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.

Question

A continuous foundation is shown in Figure 12.7. If the load eccentricity is 0.5 ft, determine the ultimate load, [latex]Q_{ult}[/latex], per unit length of the foundation.

Accepted Answer

With c′ = 0, Eq. (12.20) gives

[latex]q_u^\prime = q N_qF_{qs}F_{qd} F_{qi} +\frac{1}{2} \gamma B^\prime N_\gamma F_{\gamma s}F_{\gamma d}F_{\gamma i}[/latex]

q = (110)(4) = 440 lb/ft²

For φ′ = 35°, 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_q[/latex] = 33.3 and [latex]N_\gamma[/latex]= 48.03
B′ = 6 - (2)(0.5) = 5 ft

Because it is a strip foundation, B′/L′ is zero. Hence [latex]F_{qs}[/latex] = 1 and [latex]F_{\gamma s}[/latex] = 1, and
[latex]F_{qi} = F_{\gamma i} = 1[/latex]
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_{qd} = 1 + 2 an \phi^\prime (1 - \sin \phi^\prime)^2\frac{ D_f}{B} = 1 + 0.255\left(\frac{4}{6} ight)=[/latex]1.17
[latex]F_{\gamma d} =[/latex] 1
[latex]q_u^\prime =(440)(33.3)(1)(1.17)(1) +\left(\frac{1}{2} ight) (110)(5)(48.03)(1)(1)(1)=[/latex] 30,351 lb/ft²
Hence,
[latex]Q_{ult} = (B^\prime)(1)(q_u^\prime)=(5)(1)(30,351)=[/latex] 151,755 lb/ft = 75.88 ton/ft

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

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