A circle of diameter d 9 in is scribed on an unstressed alu

A circle of diameter d = 9 in is scribed on an unstressed aluminimum plate of thickness t = 0.75 in. Forces acting in the plane of the plate later cause normal stresses sigma_x = 12 ksi and theta_t = 20 ksi. For E = 10 middot 10^6 psi and v = 1/3, determine the change in (a) the length of the diameter AB, (b) the length of the diameter CD, (c) the thickness of the plate, and (d) the volume of the plate.

Solution

tehe longation is given by

axis y

20000/(10*10^6)= 0.002 =L/9

L=0.018

so the diameter AB is now 9.018 in

axis Z

12000/(10*10^6)= 0.0012 = L/9

L=0.0108 in in DC

so the new leng in DC is 9.0108 in

e=(1-2v)/(10*10^6) (20000+12000) = 0.001067

so the new volume is

15*15*0.75 (inicial) + (0.001067)(15*15*0.75) = 168.93 in^3

the new leng in z is

15+15* 0.0012 = 15.018 in

the new leng in y is

15+15*0.002=15.03 in

so the thickness is

15.03*15.018*t=168.93

0.7484 in