Diagram, Analysis More details. Words: ,; Pages: ISBN: S Srinath, L. S Varghese, P. C Varghese, P. C Devdas Menon At the bottom you will have the free moment diagram M0 x by EI.. Distinctive coverage of Earthquake resistant design, Structural Systems, Green Analysis for Design Moments in Continuous Systems. Advanced Structural Analysis Prof. Advanced Structural Analysis. Department of Civil Engineering. Indian Institu This file you can free download Narendra Pyla. 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.. Pdf is not opening. So, another way to look at it is, in this manner, here it is very clear which is a child and which is a parent.
A space frame is also very difficult to compute, to analyze. Why is it difficult to analyze? Six degrees there. Because it has six degrees? Because it has? Six degrees of freedom? No, that is not the right answer. It is highly indeterminate. You need to solve many simultaneous equations to crack the problem. So, traditionally you try to simplify. So, what should you do is, you break it up into plane frames and you make some assumptions.
The assumption you make is that the frames in the transverse direction and the frames in the longitudinal direction really do not interact. They will interact, for example, if the building twists, but if it does not, if it is a kind of regular symmetric structure, we can make this idealization and if it is a long building, almost all the plane frames are identical, so you need to analyze only one of them or may be two of them, an intermediate frame and end frame.
Refer Slide Time: So, it become much simpler. This is how it is handled and that is a plane frame element. So, under gravity loads for example, a plane frame will look like this and even this is a little difficult to do. And traditionally, the concept of substitute frames have been used to simplify the analysis. So you could take out one floor separately and assume the columns to be fixed at top and bottom. So, you are actually separating out that frame. You are substituting the small frame for the big frame, there will be errors, but the order of those errors will be not significant.
You have to be careful when you do this idealization. You cannot do it when the frame is un-symmetric or when the frame is subjected to sway.
Otherwise, you could do this and if you want to take some more shortcuts you could actually take out one beam. You could reduce this to a continuous beam and say that not much moments get transmitted to the columns, which is true for interior beams because the moment on the left side is more or less equal to moment on the right side. You could reduce this to a continuous beam and even further, you could take out just one beam.
When you take out one beam from the whole big structure, you must recognize that it is not really a simply supported beam because you get some partial fixity at the two ends and so, it is appropriate to put a rotational spring at the two ends and design for some moments. So, this, we know that the bending moment diagram will look like that, the shear force diagram will look like this.
The exact numbers are something that you need to find out. But a good guess is good to start off when you do modeling of the beam. For example, if the total load is W and the span is L, it is a general rule of thumb that the bending moments due to gravity load will be in the order of plus or minus WL by If you design for it, it is good enough, but of course, it is not exactly, so and you can do it more accurately. Refer Slide Time: The grillage looks like this. It is all in one floor and as I mentioned earlier, the additional internal force you get is a twisting moment, which is what makes it different from a normal beam.
So, grid element is a beam which also has torsion and if the torsional stiffness is negligible, for example, if you are dealing with reinforced concrete and this kind of torsion is called secondary torsion or compatibility torsion. A concrete to concrete connection for torsion is not very strong. When this beam bends, it is going to have a flexural rotation at this end because it is integrally connected to this beam, that beam is also going to rotate but, what is a flexural rotation for one beam is a angle of twist for the other beam and how much moment is transmitted at that junction depends on the torsional stiffness of this connecting beam in addition to the flexural stiffness of this beam.
Concrete cracks rather easily under torsion and the twisting moment is nothing but the torsional stiffness multiplied by the angle of twist But if we have low torsional stiffness and the torsional stiffness can degrade up to 15 percent in normal buildings especially under high loads, then you practically get no twisting moment.
Often, you can assume that there is no moment transfer and this is allowed by many codes. You can treat the beam as being simply supported. So, if you make that assumption, the grid element reduces to a beam element. Refer Slide Time: Now, we come to the topic of Static Indeterminacy and it is important to realize that although we use words like statically indeterminate structures, as far the structure is concerned, it is more appropriate to talk in terms of kinematics rather than statics.
The indeterminacy is the problem that the analyst faces, it is no problem for the structure. The analyst find it difficult to analyze a structure using simple equations of static equilibrium. So the statics is indeterminate. Statics refers to the force field. Why does it happen?
It happened so because you provided more constraints and the absolute minimum required to keep the structure stable. So, we use a word called over rigid or over constrained to describe a structure which is excessively constrained and to the extent to which it is constrained excessively, you have an order of a degree of static indeterminacy.
For example, you are familiar with this. This is a simple truss. And at every joint in a plane truss you can have two forces that can act.
You have two equations of equilibrium, if we use the method of joints. That is how this equation holds good.
Now, here you still satisfy m plus r equal to 2 j, but the structure is unstable because you have not located the members properly. You do not have triangulation here and that segment on the right can deform and it is important to know how it will move. It is unstable. Those elements cannot taken any shear, but if you provide an extra diagonal element, then you have a degree of static indeterminacy equal to 2. We have studied all this, it is just a simple introduction.
So, here you have an example of an unstable structure in the middle, a just rigid structure whose degree of static indeterminacy is 0 and on the right you have an over rigid structure whose degree of static indeterminacy is 2. Refer Slide Time: Now, here is something that we will really need to understand it gives us a frame work to do matrix analysis of structures.
Take a look at this simple plane truss which satisfies m plus r equal to 2j. You have 11 members and you have 7 joints, each joint has 2 degrees of freedom so you have 14 degrees of freedom. Let us first see in what all ways we can apply forces. You can apply F1 and F2 orthogonally.
If you treat this as a simply supported system, at the extreme left support, you have a vertical reaction and horizontal reaction, those are labeled F12 and F13 and at the right side you have F14 acting upward or downward. Is it clear? So, these arrows show joint forces, some of those joint forces are support reactions and some of those joint forces are potential loads and with this kind of description you have a complete description of the external forces on a structure, some of which could be unknown reactions.
So, here I have shown a free body of that structure, but the force field in a structure also includes internal forces and there are 11 bars in that structure. Each one has an unknown axial force which could be tension or compression. So, if we take out for example, bar number 11, if you take a free body of that bar, I have shown an axial tension in that bar.
The internal forces marked as N11, N stands for normal force and here is another picture which shows the possible displacements in this structure and we will use the same numbering system. If you have F1 and F2 at that top corner joint toward the left and right, we have the same labeling system for displacements D1 and D2 So, 1 and 2 are sometimes referred to as coordinates.
They are kind of unit vectors, because they identify both the location and the direction of either a force or a displacement. But they are all external. The forces are external to the structure, so are the displacements. They act at the joints. So, you can have 14 values of displacements.
Now, let us look at that same bar and see it has displaced in a certain manner. You will find that the movements at the two extreme joints which we have labeled here, D9, D10, D5, and D6 will decide, will dictate how much that bar has deformed. So, this is simple geometry. This is called compatibility. So, the bar, let us say it has elongated and we label the elongation e11, e standing for So, if you look at the displacement field, the displacement field is made up of joint displacements which I have labeled Dj and member elongations which I have labelled ei, j can vary from 1 to 14 in this case and i can vary from 1 to 11 and conjugate with this definition you have Fj, the joint forces, j can vary from 1 to 14 and Ni, the internal axial forces, i can vary from 1 to So, this is a kind of simple frame work which is useful for us.
Important thing to note is all the forces must satisfy equilibrium and all the displacements must satisfy compatibility. Refer Slide Time: Now, you are familiar with static indeterminacy. It is important to know this new term called Kinematic Indeterminacy. It is very simple. Our real task in structure analysis to know everything about the force field and the displacement field right?
In the force field, you saw earlier there were some knowns and some unknowns. The knowns are the loads that are applied and the unknowns are the reactions and the internal forces. That is a force field. As far the displacement field is concerned, usually everything is unknown. If we wish to we should be able to calculate everything and that is a whole idea of doing structure analysis. Now, kinematics is concerned with movements and statics is concerned with forces.
So, in a truss, the movements are joints movements which are translations in the x and y directions in this example and the elongations in the different members. So, we also found that it is enough to talk about joint displacement because once you know the joint displacements, you also know the bar elongation because there is a compatibility which ties the bar elongations to the joint displacement.
So indeterminacy in a kinematic sense in a truss is related to only joint displacements. Now, in this problem you have 7 joints and in each joint you can have two independent degrees of freedom, that is in this case a horizontal and a vertical translation and so, the degree of Kinematic Indeterminacy you say here is 14 except at those locations, where you provided supports, because the support ensures that there is no translation.
So, in this example, if it is simply supported one would say that D12, D13 and D14 are known to be 0. So, your indeterminacy now, kinematic indeterminacy reduces from 14 to 11, 14 minus 3 because you know the displacements there are 0, is So, a truss is a nice example of a structure, this particular truss which is statically determinante ns is equal to 0. Degree of static indeterminacy, if you make it simple supported 0, you could also make it into a cantilever. Does not matter. But, the degree of kinematic indeterminacy is huge.
That is why if you were to solve this problem manually, you would not try the displacement method, you will try the force method of analysis. So, this is Kinematic Indeterminacy and you could define it as a total number of degrees of freedom, which are independent displacement coordinates at the various joints in a skeletal structure. Refer Slide Time: Let us shift from the truss to a beam element. So, I am isolating a beam element AB and you can see clearly that at the joints there are four movements possible and they are independent.
D1 refers to a vertical translation at A, D2 refers to a slope rotation at the same location A, D3 refers to a vertical translation at B and D4 refers to a rotation at B and I have chosen some directions.
You could change those directions. I have shown upward is positive and anticlockwise is positive. So, here you would say, this particular beam element has four degrees of freedom. You can erase all the four degrees of freedom as you would in a fixed beam, fixed-fixed beam and that is interesting. The fixed beam is kinematically determinate. It is strange.
That which is kinematically determinate is statically indeterminate. The cantilever has a degree of Kinematic Indeterminacy of 2 because on one end you do not allow movements, on the other end you can have a translation and a rotation. Does it make sense? If you have a continues beam, then you can see that there are four movements possible, three rotations and one translation.
If you have a plane frame element, you have two additional degrees of freedom compared to a beam element because the two ends can move, 3 and 4 can move and it is possible for that member to elongate. One assumption we often make is that axial deformations are negligible. If you make that assumption then 1 and 4 are not independent. D4 and D1 may be interrelated. But, otherwise you can have 6 degrees of freedom.
So, if you take a portal frame with the bottom of the columns as fully fixed, you have 6 degrees of freedom. The kinematic indeterminacy is 6. I have labeled them as 1, 2, 3, 4, 5 and 6. But, if you ignore the axial deformation, that 6 reduces to 3 because the two columns will not change their lengths, so there is no vertical movement possible and the sway is the same for the entire frame. That means, if that beam element will move horizontally, the left end and the right end will move identically.
Does it makes sense?
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