Stiffness Method in Structural Analysis

I see your doubt. I learnt this method through a different view. I used to, instead of doing as you stated, replace one of the beam support by a unitary load, and then solve the system as a isostatic one. Then, go back and replace all the other stuff with the results you achieved using the unitary load, because all will be proportional to that unity.

The method you mentioned is quite the same thing, but done in another fashion. It works, because it introduces another set of equations when you didn't have it, suposing unitary displacements or loads to fill the gaps. All a matter of building the equation system with something that you know, and calculating everything else proportionally.

Just like: I don't know the support load in the middle, but if it were equal to 1, so the other reactions would be R1 and R2. So, if the middle support is actually 10 after calculations, everything is just set to get to the other values.

The example you have provided may be solved in other ways. The horizontal component of F is taken by the hinge at A so a portion of the left beam is in compression. You could remove the support at B and calculate the deflection at B, then calculate the force required to push it back to zero.

Or you could remove the support at C, calculate the deflection at C, then calculate the force required to pull it back to zero...pretty easy stuff.

But you are asking about the stiffness method. The stiffness method is primarily used in situations where the number of members is large and computer methods are needed because they cannot easily be solved by hand methods.

The basic premise of the stiffness method is consistent with the principles which you have already learned in elementary structural theory, but it takes a bit of study to see what is going on. I did not find it easy at first either because I was not familiar with matrix algebra.

I do not understand your difficulty. If you can define the problem more clearly, I would be pleased to help.

Well, all forces eventually resolve into shear, but their transmission to the loaded point may be by torsion. All the torsion forces and linear forces must resolve to zero, or the problem will begin to rotate or move at increasing speed due to the remaining force. Like a jet plane, with pitch, roll, and yaw rotations, and sideslip, lift, and forward (reverse) translation displacements, all 6 sets of forces must balance out.

If you do not let the bending forces rotate the members, you get into a lot of complex forces that complicate the problem, so we ignore them to look at the larger forces more simply, but in real life (welds or multiple bolts or pins preventing rotation) they might show up as bent metal!

With the computer, you can solve lots of simultaneous equations, so the equation writers can be less shy, but it makes for hard reading.

In the illustration, the horizontal beam is tied to the page at the left end, and seems to float on two pivots, but it is not clear to me if the right end is captured by the pivot, so the load is cantilevered to tension on that pivot. If the round object is a free roller, deformation of the beam makes the right end float off the roller.

If rotation is not allowed at just the left end, when the clockwise torsion from the force through the leverage of the horizontal member arrives at the left end anchor, it is combined with the linear (shear) forces to form the total vector of stress on each part of the beam at that end. The top of the beam will be in tension, and the bottom in compression, due to the torsion. These force vectors have to be added to the shear (linear) forces to determine the magnitude of the stress at every point.

Well, I got an A in that course, but now I am a EE programmer, so I hope my simpler understanding helps.

Well, I hope so too. But it sure doesn't help me. I haven't the faintest idea of what you said. To tell the truth, I don't believe you know what you are talking about. If you think you do know what you are talking about, please let us know.

Well, I had hope -- that's why I posted. Are you Guest, the original inquiring person, or just another befuddled specatator? Helping him is the original objective.

It is hard to find the simple model equations in the equations of more complex but realistic models. Basic, simplified analysis starts with impossible things like the drawing in the URL in the original post: inflexible beams attached at one point and on rollers at all other points, so there is no flex and only one point of tension. In that drawing, all the force's horizontal component pushes left on the top of the triangle, and the vertical component is split proportionately by distance to the triangle and the middle, roller support.

Real beams are attached with joints of limited flexibility, they flex and put complex strains on the entire attached structure. When beams flex, they pull the two attachment points together, producing tension, so they stretch, and they put torsion on all attachment points, either directly on adjacent attachments for the force producing the flex, or through cantilevered forces (when the first span bends down, any adjacent spans of the same beam bend up, absorbing part of the torsion not absorbed by the intermediate attachment, and then any farther out spans of that beam bend down, absorbing part of the torsion not absorbed by the closer attachements and closer span cantilevered flexure, etc.). You can observe this on a deck plank.

I guess I will have to admit to being another befuddled spectator. I wrote a 2D frame program about ten years ago which I have used from time to time but the theory is not all that fresh in my mind now. The software available today is much more user friendly than mine was.

Hi

without compliant middle and end you will not get any forces???

What is usually done if only the beam and not the bearings are elastic:

Step 1:

Fix the beam at two points only and calculate the forces, linear and angular deflections.

This to be done at the two remaining fix-points and at the omitted fix-point.

Step 2:

Introduce an auxiliary force at the omitted fix-point that will force the beam to zero linear deflection at this omitted fix-point. Calculate the angular deflection at the 3 fix-points that are caused by this auxiliary force. And the forces on the 2 fix-points.

Step 3: Add up angular deflections and forces.

Here it should be clear that the forces and angular movements that are introduced by the auxiliary force have to be added.

(I prefer Castigliano's method)

RHABE

I think the question is totally misunderstood it had allmost nothing to do with the picture I posted. I thought I made myself clear in that what I was wondering about is why the stiffness method works, it is a general question and I am not asking you to help me find the forces in that picture.

I will try to clear up what I asked about and ask again:

Lets consider a frame structure in 2D in the plane, to make it easier at the joints the only degree of freedom is rotation(which was in the picture earlier, that is the reason why I posted it). The stiffness method says that in order to find the internal moments we first look at the frame as completely rigid, and then calculate the fixed end moments. The method then says that we shall look at the joints and at the fixed-end moments that comes to each joint. If the sum of the moments in each joint is not zero the joints aren't in equilibrium, this I get. What the method then says is that the joints have to rotate(since the sum of the fixed-end moments are not zero), now comes the tricky part: It says that the joints has to rotate in a way so that the sum of the fixed end moments and the "rotation moments" are zero.

In this case the rotation moments (for each element) are calculated from this stiffness matrix(left end A right end B)

Ma 4EI/L 2EI/L angle a

= *

Mb 2EI/L 4EI/L angle b

Finally I now come to what I dont get. How can we justify that the sum of the rotation moments and the fixed end moment over each joint has to be zero? It is easy to say that this is just because of equlibrium. But I still don't get it, because I don't see where the rotation moments come from, are they virtual moments that we just put on, that has to be there so we can get equilibrium for the joints?

It is a very theortical question,b ut I hope you can help. If CR4 can't help ,no one can!=)

Finally I now come to what I dont get. How can we justify that the sum of the rotation moments and the fixed end moment over each joint has to be zero? It is easy to say that this is just because of equlibrium. But I still don't get it, because I don't see where the rotation moments come from, are they virtual moments that we just put on, that has to be there so we can get equilibrium for the joints?

The sum of the rotation moments and the fixed end moment at each joint is usually not zero. In the case you posted, it had to be zero because you specified a pin at each joint.

The rotation moments come from a vector multiplication, M = K* theta where M, K and theta are all vector quantities. K is the stiffness matrix which is a property of the structure irrespective of what loads you apply. If you apply a moment Ma at joint A and you define joint A to be pinned, it will turn out that Ma' from the vector multiplication will be equal and opposite to Ma so that their sum equals zero.

If you had specified a spring with rotational stiffness at joint A, the sum of the two would not be zero. Its value would depend on the rotational stiffness of the spring.

You seem like a very insightful guy when it comes to structural analysis Bruce, but in my book it says that the sum of the rotation and the fixed-end moments are zero.

And I have obtained correct results in my calculations by using this fact aswell. I am talking about fromes. I don't understand why you are talking about something that is pin-connected. In a frame different elements comes into each joint, but joints arent pin connected?

Sorry, I'm a bit rusty on this stuff and trying to read through my old textbooks hastily has not been helpful. The following link appears to be a fairly simple example of a two span continuous beam:

http://www.engr.uky.edu/~gebland/CE%20582/Summary%20-%20Stiffness%20Analysis.pdf

I have not gone completely through it yet, but can you follow what the writer is doing and explain where you are having a problem? That way we will both be working with the same reference.

I think I know the problem you are having. In the current problem (using this link), the structure has one degree of freedom, namely rotation at node 2.

Node 2 is first held against rotation. The fixed end moment (FEM) of Beam 2-3 is acting clockwise on Node 2. Node 2 has an unbalanced moment of qL²/12.

Node 2 is now released and the unbalanced moment is distributed to all of the members meeting at Node 2 in accordance with their stiffness. In this case there are only two members.

If Beam 1-2 had the same load, it would exert an equal and opposite moment on joint 2 so the unbalanced moment would be zero and the final moments on each beam would be equal to the FEM.

Imagine 10 beams arranged like the spokes of a wheel, each fixed at the far end, all meeting at node 2 and all of equal stiffness. Let us say that Beam 1-2 is loaded and all the rest are unloaded. When Node 2 is held against rotation, the unbalanced moment is the FEM of Beam 1-2. Say that is 100 '# (positive because it is counterclockwise).

When Node 2 is released, each beam takes 100/10 = 10'#, its share of the unbalanced moment. For nine beams, the moment is 10 at node 2 and 5 at the far end. For Beam 1-2, the moment is 100 + 5 = 105 at Node 1 and -100 + 10 = -90 at Node 2.

Note that 9*10 - 90 = 0 (sum of moments of all beams at Node 2).

One word: Timoshenko

(Theory of Structures - Chapter 10, Matrix Methods in Structural Analysis)