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Thermal Expansion with bending bars

  1. Aug 22, 2007 #1
    Well I'm having problems trying to understand this problem with a bending bar in my text book. Since we barely touched the subject of thermal expansion in class (there were no examples on this kind of problem) I'm a little confused. The hint in the problem makes no sense to me and gets me no where and I've looked all over for something to give me any clues on it and unfortunatly the search has left me empty handed.

    The Original Problem

    I've attached a picture of the original problem incase there are things I leave out on accident.

    Problem: A metal bar is 1.75 meters long with a coefficent of thermal expansion of
    1.34 X 10^(-5). It is rigidly held between two fixed beams. When the temperature rises, the metal bar takes on the arc of a circle (see attachment). What is the radius of the curvature of the circle when the temperature rises by 40 degrees celcius?

    Hint the problem gave me

    use small angle approximation sine theta = theta - theta^3/6

    Here is what I came up with

    At first I thought the problem wasn't going to be to bad and immediately calculated the dL and the resulting L'

    using the following

    a*L = dL/dT

    (1.35 * 10^(-5))*(1.75) = dL/40

    dL = 0.000938

    L + dL = L'

    1.75 + 0.000938 = 1.75094

    After that I tried going to the chapter where the hint came from to make more sense out of it and figure out how to use it to solve the problem. Unfortunatly it didn't help me and the hint made no sense becuase I couldn't think of any angles to use. Basically I need help figuring out what to do next.

    Any and all help is greatly appreciated
     

    Attached Files:

  2. jcsd
  3. Aug 22, 2007 #2
    Think of a circle, where the original, straight bar is a chord, and the new bar is part of the circumference, spanned by the chord. Now it's some circle geometry, to find the radius. Let's make some variables: radius is R, the angle subtended by the chord is theta...

    Relate chord length:
    [tex]2 R \sin(\theta/2) = L \approx 2 R \left(\frac{\theta}{2} - \frac{\theta^3}{24}\right) = R\theta\left(1 - \frac{\theta^2}{12}\right)[/tex]

    Relate circumference segment:
    [tex]R \theta = L + dL[/tex]

    Combine:
    [tex]\frac{L}{L+dL} = 1 - \frac{dL}{L+dL} \approx 1 - \frac{\theta^2}{12}[/tex]

    Solve for theta, and substitute back into the 2nd equation to find R.
     
  4. Aug 22, 2007 #3
    thanks alot! I think I understand it now.

    Well after reading over it I'm not quite sure where the theta cubed / 24 and 1 - theta squared / 12 came from but I recognize it from somewhere.....ok I get it nevermind.
     
    Last edited: Aug 22, 2007
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