How Do You Calculate Young's Modulus from Tensile Stress and Strain Data?

AI Thread Summary
Young's modulus can be calculated using the formula E = σ/ε, where σ is the tensile stress and ε is the strain. In this case, the tensile stress is 280 MN/m², and the alloy bar's length increases by 0.35%, which translates to a strain of 0.0035. To find Young's modulus, substitute the values into the equation: E = 280 MN/m² / 0.0035. The discussion emphasizes the relationship between stress, strain, and the calculation of Young's modulus, clarifying that strain is the fractional increase in length. Understanding these concepts is essential for accurately determining material properties.
anthonyk2013
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A light alloy bar is observed to increase in length by .35% when subjected to a tenstile stress of 280MN/m^2.
Calculate youngs modulus for the material.

σ=280MN/M^2
x=.35%
ε=?
E=?
L=?

I have too many unknowns where do I start?
 
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What's the definition of Young's modulus? How is it calculated?
 
SteamKing said:
What's the definition of Young's modulus? How is it calculated?

E=σ/ε

to find ε=x/L so think I might have to work of .35% but not sure how.
 
What does the 0.35% represent?
 
The length of the alloy has increased by .35%
 
Last edited:
And how would this increase be related to the strain?
 
Is strain not increase in length/original length ?
 
Yes.
 
I have the increase in length I don't have strain and I don't have original length.
Is it possible to work out length using only .35%?
 
  • #10
Well, let's start with the basics. Assume the original length of the test specimen was L. After the stress was applied, what is the length of the test specimen? What is the strain due to the applied stress?
 
  • #11
Strain=.35/L
 
  • #12
anthonyk2013 said:
Strain=.35/L

From the OP:

"A light alloy bar is observed to increase in length by .35% when subjected to a tenstile stress of 280MN/m^2."

Would you like to revise your answer?
 
  • #13
Sorry SK but I'm lost now.
 
  • #14
Stress and strain are proportional according to hooks law?
 
  • #15
Yes, but you are avoiding the question about the strain of the test piece after it has the stress applied.

If the length of the test piece increases by 0.35%, what is its new length in terms of its original length?
 
  • #16
Sorry SK but I'm still lost. Do I need to convert .35 from a % to measurement by dividing by 100.
 
  • #17
Look, if something increases in length by 25%, what is its new length compared to its original length?
 
  • #18
If you have D dollars in your account to start with, and you earn 0.35% per year interest (compounded annually), how much money do you have in your account after 1 year? How much has the account value increased?
 
  • #19
D+.35%?
 
  • #20
anthonyk2013 said:
D+.35%?

.35% is not an amount of money. To verify this, try spending .35% at your local store.

The amount of money you would have after 1 year would be (D+0.0035D), and the amount of money you would have earned would be 0.0035D dollars.
 
  • #21
Divide by 100 to get the value. I won't forget that too soon. Sorry guys elementary stuff.

So X=.0035
 
  • #22
anthonyk2013 said:
Divide by 100 to get the value. I won't forget that too soon. Sorry guys elementary stuff.

So X=.0035
Actually, the bar getting larger by 0.35% means that ε = 0.0035.
 
  • #23
E=280/.0035

So if you multiply strain by 100 you get the percentage it increases by?
 
  • #24
Yes. The strain is the fraction it increases by. To get the % increase in length, you multiply the strain by 100%.
 
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