Question

an aluminum bar 125mm (5in) long and having a square cross section 16.5mm (.65in) on an edge is pulled in tension with a load of 66,700 N (15000lb) and experiences an elongation of .43mm (1.7*10^-2 in ). assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.

Answer 1

Answer:

$E = 71,22 MPa$

Explanation:

According to hookes law:

elongation = $e = \frac{dL}{L} = \frac{F}{A*E}$

Where F is the for aplied, A the cross section area, and E de modulus of elasticity. Solving for E:

$E = \frac{F+L}{A*dL}$

here $A= (0,0165m)^{2} = 2,7225*10^{-4} m2$

$L = 0,125m\\dL = 0,00043 m\\F = 66700 N$

$E = \frac{66700N+0,125m}{2,7225*10^{-4} m2 * 0,00043 m}$

$E = 71,22 MPa$