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Section Modulus, major & minor axis

  1. Jan 20, 2009 #1
    1. The problem statement, all variables and given/known data

    I have attached a figure showing beam bending around 2 axis. I need to calculate section modulus Sx and Sy but seem to be getting the major and minor axis confused. I guess the problem comes in when I go to select which term is squared in my equation below.

    2. Relevant equations

    S = 1/6 * b * h^2 for rectangular cross section

    3. The attempt at a solution

    Sx = (70*750^2)/6 = 6.56E+6

    Sy = (750*70^2)/6 = 6.13E+5

    If this is correct, could you please help clarify why b and h are what they are in the section modulus equation for 1) major axis and 2) minor axis.

    Thanks.
     

    Attached Files:

    Last edited: Jan 20, 2009
  2. jcsd
  3. Jan 20, 2009 #2

    PhanthomJay

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    Yes, you have it correct. The major axis (X-X) bending section modulus is bh^2/6, and the minor (Y-Y) axis modulus is hb^2/6. What you have is a cantilever beam, 70 units x 750 units in cross section, and 320 units in length. The Fy force creates bending moments about the X axis, called the major axis because the section modulus is greater about that axis, and the Fx force creates bending moments about the (minor) Y axis. Visualize that under the Fy force, the beam is strong (large S_x) because it is 750 units deep, whereas under the Fx force, the beam is weak (small S-y), because it is only 70 units deep. Did I answer your question?
     
  4. Jan 20, 2009 #3
    Yes, thank you for your help. I don't know if you would also be able to help with this question, but how would you go about calculating stress in a fillet weld at the base of the cantilever due to this bending?

    I found some information online about finding where you find the throat of the weld by width / sqrt(2) but then how do you use this with the bending (moment)? Do you use a moment diagram / shear diagram to take max shear and divide by weld throat * weld length?
     
    Last edited: Jan 20, 2009
  5. Jan 21, 2009 #4

    PhanthomJay

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    You should probably go for a full penetration double bevel butt (tee) weld using the appropriate electrode and be done with it, but if you want practice in designing fillet welds to take shear and bending stresses, here's a site you can check: http://www.roymech.co.uk/Useful_Tables/Form/Weld_strength.html
    I used to design welds of all types 25 years ago, but I've since moved on to just specifying the loadings on the welds and have the designers do the calcs, so I've gotten away from the specifics of weld design (I've earned it!).
     
  6. Jan 21, 2009 #5
    Thanks PhanthomJay, this is exactly what I am looking for.
     
  7. Jan 21, 2009 #6
    Just to clear up what I read on the site,

    basically I take .707*weld leg and multiply it by the weld length to get a unit Area

    then calculate the moment of inertia for the weld as a line: 1/12 L^3

    Tbending = M.y/I u
    Tshear = P /A
    Tresultant = Sqrt (τ b2 + τ s2 )

    then plug into the above equations, and Presto! I can compare Tresultant to yield of material.
     
  8. Jan 21, 2009 #7

    PhanthomJay

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    I don't think that is quite correct; once you get T_resultant, that is the stress for a 1 inch weld; then you must divide T_resultant by the allowable weld shear stress, to get the required weld thickness. Note also that the allowable filet weld stresses are the shear allowables, I think it's 0.3*(nominal tensile strength of weld metal), but not to exceed 0.4*(F_yield) of base metal.
     
    Last edited: Jan 21, 2009
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