Shear flow for indeterminate beam

[StrEng007]

I got a question that's come up regarding the theory behind how shear flow works. Considering a simple span beam with uniform loading. I've always used the following to explain shear flow.
The chord force of the beam is the area under the shear flow diagram. Said in a different way, shear flow is the accumulation of shear stress that results in the chord force due to moment.

Equations:
V = wL/2
M= wL²/8
Shear Flow = VQ/I
Chord force in the beam section = the area under the shear flow diagram = (VQ/I) x L/4

Putting some numbers behind it. Take a 8"x12" rectangular cross section:
I = 1152 in^4
Q = 144 in^3

w = 100 lb/ft, L= 10ft
M = 1250 lb-ft or 15,000 lb-in
V = 500 lb

Chord force illustrated by shear flow:
Shear Flow = 62.5 lb/in
Chord Force = 62.5 lb/in x (120 in/4) = 1,875 lb

Chord force illustrated by bending stress:
Bending Stress = 15,000 lb-in (6 in)/ 1152 in^4 = 78.125 psi
Chord Force = 78.125 psi x (8in)(6in)x1/2 = 1,875 lb

However, this theory doesn't work when applied to a single span beam with fixed end conditions. The shear at the end of this beam would be the same as the simple span beam shown above. However, we know that using the area under this shear flow diagram would NOT be equal to the maximum chord force produced by the moment.

So the question is, when you have a fixed end single span beam, is shear flow calculated in the same exact manner? If so, how do you explain how to derive the chord forces in relationship to the shear flow force?

 

[KootK]

So the question is, when you have a fixed end single span beam, is shear flow calculated in the same exact manner?

Yes in the sense that VQ/It will still produce an appropriate result.

If so, how do you explain how to derive the chord forces in relationship to the shear flow force?

It requires adding a bit of nuance to your initial procedure to make it more general in nature.

The CHANGE in chord force of the beam is the area under the shear flow diagram. THE MAGNITUDE OF CHORD FORCE AT ANY LOCATION IS THIS CHANGE ADDDED TO THE STARTING VALUE AT THE SUPPORTS WHICH, IN THE CASE OF NON-SIMPLE SPAN BEAMS, MAY WELL BE NON-ZERO.

I suspect that the main thing that you are missing is giving proper account of the chord forces present at the end of your fixed ended beam.

 

[StrEng007]

THE MAGNITUDE OF CHORD FORCE AT ANY LOCATION IS THIS CHANGE ADDDED TO THE STARTING VALUE AT THE SUPPORTS WHICH, IN THE CASE OF NON-SIMPLE SPAN BEAMS, MAY WELL BE NON-ZERO.

Oh, yes that makes a lot of sense. Hence the reason why the overall magnitude of moment is always WL²/8. It just happens that with fixed ends, some negative moment is pulled out of that overall magnitude (i.e. wl²/12 + wl²/24 = wl²/8).

So going back to what you said, let's say I made a built up steel beam from two separate identical square shapes. Regardless of the beam being fixed or simple span, to get the composite action all I need to do is design the weld between the two beams using the shear flow VQ/I. It doesn't matter that the beam's ends "start" with an internal moment?

 

[KootK]

It doesn't matter that the beam's ends "start" with an internal moment?

That's right. Again, for theoretical nuance, we'll acknowledge that the beam fixity that produces the starting chord forces implies a great deal of shear flow. It's just shear flow that occurs outside of the span being studied. Consider the shear diagram on an encastre version of such a beam.

 

[human909]

I think I might include that drawing on the next set of calculations I submit and see whether the building surveyor reacts. I especially love the description "So much f'ing shear!".

(They probably won't, down here nobody even looks at the calculations submitted. Oddly enough the only times I am even asked for them is for domestic jobs. I can and do design sprawling industrial plants and never get asked for anything more than a signature.)

 

[rb1957]

I think you're mixing things up.

in a beam with a uniform load, shear is constantly, linearly, increasing (due to the distributed load), the moment is increasing parabolically.
the local section shear (VQ/I) is proportional to the local shear which of course includes the end reaction support force.

I would rather say that the shear in the beam is the area under the distributed load - the end reaction.