Calculate Temperature for Closed Expansion Joint on Steel Bridge

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SUMMARY

The discussion focuses on calculating the temperature required to close an 18.0 cm expansion joint on a 215 m steel bridge at an initial temperature of -15.0 degrees Celsius. The equation used is ΔL = L(α)ΔT, where ΔL is the change in length, L is the original length, α is the thermal expansion coefficient, and ΔT is the temperature change. The calculation shows that to achieve an expansion of 18 cm, the final temperature must reach approximately 54.8 degrees Celsius. This conclusion is confirmed by the participants in the discussion.

PREREQUISITES
  • Understanding of thermal expansion principles
  • Familiarity with the equation ΔL = L(α)ΔT
  • Knowledge of the thermal expansion coefficient for steel
  • Basic algebra skills for solving equations
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  • Research the thermal expansion coefficient for various materials, particularly steel
  • Learn how to apply the thermal expansion formula in practical engineering scenarios
  • Explore the effects of temperature changes on structural integrity in civil engineering
  • Investigate methods for measuring and monitoring thermal expansion in large structures
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Civil engineers, structural engineers, and anyone involved in the design and maintenance of large steel structures, particularly bridges, will benefit from this discussion.

lilkrazyrae
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A 215.m steel bridge has an expansion joint of 18.0cm at one end at -15.0 degrees C. At what temperature would the joint be closed?

I would use the equation Change in L=L(alpha)T

But would the L equal zero since the joint is closed?
 
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Yes sorry I forgot the m. So are there two answers? And the bridge would have to move from 18cm to 215m for the second part that seems improbable.
 
No, the bridge has length of 215 m when its temperature is -15°C.

If the temperature of the bridge increases, the length will increase (thermal expansion).

One is asked to find the temperature at which the bridge will expand 18 cm to close the gap, in which case the bridge length will be 215.18 cm.

The thermal strain is the ratio of length after expansion to the length before expansion, i.e. 215.18/215 = 1 + \epsilon, where \epsilon is the thermal strain.

Look at the link I provided - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thexp.html#c2

Let L = 215 m and \Delta{L} = 0.18 m (or 18 cm), and then find the temperature difference \Delta{T} assuming you know the thermal expansion coefficient \alpha.

See also the plate below - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thexp.html#c3
 
ok so .18=(215m)(12*10^-6)(T(f)-(-15)) and 69.767=T(f) +15 Finally T(f)=54.8 degrees C. Is this correct??
 
Correct. :smile:
 

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