Chapter 14
14.1 Pp = Kpγ H 2 ; Kp = Kp(δ′ = 0)R. With φ′ = 30º, θ = 15º, and α = 0, the value of
Kp(δ′ = 0) = 2.34 (Table 14.2). With θ = 15º, δ′ = 18º, δ′/φ′ = 18/30 = 0.6, the value
of R is 1.55 (Table 14.3). So,
1
Pp = ( 2.34 × 1.55)(17.8)(5) 2 ≈ 807 kN/m
2
14.2 Pp = Kpγ H 2. From Figure 14.5, for φ′ = 35º and δ′ = 23.33º, δ′/φ′ = 0.666, the
value of Kp is about 6.2. So,
1
Pp = (6.2)(112)(14) 2 = 68,051 lb/ft
2
14.3 Pp = Kpγ H 2. From Figure 14.4, for φ′ = 30º and δ′ = 20º, the value of Kp is
about 5.5. So,
1
Pp = (5.5)(19)( 4) 2 = 836 kN/m
2
14.4 From Table 14.2, for φ′ = 30º and θ = 0, the value of Kp(δ′ = 0) = 3.0. For θ = 0 and
δ′/φ′ = 20/30 = 0.667, the value of R is about 1.75. So
1
Pp = (3 × 1.75)(19)( 4) 2 ≈ 798 kN/m
2
14.5 From Equation (14.15): Pp = γ H 2Kp
α 10
Step 1: = = 0.333
φ ′ 30
121
© 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.
, Step 2: From Figure 14.7: K p (δ ′=φ ′) ≈ 8.5
δ ′ 18
Step 3: = = 0.6
φ ′ 30
Step 4: From Table 14.4: R′ = 0.811
Step 5: K p = ( R′)[K p (δ ′=φ ′) ] = (0.811)(8.5) = 6.893
1
Pp = (16.8)( 4.75) 2 (6.893) ≈ 1306 kN/m
2
⎡1 ⎤ 1
14.6 Eq. (14.16): Ppe = ⎢ γ H 2 K pγ ( e ) ⎥
⎣2 ⎦ cos δ ′
For kv = 0, kh = 0.25, δ′/φ′ = 15/30 = 0.5, the value of Kpγ (e) ≈ 3.95.
[
Ppe = (0.5)(118)(15) 2 (3.95) ] cos115 = 54,286 lb/ft
2.75 m Pa
14.7 na = = 0.5 . φ′ = 40º; δ′ = 15º. Table 14.5: = 0.216
5 .5 m 0.5γH 2
Pa = (0.216)(0.5)(15.8)(5.5)2 = 51.61 kN/m
6.5 ft c′ 255
14.8 na = ≈ 0 .3 ; = = 0.1
21 ft γH (121)(21)
Pa
From Table 14.6 for φ′ = 25º and δ′ = 15º, = 0.085 .
0.5γH 2
Pa = (0.085)(0.5)(121)(21)2 ≈ 2268 lb/ft
122
© 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.
14.1 Pp = Kpγ H 2 ; Kp = Kp(δ′ = 0)R. With φ′ = 30º, θ = 15º, and α = 0, the value of
Kp(δ′ = 0) = 2.34 (Table 14.2). With θ = 15º, δ′ = 18º, δ′/φ′ = 18/30 = 0.6, the value
of R is 1.55 (Table 14.3). So,
1
Pp = ( 2.34 × 1.55)(17.8)(5) 2 ≈ 807 kN/m
2
14.2 Pp = Kpγ H 2. From Figure 14.5, for φ′ = 35º and δ′ = 23.33º, δ′/φ′ = 0.666, the
value of Kp is about 6.2. So,
1
Pp = (6.2)(112)(14) 2 = 68,051 lb/ft
2
14.3 Pp = Kpγ H 2. From Figure 14.4, for φ′ = 30º and δ′ = 20º, the value of Kp is
about 5.5. So,
1
Pp = (5.5)(19)( 4) 2 = 836 kN/m
2
14.4 From Table 14.2, for φ′ = 30º and θ = 0, the value of Kp(δ′ = 0) = 3.0. For θ = 0 and
δ′/φ′ = 20/30 = 0.667, the value of R is about 1.75. So
1
Pp = (3 × 1.75)(19)( 4) 2 ≈ 798 kN/m
2
14.5 From Equation (14.15): Pp = γ H 2Kp
α 10
Step 1: = = 0.333
φ ′ 30
121
© 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.
, Step 2: From Figure 14.7: K p (δ ′=φ ′) ≈ 8.5
δ ′ 18
Step 3: = = 0.6
φ ′ 30
Step 4: From Table 14.4: R′ = 0.811
Step 5: K p = ( R′)[K p (δ ′=φ ′) ] = (0.811)(8.5) = 6.893
1
Pp = (16.8)( 4.75) 2 (6.893) ≈ 1306 kN/m
2
⎡1 ⎤ 1
14.6 Eq. (14.16): Ppe = ⎢ γ H 2 K pγ ( e ) ⎥
⎣2 ⎦ cos δ ′
For kv = 0, kh = 0.25, δ′/φ′ = 15/30 = 0.5, the value of Kpγ (e) ≈ 3.95.
[
Ppe = (0.5)(118)(15) 2 (3.95) ] cos115 = 54,286 lb/ft
2.75 m Pa
14.7 na = = 0.5 . φ′ = 40º; δ′ = 15º. Table 14.5: = 0.216
5 .5 m 0.5γH 2
Pa = (0.216)(0.5)(15.8)(5.5)2 = 51.61 kN/m
6.5 ft c′ 255
14.8 na = ≈ 0 .3 ; = = 0.1
21 ft γH (121)(21)
Pa
From Table 14.6 for φ′ = 25º and δ′ = 15º, = 0.085 .
0.5γH 2
Pa = (0.085)(0.5)(121)(21)2 ≈ 2268 lb/ft
122
© 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.
