Fixed Beams and end moments

  • Thread starter Urmi Roy
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  • #26
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I've been doing a lot of homework on this....
...I re-read the part on shear stresses on beams from my book and I think the most important part would be where it says "the ementary normal and shearing forces exerted on a given transverse section of a prismatic beam with a vertical plane of symmetry are equivalent to the bending couple M and the shearing force V" (where V and M would be the Vc and Mc on the diagrams you attached).

So now I understand that the moment Mc as per your diagram is not caused directly by an applied load...but its a result of the distribution of shear stresses produced when there is an applied load.

Now, coming to the second part of my post#24, I wrote:
"....how Ma and Mb in a fixed beam can be different values if there is indeed a pair of equal and opposite forces (denoted by 'R' in the diagram that I took from the book- post #15 on the other thread) that is responsible for them... "


...now, in my book, they have an equation that goes Mb-Ma=RL (referring to my earlier diagram showing the two forces 'R'...L is the length of the beam).....also, I found that in case of symmetrical loading, the end moments (Ma and Mb)are equal but this leads to the fact that R=0.....so I guess that 'R' is not the only fixed end reaction that causes the bending moments.....so in the case of symmetrical loading(say with load at centre point), there is also some other force that produces the end moments Ma and Mb....is this even remotely correct?
 
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  • #27
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You are getting there, hopefully some things are becoming clearer.

I will try to find some time over the weekend to post some more diagrams.

What are you studying, by the way?
 
  • #28
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Hi Studiot, thanks for your time :-)

I'm currently doing second year of my Bachelor's degree in Mechanical Engineering in India.

Coming back, if you think what I've understood is okay, could you give me an idea of what/where the "other force that produces the end moments Ma and Mb" are?...I'm referring to the part where I said -

...I found that in case of symmetrical loading, the end moments (Ma and Mb)are equal but this leads to the fact that R=0[/B].....so I guess that 'R' is not the only fixed end reaction that causes the bending moments.....so in the case of symmetrical loading(say with load at centre point), there is also some other force that produces the end moments Ma and Mb....[/B]
 
  • #29
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Mechanical Engineering, that's a good subject, good luck.

I will try to bear it in mind when making examples.

First to answer this question about your book adding a pair of opposing forces to make a couple.

I think this is probably what they are talking about, which is different from the method and discussion I have been pursuing.

Fig1
Shows a beam with a bunch of loads L1 etc and supports S1 etc.

Fig2
It is a fundamental theorem of mechanics that any bunch of (not necessarily parallel) forces can be replaced by a single resultant.
I have shown this as R.
The resultant acts at a particular line as shown and at a distance a from our test section C-C.
So it exterts a moment Ra at CC.

Fig3
This is equivalent to adding two equal and opposite forces equal to R at C forces as we did before.


Now to move on to The relationship between shear and moment.
In all the following figures we are back to considering the left hand free body as before.

Fig4
A beam with a single load L.

Fig5
We are agreed that the reaction at S1 plus the load L tends to cause the beams to shear as shown.

Fig 6
Your book probably shows the shear and moment as in this diagram. Before I only considered the left hand sections, but of course equal and opposite forces act on the right hand face of the break.

Fig7
I have swopped things around a bit so that my free body now has two cut faces, a left hand and a right hand face.
I have also shown the forces acting on each of these faces.

Fig 8
Dropping the moments and considering only the shear forces, note they are in force equilibrium, but form couple.
So they are not in moment equilibrium.

Fig9
I have added two horizontal forces which form an opposing couple.
Now the free body is in moment equilibrium as well.
This is how the horizontal forces are generated in elements of the beam.
Note this is a local effect on each small element of the beam. The actual value of these effects vary with the position of the element in the beam.

Figs 10 and 11
Back to the beams. The local effects add up to the (hopefully) familiar compression and tension forces an the top and bottom halves of the beam (left hand section).

go well
 
  • #30
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Hi Studiot...but where are the diagrams....
 
  • #31
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Ooops, sorry.

That's what I like about a good student. You just cough politely when the lecturer make howler, rather than rolling about laughing in the aisles.
 

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  • #32
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Hi Studiot...I haven't finished reading/understanding your last post yet...just 2 queries regarding the diagrams on the way...

1. I notice that in figure 3, where you added the two forces at end C, you missed out writing Vc, and you said 'add Mc'....so is the absence of 'Vc' in the labelling something important in the overall concept?

2. In my book, they bring in the two 'R' forces separately....while not considering the loads or support reactions...it's like they split the analysis into two parts- one considering the fixed beam as a simply supported beam(first part of my diagram) along with the loads and supports..and the next part considering the fixed beam only with thr 'R' forces and Ma and Mb..

(Also, unlike in your diagrams, the 'R' in my book become 0 (zero) when there's symmetrical loading..)
 
  • #33
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To answer both 1 & 2

I was not trying for a free body analysis I was just speculating how the pair of opposing forces might have arisen in your book which I haven't seen to understand the context.
So yes, if you actually cut the beam at CC and created the free body, then Vc would act there along with the moment as detailed before.

But the real point was that I wondered if the R in you book was a resultant as I showed.
 
  • #34
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Hmm....well, R in my book is certainly not resultant...it even becomes zero in the situation that I mentioned in post #26....

..it is mentioned specifically, however, that R is responsible for causing the end moments...
....and that if there is a greater applied load ,say 2W nearer to end B and there is another applied load W nearer to A, then R would act upward on end B and its pair(with which it forms the couple) acts downward at end A...this is all not considering the support reactions.....thereby the net vertical force being zero....

.....so does this approach seem invalid to you...should I just skip it?
 
  • #35
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...Btw, when we say that a fixed beam is a 'statically inderterminate' beam of third degree, do we mean that there are certain forces that we just can't find out without a lot of trouble....I mean then there'd be vertical forces other than the loads and supports..no?.....
 
  • #36
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I should skip it for now, perhaps it will become clear later. You already have enough to be going on with.

With these last few posts I have tried to give you a 'look ahead' since you were interested how the moment arises and how it is inter related with the shear.

Don't try to go into too much detail on this as this was for understanding purposes only. It is not wrong it is just that reality is more complicated as you will find out when you start to study shear distribution.

I hope what I have shown you will give you an insight and a good start into the more difficult stuff that the more complciated analysis entails.

Indeterminate beams have too many supports to solve by just the application of the six conmditions of equilibrium at every point (6 = three force and three moment equations in 3 dimensions)

go well
 
  • #37
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Smooth explanation, Studiot :-)

comparing and correlating what you said with the chapter on bending stress,

- In addition to the shear stresses that you explained to cause the tensile and compressive stresses, the bending stresses due to applied couples (like what is usually explained in chapters relating to pure bending) also contribute to the tensile/compressive stress as shown in your diag. ...right?

- Could we say that the shear stresses(denoted by tau) are caused exclusively due to shear forces (like L in your diag.) and the bending stresses are exclusively due to applied couples?

-the shear stresses that you described vary along the length of the beam whereas the bending stresses don't(since they're caused due to couples)?

Also, please say something related to #35
(this was my last set of doubts, promise !!)
 
  • #38
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Could we say that the shear stresses(denoted by tau) are caused exclusively due to shear forces (like L in your diag.) and the bending stresses are exclusively due to applied couples?
Why is this calssification so important to you.
It is very difficult to get all embracing statements correct.

Forces & stresses

There is always a force (normal or shear) associated with a stress so the stress is the force divided by the area. Alternatively the force is sum of all the stresses taken over the whole area.

However it is not true to say that a shear or normal (also called axial or direct) stress is only associated with a shear or axial stress respectively.

You will eventually come to study stress at a point where you will find the the orientation of the area also plays a part as to the nature of the stress.
If you wish to look ahead look up 'principal stresses' in your book.

In general, both the horizontal and vertical shear stresses vary along a beam, as do the moments with which they arre intimately associated in the bending action.

Also, please say something related to #35
I already have at the end my last post.

The 'degree of indeterminancy' is just a fancy classification for lecturers to look good with.
Either a structure is statically determinate, in which case it can be solved as previously indicated.
Or it is indeterminate, in which case you have to get extra equations form somewhere other than the equations of equilibrium.
This somewhere is usually the equations of geometrical compatibility eg slope=0 somewhere or the deflection =0 at a support.

go well
 
  • #39
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Thanks Studiot...I've been going over your posts and things are really starting to make sense:smile: ...All the info you gave will help me in the next chapter on columns and struts too.

(Congrats on your 2000th post,btw!!)
 
  • #40
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Hi Studiot and everyone else!

I've been doing some slightly complicated things in relation to mechanics of solids in the last one year or so. I have a doubt about how a horizontal/axial force along a beam can cause bending moment in it.

The situation is that I have a shaft mounted on bearings at both ends (treated like a simply supported beam). I have a pulley rotating on it, as a part of a transmission system.

The pulley is inclined to the horizontal axis. Due to this inclination, the two forces of tension on both sides of the pulley form a net inclined force downwards.
These forces are resolved into vertical and horizontal components.

In a book of mine, they draw separate Bending moment diagrams due to the vertical components and horizontal components.

But I don't understand why the horizontal component should cause bending moment at all!
 
  • #41
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Hello Urmi, is this a new term?

Remember the inclined plane from elementary mechanics.

If you resolve parallel and normal to the inclined plane the equations are simpler than if you resolve horizontally and vertically.

Similarly with your shaft, if you resolve horizontally and vertically each will have a component both parallel to and normal to the shaft.

In both cases the normal component provides the bending moment.
 
  • #42
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Hi Studiot,

Yes, this is a new term...and I'm on my final year now!

Well, as you said, and as per my diagram, the bending moment is due to Fn, and not Fp.

I also figured out the book is resolving the tension forces on the pulley in the horizontal and vertical planes.

I still can't visualize, though how the pulleys must be arranged in order for the tension forces to be creating bending moments in the horizontal plane. Could you think of how this is possible?
 

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  • #43
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I still can't visualize, though how the pulleys must be arranged in order for the tension forces to be creating bending moments in the horizontal plane. Could you think of how this is possible?
I think I'd have to see exactly what the book says.
 

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