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A metal is deformed in a tension test into its plastic region. The starting specimen had a gage length = 2.0 in and an area .50 sq. in. At one point in the tensile test, the gage length = 2.5 in and the corresponding engineering stress = 24000 psi; at another point in the test prior to necking, the gage length = 3.2 in and the corresponding engineering stress = 28000 psi. Determine the strength coefficient and strain-hardening exponent.

To summarize,

L0 = 2.0 in

A0 = .50 sq. in.

L1 = 2.5 in engineering stress = 24000 psi

L2 = 3.2 in engineering stress = 28000 psi

Here's my solution:

*At the first point given, we can find engineering strain by e = (L1-L0)/L0 = (2.5 - 2.0)/2.0 = .25

-We can then use the engineering stress and the engineering strain to find the true stress by the relationship:

true stress = engineering stress (1+engineering strain) = 24000(1+.25) = 30000 psi

-The true strain is given by ln (L1/L0) = ln (2.5/2.0) = .22

* For the second point, engineering strain = (L2-L0)/L0 = (3.2-2.0)/2.0

= .6

- Hence, true stress = engineering stress (1+engineering strain) =

28000(1+.6) = 44800 psi

- True strain is ln (L2/L0) = ln (3.2/2.0) = ln 1.6 = .47

*So we now have two points on our true stress-strain curve: (.22, 30000) and (.47, 44800).

The book plots these points using a log scale and then finds the slope to determine the strain hardening exponent. So n = (log 44800 - log 30000)/(log .47 - log .22) = .52

Does this seem right...I know n for most metals is between 0 and .5, so is .52 a reasonable result???

*The strength coefficient, K, equals the value of true stress at a true strain value equal to 1.

Using the point slope equation with the x-coordinate (true strain) equal to 1, I get:

y - y1 = m(x-x1)

log y - log 30000 = .52 (log x - log .22)

log y - log 30000 = .52 (1 - log .22)

log y - log 30000 = .8619402

y = 218304 psi

K = 218304 psi

Does this seem correct? It seems a little large!

Thanks for reading! Any help much appreciated!

I apologize for