Why are moments not included in free body diagrams?

[per persson]

Homework Statement

free body diagram

Relevant Equations

static equilibrium equations

Many times I've seen that the moments in the free body diagram are not included for example for an hinge or fixed beam.

I asked the same question here https://physics.stackexchange.com/q...dy-diagram?noredirect=1#comment1840797_818418

But I don't understand the answers.

 

[Lnewqban]

The free body diagram represents all the external forces and moments acting on a body.

Those external forces could induce an internal moment, as a reaction, which should not be represented in the FBD.

In similar way (but less common), any external moment could induce an internal pair of forces, as a reaction, which should not be represented in the FBD.

This site shows good examples:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/5-7-drawing-free-body-diagrams/

The case of the hinge is different, by definition, you will never get a reacting moment at a hinge, which offers no resistance to rotation.

 

[per persson]

How am I supposed to know when to include the moments or not? For example for a fixed beam in a wall in 2d. In the free body diagram in my book the moment where the beam is fixed in the wall is neglected.

 

[kuruman]

Is this what you have in mind? Shown on the right is the FBD of a tipping plank at an angle on a horizontal floor and rotating clockwise as it falls. The moment is shown in red and is directed into the screen.

For 2D situations like this, moments are either into or out of the screen. If there are more than one, they must all be computed about the same reference point so stacking them all at the same point in the diagram is not really informative.

 

[kuruman]

In the free body diagram in my book the moment where the beam is fixed in the wall is neglected.

One is free to choose the reference point about which to calculate moments. If a force is applied at that point, then the moment generated by that force is zero and can be ignored.

You can also see in the diagram in post #4 that the moments generated by $F_x$ and $F_y$ are zero because I chose to calculate moments about the point of contact of the plank with the floor.

 

[per persson]

One is free to choose the reference point about which to calculate moments. If a force is applied at that point, then the moment generated by that force is zero and can be ignored.

You can also see in the diagram in post #4 that the moments generated by $F_x$ and $F_y$ are zero because I chose to calculate moments about the point of contact of the plank with the floor.

I'm not talking about the moment caused by the forces. I'm talking about moments which are put into free body diagrams. For example with a beam fixed into a wall. If drawing a free body diagram of the beam there should be a moment at the point where the beam is fixed to the wall. I've seen sometimes the moment isn't included and sometimes it is.

 

[Lnewqban]

How am I supposed to know when to include the moments or not? For example for a fixed beam in a wall in 2d. In the free body diagram in my book the moment where the beam is fixed in the wall is neglected.

Your book represents a system first: wall, beam, forces.
No internal forces and moments to that system are represented there.

Then, the book shows the FBD of the beam (alone in the middle of the air).
In order to remain in balance, the external forces acting on the beam must be compensated by internal forces and moments.
Because of that, the FBD must show a moment where the wall was.

You have to ask yourself "which are the external forces and moments acting on this body while isolated from anything else?"
Then, you need to figure what internal forces must exist for keeping the isolated body in balance.

 

[haruspex]

How am I supposed to know when to include the moments or not? For example for a fixed beam in a wall in 2d. In the free body diagram in my book the moment where the beam is fixed in the wall is neglected.

Does it show forces from the wall on the beam?
Can you post the image?

 

[DaveC426913]

Correct me if I'm wrong here, but moments are derived. There won't be any moments unless they are driven by other forces which will be listed in the diagram. And they can all be calculated from those primary forces, so you have everything you need.

So, one diagram might have the primary forces listed, while another might do some of the work for you by showing these secondary, derived moments.

 

[haruspex]

Correct me if I'm wrong here, but moments are derived.

It is common for engineering diagrams, especially involving beams, to show some external force pairs as pure torques. Exactly how the forces are distributed is irrelevant.

 

[haruspex]

A beam is fixed to the wall at O. And this is its free body diagram.

Then I agree with you, the diagram is deficient. With the forces shown, it is assuming a free joint at the wall.

 

[DaveC426913]

It is common for engineering diagrams, especially involving beams, to show some external force pairs as pure torques. Exactly how the forces are distributed is irrelevant.

Yes. My point is that such additions are for convenience. They're not strictly necessary, since part of the problem is deriving them from a basis of of FBD. For engineering, it would be useful to have those done ahead of time. Am I right?

For the OP, it's not wrong either way. Including them would be up to what the prof expects I guess.

 

[per persson]

Then I agree with you, the diagram is deficient. With the forces shown, it is assuming a free joint at the wall.

sorry I got confused it is not that picture it is this

At O and A there are hinges but no moment

 

[haruspex]

My point is that such additions are for convenience.

Yes, but if you don’t show it as a pure torque then you must show an equivalent pair of forces. The diagram in post #9 shows neither.

 

[haruspex]

sorry I got confused it is not that picture it is thisView attachment 346958
At O and A there are hinges but no moment

If it is a free hinge (no frictional torque) then there is no moment to show.

 

[per persson]

If it is a free hinge (no frictional torque) then there is no moment to show.

it says that the door can rotate around the axis OA. How can I determine if it is a free hinge or not?

 

[haruspex]

it says that the door can rotate around the axis OA. How can I determine if it is a free hinge or not?

It is clearly shown as a hinge, and hinges are designed to have very little friction. So assume free unless told otherwise.

 

[per persson]

It is clearly shown as a hinge, and hinges are designed to have very little friction. So assume free unless told otherwise.

But there is rotation around one axes the y axis. But no rotation about the x and z axes. Should there not be 2 moments M_z and M_x because of this?

 

[Lnewqban]

But there is rotation around one axes the y axis. But no rotation about the x and z axes. Should there not be 2 moments M_z and M_x because of this?

There is a moment about the x axis and another about the z axis, both caused by the asymmetric rope BD and the weight of the platform.
The result are reaction forces Ra and Ro at the hinges, which are not able to slide over the wall or to penetrate it.

 

[haruspex]

But there is rotation around one axes the y axis. But no rotation about the x and z axes. Should there not be 2 moments M_z and M_x because of this?

The image showing forces is rather fuzzy and dark, but it looks like the two reaction forces there are in arbitrary directions.
Yes, there could also be moments from those hinges about the x and z axes, but that makes the set-up statically indeterminate. E.g., what if the hinges don't line up perfectly? They would fight each other in a way that depends on how tight the screws are.
Even if the hinges were universal joints, thereby eliminating such moments, the system would be stable. Perfectly installed, the local moments would be negligible. It is therefore reasonable to ignore them.

 

[per persson]

There is a moment about the x axis and another about the z axis, both caused by the asymmetric rope BD and the weight of the platform.
The result are reaction forces Ra and Ro at the hinges, which are not able to slide over the wall or to penetrate it.

View attachment 346967

Why are there no moments at hinges in the FBD?

 

[DaveC426913]

Why are there no moments at hinges in the FBD?

Because they are free-moving, have negligible friction and have no associated forces?

 

[haruspex]

Because they are free-moving, have negligible friction and have no associated forces?

As is made clear in post #18, @per persson means moments about the x and z axes, where the axle of the hinges is the y axis.
Indeed, if extremely strong, one hinge would do. In that case there would be moments about the other axes.
With two hinges, there could still be such other moments, but if properly installed they will be negligible.

 

[Lnewqban]

Why are there no moments at hinges in the FBD?

Respect to which axis?
When we talk about moments, it is important to define the axis of reference.

The posted picture shows a situation that may be too complicated for you to fully understand at this moment.
We could try to explain the need or not to represent moments while using a simpler example perhaps.

 

[per persson]

I think I got it.

 

[Lnewqban]

I think I got it.

Could you explain it to us?

 

[DaveC426913]

I think I got it.

"All long help threads should have a sticky globally-editable post at the top saying 'DEAR PEOPLE FROM THE FUTURE: Here's what we've figured out so far ...'"

 

[per persson]

Could you explain it to us?

So the hinges prevent each other from rotating. So when drawing a free body diagram I have to take into account all constraints. So for a case where there is a horizontal beam in 2d that is fixed into a wall at each end there is not moment at these points in the FBD. But if the beam instead is fixed into one wall and the other end is free then in the FBD there will be a moment.

 

[haruspex]

where there is a horizontal beam in 2d that is fixed into a wall at each end there is not moment at these points

As I wrote, that's not quite true. Ideally there is none, but in the real world there's imprecision and flexing.
As you load the beam it flexes slightly, leading to a torque at each wall.

 

[Lnewqban]

So the hinges prevent each other from rotating. So when drawing a free body diagram I have to take into account all constraints. So for a case where there is a horizontal beam in 2d that is fixed into a wall at each end there is not moment at these points in the FBD. But if the beam instead is fixed into one wall and the other end is free then in the FBD there will be a moment.

Please, see:
https://pressbooks.library.upei.ca/statics/chapter/reaction-forces/

https://en.wikipedia.org/wiki/Cantilever

 

[per persson]

As I wrote, that's not quite true. Ideally there is none, but in the real world there's imprecision and flexing.
As you load the beam it flexes slightly, leading to a torque at each wall.

I remember yes.

Please, see:
https://pressbooks.library.upei.ca/statics/chapter/reaction-forces/

https://en.wikipedia.org/wiki/Cantilever

I think I understand moment better know. Consider a beam that is fixed to the wall. The gravitational force on the beam causes a moment and where it is fixed to the wall prevents the rotation of the beam thus in a free body diagram we draw a moment there. But if beam is fixed at both ends the gravitational force does not cause a moment. It is the reactionary forces at the walls that prevents rotation. Am I on the right track?

 

[haruspex]

Consider a beam that is fixed to the wall. The gravitational force on the beam causes a moment and where it is fixed to the wall prevents the rotation of the beam thus in a free body diagram we draw a moment there. But if beam is fixed at both ends the gravitational force does not cause a moment. It is the reactionary forces at the walls that prevents rotation. Am I on the right track?

There still could be, and in general will be, both a net force and a torque at each end, but the situation is ‘statically indeterminate', i.e. there is a continuum of possible combinations.
If we start the beam simply laid across two supports, then apply whatever loads, there will be just a force at each end, no torques. If we now encase the ends rigidly, there's no reason for any torque to arise.
Starting again, encase the ends before applying the loads. Now, flexion in the beam will lead to a torque at each end.