Chapter 12
12.1 a. c′ = 0. From Eq. (12.3): τf = σ′ tan φ′
300
τ= 2
= 75 kN/m 2
(1000)(0.063)
So, 75 = 105 tan φ′
⎛ 75 ⎞
φ ′ = tan −1 ⎜ ⎟ = 35.5°
⎝ 105 ⎠
b. For σ′ = 180 kN/m2, τf = 180 tan 35.5° = 128.39 kN/m2
Shear force, S = (128.39)(1000)(0.063) 2 = 509.5 N
12.2 The point O (180, 128.4) represents the failure stress conditions on the Mohr-
Coulomb failure envelope. The perpendicular line OC to the failure envelope
determines the center, C of the Mohr’s circle. With the center at C, and the radius
as OC, the Mohr’s circle is drawn by trial and error such that the circle is tangent
to the failure envelope at O. From the graph,
99
© 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.
, a. σ 3′ ≈ 115 kN/m 2 ; σ 1′ ≈ 420 kN/m 2
b. The horizontal line OP drawn from O determines the pole P. Therefore, the
orientation or the major principal plane with the horizontal is given by the
angle α ≈ 65°.
12.3 For σ′ = 28 lb/in2, τf = 28 tan 33° = 18.18 lb/in2
Shear force, S = (18.18)(2.5) 2 = 113.65 lb
⎛π ⎞
12.4 Area of specimen A = ⎜ ⎟( 2) 2 = 3.14 in.2
⎝4⎠
N S ⎛τ f ⎞
Test Normal σ′= Shear τf = φ ′ = tan −1 ⎜⎜ ⎟⎟
No. force N A force S A ⎝σ′⎠
(lb) (lb/in.2) (lb) (lb/in.2) (deg)
1 15 4.77 5.25 1.67 19.29
2 30 9.55 10.5 3.34 19.27
3 48 15.28 16.8 5.35 19.29
4 83 26.43 29.8 9.5 19.77
A graph of τf vs. σ′ will yield φ′ = 19.5º.
⎛π ⎞
12.5 Area of specimen A = ⎜ ⎟(0.05) 2 = 0.00196 m 2
⎝4⎠
N S ⎛τ f ⎞
Test Normal σ′ = Shear τf = φ ′ = tan −1 ⎜⎜ ⎟⎟
No. force N A force S A ⎝σ′⎠
(N) (N/m2) (N) (N/m2) (deg)
1 250 79.6 139 44.26 29.07
2 375 119.4 209 66.56 29.13
3 450 143.3 250 79.61 29.05
4 540 171.9 300 95.54 29.06
A graph of τf vs. σ′ will yield φ′ ≈ 29º.
⎛ φ′ ⎞
12.6 c′ = 0. From Eq. (12.8): σ 1′ = σ 3′ tan 2 ⎜ 45 + ⎟; φ ′ = 30°
⎝ 2⎠
100
© 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.
12.1 a. c′ = 0. From Eq. (12.3): τf = σ′ tan φ′
300
τ= 2
= 75 kN/m 2
(1000)(0.063)
So, 75 = 105 tan φ′
⎛ 75 ⎞
φ ′ = tan −1 ⎜ ⎟ = 35.5°
⎝ 105 ⎠
b. For σ′ = 180 kN/m2, τf = 180 tan 35.5° = 128.39 kN/m2
Shear force, S = (128.39)(1000)(0.063) 2 = 509.5 N
12.2 The point O (180, 128.4) represents the failure stress conditions on the Mohr-
Coulomb failure envelope. The perpendicular line OC to the failure envelope
determines the center, C of the Mohr’s circle. With the center at C, and the radius
as OC, the Mohr’s circle is drawn by trial and error such that the circle is tangent
to the failure envelope at O. From the graph,
99
© 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.
, a. σ 3′ ≈ 115 kN/m 2 ; σ 1′ ≈ 420 kN/m 2
b. The horizontal line OP drawn from O determines the pole P. Therefore, the
orientation or the major principal plane with the horizontal is given by the
angle α ≈ 65°.
12.3 For σ′ = 28 lb/in2, τf = 28 tan 33° = 18.18 lb/in2
Shear force, S = (18.18)(2.5) 2 = 113.65 lb
⎛π ⎞
12.4 Area of specimen A = ⎜ ⎟( 2) 2 = 3.14 in.2
⎝4⎠
N S ⎛τ f ⎞
Test Normal σ′= Shear τf = φ ′ = tan −1 ⎜⎜ ⎟⎟
No. force N A force S A ⎝σ′⎠
(lb) (lb/in.2) (lb) (lb/in.2) (deg)
1 15 4.77 5.25 1.67 19.29
2 30 9.55 10.5 3.34 19.27
3 48 15.28 16.8 5.35 19.29
4 83 26.43 29.8 9.5 19.77
A graph of τf vs. σ′ will yield φ′ = 19.5º.
⎛π ⎞
12.5 Area of specimen A = ⎜ ⎟(0.05) 2 = 0.00196 m 2
⎝4⎠
N S ⎛τ f ⎞
Test Normal σ′ = Shear τf = φ ′ = tan −1 ⎜⎜ ⎟⎟
No. force N A force S A ⎝σ′⎠
(N) (N/m2) (N) (N/m2) (deg)
1 250 79.6 139 44.26 29.07
2 375 119.4 209 66.56 29.13
3 450 143.3 250 79.61 29.05
4 540 171.9 300 95.54 29.06
A graph of τf vs. σ′ will yield φ′ ≈ 29º.
⎛ φ′ ⎞
12.6 c′ = 0. From Eq. (12.8): σ 1′ = σ 3′ tan 2 ⎜ 45 + ⎟; φ ′ = 30°
⎝ 2⎠
100
© 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.
