# Mod-06 Lec-35 Matrix Analysis of Plane and Space Frames

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### Mod-06 Lec-35 Matrix Analysis of Plane and Space Frames Good morning, this is lecture number 35 We are still with module 6 – Matrix Analysis of Plane and Space frames If you recall, in the last class we had covered the conventional stiffness method. So, in this session, we will look at the reduced stiffness method, as applied to plane frame elements This is covered in the chapter on Plane and Space Frames in the book on Advanced Structural Analysis So, I keep showing you these maps, because I want you to see how the system of analysis is the same; the structure is changing, but the methodology is not changing. We saw how well it worked for the simplest type of axial element; then, we worked with plane trusses; then we worked with space trusses; then we worked with beams; then with grids, and now, with plane frames, and in the next class, or the class after that, with space frame So, we are covering all kinds of skeletal structures, and you can clearly see there are two broad methods: There is a stiffness method which is preferred for programming compared to the flexibility method. In the stiffness method itself, you have the conventional stiffness method; you have a simplified formulation called, the reduced element stiffness method, and that is a method that we are going to discuss in this session If you recall, this is the 6 degree of freedom plane frame element that we used for the conventional stiffness method. This is a large matrix and it is a singular matrix. What is the rank of this matrix? 3, and one way of understanding why the rank is 3, is because you have, mathematically, you have 3 dependent rows or columns, but physically, what does it mean? Physically For a stiffness matrix to be non-singular, what you have to make it is, you have to make the element stable Now, a singular stiffness matrix still works in a global scenario because your structure is stable. When you assemble the structure, stiffness matrix – the k a, a is non-singular, but here, you can begin with a non-singular element stiffness matrix by giving how many restrains? 3; then only you have stability, and you can choose your type of restraint We have been assuming that the simply supported condition is convenient, and we will stick to that So, we are now going to use a 3 degree of freedom system, and it is very easy to write down; at this stage, you should find it very easy to write down the element stiffness matrix You can do it from first principles You can also do it from the 6 by 6 element stiffness matrix. how Just talk of the irrelevant rows and columns; there are 3 dependent rows and columns. If you make it simply supported, you will find that the first row is important; the second is not because we do not want the shear degrees; just delete the shear degrees of freedom, and you do not need 2 axial degrees of freedom. So, that is how, this reduces to this element. It is very easy to derive; very easy to remember. E A by L, axial stiffness 4 EI by L, 2 EI by L, 2 E I by L, 4 EI by L; is it clear? It is actually a combination of your axial element and your beam element right You can also derive from first principles You apply a unit displacement, one at a time, and you can generate these. This diagram will be very familiar to you now. It is not difficult; you can generate the element stiffness matrix Now, we also need that displacement, the T D matrix that is the displacement transformation matrix. Well, we are familiar with this slide because we used this when we dealt with plane trusses. The plane frame is advancement on the plane truss because in a plane truss, you have 4 degrees of freedom in the conventional system. So, we are familiar with this transformation minus cos theta, minus sin theta, cos theta, sin theta. And you will recall there are two ways of deriving this. This is the kinematic way, but there is also an easy static way, where you get the T D transpose matrix. So, you are familiar with this You are also familiar with what you do for a continuous beam. And if you have chord rotations, relative supports settlements, then you have to use this chord rotation which is given by 1 by L, and clockwise chord rotations are treated as negative, but the equivalent beam and flexural rotations which is 1 by L, will turn out to be positive. So, you are familiar with these two; if you put them together, then you get what you need to do for a plane frame element with 3 degrees of freedom You can easily work this out from first principles What you need to recognize is – if you have any of those translations, you have to find out what is the elongation you get in the element; right so, it is either cos theta or sin theta. If you get an extension, give a positive sign; if you get a contraction, give a negative sign; it is very easy. So, you can generate this by pushing one at a time. I have shown here, the translation effects So, rotational effect is very straight forward because there is no transformation required when you have a rotation because it is same; you get 1 So, let us just look at this. Let us look at the first one; this first column in your T D matrix corresponds to D 1 equal to 1 in your structure. So, if you have D 1 equal to 1, you need to look at this picture. If you apply D 1 equal to 1, that element undergoes a contraction of cos theta; so, it has got a minus sign; that is why, we wrote minus c I; c stands for cos theta At the same time, you get a chord rotation Have you noticed? You get a chord rotation The chord rotation is anticlockwise; the value of the chord rotation is 1 by… not 1; it is sin theta by l. right And so, you get equivalent flexural rotation which will be clockwise, and that is why, you get minus s i by l; minus s i by l is it clear? Any doubts on this? From first principles, you can generate this; alternatively, you can use a force approach and generate this, and find the T D transpose Did you understand what we did in the first column? Yes. This corresponds to the first degree, the axial degree. These two correspond to the rotational degree; so, if you have a chord rotation 1 sin theta by l clockwise, you will get equivalent flexural rotations minus s by l; is it clear? Like this, you can work out for the second degree of freedom If you take the third degree of freedom, it is a rotation D 3 equal to 1; that does not need any transformation here. So, corresponding to D 2 star, you get 1. So, 0 1 0; does it make sense? That is all. Once you have got the physics in this, you got it. This is the actually not difficult to do, once you realize that it is just a superposition of the plane truss element and the beam element. We did the same in the beam element. We had 1 by L because you did not have sin theta cos theta, but now, your plane frame element can be oriented in any direction, and not necessarily align with the global x axes; is it clear? ok So, if you have got this, then we can go ahead This slide should also be familiar to you These are the shortcuts you will take when you want to avoid considering axial deformations Remember, you can convert; you have to find out the sway degrees of freedom, the minimum sway degrees of freedom, and convert them to chord rotations. right We had done this when we did the slope deflection method, remember So, we will invoke this concept when we want   matches with 6. So, 3 and 6 get those numbers, but then, you have a degree of freedom here, 2 going up, and 5 going up. That is why you put 74.04; is it clear? And it is positive; both are positive; is it clear? That is why you need to calculate those fixed end forces. Because once you look at the global coordinates, you are just looking Do I get any contribution from the elements? If I do, put them all together; does it make sense? It makes absolute sense totally, rational logical method Only thing, this needs a little input from your side. It is not mechanical, the way the conventional stiffness method is. Conventional stiffness method, you can program it and just forget about it; do not even look at the physics of the problem, where you have to, but that is what makes it interesting. This is good, for human beings should do it; that is good for machines. ok So, as far as the support reactions are concerned, there is nothing because elements 1 and 3 are where you have fixity. You have null vector that this is done by inspection. So, you got F F A, you got F F R, and you got it by looking at each element separately. Clear? Can we proceed? Is this clear to you? ok What is the next step? No. We still need resultant loads which is actually the same as this because you do not you have a 50 kilo newton acting at the third coordinate. So, you you do this and you draw, sketch, and the same. Now, this is a structure that I am going to analyze and superpose this result with the fixed end force values that I get. This structure is loaded with equivalent joint loads and this is identical to my original structure. I have got rid of the concentrated load acting in the middle of the element 2. I have replaced it with the equivalent end forces; both vertical forces end moment; got it? And what is guaranteed? What is guaranteed is the D A vector will be the same, and that is the beauty of the equivalents. right ok And also, note in this problem, you have a 10 mm settlement. Now, I have given you an assignment. The last next assignment where you get two problems only to do; one is a simple problem for conventional stiffness method, but I have thrown in a bit of temperature because we discussed that in the last class Second is of a funny shape frame; a shape frame like that fixed here, hinged here. So, you should take advantage of that hinge and with the support settlement and U D l on that beam. So, this is kind of suspended from above, and this is resting on the ground below; it is an interesting problem; try it ok Element stiffness matrices, you know, the formula you know the EI by L. So, this is child’s play. You can do this EA by L is also known. So, for the first and third elements, it is going to be identical because they are identical columns. Only for the second element, it is easy to write always EA by L 0 0 and the rest is 4 EI by L 2, EI by L; very easy to write down; mechanically you can do this What do you do next? You generate the structure stiffness matrix. You do it in two phases; first you do this k i T D i for the three elements; all these can be done by matrix multiplication. If you are doing manually also, it is not difficult. It is only a 3 by 3, 3 by 6; then, you add up all the contributions You do not go have worry about slotting here because you got the T D, and you can generate it It is a big matrix you get; 12 by 2, and low end behold; this is identical to what we got by the conventional stiffness method, but the operations involved much less effort because you dealt with much smaller matrices. You did not have to worry about the slotting also So, it is a much more powerful method, and it is including the effect of axial deformations, and you are getting exactly the same results So, the next step should give you also the same result. You will get the same deflections and you will get the same rotations. So, you can do that, and we got the same solution as we got earlier A support reaction also is identical. So, these two steps are common to both reduced element stiffness method and the conventional stiffness method. And so, you can check equilibrium, find out your reactions, make sure everything is okay; find out your member forces Now, this is where there is a slight difference from the previous method. What is the difference? You have only three degrees of freedom. In the conventional stiffness method, you got everything; you got the member end; moments member end; axial forces member end shear forces. Now, you got only three; the rest you got to figure out yourself, which is which is ok So the member end shear forces are not directly obtainable as in conventional stiffness method, but can be easily computed from the free bodies, applying equilibrium equations. So, let see how to do that. So, this is what you get from those vectors. You got the two end moments for each of the elements and you got the axial forces. If it is plus, it means extension; if is minus, it means it is compression Now, the rest of it, you can get from equilibrium; isn’t it? You take the second element; you can get the vertical shear force; so, you do that. That is is one step away; that is it; you are ready; bending moment diagram is the same; exactly, what we got earlier Now, is a interesting step; why should we work so hard with even that matrix? What is the advantage of ignoring axial deformations? The plane element reduces to a beam element So, you can work with beam elements, throw away the axial deformation, and one way to check the answer is put EA tending to infinity in the problem because then it becomes actually rigid and you should get the same result, but your effort required is much less in this method So, if axial deformations are ignored, the kinematic indeterminacy itself reduces from 6 to 3; that is a big reduction. Why does it reduce from 6 to 3? because the columns A B and C D will not change in length. So, you got rid of two vertical degrees of freedom at B and C, and BC also will not change in length. So, you have only one sway degree of freedom. You can choose either the left end or the right end; the choice is yours The deflection will be the same. This is what we did in slope deflection method and moment distribution method right So, it is a massive reduction of effort, and let us not waste time in calculating reactions through these techniques. So, we do not even put global coordinates for reactions. So, from a 12 degree of freedom model and now three degree of freedom model of the structure level, it is a tremendous saving in effort So, the two degree of freedom beam elements can be used in place of three degree of freedom frame elements. So, we throw away the axial degree of freedom; that is it So, you have 1 star, 2 star, for each of the three elements, and the restrain coordinates are not shown as they cannot be included in the simplified analysis. The reason is – once you are going for chord rotation way of dealing with sway, then do not bring in do not bring in those restrain degrees of freedom; which means, now, how do you deal with the 10 mm support settlement at D? Do you understand? Now, how did you do it in slope deflection method? Let us say, the portal frame D goes down by 10 mm; earlier you could handle it because in the D R vector, you could fit it in. Now, what do you do in the fixed end force? You have to handle it because if D goes down by 10 mm, C also will also go down exactly by 10 mm. If CD is not going to change in length, BC will undergo chord rotation, and that can you can get the fixed end moments So, that is the clever way of doing it, and that is what we will do But first, can you write down the T D A matrix? Do not worry about T D R. There is no T D R here. Can you write down the T D A matrix for the three elements? What is the size of each of them? What is the size of each of them? 2 by 3 2 by 3 right 2 by 3. So, write them down So, this you should do; no excuse. This is simple. Write down the 2 by 3 T d A matrices for the three elements. Well, they are identical for elements 1 and 3, they have the same Yes sir They are going to behave identically; is it not? Yes sir The start node is the same at the bottom So, reduces your effort considerably. Write    