Calculate Temperature for Closed Expansion Joint on Steel Bridge

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Homework Help Overview

The discussion revolves around calculating the temperature at which a closed expansion joint on a steel bridge would be reached, given an initial temperature and the dimensions of the bridge and joint. The subject area includes thermal expansion and its implications on structural engineering.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between temperature change and the resulting length change of the bridge due to thermal expansion. Questions arise regarding the initial conditions and the implications of the joint being closed.

Discussion Status

The discussion has progressed with participants clarifying the parameters of the problem and attempting to derive a formula to find the final temperature. Some guidance has been provided regarding the use of thermal expansion coefficients and the relationship between length and temperature change.

Contextual Notes

There is a focus on the assumptions regarding the initial length of the bridge and the temperature at which the joint closes. Participants are also considering the implications of the thermal expansion coefficient in their calculations.

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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