Question 7.4: Reconsider the tension test in example problem 5.3. Let n = ......
Reconsider the tension test in example problem 5.3. Let n = 0.226 and m = 0.012 (typical values for low carbon steel) and let f = 0.98 as in the previous problem. Calculate the strain, \varepsilon _{a}, in the region with the larger diameter if the bar necks to a 25% reduction of area (\varepsilon _{b} = ln(4/3) = 0.288.)
Substituting into Equation (7.12),
\int_{0}^{\varepsilon_{a}}{\exp(-\varepsilon_{a}/m) \varepsilon^{n/m}_{a} d\varepsilon_{a} } =f^{1/m}\int_{0}^{\varepsilon_{b}}{\exp(-\varepsilon_{b}/m) \varepsilon^{n/m}_{b} d\varepsilon_{b} }
or
\int_{0}^{\varepsilon_{a}}{\exp(-83.33 \varepsilon_{a}) \varepsilon^{18.86}_{a} d\varepsilon_{a} } =0.1587 \int_{0}^{\varepsilon_{b}}{\exp(-83.33 \varepsilon_{b}) \varepsilon^{18.83}_{b} d\varepsilon_{b} } .
Numerical integration of the right-hand side gives 0.02185. The left-hand side has the same value when \varepsilon _{a} = 0.2625. The loss of elongation in the thicker section is much less than that in Example Problem 5.3, where it was found that \varepsilon _{a} = 0.195 with n = 0.226 and m = 0.