Uncategorized Good morning. This is session number 34, module 6 – matrix analysis of plane and space frames If you recall, we had just started the application of the conventional stiffness method to plane frames. This is covered in the chapter on matrix analysis of plane and space frames in the book on Advanced Structural Analysis As you know, there are three broad methods we studied. We are looking at the first of them – Conventional Stiffness Method. We will now look at problems where you have internal hinges in frames We have already learnt how to deal with these in beams, so it is basically the same. I am just showing you some of the old slides on how to deal with an internal hinge. There is no transfer of bending moment, across the hinge, and no compatibility requirement regarding rotation. So we introduce a moment release and we introduce a clamp If you recall, we need to modify the element stiffness matrices. Essentially, one of those rows and columns will become full of 0s, corresponding to the release degree of freedom If you have a hinge at the start node, it is a second row and second column that gets released. Not only that, from 4 EI by L, it goes to 3 EI by L, the flexural stiffness, and 6 EI by L square becomes 3 EI by L square, etc,. We have done this earlier. If you have hinges at both the ends, the second and fourth rows and columns will become 0s. We are just refreshing something that we did earlier We also know that we need to modify the fixed-end forces, where you have a moment release. Essentially, the fixed beam will become like a propped cantilever and you need to modify the fixed-end moments Finally, we said that we need to deal with the fact that we will end up with zero stiffness in corresponding to the degree of freedom, where we have the moment release. Because how we deal with the moment release is – we are going to shifts that global coordinates from active category to restrained category You do not need to inverse the kAA matrix with the 0. Diagonal element in it moment you have a release. It will have a 0 diagonal element. So we conveniently pass it on to the restrained coordinate and it does not really matter. We do this by means of an imaginary clamp. We have done this for the beam. We are just repeating this idea for a frame If you remember, we did this problem in the last class – portal frame with a concentrated load. The only change we have now introduced is we bring an internal hinge at a beam column joint. In this case, joint C How do we deal with the same problem? Can we make use of some of the work we have done earlier. Can we use the same stiffness matrices? Partly yes. For A B Let us go through this problem. The procedure is exactly the same as we did earlier, except that we need to make modifications for the internal hinge. If you recall in this slide, we have looked at the six global coordinates and the six restrained coordinates. Just reproducing a slide which we covered in the last lecture But there are going to be changes now. Firstly, you have the internal hinge, at that hinge at C. We are going to convert one of those degrees of freedom. Which degree of freedom? What is the number? Six. We convert it from active to restrained, and we will introduce a clamp. We do all that. Is it clear? We do it so that and that degree of freedom goes to the restrained category. Nothing else changes So, how do we deal with this problem, with the internal hinge? Do we need to make any    exactly the same active displacements The same DA vector, the same deflections at B and C vertical and horizontal and the same rotations at B and C, and at A and D. You have to superimpose these results with the results that you get in the primary structure, when you have that UDL So, next you get the element and structure stiffness matrices, standard formulation for a plane frame element, you just have to plug in the values of EA by L EI by L, and it all falls into place. Generate this for all the three elements What do you do next? you have to generate the structure stiffness matrix. how do you do that? TD transpose k* TD Which you can do it in two stages? You can first do k* TD and then do the TD transpose or vice versa. It is best to program all these and let MATLAB do it. Just check, after you get the results, that you have a symmetric matrix and that your diagonal elements are all positive. You will get some null some 0s somewhere. If you number it more intelligently, it will fall into a nice banded matrix with minimum half band width for our convenience in storage and solution I hope, up to this stage, you know how to do it. If you write the full k matrix, you cannot fit it into that slide because the size is 12 by 12. It makes sense to do it – kAA kAR kRA kRR. Even here, it helps to draw partitions as we have done it here, because 1 2 3 refers to the joint B. So, I have drawn a partition for that – 4 5 6 refers to the joint C, and the last two refer to the rotational coordinates at A and D. Is it clear? You must have your own system of identifying for convenience because you need to interpret and you need to know why those zeroes are where there are. They should make some kind of sense Next, you generate all your matrices – kRR kRR is 4 by 4, and kAA is 8 by 8. Then, you solve this. Here there are no support settlements Solve these two equations. First, you find the deflections. When you look at that kind of deflection vector really, you can make sense mostly only after translations The order of magnitude of the translations should make some sense in a real structure So, I mark them in yellow color, the once that are significant; the once that are almost 0, I have not marked them. And you have to draw a sketch The sketch would look like this. Imagine, you have these two hinged frame, bent frame with a UDL on B C. This is how it is going to deflect. And the deflections are very small – 0.093 mm to the right. That is the sway at B, and vertical deflection of 0.278 mm, which is extremely small. So you got this? Then, you can find your support reactions solving the second equation, interpret those results, check equilibrium. The least you can do is check force equilibrium, so sigma Fy should add up to 0. Those two vertical reactions should add up to total load of 120 kilo Newton and the horizontal reactions must cancel out. It makes sense Last step is to find the member forces by using the TD matrix and draw the free body diagrams, and draw the bending moment diagram, draw the axial force diagram, shear force diagram All this can be done. And compare these results with the results we got earlier by the slope deflection method, not moment distribution Moment distribution is not good when you have sway. You will find that they are almost identical This is more accurate because you are accounting for axial deformation. We will do the same problem by the reduced element stiffness method, where we ignore axial deformations and compare the results Here is one last problem. A really interesting problem. Remember at the end of the last class, we also said that let us look at temperature loading. This is more tricky because the legs are inclined. The same frame. Let us do this slowly so that you get it. Same frame with sloping legs. We are now talking of a fall in temperature by 40 degree Celsius. Coefficient of thermal expansion is given. How do you solve this problem? There are no loads – no direct loads, indirect loading Yes, tell me step by step, how to do this problem? Find the elongations first What are yes first Find the elongations and then Elongations you have to speak very clear change in length Find the free change in length Yes sir For each element, which is L, alpha delta T, then what? And then fixed end forces for this fixed end forces It is more right to say that this structure is kinematically indeterminate. So, the primary structure is one and which all the degrees of freedom are restrained. In the restrained structure, if you allow this temperature change to take place, you will end up with what kind of forces? compression Obviously tension, because if you take any element it wants to shrink. It is prevented from shrinking you will get tension Will you get bending moments and shear forces? Not in the primary structure, but in the final structure, you should get. You will be in for lot of interesting results. So let us look at it slowly First, as you rightly said, we will find the fixed end forces in the primary structure and that is simply the axial stiffness ki Nif is ki into E naught i. We put a minus sign in general because normally you have a rise in temperature and E naught is an extension But in this case, E naught will turn negative, so when you substitute the results you should get a tension Now, it is interesting that ki e is EA by L and E naught i is L alpha delta T. L cancels out. It is interesting. Am I right? So for all the three elements, you will get the same axial force. Yes or no? That is what logic says. You get the same axial force for all the three elements. Can you work out how much that force is it? Small force or a big force? Plug in those values Let us go back. All the members have cross section 300 by 300, so the area is 0.09 meter squared, material has elastic modulus of 2.5 into 10 raise to 7 kilo Newton by meter square With EA and L, can you work out what that force is? That is it. EA alpha delta T, where delta T is minus 40 degree Celsius, you get a huge value 990 990 is not a small value. It is a huge axial force. Do you think the final force will be so large? How much you think? See this is what you would get. If you were to arrest those degrees of freedom, prevent that cooling from manifestly in the primary structure, you get a huge force of 990 kilo Newton But had this structure, for example, being simply support, what will be the force in all those members? 0. So it is actually not   