Respuesta :
Answer:
The value of poisson's ratio of the material = 0.28
Explanation:
Given data
Force F = 15700 N
Diameter D = 8 mm
Change in diameter ΔD = 0.005 mm
Modulus of elasticity E = 140 × [tex]10^{3}[/tex] M Pa.
Stress induced in the shaft due to applied force
[tex]\sigma = \frac{F}{A}[/tex]
Area A = [tex]\frac{\pi}{4} d^{2}[/tex]
⇒ A = [tex]\frac{\pi}{4} 8^{2}[/tex] = 50.24 [tex]mm^{2}[/tex]
Now stress [tex]\sigma = \frac{15700}{50.24}[/tex]
⇒ Stress [tex]\sigma[/tex] = 312.5 M pa ------ (1)
Now strain in the element is given by
∈ [tex]= \frac{\sigma}{E}[/tex]
⇒ ∈ = [tex]\frac{312.5}{140000}[/tex]
⇒ Strain ∈ = 2.232 × [tex]10^{-3}[/tex]
We know that poisson's ratio is given by
⇒ [tex]\mu =[/tex] [tex]\frac{\frac{change in diameter}{original diameter} }{strain}[/tex]
Change in diameter ΔD = 0.005 mm
Diameter D = 8 mm
Strain ∈ = 2.232 × [tex]10^{-3}[/tex]
⇒ [tex]\mu = \frac{\frac{0.005}{8} }{0.002232}[/tex]
⇒ [tex]\mu =[/tex] 0.28
This is the value of poisson's ratio of the material.
