What is the diameter of the cylindrical rod

AI Thread Summary
To determine the diameter of a cylindrical copper rod subjected to a load of 6660N with an elongation of 0.5mm, Young's modulus (E) is used rather than yield strength. The relevant formulas include stress, strain, and the area of the rod. After calculations, the diameter was found to be approximately 7.65 mm, correcting earlier miscalculations. The discussion emphasizes the importance of correctly applying the square root in the formula. The final diameter of the rod is confirmed as 7.65 mm.
Perodamh
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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
 
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You don't really need the yield strength to solve this problem. Young's modulus is all you need.
 
phyzguy said:
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?
 
Perodamh said:
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.
 
phyzguy said:
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
 
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
 
Perodamh said:
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?
 
haruspex said:
Seems too small. Did you forget to take the square root?
phyzguy said:
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
 
Perodamh said:
I skipped the square root part so uhm 0.765 * 10^-2

That's what I got. 7.65 mm.
 
  • #10
phyzguy said:
That's what I got. 7.65 mm.
Neat, thanks so much
 
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