Question

# Due to the operational conditions and temperature gradient, a 70 mm diameter, 1.55 m long steel...

Due to the operational conditions and temperature gradient, a 70 mm diameter, 1.55 m long
steel rod within a machine assembly is subjected to a combination of tensile loading of 150 N and
distribution of forces on the body of the rod causing it to change its dimensions by increasing 0.5
mm in length at both ends and decreasing by 0.0125 mm at the diameter

(h) Poisson’s Ratio, v, may be determined by dividing the negative value of the lateral strain by
the axial strain. What is Poisson’s Ratio for the rod material? Is this an acceptable amount for
the rod?
(i) The Shear Modulus, G, of a material may be determined by the following equation: G = E/
2(1 + ν), where E is the Modulus of Elasticity, and ν, is Poisson’s Ratio. If the Modulus of
Elasticity for the rod material is E = 195 GPa, what is the value of the Shear Modulus?

h)

lateral strain of the rod, lateral = change in diameter / diameter

as given, change in diameter, D = -0.0125 mm

Diameter of the rod, D = 70 mm

so, lateral = -0.0125 / 70 = -1.785 * 10-4

Now, longitudinal strain,  longitudinal = change in the length / length of the rod

change in the length ,  L = 0.5 mm

Actual length of the rod, L = 1.55 m = 1550 mm

so,  longitudinal = 0.5 / 1550 = 3.2 * 10-4

Now, poisson's Ratio of the rod material , V = -  lateral / longitudinal

= -  -1.785 * 10-4 / 3.2 * 10-4 = 0.553

Value of poisson Ratio should be vary between 0.0 to 0.5, as here V = 0.553 , so it is not an acceptable amount.

i)

as G = E / 2(1+ 2V)

= 195 / 2(1+2*0.553)

so, Shear modulus, G = 46.28 GPa

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