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Shear stress question

  1. May 12, 2012 #1
    1. The problem statement, all variables and given/known data

    26eyH.png

    2. Relevant equations

    Mx=200Nm

    My=300Nm

    Mz=600Nm


    Normal stress caused by bending moment at A = Mc/I = (300*0.02)/((pi*0.02^4)/4)= 47.7MPa. This answer is correct.




    Shear stress by Torsional Moment=Tc/J

    Polar moment of inertia J=(pi/2)*c^4

    So I did (600*0.02)/((pi*0.02^4)/2)= 47.7MPa Is this correct?

    &

    VQ/It=1000*((4*0.02)/3*pi)*((pi*0.02^2))/(1.257*10^-7*0.04)=10.47 MPa Is this correct?

    One of the stress calculations is wrong beacuse when added that should equal 48.8 MPa

    So what I'm I doing wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 12, 2012 #2

    PhanthomJay

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    it appears you calculated Q incorrectly, ormade a math error, one or the other. The Q of a semicircle about its base is 2r^3/3
     
  4. May 13, 2012 #3
    Is Q not =y bar prime * A prime = 4r/3pi * (pi*r^2)/2 ?
     
  5. May 14, 2012 #4

    PhanthomJay

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    certainly, which simplifies to 2r^3/3. So you have made a math error...you
    forgot to divide by 2 when determing area of semicircle.....and some other calculation error..try again and you should get the correct shear stress as 1.06 MPa
     
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