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Temperature and materials

  1. Nov 23, 2009 #1
    how do i go about solving this
    Capture.JPG

    im not sure ive learned this material yet but i have learned defrmations and deflections,
    i need to somehow find the force applied to the bars due to their exopansion and the restraint.

    what is [tex]\alpha[/tex] exactly? the increase in length for each degree F??
     
  2. jcsd
  3. Nov 23, 2009 #2

    mgb_phys

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    Yes it's the fraction increase in length with temperature.
     
  4. Nov 23, 2009 #3
    line the strain per degree?
    can i use superposition here, find the strain on each material annd then add them? how will i take into account the frame if the total deflection comes more than 0.02in
     
  5. Nov 23, 2009 #4

    mgb_phys

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    Yes, you assume the base doesn't change an the two parts each change in length by whatever factor.
    Then you have two bars that are under a certain strain (they want to be longer than they are) you then use Young's modulus to fin the stress and so the force
     
  6. Nov 24, 2009 #5
    so then can i say that
    εxx=α*T
    εyy=0
    εzz=0

    for aluminium
    εxx=α*T=2.56*10-3

    for bronze
    εxx=α*T=2.02*10-3

    now i have all the strains for both materials
    from a table i get the poisson ratios
    aluminuim->0.33
    bronze->0.34

    (aluminium) σxx= 37.93*103Pa
    (Bronze) σxx= 59.1*103Pa


    now what i tried here was Fx=σxx*A but that doesnt come right, how do i tak the interaction between the 2 materials into account?? i think that may be where im going wrong?
     
    Last edited: Nov 24, 2009
  7. Nov 24, 2009 #6

    mgb_phys

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    so calculate how long each bar would expand to if it was free.
    then if this is more than the 0.2" extra space - you have two bars with a total length of X compressed into 27.2" from the youngs moduluses work out how much each bar would be compressed ad there fore the stress and strain.
    Note the bars are different materials so the stronger one will squash the weaker one
     
  8. Nov 24, 2009 #7
    for Bronze
    εxx=α*T=0.00202
    A=2.5
    E=15*106
    v=0.34

    according to hookes law

    σxx=[tex]\frac{E(1-v)εxx}{(1+v)(1-2v)}[/tex]

    =[tex]\frac{15*106(1-0.34)*0.00202}{(1+0.34)(1-2*0.34)}[/tex]=46637.12687

    Fx=σxx/A=46637.12687/2.5
    Fx=18654.85075

    for aluminium
    εxx=α*T=0.00256
    A=3
    E=10*106
    v=0.33

    according to hookes law

    σxx=[tex]\frac{E(1-v)εxx}{(1+v)(1-2v)}[/tex]

    =[tex]\frac{10*106(1-0.33)*0.00256}{(1+0.33)(1-2*0.33)}[/tex]=37930.11942

    Fx=σxx/A=37930.11942/3
    Fx=12643.37314


    FxBronze-FxAl= 6011.47
     
    Last edited: Nov 24, 2009
  9. Nov 24, 2009 #8
    another way i looked at it was through the deflection, now since i only have force on the x axis

    delta=εxx*L
    =σxx*L/E
    =(F*L)/(A*E)

    now i know that the total deflection is 0.02

    0.02=(F*15)/(3*10^7)+(F*12)/(2.5*15*10^6)

    F=24390.2439

    i just cant get anywhere near the correct answer, obviously i am doing something fundamentally wrong
     
  10. Nov 24, 2009 #9

    mgb_phys

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    Bronze a=10.1e-6, original length = 12"

    Al a=12.8e-6 orig = 15"

    so after 200f,
    Al = 200*10.1e-6 * 15 = 0.0303 extra, so 15.03 total
    Br = 200*12.8e-6 *12 = 0.0307 extra, so 12.03 total

    total extra length = 0.061 an we only have 0.02 so we have 0.041 extra compression providing the stress

    The force in both bars must be the same - otherwise they would move!
    So we have an unknown expansion of each but a known total expansion.
    We know the total strain of the two bars
    From the strain and E (and A) we can find the stress and so the force.

    Should just be a couple of simultaneous equations
     
  11. Nov 24, 2009 #10
    εtot=(0.02) / 27
    =7.407407e-4

    σ=ε*E
    F=ε*E*A

    F=al*10e6*3
    F=br*15e6*2.5
    al*15+br*12=0.02

    after solving i get
    F=24390.2439
    al=8.13e-4
    br=5.65e-4


    still no good

    i can see where this might be wrong, nowhere here do i take into account the amount that each material expands.
    but i have no idea how to fix it
     
    Last edited: Nov 24, 2009
  12. Nov 25, 2009 #11
    i think i found a mistake, in the 3rd equation

    F=al*10e6*3
    F=br*15e6*2.5
    al*15+br*12=0.041

    F=50 000
    al=1/600
    br=1/750


    delta=al*L

    =15/600=0.025


    still not the right answers
     
  13. Nov 25, 2009 #12
    i think i might finally have it (again)

    the strain i found in my last post was the strain that the force f applied, but i cannot find the deflection from this

    what i think i need to do is find some kind of effective strain which is the difference between the strain had there been no restriction and the strain i just found ??
     
  14. Nov 25, 2009 #13
    got it! thanks for all the help
     
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