What is the diameter of the cylindrical rod

[Perodamh]

Homework Statement

A cylindrical rod of copper (E = 110GPa) having a yield strength of 240MPa is to be subjected to a load of 6660N. If the length of the rod is 380mm, what must be the diameter to allow an elongation of 0.5mm.

Homework Equations

E = stress/ strain
stress = Force/Area ; Area = pi*(diameter)/4
strain = elongation/original length
E= young's modulus = 110GPa = 110 * 10^9Pa
yield strength = 240MPa = 240 * 10^6Pa
Force = 6660N
elongation = 0.5mm
original length = 380mm

The Attempt at a Solution

I'm trying to understand relation between yield strength and young modulus.So far all i got from google is that it is the max stress a body goes through before plastic deformation occurs, but how do i plug that in anywhere? Thanks

 

[phyzguy]

You don't really need the yield strength to solve this problem. Young's modulus is all you need.

 

[Perodamh]

You don't really need the yield strength to solve this problem. Young's modulus is all you need.

Thanks, was wondering why it was there, I tried another go at the problem, by plugging in values and got (0.586 * 10^-4)m. Don't know how correct that is?

 

[phyzguy]

Thanks, was wondering why it was there, I tried another go at the problem, by plugging in values and got (0.586 * 10^-4)m. Don't know how correct that is?

Show us your work so we can see how you got there. then we can make meaningful comments.

 

[Perodamh]

Show us your work so we can see how you got there. then we can make meaningful comments.

E = stress / strain
stress = F / A
Area = pi*(diameter)^2/4
strain= elongation/original length
making d subject of formula = sqrt(4*F*L/(E*pi*elongation))
plugging in values after conversion to S.I units gave 58617.25 * 10^-9m which is same as 0.586 * 10^-4m

 

[phyzguy]

Well, I agree with this, "making d subject of formula = sqrt(4*F*L/(E*pi*elongation))", but that's not the answer I got, so one of us made a mistake. I suggest you check your numbers

 

[haruspex]

E = stress / strain
stress = F / A
Area = pi*(diameter)^2/4
strain= elongation/original length
making d subject of formula = sqrt(4*F*L/(E*pi*elongation))
plugging in values after conversion to S.I units gave 58617.25 * 10^-9m which is same as 0.586 * 10^-4m

Seems too small. Did you forget to take the square root?

 

[Perodamh]

Seems too small. Did you forget to take the square root?

Well, I agree with this, "making d subject of formula = sqrt(4*F*L/(E*pi*elongation))", but that's not the answer I got, so one of us made a mistake. I suggest you check your numbers

I skipped the square root part so uhm 0.765 * 10^-2

 

[phyzguy]

I skipped the square root part so uhm 0.765 * 10^-2

That's what I got. 7.65 mm.

 

[Perodamh]

That's what I got. 7.65 mm.

Neat, thanks so much