Answer:
(a) stress in the post is [tex]\sigma =[/tex] 816.4 M pa
(b). strain in the post is ∈ = 0.004082
(c). strain is dL = 0.01347 m
Explanation:
Given data
Force = mg = 9900 × 9.81 = 97119 N
Diameter (D) = 0.39 m
Radius = 0.195 m
Length = 3.3 m
(a) stress in the post is given by
[tex]\sigma = \frac{F}{A}[/tex]
Area (A) = [tex]\pi r^{2}[/tex]
A = 3.14 × [tex]0.195^{2}[/tex]
A = 0.119 [tex]m^{2}[/tex]
Now stress
[tex]\sigma = \frac{97119}{0.119}[/tex]
[tex]\sigma =[/tex] 816.4 M pa
(b). strain in the post is given by
∈ = [tex]\frac{\sigma }{E}[/tex]
E = 2 × [tex]10^{5}[/tex] M pa
Now strain
∈ = [tex]\frac{816.4}{2(10^{5} )}[/tex]
∈ = 0.004082
(c). We know that strain is the ratio of change in length to original length.
∈ = [tex]\frac{dL}{L}[/tex]
dL = ∈ × L
dL = 0.004082 × 3.3
dL = 0.01347 m
