Question

(a)
Two separate direct shear tests were conducted on soil samples with the shear box having dimensions of 80 mm × 80 mm on plan view. Assuming this cross-sectional area remains unchanged, determine the cohesion intercept (c’) and friction angle (φ’) of the soil based on the following results obtained at failure.
Direct shear test results
Test Normal force (N) Shear force (N)
1 210 257
2 360 372
(b)
A CU triaxial test was conducted on a specimen of sand (assume c’=0). The cell pressure was 450 kPa, and the back pressure was 200 kPa. During the compression stage, the pore water pressure became 263 kPa at failure. Also, when the specimen failed, the applied axial force was 880 N, and the cross-sectional area of the specimen was 24.35 cm2. Using such information, estimate the friction angle (φ’) of the sand.

Answer 1

Normal Force	Shear force	Area	Normal stress, σ = Normal force/ area	Shear stress, τ = Shear force /area
N	N	mm^2	kPa	kPa
210	257	6400	32.81	40.16
360	372	6400	56.25	58.13

τ = 0.7667*σ + 15 Comparing with τ = tanφ'*σ + c'

c' = 15 kPa

tan φ' = 0.7667

φ' = 37.47 degree

b.

σ3 = 450 kPa

σd = σ1-σ3 = 880/24.35/10000*1000 = 361.4 kPa

σ 1 = 361.4+450 = 811.4 kPa

σ1 = σ3*tan(45+φ'/2)2   + 2*c*tan(45+φ/2)

c= 0

σ1 = σ3*tan(45+φ'/2)2  

811.4 = 450 *tan(45+φ'/2)2

φ' = 16.65 degree

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