Chapter 13
13.1 — 13.4 K o = (1 − sin φ ′)(OCR ) sin φ ′
Problem φ′ (deg) Ko Po = 12 K o γH 2 z = H /3
13.1 35 0.634 143.44 kN/m 1.67 m
13.2 33 0.627 8,607.1 lb/ft 5.67 ft
13.3 29 0.515 176.13 kN/m 2m
13.4 40 0.463 8,625.7 lb/ft 6 ft
13.5 — 13.8 K a = tan 2 (45 − φ ′ 2 )
Problem φ′ (deg) Ka σ a′ ( z = H ) = K a γ H Pa = 12 K aγ H 2 z = H /3
(0.307)(110)(14) ()(0.307)(110)(14)2
13.5 32 0.307 4.66 ft
= 472.7 lb/ft2 = 3309.4 lb/ft
(0.361)(99)(22) ()(0.361)(99)(22)2
13.6 28 0.361 2 7.33 ft
= 786.2 lb/ft = 8648.8 lb/ft
(0.248)(17.6)(5) ()(0.248)(17.6)(5)2
13.7 37 0.248 2 1.67 m
= 21.8 kN/m = 54.56 kN/m
(0.207)(19.5)(9) ()(0.207)(19.5)(9)2
13.8 41 0.207 3m
= 36.32 kN/m2 = 163.47 kN/m
Note: 1. Pressure distribution is similar to that shown in Figure 13.11a., i.e.,
σ a′ = 0 at z = 0 and σ a′ = K a γH at z = H
2. z = distance measured from the bottom of the wall
13.9 — 13.12 K p = tan2 (45 + φ′ 2)
Problem φ′ (deg) Kp σ p′ ( z = H ) = K p γ H Pp = 12 K pγ H 2 z = H /3
()(3.254)( 117)(11)2
(3.254)(117)(11)
13.9 32 3.254 3.67 ft
= 4187.9 lb/ft2
= 23,033 lb/ft
()(4.203)(101)(16)2
(4.203)(101)(16)
13.10 38 4.203 5.33 ft
= 6792 lb/ft2
= 54,336 lb/ft
()(3)(16.6)(7)2
(3)(16.6)(7)
13.11 30 3.0 2.33 m
= 348.6 kN/m2
= 1220.1 kN/m
()(2.662)(20.5)(12)2
(2.662)(20.5)(12)
13.12 27 2.662 4m
= 654.8 kN/m2
= 3929.1 kN/m
Note: 1. σ′p( z=0) = 0; triangular pressure distribution
2. z = distance measured from the bottom of the wall
109
© 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.
, ⎛ φ′ ⎞ ⎛ 28 ⎞
13.13 K a = tan 2 ⎜ 45 − ⎟ = tan 2 ⎜ 45 − ⎟ = 0.361. Refer to the figure.
⎝ 2⎠ ⎝ 2⎠
z = 0 ft: σ a′ = σ′a Ka = 0; u = 0
z = 7 ft: σ a′ = (103)(7)(0.361) = 260.28 lb/ft2; u = 0
z = 14 ft: σ a′ = [(103)(7) + (127 – 62.4)(7)](0.361) = 423.5 lb/ft2
u = (62.4)(7) = 436.8 lb/ft2
Area No. Area
1 ()(7)(260.28) = 910.98 = 910.98
2 (260.28)(7) = 1,821.96
3 ()(7)(423.5 – 260.28) = 571.27
4 ()(7)(436.8) = 1,528.8
Σ4,833 lb/ft
Resultant: Taking the moment about the bottom of the wall,
⎡ ⎛ 7⎞ ⎛7⎞ ⎛7⎞ ⎛ 7 ⎞⎤
⎢(910.98)⎜ 7 + 3 ⎟ + (1821.96)⎜ 2 ⎟ + (571.27)⎜ 3 ⎟ + (1528.8)⎜ 3 ⎟⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦
z= ⎣
4833
= 4.09 ft
110
© 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.
13.1 — 13.4 K o = (1 − sin φ ′)(OCR ) sin φ ′
Problem φ′ (deg) Ko Po = 12 K o γH 2 z = H /3
13.1 35 0.634 143.44 kN/m 1.67 m
13.2 33 0.627 8,607.1 lb/ft 5.67 ft
13.3 29 0.515 176.13 kN/m 2m
13.4 40 0.463 8,625.7 lb/ft 6 ft
13.5 — 13.8 K a = tan 2 (45 − φ ′ 2 )
Problem φ′ (deg) Ka σ a′ ( z = H ) = K a γ H Pa = 12 K aγ H 2 z = H /3
(0.307)(110)(14) ()(0.307)(110)(14)2
13.5 32 0.307 4.66 ft
= 472.7 lb/ft2 = 3309.4 lb/ft
(0.361)(99)(22) ()(0.361)(99)(22)2
13.6 28 0.361 2 7.33 ft
= 786.2 lb/ft = 8648.8 lb/ft
(0.248)(17.6)(5) ()(0.248)(17.6)(5)2
13.7 37 0.248 2 1.67 m
= 21.8 kN/m = 54.56 kN/m
(0.207)(19.5)(9) ()(0.207)(19.5)(9)2
13.8 41 0.207 3m
= 36.32 kN/m2 = 163.47 kN/m
Note: 1. Pressure distribution is similar to that shown in Figure 13.11a., i.e.,
σ a′ = 0 at z = 0 and σ a′ = K a γH at z = H
2. z = distance measured from the bottom of the wall
13.9 — 13.12 K p = tan2 (45 + φ′ 2)
Problem φ′ (deg) Kp σ p′ ( z = H ) = K p γ H Pp = 12 K pγ H 2 z = H /3
()(3.254)( 117)(11)2
(3.254)(117)(11)
13.9 32 3.254 3.67 ft
= 4187.9 lb/ft2
= 23,033 lb/ft
()(4.203)(101)(16)2
(4.203)(101)(16)
13.10 38 4.203 5.33 ft
= 6792 lb/ft2
= 54,336 lb/ft
()(3)(16.6)(7)2
(3)(16.6)(7)
13.11 30 3.0 2.33 m
= 348.6 kN/m2
= 1220.1 kN/m
()(2.662)(20.5)(12)2
(2.662)(20.5)(12)
13.12 27 2.662 4m
= 654.8 kN/m2
= 3929.1 kN/m
Note: 1. σ′p( z=0) = 0; triangular pressure distribution
2. z = distance measured from the bottom of the wall
109
© 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.
, ⎛ φ′ ⎞ ⎛ 28 ⎞
13.13 K a = tan 2 ⎜ 45 − ⎟ = tan 2 ⎜ 45 − ⎟ = 0.361. Refer to the figure.
⎝ 2⎠ ⎝ 2⎠
z = 0 ft: σ a′ = σ′a Ka = 0; u = 0
z = 7 ft: σ a′ = (103)(7)(0.361) = 260.28 lb/ft2; u = 0
z = 14 ft: σ a′ = [(103)(7) + (127 – 62.4)(7)](0.361) = 423.5 lb/ft2
u = (62.4)(7) = 436.8 lb/ft2
Area No. Area
1 ()(7)(260.28) = 910.98 = 910.98
2 (260.28)(7) = 1,821.96
3 ()(7)(423.5 – 260.28) = 571.27
4 ()(7)(436.8) = 1,528.8
Σ4,833 lb/ft
Resultant: Taking the moment about the bottom of the wall,
⎡ ⎛ 7⎞ ⎛7⎞ ⎛7⎞ ⎛ 7 ⎞⎤
⎢(910.98)⎜ 7 + 3 ⎟ + (1821.96)⎜ 2 ⎟ + (571.27)⎜ 3 ⎟ + (1528.8)⎜ 3 ⎟⎥
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦
z= ⎣
4833
= 4.09 ft
110
© 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.
