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An aluminium cylinder 10 cm long, with a...

An aluminium cylinder 10 cm long, with a cross-sectional area of `20cm^(2)`, is to be 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 if exerts on each wall, assuming that the walls are perfectly rigid and a constant distance apart.

Text Solution

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For aluminum, `Y=7.0xx10^10 Pa` and prop `=2.4xx10^-5k^-1`. The temperture change is `DeltaT`,
`DeltaT=5.1 C^(@)`=5.1 K
Stress=`F/A=-YpropDeltaT`
`=-(0.70xx10^(11)Pa)(2.4xx10^(-5)k^(-1))`(5.1K)
Stress =-`8.6xx10^6Pa`
The total force F is the cross-sectional times the stress
`F=((F)/(A))A`
`=(20xx10^(-4)m^(2))(-8.6xx10^(6)Pa)`
`=-1.7xx10^4N`
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