Answer:
Shearing strain will be 0.1039 radian
Explanation:
We have given change in length 
Length of the pad L = 1.15 inch
We have to find the shearing strain
Shearing strain is given by
Shearing strain is always in radian so we have to change angle in radian
So 
Its safe because it isn't something with electricity
Answer:
a) P = 86720 N
b) L = 131.2983 mm
Explanation:
σ = 271 MPa = 271*10⁶ Pa
E = 119 GPa = 119*10⁹ Pa
A = 320 mm² = (320 mm²)(1 m² / 10⁶ mm²) = 3.2*10⁻⁴ m²
a) P = ?
We can apply the equation
σ = P / A ⇒ P = σ*A = (271*10⁶ Pa)(3.2*10⁻⁴ m²) = 86720 N
b) L₀ = 131 mm = 0.131 m
We can get ΔL applying the following formula (Hooke's Law):
ΔL = (P*L₀) / (A*E) ⇒ ΔL = (86720 N*0.131 m) / (3.2*10⁻⁴ m²*119*10⁹ Pa)
⇒ ΔL = 2.9832*10⁻⁴ m = 0.2983 mm
Finally we obtain
L = L₀ + ΔL = 131 mm + 0.2983 mm = 131.2983 mm
C, because a narrow structure evacuation below surface ground isn’t the best and a structure holding forces and isn’t to do with the question at all and d doesn’t matter if there include away or not
