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Shear Flow 15

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Stillerz

Structural
Mar 27, 2008
298
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?
 
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Horizontal shear in a beam or plate girder is mobilized by bending, so horizontal shear is not 0 at the center as would be for vertical shear. It is actually at a maximum with the maximum moment. The shear is created by the tension/compression differential in bending.
 
You should not design this weld for MC/I. The entire section resists the moment if the welds are adequate to transfer the shear. I remember a demonstration in statics class where a load is applied to a stack of unattached plates and to a stack of plates attached together. The unattached plates slip relative to each other resulting in a large deflection. The attached plates are much stiffer. The attachment is meant to prevent the slip and allows I to be calculated based on the whole section, not just the sum of the I's of the individual plates. At the middle of the beam you describe, the shear has already flowed into the flanges. Theoretically, no weld is needed at that infinitesimal point.
 
Ron- This was my thought as well.
Maybe I am misunderstanding my texts.
How is the horizontal shear calculated?

VQ/I is the horizontal shear due to vertical shear forces on the beam. internal equilibrium states that this is true. Horizontal shear forces due result from vertical.
 
Ron-
What is the horizontal shear force in the center of the beam I described in the original post...i.e., where the moment is max?
 
Graybeach-
Your explanation seems perfectly logical and agrees with what the texts say.
 
Here are my thoughts.

The horizontal shear stresses are a result of the differential moment. If you think about a simply supported beam with a uniform load, the moment diagram takes on a parabolic shape. It increases rapidly from 0 at the ends, and levels off to a maximum at the center. As you get closer to the center, the differential moment decreases which, in turn, causes a decrease in the horizontal shear flow. This continues to the point where the moment is at a maximum (the slope is 0) in which case the differential moment is 0, hence your horizontal shear flow is 0.

This matches what you will get if you plot the horizontal shear flow of a uniformly loaded, simply supported beam. The horizontal shear flow will be a maximum at the ends (where the slope of the moment diagram, or the differential moments, are highest) and taper down to 0 at the center (where the slope of the moment diagram, or the differential moments, are zero).
 
Can you explain what you are calling "differential moments"?
 
Am I wrong in saying that horizontal shear forces in a beam are the direct result of vertical shear forces being applied to the beam?
Internal equilibrium in the beam would dictate that.
For the case of the simply supported beam with uniform loading, the shear at the support is max and moment is ZERO, and, horizontal shear is VQ/I.
So how can the horizontal shear exist where no moment does if what you guys are saying is correct?
 
As you increnentally walk along any beam from point A to point B, it is the difference in the moment at A and the moment at B due to the applied vertical load.

Mike McCann
MMC Engineering
 
Stillerz.

Taking your simply supported beam idea - at midspan, the moment is maximum and vertical shear is zero.

At this point, the horizontal shear between the flanges and webs is zero (i.e. VQ/I = 0 because V=0). As you move outward towards the supports, the vertical shear V increases and so does the horizontal shear.

What is basically happening is that as you move from the support towards midspan, the shearing stresses are gradually transfered into longitudinal stresses in the flanges through the horizontal shear at the flange/web interface.

At midspan, there is Mc/I stress in the weld, but it is NOT a shear through the weld but rather a axial force in the cross section of the total weld itself.
 
Yes.

I'll use a simple working example to make it easier to talk about. Let's talk about a simply supported, 20' long beam with a uniform load of 1klf. Let's look at the "differential element" from 0' - 1'. The moment at 0' = 0, and the moment at 1' = 9.5k-ft. This would be the differential moment for that element (9.5-0 = 9.5)

Now let's look at the element from 9' - 10'. The moment at 9' = 49.5k-ft, and the moment at 10' = 50k=ft. The differential moment for that element is only 0.5k-ft.

Now, if you take infinitesimally small elements (instead of the 1' elements discussed above) what you will find is that the differential moment is equal to the slope of the moment diagram. This makes sense if you think about it from a mathematical point of view - dM=slope of the moment function.

Now, with the above talked through, you should be able to see that the differential moment at the midspan (point of maximum moment) is zero, therefore the horizontal shear flow is zero.

Now just for a little more practical look at the situation. If you imagine the traditional picture that is used to illustrate shear flow (It was mentioned above), of several elements lying on top of each other with no connection between them. You load the system and they deflect together, but they also slide with respect to each other. That sliding is a maximum at the ends, and is 0 in the middle. This also supports the fact that horizontal shear flow is 0 where the slope of the moment diagram is 0, or decreases as the moment decreases.
 
JAE-
This is precisely what I was picturing.

As you say, at midspan, there is no vertical shear force on the weld, however, if were to calculate a stress on the weld cross-section by MC/I (c being = to the distance to the centroid of the weld cross section from the n.a.), I could then determine the longitudinal force on that weld. Correct?
This would be a longitudinal force that results from bending, no? But since the force in the surrounding material (flange, web) would have only infintesimally higher and lower (varying by distance from the n.a. only ever so slightly), the weld really sees no stress in terms of being a connection between web & flange.

agreed?
 
Shear is the derivative of moment along length and is at maximum for ordinary beams near supports; the rate of variation of moments is then also at maximum there, at ends, and so is horizontal shear in web-flange welds or vertical sections in flanges. Remember the old practice of setting rigid welded stubs at ends of top flanges of composite beams, that proceeds just of these considerations.
 
I should make clear that my point here is trying to justify to myself why the longitudal force in the weld due to bending stresses can be ignored.
 
The reason that the flange-web weld is critical at the mid point of the beam is that the compressive load on the flange is maximum at that point. The purpose of the web is to support the flange to keep it from buckling out-of plane. Although the theoretical loads are zero the effect on "I" is what is important to prevent column buckling.
 
I think this is the first time I have ever said--

in my opinion JAE is wrong.

The welds which stitch together a flange and a web transfer longitudinal force from web to flange. This is what we call shear flow. The shear flow is maximum at the ends, and zero at midspan. By the time you reach midspan, all of the flange force has developed in the flange, because of all the shear flow which has been transferring to the flange. So Mc/I defines the stress in the flange at midspan, but not the force in the weld--because that force is already in the flange.

I am really just repeating what graybeach said.

DaveAtkins
 
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