Question
Download Solution PDFA shear box test was performed to give the following results for a cohesive soil sample.
|
Result: |
(1) |
(2) |
|
Normal stress σ (kN/m2) |
150 |
250 |
|
Shear stress at failure τ (kN/m2) |
110 |
120 |
The value of c and tan ϕ are;
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Coulomb's equation,
τ = C + σ tan ϕ
Where,
τ = Shear strength of soil
C = Apparent cohesion
σ = Normal stress on the plane of rupture
ϕ = Angle of internal friction
Calculation:
Given,
σ1 = 150 kN/m2 , σ2 = 250 kN/m2
τ1 = 110 kN/m2 , τ2 = 120 kN/m2
We know,
τ = C + σ tan ϕ
When, σ1 = 150 kN/m2 and τ1 = 110 kN/m2
110 = C + 150 tan ϕ........(1)
When, σ1 = 250 kN/m2 and τ1 = 120 kN/m2
120 = C + 250 tan ϕ........(2)
Equation (2) - (1)
10 = 100 tan ϕ
tan ϕ = 0.1
Put tan ϕ in equation (1),
110 = C + 150 × 0.1
C = 95 kN/m2.
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