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Homework Help: Finding Reactions In Simply Supported Beams

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Hope you can access this link for question http://s353.photobucket.com/albums/r387/james_nikko/?action=view&current=img002.jpg

    Supports are at 1.5m and 4.5m

    2. Relevant equations

    3. The attempt at a solution
    Ok so i have attempted the problem using the following
    (5x1.5) + (5x3) + (5x6) = (Rb x 4.5)

    Therefore giving:
    (Rb x 4.5) = 52.5
    Rb= 11.67kN

    Ra = (5+5+5) - Rb
    Ra = 3.33kN
    Last edited: Oct 29, 2009
  2. jcsd
  3. Oct 29, 2009 #2


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    When the beam is in equilibrium, you can sum moments of a force about any point and set the sum equal to zero, but you must compute the moments of each force about that same chosen point, and then watch your plus and minus signs (clockwise vs. counterclockwise). Although you can choose any point, it is convenient to choose a point on the beam at one of the reaction supports. Try again to sum moments about R_a, and watch your plus and minus signs. Note from the symmetry of the loading and beam, the values of the end reactions, as you gain more experrence, should pop right out at you.
  4. Oct 29, 2009 #3
    ok so my first force which is before point Ra on an overhang should be negative whereas the other two forces are positive? assuming i use a negative reaction at b of course. therefore formula would be, -(5x1.5) + (5x3) + (5x6) - (Rbx4.5)?
  5. Oct 30, 2009 #4


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    You are not summing moments correctly. All perpendicular distances must be measured between the force and the point (Ra) about which you are summing moments.

    So I'll start you off, it's -(5 x 1.5) + (5 x 1.5) + (5 x ___) -(Rb x ___) = 0. Fill in the blanks and solve for Rb.
  6. Oct 30, 2009 #5
    sorry i was working from memory :) formula should be
    -(1.5 x 5) + (5 x 1.5) + (5 x 4.5) - (Rb x 3)
    My major problem was that i wasn't using the force to the left of reaction a as a - number.
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