# Beams subject to distributed load/centroids

In summary, you are trying to find the reaction at point B for a beam in equilibrium by using the moment equation and the force equation. You are having a hard time figuring out what the value for a is that would bring the reaction at B to a minimum.
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.

Last edited:
Could you post your derivation of Ma?

SteamKing said:
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.

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:
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.

Have you tried drawing the graph of By v a?

## 1. What is a distributed load on a beam?

A distributed load on a beam is a type of external force that is applied along the length of the beam, rather than at a single point. This load is evenly spread out over the entire length of the beam, and is typically represented by a linear function.

## 2. How does a distributed load affect the bending of a beam?

A distributed load can cause a beam to bend, as the load creates a moment along the length of the beam. The magnitude of the bending moment will vary along the length of the beam, depending on the distribution of the load.

## 3. What is the centroid of a beam?

The centroid of a beam is the point at which the entire weight of the beam can be considered to act. It is the geometric center of the cross section of the beam, and is important in understanding the distribution of forces and moments on the beam.

## 4. How is the centroid of a beam calculated?

The centroid of a beam can be calculated by dividing the moment of inertia of the cross section by the area of the cross section. This calculation can be complex for irregularly shaped beams, but for simple shapes like rectangles or circles, the centroid can be easily determined.

## 5. What is the difference between a point load and a distributed load on a beam?

A point load is an external force applied at a specific point on a beam, while a distributed load is applied along the entire length of the beam. Point loads can cause the beam to bend and shear, while distributed loads can cause bending, shear, and deflection.

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