Cantilever Hinging 1

[Quence]

In beams in between columns, you use Moment (left) + Moment (right) / Length to get the Vu.. but in case of cantilever where there is only one side fixed at the column.. what formula should be used?

If it is still Moment (left) + Moment (right) / Length, how could it work when one side of the cantilever is free (can't form hinge)?

 

[dhengr]

Quence:
Why don’t you draw the shear and moment diagrams for the two different beams, and see what sense you can make out of those, how you draw/determine those diagrams and what is really means? Maybe dig out your Statics and Engineering Mechanics textbooks and do a little self-study. You did have these courses in school, didn’t you? These are pretty elementary engineering topics. Start thinking in terms of how do I do this, why, what is the meaning of the various thought steps in the process, rather than just asking what formula do I use?

 

[Quence]

see attachment...

I'm only concerned with the seismic hinging/rebars yielding and plastic rotations and not the normal shear and moment diagram..

Statics shows that for equilibrium Vpr * ln = M(+) + M(-)

so Vpr = (M(+) + M (-))/ln

in the case of a cantilever.. there is only one side that is hinging (rebar yielding and plastic rotation).. so logically it should be

V = M(+) / ln

however, it seems the formula is similar to the former where the beam has 2 fixed column support...

May I know where I err in the analysis? In the cantilever, there is no opposite side hinging but why both still show up in the formula?

 

[BridgeSmith]

Forget the formula. It's simple statics for the cantilever - shear = load applied, moment = shear X distance from load to reaction.

 

[Quence]

I wasnt talking about the gravity loads which has its own moment and shear. I was referring to probable moments strength andcapacity during seismic when the beam ends encounter plastic rotation and rebars yielding. So my subject should have been "Cantilever Plastic Rotations". Does a cantilever have the same (Mpr (+) + Mpr (-))/ln reactions as.that of beams between columns? But how could the other free end encounters plastic rotations since its rebars were not fixed to a column?

 

[BridgeSmith]

There cannot be plastic rotations at the end of the cantilever, nor moments. You'll have to quantify the restraint force on the cantilever and use statics to determine shear and moment.

 

[Quence]

If there is no restrain in the free end but only connected to a column at one end. Then vpr should only be M(-) / ln? I couldnt find the reference about this in any textbook. They always use beam in between columns in determining mpr and vpr. Does anyone have any idea and can point to any reference about this?

 

[BridgeSmith]

If there's no restraint, there's only the inertial load of the mass of the cantilever itself, which shouldn't be enough to cause plastic hinging. It seems you're trying to calculate a moment without the load applied to produce it.

 

[Quence]

Are you familiar with capacity design in seismic? It only depends on the section properties and yielding of the rebars. The moments and shear would then be added to the factored gravity loads. I'm only concerned with the contribution of the moments and shears by the plastic rotations and hinging. In a normal beam moment connected to 2 columns. You calculate the Mpr of the negative and positive regions and multiply it by the length. Then you add this Vpr to the factored Vu of the gravity loads (produced by say point loads at the free end). I'm aware of this. Im just concerned of the computations of the plastic Mpr and Vpr. In the case of Cantilever, How is Vpr determined? is it Mpr (-) / ln only? The result would be added to gravity loads Vu and let's not concerned this for now.

 

[BAretired]

Actually, you could do the same thing for a cantilever.

For a simple span beam, when you apply moments at each end, say M1 and M2 (positive when clockwise) then V = (M1+M2)/L.

For a cantilever, if you apply M1 at the free end, then for equilibrium, the moment at the restrained end must be M2 = -M1. If M1 is clockwise, then M2 is equal in magnitude but counter-clockwise. The moment is constant throughout the length of the cantilever and the shear V = (M1+M2)/L = (M1-M1)/L = 0.

The shear in the cantilever is zero.

BA

 

[Quence]

But how could you have Mpr at the free end when its reinforcement is not yielding (no plastic rotation). Hence Mpr at the free end should be 0. Therefore you need to consider only the Mpr at the support and Vpr at support must be Mpr/ Ln. No?

 

[Toby43]

Assuming the cantilever projects from a moment frame, it has no stiffness to contribute in resisting drift, thus doesn't develop moments from drift. Thus there is no need to provide "Seismic capacity design" If the load can not shed away from cantilever (I assume this as this is fundamental principle of a cantilever, otherwise it would be something else) then moment failure and shear failure would not have a different outcome (gross change in shape), so again capacity design would not be applicable.

Toby

 

[BridgeSmith]

"Are you familiar with capacity design in seismic?"

I am, and as Toby said "...capacity design would not be applicable." I didn't realize that's what you were trying to apply. A truly free cantilever, without loads or restraint cannot develop any substantial shear or moment (only small inertial forces from dynamic movement which would require a crazy complicated dynamic model to even estimate with any accuracy).

 

[Quence]

No. Let me elaborate. Most applications of cantilever were overhang like second floor and up of buildings where the front extend two to three meters. There is always a lot of torsion introduced. Now torsion can eat up the capacity of stirrups. If there is seismic movement, the Vu could rise up. What if the loading makes the cantilever finally reaches plastic strain (bars yielding at support). The stirrups capacity would be almost used up by the torsion and shear of the seismic load. Now if there is additional Vpr from plastic rotations. Then the stirrups can yield.
Here is the weird part. Im manually verifying each output of Etabs version 9. In the cantilever output. The program add up the top and bottom steel of the cantilever to give such a huge Vpr amount (using still the equation M(+) + M (-) /Ln. Can anyone try it. Is the program faulty?

 

[jec67]

Looking at the sketch you submitted for the case of a beam supported at both ends, it appears the moments at the ends are due to the fixity of the joints at the columns and the lateral deflection of the frame. The moments then result in equal and opposite shears at the beam ends.

For a cantilever, how are you getting a moment at the free end? It would either be due to a moment applied at the free end or a vertical load on the cantilever. Am I missing something?

 

[Quence]

You can see this in many buildings in the streets where the second floor and up extend two meters forward..the cantilever beams at the sides carry another secondary beam at the free ends perpendicular to them and the secondary beam is carrying full wall. Then you will hav so much torsion at the cantilever beams and heavy load. When the cantilever yields. There will be no Vpr produced? Try it on etabs. It will produce Vpr. The program seems faulty (?)

 

[BridgeSmith]

"Is the program faulty?"

Doubtful.

"The program add up the top and bottom steel of the cantilever to give such a huge Vpr amount..."

If the cantilever is loaded beyond yielding of the longitudinal reinforcement, then the shear will be huge, especially if the cantilever is relatively short.

 

[jec67]

I am completely confused. Your last post talks about the wall loads at the end of the cantilever, but your third post from the start of this string states your are not concerned about gravity loads. Which is it?

 

[jec67]

Assuming the beam yields, then you have a hinge (i.e. rotation) and an unstable system. This likely explains why the program reports large shear.

Imagine a beam simply supported at one end and apply a load at the other end. What do you have for a shear if you run the numbers by hand? The system is unstable until the beam rotates to a vertical orientation. Then you can load it in tension (assuming you have restraint in that direction).

 

[BridgeSmith]

If you're getting a large shear in your cantilever under seismic without applying the gravity loads, then it's not modeled as a cantilever - it's a fixed-pinned beam and you're rotating the fixed end, right? Do you actually have complete restraint against translation (the end of the 'cantilever' cannot move up or down) up to the force necessary to fail the beam, or does the restraint fail first?