An unconfined compression test yielded a strength of 0-1N-mm2- If the failure plane is inclined at 50- to the horizontal- what are the values of the shear strength parameters-

Given σ _1=0.1 N/mm^2 Inclination of failure plane with horizontal =50^∘ 45+ϕ2=50^∘ ϕ = 10^∘ We know that σ _1=σ _3 ( 45+ϕ2 )+2c tan ( 45+ϕ2 ) For UCS te ...

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Soil Mechanics

Unconfined Compression Test

An unconfined...

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An unconfined compression test yielded a strength of 0.1 $N / m m^{2}$ . If the failure plane is inclined at $50^{\circ}$ to the horizontal, what are the values of the shear strength parameters?
0.042

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Solution

The correct option is A 0.042
Given

$σ_{1} = 0.1 N / m m^{2}$

Inclination of failure plane with horizontal
$= 50^{\circ}$

$45 + \frac{ϕ}{2} = 50^{\circ}$

$ϕ = 10^{\circ}$

We know that

$σ_{1} = σ_{3} (45 + \frac{ϕ}{2}) + 2 c t a n (45 + \frac{ϕ}{2})$

For UCS test,
$σ_{3} = 0$

$σ_{1} = 2 c t a n (45 + \frac{ϕ}{2})$

$0.1 = 2 c t a n 50^{\circ}$

$\Rightarrow c = 0.042 N / m m^{2}$

$c = 0.042 N / m m^{2}; ϕ = 10^{\circ}$

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An unconfined compression test yielded a strength of 0.1N/mm2. If the failure plane is inclined at 50∘ to the horizontal, what are the values of the shear strength parameters? 0.042

The correct option is A 0.042Given σ1=0.1N/mm2 Inclination of failure plane with horizontal =50∘ 45+ϕ2=50∘ ϕ=10∘ We know that σ1=σ3(45+ϕ2)+2ctan(45+ϕ2) For UCS test, σ3=0 σ1=2c tan(45+ϕ2) 0.1=2c tan50∘ ⇒c=0.042N/mm2 c=0.042N/mm2; ϕ=10∘

An unconfined compression test yielded a strength of 0.1 $N / m m^{2}$ . If the failure plane is inclined at $50^{\circ}$ to the horizontal, what are the values of the shear strength parameters?
0.042