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Problem with how we solved Temp stress question

  1. Dec 2, 2013 #1
    We solved a problem today in class calculating stress in a compound tube. I haven't posted the full question because I don't think I need to.

    There are two areas that im not sure how we found them, EX 1: -60*10^6
    how does 10^-6 and 10^9 become 10^6

    and Ex2: 6.7*10^-5
    and here how does10^-6 and 10^3 become 10^-5

    sure its simple enough but I cant figure it out.

    Hope the image is not confusing the first term on ex2 is Δx not Δ*
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

    Last edited: Dec 2, 2013
  2. jcsd
  3. Dec 2, 2013 #2
    Your attachment doesn't make any sense (to me). Please state the two problems that were solved.
     
  4. Dec 3, 2013 #3
    Here is the problem in full both are in there highlighted in light blue

    A compound tube is formed by a stainless outer tube of 50mm diameter and 47mm inside diameter, together with a concentric mild steel inner tube of wall thickness 6mm.The radical clearance between the inner and outer tubes is 2mm.
    The two tubes are welded together at their ends, the compound tube being free to expand when heated. Calculate the stress in each tube due to a temperature rise of 50degC and the final extension of the tube length is 500mm.
    s-stainless steel:E=175Gpa and coefficient of linear expansion α=18*10^-6 degC
    m-mild steel :E=200Gpa and coefficient of linear expansion α=12*10^-6 degC

    Im not sure if this α is the correct symbol for coefficient of linear expansion. for the porpose of making solution easier to write ill use it.

    αmΔT-αsΔT=σs/Es-σm/Em

    (αm-αs)ΔT=σs/Es-σm/Em

    (12*10^-6-18*10^-6)50=σ/175*10^9-σ/220*10^9

    (-6*10^-6)50*200*10^9=200σs/175-σm

    σs*As=-σm*Am

    Area s =228.6mm^2
    Area m=697.4mm^2

    -σm=σmAs/Am

    -σm=σs*228.6/697.4

    -σm=σs*0.328

    (-6*10^-6)50*200*10^9=200σs/175-σm
    sub in .328σs for σm

    (6*10^-6)50*200*10^9=1.14σs+0.328σs

    -60*10^6 =(1.14=0.328)σs

    σs=-40.87*10^6


    Stress σm
    -σm=0.328σs
    -σm=.0328*(-40.87)=13.4Mpa

    Extension on mild steel tube
    ΔX=strain*L
    (αmΔT=σm/Em)L

    ΔX=(12*10^-6*50+13.4/200*10^3)500

    ΔX=(600*10^-6+6.7*10^-5)500

    (667*10^-6)500

    ΔX=0.33mm
     
  5. Dec 3, 2013 #4
    Everything looks fine up to the calculation of the length change. That should be a 10-3, not a 103.
     
  6. Dec 3, 2013 #5
    Is it this line ΔX=(12*10^-6*50+13.4/200*10^3)500

    should be ΔX=(12*10^-6*50+13.4/200*10^-3)500


    I don't understand how we got 10^-5 in the next line.
     
  7. Dec 3, 2013 #6
    Yes. That's the line. Just be careful how you express things.

    [tex]\frac{13.4}{200}\times10^{-3}=6.7\times10^{-5}[/tex]
     
  8. Dec 3, 2013 #7
    I know this is very basic but I have been out of the education system along time.

    13.4/200=.067*10^-3, to bring the decimal after 6 I move it two places which adds -3+-2=-5 and that where 10^-5 comes from.
     
  9. Dec 3, 2013 #8
    For someone who has been out of the education system a long time, you did very well on the hard part of this problem. Then you got tripped up by the easy part: the arithmetic.
     
  10. Dec 3, 2013 #9
    That my problem in both maths and physics, the basics. it takes up time but I don't mind once I get solution finished correctly. am I right in how I got to 10^-5? -2+-3=-5
     
  11. Dec 3, 2013 #10
    Yes.
     
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