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2 answers:
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Answer:
Bending stress at point 3.96 is \sigma_b = 1.37 psi
Explanation:
Given data:
Bending Moment M is 4.176 ft-lb = 50.12 in- lb
moment of inertia I = 144 inc^4
y = 3.96 in
putting all value to get bending stress
Bending stress at point 3.96 is = 1.37 psi
Rubber block is not shown. I have attached an image of it.
Answer:
A) ε_x = 0.0075
B) ε_y = 0.00375
C) γ_xy = 0.0122 rad
Explanation:
We are given;
δ = 0.03 in
L = 4 in
ν_r = 0.5
θ = 89.3° = 89.3π/180 rad
Let's calculate ε_x in the direction of axis x
Thus, ε_x = δ/L = 0.03/4 = 0.0075
Let's calculate ε_y in the direction of axis y;
ε_y = v•ε_x = 0.5 x 0.0075 = 0.00375
Now, shear strain is angle between π/2 rad surfaces at that point.
Thus,
γ_xy = π/2 - θ = π/2 - 89.3π/180
γ_xy = π(0.003889) = 0.0122 rad
