Gusset Design - basic help needed.

If I understand you correctly, you have calculated the stresses (or, if you would, the allowable moment) in the channel at the point of attachment to the gusset.

It's a good start. If your channel fails there, your system fails (you knoew that). If your channel passes, then even though it's possible that the gusseted portion will be sound, it may still fail, thereby failing your system again.

By adding gussets, you have certainly increaed the allowable bending moment at the true base of the channel; but by how much I can't tell. By enough? Don't know.

Deflection will be the sum of the deflection you calculate from the point where the gusset ends on the channel, plus the deflection at the end of the gusseted portion of the channel (see "tapered beam" theory), plus the efect of angular deflection at that second mentioned point.

..."By adding gussets, you have certainly increaed the allowable bending moment at the true base of the channel; but by how much I can't tell. By enough? Don't know."...

This sums up my take on the design, which is why I'm asking for some help. This image will help me illustrate what I'm attempting to explain.



Since I have assumed zero deflection from A-B, I have used the following deflection equations to solve for load P2:

(Deflection at B due to load P1) + (Deflection at B due to load P2) = 0

Moment: Then I have summed moments about A to find the 'new moment'. It seems like this method would work if I understood how to check the gusset for it's maximum deflection due to load P1 transferred through the channel.

Deflection: Here I have simply taken the deflection at C due to load P1 over the distance B-C (since I assumed earlier that deflection at B was zero). Again not sure if I can do this as it largely depends on the gusset section not deflection, or me knowing how much it will deflect.

I need to know how to design the gusset.

Thanks again.
G.

Well, first off, P2 (the reaction at B from P1) equals P1.

If you "cut" the beam just left of B, the forces remaining at B would equal P1, plus you'd have an imposed moment of P1*(C-B).

If you note, P1*(C-B) + P1*B = P1*C; which is correct, since the moment at the root of your weldment stays the same no matter where you make your "imaginary" cut. (For simplicity, I state the length as "C"; some may insist that it should properly be "C-A"; tough on them).

Step 1: Check for M = P1*(A-B) at point B. This is the cantilever moment in the beam proper.

Step 1.5: You will have some stress concentration going on at point "B" due to the change in stiffness. I can't tell you what that is (because I don't know).

Step 2: Check the Section Modulus of the compound beam at A; sheck it against M = P1*C.

The big trick is determining the section modulus at "A". The smaller trick is determining the stress concentration (if any) at "B".

You'll probably see a bit less deflection at "C" from the gusseted length (B-A), but it will not be zero.

Hmm... I guess I'm more lost than I thought.

I don't see how P2 = P1, if it did then there wouldn't be any reaction at A. Maybe my assumption that the gusset could be replaced by a load resulting in 0 deflection (similar to solving a propped cantilever beam) at that point is invalid.

I never stated that the deflection at C would be 0.

Thanks for your suggestions anyway, I'll leave this one to our PE I guess.

I think I meant "Deflection at 'B'." Sorry.

Take a couple of linear bearings on a vertical rail--enough so that the moment you're imposing is esily handled.

Now, assume that they're frictionless. Attach a horizontal broomstick (of length "C") coming off the bearing carrier. Attach a spring-scale to the top of the carrier.

Apply a force P1 at the far end of the broomstick. You'll see the same force reacting at teh spring scale. Now move the force closer in...to point "B" perhaps? Again, you'll see the same force (P1) acting on the spring scale. A cantilevered beam will have a constant shear across its length. That's what you've got.

It's great to see people trying, though!

Can I ask a question about that last explanation -

In your example what you stated would be valid because there is no reaction in the vertical direction since you're using linear bearings, but in the problem above I would think that if you sum forces in the vertical direction it would look something like:

Fy = 0 = -P1 + P2 + Ra

P1 is the external load, P2 is the 'reaction' of the gusset (I guess this would technically act through the centroid of the gusset), and Ra is the reaction at A (base of the channel). If P1 & P2 are of equal magnitude then the equilibrium of the channel is invalidated is it not?

Maybe where I'm going wrong with this is P2 AND Ra. Now that I'm looking at it, I guess there is really only one force there and it should act through the centroid of the gusset (shouldn't it?). I guess if I used a knee brace instead of a gusset that would be more similar to the way I've been trying to solve this, not that it helps me understand a way to effectively determine what size/shape gusset I need to resist the original bending moment.

I'll start tomorrow with a clear head. Thanks again.

What boothby is trying to describe is that the vertical shear is constant over the entire length of a weightless cantilever beam that's loaded by a concentrated force at its extreme end. It's simplest if you determine the shear and moment for any given point straight from the applied force and its distance from the point in question and ignore any intermediate points that you've already solved for.

But you can work from one point to the next (it's just more work and bookkeeping than necessary).

When you cut off the left part of the beam and analyze it as a free body you have to insert a vertical force reaction of "P2" and a clockwise bending moment reaction of magnitude P2*BC at point B for reasons of equilibrium. These reactions replace whatever it is that segment AB must be doing to support BC and hence do not represent loads applied to segment AB. Actually, we can ignore AB for the moment; so just cover that portion of the beam over with something opaque. You can solve portion BC.

For equilibrium at point B, the reactions computed above are equal and opposite to the internal member loads within BC at point B. So when you then go to evaluate segment AB as a separate free body, the internal member BC loads at point B become the loads that are applied to member AB (equal to P1 and P1*BC in this example). As far as a free body analysis of AB is concerned, they replace member BC and its end load P1, so cover up P1 and segment BC. At point A you'll have a vertical reaction that's equal to P2 (equal to -P1) and a moment equal to P1*BC plus -P2*AB. Simplifying the moment gives P1*BC + P1*AB, or P1*(BC+AB), or P1*AC.

If you're summing all the forces across the two beam segments, you'd need to look at both sides of the cut through point B, and it would be -P1 + P2 - P2 + Ra = 0.

As far as the location of Ra is concerned, I'm assuming that the gusset would be aligned with the web of the channel, so Ra would lie along this line. But since this is a channel there would also be some amount of torsion induced as a consequence of the unsymmetrical cross section shape (think end view; the line of the load does not pass through the shear center of the beam).

Norm

For Gusset Design, Look in Blodgett's Design of Welded Structures, Chapter 5.3 & Check for Buckling (i.e.,5th ed Roark T.35 c.1d)

The stress issue in the channel is one of bending, more than shear, but you can check for both. The gusset adds to the moment of inertia of the channel, thus increasing its capacity. The trick to this is either using an equation to determine the length of the gusset or just doing it by trial and error. To cut the brain damage, use trial and error. You'll probably get it in 2 or 3 tries.

Draw the moment diagram accurately without the gusset. Determine the section modulus required at the fixed end and for every 3 or 4 inches from the fixed end. Now determine the difference in section modulus needed to make it work within the allowable stress range at each of these points. Once you know the difference, you can then compute the thickness,width and length of your gusset. Use transposition equation to compute the new "I", then an average "c" value to get the "S".