An aluminium cylinder 10 cm long. As a spacer between two steel walls. At 12^(@)C it juct slips in between the walls. When it warms to 23^(@)C calculate the stress in the chlinder and the total force it exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.

An aluminium cylinder 10 cm long. As a spacer between two steel walls.

1.	An aluminium cylinder 10 cm long. As a spacer between two steel walls. At 12^(@)C it juct slips in between the walls. When it warms to 23^(@)C calculate the stress in the chlinder and the total force it exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.
Answer» Solution :GIVEN, For aluminium `Y=7.0xx10^(-10)` PA and `alpha = 2.4 xx 10^(-5) K ^(-1)` Change in temperature, `DeltaT = 5.1^(@)C = 5.1 K ` Area A `= 20 cm^(2) = 20xx10^(-4) m^(2)` `therefore STRESS=(F)/(A)=YalphaDeltaT=(0.70xx10^(11)Pa)(2.4xx10^(-5)K^(-1))(5.1K)` `= 1.68xx5.1 xx 10^(6)= 8.6 xx 10^(6)` Pa The total forcr F is the cross-seactional times thje stress. `F= "stress xx area"` `=(8.6xx10^(6)Pa)(20xx10^(-4)m^(2))=1.7 xx10^(4)N`