Force required to reduce the diameter

1. Aug 31, 2013

oxon88

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

force = Stress x Area

3. 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

2. Sep 2, 2013

oxon88

can anyone check my workings please?

3. Sep 2, 2013

4. Sep 2, 2013

Many Thanks.

5. Dec 3, 2013

6. Dec 3, 2013

oxon88

yes. what answer did you get?

7. Dec 3, 2013

Big Jock

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

8. Dec 3, 2013

oxon88

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

9. Dec 3, 2013

Big Jock

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.

10. Dec 3, 2013

PhanthomJay

Values should be rounded to the lowest number of significant figures in the given values. All such values have only 1 significant figure. Therefore, the answer is good only to one significant figure.