I agree with that someone. Roller can only introduce a load perpandicular to the support surface.

Who you argue with, your spouse? If so, you are right :)

The simplest example of a roller support is an unlocked car tire, it rolls until you hit the break, at that moment (with break applied), it becomes a pin support. Both roller and pin supports do not have moment capacity, as they will rotate about the contact point with the beam they support. However, although no moment capacity at the supports, a simply support beam will fail by moment, why, where is the moment acting on to cause failure? I sincerely hope you know the answer.

Whoever you were arguing with is right - for 2D analysis pin is reaction in the x and y, roller is reaction in y, and fixed is moment about z and reaction in x and y. This assumes x is parallel to the support, y is perpendicular, and z is orthogonal to x and y.

Of course this is statics at its most basic level. In reality you have to look at it in 3 dimensions, with reaction forces in all three axes and reaction moments about all three axes. Then you have partial fixity, spring supports, etc.

A 2D roller is limited to a single reaction normal to the bearing surface including, but not limited to, a horizontal surface. A 2D roller allows two degrees of freedom, a reaction parallel to the bearing surface and a moment.

The concept should not cause confusion...Fixed, Pinned, Roller = 0, 1 and 2 degrees of freedom respectively.

BA

All

Thank you for your responses, I really appreciate it.

Just a follow on question.

The attached drawing (Picture 1) shows a member that support a UDL of 200kN.

The member sits directly on 3 x supports, to me I would describe these supports as roller supports?

If this is correct then they have no horizontal reaction forces which I am happy with.

However since it is a roller support it has no moment reaction but I struggle to visualize this, the flat underside of the member sits directly on a flat horizontal supports - would it be correct to say that the the member rotates about the edge of the supports as shown in picture 2 (forgive my poor sketch, I have highlighted in red circles the rotation points).

https://files.engineering.com/getfile.aspx?folder=...

https://files.engineering.com/getfile.aspx?folder=...

Thank you.

• https://files.engineering.com/getfile.aspx?folder=2700eb9d-16f2-4632-a7d3-7

Let's start with the problem declaration: A member with 200kN/m UDL (Uniform Distributed Load), over 3 support beams. What type of support is typically assumed? The answer can be either one below:

1) All 3 are pin support - this is correct if the member is connected to the support beams through positive fastening.
2) All 3 are roller support - this is acceptable for gravity load only, provides the member is not positively connected to the support beams, and having frictionless medium in between the member and the supports(note this type of support specification violates the rule regarding global stability - the member can move horizontally, but will not affect the analytical results).

Regarding your second question: Theoretically the member will rotate about the centroidal axes. Your sketch depicts the deflect shape of the member with 3 roller support - the supports do not rotate with the member. For the member support on 3 pin supports, at least the outer two supports will rotate with the member, because of the drag force from the connections and shear friction.

Quote (r13, BAretired in red)

Let's start with the problem declaration: A member with 200kN/m UDL (Uniform Distributed Load), over 3 support beams. What type of support is typically assumed? The answer can be either one below:

I think the OP meant 200kN total load. Makes a big difference to reactions, but it doesn't really affect the answer.

1) All 3 are pin support - this is correct if the member is connected to the support beams through positive fastening.
Three pin supports increase the indeterminacy of the structure by two degrees of freedom and should never be used with a computer program as it will yield incorrect answers. The correct model is one pin and two rollers.
2) All 3 are roller support - this is acceptable for gravity load only, provides the member is not positively connected to the support beams, and having frictionless medium in between the member and the supports(note this type of support specification violates the rule regarding global stability - the member can move horizontally, but will not affect the analytical results).
Three roller supports produce an unstable structure, even in the absence of horizontal applied forces. Should be one pin and 2 rollers.

Regarding your second question: Theoretically the member will rotate about the centroidal axes. Your sketch depicts the deflect shape of the member with 3 roller support - the supports do not rotate with the member. For the member support on 3 pin supports, at least the outer two supports will rotate with the member, because of the drag force from the connections and shear friction.

Pins and rollers are theoretical concepts. They are deemed to be knife edge supports which rotate freely. Real structures do not have knife edge supports. If a support is flat on top, then the beam would tend to bear closer to the edge when it rotates, not precisely at the edge as the bearing area has a finite length.

BA

BA,

Quote (OP)

The attached drawing (Picture 1) shows a member that support a UDL of 200kN.

I just wanted to point out the load description (UDL) and the units of the load (kN) do not agree with each other.

Regarding to the 3 rollers system, it is unstable, so theoretically there is no solution, and in reality, no one shall do it. But strictly speaking, in a limit equilibrium situation without any cause for lateral movement, the vertical reaction under each support is identical to that obtained for a stable system.

The image below compares results from the 3 pins system, and the system with 1 pin and 2 rollers. No warning was issued from the program.

Quote (r13, BA in red)

I just wanted to point out the load description (UDL) and the units of the load (kN) do not agree with each other.

A uniformly distributed load (UDL) can be expressed in kN or kN/m, depending on whether you are referring to W or w (W = w*l).

Regarding to the 3 rollers system, it is unstable, so theoretically there is no solution, and in reality, no one shall do it. But strictly speaking, in a limit equilibrium situation without any cause for lateral movement, the vertical reaction under each support is identical to that obtained for a stable system.

I suppose one could say that the system is in neutral equilibrium, where the slightest lateral force would set the entire beam in motion, but in reality, the structure should be considered unstable. Some software (perhaps not all) would issue a warning.

The image below compares results from the 3 pins system, and the system with 1 pin and 2 rollers. No warning was issued from the program.

I would not expect a warning to be issued for those conditions as they are both stable systems; but you will find different results for shear and bending moment if the supports are placed at the bottom of the beam. With three rollers you should get a warning.

BA

Additionally, with respect to the analysis r13 presents above, I'd expect the 3 pin case you present above should have horizontal reactions to maintain equilibrium as the beam cannot shorten under the flexural loading if restrained horizontally at all supports. You get some catenary action and tension in the beams. This is why the beanding/shear/axial/deflections will all be different as stated by BAretired. Not sure why the 3 pin analysis isn't showing an x direction reaction on the N1 & N3 supports?

Everything BAretired has added in his last 2 posts is correct here

Quote (Agent666)

Not sure why the 3 pin analysis isn't showing an x direction reaction on the N1 & N3 supports?

Most likely, the beam was modeled as being rigid or very stiff. No bending = No deflection = no change in the length of the beam = no horizontal forces.

Rod Smith, P.E., The artist formerly known as HotRod10

To BA,

To be precise and clear, the conventional description of the loading on the OP's example should be:
- ..the beam has a total load capacity of 200kN, that is uniformly distributed over..., or
- ..the beam supports an UDL of XX kN/m in metric system, or XX plf/klf in the US system.

TO Agent666,

You have raised an important question, though it is not in the OP's query.

I think in general purpose structural analysis, the program ignores the secondary effect, that produces horizontal reaction at pin support. Evident by another examples below - left end fixed, right end on roller, or pin support. A person familiar with structural programming should be able to point out why, or RISA has committed a mistake.



We know in reality, secondary effect shall be included to obtain true reactions afford by the pin support, that is important in design of the connection. Also, in reality, for beam over multiple supports, imo, should be modelled as beam supported by multiple pin supports, unless there is clearly no physical horizontal restrain at the interface, which is rare.

I guess the OP is in the elementary level of structural analysis and design. Wish he/she has knowledgeable mentor around, and study more.

I don't know whether all software ignores the secondary effect of a continuous beam, pinned at every support, but I do know that it makes a difference for a simple or continuous truss. The bottom chord of a simple truss is in tension throughout its length, so it must strain accordingly. If both ends are pinned, it can't change length, so pinning both ends of a simple truss is wrong and will lead to incorrect results for member forces.

That has been demonstrated in a thread fairly recently (I forget the name of the thread offhand). Simply extend the rule to a beam, whether simple or continuous. Use at most one pinned support. It doesn't matter which support is pinned. The remainder should be rollers if you want to be sure the results are accurate.

In the case of cable, it's a different story. The cable has an initial sag and the supports must be able to resist the horizontal component of the cable.

BA

Quote (r13)

Also, in reality, for beam over multiple supports, imo, should be modelled as beam supported by multiple pin supports, unless there is clearly no physical horizontal restrain at the interface, which is rare.

That statement is completely wrong, hence very bad advice to give to a young engineer reading this thread. It may be true for some software, I don't know, but it is theoretically incorrect. If the pin supports are placed at the bottom of the beam, where I would expect them to be, all pins are located h/2 below the beam's neutral axis, where h is the height of beam.

In the case of a simple span, the bottom fibres are in tension across the full span. If the ends are pinned, how can that be? A pin at each end means there is no change in length.

BA

I'll let the sketch speak itself. [Sketch was withdrawn and revised]

r13,

Your sketch speaks with forked tongue.

When pin supports are placed at the bottom of the beam, they must resist a horizontal reaction. If there is one pin and the remainder rollers, there can be no horizontal reaction. See simple span at the left in sketch below for support location. The bearing point of a pin or roller is not at the neutral axis of the beam. It is usually at the bottom as you show on Sections 1 and 2.

Try relocating your pin supports h/2 below the neutral axis by using a rigid extension from the neutral axis down to each pin and let us know if you get horizontal reactions.

BA

Unless the beam has a relatively (compared to the load) very large moment of inertia, having more than one pinned connection in the structural model will produce different results for the reactions than if there is only one pinned support. Unless the supports are really very rigid (and it's difficult to imagine how that would ever be accomplished in a real structure), modeling multiple pinned supports will result in model that does not come close to reflecting reality.

Rod Smith, P.E., The artist formerly known as HotRod10

Quote:

I'll let the sketch speak itself.

Unfortunately it's telling lies, that picture is wrong in the sense that if you apply a concentrated horizontal load to the end of the beam at the support location as shown in the bottom picture with 4 pins, all of the H load is resisted by the first pin supoort. The beam doesn't distribute the load to all four pin supports as implied.

You don't need a computer analysis, just some basic understanding of statics (usually 1st year stuff at University) to follow the logic.

Quote:

modeling multiple pinned supports will result in model that does not come close to reflecting reality.

Exactly.

Thanks for the responses and your opinions. Yes, the horizontal load was drawn incorrectly. In order to generate the support reactions, it shall be located on the beam as member force.

Something horizontal, I suspect...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

Quote (dik)

Something horizontal, I suspect...

If it's a roller, there can't be something horizontal or it would roll.

BA

r13,

You have not responded to my suggestion that pin and roller supports are at the bottom of the beam, not at the neutral axis.

For a simple span beam with uniformly distributed load, the end rotations can easily be calculated. If the rotation at each end is θ, the total strain of the bottom fibre is 2yθ where y is the distance between the neutral axis and the bottom of beam. If a pin support is used at each end, the total strain must be zero.

I suggested a way you can check that with your software, but you have simply ignored the suggestion. You seem to be more interested in clouding the issue than seeking the truth.

BA

I don't know what exactly you don't understand, so no comment. I've suggested many times to get the free copy of Risa2D, so you can use it to carry out your suggestion. I don't think we can discuss in a civilized manner, why drag on!

r13,

I don't know why you can't discuss this in a civilized manner. And I don't need software to carry out my suggestion!

BA

Quote (r13)

We know in reality, secondary effect shall be included to obtain true reactions afford by the pin support, that is important in design of the connection. Also, in reality, for beam over multiple supports, imo, should be modelled as beam supported by multiple pin supports, unless there is clearly no physical horizontal restrain at the interface, which is rare.

With all due respect and in a civilized manner:

What should be included in secondary effects? How about temperature change? Is that a secondary effect? If multiple pin supports are modelled, the beams cannot change in length, causing axial forces throughout the beam and horizontal reactions at some supports.

Continuous beams do not always have the same cross section in every span. Even when they do, placing pin supports at the bottom of a beam, instead of at the neutral axis, causes horizontal support reactions which alter moments and axial loads throughout the beam.

This can be avoided by using one pin support, and the rest roller supports. Alternatively, by using horizontal roller supports at every location and adding one vertical roller support anywhere to provide stability to the structure. That would be my recommendation, contrary to the red text in the above quote.

BA

Maybe I am looking things differently, or simply wrong. But I have no intention to lie, and mislead less experienced.

Quote (If it's a roller, there can't be something horizontal or it would roll.)

Exactly, BART...

Rather than think climate change and the corona virus as science, think of it as the wrath of God. Feel any better?

-Dik

r13, what are you trying to show. You have pin supports deflecting, rollers sitting still. It does not make sense at all.

Honestly unsure where you are trying to go with your argument, are you just taking the piss and winding everyone up on purpose by posting the incorrect information repeatedly? If so you should stop, it isn't adding to the original intent of the post.

It was a view of a floor framing plan, with indicative support conditions at the beam, so not to waste my energy to draw cross section views. Please be more understanding before offering your harsh comment/opinion. I agree that this back and forth shall be stopped a long while ago, if not called repeatedly to respond. I will stop here then.

Quote (r13)

Maybe I am looking things differently, or simply wrong. But I have no intention to lie, and mislead less experienced.

Maybe you are looking at things differently. I do not believe, and do not suggest that you are intending to mislead anyone. However, I am not following your latest diagrams.

Below is a sketch of a two span continuous beam. The upper diagram has three pin supports located at the bottom of the beam. The lower diagram has one pin support and two rollers. The loading is the same for both diagrams, namely a UDL over both spans and a couple of sloping forces on the right hand span represented by sloping arrows.

An analysis of these beams will yield different results for shear, bending moment and axial load because the boundary conditions are different. We can discuss these differences in more detail if anyone wishes. The second diagram is the arrangement I suggested, but the location of the pin support can be moved to either point b or c if preferred.



BA

Interesting point, below is the two supporting schemes and results of the beam with inclined load (V=1, H=1). The upper beam is supported by 3 pins, the lower is pin-roller- roller. It seems the differences are the support reactions (horiz.), member axial forces, and lateral displacement and rotation.

r13,

Your program is representing the beam as a single line with supports at the same level. In effect, that means your supports are at the neutral axis of the beam. You need to drop the pin and roller supports by half the beam depth to see a difference in the shear and moment diagrams. This means adding nodes N4, 5 and 6 at an elevation one foot (or half the beam height) below N1, 2 and 3 with three infinitely stiff vertical members fixed to the beam from N1 to N4, N2 to N5 and N3 to N6.

BA

BA,

Your suggestion/request is beyond the realm of elementary/general structural analysis, that considers all applied and resultant forces are acting on/about the centroid of the member; similarly, the support is located at the intercept of its axes and the centroid of the connected member, not the extreme fiber of the physical beam. You may try yourself the model as you suggested, and post the findings. I am eager to learn the difference, but don't know how to properly handle the modeling in a general purpose structural analysis program such as Risa2D. Sorry.

Note: There is a "member offset" commend in STAADPro, maybe that is something you are interested in. Haven't use the program for a longtime, so really can't provide information on how it works.

Quote (r13)

Your suggestion/request is beyond the realm of elementary/general structural analysis, that considers all applied and resultant forces are acting on/about the centroid of the member; similarly, the support is located at the intercept of its axes and the centroid of the connected member, not the extreme fiber of the physical beam. You may try yourself the model as you suggested, and post the findings. I am eager to learn the difference, but don't know how to properly handle the modeling in a general purpose structural analysis program such as Risa2D. Sorry.

The simplest frame is a single portal frame which can easily be analyzed by any 2D frame program. Multiple bay frames can also be analyzed by the same software.

When a single span beam is analyzed, it is usual to consider supports to be one pin and one roller. There is a good reason for that. If two pins are chosen as supports, the assumption of all forces acting at the centroid of the member cannot be achieved in a real structure because the pins are usually located at the bottom of the beam rather than at the neutral axis.

I am proposing to consider a single beam as a portal frame as shown below where the distance from the neutral axis down to the pin is 'h'. I will assume that I₂/I₁ = 1, although it should be much smaller.

k = h/L and N = 2h/L + 3
Mb = Mc = -wL^2/4N = WL/4N where W = wL
Ha = Mb/h = WL/4hN

If L = 20'-0" and h = 1'-0",
then N = 5 and Mb = Mc = WL/(4*5) = WL/20
and Ha = WL/20*1 = W

The assumption of pin supports, using this method, produces significant end moments on the beam as well as a substantial axial compression.

If one of the pin supports is changed to a roller, the end moments and the horizontal force Ha both vanish.

Quote (r13)

Note: There is a "member offset" commend in STAADPro, maybe that is something you are interested in. Haven't use the program for a longtime, so really can't provide information on how it works.

That is interesting and may be another good approach to the problem. I do not have STAADPro, however and can't justify purchasing it under the present circumstances.

BA

Quote (BA)

...because the pins are usually located at the bottom of the beam rather than at the neutral axis.

This is the mean difference of us. I do recognize your concern. In practice, we model according to member centroidal axis, and put support on the same level. If the perceived effect (I call it secondary effect) is large, we perform hand calculation (ie. M_a = H*d/2), or use member offset comment, and impose the load on the support. It is just something needs to be designed for, because in real structure the supports are most likely connected to the member above.

The structural models below show the typical way of modelling.

FYI, here is the instruction of STAADPro "Member Offset Command". You can browse other topics too. Link

If, in the expression in my last post, I₁ is taken as infinite:

then k = 0, N = 3
and Mb = Mc = WL/12 or wL²/12 (a fully fixed end)
Ha = Mb/h = WL/12h

So the elevation of pin supports makes a huge difference in the result.
That is why we use pin and roller for a simple beam, rather than two pins.

BA