Determine the safety factors for the cantilever retaining wall shown in figure 14.16 with respect to sliding and overturning, using Coulomb’s active earth pressures. Assume δ′ is 24° and the unit weight of concrete is 24 kN/m³. How would you improve the stability of this wall?

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Accepted Answer

From Table 11.4 for α = 10°, φ′ = 36° and δ′ = 24°,

α (deg)	φ′ (deg)	θ (deg)
α (deg)	φ′ (deg)	0	5	10	15	20	25
0	28	0.3213	0.3588	0.4007	0.4481	0.5026	0.5662
	29	0.3091	0.3467	0.3886	0.4362	0.4908	0.5547
	30	0.2973	0.3349	0.3769	0.4245	0.4794	0.5435
	31	0.286	0.3235	0.3655	0.4133	0.4682	0.5326
	32	0.275	0.3125	0.3545	0.4023	0.4574	0.522
	33	0.2645	0.3019	0.3439	0.3917	0.4469	0.5117
	34	0.2543	0.2916	0.3335	0.3813	0.4367	0.5017
	35	0.2444	0.2816	0.3235	0.3713	0.4267	0.4919
	36	0.2349	0.2719	0.3137	0.3615	0.417	0.4824
	37	0.2257	0.2626	0.3042	0.352	0.4075	0.4732
	38	0.2168	0.2535	0.295	0.3427	0.3983	0.4641
	39	0.2082	0.2447	0.2861	0.3337	0.3894	0.4553
	40	0.1998	0.2341	0.2732	0.3179	0.3696	0.4304
	41	0.1918	0.2278	0.2689	0.3164	0.3721	0.4384
	42	0.184	0.2197	0.2606	0.308	0.3637	0.4302
5	28	0.3431	0.3845	0.4311	0.4843	0.5461	0.619
	29	0.3295	0.3709	0.4175	0.4707	0.5325	0.6056
	30	0.3165	0.3578	0.4043	0.4575	0.5194	0.5926
	31	0.3039	0.3451	0.3916	0.4447	0.5067	0.58
	32	0.2919	0.3329	0.3792	0.4324	0.4943	0.5677
	33	0.2803	0.3211	0.3673	0.4204	0.4823	0.5558
	34	0.2691	0.3097	0.3558	0.4088	0.4707	0.5443
	35	0.2583	0.2987	0.3446	0.3975	0.4594	0.533
	36	0.2479	0.2881	0.3338	0.3866	0.4484	0.5221
	37	0.2379	0.2778	0.3233	0.3759	0.4377	0.5115
	38	0.2282	0.2679	0.3131	0.3656	0.4273	0.5012
	39	0.2188	0.2582	0.3033	0.3556	0.4172	0.4911
	40	0.2098	0.2489	0.2937	0.3458	0.4074	0.4813
	41	0.2011	0.2398	0.2844	0.3363	0.3978	0.4718
	42	0.1927	0.2311	0.2753	0.3271	0.3884	0.4625
10	28	0.3702	0.4164	0.4686	0.5287	0.5992	0.6834
	29	0.3548	0.4007	0.4528	0.5128	0.5831	0.6672
	30	0.34	0.3857	0.4376	0.4974	0.5676	0.6516
	31	0.3259	0.3713	0.423	0.4826	0.5526	0.6365
	32	0.3123	0.3575	0.4089	0.4683	0.5382	0.6219
	33	0.2993	0.3442	0.3953	0.4545	0.5242	0.6078
	34	0.2868	0.3314	0.3822	0.4412	0.5107	0.5942
	35	0.2748	0.319	0.3696	0.4283	0.4976	0.581
	36	0.2633	0.3072	0.3574	0.4158	0.4849	0.5682
	37	0.2522	0.2957	0.3456	0.4037	0.4726	0.5558
	38	0.2415	0.2846	0.3342	0.392	0.4607	0.5437
	39	0.2313	0.274	0.3231	0.3807	0.4491	0.5321
	40	0.2214	0.2636	0.3125	0.3697	0.4379	0.5207
	41	0.2119	0.2537	0.3021	0.359	0.427	0.5097
	42	0.2027	0.2441	0.2921	0.3487	0.4164	0.499
15	28	0.4065	0.4585	0.5179	0.5868	0.6685	0.767
	29	0.3881	0.4397	0.4987	0.5672	0.6483	0.7463
	30	0.3707	0.4219	0.4804	0.5484	0.6291	0.7265
	31	0.3541	0.4049	0.4629	0.5305	0.6106	0.7076
	32	0.3384	0.3887	0.4462	0.5133	0.593	0.6895
	33	0.3234	0.3732	0.4303	0.4969	0.5761	0.6721
	34	0.3091	0.3583	0.415	0.4811	0.5598	0.6554
	35	0.2954	0.3442	0.4003	0.4659	0.5442	0.6393
	36	0.2823	0.3306	0.3862	0.4513	0.5291	0.6238
	37	0.2698	0.3175	0.3726	0.4373	0.5146	0.6089
	38	0.2578	0.305	0.3595	0.4237	0.5006	0.5945
	39	0.2463	0.2929	0.347	0.4106	0.4871	0.5805
	40	0.2353	0.2813	0.3348	0.398	0.474	0.5671
	41	0.2247	0.2702	0.3231	0.3858	0.4613	0.5541
	42	0.2146	0.2594	0.3118	0.374	0.4491	0.5415
20	28	0.4602	0.5205	0.59	0.6714	0.7689	0.888
	29	0.4364	0.4958	0.5642	0.6445	0.7406	0.8581
	30	0.4142	0.4728	0.5403	0.6195	0.7144	0.8303
	31	0.3935	0.4513	0.5179	0.5961	0.6898	0.8043
	32	0.3742	0.4311	0.4968	0.5741	0.6666	0.7799
	33	0.3559	0.4121	0.4769	0.5532	0.6448	0.7569
	34	0.3388	0.3941	0.4581	0.5335	0.6241	0.7351
	35	0.3225	0.3771	0.4402	0.5148	0.6044	0.7144
	36	0.3071	0.3609	0.4233	0.4969	0.5856	0.6947
	37	0.2925	0.3455	0.4071	0.4799	0.5677	0.6759
	38	0.2787	0.3308	0.3916	0.4636	0.5506	0.6579
	39	0.2654	0.3168	0.3768	0.448	0.5342	0.6407
	40	0.2529	0.3034	0.3626	0.4331	0.5185	0.6242
	41	0.2408	0.2906	0.349	0.4187	0.5033	0.6083
	42	0.2294	0.2784	0.336	0.4049	0.4888	0.593

Coulomb’s active earth pressure coefficient is given by [latex]K_a =[/latex] 0.2633
[latex]P_a =\frac{1}{2}\gamma H^2K_a =\frac{1}{2}(18.5)(6.0)^2(0.2633) =[/latex] 87.7 kN/m
[latex]P_k = P_a \cos 24 = [/latex]80.1 kN/m
[latex]P_v = P_a \sin 24 =[/latex] 35.7 kN/m

Section	Weight (kN/m)	Moment arm from C (m)	Moment about C (kN⋅m/m)
1	[latex](0.5)(0.5)(5.5)(24.0)=33.0[/latex]	1.83	60.4
2	[latex](0.5)(5.5)(24.0)=66.0[/latex]	2.25	148.5
3	[latex](0.5)(3.5)(24.0)=42.0[/latex]	1.75	73.5
	[latex]P_v=35.7[/latex]	2.5	89.3
	[latex]\sum V =176.7[/latex]		[latex]\sum M_R=371.7[/latex]

Stability with Respect to Overturning
Overturning moment [latex]M_O = P_h\frac{H^\prime}{3} = 80.1 imes 2.0 =[/latex] 160.2 kN⋅m/m
Resisting moment [latex]\sum M_R =[/latex] 371.7 kN⋅m/m
Therefore,
[latex]FS_{(overturning)} =\frac{\sum M_R}{M_O}=\frac{371.7}{160.2} =[/latex] 2.32
Stability with Respect Sliding
For φ′ = 36° using Eq. 11.28, [latex]K_p = an^2\left(45+\frac{\phi^\prime}{2} ight)=[/latex] 3.852
[latex]P_p =\frac{1}{2} \gamma D^2K_p =\frac{1}{2}(18.5)(1.0)^2(3.852) =[/latex] 35.6 kN/m
Taking δ′ as 2/3 φ′, δ′ = 24°
[latex]FS_{(sliding)} = \frac{(\sum V) an \delta^\prime + P_p}{P_h}=\frac{176.7 imes an 24 + 35.6}{80.1} =[/latex] 1.43
The factor of safety against overturning is adequate, but the one against sliding is not. Increasing the width of the base slab, provision of a key, or a dead-man anchor will increase the safety factors.
In this case, provision of a small key at the base of the wall will improve the factor of safety with respect to sliding.

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