total here, which is characteristic of a c'–phi' soil with a modest friction angle and a narrow footing.
5. Part (b): re-establish the water-table effect for a table 0.5 m below the base. With the water table now at 2.0 m, the whole of the founding depth is above it, so the surcharge is entirely moist soil, $$q=\gamma D_f=(17.375)(1.5)=26.063\ \text{kPa}$$ The water table now lies a distance d = 0.5 m below the base, which is less than B = 1.5 m, so it still intrudes into the failure wedge and an averaged unit weight is used in the width term, $$\bar{\gamma}=\gamma'+\frac{d}{B}(\gamma-\gamma')=9.81+\frac{0.5}{1.5}(17.375-9.81)=12.332\ \text{kN/m}^{3}$$
6. Part (b): apply the Terzaghi square equation. The square form carries the empirical shape constants 1.3 on the cohesion term and 0.4 on the width term, $$q_u=1.3c'N_c+qN_q+0.4\bar{\gamma}BN_\gamma$$ $$q_u=816.7+331.5+61.7=1209.9\ \text{kPa}$$ $$\boxed{q_{all}=\frac{1209.9}{3}=403.3\ \text{kPa}}$$ The square footing gains 24 per cent over the strip, of which most comes from the 1.3 factor on the large cohesion term and the rest from the drier surcharge and the heavier averaged soil beneath the base.
7. Part (c): assemble the general bearing-capacity factors. The general equation replaces Terzaghi's factors with the Reissner–Prandtl–Vesic set, $$N_q=e^{\pi\tan\phi'}\tan^{2}\!\left(45^\circ+\tfrac{\phi'}{2}\right)=10.662,\quad N_c=(N_q-1)\cot\phi'=20.721,\quad N_\gamma=2(N_q+1)\tan\phi'=10.876$$ These are smaller than Terzaghi's because the general solution assumes a smooth base and does not carry Terzaghi's rigid soil wedge.
8. Part (c): evaluate the shape, depth and inclination factors. For a square footing, B/L = 1, and with D_f/B = 1.0, $$F_{cs}=1+\frac{B}{L}\frac{N_q}{N_c}=1.5146,\qquad F_{qs}=1+\frac{B}{L}\tan\phi'=1.4663,\qquad F_{\gamma s}=1-0.4\frac{B}{L}=0.60$$ $$F_{qd}=1+2\tan\phi'(1-\sin\phi')^{2}\frac{D_f}{B}=1.3109,\qquad F_{cd}=F_{qd}-\frac{1-F_{qd}}{N_c\tan\phi'}=1.3431,\qquad F_{\gamma d}=1$$ The inclination of the resultant, beta = 15 degrees to the vertical, enters through $$F_{ci}=F_{qi}=\left(1-\frac{\beta}{90^\circ}\right)^{2}=0.6944,\qquad F_{\gamma i}=\left(1-\frac{\beta}{\phi'}\right)^{2}=\left(1-\frac{15}{25}\right)^{2}=0.16$$ Note how severe the width-term reduction is: beta is 60 per cent of phi', so F_{gamma i} falls to 0.16 and the width term is all but eliminated.
9. Part (c): assemble the general bearing-capacity equation. Retaining the part (b) water-table condition (q = 26.06 kPa and gamma-bar = 12.33 kN/m³), $$q_u=c'N_cF_{cs}F_{cd}F_{ci}+qN_qF_{qs}F_{qd}F_{qi}+\tfrac{1}{2}\bar{\gamma}BN_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i}$$ $$q_u=731.8+370.9+9.7=1112.4\ \text{kPa}$$ $$\boxed{q_{all}=\frac{1112.4}{3}=370.8\ \text{kPa}}$$
10. Interpret the effect of the inclination. Repeating step 9 with beta = 0 gives q_u = 1648.2 kPa, so tilting the resultant by 15 degrees costs 32.5 per cent of the vertical bearing capacity. The allowable vertical load on the 1.5 m square footing is Q = q_allA = (370.8)(2.25) = 834 kN, and the corresponding inclined resultant is that value divided by cos 15 degrees, 864 kN. A separate check on sliding along the base would also be required in practice, since the horizontal component is 224 kN.