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Natasha_Volkova [10]
3 years ago
6

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.
Engineering
1 answer:
Alenkasestr [34]3 years ago
6 0

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

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