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I-beam Shear Force calculation

  1. Nov 16, 2008 #1
    So I'm taking a Strengths of Materials class right now (aka Mechanics of Materials) and my teacher absolutely sucks. We ask him questions but he doesn't have answers, so in turn I have no idea what is going on and why the book does what it does.

    Right now we're working with I-beams and some of the problems deal with Shear Force (not to be confused with Shear stress). The question says "If the wide flanged beam is subjected to a shear of V = 30 kN, determine the shear force resisted by the web of the beam." Then it gives the dimensions and all that. I've asked my teacher to do this problem but he just copies from the solutions manual and doesn't explain anything.

    So what the solution did was find the things needed for the equation T(shear) = VQ/It. It first finds I, which I can figure out, but I don't get how it calculates Q. It takes the height from the Neutral Axis to the top (.155 in this case), then adds an arbitrary y value. It divides this value by 2, then multiplies by (.155 - y)*(.2) ...where .2 is the length of the top of the I-beam. And that's it. Usually when Q is calculated (that I've seen) it's the distance of the centroid of the part to the NA, multiplied by that parts area. (Part being...that you can break I-beams into the horizontal part and the vertical part). But I don't understand why in this problem it seems to be taking the whole area (of a box instead of the top half of the I-beam) and multiplying it by (.155 + y)/2. Where does this value come from?

    Then in a DIFFERENT problem reading, "If the wide flange beam is subjected to a shear of V = 25 kip, determine the shear force resisted by the web of the beam." I is calculated the same way...but Q is totally different than the previous problem. It calculates the horizontal part like I thought, but then it adds what I presume is the vertical part, making it look like:

    (.5y + 2.5)*(5-y)*(1)

    Where the total height of the vertical part is 10 inches (thus the 2.5 is 1/4 of 10 [half away from the NA]) and 5-y I guess is the arbitrary height of the "point". I have no idea where the .5 comes from (except maybe half of the thickness of the horizontal part, which is 1 in), but why is it multiplied by y? And why is it added to the horizontal part whereas in the last problem it was just the Q of what seemed to be the whole box?

    Sorry if this is terribly confusing, but I don't know how else to explain it without using pictures. I'm just confused what their method of calculating Q is and why it's different between the two seemingly same problems. Thank you for any help in advance and let me know if you need me to clear some things up! ^_^
  2. jcsd
  3. Apr 13, 2011 #2
    I know this is really old but I have exactly the same issue! I'll be asking my lecturer within the week though. I think it has to do with the thickness value used though...

  4. Apr 14, 2011 #3
  5. Apr 14, 2011 #4
    Thanks. Its just that I had exactly the same question as the OP which I thought was interesting. Anywyas I figured it was just a different method of calculation.

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