Answer:
[tex]T = 116 F[/tex]
average normal stress = 11.6 ksi
Explanation:
Data:
First, we calculate the thermal expansion required to close the gap:
[tex]\delta _{T} = \alpha _{al}\delta TL + \alpha _{cu}\deta TL_{cu}\\ 0.008 = 13 (10^{-6})(T_{2} - 60)(8) + 9.4 (10^{-6}) (T_{2} - 60) (4)\\ T_{2} = 116 F[/tex]
Next, we determine the resultant compatibility:
[tex]0.008 = (\delta _{Tcu})-(\delta _{Tal})-(\delta_{Fal})\\ 0.008 = 9.4 (10^{-6}) (200 - 60)(4)- \frac{F(4)}{\frac{\pi }{4}*(125^{2})(18)(10^{3}) } + 13 (10^{-6}) (200 - 60)(8) - \frac{F(8)}{\frac{\pi }{4}(1.25^{2})(10)(10x^{3}) } \\F = 14.195 kip[/tex]
The average normal stress:
[tex]\sigma _{al} = sigma_{cu} = \frac{F}{A} = 11.6 ksi[/tex]
