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Homework Help: Maximum Bending Stress in Beam

  1. May 2, 2010 #1
    1. The problem statement, all variables and given/known data
    Note the attachment. I scanned it out of my book and I drew a FBD on it.


    2. Relevant equations



    3. The attempt at a solution

    Ok so i havent gotten very far at all with this one because it just doesnt seem to have enough information.

    So first I summed the forces in the y direction.

    Fy = 0 = 9k - RB.... RB = 9k

    Then I found the Moment @ B

    MB = 9k(192) = 1728k

    and thats about as far as I got. It seemed odd that there was no information given about the forces or distances at point A.

    so I know that the Bending stress = -My/I

    but how do I find the y. Im assuming it would come from k = 1/p but I do not know any of those quantities. Same with I = integral y2dA. Can I get a hint as to my next step
     

    Attached Files:

  2. jcsd
  3. May 2, 2010 #2

    PhanthomJay

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    There is a force at A (and B), but the max bending moment at B (in inch-kips) is the same on either side of support B, so you don't need the force at A to calculate it. You should calculate the moment of Inertia , I, of the wide flanged beam, using the parallel axis theorem. Ignore the small fillets where the web connects to the flanges.
     
  4. May 3, 2010 #3
    Ok so find I i used an equation from wikipedia

    http://en.wikipedia.org/wiki/Second_moment_of_area#Parallel_axis_theorem

    I calculated it for Ix & Iy, but it my book it doesnt specify which ones to use. do i add them together or did i use the wrong formulas

    Ix = 1995 & Iy = 75

    my book says for doubly symmetric shapes... [tex]\sigma[/tex]max = M/S where S = I/c

    where do I go from here
     
  5. May 3, 2010 #4

    PhanthomJay

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    The beam is bending about the strong axis of the beam, not the weak axis, so you just need to use the I that applies, not both. You don't add the moments of inertias. Don't forget your units of I (in^4). I assume you understand the value to use for 'c' ?
     
  6. May 3, 2010 #5
    Ok so i used the Iy because that is the direction of movment = 75 in^4

    and c would just be half of the height because it is a symetric beam

    S = I/c = 75in^4/11in = 6.81in^3

    [tex]\sigma[/tex] = M/S = 1725kip*in/6.81in^3 = 254 kips/in^2

    is this correct
     
  7. May 3, 2010 #6

    PhanthomJay

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    You've got your Ix and Iy mixed up. But you managed to get the correct c distance.
     
  8. May 3, 2010 #7
    ok I missunderstood

    S = 1995/11 = 181.4 in^3

    Stress max = M/S = 1728kips*in / 181.4 in^3 = 9.5 is kips/in^2 = psi
     
  9. May 3, 2010 #8

    PhanthomJay

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    yes, that's 9.5 kips/in^2, or 9.5 ksi, or 9500 psi.
     
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