Why Does a Triangular Load on a Beam Require Multiple Moment Calculations?

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In summary, the conversation discusses a problem involving a triangular distributed load on a simply supported beam. The triangular load is equivalent to a concentrated load at the centroid, and to find the unknown moment at one of the supports, the moments must be added together and set equal to zero. The confusion arises from the inclusion of a moment at the point where the triangular load tapers off, which is not accounted for in the moment expression. A sketch of the problem is requested for clarification.
  • #1
nonlocalworlds
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I have a problem that shows a triangular distributed load on a beam (studying for NCEES civil engineering exam). At one end of the triangle we have a force magnitude level of "w" and the other end is labeled "wL/6."

They tell me that a triangular load is equivalent to a concentrated load of (1/2)wL acting at the centroid.

My load is on a simply supported beam on either end. To get the 'correct' answer the moments about one of the simple supports are added together and set equal to zero (non-rotating system). There is an unknown moment on the opposite simple support (that the problem is trying to get you figure out), but then my confusion is that there appears to be two counteracting forces, a moment at the centroid of the triangular load as well as another moment where it tapers off (the triangular load is shown at a maximum w on top of the simple support whose counteracting moment I am trying to find, and tapers to wL/6 at a distance L from the unknown simple support, with the final simple support being 2L away from the unknown location).

I do not understand why all of the load is not accounted with the moment expression for the concentrated load at the centroid of the triangle.

Apparently the correct expression is:

X*2L (unknown moment to calculate) - (wL/6)L (a moment at the point the triangle tapers whose inclusion I don't understand) - (wL/2)(5/3L) (this part is fine, the concentrated load at the centroid) = 0.
 
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  • #2
I am confused as well by your description of the triangular load. Try not to use L for the distance to the start of the load or from the end of the load to the support. Generally, L is used to indicate the total length of the beam.

If you could post a sketch of the problem, that would save quite a bit of hand-waving explanation.
 
  • #3
sketch
 

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  • #4
The beam in question has a concentrated load of ##w_0L/6## located at x = L from point A. (Total length of the beam = 2L). This concentrated load is not related to the triangular load distributed between 0 < x < L, with ##w_0## @ x = 0 as its load magnitude. The equivalent concentrated load would be ##w_0L/2## located @ x = L/3 from point A.
 

Related to Why Does a Triangular Load on a Beam Require Multiple Moment Calculations?

What is a distributed load?

A distributed load is a type of external force that is applied evenly over a certain area rather than at a single point. This can include loads like weight, pressure, or tension that are spread out over a surface or structure.

How do you calculate the magnitude of a distributed load?

The magnitude of a distributed load can be calculated by multiplying the force or weight per unit area by the area over which the load is being applied. This will give you the total force being applied to the surface or structure.

What is the difference between a uniform and a non-uniform distributed load?

A uniform distributed load is one in which the magnitude of the load is the same at every point on the surface or structure. A non-uniform distributed load, on the other hand, varies in magnitude at different points along the surface or structure.

What types of structures are affected by distributed loads?

Distributed loads can affect a variety of structures, including beams, bridges, roofs, and floors. These loads can cause stress and deformation in the structure, so it is important to consider them in the design and analysis process.

How can distributed loads be managed or reduced?

There are several ways to manage or reduce distributed loads, including distributing the load over a larger area, using stronger and more durable materials, and implementing structural reinforcements. It is also important to regularly inspect and maintain structures to ensure they can withstand distributed loads over time.

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