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Finding the shear centre of a non-homogeneous cross section cross

  1. May 18, 2010 #1
    for the following cross section
    Capture.JPG
    the ratio Eb/Ep=30

    find the shear centre of the cross section
    -------------------------------------------
    first of all i need to find the equivalent cross section, which will be the same except that the horizontal parts will be of length n*L=30*1=30cm

    since this cross section has an axis of symmetry, i know that the shear centre passes through that axis, now all i need to find is the distance "eo"

    i know that i can move the force to the shear centre and the cross section must feel the same moment, i calculated the moment about a point that passes through the Vertical portion so that only the sums of the horizontal shearing stresses have an effect on the moment,

    i know that Q1y is the 1st area moment of each of the horizantal portions,

    Q1y=(0.15*s)*1.075 =0.16125*s

    I=10.6cm^4

    shearing stress=[tex]\int[/tex][tex]\int[/tex](V*Q/I/b)da*2.15 where 2.15 is the distance between the horizontal forces

    V*e=[tex]\int[/tex][tex]\int[/tex](V*Q/I/b)da*2.15

    e=2.15/(I*0.15)*[tex]\int[/tex]dt[tex]\int[/tex]Qds

    e=[2.15/(I*0.15)]*0.15*[tex]\int[/tex](0.16125*s)ds -->from 0 to 30



    [2.15/(10.6*0.15)]*0.15*0.16125*302/2=14.7cm

    the correct answer is somehow meant to be 0.408cm, can anyone see where i have gone wrong??
     
  2. jcsd
  3. May 18, 2010 #2
    your line: "shearing stress=LaTeX Code: \\int LaTeX Code: \\int (V*Q/I/b)da*2.15 where 2.15 is the distance between the horizontal forces"

    comment: I'm not sure what all those integral signs are doing. I thought shearing stress was just V*Q/I/b. (That is, the result of integration). OK, so you multiply by area to get force, and then by 2.15 to obtain moment. So should your "shearing stress" read as "moment"?
     
  4. May 19, 2010 #3
    o yes, thats correct, meant to be moment, but it still wont change the answer,

    im beginning to think that the way i expanded the cross section was incorrect, i just made the 2 horizontals 30 times longer ( to the right) maybe i was meant to expand them to both sides, 14.5 to each side,

    doing this would still not give me a correct answer, but much closer,
    by doing this i get 0.49cm

    is it possible that the answer i was given in the textbook (0.408) is incorrect,
    could someone please check this for me
     
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