Chapter 16
16.1 φ′ = 32º; Nc = 44.14; Nq = 28.52; Nγ = 26.87 (Table 16.1)
qu 1 ⎛ 1 ⎞
qall = = ⎜ c ′N c + qN q + γBN γ ⎟
Fs 3 ⎝ 2 ⎠
1⎡ 1 ⎤
= ⎢ ( 21)( 44.14) + (1)(17.5)( 28.52) + (17.5)(1.5)( 26.87)⎥
3⎣ 2 ⎦
= 593 kN/m2
16.2 φ′ = 24º; Nc = 23.36; Nq = 11.40; Nγ = 7.08 (Table 16.1)
qu 1 ⎛ 1 ⎞
qall = = ⎜ c ′N c + q N q + γ BNγ ⎟
Fs 4 ⎝ 2 ⎠
1⎡ 1 ⎤
= ⎢ (1500)( 23.36) + (118)( 4)(11.40) + (6)(118)(7.08)⎥
4⎣ 2 ⎦
= 10,732 lb/ft 2
16.3 φ′ = 0º; Nc = 5.7; Nq = 1; Nγ = 0 (Table 16.1)
= (cu Nc + qN q ) = [(37)(5.7) + (19.5)(0.75)(1)] = 37.58 kN/m 2
qu 1 1
qall =
Fs 6 6
16.4 For a continuous foundation with vertical loading, all inclination factors and
shape factors are equal to one. So,
qu 1 ⎛ 1 ⎞
q all = = ⎜ c ′N c λcd + qN q λ qd + γBN γ λγd ⎟
Fs Fs ⎝ 2 ⎠
φ′ = 32º; Nc = 35.49; Nq = 23.18; Nγ = 30.22 (Table 16.2)
⎛ Df ⎞
⎟⎟ = 1 + 0.4⎛⎜ ⎞⎟ = 1.266
1
λcd = 1 + 0.4⎜⎜
⎝ B ⎠ ⎝ 1.5 ⎠
141
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, Df 2⎛ 1 ⎞
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 = 1 + 2 tan 32(1 − sin 32) ⎜ ⎟ = 1.184
B ⎝ 1.5 ⎠
λ γd = 1
⎡(21)(35.49)(1.266) + (17.5)(1)(23.18)(1.184)⎤
1⎢ ⎥ = 606.8 kN/m 2
q all = 1
3⎢ + (17.5)(1.5)(30.22)(1) ⎥
⎣ 2 ⎦
16.5 φ′ = 24º; Nc = 19.32; Nq = 9.60; Nγ = 9.44
⎛4⎞
λcd = 1 + 0.4⎜ ⎟ = 1.266
⎝6⎠
Df 2⎛ 4⎞
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 = 1 + 2 tan 24(1 − sin 24) ⎜ ⎟ = 1.209
B ⎝6⎠
λ γd = 1
⎡(1500)(19.32)(1.266) + ( 4)(118)(9.60)(1.209)⎤
1⎢ ⎥
qall = ⎢ 1 ⎥ = 11,377 lb/ft 2
4 + (118)(6)(9.44)(1)
⎢⎣ 2 ⎥⎦
1 ⎛ 1 ⎞
16.6 q all = ⎜ c ′N c λcd + qN q λ qd + γBN γ λγd ⎟
Fs ⎝ 2 ⎠
φ′ = 0º; Nc = 5.14; Nq = 1.0; Nγ = 0
⎛ Df ⎞
⎟⎟ = 1 + 0.4⎛⎜
0.75 ⎞
λcd = 1 + 0.4⎜⎜ ⎟ = 1.12
⎝ B ⎠ ⎝ 2.5 ⎠
Df
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 =1
B
λ γd = 1
1
q all = [(37)(5.14)(1.12) + (0.75)(19.5) + 0] = 37.94 kN/m 2
6
142
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16.1 φ′ = 32º; Nc = 44.14; Nq = 28.52; Nγ = 26.87 (Table 16.1)
qu 1 ⎛ 1 ⎞
qall = = ⎜ c ′N c + qN q + γBN γ ⎟
Fs 3 ⎝ 2 ⎠
1⎡ 1 ⎤
= ⎢ ( 21)( 44.14) + (1)(17.5)( 28.52) + (17.5)(1.5)( 26.87)⎥
3⎣ 2 ⎦
= 593 kN/m2
16.2 φ′ = 24º; Nc = 23.36; Nq = 11.40; Nγ = 7.08 (Table 16.1)
qu 1 ⎛ 1 ⎞
qall = = ⎜ c ′N c + q N q + γ BNγ ⎟
Fs 4 ⎝ 2 ⎠
1⎡ 1 ⎤
= ⎢ (1500)( 23.36) + (118)( 4)(11.40) + (6)(118)(7.08)⎥
4⎣ 2 ⎦
= 10,732 lb/ft 2
16.3 φ′ = 0º; Nc = 5.7; Nq = 1; Nγ = 0 (Table 16.1)
= (cu Nc + qN q ) = [(37)(5.7) + (19.5)(0.75)(1)] = 37.58 kN/m 2
qu 1 1
qall =
Fs 6 6
16.4 For a continuous foundation with vertical loading, all inclination factors and
shape factors are equal to one. So,
qu 1 ⎛ 1 ⎞
q all = = ⎜ c ′N c λcd + qN q λ qd + γBN γ λγd ⎟
Fs Fs ⎝ 2 ⎠
φ′ = 32º; Nc = 35.49; Nq = 23.18; Nγ = 30.22 (Table 16.2)
⎛ Df ⎞
⎟⎟ = 1 + 0.4⎛⎜ ⎞⎟ = 1.266
1
λcd = 1 + 0.4⎜⎜
⎝ B ⎠ ⎝ 1.5 ⎠
141
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
, Df 2⎛ 1 ⎞
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 = 1 + 2 tan 32(1 − sin 32) ⎜ ⎟ = 1.184
B ⎝ 1.5 ⎠
λ γd = 1
⎡(21)(35.49)(1.266) + (17.5)(1)(23.18)(1.184)⎤
1⎢ ⎥ = 606.8 kN/m 2
q all = 1
3⎢ + (17.5)(1.5)(30.22)(1) ⎥
⎣ 2 ⎦
16.5 φ′ = 24º; Nc = 19.32; Nq = 9.60; Nγ = 9.44
⎛4⎞
λcd = 1 + 0.4⎜ ⎟ = 1.266
⎝6⎠
Df 2⎛ 4⎞
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 = 1 + 2 tan 24(1 − sin 24) ⎜ ⎟ = 1.209
B ⎝6⎠
λ γd = 1
⎡(1500)(19.32)(1.266) + ( 4)(118)(9.60)(1.209)⎤
1⎢ ⎥
qall = ⎢ 1 ⎥ = 11,377 lb/ft 2
4 + (118)(6)(9.44)(1)
⎢⎣ 2 ⎥⎦
1 ⎛ 1 ⎞
16.6 q all = ⎜ c ′N c λcd + qN q λ qd + γBN γ λγd ⎟
Fs ⎝ 2 ⎠
φ′ = 0º; Nc = 5.14; Nq = 1.0; Nγ = 0
⎛ Df ⎞
⎟⎟ = 1 + 0.4⎛⎜
0.75 ⎞
λcd = 1 + 0.4⎜⎜ ⎟ = 1.12
⎝ B ⎠ ⎝ 2.5 ⎠
Df
λqd = 1 + 2 tan φ ′(1 − sin φ ′)2 =1
B
λ γd = 1
1
q all = [(37)(5.14)(1.12) + (0.75)(19.5) + 0] = 37.94 kN/m 2
6
142
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