Answer to Question #350406 in General Chemistry for justin

Question

A cylindrical specimen of cold-worked copper has a ductility (%EL) of 15%. If its cold-worked radius is 6.4 mm, what was its radius before deformation?

Answer 1

The equation for a cylindrical specimen for percent cold work is a given by:

$CW= \bigg( \cfrac {r_0² π-r_d²π}{r_0²π}\bigg)*100$

However we are interested in the radius before deformation r₀ . We can therefore rearrange the above equation to:

$r_0=\cfrac {r_d}{\sqrt{1-\frac{CW}{100}}}$

15%EL correspond to 20% CW for copper.Thus,

$r_0=\cfrac {6.4}{\sqrt{1-\frac{20}{100}}}=\bold{ 7.155}$

Answer: 7.155 mm