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Find n and K value? (sheet metal question!)

  • Thread starter emlekarc
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  • #1
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Thread moved from the technical forums, so no HH Template is shown.
Hello! Can someone please tell me where I'm messing up in this problem?

At 4% and 8% elongation, the loads on a tensile test-piece of half-hard aluminium alloy are 1.59 kN and 1.66 kN respectively. The test-piece has an initial width of 10 mm, thickness of 1.4 mm and gauge length of 50 mm. Determine the K and n values.

So, first I calculated the engineering stress at the second point:
1.66/(10*1.4) = .11857

Then the engineering strain at that point: (54-50)/50 = .08

The true stress is then: .11857(1+.08) = .128056

The true strain is then: ln(1+.08) = .07696

For for the first point, I did the same thing:
Engineering Stress: 1.59/(10*1.4) = .011357
Engineering Strain: (52-50)/50 = 0.4
True stress: .11357(1+.04) = .0118
True strain: ln(1+.04)=.03922

Then, I tried to calculate the n-value by:
ln(.128056)-ln(.0118) / ln(.07696)-ln(.03922) = 3.5371
ln(true stress)/ln(true strain)

The book says the answer is supposed to be 0.12, but I'm not sure what I'm doing wrong?

Thank you VERY much for your help!
 

Answers and Replies

  • #2
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Your engineering stress for the first point is off by a factor of 10.
 
  • #3
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Your engineering stress for the first point is off by a factor of 10.
Thank you!! I didn't even realize I did that! I really appreciate it!!
 

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