# How Do Horizontally Oriented Gussets Perform in Cantilever Bending?

• lordvon
In summary: Sorry, I thought my question was clear. I am only concerned with the gusset (point #3). The other points are more straightforward. Also, I only asked about the horizontal orientation in my original question.
lordvon
TL;DR Summary
Do gussets (horizontally oriented) perform well in cantilever bending?
Do gussets perform well in cantilever bending when horizontally oriented, as opposed to vertically oriented? I assume vertical orientation performs well in bending, in which you can calculate the area moment of inertia by adding that of the gusset cross sections individually, and the joint helps with geometric stability. But how would you calculate the effective area moment of inertia for bending when the gussets are horizontal?

Could you show us a drawing of beam and gussets?
Gussets don’t do as much for increasing area moment of inertia as for increasing welding area.

Lnewqban said:
Could you show us a drawing of beam and gussets?
Gussets don’t do as much for increasing area moment of inertia as for increasing welding area.
Sure, here is what i mean (no welding, just bolts/screws+plates):

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Lnewqban
There is no simple "one answer fits all" answer to your question. You need to analyze the entire joint. A simplified outline is as follows for the vertical orientation:

1) Bending strength of the horizontal beam at the point of highest moment.
2) The horizontal beam bolted joint. Note that this involves multiple calculations for shear, crushing, tensile failure, slippage, etc.
3) The gusset at the point of highest stress.
4) Bending strength of the vertical beam at the point of highest moment.
5) The vertical beam bolted joint. Note that this involves multiple calculations for gusset shear, bolt shear, local crushing, tensile failure, slippage, etc.

The strength of the assembly is whichever of the above is the weakest. A similar analysis applies to the horizontal orientation.

A complete description of how to do all of this would involve writing a book. Fortunately, there are multiple books available that do exactly that. Search Amazon books for structural steel design to find a selection.

jrmichler said:
There is no simple "one answer fits all" answer to your question. You need to analyze the entire joint. A simplified outline is as follows for the vertical orientation:

1) Bending strength of the horizontal beam at the point of highest moment.
2) The horizontal beam bolted joint. Note that this involves multiple calculations for shear, crushing, tensile failure, slippage, etc.
3) The gusset at the point of highest stress.
4) Bending strength of the vertical beam at the point of highest moment.
5) The vertical beam bolted joint. Note that this involves multiple calculations for gusset shear, bolt shear, local crushing, tensile failure, slippage, etc.

The strength of the assembly is whichever of the above is the weakest. A similar analysis applies to the horizontal orientation.

A complete description of how to do all of this would involve writing a book. Fortunately, there are multiple books available that do exactly that. Search Amazon books for structural steel design to find a selection.
Sorry, I thought my question was clear. I am only concerned with the gusset (point #3). The other points are more straightforward. Also, I only asked about the horizontal orientation in my original question. I only show the vertical just to clarify what I mean by horizontal orientation.

lordvon said:
But how would you calculate the effective area moment of inertia for bending when the gussets are horizontal?
If I understood you correctly, you would model your gussets as an I-beam where a = 0:

Ix = (a h3 / 12) + (b / 12) (H3 - h3)

Iy = (a3 h / 12) + (b3 / 12) (H - h)

The width b of your "I-beam" is the smallest value within the gap between your bolted beams.

Ix is when they are placed horizontally and Iy is when they are placed vertically.

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jack action said:
If I understood you correctly, you would model your gussets as an I-beam where a = 0:

Ix = (a h3 / 12) + (b / 12) (H3 - h3)

Iy = (a3 h / 12) + (b3 / 12) (H - h)

The width b of your "I-beam" is the smallest value within the gap between your bolted beams.

Ix is when they are placed horizontally and Iy is when they are placed vertically.
Thanks @jack action, this is what I hoped it would be, but it seems too good to be true. The reason I say that is that the gap between the beams containing only gusset cross section differs from the I-beam in that it is missing the vertical center flange / web. So is it really as strong as an I-beam, or is it as weak as 2 individual thin plates, or something in between?

Where two narrow members are joined at right angles, the size and the distance between the pairs of bolts that handle the moment is limited. Using bigger bolts would reduce the section of the members, and so weakens the structure. It is necessary to use smaller bolts, with greater separation, on the neutral axis of the members to be joined.
There are two simple ways to maintain the strength, while handling the moment.
1. Use a diagonal or asymmetric triangle with three bolts. That joint does not keep members co-planar, so is uneven, and must be deformed and stressed during assembly.
2. Use a symmetric gusset with four bolts. That makes a better flat base, while keeping the structural members co-planar.

Baluncore said:
Where two narrow members are joined at right angles, the size and the distance between the pairs of bolts that handle the moment is limited. Using bigger bolts would reduce the section of the members, and so weakens the structure. It is necessary to use smaller bolts, with greater separation, on the neutral axis of the members to be joined.
There are two simple ways to maintain the strength, while handling the moment.
1. Use a diagonal or asymmetric triangle with three bolts. That joint does not keep members co-planar, so is uneven, and must be deformed and stressed during assembly.
2. Use a symmetric gusset with four bolts. That makes a better flat base, while keeping the structural members co-planar.
Isnt the weakest point in between the beams, where there is only gusset plate in the cross-section?

lordvon said:
Isnt the weakest point in between the beams, where there is only gusset plate in the cross-section?
I hope not.
The gusset plate does not need to be thicker than the member, as it is wider.
The weakest point will be where a bolthole in the member, reduces the section of the member.

Holes should be punched on the neutral bending axis. But there is no neutral axis under tension, so the member section will be reduced. Design of the gusset will come down to balancing shear of the bolts, damage to the sides of the holes, and the remaining material member strength under tension.

Lnewqban said:
yes, this is what i am trying to get at. the gusset plates could be acting as mere thin plates (that might buckle) in bending, which would be by far the greatest weak point @Baluncore

lordvon said:
So is it really as strong as an I-beam, or is it as weak as 2 individual thin plates, or something in between?
Whether you look at it as 2 plates or an I-beam with no web, the geometry is exactly the same. If the plates are thin, it is weak, and if they are thick, it is stronger.

https://en.wikipedia.org/wiki/I-beam said:
The horizontal elements of the I are flanges, and the vertical element is the "web". [...]

The web resists shear forces, while the flanges resist most of the bending moment experienced by the beam. The Euler–Bernoulli beam equation shows that the I-shaped section is a very efficient form for carrying both bending and shear loads in the plane of the web. On the other hand, the cross-section has a reduced capacity in the transverse direction, and is also inefficient in carrying torsion, for which hollow structural sections are often preferred.
So, used horizontally, it is an I-beam and when used vertically, it is an H-beam (the "transverse direction" in the previous quote).

jack action said:
Whether you look at it as 2 plates or an I-beam with no web, the geometry is exactly the same. If the plates are thin, it is weak, and if they are thick, it is stronger.So, used horizontally, it is an I-beam and when used vertically, it is an H-beam (the "transverse direction" in the previous quote).
Unfortunately, that is incorrect. There must be a transfer of longitudinal shear through the section, and in the case of separated plates, that cannot happen. In the I-beam it does happen. If the two thin plates are indeed to be treated separately (this is part of my question), you must use the area moment of inertia calculation for thin/short rectangles (bh^3/12), and not that of the I-beam.

lordvon said:
the gusset plates could be acting as mere thin plates (that might buckle) in bending, which would be by far the greatest weak point
Then the gusset plate thickness selected is clearly insufficient for that application.

Baluncore said:
Then the gusset plate thickness selected is clearly insufficient for that application.
well, the question really is how do you calculate the stress in these plates (or geometric instability / buckling limit). so that one can determine the right size/thickness of plate. we can be certain though that the gusset plates cannot be treated as an i-beam.

lordvon said:
well, the question really is how do you calculate the stress in these plates
The plates here attach cross members that tie between two chords. The plates are handling mainly moments in the plane of the horizontal structure.

Two gusset plates can be used to join I-beam RSJs, or universal columns at right angles. Column stability under compression can be avoided. Since the gussets have a minimum span, they do not require a web, the gusset plate only needs to have the same section as the flange to which it is bolted.

A similar but different method will be used to join end-to-end chord sections. That will require the web also be plated, to prevent buckling of the web, due to shear between the flanges.

lordvon said:
Unfortunately, that is incorrect. There must be a transfer of longitudinal shear through the section, and in the case of separated plates, that cannot happen. In the I-beam it does happen. If the two thin plates are indeed to be treated separately (this is part of my question), you must use the area moment of inertia calculation for thin/short rectangles (bh^3/12), and not that of the I-beam.
There is still shear going through the plates. It is just not very efficient.

As I said in my previous post:

Ix = (a h3 / 12) + (b / 12) (H3 - h3)

where a = 0, thus Ix = (b / 12) (H3 - h3).

The geometry is the same.

jack action said:
There is still shear going through the plates. It is just not very efficient.

As I said in my previous post:

Ix = (a h3 / 12) + (b / 12) (H3 - h3)

where a = 0, thus Ix = (b / 12) (H3 - h3).

The geometry is the same.
No, that is still incorrect. That formula implies longitudinal shear going through a cross section that is H in height. But, for disparate plate cross sections of height h, the longitudinal shear is only going through a cross section of height h, in 2 plates. Do you understand what I am saying?

From post #6, h is not the thickness of the plates but the distance between them. The shear goes through a section that is (H - h) in height, which is the thickness of the 2 plates.

Baluncore said:
The plates here attach cross members that tie between two chords. The plates are handling mainly moments in the plane of the horizontal structure.

Two gusset plates can be used to join I-beam RSJs, or universal columns at right angles. Column stability under compression can be avoided. Since the gussets have a minimum span, they do not require a web, the gusset plate only needs to have the same section as the flange to which it is bolted.

A similar but different method will be used to join end-to-end chord sections. That will require the web also be plated, to prevent buckling of the web, due to shear between the flanges.

When you say "do not require a web" are you saying the gussets arent the weak point? If not, wouldn't the web (if it could somehow be applied) make the strength of the joint more similar to the beam?

jack action said:
From post #6, h is not the thickness of the plates but the distance between them. The shear goes through a section that is (H - h) in height, which is the thickness of the 2 plates.
oh, my mistake, I meant thickness of one plate. but, my point still stands. e.g. (H^3-h^3) is much greater than 2 * thickness^3.

It's the area bh³ subtracted from the area bH³, like for a hollow rectangle, except the inner "empty" rectangle is the same width as the outside rectangle.

lordvon said:
When you say "do not require a web" are you saying the gussets arent the weak point? If not, wouldn't the web (if it could somehow be applied) make the strength of the joint more similar to the beam?
Any weakness is in the discontinuity of the web. The flanges do the work, the web keeps the flanges separated by a fixed distance.
The strength is being considered, but the direction of the strength is not. There is no advantage in having strength in a useless direction.
Where a 'T' is formed from two RSJs, the 'T'-cross will twist axially before the web of the 'T'-stem fails due to compression or tension in the flanges or gussets.

jack action said:
It's the area bh³ subtracted from the area bH³, like for a hollow rectangle, except the inner "empty" rectangle is the same width as the outside rectangle.

haha, i was going to cite this example to convince you of the opposite, but somehow you are further convinced of what i believe is the incorrect view. why dont you try this out in an experiment? Get a tube, then get 2 plates and set them apart a distance such that the formula you are using gives an equal area moment of inertia. Which one fails much earlier?

Baluncore said:
Any weakness is in the discontinuity of the web. The flanges do the work, the web keeps the flanges separated by a fixed distance.
The strength is being considered, but the direction of the strength is not. There is no advantage in having strength in a useless direction.
Where a 'T' is formed from two RSJs, the 'T'-cross will twist axially before the web of the 'T'-stem fails due to compression or tension in the flanges or gussets.
i think you do not understand that longitudinal shear (by which the moments manifest in a cross section) cannot be transmitted between 2 separate plates.

lordvon said:
i think you do not understand that longitudinal shear (by which the moments manifest in a cross section) cannot be transmitted between 2 separate plates.
I think you are analysing joint competence in modes that are not important.
Maybe you are thinking of a different structure to me.
Without the bigger picture, discussion of the joints is meaningless.

Baluncore said:
I think you are analysing joint competence in modes that are not important.
Maybe you are thinking of a different structure to me.
Without the bigger picture, discussion of the joints is meaningless.
hmm? joints can definitely be discussed in isolation. how can the weakest link be "not important"?

If I understand you correctly, this is where I think you misunderstand:
lordvon said:
But, for disparate plate cross sections of height h, the longitudinal shear is only going through a cross section of height h, in 2 plates. Do you understand what I am saying?
No, the longitudinal shear is going through a cross-section of height h (the thickness) at a certain distance from the centroid of the area formed by both plates (even if they are not connected together).

A single plate (compression at the bottom, tension on top) is not the same as 2 plates apart (compression in bottom plate, tension in top plate).

lordvon said:
hmm? joints can definitely be discussed in isolation.
Not sensibly.
The joint would evaporate, due to the lack of external load.
Or become really heavy because of fear, uncertainty and doubt, about the unspecified possible externals.

lordvon said:
... how can the weakest link be "not important"?
The weakest link may be weak in a direction that is not the limiting factor.

lordvon said:
Sure, here is what i mean (no welding, just bolts/screws+plates):
In post #3 you show a T junction.
By “vertical orientation”, do you mean “plan view”?
By “horizontal orientation”, do you mean “end elevation”?
If the members are RSJs, or I-beams, then how do you propose to put the nuts and bolts through the flange with the web in the way?

jack action said:
If I understand you correctly, this is where I think you misunderstand:

No, the longitudinal shear is going through a cross-section of height h (the thickness) at a certain distance from the centroid of the area formed by both plates (even if they are not connected together).

A single plate (compression at the bottom, tension on top) is not the same as 2 plates apart (compression in bottom plate, tension in top plate).

I actually agreed with you at one point. But I did the experiment I described to you in the previous post and got the result that leads to my belief now. Also I believe that the longitudinal shear has to be transmitted throughout the cross-section somehow. But I am happy to be proven wrong.

Baluncore said:
Not sensibly.
The joint would evaporate, due to the lack of external load.
Or become really heavy because of fear, uncertainty and doubt, about the unspecified possible externals.The weakest link may be weak in a direction that is not the limiting factor.
well, agree to disagree then. thanks for the input.

lordvon said:
I actually agreed with you at one point. But I did the experiment I described to you in the previous post and got the result that leads to my belief now. Also I believe that the longitudinal shear has to be transmitted throughout the cross-section somehow. But I am happy to be proven wrong.
I think you are referring to buckling which is another mode of failure. For example, if you take a square tubing and replace it with a pile of thin plates forming the exact same shape, a force acting in the plane perpendicular to the plates may induce buckling on the compressed exterior plate. If it fails, then the effective area of the whole beam is reduced which may lead to a chain reaction of failures one after another.

jack action said:
I think you are referring to buckling which is another mode of failure. For example, if you take a square tubing and replace it with a pile of thin plates forming the exact same shape, a force acting in the plane perpendicular to the plates may induce buckling on the compressed exterior plate. If it fails, then the effective area of the whole beam is reduced which may lead to a chain reaction of failures one after another.
yes i am familiar with buckling, and i mentioned it here in earlier posts. the failure i observed was not due to buckling.

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