Chapter 11
1 − μs2
11.1 Se( flexible, center ) = Δσ (αB′) IsI f
Es
355
Δσ = = 59.16 kN/m 2
( 2)( 3)
2
Given: α = 4; B′ = = 1; μs = 0.35; Es = 13,500 kN/m2
2
1 − 2μs
I s = F1 + F2
1 − μs
L 3 H 4
m′ = = = 1.5; n′ = = =4
B 2 ⎛ B⎞ ⎛2⎞
⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝2⎠
From Table 11.1, F1 = 0.454; from Table 11.2, F2 = 0.054.
1 − ( 2)(0.4)
I s = 0.454 + (0.054) = 0.472
1 − 0 .4
Df 1.5 L
Also, with = = 0.75 and = 1.5, Table 11.3 gives If = 0.765. Hence,
B 2 B
⎛ 1 − 0.4 2 ⎞
S e ( flexible, center ) = (59.16)[(4)(1)]⎜⎜ ⎟⎟(0.472)(0.765) = 0.0053 m = 5.3 mm
⎝ 13500 ⎠
Se(rigid) = (0.93)(5.3) ≈ 4.93 mm
1 − μ s2
11.2 As in Problem 11.1, S e ( rigid ) = 0.93Δσ (αB ′) IsI f
Es
∑ E s ( i ) Δz ( 3000 )( 6) + (1100 )(8) + (8500 )(10 )
Es = = = 4658 lb/in 2 = 670,752 lb/ft 2
z 24
Given: B = L = 6 ft; μs = 0.3; α = 4
83
© 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.
, 100000
Δσ = = 2778 lb/ft 2
( 6)( 6)
B 6
B′ = = = 3 ft.
2 2
1 − 2μs
I s = F1 + F2
1 − μs
L H 24
m′ = = 1; n′ = = =8
B ⎛ B⎞ ⎛6⎞
⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝2⎠
From Table 11.1, F1 = 0.482; from Table 11.2, F2 = 0.02
1 − ( 2)(0.3)
I s = 0.482 + (0.02) = 0.493
1 − 0 .3
Df 3
Also, = = 0.5. From Table 11.3, If ≈ 0.77. So,
B 6
⎛ 1 − 0.3 2 ⎞
S e ( rigid ) = (0.93)(2778)(4 × 3)⎜⎜ ⎟⎟(0.493)(0.77) = 0.016 ft ≈ 0.2 in
⎝ 670752 ⎠
11.3 a. The plot of e vs. σ′ is shown.
84
© 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.
, b. σ c′ = 170 kN/m2
e1 − e2 0.755 − 0.65
c. C c = = ≈ 0.35
⎛ σ 2′ ⎞ ⎛ 16 ⎞
log⎜⎜ ⎟⎟ log⎜ ⎟
⎝ σ 1′ ⎠ ⎝8⎠
e1 − e2 0.658 − 0.65
Cs = = = 0.026
⎛ σ 2′ ⎞ ⎛ 16 ⎞
log⎜⎜ ⎟⎟ log⎜ ⎟
σ
⎝ 1⎠ ′ ⎝8⎠
C s 0.026
= = 0.074
Cc 0.35
Ws 12 g
11.4 a. Height of solids: H s = = = 0.152 cm ≈ 0.06 in.
AG sγ w (4.91)(2.5 4) 2 ( 2.49 )(1)
Change in Final Hv
σ′ dial reading height, H Hs Hv = H – Hs e=
Hs
(ton/ft2) (in.) (in.) (in.) (in.)
0.063 0.0112 0.8013 0.06 0.7413 12.35
0.125 0.0059 0.7954 0.06 0.7354 12.25
0.250 0.0124 0.7830 0.06 0.723 12.05
0.500 0.0222 0.7608 0.06 0.7008 11.68
1.000 0.0324 0.7284 0.06 0.6684 11.14
2.000 0.0886 0.6398 0.06 0.5798 9.66
4.000 0.2105 0.4293 0.06 0.3693 6.15
The e-log σ′ graph is plotted on the following page.
85
© 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.
1 − μs2
11.1 Se( flexible, center ) = Δσ (αB′) IsI f
Es
355
Δσ = = 59.16 kN/m 2
( 2)( 3)
2
Given: α = 4; B′ = = 1; μs = 0.35; Es = 13,500 kN/m2
2
1 − 2μs
I s = F1 + F2
1 − μs
L 3 H 4
m′ = = = 1.5; n′ = = =4
B 2 ⎛ B⎞ ⎛2⎞
⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝2⎠
From Table 11.1, F1 = 0.454; from Table 11.2, F2 = 0.054.
1 − ( 2)(0.4)
I s = 0.454 + (0.054) = 0.472
1 − 0 .4
Df 1.5 L
Also, with = = 0.75 and = 1.5, Table 11.3 gives If = 0.765. Hence,
B 2 B
⎛ 1 − 0.4 2 ⎞
S e ( flexible, center ) = (59.16)[(4)(1)]⎜⎜ ⎟⎟(0.472)(0.765) = 0.0053 m = 5.3 mm
⎝ 13500 ⎠
Se(rigid) = (0.93)(5.3) ≈ 4.93 mm
1 − μ s2
11.2 As in Problem 11.1, S e ( rigid ) = 0.93Δσ (αB ′) IsI f
Es
∑ E s ( i ) Δz ( 3000 )( 6) + (1100 )(8) + (8500 )(10 )
Es = = = 4658 lb/in 2 = 670,752 lb/ft 2
z 24
Given: B = L = 6 ft; μs = 0.3; α = 4
83
© 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.
, 100000
Δσ = = 2778 lb/ft 2
( 6)( 6)
B 6
B′ = = = 3 ft.
2 2
1 − 2μs
I s = F1 + F2
1 − μs
L H 24
m′ = = 1; n′ = = =8
B ⎛ B⎞ ⎛6⎞
⎜ ⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝2⎠
From Table 11.1, F1 = 0.482; from Table 11.2, F2 = 0.02
1 − ( 2)(0.3)
I s = 0.482 + (0.02) = 0.493
1 − 0 .3
Df 3
Also, = = 0.5. From Table 11.3, If ≈ 0.77. So,
B 6
⎛ 1 − 0.3 2 ⎞
S e ( rigid ) = (0.93)(2778)(4 × 3)⎜⎜ ⎟⎟(0.493)(0.77) = 0.016 ft ≈ 0.2 in
⎝ 670752 ⎠
11.3 a. The plot of e vs. σ′ is shown.
84
© 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.
, b. σ c′ = 170 kN/m2
e1 − e2 0.755 − 0.65
c. C c = = ≈ 0.35
⎛ σ 2′ ⎞ ⎛ 16 ⎞
log⎜⎜ ⎟⎟ log⎜ ⎟
⎝ σ 1′ ⎠ ⎝8⎠
e1 − e2 0.658 − 0.65
Cs = = = 0.026
⎛ σ 2′ ⎞ ⎛ 16 ⎞
log⎜⎜ ⎟⎟ log⎜ ⎟
σ
⎝ 1⎠ ′ ⎝8⎠
C s 0.026
= = 0.074
Cc 0.35
Ws 12 g
11.4 a. Height of solids: H s = = = 0.152 cm ≈ 0.06 in.
AG sγ w (4.91)(2.5 4) 2 ( 2.49 )(1)
Change in Final Hv
σ′ dial reading height, H Hs Hv = H – Hs e=
Hs
(ton/ft2) (in.) (in.) (in.) (in.)
0.063 0.0112 0.8013 0.06 0.7413 12.35
0.125 0.0059 0.7954 0.06 0.7354 12.25
0.250 0.0124 0.7830 0.06 0.723 12.05
0.500 0.0222 0.7608 0.06 0.7008 11.68
1.000 0.0324 0.7284 0.06 0.6684 11.14
2.000 0.0886 0.6398 0.06 0.5798 9.66
4.000 0.2105 0.4293 0.06 0.3693 6.15
The e-log σ′ graph is plotted on the following page.
85
© 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.
