Maximum reduction per pass for a strain-hardening material

Obtain an expression for the maximum reduction per pass for a material with a true- stress– true-strain curve of $\sigma =K\epsilon ^{n}$ . Ignore friction and redundant work.

Question

Maximum reduction per pass for a strain-hardening material

Obtain an expression for the maximum reduction per pass for a material with a true- stress– true-strain curve of [latex]\sigma =K\epsilon ^{n}[/latex] . Ignore friction and redundant work.

Accepted Answer

From Eq. (6.62),

[latex]\sigma d=Y\ln \left\lgroup\frac{A_{\omicron }}{A_{f}} \right\rgroup.[/latex] (6.62)

we have

[latex]\sigma _{d}= \overline{Y}\ln \left\lgroup\frac{A_{\omicron }}{A_{f}} \right\rgroup =\overline{Y} \epsilon _{1} ,[/latex]

where

[latex]\overline{Y}=\frac{K\epsilon ^{n}_{1}}{n+1} [/latex]

and [latex]\sigma _{d}[/latex], for this problem, can have a maximum value equal to the flow stress at [latex] \epsilon_{1}[/latex], or

[latex]\sigma _{d} =Y_{f}=K\epsilon^{n}_{1}. [/latex]

We can now write Eq. (6.72) as

[latex]\sigma _{d}=Y\ln \left\lgroup\frac{A_{\omicron }}{A_{f}} \right\rgroup=Y,[/latex] (6.72)

[latex]K \epsilon ^{n}_{1} =\frac{K \epsilon ^{n}_{1}}{n+1} \epsilon _{1} ,[/latex]

or

[latex]\epsilon _{1}=n+1[/latex]

With [latex]\epsilon _{1}= ln(A_{o}/A_{f}) [/latex] and a maximum reduction of [latex](A_{o} - A_{f})/A_{o}[/latex], these expressions reduce to

Maximum reduction per pass = [latex]1 – e^{-(n + 1)}[/latex] (6.74)

Note that when [latex] n = 0 [/latex] (that is, a perfectly plastic material), this expression reduces to Eq. (6.73),

Maximum reduction per pass = [latex]\frac{A_{\omicron }-A_{f}}{A_{\omicron }} = 1-\frac{1}{e} =0.63 = 63 [/latex]%. (6.73)

and that as [latex]n [/latex] increases, the maximum reduction per pass increases.

Question 6.8: Maximum reduction per pass for a strain-hardening material O...

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