Bending Moment of a Cantilever Beam

[BananaJelly]

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.

 

[SteamKing]

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.

 

[BananaJelly]

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.

 

[SteamKing]

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.

 

[BananaJelly]

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.

 

[SteamKing]

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.

 

[BananaJelly]

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.

 

[SteamKing]

For the beam to be in static equilibrium, there's always a reaction at the wall.

 

[BananaJelly]

I meant there's no vertical reaction. Or am I wrong?

 

[SteamKing]

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?

 

[BananaJelly]

I don't know. But based on what I looked up, there would be a moment, vertical force, and a horizontal force.

 

[SteamKing]

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.)

 

[BananaJelly]

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.

 

[SteamKing]

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.

 

[BananaJelly]

Yeah, thanks. My intuition regarding mechanical engineering and forces is lacking.