Stress-Strain question -- The maximum elastic elongation of a steel sample

• Engineering
menotu3169
Homework Statement:
A cylindrical steel specimen 350mm long is to be subjected to a tensile load of 10KN. The modulus of elasticity is 207GPA, Yield Strength is 415 MPa, and Poisson's ratio is 0.30
Determine
a)The maximum elastic elongation
b)the reduction in diameter
Relevant Equations:
stress = Force / Area
strain = Stress / Modulus of Elasticity
Poisson's ratio = transverse stress / longitudinal stress
I was able to calculate a), and got 0.7mm

Homework Helper
Gold Member
It's been awhile since done these types of calculations, but I believe you need cross sectional area. I assume that you have this information since you were able to calculate the elongation.

menotu3169 and Lnewqban
menotu3169
It's been awhile since done these types of calculations, but I believe you need cross sectional area. I assume that you have this information since you were able to calculate the elongation.
I was not given the area or the diameter, which is why I am having trouble with the question

Homework Helper
Gold Member
Could it be percentage?

Homework Helper
Gold Member
I was not given the area or the diameter, which is why I am having trouble with the question
What formulas did you use to get the 0.7mm ?

Gold Member
Poisson's ratio is usually defined as the ratio of strains, not stresses (although this shouldn't matter if you're in the elastic regime). I think the tensile load of 10 kN that OP listed is a red herring. The yield strength and the elastic modulus allow you to calculate the maximum strain (elongation) of the material before plastic deformation sets in. This strain will be a percentage elongation which can be used to find a length (which the OP seems to have done correctly). The Poisson's ratio is then just the ratio of transverse strain to axial strain, with a negative sign out front because most materials shrink transversely when you pull them axially.

scottdave, menotu3169, Chestermiller and 2 others
menotu3169
Spoke with the professor and got the solution. Seems like it is just a poorly worded question. The question was worded exactly as I have it in the original post.
Solution:
A=10000/415=24.09 mm2
d=5.54 mm
change in length = 415/207000*350=0.7 mm
change in diameter = 415/207000*(-0.3)*5.54=-0.003 mm

This seems to indicate that the Yield Strength corresponds to the applied force of 10kN, which in my opinion is not obvious from the question.

Thanks everyone for all the help!

scottdave and Lnewqban