Sheet metal of two different alloys is produced by rolling to 0.1 mm thickness. You must select one of these alloys for a sheet forming operation. To make the selection, tensile samples are cut from each alloy at 0o , 45o , and 90o to the rolling direction. The initial gauge section of each tensile sample is 0.75 mm width. After subjecting all samples to the same tensile load, th

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Preciousorekha1 Preciousorekha1 · Answer 1 · 2020-04-29T17:12:37+02:00

Answer: Alloy A is best chosen for the sheet forming operation

Explanation:

taking both alloy A and B into consideration;

For alloy A;

the true strain ration, R₀ = width strain / thickness strain

R₀ = in (0.65/0.75) / in (0.092/0.1) = 1.716

at 45⁰ and 90⁰, the true strain ratio becomes;

R45⁰ = in (0.63/0.75) / in (0.093/0.1) = 2.4025

R90⁰ = in (0.67/0.75) / in (0.087/0.1) = 0.8099

where R(avg) = (R₀+2Ras+Rao) / A = (1.716+2(2.4025)+0.8099) = 3.0340

R(avg) = 3.0340

For alloy B;

R₀ = in (0.7/0.75) / in (0.082/0.1) = 0.3477

R45⁰ = in (0.69/0.75) / in (0.083/0.1) = 0.4475

R90⁰ = in (0.71/0.75) / in (0.078/0.1) = 0.2206

R(avg) = (R₀+2Ras+Rao) / A = (0.3477+2(0.4475)+0.2206) = 0.365825

R(avg) = 0.365825

comparing both we have that,

R(avg) for allow A > R(avg) for allow B

∴ Alloy A is the best to be selected for sheet formation operation.

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