Magnitude of force within thermally expanding restrained bar

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SUMMARY

The discussion centers on calculating the internal force developed in a steel rod due to thermal expansion when subjected to a temperature increase of 41.0 K. The rod has a length of 1.55 m and a cross-sectional area of 4.52 cm². Using the coefficient of linear expansion (α = 1.17×10-5 1/K) and Young's modulus (E = 200 GPa), the formula F = EA(ΔL/L) is applied. The correct internal force is determined to be approximately 43,364 N, contrasting with an incorrect calculation of 105,768 N.

PREREQUISITES
  • Understanding of thermal expansion principles
  • Familiarity with Young's modulus and its application
  • Basic knowledge of unit conversions (GPa to Pa, cm² to m²)
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Review the derivation of the thermal expansion formula ΔL = L*α*ΔT
  • Study the application of Young's modulus in material science
  • Practice unit conversions relevant to engineering calculations
  • Explore error analysis techniques in mechanical calculations
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Mechanical engineers, materials scientists, and students studying thermodynamics or mechanics of materials will benefit from this discussion.

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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