chandra_324_hw7_soln

chandra_324_hw7_soln - Solution for ME324 HW#7 1 SOLUTION...

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Solution for ME324 HW #7 1 SOLUTION This is a matter of the ductility of the material, particularly the reduction in area as depicted by Eq. and . Thus, hard material conditions mean lower tensile reduction and therefore higher R/T ratios. In other words, for a constant sheet thick- ness T, the bending radius R has to be larger for better bendability. 2 SOLUTION Higher R value means the material has better capability of undergoing reduction in width, yet resisting thinning in the cup wall; thus, it gives better deep drawability. 3 SOLUTION To determine the true strains, we refer to Eq. which gives the relationship required for the experimental data agreement, and obtain the tensile reduction of area (percent reduction of area in a tension test). After determining , we refer to the first equation on that page and find the true strain . Thus, (1) For , , and (2)For , , and (3)For , , and 4 SOLUTION In this problem, we use Eq. where, we know for Ti-5 Al- 2.5Sn titanium alloy, and noting that ., . and .
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This note was uploaded on 10/13/2008 for the course ME 324 taught by Professor Boylan during the Spring '08 term at Iowa State.

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chandra_324_hw7_soln - Solution for ME324 HW#7 1 SOLUTION...

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