Acircular bar is 800mm in length and 32mm in diameter. the bar is made from a material with a modulus of elasticity e = 150 gpa and poisson’s ratio of 0.27. when a force is applied to the bar, it extends by 0.7mm. what is the change in diameter and the force applied?

For any material if ∈ is the axial strain then the lateral strain is given by -μ∈ is the lateral strain in the object

Where,

μ is the poisson's ratio of the material

The longitudinal strain is calculated as follows

Thus the lateral strain becomes

now by definition of lateral strain we have

By hookes law the stress developed due to the given strain is given by

Applying values we get

Thus the force is calculated as

twist boundary

explanation:

answer:

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