# 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