Acylindrical specimen of some metal alloy 11.1 mm (0.4370 in.) in diameter is stressed elastically in tension. a force of 14000 n (3147 lbf) produces a reduction in specimen diameter of 7 × 10-3 mm (2.756 × 10-4 compute poisson's ratio for this material if its elastic modulus is 100 gpa (14.5 × 106 psi).

To develop this problem it is necessary to apply the concepts related to the uni-axial deflection of bodies.

From the expression of Hooke's law we have to

Where,

E= Young's modulus

The strain

And substituting P/A for stress and for strain gives that

Where,

P = Force

A = Area

L = Length

Therefore this can be re-arranged to give

If we want to calculate the deformation per unit area then we can also rewrite the equation as

Replacing with our values we have to

Therefore the posion ratio would be

Therefore the Poisson's ratio for this material is 0.4238

answer: he got 23

explanation:

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.