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## Homework Statement

Mild steel bar 40 mm diameter and 100 mm long, subjected to a tensile force along its axis (through central length)

Youngs Modulus = 200GN m^-2

Poisson's ratio = 0.3

Calculate the force required to reduce the diameter to 39.99mm

(there is also a picture saying Use the x- y co-ordinate system above). I shall upload that momentarily if needed?

## Homework Equations

Poissons ratio = Transverse strain = - εt / εl

Diameter = 0.04m

Lo = 0.1m

E = 200GN m^-2 = 200x10^9 Pascals (Pa)

V = 0.3

Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter

Axial strain = Transverse strain / V

Δd = εtd

ε = εt / -V

ε = σ / E

σ = F/A

## The Attempt at a Solution

So here's what I did and hoping it is right?

Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter

(0.3999 - 0.4000) / 0.4000 = -0.00025 Transverse strain (εt)

Axial strain = Transverse strain / V

-0.00025 / 0.3 = .00083333333 Axial strain

I confirmed these by dividing Transverse strain by Axial strain = -0.3

Since decrease in diameter = transverse strain x Original Diameter then:

Δd = εtd = -0.00025 x .40 = 0.1mm which is correct.

As εt = -Vε Then:

-0.00025 = -0.3 x ε

∴

ε = εt / -V

ε = 0.00083333333

ε = σ / E

σ is unknown ∴

σ = ε x E

σ = 16666666 Pa (stress value)

σ = F / A

∴

Force = σ x A (Area)

= 16666666 Pa x 0.00125663706

= 20943.95019 Nm

I think here I can bring the figure down a thousand to 20.944 KNm?

Thank guys for ANY help! This took me ages!!