Question

The two cylindrical rod segments are fixed to the rigid walls such that there is a gap of 0.01 in. between them when t1 = 60°f. each rod has a diameter of 1.25 in. determine the average normal stress in each rod if t2 = 400°f, and also calculate the new length of the aluminum segment.

Answer 1

Answer:

$T = 116 F$

average normal stress = 11.6 ksi

Explanation:

Data:

First, we calculate the thermal expansion required to close the gap:

$\delta _{T} = \alpha _{al}\delta TL + \alpha _{cu}\deta TL_{cu}\\ 0.008 = 13 (10^{-6})(T_{2} - 60)(8) + 9.4 (10^{-6}) (T_{2} - 60) (4)\\ T_{2} = 116 F$

Next, we determine the resultant compatibility:

$0.008 = (\delta _{Tcu})-(\delta _{Tal})-(\delta_{Fal})\\ 0.008 = 9.4 (10^{-6}) (200 - 60)(4)- \frac{F(4)}{\frac{\pi }{4}*(125^{2})(18)(10^{3}) } + 13 (10^{-6}) (200 - 60)(8) - \frac{F(8)}{\frac{\pi }{4}(1.25^{2})(10)(10x^{3}) } \\F = 14.195 kip$

The average normal stress:

$\sigma _{al} = sigma_{cu} = \frac{F}{A} = 11.6 ksi$

Answer 2

Hmmm a b cd hi ugdff