Chapter 15 Lateral Earth Pressure, Curved Failure Surface

Chapter 15

15.1 Eq. (15.15):

c′ tan φ ′
Fs = +
γ H cos β tan β
2
tan β

31 tan 28
2.75 = 2
+
(17.8)( H )(cos 25)(tan 25) tan 25
4.546
2.75 = + 1.14
H

H = 2.82 m


15.2 Eq. (15.16):

c′ 1 300 1
H cr = = = 17.97 ft
γ cos β (tan β − tan φ ′) 115 (cos 30)(tan 30 − tan 21)
2 2




15.3 γ′ = 19.2 – 9.81 = 9.39 kN/m3

Eq. (15.28):

c′ γ ′ tan φ ′ 46 9.39 tan 22
Fs = + = +
γ sat H cos β tan β γ sat tan β (19.2)(11)(cos 18) (tan 18) 19.2 tan 18
2 2


= 0.741 + 0.608 ≈ 1.35


(G s + e)γ w (2.73 + 0.69)(62.4)
15.4 γ sat = = = 126.27 lb/ft 3
1+ e 1 + 0.69

γ′ = 126.27 – 62.4 = 63.87 lb/ft3




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, c′ γ ′ tan φ ′
Fs = +
γ sat H cos β tan β γ sat tan β
2


1000 63.87 tan 18
= 2
+
(126.27)(27)(cos 28)(tan 28) 126.27 tan 28
= 0.707 + 0.309 ≈ 1.016


15.5 a. Eq. (15.15):

c′ tan φ ′
Fs = +
γ H cos β tan β
2
tan β
21 tan 26
Fs = +
⎡ (1950)(9.81) ⎤ tan 18
⎢⎣ ⎥ (5)(cos 2 18)(tan 18)
1000 ⎦
≈ 2.25

b. Eq. (15.15):

c′ tan φ ′
Fs = +
γ H cos β tan β
2
tan β
21 tan 26
1.75 = +
⎡ (1950)(9.81) ⎤ 2 tan 27
⎢⎣ 1000 ⎥⎦( H )(cos 27)(tan 27)

2.713
1.75 = + 0.957
H

H = 3.42 m


15.6 Consider a 1-ft length of the wedge ABC




128
© 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.