Consider a shallow foundation with the following:
● Foundation: B = 3 ft
L = 5 ft
$D_f$ = 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.

Question

Consider a shallow foundation with the following:
● Foundation: B = 3 ft
L = 5 ft
[latex]D_f[/latex] = 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.

Accepted Answer

From Eq. (12.7), noting that [latex]F_{ci} = F_{qi} = F_{\gamma i} = 1[/latex]
[latex]q_u = c^\prime N_cF_{cs}F_{cd} + qN_qF_{qs}F_{qd} +\frac{1}{2} \gamma BN_\gamma F_{\gamma s}F_{\gamma q}[/latex]
For φ′ = 25°, 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[/latex] = 20.72
[latex]N_q[/latex] = 10.66
[latex]N_\gamma[/latex] = 10.88
[latex]q = \gamma D_f = (110)(3) =[/latex] 330 lb/ft²
From Table 12.2,

Factor	Relationship	Source
Shape*	[latex]F_{cs} = 1 +\frac{B}{L}\frac{ N_q}{ N_c}[/latex]	De Beer (1970)
	[latex]F_{qs} = 1 +\frac{B}{L} an \phi^\prime[/latex]
	[latex]F_{\gamma s} = 1 - 0.4\frac{B}{L}[/latex]
	where L = length of the foundation (L > B)
Depth[latex]{}^†[/latex]	Condition (a): [latex]Df/B \leq 1[/latex]	Hansen (1970)
	[latex]F_{cd} = 1 + 0.4\frac{D_f}{B}[/latex]
	[latex]F_{qd}=1+2 an\phi^\prime(1-\sin \phi^\prime)^2\frac{D_f}{B}[/latex]
	[latex]F_{\gamma d} = 1[/latex]
	Condition (b): [latex]D_f /B \gt 1[/latex]
	[latex]F_{cd} = 1 +( 0.4) an^{-1}\left(\frac{D_f}{B} ight)[/latex]
	[latex]F_{qd}=1+2 an\phi^\prime(1-\sin \phi^\prime)^2 an^{-1}\left(\frac{D_f}{B} ight)[/latex]
	[latex]F_{\gamma d} = 1[/latex]
Inclination	[latex]F_{ci}=F_{qi}=\left(1-\frac{\beta^\circ}{90^\circ} ight)^2[/latex]	Meyerhof (1963); Hanna and Meyerhof (1981)
	[latex]F_{\gamma i}=\left(1-\frac{\beta}{\phi^\prime} ight)^2[/latex]
	where β = inclination of the load on the foundation with respect to the vertical

[latex]F_{cs} = 1 +\frac{B N_q}{L N_c}= 1 + \left(\frac{3}{5} ight)\left(\frac{10.66}{20.72} ight) =[/latex] 1.309
[latex]F_{qs} = 1 +\frac{B}{L} an \phi^\prime=1+\left(\frac{3}{5} ight) an 25 =[/latex] 1.28
[latex]F_{\gamma s} = 1 - 0.4\frac{B}{L}=1-(0.4)\left(\frac{3}{5} ight)=[/latex] 0.76
[latex]F_{cd} = 1 + 0.4\frac{D_f}{B}= 1 + (0.4)\left(\frac{3}{3} ight)=[/latex] 1.4
[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\left(\frac{3}{3} ight) =[/latex] 1.311
[latex]F_{\gamma d} = 1[/latex]
[latex]q_u = (400)(20.72)(1.309)(1.4) + (330)(10.66)(1.28)(1.311) +\frac{1}{2}(110)(3)(10.88)(0.76)(1)[/latex]
= 15,188.6 + 5903.1 + 1364.4 = 22,456.1 lb/ft²
[latex]q_{all} =\frac{q_u}{FS} = \frac{22,456.1}{4}[/latex] ≈ 5614 lb
[latex]Q_{all} = (q_{all})(BL) = (5614)(3 imes 5) =[/latex] 84,210 lb≈ 84.2 kip

Question 12.1: Consider a shallow foundation with the following: ● Foundati...

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