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So, first is the type of indeterminacy, which you answered. As you know very well, in force methods, they are of the nature of compatibility equations; in displacement methods, they will be of the nature of equilibrium equations.

You need to express the displacements in terms of forces. You can do it in two ways: So, these are broadly the.

As I mentioned, in the first three lectures we will focus on force methods and in the next four lectures we will focus on displacement methods. You've reached the end of this preview. Share this link with a friend: Other Related Materials pages. You can identify your start nodes and end nodes for the three elements. You could draw the three elements separately or you could choose to draw a typical element and you can generate the matrix.

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The Ti matrix and the TD matrix actually are similar, except that the size changes. So, we are familiar with the Ti matrix.

You have to still rope in the cos theta, sin theta, and 1. So, the method is the same. Only thing that you have to do it with your eyes open and fill up these matrices.

Refer Slide Time: It is convenient to separate out the active degrees of freedom from the restrained.

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So actually the TD matrix for each element will have a size 6 into 12, which you can partition into 6 into 8 for TDA, active degrees of freedom, and 6 into 4 for restrained degrees of freedom. Refer Slide Time: Most of these elements in the matrix will be 0s, except where you have the degree of freedom, which is active.

Let us see, take the first column in these two matrices for TD1 and TD2. The first column actually corresponds to D1 equal to 1.

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Refer Slide Time: If you apply D1 equal to 1, it is going to affect elements 1 and 2. It is at the tail end of element 1 and at the beginning of element 2. So basically, you have to use this transformation matrices and fill up at the appropriate column. You see the tail end gets filled in element 1 and the start end gets filled in the element 2. It is only 1. Is it clear?

I hope you know how to fill up this. This is actually similar to what you would get, if you did the Ti formulation, except that you have to keep track.

Now on the top row, I have written 1 2 3 4, all the way to 12, but you need not do that in this case. It is obvious it has to be a sequential form from 1 to I hope you know how to do this.

We have done it for a truss, you have done it for a beam, it is not a big deal to extend that to a plane frame. Refer Slide Time: Similarly, for the third element, you can fill up.

Now the fixed end forces. It is very easy to calculate in this case. Of the three elements, only the second element is loaded.

It has a UDL. What do you do next? You convert it to? You want to get it for the whole structure.

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You want the FfA and FfR vectors. How do you do that? TD transpose. Refer Slide Time: TD transpose, so that contragradient principle works. And you can sum up, do for one element at a time and sum it up for the three elements. This is something you can work out for all the elements -- you get FfA and FfR.

You need not write down those numbers 1 2 3 4 all the way to 12, because it is obvious. Those numbers you need to do when you do the Ti matrix. When you do TD, it has to fall into this numbering scheme. So you get FfA and FfR.

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You have got the load vector. Refer Slide Time: You can draw a sketch. Look at this. This is your resultant load vector FA minus FfA. FA is a null vector because there are no nodal loads in this particular problem. Put a negative sign to it and interpret those results. You will find that the original problem had a UDL.

What you are having here will give you 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.

Refer Slide Time: 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. Internal Forces in Frames 9.

Deflection in Beams: Conventional Methods Principle of Virtual Forces Introduction to Force Methods Method of Consistent Deformations Theorem of Least Work Column Analogy Methods Introduction to Displacement Methods Slope-Deflection Methods TD is the displacement transformation matrix.

I have marked them there in green color, and you have 4 restrained degrees of freedom. L cancels out. About Narosa. Load 3. First, with regard to the type of indeterminacy - static indeterminacy, which means you are talking about unknown elements in the force field; that is part of the force methods.

The methods of consistent deformations and Theorem of least work, which you have been introduced to.

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