An aluminum alloy is characterized by the following engineer

An aluminum alloy is characterized by the following engineering stress-strain behavior. Determine the modulus of elasticity. the yield stress and the tensile strength of the alloy. A specimen of the same alloy (diameter-10 mm, length = 120) is subjected to a tensile force of 60.000 N. Compute the final length after the force is release, Compute the final length after a force of 120.000 N is released.

Solution

solution:

1) from graph above we can get value of modulus of elasticity as it is slope of stress strain curve within elastic limit

hence for elastic limit

stress=200*10^3 psi=1378953333N/m2

strain=.006875

hence

modulus of elasticity=E=stress/strain=1378953333/.006875=200.57*10^9 N/m2

2) elongation of rod of

d=10 mm nad L=120 mm

F=60000N

is given by

dl=F*L/A*E

dl=60000*.12*4/pi*(10*10^-3)^2*200.57*10^9=.457 mm

hence final length is l=L+dl=120+.457=120.457 mm

3)for F=120000 N

dl=F*L/A*E

dl=120000*.12*4/pi*(10*10^-3)^2*200.57*10^9=.9141 mm

hence final length

l2=L+dl=120+.9141=120.9141mm

4) yield stress is obtain at which stress gives certain deformation from graph we get

maximum stess sustain is of 2000 MPA

hence tensile strength=2000 MPA

yield stress=1400 MPA from graph