Consider a 3-m-high (H) retaining wall with a vertical back (θ = 0°) and a horizontal granular backfill. Given: γ = 15.7 kN/m³, δ′ = 15°, and $\phi ^\prime$ = 30°. Estimate the passive force, $P_p$ , by using
a. Coulomb’s theory
b. Terzaghi and Peck’s wedge theory
c. Shields and Tolunay’s solution (method of slices)
d. Zhu and Qian’s solution (method of triangular slices)

Question

Consider a 3-m-high (H) retaining wall with a vertical back (θ = 0°) and a horizontal granular backfill. Given: γ = 15.7 kN/m³, δ′ = 15°, and [latex]\phi ^\prime [/latex] = 30°. Estimate the passive force, [latex]P_p[/latex], by using
a. Coulomb’s theory
b. Terzaghi and Peck’s wedge theory
c. Shields and Tolunay’s solution (method of slices)
d. Zhu and Qian’s solution (method of triangular slices)

Accepted Answer

Part a
From Eq. (13.70),
[latex]P_p=\frac{1}{2}K_p\gamma H^2[/latex]
From Table 13.9,

[latex]K_p=\frac{\cos ^2(\phi ^\prime + heta )}{\cos ^2\cos (\delta^\prime - heta )\left[1-\sqrt{\frac{\sin (\phi ^\prime +\delta ^\prime )\sin (\phi ^\prime +\alpha )}{\cos (\delta ^\prime - heta \cos (\alpha - heta ))} } ight]^2 } [/latex] (13.71)

Table 13.9 Values of [latex]K_p[/latex] [Eq. 13.71] for θ = 0°, α = 0°
↓[latex]\phi ^\prime [/latex] (deg)	δ′ (deg)→
↓[latex]\phi ^\prime [/latex] (deg)	0	5	10	15	20
15	1.698	1.900	2.130	2.405	2.735
20	2.040	2.313	2.636	3.030	3.525
25	2.464	2.830	3.286	3.855	4.597
30	3.000	3.506	4.143	4.977	6.105
35	3.690	4.390	5.310	6.854	8.324
40	4.600	5.590	6.946	8.870	11.772

for [latex]\phi ^\prime [/latex] = 30° and δ′ = 15°, the value of [latex]K_p[/latex] is 4.977. Thus,
[latex]P = (\frac{1}{2})(4.977)(15.7)(3)^2 =[/latex] 351.6 kN/m
Part b
From Fig. 14.4, for [latex]\phi ^\prime [/latex] = 30° and δ′ = 15°, the value of [latex]K_p[/latex] is about 4.53. Thus,
[latex]P_p= (\frac{1}{2})(4.53)(15.7)(3)^2 =[/latex] 320 kN/m
Part c
[latex]P_p=\frac{1}{2}K_p\gamma H^2[/latex]
From Figure 14.5, for [latex]\phi ^\prime [/latex] = 30° and δ′ = 15° (i.e [latex]\frac{\delta^\prime}{\phi^\prime}=0.5[/latex]) the value of [latex]K_p[/latex] is 4.13. Hence,
[latex]P_p= \left(\frac{1}{2} ight)(4.13)(15.7)(3)^2 =[/latex] 292 kN/m
Part d
From Eq. (14.12),
[latex]K_p = K_{p(\delta^\prime= 0)}R[/latex]
For [latex]\phi ^\prime [/latex] = 30° and θ = 0, [latex]K_{p(\delta^\prime= 0)}[/latex] is equal to 3.0 (Table 14.1).

Table 14.1 Variation of [latex]K_{p(\delta^\prime= 0)}[/latex] [see Eq. (14.12) and Figure 14.3a]*
[latex]\phi ^\prime [/latex] (deg)	θ (deg)
[latex]\phi ^\prime [/latex] (deg)	30	25	20	15	10	5	0
20	1.70	1.69	1.72	1.77	1.83	1.92	2.04
21	1.74	1.73	1.76	1.81	1.89	1.99	2.12
22	1.77	1.77	1.80	1.87	1.95	2.06	2.20
23	1.81	1.81	1.85	1.92	2.01	2.13	2.28
24	1.84	1.85	1.90	1.97	2.07	2.21	2.37
25	1.88	1.89	1.95	2.03	2.14	2.28	2.46
26	1.91	1.93	1.99	2.09	2.21	2.36	2.56
27	1.95	1.98	2.05	2.15	2.28	2.45	2.66
28	1.99	2.02	2.10	2.21	2.35	2.54	2.77
29	2.03	2.07	2.15	2.27	2.43	2.63	2.88
30	2.07	2.11	2.21	2.34	2.51	2.73	3.00
31	2.11	2.16	2.27	2.41	2.60	2.83	3.12
32	2.15	2.21	2.33	2.48	2.68	2.93	3.25
33	2.20	2.26	2.39	2.56	2.77	3.04	3.39
34	2.24	2.32	2.45	2.64	2.87	3.16	3.53
35	2.29	2.37	2.52	2.72	2.97	3.28	3.68
36	2.33	2.43	2.59	2.80	3.07	3.41	3.84
37	2.38	2.49	2.66	2.89	3.18	3.55	4.01
38	2.43	2.55	2.73	2.98	3.29	3.69	4.19
39	2.48	2.61	2.81	3.07	3.41	3.84	4.38
40	2.53	2.67	2.89	3.17	3.53	4.00	4.59
41	2.59	2.74	2.97	3.27	3.66	4.16	4.80
42	2.64	2.80	3.05	3.38	3.80	4.34	5.03
43	2.70	2.88	3.14	3.49	3.94	4.52	5.27
44	2.76	2.94	3.23	3.61	4.09	4.72	5.53
45	2.82	3.02	3.32	3.73	4.25	4.92	5.80

Again, for θ = 0 and δ′/[latex]\phi ^\prime [/latex] = 0.5, the value of R is about 1.52. Thus, [latex]K_p[/latex] = (3)(1.52) = 4.56.
[latex]P_p= (\frac{1}{2})[/latex](4.56(15.7)(3)² = 322 kN/m

Question 14.1: Consider a 3-m-high (H) retaining wall with a vertical back ...

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