Hello friends, welcome to DCBA and this particular video lecture is about stiffness, what is stiffness, how to form stiffness matrices. It is basically about the basics of stiffness matrices and detailed video lectures will be posted further on. This video lecture also enables you to learn the forces and the interaction between the forces and the reaction when a body gets displaced. So let's get started with what is stiffness.
So stiffness is a force at a moment required to produce a unit deflection or rotation as simple as it sounds it is a force or moment that produces a unit deflection or a unit rotation let us see an example to it so i've considered this beam and actually loaded beam as you can see here and it is loaded by a force p kilo newton so as obvious when i load it with a force p kilo newton it is going to get deflected or it is going to get elongated here and this is it is elongated by an amount delta and as you might know that the equation for delta or the equation for deflection due to an axial load is given as delta is equal to pl upon a where p is the applied force l is the length of the member a is the cross-sectional area and e is the young's modulus what is stiffness is given by this basic definition that it is a force or moment required to produce unit deflection so if i substitute this delta equal to 1 what i get 1 equal to pl by a and stiffness is a force or moment so if i manipulate this equation just to get in terms of p so p will be equal to ae by l and this is the force that is required to produce a unit deflection or a rotation the rotation does not come into this picture this particular example but a unit deflection so this is basically my stiffness stiffness for this member you so i hope you got it and you got the fair idea about what is stiffness and suppose if i give a anti-clockwise rotation at point b then it gets deflected like this okay so if you can see that this point b has been rotated in anti-clockwise direction and my whole beam has been deflected this way and this is theta b okay so for as you know for stiffness to find stiffness we want our unit rotation So we want this theta be equal to 1. So first of all the moment that causes this rotation is given as 4EI theta B by L. Okay you have to remember these formulas. These formulas have been derived from extensive derivations that cannot be covered right now.
I'll be posting those derivations further on in some other video tutorials but you for now first difference matrices you need to remember these formulas. So 4EI theta B by L causes the moment that causes this theta b equal to 1 or theta b basically because i have not substituted here theta b as 1 and because of this deflection obviously there will be reactions formed at a and b so first of all a moment at a is generated and it is anticlockwise direction again i'll tell you how it is anticlockwise direction but the magnitude of this moment is given as 2 ei theta b by m it is basically half of this moment here so this is a force which is shown in red and reactions have been shown in blue so a force of anti-clockwise moment or a moment of anti-clockwise nature is given at point b and a reaction of anti-clockwise nature has been formed at a so why it is anti-clockwise basically so if you could see the deflection profile of this beam then this member here which is connected at a is trying to get deflected in clockwise manner if you see that this line shows the original shape of the beam which was horizontal and this uh member here has been deflected in clockwise direction to resist this reflection here that is resist the clockwise deflection we require an anti-clockwise reaction and the magnitude of the reaction is given as 2 e i theta b by n and along with these moments here there will be also vertical reactions formed at both the ends and the magnitude of this reaction is 68 theta p by l square and don't worry about the directions i'll explain you how the directions are formed the direction of reaction at point a depends on the direction of moment at point b and the direction of reaction at point b depends on the direction of moment at point a so let us talk about this reaction first why it is upward in nature so as i said that it depends on the direction of moment applied at point B and the direction of moment at point B is anticlockwise so this direction should be such that it causes a clockwise moment at B so if you see this is acting in upwards direction and if i take the center of rotation at point b then this is going to cause a clockwise moment at point b okay so it should basically be opposite to what is applied at point b if my applied moment was clockwise then this direction should have been downward so that it would have caused an anti-clockwise moment again here why is this downward as it depends on the direction of moment at point a and it is acting in the anticlockwise direction so this should be such that it causes a clockwise moment at point a so if this is downwards and if i fix my center of rotation at point a then this causes a clockwise movement here so i hope you got how the directions of these reactions are fixed and when we substitute theta b equal to 1 as we require for stiffness These are all the stiffnesses that we require when we substitute theta b equal to 1 in all these equations. These are the stiffnesses that we require. Now we will see a similar example to this and again a unit rotation at end. But this time we will give a clockwise moment.
In the previous example it was anti-clockwise moment. Let us see how the reactions and forces interact when we give a clockwise moment at point B. As you can see it has been deflected in the clockwise manner and causing a rotation of theta b and if you remember then the moment that causes this angle theta b is given as 4 e i theta b by l and the reaction is given as 2 e i theta b by l and these are in 6 e i theta b by l square so i hope now you must be understanding how these reactions direction of these reactions have been formed so why it is clockwise here because the member at a is trying to get rotated or trying to get deflected in anti-clockwise manner and to resist this anti-clockwise deflection we require a clockwise moment okay so if you if you are able to draw this deflected profile correctly then half of your problem is solved there itself and how to draw this deflected profile correctly so when i rotate this joint at point b then you should always notice that this angle is always a 90 degree angle okay you cannot just form a hinge here okay these are always uh fixed points so this will cause a 90 degree angle and when you rotate your beam at 90 degree angle in this manner then this is how your deflected shape forms okay and hence it is clockwise in nature to resist this anti-clockwise turning of the beam and why is this downward reaction as i told you previously that this depends on the direction of this moment and it should cause a anti-clockwise movement now because here the direction of the moment is clockwise hence 6 ci theta b by l square when rotated about point b it will cause an anti-clockwise why is this direction upwards because it should cause a anti-clockwise moment again because as you can see this is clockwise moment so when this reaction is rotated about point a it will cause an anti-clockwise and when I substitute theta be equal to 1 we get 4 e I by L so we have I'll 6 e I by L square and 6 e I by L square and these are the required stiffnesses when my beam gets rotated this way and these are the required directions of the reaction so i hope you got how the forces and reactions interact with each other everything is based on how your beam gets deflected so deflected profile is very important again let us check a similar example but this time giving rotation at point a and we are giving a clockwise rotation at point a and if you can see the deflected shape now becomes like this again causing a 90 degrees angle here maintaining a 90 degrees angle here and causing a deflected profile to be this way so let us find how the reactions are formed again the magnitude of this moment this clockwise moment is 4 e i theta e by l and the reaction formed is clockwise because the beam which is connected to b is trying to get rotated in anti-clockwise manner this is our original axis of the beam and it has been rotated in an anti-clockwise manner thus to resist this we have to give a clockwise moment at or a clockwise reaction at point b and the magnitude of this reaction is given as 2 e i theta a by l and let us check again the reactions upwards reactions vertical reactions this is downwards this is upwards and the magnitude of both the reaction is given as 6a theta a by l square why is this downwards because it should cause a anti-clockwise moment at point b and this should cause a anti-clockwise moment at point a that is a moment that is opposite to these first of all this is applied moment and this reaction so these reactions should cause opposite moments at both the joints and these are the required stiffnesses when we substitute theta a equal to 1 in all the equation now let us check what happens when we give a unit deflection at an end so i am giving a unit deflection at point a and this is going upwards okay and this will take my beam at a upwards causing a deflected profile like this and to cause a deflection in of delta in upwards or downward direction the magnitude of this force has to be 12 ei delta by l cube again you need to remember this formula and from this deflected shape you must now have already been able to know the direction of how the moments are going to form since this beam is rotated in a clockwise manner this was the original axis of the beam if you see just a parallel line to this parallel line to this at point a and this member has been rotated in clockwise direction hence a moment of anti-clockwise nature is formed and the magnitude of this moment is given as 6c and delta by l square again this moment depending on the delta value and the reaction the nature of the reaction formed here is again of anti-clockwise nature as even this beam has tried to rotate in clockwise moment clockwise nature and again this reaction is of 6 e i delta by l square and there is also a vertical reaction of downwards direction the directions of these both these are again similar to the previous example that they should cause a clockwise moment here and that this depends on the direction of the moment here and this depends on the direction of the moment here but this particular value is not a reaction but it is a force It is a force that caused this deflection.
So this is already fixed and it will also give you a check to see that whether your moment direction is correct or not. So this should cause a clockwise moment here. When I apply this force at point A, it should cause a clockwise moment here and if it is causing a clockwise moment here, then the reaction should be anticlockwise. Okay, and this should cause a clockwise moment at A.
and again the reaction should be anticlockwise and when we substitute delta equal to one we get the stiffness values at both the joints that is 6 e i by l square 12 e i by l cube 6 e i by l square 12 e i by l cube and these are the reaction directions that you need to know now let us check what happens when we give a unit deflection at again point a but in the downward direction and it gets deflected downwards and again the magnitude of this force is given as 12 e i delta by l cube and the direction of these moments as you might have guessed already that this should be of clockwise nature because the beam has been deflected in an anti-clockwise manner this being an original axis a straight line and my beam element has been deflected in the anti-clockwise manner thus the moment is of clockwise nature resistive moment of clockwise nature and the magnitude is 6EI delta by L square. Again this beam has been reflected in the anti-clockwise manner. The resistive moment here should be of clockwise nature and magnitude is 6EI delta by L square.
The reactions upwards and downwards, they should be upwards as you might already know that it should cause an anti-clockwise moment at point A and this should cause an anti-clockwise moment at point B. and this is how these reaction directions are determined this is a force which is already predetermined or it was basically it was applied and because of this force our beam got deflected this way so i hope you got this and now finally we will check what happens when we apply an actual deflection so let us apply an actual force at point a the magnitude of this force is given as a e delta by l the magnitude of this force is given as a e delta by l and the reaction is as you can see it has to be opposite in nature because if this is acting in the right direction this has to be in the left direction and it has to be equal to the force applied uh and when we cause an actual deflection there is no moments formed here so this itself is the stiffness value and basically if you remember then this was our first example that we saw uh when we caused an actual deflection and we proved that the stiffness value is ae by l when we substitute delta equal to one so this is basically the same kind of problem in which we consider an axial deflection of force acting actually and a reaction generating action now let us check the sign conventions that we have to follow while applying forces as you can see here we have got our three axes in cartesian coordinates x y and z and the force is acting along these three axes are considered to be positive okay x-axis going in the right direction y-axis going upwards and z-axis coming towards you okay so these are considered as positive forces and the moments are shown basically by double arrows this is a force and if i show double arrows then it is a moment and how to find the direction of this moment then the direction of this moment is given by the right hand thumb rule where my thumb is in the direction of these double arrows and the curl of the fingers will tell you the direction of the moment so in this particular case it is anti-clockwise along the y-axis this is positive in the x direction again this moment is positive and again if we apply the right hand thumb rule then a moment which is anti-clockwise along the x-axis is positive and in the z direction again applying the right hand thumb rule this moment is positive so if you see the all the anti-clockwise moments along the particular axis are positive and all the forces acting in the particular axis are positive so these are the sign conventions that we need to follow by solving the stiffness matrix or while generating a stiffness matrix so i hope you understood the basics of stiffness and stiffness matrix is basically a very important application not only for solving problems but it is it can also be applied in computers as as computer is more easily able to solve matrices rather than solving equations so whenever matrices come into picture you can easily code them and you can form a program a small program using stiffness matrix which will make your work very very less the stiffness matrix is very important for you to learn so i hope this video was fruitful to you and you have learned the basics of stiffness matrix you will see video tutorials about how to generate stiffness matrix for a beam element or for a plane frame element in the future videos Please like, share and subscribe these videos as it gives me a lot of motivation to make more such videos for you. Thank you so much.