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Homework Help: Shear stress in a thin profile

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello guys, I have to calculate the shear stress in this figure.
    The sizes are: a = 250mm, b = 100mm, c = 250mm, d = 150mm and t = 10mm.
    The shear force working on it is Vcd = 39802.01 N (calculated before)
    The center of the construction Nc is at 157.95 mm (from the bottom). (calculated before)
    The moment of inertia (I don't know if this is the correct english term) I = 147258143.9 mm4. (calculated before)

    The things I already calculated have been checked and approved.
    The calculating of Tau1 is also correct but after that one it goes wrong...

    2. Relevant equations
    Taui =(V*Si)/(b*I)
    Si = Ai * zi, in which z is the distance from the center of gravity of that part till the center Nc.

    3. The attempt at a solution
    I am not english from origin so my answer isn't in english but because it's math/physics you'll probably understand :). Otherwise I'll explain what I did.

    I ordered the different parts as followed: 1 is the vertical top part. 2 is the large horizontal part (with length a). 3 are the two vertical pieces and 4 are the horizontal pieces at the bottom.

    Tau1 = (V*S1)/(b*I) = (39802.01*S1)/(10 * I)
    A1 = 150*10 = 1500 mm2
    z1 = 400 - Nc - 0.5*150 = 167 mm
    S1 = 1500*167 = 250568 mm3 (numbers may differ because I rounded some numbers)
    Tau1 = 6.77 N/mm2

    Tau2 in the middle of the horizontal part is half of Tau1. So we start at:
    Tau2 = 1/2 * 6.77 N/mm2 = 3.39 N/mm2

    So from here on I am going to make some mistakes.

    Tau2 at the end of the horizontal part:
    Tau2= (V*S2)/(b*I) = (39802.01*S2)/(10 * I)
    A2 = 150*10+250*10 = 4000 mm2
    N1,2 = (d*t*0.5d+250*t*d)/(d*t+a*t) = 121.88 mm (from the top)
    z2 = 400 - Nc - N1,2 = 120 mm (rounded)
    S2 = 480681 mm3 (numbers may differ because I rounded some numbers)
    Tau2 = (V*S2)/(b*I*4)
    I added the 4 because the shear force splits in 2 ways and the area I calculated has to be divided in 2.
    Tau2 = 3.25 N/mm2

    Apparantly my Tau2 should be twice as high. I got calculations for part 3 too but you can guess, those are also wrong. Is there somebody who can help me with this?

    Thanks in advance!

    PS: sorry if I am a bit unclear with my notations

    Attached Files:

  2. jcsd
  3. Nov 30, 2015 #2
    I forgot to add my second page.

    My paperwork might be a bit unreadable so that's why I started to write it out. If you can read it however all my work is shown on those 2 handwritten pages.

    Attached Files:

  4. Nov 30, 2015 #3
    Please put yourself in our place. We have no idea what the loading on this object is. Please describe the loading.
  5. Nov 30, 2015 #4
    Sorry I totally forgot haha. The thing you are looking at is a cross section of a beam which has a shear force in it. The loading is the shear force which works downwards with 39.80201 kN or 39802.01 N.
    The load is centered so it works downwards in the middle (but it doesn't change in the formula so you can assume it constant at 39.8 kN)
    Does this make it better understandable?
    Last edited: Nov 30, 2015
  6. Nov 30, 2015 #5
    Yes. It is much more understandable now. But I still must be missing something. Shouldn't the shear stress simply just be the shear force divided by the cross sectional area of the beam?

  7. Nov 30, 2015 #6


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    Staff Emeritus
    Science Advisor
    Homework Helper

    As far as calculating τ2, the way you have set up the calculation is incorrect. Using the formula τ = VS / (I b) allows you to calculate the shear stress for a certain distance y above the centroid of the section. You can't split one wide horizontal member into smaller pieces.

    For your calculation of τ2, b = 250 mm and S is the first moment for the whole 250 x 10 mm piece.

    To obtain more accurate shear stress calculations in a wide horizontal piece, the shear flow method is used when analyzing complicated thin-walled sections like this one.

    No, that only gives an "average" shear stress. It is a very inaccurate method for calculating the shear stress for thin-walled beams composed of a lot of little pieces parts.
  8. Nov 30, 2015 #7
    Thanks SK. I am totally unfamiliar with all this, but it looks like it is now in your competent hands.

  9. Dec 1, 2015 #8
    Thanks SteamKing!
    I'll try to solve it, if I get the right answers I'll post my findings. And otherwise if I get stuck I'll just be asking new questions ;).
    Thanks Chestermiller for the input!
    Last edited: Dec 1, 2015
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