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Answer :
Answer:
a) 0.4393 mm
b) -0.141 mm
Explanation:
Cylindrical bar : Diameter = 20.3 mm , Length = 205 mm
Force that deforms bar of metal = 46400 N
elastic modulus = 66.9 GPa
Poisson's ratio ( u ) = 0.32
A) Determine the amount by which this specimen will elongate in the direction of applied stress
dl = [tex]\frac{P*L}{A*E}[/tex]
where P = 46400 N
L = 205 mm
A = [tex]\frac{\pi }{4} * 20.3^2[/tex] = 323.65
E = 66.9 GPa
dl = ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 ) = 0.4393 mm
B) determine the change in diameter of the specimen
change in diameter( compressed due to elongation in length )
= - u * dl
= - 0.32 * 0.4393
= -0.141 mm
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