# Mod-06 Lec-36 Matrix Analysis of Plane and Space Frames

Good Morning This is lecture number 36, module 6, Matrix Analysis of Plane and Space Frames. If you recall, in the last session, we covered the reduced stiffness method; We will continue with that method, and also complete the flexibility method, as applied to plane frames This is covered in the chapter on plane and space frames, in the book on Advanced Structural Analysis So, this is the reduced element stiffness method You know that the size of the plane frame element stiffness matrix is 3 by 3. We have done a few problems This is how you do the transformation the T D matrix in the reduce element stiffness method Now, let us look at a problem with sloping legs, which is actually a complicated type of problem, when you apply the reduced stiffness method. So, you have to intelligently identify the sway degrees of freedom; we have discussed this earlier. And you need to express the chord rotations in terms of the identified sway degree of freedom So, let us look at this problem which we have solved earlier by the more regress method So, let us solve this by the reduced element stiffness method and let us a take advantage of the fact that we can ignore axial deformations We have solved this problem earlier by the slope deflection method. What is the degree of indeterminacy, kinematic indeterminacy? Shortest needed degree of kinematic indeterminacy refers to the active global coordinates. The absolute minimum when you have when you do not have axial deformation 8, 5, 8 Ok Five At the supports A and D, do you need to have anything? No, No No. We do not even bother about finding reactions We do the minimum work. with So, at B and C, how many do we have? At each of them You have a rotation at B; you have a rotation at C Vertical reaction And you have just one sway degree of freedom; that is what we did in slope deflection method; remember – theta B, theta C, and delta BC Correct, but we have done this problem by the conventional stiffness method. There the degree of kinematic indeterminacy was – at B and C, you had each 3 degrees of freedom; 3 plus 3; you also had a rotation at A and D; it was 8 So, we have, we are reducing 8 to 3, but we are ignoring axial deformation; that is a major reduction. The method of solving is, as we have discussed earlier, the steps are the same plus we take advantage of the fact that, we limit the considerations only to active degrees of freedom because we can always get the support reactions at the end from the free bodies So, let us do this problem. So, I have got 3 degrees of freedom; global coordinates 1, 2, 3; that is all I need. Local coordinates, again, I take advantage of the fact that the there is a hinge at A and D So, my 2 degrees of freedom B model will now become 1 degree of freedom model. What I have achieved is massive reduction in the quantum of work A, I am not going to use a plane frame element; I am using a beam element. A plane frame element has 3 degrees of freedom in the reduced formulation, but I am not worried about the axial degree of freedom. So, the 3 becomes 2 The plane frame element becomes a beam element and I am using that formulation for element number 2 because there is a fixed end force coming in there. There is an intermediate

load, and B and C are not hinged. But for elements 1 and 3, I take advantage of the fact that, I have a hinged support at the ends A and D where, the bending moments are 0. And so, you know, I can take advantage of a reduced stiffness. What is the element stiffness for that flexural stiffness? 3EI by L. So, I just need 1 degree of freedom which I marked as 1 star; is it clear? So, with this, I can proceed. I first need to write down my T D matrix T D A matrix Will you try that? To do that, you have to be very careful. When you are dealing with sloping legs, you have to correctly express the chord rotations in terms of the sway. So, the displacement D 1 corresponds to the delta in this figure right The figure at the bottom is a generic figure So, you have chord rotations of delta by the height for the elements 1 and 3. right And for the element 2, you have to work this out using trigonometry; it turns out to be as shown in that formula delta tan alpha plus tan beta by the span L. I am not going through this all over again because we did exactly this, when we did the slope deflection method So, if you plug in the values of tan alpha and tan beta because this the dimensions are shown there, you will get the chord rotation for the element 2 as plus 27 by 36 times the sway D 1 Just plug in the values of tan alpha tan beta; this, the span of that element BC is 3 meters So, you will get this. Shall we proceed? We have done this earlier. It is just a repetition We put a positive sign because you are getting an anticlockwise chord rotation for element 2, but please note – for elements 1 and 3, the chord rotation is clockwise; clockwise; chord rotation is clockwise. So, and it is given by delta by the respective heights Now, can we write down the T D matrix. We can So, the T D A matrix for the three elements are as shown there. If you apply D 1 equal to 1, then for the first element, the chord rotation is 1 by 4; it is a clockwise chord rotation. You get equivalent anticlockwise end rotations and it will be plus 1 by 4 because anticlockwise is assumed to be positive in the sign convention The same is true for the element 3. So, you get 1 by 3 because the height is 3 meters As far as the element 2 is concerned, it is 27 by 36. right It is We just worked it out know 5 2; it is 27 by 36, if you expand it, sorry 17 by 36, it turns out to be 0.4722 But it is a clockwise rotation. So, your equivalent end rotation will be, sorry, it is an anticlockwise chord rotation. You are you get clockwise equivalent end rotations. So, it turns out to be minus. This is the only difficult part; that first column in your T D matrix has to be very very carefully written because it is totally dependent on your correct assessment of the chord rotations; both magnitude and direction The others are simple. this If you put D 2 equal to 1, we are now talking about the second column in the T D matrix. Then, you will find that you get it will effect only elements 1 and 2. The right end of element 1 and the left end of element 2. So, you get… and element 3 is [un effective] So, you get 1 1 0 0, and similarly, for D 3; is it clear? This is easy to generate. Can I proceed? You got your T D matrix. Now, you have to carefully write down your fixed end forces We did this earlier, and remember, we just calculated the fixed end moments. But here, you have to be careful because you have only 3 degrees of freedom D 1, D 2, D 3, and you do not have those vertical forces which we put on the last occasion. right There are