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An aluminum cylinder 10 cm long with a c...

An aluminum cylinder `10` cm long with a cross section area of `20cm^2` is used as a spacer between two steel walls. At `17.2^(@)C` it just slips in between the walls. When it warms to `22.3^(@)C` calculate the stress in the cylinder and the total force it exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.

Text Solution

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For aluminium, `Y=7.0xx10^(10)Pa and alpha=2.4xx10^(-5)K^(-1)`. The temperature change is `DeltaT,DeltaT=5.1C^(2)=5.1K`

Stress`=(F)/(A)=-YalphaDeltaT`
`=-(0.70xx10^(11)Pa)(2.4xx10^(-5)K^(-1))(5.1K)`
The total force F is the cross-sectional times the stress.
`F=((F)/(A))A`
`=(20xx10^(-4)m^(2))(-8.6xx10^(6)Pa)`
`=-1.7xx10^(4)N`
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