Find n and K value? (sheet metal question)

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The discussion revolves around calculating the K and n values for a tensile test of half-hard aluminum alloy based on given elongation and load data. The user initially calculates engineering stress and strain but makes an error in the engineering stress for the first point, which is off by a factor of ten. This mistake leads to incorrect true stress and strain calculations, affecting the final n-value computation. After receiving feedback about the error, the user acknowledges the mistake and expresses gratitude for the clarification. Accurate calculations are essential for determining material properties in engineering contexts.
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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!
 
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Your engineering stress for the first point is off by a factor of 10.
 
Chestermiller said:
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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