A copper rod is placed in an aluminum sleeve of length L = 8 in. The rod is 0.00

Question

A copper rod is placed in an aluminum sleeve of length L = 8 in. The rod is 0.005 in. longer than the sleeve. If a load of P = 71044 lb is applied to the bearing plate as shown, the gap between sleeve and rod is closed because of deformation. Determine the normal stress in the copper rod. The copper has a cross sectional area of 3 square inches and an elastic modulus of 17 x 10^6 psi. The aluminum has a cross sectional area of 2 square inches and an elastic modulus of 10 x 10^6 psi

Answer is between 18 - 25 ksi. Please show work

A copper rod is placed in an aluminum sleeve of length L = 8 in. The rod is 0.005 in. longer than the sleeve. If a load of P = 71044 lb is applied to the bearing plate as shown, the gap between sleeve and rod is closed because of deformation. Determine the normal stress in the copper rod. The copper has a cross sectional area of 3 square inches and an elastic modulus of 17 times 10^6 psi. The aluminum has a cross sectional area of 2 square inches and an elastic modulus of 10 times 10^6 psi

Explanation / Answer

deflection due to force P1 on copper rod is

del1=P1*(8.005)/(17*10^6*3)

deflection due to force P2 on aluminium is

del2=P2*8/(10*10^6*2)

we know P1+P2=71044 -------------i

and del1=0.005+delta2

so

P1*(8.005)/(17*10^6*3)=P2*8/(10*10^6*2)+0.005------------------ii

from equation i and ii we get

P1=60000 lb

P2=11044 lb

so stress=60000/3=20000psi=20ksi

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