A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ௫ = 2.75*10-6, determine the modulus of elasticity E and the change in the rod’s diameter. Let Poisson’s ratio ν = 0.23.

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

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πex

E =

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 = (бy - v (бx + бz)) = бx

= =

Finally

ey = Δr / r

Δr = ey * r = 10 *

Δd = 2Δr =

Explanation: