Beams subject to distributed load/centroids

[adam199]

Picture of the problem:

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.

 

[SteamKing]

Could you post your derivation of Ma?

 

[adam199]

Could you post your derivation of Ma?

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.

 

[adam199]

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.

 

[SteamKing]

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.

 

[pongo38]

Have you tried drawing the graph of By v a?