Shear Flow 15

[Stillerz]

I am sure most here are familiar with the concept of shear flow as it related to horizontal shear stresses in beams.
When designing a I-Shaped plate girder, most references, if not all, will design the weld between the flange and web using the shear formula VQ/I to determine the force on the welds.
My question is, isn't there bending stress on the weld as well in the form of MC/I?
If one had a simply support girder with a uniformly distributed load, the shear at the center of the beam would be = zero and the moment at a maximum. This would imply that no weld would be needed at the center, yet this is the section where having the entire cross-section engaged in bending is most critical.
What am I missing here?

 

[ishvaaag]

The thing with many of the good researchers of the XIX and XX century is that they have high respect of the mathematical basis of mechanics, and not as many professors to get the bunch of them not to be brilliant. I lacking much the mathematics I would love to know, appreciate very much the overspread reach of those people able to adjudicate symbolical models for everything subject to their consideration at ease. It should also be matter of reflection to see how much the mathematical advances themselves were tightly tied to the advances in mechanics always targeted to science of course but then immediately to engineering. Said another way, at those societies these gents got intelligent numbers of them to be fully armed to mathematically describe phenomenons. I don't think now that is true to the same extent, if only by overtextension of the related "trades".

 

[miecz]

ishvaaag-

Nice try, but that was more than 30 words.

 

[Stillerz]

HAHAHAHA-
Classic Miecz!!

 

[ishvaaag]

You know, I can't count, today was giving 70 euros to pay 21.

 

[paddingtongreen]

I always found Timoshenko complicated.
That website I posted tells about the same story as Timoshenko, but in simpler terms.

In working stress, I'm sure you know the normal shear distribution in beams, zero at the extreme fibres and maximum in the middle. The complimentary shear stress is longitudinal and in proportion to the normal shear stress. That is what the welds have to resist. At the ends, there is maximum shear, normal and complementary, and minimum compression or tension due to bending. At the point of zero shear, there is no shear but maximum compression/tension due to bending. Besides the two are not additive.

Sorry for the pedantry.

Michael.
Timing has a lot to do with the outcome of a rain dance.

 

[hemis]

Dear All

In the above analysis for calculating the horizontal shear, in the equation MC/I, the parameter C approaches zero towards the N.A. and maximum at the flanges. This assumes the beam is subjected to pure flexure. Timoshenko titles this expression as a the "flexure formula" and gives good examples of the theory (Gere and Timoshenko: Mechanics of Materials: Third ed SI: pp260). In pure bending, shear stresses on the N.A. are zero so a weld not required but try to make the beam stay together. Other factors arise if the beam is subject to non-uniform bending.

 

[apsix]

The equation MC/I calculates flexural stress, not horizontal shear.
Horizontal shear is at a maximum at the NA, not zero.

 

[BAretired]

A beam in pure bending has no shear. So the shear stress, VQ/Ib is zero at all points along the span. The maximum principal shear at any section, however is f[sub]b[/sub]/2. Draw the Mohr's circle to confirm.



BA

 

[apsix]

I did think about that (a little bit) but couldn't think how you could get bending into a beam without shear.

Is it possible, and how?

 

[BAretired]

Yes, it is possible. Place an equal and opposite bending moment at each end of the beam. You have a case of pure bending throughout the span with no end reactions and a deflected shape in the form of a circular arc with radius of EI/M.

BA

 

[apsix]

You're right! Thanks.

I can't yet visualise how a beam can be bent, but have zero shear flow.

 

[BAretired]

Sleep on it. It will come to you.

BA

 

[hemis]

Dear apsix

That was a pretty dumb sounding answer I gave for shear flow when I said a weld would not be needed. I saw the problem clearly when I wrote it.

I assumed that if the beam were a series of plates, like a cart-spring, pinch the middle tight beneath the load (axle) and the plates slide away from the pinch - zero stain beneath the axle.

If the plates were pinched at each end and torqued equally with the centre of the moment on the line of the N.A., then the plates would slide over each other but remain static at the N.A.

Maybe

 

[rtmote]

Fascinating discussion.

My tuppence worth...I got involved in this when trying to understand Staad country code calculation to the French Code. I was finding beams that passed every other country code checks without issue, fail under the french code. It came down to Von Mises and the programming assumption of a single shear centre, that was producing misleading results and I see that in this thread regarding the classical view. If the check was done by hand, it worked fine.

Basically, an open thin-walled section like a deep I-beam, involving Von Mises,should be modelled with two shear centres. Imagine a thread throughout the length of the beam, at the top and bottom of the web/flange intersections. Without stiffeners and loading our uniformly distributed beam: at midspan it is possible that the shear centres which must be complementary can behave very differently and lead to flange or web buckling as further loading is applied. The top thread is in compression axially and may experience different shear flow behaviour as it becomes unstable. At the centreline of the web there is no shear flow. I think of castellated beams. At the top and bottom there is potential for separate regions of horizontal shear flow as the subsections are mobilised around their respective shear centre, with its own polar moment of inertia value on the compression shear centre. The top flange/shear centre can become unstable and we cannot use the I-value for the full section.

So what is the horizontal or vertical shear flow in this situation? The weld is part of a surface area under torsional stress.

The classic formulae work for the most part but do not always apply to all details. Was that gobbleydook?


Robert Mote

http://www.motagg.com

 

[miecz]

rtmote-

At the centreline of the web there is no shear flow.

Have to disagree with you there. As apsix recently pointed out, for a normal beam (no special cases), shear (horizontal and vertical shear stress for any element are equal) is highest at the neutral axis, or the centerline, of the web.

A castellated beam does not cut out the entire web, and is generally used for beams with high bending moment and low shear. Those remaining sections of web are working hard.

 

[hemis]

Therefore:
If a beam was under pure flexure (ignoring self-weight and other gravity effects), is there any horizontal shear stress along the N.A.?

 

[ishvaaag]

No. Moment=constant, Shear=0. Each fiber is compressed or elongated as to satisfy the plane-remains-plane in the same radians per unit length and no shear develops between fibers, along any height of the constant section.

 

[shin25]

Simply talking, moment is the sum of area of the shear diagram, therefore, using the max moment for a simply supported beam at the mid span to calculate the horizontal shear flow at a given plane should give the total horizontal shear in that plane up to the section considered. This methodology is simple to use, however, since it's the total shear does not given you the higher shear flow that should occur near the support.

 

[Compositepro]

This is an interesting discussion but I'm starting to wonder what is the point that is being made. An example that I have often found useful is comparing a 2x4 to a phone book. A phone book cannot carry shear loads between the pages. If you bend a phone book the pages slide but stay the same length. The example presented above of a beam in pure bending where there is no shear is interesting but what is the point? The welds are still absolutely critical to prevent buckling of the flanges and web. Try bending a phone book in pure bending. The pages in compression will buckle.

 

[BAretired]

Let's get back to the original questions:

(1) When designing an I-Shaped plate girder, most references, if not all, will design the weld between the flange and web using the shear formula VQ/I to determine the force on the welds.
(2) My question is, isn't there bending stress on the weld as well in the form of MC/I?
(3) If one had a simply support girder with a uniformly distributed load, the shear at the center of the beam would be = zero and the moment at a maximum. This would imply that no weld would be needed at the center, yet this is the section where having the entire cross-section engaged in bending is most critical.
4 What am I missing here?

The answers are:

(1) The horizontal shear is VQ/I per unit length which must be transferred by the weld from web to flange.

(2) There is bending stress in the weld, Fb = Mc/I which acts over the cross sectional area of the weld.

(3) No weld is needed at midspan for the purpose of transferring horizontal shear, but weld may be needed to prevent the web from buckling sideways.

(4) I'm not sure that you are missing anything. If the entire cross section is engaged an inch or two each side of midspan, you do not need any weld at midspan to engage it for that short length. The only purpose of the weld at the point of zero shear would be to prevent the web from moving away from its position at the center of the flange.

BA