Calculating Young's Modulus for CopperWhat is the Young's Modulus for copper?

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SUMMARY

The Young's Modulus for copper is calculated using the formula Y = Stress/Strain, resulting in a value of 4 x 109 Pa when stress is 1.4 x 108 Pa and strain is 0.035. To determine the tensile force required to reach the elastic limit of a copper wire with a cross-sectional area of 1 mm2, the formula F = (A x e x Y) / l is applied, yielding a force of 1.4 x 105 N when the length is 1 m and extension is 0.035 m. Proper unit conversions are crucial, particularly converting mm2 to m2 for accurate calculations.

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  • Understanding of Young's Modulus and its significance in material science.
  • Familiarity with stress and strain concepts in physics.
  • Basic knowledge of unit conversions, particularly between metric units.
  • Proficiency in algebraic manipulation of formulas related to tensile strength.
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pmh118
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Got exams in January so just want to make sure I am correct the questions read exactly like this

a) The sample of copper wire reached its elastic limit when the stress is 1.4x10^8 Pa and the strain is 0.035 Calculate the Young's Modulus for copper.

My attempt is Y= Stress/Strain => Y= 140000000/0.035 => 4,000,000,000 => 4x10^9

Hope this part is correct

b) What tensil force would be needed to reach the elastic limit of the copper wire of cross sectional area 1mm sq

F = Force A= Area e= extention Y= Young's Modulus l = lenght

My attempt is F= (A x e x Y) / l => F= (0.001 x 0.035 x 4,000,000,000) / 1 => 140,000 => 1.4x10^5

I took my length as 1m and extention was 0.035m from the original strain = 0.035

not sure if I am right as I have no contact with my lecturures until my exam I ask if somebody could tell me if my work is right or wrong and where I have gone wrong if so

Thx Paul
 
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pmh118 said:
Got exams in January so just want to make sure I am correct the questions read exactly like this

a) The sample of copper wire reached its elastic limit when the stress is 1.4x10^8 Pa and the strain is 0.035 Calculate the Young's Modulus for copper.

My attempt is Y= Stress/Strain => Y= 140000000/0.035 => 4,000,000,000 => 4x10^9

Hope this part is correct
yes, but don't forget the units!
b) What tensil force would be needed to reach the elastic limit of the copper wire of cross sectional area 1mm sq

F = Force A= Area e= extention Y= Young's Modulus l = lenght

My attempt is F= (A x e x Y) / l => F= (0.001 x 0.035 x 4,000,000,000) / 1 => 140,000 => 1.4x10^5

I took my length as 1m and extention was 0.035m from the original strain = 0.035

not sure if I am right as I have no contact with my lecturures until my exam I ask if somebody could tell me if my work is right or wrong and where I have gone wrong if so

Thx Paul
Why go through all this when Tensile Stress = Tensile Force/Cross Section Area?

Units?

Check conversion from mm^2 to m^2.
 
Ah yes I see that F= 140000000 x 0.001

so my answer with units should be F= 1.4x10^5 Pa ?
 
While I like the SI system in Physics, I don't like it at all in Engineering. Partly because I'm from the USA, and partly because I can't handle all those darn zero's before or after the decimal point. Regardless, 1 mm2 is 0.000 001 m2, or 1 X 10-6 m2. Please confirm.
 
Ah thanks my conversions need a lot of work but now i see 1mm^2 = 1/(1000x1000)

1x10^-6 so my new equation is F = 140000000 x 0.000001 => 140 Pa

Thx all for helping me
 
A Pa is a stress unit (1 Pa = 1 N/m2). Forces are measured in Newtons. Are you from the contiguous 48?
 
Ah sorry I am being dumb today, I was thinking a Newton was a Pascal, think maybe I should take a break from this study for a day ;p
and nope not from the contiguous 48 as much as I would love to live there and study at MIT
I am English just got back into studying I am a Carpenter by trade and looking for a new direction been accepted into University if I get this Maths and Physics courses mastered :)
 
How do you buy apples and bananas, by the pound, kilo, Newtons, or is it stones?:wink:
 
Well tbh I by per bag :p but when I was just a lad I worked on a fruit stall selling them in lbs but then it all went metric so is kilo now :)
 
  • #10
:cool:Cool. The feds have been trying to get us to convert to metric for the past 40 years...we just ignore them... we buy them by the pound...

Best of luck in your studies!

Happy New Year!:smile:
 
  • #11
Haha change over its for the best :) thanks for all the help and the nice chat
All the best for new year
 

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