Choose the person who best completes each statement.
2 answers:
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Answer:
γ =0.01, P=248 kN
Explanation:
Given Data:
displacement = 2mm ;
height = 200mm ;
l = 400mm ;
w = 100 ;
G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²
a. Average Shear Strain:
The average shear strain can be determined by dividing the total displacement of plate by height
γ = displacement / total height
= 2/200 = 0.01
b. Force P on upper plate:
Now, as we know that force per unit area equals to stress
τ = P/A
Also, τ = Gγ
By comapring both equations, we get
P/A = Gγ ------------ eq(1)
First we need to calculate total area,
A = l*w = 400 * 100= 4*10^4mm²
By putting the values in equation 1, we get
P/40000 = 620 * 0.01
P = 248000 N or 2.48 *10^5 N or 248 kN
Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, = 0.51
Void ratio in the loosest state, = 0.87
Now,
Dry density,
= 17.14 kN/m³
Also,
here, G = Specific gravity = 2.7 for sand
or
e = 0.545
Relative density =
=
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%