Answer:
a. 0.9263
b. 0.4872
c. 13.83kN/m
Explanation:
moisture content (ω) = 0.32
void ratio (e) = 0.95
specific gravity () = 2.75
the degree of satruation (S) = =0.32×2.75/0.95 = 0.9263
b. porosity (n) = = 0.95/(0.95 + 1)= 0.4872
c. dry unit weight (γ) =
taking specific unit weight of water (V)= 9.81kN/m
γ = 2.75 × 1000/(1 + 0.95) = 13.83kN/m
Answer:
Maximum shear stress= 5.66 ksi
Maximum tensile stress= 5.66 ksi
Maximum compressive stress=-5.66 ksi
Maximum shear strain=0.000943
Maximum tensile strain= 0.0004715
Maximum compressive strain= -0.0004715
Explanation:
For acircular bar, the maximum shear stress will be given by
where d is the diameter and T is torque.
By substituting 30 kip-in for torque and 3 in for d then
Maximum shear stress=
Also, the maximum tensile and compressive stresses will be 5.66 ksi and -5.66 ksi respectively.
The maximum shear strain will be given by stress divided by modulus of elasticity, in this case 6000 G
Maximum shear strain will be
The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence
Maximum compressive strain will be