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3 years ago

6

Undisturbed samples from a normally consolidated clay layer were collected during a subsoil exploration. Drained triaxial tests

showed that the effective friction angle was φ’ = 25°. The unconfined compressive strength, qu, of a similar specimen was found to be 133 kN/m2. Find the pore pressure at failure for the unconfined compression test.

1 answer:

kipiarov [429]3 years ago

3 0

Answer:

pre water pressure $= \sigma_1 = 208.76 kN/m2$

Explanation:

given data:

frictional angle 25°

unconfined xompressive strength qu = 133 kN/m2

for clay, unconfined compressive strenth is cu

$cu =\frac{qu}{2}$

cu = 133/2 =66.5 kN/m2

cell pressure $\sigma_3$ is zero in uncomfined compressive strenth test

from principle effective stress we have

$\sigma_1 = \sigma_2 tan^{2}(45+ \frac{\theta}{2}) +2cu tan(45+ \frac{\theta}{2})$

$\sigma_1 = 2cu tan(45+ \frac{\theta}{2})$

$\sigma_1 = 2*66.5 tan(45+ \frac{25}{2})$

$\sigma_1 = 208.76 kN/m2$

as this test is quick test i.e. undrained test therefore

pre water pressure $= \sigma_1 = 208.76 kN/m2$

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A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.

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Answer:

Explanation:

Given a ball of mass m= 0.5kg

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Answer:

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Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

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(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

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Put the value into the formula

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(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

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Square root of (4^2 + 4^2) = 4*squareRoot(2)

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