Bending Moment of a Cantilever Beam

In summary, the problem involves a cantilever beam attached to a wall with a force applied at a 30 degree angle. The beam is made of steel with specific properties and a mass per unit length. The solution involves calculating the maximum bending moment in the beam due to its own mass. This is done by replacing the distributed load with an equivalent point load and using the concept of centroids. The process is similar to calculating the reactions on a beam with a triangular load. The solution also involves drawing shear force and bending moment diagrams.
  • #1
BananaJelly
8
0
This is a problem from Michael Lindeburg's FE Review Manual. It is in Chapter 20, Number 11 if you have the book.

PROBLEM:
There is a picture of a cantilever 2 m in length attached to a wall. There is a force applied at a 30 degree angle from the horizontal at the free end of the beam.

"The beam is manufactured from steel with a modulus of elasticity of 210 GPa. The beam's cross-sectional area is 37.9 cm^2; its moment of inertia is 2880 cm^4. The beam has a mass of 45.9 kg/m. A 6000 N force is applied at the top of the beam, at an angle of 30 degrees from the horizontal. Neglect buckling.

What is the approximate maximum bending moment in the beam?"


SOLUTION:
I have the solutions, but I cannot understand the first part of it:

The moment due to the beam's own mass is

M1 = (0.5)wL^2 = (0.5)(45.9 kg/m)(9.81 m/s^2)(2 m)^2

I do not understand where they got the "(0.5)" from. If it is moment, shouldn't the equation just be M = F L = mgL = (wL)(g)(L) = wgL^2 without the 0.5 in front?

There's more to the solution, but I am stuck at this one part regarding the moment due to the beam's mass.
 
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  • #2
Think about how the mass of the beam is distributed along the length.

You can always draw the shear force and ending moment diagram of the beam if that helps.
 
  • #3
I assumed the mass was distributed uniformly along the beam.

I don't know how I would draw the shear force and moment diagram in this case, since the force is applied at an angle.

I'm really just trying to understand why you have to divide mgL by half to solve for the moment of the beam due to its mass. Does it have anything to do with the assumption that the force of gravity is acting at the center of the beam? Is it really just as simple as that? Also, I imagined the 45.9 kg/m as more of a load along the entire beam.
 
  • #4
Your question was why the bending moment due to the beam's mass had a factor of 0.5 in it. The force of gravity acting on the mass of the beam produces a distributed load like any other distributed load.

For analyzing the statics of the beam, a uniformly distributed load can be replaced by an equivalent point load. You should review how to do this for other types of distributed loading, like a triangular load.

The force applied at the free end of the beam makes no contribution to this moment. Just because it is applied at an angle should produce no great impediment to analysis. After all, such forces can always be broken down into components acting vertically and horizontally.
 
  • #5
Okay. That's what I thought, but I didn't understand the concept behind it. So if it is a triangular load, then it would be M = (1/3)*(1/2)wL or a M = (2/3)*(1/2)wL with w = mg [=] Newtons depending on how the triangular load is distributed, correct?

I will need to look up how to draw a stress diagram for something attached to a wall.
 
  • #6
BananaJelly said:
Okay. That's what I thought, but I didn't understand the concept behind it. So if it is a triangular load, then it would be M = (1/3)*(1/2)wL or a M = (2/3)*(1/2)wL with w = mg [=] Newtons depending on how the triangular load is distributed, correct?

Correct. You should probably review centroids and how to calculate them, and learn the formulas for simple shapes like a rectangle and a triangle.

The concept behind it is simple: as long as you replace the distributed load with its equivalent point load value, and the point load is located at the centroid of the distributed load, you have not changed the values of the reactions on the beam.

This article shows how to replace an offset force, with a force and a couple:

http://emweb.unl.edu/NEGAHBAN/EM223/note8/note8.htm

A similar procedure is followed for a distributed load.

I will need to look up how to draw a stress diagram for something attached to a wall.

IDK what you mean here. Your review should cover how to construct the shear force and bending moment diagrams for a beam, given point loads and distributed loads.
 
  • #7
Thanks!

It covers it a little bit, but I mainly learned how to draw them from a few youtube videos. The review manual lacks concepts. For the case with the wall, I was thinking that there was no reaction at the wall, so I wasn't sure how to approach it. But since there is a load, I guess you just start with that or start it at the applied Force. Anyway, thanks again! I'll look into it more.
 
  • #8
For the beam to be in static equilibrium, there's always a reaction at the wall.
 
  • #9
I meant there's no vertical reaction. Or am I wrong?
 
  • #10
If you draw a free body diagram for the loaded cantilever beam, what do you replace the wall with and still keep the beam from translating and rotating?
 
  • #11
I don't know. But based on what I looked up, there would be a moment, vertical force, and a horizontal force.
 
  • #12
BananaJelly said:
I don't know. But based on what I looked up, there would be a moment, vertical force, and a horizontal force.

Those are also known as reactions.

(BTW, I think you are talking about the original problem now, if I'm not correct.)
 
  • #13
Sort of. But I'm also just talking about a general cantilever fixed on a wall at one end, and the forces or reactions at the wall end.
 
  • #14
That's why we draw free body diagrams for analyzing beams. It helps to replace things like walls and other supports with a set of forces and/or moments sufficient to keep the beam in equilibrium.

BTW, in general, a cantilever beam will not have a horizontal reaction unless a horizontal load is applied.
 
  • #15
Yeah, thanks. My intuition regarding mechanical engineering and forces is lacking.
 

1. What is a cantilever beam?

A cantilever beam is a beam that is supported at only one end, with the other end projecting freely into space. It is commonly used in construction and engineering projects for its ability to support heavy loads without the need for additional support.

2. What is bending moment?

Bending moment is the measure of the bending or twisting force experienced by a beam. It is caused by external forces acting on the beam, such as loads or weight, and can result in the beam bending or deforming.

3. How is the bending moment of a cantilever beam calculated?

The bending moment of a cantilever beam is calculated by multiplying the applied force by the distance from the fixed end of the beam to the point where the bending moment is being measured. This calculation can be done using the formula M = F * d, where M is the bending moment, F is the applied force, and d is the distance from the fixed end.

4. What factors affect the bending moment of a cantilever beam?

The bending moment of a cantilever beam is affected by several factors, including the magnitude and direction of the applied force, the distance from the fixed end of the beam to the point of measurement, and the properties of the beam such as its material, shape, and cross-sectional area.

5. How can the bending moment of a cantilever beam be reduced?

The bending moment of a cantilever beam can be reduced by increasing the distance from the fixed end to the point of measurement, using a stronger or stiffer material for the beam, or by adding additional support or reinforcement to the beam. Proper design and placement of loads can also help to minimize bending moment on a cantilever beam.

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