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## Homework Statement

A mild steel bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis.

Young’s modulus of elasticity for mild steel = 200 GN m–2.

Poisson’s ratio is 0.3.

Calculate the force (F) required to reduce the diameter to 39.99 mm.

Use the x–y coordinate system as shown above.

## Homework Equations

Poissson's Ratio = - (transverse strain / axial strain)

force = Stress x Area

## The Attempt at a Solution

transverse strain = (39.99 - 40) / 40 = -0.25x10^-3

axial strain = - (-0.25x10^-3 / 0.3) = 833.333 x 10^-6

axial stress = (833.333 x 10^-6) x (200 x 10^9) = 166.666 x 10^6

force (F) = 166.666x10^6 x (0.25∏ x 0.04^2) =

**209439.51 N**