SSC JE Mechanical Engineering 2018 Set 1 Paper

Question : 119 of 200

Marks: +1, -0

Calculate the value of modulus of rigidity (N/mm

^{2}

. if the Poisson’s ratio is 0.25 and modulus of elasticity for the material is 200 N/mm

^{2}

Solution:

Poisson ratio

(1/mrµ)=0.25

Modulus of elasticity E = 200 N/m

^{2}

.
From relation E = 2 c(1 + µ)
200 = 2 × c(1 + 0.25)

⇒ c=100/1.25

[c=80\N\/\mm^2]

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