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A consolidated-drained triaxial test is carried out on a sand specimen that is subjected to 80 kN/m2 confining pressure. The vertical deviator stress was increased slowly such that there is no built-up of pore water pressure within the specimen. The specimen failed when the addition axial stress reached 240 kN/m2. Find the friction angle of the sand. If another identical sand specimen was subjected to 150 kN/m2 confining pressure, what would be the deviator stress at failure.

Sagot :

nuhulawal20

Answer:

a) the friction angle of the sand is 36.87°

b) the deviator stress at failure is 450 kN/m³

Explanation:

Given the data in the question;

For a consolidated drained test

The effective major principle stresses

σ₃ = σ₃' = 80 kN/m²

and

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 80 kN/m² + 240 kN/m²  = 320 kN/m²

now

a) friction angle of the sand

σ₁' = σ₃'tan²( 45° + Ф/2' ) + 2c' tan( 45° + Ф/2 )

for sand; c' = 0

so

σ₁' = σ₃'tan²( 45° + Ф/2' )

we substitute

320 = 80 tan²( 45° + Ф/2' )

Ф' = 2 × [ tan⁻¹ (√[tex]\frac{320}{80}[/tex]) - 45° ]

Ф' = 2 × [ 63.4349° - 45° ]

Ф' = 2 × 18.4349

Ф' = 36.87°

Therefore,  the friction angle of the sand is 36.87°

b)  deviator stress

σ₁' = σ₃'tan²( 45° + Ф/2' )

σ₁' = σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )

σ₃' + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = σ₃'tan²( 45° + Ф/2' )   3.597

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 150tan²( 45° + 36.87°/2 )

150 + (Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex] = 600 - 150

(Δσ[tex]_{d}[/tex])[tex]_{f}[/tex]  = 450 kN/m³

Therefore, the deviator stress at failure is 450 kN/m³

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