1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating support reactions in a beam

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
    b515aw.jpg

    Trying to calculate the support reactions here.


    2. Relevant equations
    ƩFy = 0
    ƩMA = 0


    3. The attempt at a solution

    Well I've come up with the first equation (I assume two are needed to solve for both support reactions from the above equations).

    ƩFy = 0

    RA + RB (up) = 10kN + 10kN + 4(10kN) (down)

    RA + RB = 60kN

    Not too sure how to go about calculating the sum of the moments in this case. But here's an attempt, I may be completely wrong! But then again that's why I'm here.

    ƩMA = 0

    Taking moments about A in a positive clockwise direction:

    (4+2+2)(10kN) + (4+2)(10kN) (clockwise+) = 4(10kN)(4 + 4/2) + 4RB (counter clockwise-)

    (8)(10kN) + (6)(10kN) - 4(10kN)(4 + 4/2) - 4RB = 0

    4/2 is the centroid of the uniformly distributed load 10kN

    Any help would be greatly appreciated!

    Thanks all.
     
    Last edited: Mar 20, 2013
  2. jcsd
  3. Mar 20, 2013 #2
    Is A the pin support and B the roller support? I dont see it on picture
     
  4. Mar 20, 2013 #3
    Yep!
     
  5. Mar 20, 2013 #4
    Your mistake is where you take the moment of the uniformly distributed load.

    You have "4(10kN)(4 + 4/2)" for the moment of the uniformly distributed load.
    You're right that the resultant force from the 10kN/m load will be 4*10 because the load is acting on the 4 meter length. This resultant force is now 40KN at the center of that length (which is 2m). So as you know, the moment will be the resultant force multiplied by the distance to point A. So you don't need to add the original distance, just where the resultant force is acting.

    Edit: Also, the moment from the resultant force acts clockwise(+) because it points down and is to the right of A
     
  6. Mar 20, 2013 #5
    Thanks Blugga!

    So the moment for the UDL is 40kN(2m), 2m being the centroid of that section? I understand now.

    Which is the moment from the resultant force?
     
  7. Mar 20, 2013 #6
    Yes it is, and it's clockwise.
     
  8. Mar 20, 2013 #7
    Great. So just to clarify, the correct summing of moments about A is:

    (8)(10kN) + (6)(10kN) + 40kN(2) - 4RB = 0

    Thank you very much
     
  9. Mar 20, 2013 #8
    That's what I got. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted