# Force required to reduce the diameter

## 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

can anyone check my workings please?

PhanthomJay
Homework Helper
Gold Member

• 1 person
Many Thanks.

yes. what answer did you get?

I got 209.36kN that's why I was wondering. I changed the dimensions of the tube into m from the start and got a different answer to your transverse strain which then saw all my answers being different all the way through. Hence why I wanted to know if yours was correct a I thought it was slightly out

looks acceptable, its pretty close to what i got. As PhanthomJay states, just round it off to 200kN

why would you round it down to 200kN? Also I commented on another question regarding tubular column which is question 2 and was waiting to hear back from that thread if you could that would be great.

PhanthomJay
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