# Is pin column-beam joint always have less moment, more deflection compared to fixed joint?

I am referring to the answer here:

It seems that on a simple structure

(The diagrams on the left the images below have fully-fixed connections, while on the right the columns are pinned connections to the beams.)

A: The deflection on beam is bigger in pinned model, but the deflection on beam/column joint is bigger in fixed model

B: The moment on column in pinned model is 0

C: The axial load on column in pinned model is bigger, because the moment is being transformed into axial load (?? is this reasoning true??)

My questions are:

1. These conclusions are true for this particular simple model, but are these three conclusions always true on any general model?
2. And why? Can it be explained in terms of loading flow and statics? Or this is just FEM behavior that we can't explain more intuitively?

Let me start by answering your second question: models such as this one, which involve only one-dimensional beam elements, are 100% analytical and can therefore always in theory be understood intuitively. There is no "FEM behavior" for such models. Sometimes the models may get complex with lots of bars and whatnot, which may make "intuitive explanations" more difficult, but the result will always be analytical.

Let's start by looking at statement B:

Now, let's take a look at beam 4 in your model (the left-most beam). More specifically, it's bending moment diagram. As you've noticed, the pinned model displays zero moment at the left-most column. This is the very definition of a hinge and is expected behavior. The moment on the beam at the central column is non-zero because the beam itself is not hinged, but the central column is hinged and therefore displays zero moment at the node.

Now, on to statement A, starting by looking at the beam's deflection:

Let's keep looking at the bending moment diagram. The beam equation tells us that

$$\dfrac{\partial^2 }{\partial x^2}\left(EI\frac{\partial^2 w}{\partial x^2}\right) = q$$

which also tells us that

$$EI\frac{\partial^2 w}{\partial x^2} = M$$

that is: bending moment (divided by stiffness $EI$) is the second derivative of deflection. From calculus, we know that the second derivative of any function described the function's curvature. So bending moment describes deflection's curvature, which describes the "acceleration" with which the beam's tangent (the first derivative of deflection, and therefore bending moment's integral) changes.

So, the more balanced a bending moment diagram is between positive and negative bending moment, the more the total "acceleration" cancels itself out, implying in smaller tangent changes, and therefore smaller deflections. So yes, a fixed node will always lead to smaller deflections

To answer the matter of the node's displacement, we first need to explain point statement C. For that we need to look at beam 4 in isolation. To do so, we need to replace the surrounding beams with elastic supports which describe their stiffness.

• The vertical supports' stiffness will be equal to the columns' axial stiffness (the node with the central column will also have a tiny addition due to the other beam's stiffness against imposed transversal displacements)
• The horizontal supports' stiffness will be equal to the columns' stiffness against imposed transversal displacements
• The rotational supports' stiffness will depend on the boundary conditions. If hinged, then the outer node will have zero stiffness and the central node will have a stiffness equal to the other beam's stiffness against imposed rotations. If fixed, then both nodes will have the columns' stiffness against imposed rotations, adding the other beam's stiffness as well for the central node.

So, basically, the only difference between the hinged and fixed cases is in the rotational stiffness (as would intuitively be expected). This increased stiffness, however, causes the node to pull in a greater proportion of all forces, thereby increasing the axial forces in your outer column and reducing them in the central column in the fixed model.

Returning to the issue of the node's deflections, they are now easy to explain. After all, in the fixed model the column suffers more axial forces, naturally increasing the vertical deflections. But it also suffers bending moment, which generates horizontal deflections as well as a tiny bit of additional vertical deflection.

• Thanks for your excellent explanation. I don't understand this The moment on the beam at the central column is non-zero because the beam itself is not hinged, but the central column is hinged, I thought if a joint is hinged, then it's hinged on both column and beam, how can it be that beam is hinged but not column? Jul 22 '16 at 2:04
• @Graviton: Well, if you just look at your own models, you'll see that the beam over the central column is clearly continuous even in the "hinged" model, which is what generates the non-zero bending moment over that support. If that weren't the case, then each span would behave independently of the other.
– Wasabi
Jul 22 '16 at 2:39
• @Graviton: If you look at my answer to the question you linked to, you'll see an image how the beam over the central column can either be hinged or not. The version to the left has a continuous beam supported by a hinged column, while the one to the right represents a node which is fully hinged.
– Wasabi
Jul 22 '16 at 2:39
• Wasabi, is there any mathematical derivation somewhere that shows that when the joint condition is hinged, it will lead to zero moment? I started from the beam equation and then applied appropriate boundary conditions, and then tried to solve the differential equation, but I couldn't get the zero moment conclusion, got lost in mathematical details :( Jul 22 '16 at 3:31
• @Graviton: I don't know what you mean... the definition of a hinged joint is to have zero moment.
– Wasabi
Jul 22 '16 at 10:10