Magnitude of force within thermally expanding restrained bar

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Homework Help Overview

The problem involves a steel rod that is thermally expanding while being restrained at both ends. Participants are tasked with determining the internal force generated due to a temperature increase, using properties such as the coefficient of linear expansion and Young's modulus for steel.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for the change in length (ΔL) and the resulting force (F) using the provided equations. There are attempts to verify unit conversions and the application of formulas.

Discussion Status

Some participants are sharing their calculated values and questioning the correctness of their results. There is a focus on checking mathematical accuracy and unit consistency, with no clear consensus on the correct answer yet.

Contextual Notes

Participants are working under the constraints of specific material properties and temperature changes, and there is an emphasis on ensuring that all units are correctly converted and applied in calculations.

AnnaWerner
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Homework Statement


A steel rod with a length of l = 1.55 m and a cross section of A = 4.52 cm^2 is held fixed at the end points of the rod. What is the size of the force developing inside the steel rod when its temperature is raised by ∆T = 41.0 K? (The coefficient of linear expansion for steel is α = 1.17×10-5 1/K, the Young modulus of steel is E = 200 GPa.)

Homework Equations


F=EA(ΔL/L)
ΔL = L*α*ΔT

The Attempt at a Solution


OK, here is what I have done:
1) Convert GPa to Pa. In my case, 200 GPa = 2*(10^11)Pa. Also, convert A from cm^2 to m^2.
2) Find ΔL: L*α*ΔT (L in m, α, and ΔT in K given).
3) Apply ΔL to F = EA(delta L/L).

Thank you for any assistance.
 
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That looks OK. Put the numbers in and see what you get
 
Well, I have tried this multiple times, but my answer consistently comes out incorrect. The units are supposed to be Newtons, correct? I'm getting 105,768 N. Seems large . . .
 
I got ΔL =7.43x10^-4m
Then F = (2x10^11 x 7.43x10^-4 x 4.52x10^-4)/1.55 which came to 43364N
Double check your maths and mine! see if you can spot a difference.
 

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