A 2.0 m wide strip foundation carries a wall load of 350 kN/m in a soil where γ = 19.0 kN/m³, c′ = 5 kN/m² and φ′ = 23°. The foundation depth is 1.5 m. Determine the factor of safety of this foundation.

Question

Accepted Answer

For φ′ = 23° from Table 12.1,

φ′	[latex]N_c[/latex]	[latex]N_q[/latex]	[latex]N_\gamma[/latex]	φ′	[latex]N_c[/latex]	[latex]N_q[/latex]	[latex]N_\gamma[/latex]
0	5.14	1.00	0.00	23	18.05	8.66	8.20
1	5.38	1.09	0.07	24	19.32	9.60	9.44
2	5.63	1.20	0.15	25	20.72	10.66	10.88
3	5.90	1.31	0.24	26	22.25	11.85	12.54
4	6.19	1.43	0.34	27	23.94	13.20	14.47
5	6.49	1.57	0.45	28	25.80	14.72	16.72
6	6.81	1.72	0.57	29	27.86	16.44	19.34
7	7.16	1.88	0.71	30	30.14	18.40	22.40
8	7.53	2.06	0.86	31	32.67	20.63	25.99
9	7.92	2.25	1.03	32	35.49	23.18	30.22
10	8.35	2.47	1.22	33	38.64	26.09	35.19
11	8.80	2.71	1.44	34	42.16	29.44	41.06
12	9.28	2.97	1.69	35	46.12	33.30	48.03
13	9.81	3.26	1.97	36	50.59	37.75	56.31
14	10.37	3.59	2.29	37	55.63	42.92	66.19
15	10.98	3.94	2.65	38	61.35	48.93	78.03
16	11.63	4.34	3.06	39	67.87	55.96	92.25
17	12.34	4.77	3.53	40	75.31	64.20	109.41
18	13.10	5.26	4.07	41	83.86	73.90	130.22
19	13.93	5.80	4.68	42	93.71	85.38	155.55
20	14.83	6.40	5.39	43	105.11	99.02	186.54
21	15.82	7.07	6.20	44	118.37	115.31	224.64
22	16.88	7.82	7.13	45	133.88	134.88	271.76

[latex]N_c = 18.05, N_q = 8.66[/latex], and [latex]N_\gamma = 8.20[/latex]. Given: B = 2.0 m and [latex]D_f =[/latex]1.5 m. For strip foundation, [latex]L\gg B[/latex]. Hence B/L ≈ 0.

Shape Factors

[latex]F_{cs} = 1 +\frac{B}{L} \frac{N_q}{N_c}[/latex]≈1

[latex]F_{qs} = 1 +\frac{B}{L} an \phi^\prime[/latex]≈1

[latex]F_{\gamma s} = 1 - 0.4\frac{B}{L}[/latex]≈1

Note that all three shape factors are 1.0 for all strip foundations.
Depth Factors

[latex]F_{cd} = 1 + 0.4\frac{D_f}{B}=1+0.4\frac{1.5}{2.0}=[/latex]1.30

[latex]F_{qd}=1+2 an\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}=1+2 an 25(1-\sin 25)^2\frac{1.5}{2} =[/latex] 1.23
[latex]F_{\gamma d} = 1.00[/latex]

Since there is no inclination in the load, all three inclination factors ([latex]F_{ci}, F_{qi}[/latex], and [latex]F_{\gamma i}[/latex]) are unity.
From Eq. (12.7), the ultimate bearing capacity is given by

[latex]q_u = c^\prime N_cF_{cs}F_{cd}F_{ci} + qN_qF_{qs}F_{qd}F_{qi} +0.5 \gamma BN_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i}[/latex]

=[latex] (5.0)(18.05)(1) (1.30)(1) + (1.5 imes 19.0)(8.66)(1)(1.23)(1) + 0.5(19.0)(2)(8.20)(1)(1)(1)[/latex]
= 576.7 kN/m²

The pressure applied on the soil is 350/2 = 175.0 kN/m². Therefore, the factor of safety is
FS = [latex]\frac{576.7}{175}[/latex] = 3.30

Question 12.2: A 2.0 m wide strip foundation carries a wall load of 350 kN/...

Related Answered Questions

A square foundation is shown in Figure 12.9. Assume that the load eccentricity e = 0.5 ft. Determine the ultimate load, Qult. ...

Verified Answer:

A 2 m × 3 m spread foundation placed at a depth of 2 m carries a vertical load of 3000 kN and moment of 300 kN⋅m, as shown in Figure 12.8, Determine the factor of safety. ...

Verified Answer:

A continuous foundation is shown in Figure 12.7. If the load eccentricity is 0.5 ft, determine the ultimate load, Qult, per unit length of the foundation. ...

Verified Answer:

Consider a shallow foundation with the following: ● Foundation: B = 3 ft L = 5 ft Df = 3 ft ● Soil: γ = 110 lb/ft³ φ′ = 25° c′ = 400 lb/ft² Determine the gross allowable load the foundation can carry. Use FS = 4. ...

Verified Answer:

Verified Answer:

A rigid shallow foundation 3 ft × 6 ft is shown in Figure 12.15. Calculate the elastic settlement at the center of the foundation. ...

Verified Answer:

Refer to Example 12.16. For the clay, cu = 270 lb/ft². If the required factor of safety against bearing capacity failure is 3, determine the depth of the foundation. ...

Verified Answer:

Refer to Figure 12.22. The mat has dimensions of 90 ft × 120 ft, and the live load and dead load on the mat are 45 × 10³ kip. The mat is placed over a layer of soft clay that has a unit weight of 120 lb/ft³. Find Df for a fully compensated foundation. ...

Verified Answer:

Determine the allowable bearing capacity of a 10 m × 15 m mat foundation, placed at a depth of 3 m in sand (N60 = 27), which can accept 40 mm settlement. ...

Verified Answer:

Determine the net allowable bearing capacity of a 10 m × 15 m raft foundation resting on a clay where cu = 80 kN/m² and Df = 3 m. Allow a factor of safety of 3. ...

Verified Answer: