# Calculate force required to reduce diameter of bar?

A mild stell bar 40 mm diameter and 100 mm long is subjected to a tensile force along its axis.
Young's modulus of elasticity = 200 GN m -2. Poisson's ratio is 0.3.
Calculate the force required to reduce the diameter to 39.99 mm

It says use the x-y coordinate system?? This i don't understand?

Thank you

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nvn
Homework Helper
bensm0: Hint 1: What is the equation for the definition of Poisson's ratio, in terms of transverse strain εy, and axial strain εx? Hint 2: What is the equation for the bar axial elongation, in terms of the applied tensile force P?

We are not allowed to give you the relevant equations for your homework. You must list relevant equations yourself, and show your work. And then someone might check your math.

Hint 3: The y axis is the bar transverse direction. The y axis is parallel to the bar diameter.

By the way, GN*m^-2 is called GPa. Always use the correct, special name for a unit. E.g., 200 GPa, not 200 GN*m^-2.

ok so i think i may have the answer:

poisson's ratio = - transverse strain/axial strain
Transverse strain = (39.99 - 40)/40 = -0.25x10^-3
axial strain = -(-0.25x10^-3/0.3) = 833.333x10^-6
Axial stress = 833.333x10^-6 x 200x10^9 = 166.6666x10^6
Force = stress x area
Force = 166.6666x10^6 x (0.25∏x0.04^2) = 5.236x10^6 N

Am i on the right track?

nvn
Homework Helper
bensm0: Nice work, except you did not check your arithmetic, and got the wrong answer.

Hi there, i am also stuck on this question along with the rest of the assignment to be honest. Bensm0 contact me on here if you want to try and tackle this togeter?

Hi there,

Ive been working on this for most of today and i have come up with the following :

Axial Strain = 8.333 x 10^-4

Axial Stress = 166.66 x 10^6

Force = Stress x Area
= (166.66 x 10^6) x (1.257 x 10^-3)
= 209491.62 N