noc18-me58 Lec 59-Free Vibration in Composite Plate: Part-I

[Music] [Music] hello welcome to advanced composites today is the fifth day of the tenth week of this course over the last four days we have been discussing about the role played by d16 and d26 as to how they influence the behaviors of plate and specifically we saw that they influenced the distribution of M X within the plate and also they lead to as effects which is the difference between the computed value and the actual value of M X as per the code so today we are going to move into a slightly different area and that relates to the dynamics of these plates and specifically we will talk about free vibrations in composite plates now when we are talking about the dynamic equilibrium plates till so far what we had discussed was or we had developed the governing equilibrium equations that was about their static equilibrium so the first equation was then an x over del X plus del NX Y over del y equals zero if the plate has to be in so this was the equilibrium of forces so we got this equation from equilibrium forces in the X direction when the plate was in static equilibrium but if the plate is in dynamic equilibrium then this right side of the equation this should be equal to mass times the acceleration of the plate and if the we are talking about acceleration and this is equilibrium in the X direction then it will be the acceleration in X direction so and so that was second thing and the second thing is that this is an X is force per unit length so essentially all what all that means is that this equation gets modified to del NX over del X plus del NX Y over del Y and that equals the area density of the plate okay so Rho a so Rho a is no excuse yeah and times the acceleration so it is del 2 u over del T square so this is not area density it is density times the thickness of the plate Rho so this is equal to density times thickness of plate so this is the equilibrium equation in the X direction similarly the equilibrium equation so this is for X direction for y direction we have a similar equation okay and for the third direction which is the combination of force and moment equilibrium we have the following equation so I am sorry this should be Delta V over del T Square so all these are mid plane displacements so these are the modified equations which have to be considered when we are talking about dynamic equilibrium of the plate next what we will do is we will develop relations for resonance frequencies of these plates which are rectangular in shape and they are simply supported so so free vibrations in simply supported plates and here we assume couple of things so we assume so if it is free vibration then there is no external force acting on the plate okay so we assume that the laminate is symmetric which means the B matrix is zero second we assume that D 1 6 is equal to D 2 6 is equal to zero and the third thing is oh so here we missed one term which is minus Q naught or Q X Y so this is this minus Q which is a function of X Y okay so going back we don't have any external forces on the plate so what does that mean that's mean that the plate has well if it vibrates it will vibrate naturally it is not vibrating because of some external agent but it has vibrating in a natural state these equations also do not include the effect of damping because if damping was there then we will also have to account for damping forces in the system so that is there and because of condition one because of one the out-of-plane solution and in plain solution these are decoupled okay out of plane and airplane solutions are decoupled so all we have to worry about is the third equation so the third equation is Del 2 MX over del X square plus 2 del 2 m XY over del X del y plus del 2 my over del Y square is equal to Rho del 2 W naught over W T square so the first thing we do is we want so our aim is to find W so the first thing is we convert this entire equation in terms of W we know already the expressions for MX MX y and my ands because D 1 6 and D 2 6 are zero so what we get is d 1 1 so this is the updated equation now to solve this equation we have what we will do is we will guess a function and we will guess it in such a way that it satisfies all the boundary conditions and also all the I am this differential equation it satisfies so we guess that W naught XY equals W naught MN sine M PI x over a sine n PI Y over B exponent J Omega T where J is equal to square root of -1 okay so this this assumed solution has a function of X and a function of time so this is a function of X by capital X I means space and this is also a function of time so excuse me this is this this is a function of space and this is a function of time so this is a process of variable separable and now we plug this equation this relation back into the governing differential equation so what we get is so this sign terms gets differentiated twice so I have four times so I end up with the same sign so I get minus D 1 1 M PI over a to the power of 4 so I will take D 1 1 a negative sign out plus 2 D 1 2 plus 2 D 6 6 M n PI square over a B the entire thing squared plus D 2 2 n PI over B whole thing to the power of 4 times w MN sign and M PI X over a sine n PI Y over B e to the power of J Omega T so remember this is Omega this is Omega this is not W and this equals on the right side row and then when I differentiate this term in with respect to time twice I get J square and I also get omega square WM n sine M PI X over a sine n PI Y over B e J Omega T okay so if you compare the left side and the right side all these terms are common so they get cancelled out so if this equation has to be true and then J square becomes negative one so if this equation has to be true then what I get is Omega square is equal to PI over four so I'll take pi/4 out and then I get D 1 1 M over a to the power of 4 plus 2 D 1 2 plus 2 D 6 6 MN over a B whole thing square plus D 2 2 + n over B whole thing to the power of 4 and on the other side I get density in the denominator so this is my expression for natural frequency so Omega is equal to 2 pi F n where and this Omega is again associated with M and n okay so this value of Omega it varies with MN n so this is what I get ok now what I say is that let a over B be the aspect ratio of the plate okay ever be so I can say that a is equal to B times R so if I substitute back in this into this equation I get Omega and what I will do is I will apply a subscript M n because each natural frequency is associated with this M and n okay so I just called it the MN 'the natural angular frequency so that equals pi over four times Rho D 1 1 M 4 over B are ok to the power of 4 plus 2 D 1 2 plus 2 to D 6 6 this is MN over be our square b square r square and this is MN whole square plus D 2 to N by B to the power of 4 so so this is B 4 R square and just to make this clear this is M the whole thing ok so I get omega MN is equal to if I take the square root of the Center entire thing and bring out be outside so I get PI square over Rho R square B Square I take everything be square as well as R square out then this is what I end up with D 1 1 M to the power of 4 plus 2 D 1 2 plus 2 D 6 6 M square n square R square plus D 2 2 n to the power of 4 R to the power of 4 the entire thing so this is my natural frequency of the system and so this is the M nth frequency and this is the angular frequency I have to calculate the cyclic frequency I just you know divide this Omega MN by 2 pi and I will get the frequency and associated with each of this natural frequency so associated with each Omega and M n so this is the frequency is a mode shape and what is that mode shape it is sine M PI x over a sine n PI Y over B okay so what it means is that if the thing is going to vibrate at angular frequency of Omega 1 1 then the shape of the plate will be such that it has one curve in the X direction and another curve in the Y direction and so on and so forth so this is the natural frequency of the system and this we can also call it as 2 pi F M n so this is another one so this is what I wanted to cover today tomorrow we will extend this discussion a little further we will also see how this behavior of the plates gets applied to isotropic plates and how do vibrations in isotropic plates as well as in orthotropic plates how do they differ based on some of the results which we have discussed till so far so that concludes our discussion and let's come let's see what gets extended in context of this discussion in the tomorrow's lecture thank you [Music] [Music]