Are Fixed Beams and End Moments Causing Deflection in Simply Supported Beams?

[Studiot]

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.

 

[Urmi Roy]

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..)

 

[Studiot]

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 V_c 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.

 

[Urmi Roy]

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?

 

[Urmi Roy]

...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?...

 

[Studiot]

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

 

[Urmi Roy]

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 !)

 

[Studiot]

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

 

[Urmi Roy]

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

(Congrats on your 2000th post,btw!)

 

[Urmi Roy]

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!

 

[Studiot]

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.

 

[Urmi Roy]

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?

 

[Studiot]

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.