The cross section of a cantilever retaining wall is shown in Figure 14.10. Calculate the factors of safety with respect to overturning, sliding, and bearing capacity. Use 150 lb/ft³ for the unit weight of concrete.

Question

Accepted Answer

Refer to Figure 14.10. Note that
[latex]H^\prime = H_1 + H_2 + H_3 = 6 an 10^\circ + 18 + 2.75 =[/latex] 21.81 ft
[latex]P_a = \frac{1}{2} \gamma_1 H^{\prime 2} Ka[/latex]
For [latex]\phi_1^\prime = 34^\circ[/latex] and α = 10°, the value of [latex]K_a[/latex] is 0.294 (Table 11.2), so
[latex]P_a =\frac{\frac{1}{2}(117)(21.81)^2(0.294)}{1000} =[/latex] 8.18 kip/ft
[latex]P_v = P_a \sin 10^\circ =[/latex] 1.42 kip/ft
[latex]P_h = P_a \cos 10^\circ =[/latex] 8.06 kip/ft
Factor of Safety Against Overturning
The following table can now be prepared for determination of the resisting moment.

Section	Weight (kip/ft)	Moment arm from C (ft)	Moment about C (kip.ft/ft)
1	[latex](1.5)(18)(0.15)=4.05[/latex]	5.75	23.29
2	[latex]\frac{1}{2}(1.0)(18)(0.15)=1.35[/latex]	[latex]4+\frac{2}{3}(1)=4.67[/latex]	6.3
3	[latex](12.5)(2.75)(0.15)=5.156[/latex]	6.25	32.23
4	[latex]\left(\frac{18+19.05}{2} ight)(6)(0.117)=13.01[/latex]	[latex]\approx 4+2.5+\frac{6}{2}=9.5[/latex]	123.6
	[latex]P_v=1.42[/latex]	12.5	17.75
	ΣV = 24.986		Σ203.17 =Σ[latex]M_R[/latex]

The overturning moment, [latex]M_O[/latex], is
[latex]M_O = P_h\frac{H^\prime}{3}=(8.06)\left(\frac{21.81}{3} ight)=[/latex] 58.6 kip.ft/ft
So
[latex]FS_{(overturning)}= \frac{\sum M_R}{M_O}=\frac{203.17}{58.6}= [/latex] 3.47 > 3 — OK
Factor of Safety Against Sliding
From Eq. (14.10),
[latex]FS_{(sliding)}=\frac{(\sum V) an (k_1 \phi_2^\prime)+B k_2 c_2^\prime+ P_p}{P_a \cos \alpha}[/latex]
Let [latex]k_1 = k_2 = 2/3[/latex] and [latex]P_p = 0[/latex]. So
[latex]FS_{(sliding)}=\frac{(\sum V) an \left[\frac{2}{3} \phi_2^\prime ight]+B\left(\frac{2}{3}c_2^\prime ight)}{P_a \cos \alpha}[/latex]
=[latex]\frac{(24.986) an \left[\frac{2}{3} (18) ight]+(12.5)\frac{2}{3}(0.9)}{8.06}[/latex]
= 1.59 > 1.5—OK
Factor of Safety Against Bearing Capacity Failure
[latex]e=\frac{B}{2}-\frac{\sum M_R -\sum M_o}{\sum V}=6.25-\frac{203.17-58.6}{24.986}[/latex]
=0.464 ft[latex]\lt \frac{B}{6}=\frac{12.5}{6}=[/latex]2.08 ft
Again, from Eqs. (14.11)
[latex] q_{toe}=\frac{\sum V}{B}\left(1+\frac{6e}{B} ight)=\frac{24.986}{12.5}\left[1+\frac{(6)(0.464)}{12.5} ight]=[/latex] 2.44 kip/ft²
The ultimate bearing capacity of the soil can be determined from Eq. (14.13):

[latex]q_u^\prime = c_2^\prime N_cF_{cd} F_{ci}+q N_qF_{qd} F_{qi} +\frac{1}{2} \gamma_2 B^\prime N_\gamma F_{\gamma d}F_{\gamma i}[/latex]

From Table 12.1 for φ′ = 18°,

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 = 13.1, N_q = 5.26[/latex], and [latex]N_\gamma = 4.07[/latex].
[latex]q = \gamma_2 D = (4)(0.107) =[/latex] 0.428 kip/ft²
[latex]B^\prime = B - 2e = 12.5 - (2)(0.464) =[/latex] 11.572 ft
[latex]F_{cd} = 1 + 0.4\left(\frac{D}{B} ight)=1+(0.4)\left(\frac{4}{12.5} ight)=[/latex]1.128
[latex]F_{qd}=1+2 an\phi_2^\prime(1-\sin \phi_2^\prime)^2\left(\frac{D}{B} ight)=1+(0.31)\left(\frac{4}{12.5} ight)=[/latex] 1.099
[latex]F_{\gamma d} = 1[/latex]
[latex]F_{ci}=F_{qi}=\left(1-\frac{\psi^\circ}{90^\circ} ight)^2[/latex]
[latex]\psi= an^{-1}\left(\frac{P_a \cos \alpha}{\sum V} ight)=\left(\frac{8.06}{24.986} ight)=[/latex]17.88°
So
[latex]F_{ci}=F_{qi}=\left(1-\frac{17.88}{90} ight)^2=[/latex] 0.642

[latex]F_{\gamma i}=\left(1+\frac{\psi}{\phi_2^\prime} ight)^2=\left(1-\frac{17.88}{18} ight)^2\approx[/latex] 0

Hence,
[latex]q_u^\prime = (0.9)(13.1)(1.128)(0.642) + (0.428)(5.26)(1.099)(0.642)+\frac{1}{2}(0.107)(11.572)(4.07)(1)(0)[/latex]
= 10.126 kip/ft²
So
[latex]FS_{(bearing capacity)} =\frac{q_u^\prime}{q_{toe}}=\frac{10.126}{2.44} =[/latex] 4.15 > 3—OK

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