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

`E = 71,22 MPa`

Step-by-step explanation:

According to hookes law:

elongation =
`e = (dL)/(L) = (F)/(A*E)`

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

`E = (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 = (66700N+0,125m)/(2,7225*10^(-4) m2 * 0,00043 m)`

`E = 71,22 MPa`