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)?

 

[jec67]

Clarifying question: Your system consists of a cantilever beam. the unsupported end of the beam has beams framing into it (spandrel beams perpendicular to the cantilever beam & supporting wall loads). Is this correct? If so, are the spandrel beams in torsion for your load case you are looking at?

 

[BAretired]

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?

No! If Mpr at the free end is zero, then Mpr at the support must also be zero. Otherwise, the cantilever is not in equilibrium. Vpr is zero throughout the cantilever.

BA

 

[Celt83]

Are these the conditions you are talking about (see linked image if it doesn't show up below):

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[BAretired]

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?

The first part of the above statement is not correct. It should read:
In beams in between columns, you use Moment (left) + Moment (right) / Length to get the change in Vu...

To get Vu, you still have to add the simple beam shear from gravity or other applied loading.

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)?

The shear and moment diagrams of a cantilever are determined strictly from the applied loading. There is no need to modify them.

BA

 

[Quence]

A picture is worth a thousand words (see attached picture first).

I drove today and saw a construction site with cantilevered overhang. This is so common that most buildings are of that kind. You mean in your places you avoid that kind of design?

It answered your many questions.

It's not a simple cantilevered beam. The arrows in red in the picture point to the cantilever beam at sides that supports front wall. So they are heavily load. Torsion is introduced because of the eccentric loading. Note there is no restraint at the free end.

To summarize. For very heavy loading that yields the longitudinal bars of the cantilevered beam, the shear only comes from the gravity shear, torsion and not from Mpr? Is this right?
For normal beams fixed in between two columns or restrained at the end. You need to address gravity shear, torsion, and Mpr.

So in essence, for normal beams fixed in between two supports. Mpr and Gravity shear is the main concern. While for cantilevered beams with huge loads at the ends. It's torsion and gravity shear that are the issues and not Mpr. Correct?

 

[Quence]

No! If Mpr at the free end is zero, then Mpr at the support must also be zero. Otherwise, the cantilever is not in equilibrium. Vpr is zero throughout the cantilever.

BA.. so Vpr occurs due to the restraints at both ends and if there are no restraint, it wont occur even if the bars at the fixed ends are yielding and having plastic rotations? But cant the load at free end be restraining the free end from upward vertical movements?

(To others pls see last message for the loading conditions. Thanks a lot for all your assistances)

 

[BAretired]

I drove today and saw a construction site with cantilevered overhang. This is so common that most buildings are of that kind. You mean in your places you avoid that kind of design?

I don't avoid it, but it is not common in my area.

It's not a simple cantilevered beam. The arrows in red in the picture point to the cantilever beam at sides that supports front wall. So they are heavily load. Torsion is introduced because of the eccentric loading. Note there is no restraint at the free end.

The cantilever on the right is more heavily loaded than the one on the left but does not have as much torsion because it is loaded from both sides. I agree there is no restraint at the free end of either beam.

To summarize. For very heavy loading that yields the longitudinal bars of the cantilevered beam, the shear only comes from the gravity shear, torsion and not from Mpr? Is this right?

Hopefully, the top longitudinal bars do not yield under service load. If they do, the cantilever will fail. Yes, you are right.

For normal beams fixed in between two columns or restrained at the end. You need to address gravity shear, torsion, and Mpr.

Yes, but in most cases, Mpr is a relatively minor consideration compared to simple span shear.

So in essence, for normal beams fixed in between two supports. Mpr and Gravity shear is the main concern. While for cantilevered beams with huge loads at the ends. It's torsion and gravity shear that are the issues and not Mpr. Correct?

For a continuous beam, simple span shear must be modified to take into account the fact that differing end moments may result in a net unbalanced moment on the beam. The main reinforcement must be designed to resist the maximum positive and negative moments.

For a cantilever, the shear force and bending moment at all points can be calculated from the loading alone. There is no correction to be made to the shear force; a cantilever is a determinate structure. Torsional shear reinforcement may be omitted if the member carrying the heavy load is sufficiently rigid.

The cantilever moment must be considered when determining shear values for the back span.

BA

 

[BAretired]

BA.. so Vpr occurs due to the restraints at both ends and if there are no restraint, it wont occur even if the bars at the fixed ends are yielding and having plastic rotations? But cant the load at free end be restraining the free end from upward vertical movements?

You cannot allow the bars at the fixed end to yield, permitting plastic rotation. For a cantilever, that is a collapse. Plastic rotation may be permissible at the ends of a continuous beam provided the load can be carried somewhere else, namely by the positive reinforcement.

Yes, load at the free end causes downward deflection; therefore it restrains upward movement.

BA

 

[Quence]

Dear BARetired. About cantilever being determinate. Wont seismic lateral forces redistribute the moments.. imagine the overhang swaying left to right. The right cantilever can have more moments at.the free ends in that instant.

 

[BAretired]

Quence,

A cantilever is determinate for any given load combination, including seismic forces. Your code will tell you what seismic forces to include in design. During a seismic event, forces will vary in magnitude and direction. Variable forces result in variable shears and moments, but to say that seismic forces "redistribute the moments" is not valid.

Moment is redistributed when one part of the structure yields and allows another part to take over. A cantilever cannot be permitted to yield because there is nothing else left to resist the load.

BA