Answer:
Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension
ex = бx/E
бx = Fx/A = Fx/π[tex]r^{2}[/tex]
Using both equation and solving for the modulus of elasticity E
E = бx/ex = Fx / π[tex]r^{2}[/tex]ex
E = [tex]\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa[/tex]
Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius
ey = [tex]\frac{1}{E}[/tex] (бy - v (бx + бz)) = [tex]-\frac{v}{E}[/tex]бx
= [tex]\frac{vFx}{Epir^{2} }[/tex] = [tex]\frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9} } = -0.63 *10^{-6}[/tex]
Finally
ey = Δr / r
Δr = ey * r = 10 * [tex]-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm[/tex]
Δd = 2Δr = [tex]-12.6 * 10^{-6} mm[/tex]
Explanation:
