The circle of diameter d 9 in is scribed on an unstressed a

The circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stress sigma_x = 12 ksi and sigma_z = 20 ksi. For E = 10 times 10^6 psi and u = 1/3, determine the change in (a) the length of diameter AB, (b) the length of diameter CD, (c) the thickness of the plate, (d) the volume of the plate

Solution

for strain produced in x direction, e_x = (xx zz) / E = (12000-0.33*20000) / 10000000 = 0.00054

so final length is 20.00054 in. so new dia.length of AB is 9.00054 in. (Assuming uniform Strain)

for strain produced in z direction, e_z= (zz xx) / E = (20000-0.33*12000) / 10000000 = 0.001604

so final length is 20.001604 in. so new dia.length of CD is 9.001064 in. (Assuming uniform Strain)

here stress is high in z direction so deflection is high in z direction.

here, thicknes will decrease. so vloume will be same as was earlier.

decrease in thickness will be = v(xx+zz) / E = -0.33(20000+12000)/10000000 = -0.001056
so final thickness will be = 0.75-0.001056 = 0.7489 in.

Final V = L*B*H = 20.00054*20.001604*0.7489 =299.59 in³