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Shear Flow for welds 1
- Thread starter struc100
- Start date
Some bemusings about this ... in any case the shear stress from shear flow seems to stay moderate or low.
I normally would specify mated faces with prepared borders and full penetration weld. These things use to be for details where to be sound is the important thing and not meaning a lot in the whole of the works. If otherwise, would have to decide.
I normally would specify mated faces with prepared borders and full penetration weld. These things use to be for details where to be sound is the important thing and not meaning a lot in the whole of the works. If otherwise, would have to decide.
BA:
I think you doubt your better judgement and knowledge too much. I think you were basically right on 8APR11, 0:03. And, I’ll take a shot at it from a slightly different angle. Of course we need to know the loads, loading orientation, torsional loading or not, and are we adding these plates under load, to assist with additional loads on the beam. But, I look at shear flow and shear stress this way: if we divide the entire section into infinitesimal horiz. sections or layers with their normal stress and multiply by the width of that element we get a normal force for that element. At any elevation above or below the N.A. of the section we want to study the glue or bonding force, actually a shear force or shear flow, btwn. two infinitesimal layers which keeps them from faying, or slipping, w.r.t. each other due to the imbalance of the sums of the normal forces above and below that level on the entire section. And, the relationship q = VQ/I, the shear flow, allows us to do this. I shouldn’t get any argument that on the OP’ers. section the shear stress will be max. at the N.A. and is tau = q/t = VQ/It, where t is the sum of the three web thicknesses.
This relationship doesn’t change as you move up and away from the N.A., but Q does, thus the shear stress decreases as you move up in the section. And, the shear stress is still tau = q/t, and the shear force in any web is (tau)(t). We have no trouble calc’g. the shear stress just below the fillets which join the WF web and flg., but the steel industry hasn’t figure out how to roll or cast a partial pen. connection btwn. the web and flg., so the shear stresses are low and the residual stresses are quite high right there. Once in awhile they come pretty close, it’s called piping and that isn’t good, it’s a slag inclusion or some such. I sure use fillets there when I can and it’s a welded connection. And the only reason I would increase the weld throat is for bearing area, bearing stresses plus the shear flow, of a concentrated load on that welded joint, or for radial stress components on that weld as in your cranked beam or my curved beam, from an earlier post. In any event I would probably design the welds in question for the shear stress around the bottom of the flg., the shear flow at that level would be divided btwn. the three webs in proportion to their thicknesses, and I would design my weld to take that shear force, and I doubt that it would take much more than a min. fillet, although I’ve not run any numbers. I think the actual shear flow is essentially as you showed it, area of section above the plane in question is (bf * tf/2) and dist. to its centroid (h/2 + tf/4) on this symmetrical section. If you plot the shear stress from the top face of the top flg. to the bottom face of the top flg. it is a long sloped, essentially straight line, from tau = zero to some value, then you get the parabolic (at least curved) shape btwn. the inner faces of the flgs., with a max. at the N.A., and then a sloped line through the bot. flg. back to zero. I’ll bet one of your ref. books shows this stress diagram, you’ve got all the same one I do, and I think I could find it for a WF or box shape, they are the same.
However big you make that weld it isn’t really going to change the pl. buckling picture appreciably, the member length and dist. btwn. the WF flanges or welds will and that’s (h). Obviously, a deeper WF, greater h will affect that pl. buckling config., as will the pl. thick. and weld size only changes edge fixity a little bit. I doubt that that pl. buckling pattern will change much with edges supported and fixed vs. hinged. Obviously, if we rotate the OP’ers. section by 90° and have vert. loads on it we will get a new set of calcs. and shear stresses. And, we may get a third set of shear flows particularly at the welds which are the subject of this lengthy dissertation if there is a torsional loading, and these may be additive, +/-.
Please don’t show me, or convince me, that I’m wrong, because I’ve got a heck of a lot of fabed. structures runnin around or standing out there based on the above approach to the problem.
I think you doubt your better judgement and knowledge too much. I think you were basically right on 8APR11, 0:03. And, I’ll take a shot at it from a slightly different angle. Of course we need to know the loads, loading orientation, torsional loading or not, and are we adding these plates under load, to assist with additional loads on the beam. But, I look at shear flow and shear stress this way: if we divide the entire section into infinitesimal horiz. sections or layers with their normal stress and multiply by the width of that element we get a normal force for that element. At any elevation above or below the N.A. of the section we want to study the glue or bonding force, actually a shear force or shear flow, btwn. two infinitesimal layers which keeps them from faying, or slipping, w.r.t. each other due to the imbalance of the sums of the normal forces above and below that level on the entire section. And, the relationship q = VQ/I, the shear flow, allows us to do this. I shouldn’t get any argument that on the OP’ers. section the shear stress will be max. at the N.A. and is tau = q/t = VQ/It, where t is the sum of the three web thicknesses.
This relationship doesn’t change as you move up and away from the N.A., but Q does, thus the shear stress decreases as you move up in the section. And, the shear stress is still tau = q/t, and the shear force in any web is (tau)(t). We have no trouble calc’g. the shear stress just below the fillets which join the WF web and flg., but the steel industry hasn’t figure out how to roll or cast a partial pen. connection btwn. the web and flg., so the shear stresses are low and the residual stresses are quite high right there. Once in awhile they come pretty close, it’s called piping and that isn’t good, it’s a slag inclusion or some such. I sure use fillets there when I can and it’s a welded connection. And the only reason I would increase the weld throat is for bearing area, bearing stresses plus the shear flow, of a concentrated load on that welded joint, or for radial stress components on that weld as in your cranked beam or my curved beam, from an earlier post. In any event I would probably design the welds in question for the shear stress around the bottom of the flg., the shear flow at that level would be divided btwn. the three webs in proportion to their thicknesses, and I would design my weld to take that shear force, and I doubt that it would take much more than a min. fillet, although I’ve not run any numbers. I think the actual shear flow is essentially as you showed it, area of section above the plane in question is (bf * tf/2) and dist. to its centroid (h/2 + tf/4) on this symmetrical section. If you plot the shear stress from the top face of the top flg. to the bottom face of the top flg. it is a long sloped, essentially straight line, from tau = zero to some value, then you get the parabolic (at least curved) shape btwn. the inner faces of the flgs., with a max. at the N.A., and then a sloped line through the bot. flg. back to zero. I’ll bet one of your ref. books shows this stress diagram, you’ve got all the same one I do, and I think I could find it for a WF or box shape, they are the same.
However big you make that weld it isn’t really going to change the pl. buckling picture appreciably, the member length and dist. btwn. the WF flanges or welds will and that’s (h). Obviously, a deeper WF, greater h will affect that pl. buckling config., as will the pl. thick. and weld size only changes edge fixity a little bit. I doubt that that pl. buckling pattern will change much with edges supported and fixed vs. hinged. Obviously, if we rotate the OP’ers. section by 90° and have vert. loads on it we will get a new set of calcs. and shear stresses. And, we may get a third set of shear flows particularly at the welds which are the subject of this lengthy dissertation if there is a torsional loading, and these may be additive, +/-.
Please don’t show me, or convince me, that I’m wrong, because I’ve got a heck of a lot of fabed. structures runnin around or standing out there based on the above approach to the problem.
SAIL3's attachment for the steel mill crane sheave beam, shows something that I have been talking about on this thread and an the Cranked Beam thread. There, BA talked about the radial components of a change in flg. direction. BA’s comments pertained to a sharp change in flg. direction, and a radial force component, and the need for a single, tension, web stiffener where the joint was opening (increasing angle btwn. the two member parts) and several compressive stiffeners and web buckling where the joint is tending to close (decreasing angle btwn. the two member parts). On the cranked beam there were stiffeners on both sides of the web.
I talked about curved beams where the transition was some form of circular curve with an inner radius to one of the flgs. and a larger outer radius to the other flg. SAIL3's sketch shows a slight variation on the flg./web welding issue, and radial stresses at the transition. Assuming his bottom flg. is in tension, the bottom curve will induce compression into the webs (a bearing stress on the weld, akin to a large point load on the top flg. of a beam) and the upper curve will induce tension into the webs and welds. And, these radial forces must be resolved from the flg. to the web through the welds btwn. them. Thus, you have the normal shear flow, plus this orthogonal shear stress in the weld, some sort of shear flow over some arc length, through the transition length, for a combined weld stress. He already shows full pen. welds so they better be O.K. for these stresses or he has web stress problems too. I might normally have used only fillets btwn. the flg. and the webs, but at the transition curves I might have a bevel cut of some length centered on each transition curve. Then I’d fill the bevels and then weld my fillets over those bevels for an increased weld throat, for the added radial forces.
I talked about curved beams where the transition was some form of circular curve with an inner radius to one of the flgs. and a larger outer radius to the other flg. SAIL3's sketch shows a slight variation on the flg./web welding issue, and radial stresses at the transition. Assuming his bottom flg. is in tension, the bottom curve will induce compression into the webs (a bearing stress on the weld, akin to a large point load on the top flg. of a beam) and the upper curve will induce tension into the webs and welds. And, these radial forces must be resolved from the flg. to the web through the welds btwn. them. Thus, you have the normal shear flow, plus this orthogonal shear stress in the weld, some sort of shear flow over some arc length, through the transition length, for a combined weld stress. He already shows full pen. welds so they better be O.K. for these stresses or he has web stress problems too. I might normally have used only fillets btwn. the flg. and the webs, but at the transition curves I might have a bevel cut of some length centered on each transition curve. Then I’d fill the bevels and then weld my fillets over those bevels for an increased weld throat, for the added radial forces.
dhengr,
Suppose the side plates had been 1/2" deeper than the WF and the fillet welds are above the top flange and below the bottom flange of the WF. What is Q? Looks like Q = 0 which means, in this case, there would be no horizontal shear in the welds.
A vertical plate applied to each side of a WF beam, is capable of carrying its own portion of gravity load in accordance with its stiffness, without reliance on the WF section, other than to prevent local buckling. VQ/I does not enter into the problem.
BA
Suppose the side plates had been 1/2" deeper than the WF and the fillet welds are above the top flange and below the bottom flange of the WF. What is Q? Looks like Q = 0 which means, in this case, there would be no horizontal shear in the welds.
A vertical plate applied to each side of a WF beam, is capable of carrying its own portion of gravity load in accordance with its stiffness, without reliance on the WF section, other than to prevent local buckling. VQ/I does not enter into the problem.
BA
This is one of those problems that you can overthink to death!
In theory, there is no shear flow at the flange edges....until you attach the plate.
Is there horizontal shear stress in the flanges? As TJ noted, if you consider this a built-up beam and neglect the web, the shear flow in the plates would be that replacing the web. Not quite that simple, but would yield a conservative answer.
Now weld the top of the plate to the top flange, leaving the bottom free. What happens when the beam is loaded? The compression in the top flange will be shared by the plates, potentially buckling them.
Now release the load and weld the bottom of the plate to the bottom flange. Reload the beam. The tension in the bottom flange is now transferred to the plates. So we have a compression/tension stress block resultant within the plates, on their face. The difference before and after becomes the effective "shear flow".
In its simplest form....
v=(VQ/I1) x (I2/I1) where I1 is the moment of inertia of the wide flange and I2 is the composite moment of inertia, and Q is for the flange only. This is a similar conservative approach as TJ.
Keep in mind, as BAretired noted, this is an inefficient way to reinforce the beam. Also, a limiting factor might well be the flange thickness...can you get enough weld on the edge?
In theory, there is no shear flow at the flange edges....until you attach the plate.
Is there horizontal shear stress in the flanges? As TJ noted, if you consider this a built-up beam and neglect the web, the shear flow in the plates would be that replacing the web. Not quite that simple, but would yield a conservative answer.
Now weld the top of the plate to the top flange, leaving the bottom free. What happens when the beam is loaded? The compression in the top flange will be shared by the plates, potentially buckling them.
Now release the load and weld the bottom of the plate to the bottom flange. Reload the beam. The tension in the bottom flange is now transferred to the plates. So we have a compression/tension stress block resultant within the plates, on their face. The difference before and after becomes the effective "shear flow".
In its simplest form....
v=(VQ/I1) x (I2/I1) where I1 is the moment of inertia of the wide flange and I2 is the composite moment of inertia, and Q is for the flange only. This is a similar conservative approach as TJ.
Keep in mind, as BAretired noted, this is an inefficient way to reinforce the beam. Also, a limiting factor might well be the flange thickness...can you get enough weld on the edge?
This proves to be an interesting dilemna, indeed.
Case 1) On one hand you have the idea of three separate bending elements (a WF and two individiual plates) that, because the neutral axes of all align, would seemingly want to take load based on their relative EI.
Case 2) On the other hand you have the idea that this thing is one big assembly, and if the side plates were full pen welded, this member doesn't know that it's supposed to have Q=0.
I'm not saying that you're wrong dh, because I believe your approach is conservative, but I would offer a couple of thoughts.
1) distributing the shear stress, in this particular member, based on relative thickness of the webs, means that the side plates take more of the load. If you look at the shear flow formula VQ/It, the shear stress should be distributed according to the inverse of the relative thicknesses. The thinnest piece should have the highest shear stress, not the thickest piece.
2) distributing the loads based on relative thickness of the webs, again resulting in the side plates taking more load than the WF, means that plane sections don't remain plane. I took an example of a W16x31 with two 3/4" side plates of equal depth. If you distribute the stresses based on relative EI, then the plates take 28.6% of the load and the WF takes 42.8% of the load. If you then look the bending stress in the plates at any height along the member you'll get identical bending stresses, which is exactly what you would expect for a pieces with equal E. If, however, you distribute the forces based on relative thickness of the web then the plates take more load than the web of the WF and the bending stresses between the flanges are no longer equal. This tells me that something funky is going on with the strain diagram between different parts of the cross-section, and makes it harder for me to believe.
I might be off with this thinking, but I don't think so.
The last thought I would offer is a condition that would provide identical bending capacity with a clear vision that there is no shear flow. Imagine these two plates are thru-bolted (as shown in the attached sketch), such there is clearly no shear flow between the parts - the bolts do nothing more than ensure all parts of the section deflect together. Please note the bolting would have to be such that local buckling of the plate wouldn't control and thatthe bolts are adequate to transfer the requird load into the plates. In this scenario, you'll have identical bending capacity to the welded solution, but you can see there is no shear flow.
Case 1) On one hand you have the idea of three separate bending elements (a WF and two individiual plates) that, because the neutral axes of all align, would seemingly want to take load based on their relative EI.
Case 2) On the other hand you have the idea that this thing is one big assembly, and if the side plates were full pen welded, this member doesn't know that it's supposed to have Q=0.
I'm not saying that you're wrong dh, because I believe your approach is conservative, but I would offer a couple of thoughts.
1) distributing the shear stress, in this particular member, based on relative thickness of the webs, means that the side plates take more of the load. If you look at the shear flow formula VQ/It, the shear stress should be distributed according to the inverse of the relative thicknesses. The thinnest piece should have the highest shear stress, not the thickest piece.
2) distributing the loads based on relative thickness of the webs, again resulting in the side plates taking more load than the WF, means that plane sections don't remain plane. I took an example of a W16x31 with two 3/4" side plates of equal depth. If you distribute the stresses based on relative EI, then the plates take 28.6% of the load and the WF takes 42.8% of the load. If you then look the bending stress in the plates at any height along the member you'll get identical bending stresses, which is exactly what you would expect for a pieces with equal E. If, however, you distribute the forces based on relative thickness of the web then the plates take more load than the web of the WF and the bending stresses between the flanges are no longer equal. This tells me that something funky is going on with the strain diagram between different parts of the cross-section, and makes it harder for me to believe.
I might be off with this thinking, but I don't think so.
The last thought I would offer is a condition that would provide identical bending capacity with a clear vision that there is no shear flow. Imagine these two plates are thru-bolted (as shown in the attached sketch), such there is clearly no shear flow between the parts - the bolts do nothing more than ensure all parts of the section deflect together. Please note the bolting would have to be such that local buckling of the plate wouldn't control and thatthe bolts are adequate to transfer the requird load into the plates. In this scenario, you'll have identical bending capacity to the welded solution, but you can see there is no shear flow.
Suppose the beam to be reinforced is W16x31 to use Lion's example. Instead of plates, suppose we reinforce the beam with a W16x31 each side of the first so that the flanges are just touching. If the load is applied equally to all three, each beam carries one third of the load and they all deflect identically.
If we weld them together at the flange tips, what changes? The welds are not doing anything. There is no shear flow.
The same argument applies to the side plates except that the welds are needed to prevent buckling.
BA
If we weld them together at the flange tips, what changes? The welds are not doing anything. There is no shear flow.
The same argument applies to the side plates except that the welds are needed to prevent buckling.
BA
As dhengr said..please don"t convince me that I'm wrong...me too, as I have many designs hanging out there based on what I convinced myself as being valid.
What caught my attention was deviding the shear into each vert. plate according to their relative thickeness. However, I agree
with the rest of your well-written post.
So going back to basics, it seems we are discussing two different
cases here.
1. side plates are welded to the fla....once one does this it
forces the bm to act as a composite section...all elements
acting together as one unit. At any given location along
the bm, all three vert. plates will deflect the same amount
vertically. As I want to see it, this shear in question is
based on the Q of the fla. or how the axial force in the
fla is transferred(or shared) among the three vert. plates.
The amount of shear each plate picks up is not based on the
relative stiffeness of each plate but by the portion of fla
it is enfluenced by.
2. Say, sides plates are just bolted to the web of the WF...
in this case also all three vert. plates will deflect the
same amount but what is different in this case is that the
sides plates will pick up the vert. shear according to
their relative stiffeness.
This ,then brings up the question of what value of not only Q but also V to use in v=VQ/I.
I want to believe that one uses the total V and a value of Q of
the fla. that is assigned to that particular vert. plate by influence.
I could very easily be out to lunch on this.
What caught my attention was deviding the shear into each vert. plate according to their relative thickeness. However, I agree
with the rest of your well-written post.
So going back to basics, it seems we are discussing two different
cases here.
1. side plates are welded to the fla....once one does this it
forces the bm to act as a composite section...all elements
acting together as one unit. At any given location along
the bm, all three vert. plates will deflect the same amount
vertically. As I want to see it, this shear in question is
based on the Q of the fla. or how the axial force in the
fla is transferred(or shared) among the three vert. plates.
The amount of shear each plate picks up is not based on the
relative stiffeness of each plate but by the portion of fla
it is enfluenced by.
2. Say, sides plates are just bolted to the web of the WF...
in this case also all three vert. plates will deflect the
same amount but what is different in this case is that the
sides plates will pick up the vert. shear according to
their relative stiffeness.
This ,then brings up the question of what value of not only Q but also V to use in v=VQ/I.
I want to believe that one uses the total V and a value of Q of
the fla. that is assigned to that particular vert. plate by influence.
I could very easily be out to lunch on this.
The formula VQ/It is for plane remain plane sections; every height on the neutral axis takes the same longitudinal stress. If the cut is horizontal, symmetrical like here, that means that the PLANE stands so, rotated on the neutral axis, and the shear flow stress at such section is also constant anywhere at the height, as per the classical plane-remain plane hypothesis; hence a distribution (per that notional theory) of the total force is on a per relative width of the cut parts of the members at the cut height. This throws one shear stress, that is what given in my posted worksheet.
So you know just under the weld the common shear flow stress at that height; I have elected to chose conservatively in such worksheet to pass the resultant of shearflow there in 1 plate through just one throat width; but more properly, the weld itself, that we could draw with its notional quarter of circle shape, is part of the member, and we might dispose the sideplates marginally short to have a section where the cut parts were two throats for the welds, and one central web at the widest part of the rounded fillets; the theory and so VQ/It also would throw one shear flow stress commoon for the two welds and the described widest part of the fillet. That is what the theory says.
Since the width of the throats is less than that of the sideplates, and the cut heights are only marginally different, and with the same disposition, for the described new setup of the welds, the central cut being the same will take relatively more shear flow force for the second case than my first; what makes my posted assumption conservative.
Shear lag, locked stresses, or any other irregular distribution of stresses at the height is not commonly described by the common theory of beams; what is not to say are not of interest, and nothing can be said about it; but you quite likely would have to go to some form of 3D FEM to get some more advanced representation of what happens at one reasonable effort these days, because available; not that closed form more perfected theories can't be advanced, but more difficult if starting from zero and by one not having dealt extensively on this kind of problem.
So you know just under the weld the common shear flow stress at that height; I have elected to chose conservatively in such worksheet to pass the resultant of shearflow there in 1 plate through just one throat width; but more properly, the weld itself, that we could draw with its notional quarter of circle shape, is part of the member, and we might dispose the sideplates marginally short to have a section where the cut parts were two throats for the welds, and one central web at the widest part of the rounded fillets; the theory and so VQ/It also would throw one shear flow stress commoon for the two welds and the described widest part of the fillet. That is what the theory says.
Since the width of the throats is less than that of the sideplates, and the cut heights are only marginally different, and with the same disposition, for the described new setup of the welds, the central cut being the same will take relatively more shear flow force for the second case than my first; what makes my posted assumption conservative.
Shear lag, locked stresses, or any other irregular distribution of stresses at the height is not commonly described by the common theory of beams; what is not to say are not of interest, and nothing can be said about it; but you quite likely would have to go to some form of 3D FEM to get some more advanced representation of what happens at one reasonable effort these days, because available; not that closed form more perfected theories can't be advanced, but more difficult if starting from zero and by one not having dealt extensively on this kind of problem.
Indeed, I have been out to lunch on this.
I believe,now, dhengr is correct in distributing this total shear flow between the various webs according to their relative stiffeness which in this case reduces down to their various thickenesses.
My error was in assuming that the distribution of shear is based on the area of influence being distribted to each resisting element.
So, calculate the total shear flow of the fla using Q of the total fla, then, distribute this total shear flow among the various webs based on their relative stiffenes.
If all the webs are of same thickeness, then devide this total shear flow by the number of webs.
Some is going to sleep well tonight!
I believe,now, dhengr is correct in distributing this total shear flow between the various webs according to their relative stiffeness which in this case reduces down to their various thickenesses.
My error was in assuming that the distribution of shear is based on the area of influence being distribted to each resisting element.
So, calculate the total shear flow of the fla using Q of the total fla, then, distribute this total shear flow among the various webs based on their relative stiffenes.
If all the webs are of same thickeness, then devide this total shear flow by the number of webs.
Some is going to sleep well tonight!
Sail3,
And some is going to toss and turn.
I agree that your method would be safe insofar as designing the welds, but if you were relying on that theory to reduce the shear in the original web, I would question it because I do not think a fillet weld is as positive a connection as the beam web plus fillets.
But I think you and dhengr are correct that a pair of flanges connected with three web plates are going to have a horizontal shear between webs and flanges.
BA
And some is going to toss and turn.
I agree that your method would be safe insofar as designing the welds, but if you were relying on that theory to reduce the shear in the original web, I would question it because I do not think a fillet weld is as positive a connection as the beam web plus fillets.
But I think you and dhengr are correct that a pair of flanges connected with three web plates are going to have a horizontal shear between webs and flanges.
BA
BA:
We agree, that Q is zero at the top surface of the top flg., and for that reason VQ/I goes to zero and makes absolutely no sense as a means of calculating that weld size. If I made the side pls. fit btwn. the t&b flgs. I doubt anyone would disagree that VQ/I at the underside of the top flg. would be divided by the sum of the three (nominal) web thicknesses and that would be the shear stress in the webs at that level, a constant value in all webs. Then this would be multiplied by the thickness of the side pl. and its weld would be designed for that shear flow, but now it’s a shear flow in kips/inch of length of weld, not thickness of web. And, in fact this is about what I said I would do to size that weld, since it seems closer to our normal practice than a very small weld, which blindly using VQ/I might suggest. I said, “In any event I would probably design the welds in question for the shear stress around the bottom of the flg., the shear flow at that level would be divided btwn. the three webs in proportion to their thicknesses, and I would design my weld to take that shear force, and I doubt that it would take much more than a min. fillet, although I’ve not run any numbers.”
And, I would assume we agree that the combined moment of inertia for this section is the sum of the moments of inertia of the WF plus that of the two side plates, it’s symmetrical. And, I agree with you that loads will be shared by the three elements in proportion to their relative stiffness, but we must connect the three element to make that load transfer take place wherever the load is applied on the combined member. If the load is applied to the top of the WF it will take about 75% and each side pl. will take half the rest, and you will have to design two sets of welds which distrib. 25% into the side pls. Lion06 picked slightly differently proportioned WF and side pls. for this example. If the load rests on your new slightly high side pls. then you will need two welds which transfer 75% of the load to the WF, and this load might be expressed in lbs./inch, a type of shear flow as relates to weld design, since that is basically the way a fillet reacts to these loads. I guess I’m using the term shear flow in about three slightly different ways here. But, 75% would lead to unrealistically large welds, so this load application wouldn’t be my first choice. I think Lion06 and Toad were debating that as a solution, but with 25%/4 as determining the weld size, and I might buy that if I could compare some real numbers.
In your example of the three WF’s side by side, you are dealing with a slightly different problem. You are saying I will divide the load between three identical type members and I would say that load will split in proportion to relative stiffness, EI, assuming you provide a means to be sure that load distrib. will happen. That’s akin to my .125/.75/.125 discussion above. That may be by welding them together at the flg. tips, probably not a good means of attachment in this case, or it may be by some other structural system which assures that the three WF’s will deflect equally under the load, but the normal stresses or the shear flows and shear stresses at any given level may not be constant or the same from WF to WF in this case. And, in fact, this probably leads to some differential shear flow btwn. each WF at the flg. tips, if you choose to weld there. The original problem was to make a built-up member which would act as an integral unit, not as a series of three beams which deflect the same amount, and carry the load.
Ishvaaag’s discussion about plane sections remaining plane, normal stresses and shear stresses at a given level being constant across the member, is exactly in keeping with the Theory of Elasticity for this symmetrical section, working in the elastic range. q = VQ/I is constant at any level, it’s a shear flow (kips/inch of thickness); and if you divide it by the sum of web thicknesses at that level you will get the shear stress at that level, tau = q/t = VQ/It (ksi). The shear force at that level if you are sizing a weld is (q/t) times (t for that web), again a shear flow (kips/inch), but now I think of this in terms of kips/inch of length because, that’s the way I think of the cap’y of a given weld throat. The shear flow and shear stress in SAIL3's cross section exactly fits this situation, 3 uniform webs of some thickness, leads to tau avg. = q/t and t is now 3(individual web thick.). If his inner web were thicker than the outer webs it would need a larger weld (q/t)*(t for that web), a shear flow used to select the needed weld throat. When we look at the WF at the underside of the top flg. I just consider the three nominal web thicknesses, q is constant at that level, but the shear stress gets funky (using Lion06's terminology, notice I don’t discuss lower joints any longer, or shorter) in the region of the WF web/flg. radii, it transitions from tau in its web to tau in the flg. and then into that long, straight sloped line as it goes to zero at the top surface of the flg. As an aside, I still have a couple buckets of shear flow left over from that last thread on this subject, but now they are all mixed up, some of them are shear flow/inch of length and some of them are shear flows/inch of web thickness, and I don’t know how to sort them out.
We agree, that Q is zero at the top surface of the top flg., and for that reason VQ/I goes to zero and makes absolutely no sense as a means of calculating that weld size. If I made the side pls. fit btwn. the t&b flgs. I doubt anyone would disagree that VQ/I at the underside of the top flg. would be divided by the sum of the three (nominal) web thicknesses and that would be the shear stress in the webs at that level, a constant value in all webs. Then this would be multiplied by the thickness of the side pl. and its weld would be designed for that shear flow, but now it’s a shear flow in kips/inch of length of weld, not thickness of web. And, in fact this is about what I said I would do to size that weld, since it seems closer to our normal practice than a very small weld, which blindly using VQ/I might suggest. I said, “In any event I would probably design the welds in question for the shear stress around the bottom of the flg., the shear flow at that level would be divided btwn. the three webs in proportion to their thicknesses, and I would design my weld to take that shear force, and I doubt that it would take much more than a min. fillet, although I’ve not run any numbers.”
And, I would assume we agree that the combined moment of inertia for this section is the sum of the moments of inertia of the WF plus that of the two side plates, it’s symmetrical. And, I agree with you that loads will be shared by the three elements in proportion to their relative stiffness, but we must connect the three element to make that load transfer take place wherever the load is applied on the combined member. If the load is applied to the top of the WF it will take about 75% and each side pl. will take half the rest, and you will have to design two sets of welds which distrib. 25% into the side pls. Lion06 picked slightly differently proportioned WF and side pls. for this example. If the load rests on your new slightly high side pls. then you will need two welds which transfer 75% of the load to the WF, and this load might be expressed in lbs./inch, a type of shear flow as relates to weld design, since that is basically the way a fillet reacts to these loads. I guess I’m using the term shear flow in about three slightly different ways here. But, 75% would lead to unrealistically large welds, so this load application wouldn’t be my first choice. I think Lion06 and Toad were debating that as a solution, but with 25%/4 as determining the weld size, and I might buy that if I could compare some real numbers.
In your example of the three WF’s side by side, you are dealing with a slightly different problem. You are saying I will divide the load between three identical type members and I would say that load will split in proportion to relative stiffness, EI, assuming you provide a means to be sure that load distrib. will happen. That’s akin to my .125/.75/.125 discussion above. That may be by welding them together at the flg. tips, probably not a good means of attachment in this case, or it may be by some other structural system which assures that the three WF’s will deflect equally under the load, but the normal stresses or the shear flows and shear stresses at any given level may not be constant or the same from WF to WF in this case. And, in fact, this probably leads to some differential shear flow btwn. each WF at the flg. tips, if you choose to weld there. The original problem was to make a built-up member which would act as an integral unit, not as a series of three beams which deflect the same amount, and carry the load.
Ishvaaag’s discussion about plane sections remaining plane, normal stresses and shear stresses at a given level being constant across the member, is exactly in keeping with the Theory of Elasticity for this symmetrical section, working in the elastic range. q = VQ/I is constant at any level, it’s a shear flow (kips/inch of thickness); and if you divide it by the sum of web thicknesses at that level you will get the shear stress at that level, tau = q/t = VQ/It (ksi). The shear force at that level if you are sizing a weld is (q/t) times (t for that web), again a shear flow (kips/inch), but now I think of this in terms of kips/inch of length because, that’s the way I think of the cap’y of a given weld throat. The shear flow and shear stress in SAIL3's cross section exactly fits this situation, 3 uniform webs of some thickness, leads to tau avg. = q/t and t is now 3(individual web thick.). If his inner web were thicker than the outer webs it would need a larger weld (q/t)*(t for that web), a shear flow used to select the needed weld throat. When we look at the WF at the underside of the top flg. I just consider the three nominal web thicknesses, q is constant at that level, but the shear stress gets funky (using Lion06's terminology, notice I don’t discuss lower joints any longer, or shorter) in the region of the WF web/flg. radii, it transitions from tau in its web to tau in the flg. and then into that long, straight sloped line as it goes to zero at the top surface of the flg. As an aside, I still have a couple buckets of shear flow left over from that last thread on this subject, but now they are all mixed up, some of them are shear flow/inch of length and some of them are shear flows/inch of web thickness, and I don’t know how to sort them out.
HereTwoLearn
Structural
I think the calcs are wrong in the original post.
If the new side plates were only connected at the neutral axis, they would just be "along for the ride" and the load in them would be proportional to I(plate) / I(WF Section).
But that is not what is happening. These plates are integral to the section because they are connected away from the neutral axis. Effectively it is a built up section with three webs instead of two.
I would calculate shear flow as:
Q = (Area of flange / 4).
y = h/2 - flange thickness/2.
I = Moment of inertia of the combined section (which is equal to I(W Beam) + I(Plates).
Guys, be careful about relying on this website for authoritative advice. Go ask a senior engineer how this should be done and go through it with him in person.
If the new side plates were only connected at the neutral axis, they would just be "along for the ride" and the load in them would be proportional to I(plate) / I(WF Section).
But that is not what is happening. These plates are integral to the section because they are connected away from the neutral axis. Effectively it is a built up section with three webs instead of two.
I would calculate shear flow as:
Q = (Area of flange / 4).
y = h/2 - flange thickness/2.
I = Moment of inertia of the combined section (which is equal to I(W Beam) + I(Plates).
Guys, be careful about relying on this website for authoritative advice. Go ask a senior engineer how this should be done and go through it with him in person.
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