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!

Beams subject to distributed load/centroids

  1. Apr 10, 2013 #1
    Picture of the problem:
    16i9ikl.jpg

    Relevant equations are the ones used to find the reactions at the supports (moment about points, forces in the x and y) and the ones used to determine the centroid.

    I'm having a hard time finding a. I found the moment of area for each triangle, with the origin at A, and the sum of the areas turned out to be ƩA=1200+600a and ƩxA=200a^2-400a+3200. I know that the coordinate X of the centroid can be found by dividing ƩxA by ƩA, and doing so brought me no closer to finding a. I then tried to set up the equations for the moment about point A, Fy, and Fx and got:

    Ma=500a^2-400a+3200-4By=0 (By is the reaction force at point B in the y-direction)

    Fx=0, Ax=0 (since all other forces, including the roller at point B, are vertical)

    Fy=Ay+By-(1200+600a)=0 (where the 1200+600a is the combined load applied to the beam, figured out from the centroid stuff earlier)

    I can't figure out how to manipulate the centroid or moment/force equations to obtain a. Any help would be much appreciated.
     
    Last edited: Apr 11, 2013
  2. jcsd
  3. Apr 11, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Could you post your derivation of Ma?
     
  4. Apr 11, 2013 #3
    Sure.

    Ma equals the moment of area cross the area for each triangle (at point A).

    So, for the first triangle on the left side, I got an area of .5(1800)a=900a (area of a triangle), and x-distance of the centroid of the triangle, which is 1/3 the height of the triangle, or a/3 in terms of the coordinate system with origin at A. I did the same for the other triangle, and got an area of .5(600)(4-a), which comes out to 1200-300a, and the x-direction for the centroid came out to a+(2/3)(4-a), which comes out to (8/3)+(a/3). To do the moment about A, I summed the x-distances to the centroids multiplied by the areas. I also accounted for the reaction at B (which is in the y-direction like the other forces). So this came out to:

    Ma=(a/3)(900a)+[(8/3)+(a/3)](1200-300a)-4B=0

    I simplified this to the Ma formula that you're inquiring about.
     
  5. Apr 11, 2013 #4
    Actually, the formula comes out to 200a^2-400a+3200-4By=0, I made an error adding up the a^2's. Still stumped on how to determine the value for a that would bring a minimum reaction at B. I tried manipulating the force and moment equations together to come up with a, but there's something I'm not seeing here.
     
    Last edited: Apr 11, 2013
  6. Apr 11, 2013 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You know from statics that the sum of the moments about A must equal zero for the beam to be in equilibrium. You have the moment equation in Post 4 which involves a and By. Set this equation = 0 and solve for By. What you want to do is find the value of a which minimizes By.
     
  7. Apr 14, 2013 #6
    Have you tried drawing the graph of By v a?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted