noc18-me58 Lec 46-Example Based On a Finite Rectangular Plate: Part-I.

[Music] [Music] hello welcome to advanced composites today is the fourth day of the ongoing week and today we will actually solve a particular problem related to finite sized plates the finite size composite plates so let's have the problem definition first so the problem is that we have a rectangular plate and we will choose our coordinate system slightly differently so that's the x axis z axis and y axis and the plate is located at the corner so the origin is not at the center of the plate but it is at the corner of the plate because it makes the mathematics slightly easy that's all so this dimension of the plate is a and in the Y direction it is B long okay and the boundary conditions are such that all the play edges are simply supported all the edges of the plate are simply supported and then finally we have another condition that the plate is seeing a normal distributed load so it seemed normal distributed load and so normal uniformly distributed distributed load and its value is q-not Newton's per meter square okay so the total load will be Q times a times B which is air a a and B is the area of the plate so this is there the other thing which we know about the plate is that lamination sequence is symmetric and it is also especially orthotropic so if it is specially orthotropic what does that mean essentially what that means is that the B matrix is zero because of symmetry and also a 1 6 a 2 6 D 1 6 d26 all of these guys are also zero so simply supported plate on all the forages it's simply supported acted upon by normal transversely distributed load the lamination sequence is symmetric and also the layers are the lamination sequences especially orthotropic so that is there so then the question is find u naught V naught W naught as functions of x and y we have to calculate that so this is the problem definition so we will look at the governing differential equations del NX over del X plus del n XY over del y equals 0 del n XY over del X plus del V over del y equals 0 and the last one 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 plus q naught equals 0 now when we look at these equations we make following observations 1 + x + y n XY only depend on they only depend on u naught and V naught because B matrix is 0 second MX my MXY only depend on w not because b matrix is zero so the solution of the third equation does not influence the solution of first two equations and the solution of the first two equations does not influence the solution of the third equations because b matrix is zero so in because of this simplification we can say that this is out of plane problem out of plane third equation is the out of plane problem and first two equations are in plane firmness out of plane and in plane problem decoupled so the solution for W naught is not related to the solution for u naught and V naught and vice-versa u naught and V naught are interlinked but the solution for W naught doesn't influence the solution for u naught and V naught and the solution for u naught and V naught does not influence the solution with W because the B matrix is zero so with this understanding so first thing what we will do is we will try to solve the out-of-plane problem so we will aim will solve the out of length problem okay so so once again this is Del 2 m X plus 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 minus Q naught and we know that M X what is MX MX is equal to d1 1 del 2 w naught over del X square minus D 1 2 so there is a minus also here del 2 W naught over del Y square and the third term is 0 because d16 is do you 0 because d 1 6 is 0 because the plate is especially your throat of orthotropic similarly my equals minus d 1 2 del 2 w naught over del X square minus D 2 2 L 2 W naught over del Y Square and the d26 term is again 0 and M X Y M X Y is minus d66 del 2 w naught over del X del Y and there is a 2 here because again d 1 6 and D 2 6 are 0 so these are the simplified terms so we put these into this equation and so in this way we will develop an expression for W so what does this become it becomes T 1 1 del 4 W naught over del X is Del X 4 plus 2 D 1 2 plus 2 D 6 6 del 4 W naught over del X square del Y square plus D 2 2 del 4 W naught over del Y 4 equals Q naught the negative sign from both sides it goes away so this is the out-of-plane equilibrium equation so this is what out of plane equilibrium equation for a symmetric laminate which is also especially orthotropic so if there is any plate whatever shape it is it will be governed by this particular equation if the plate is symmetric in lamination sequence and it's also especially orthotropic ok any plate it can be rectangle or a non rectangular and q could be a variable or constant in this case Q is Q naught which is constant uniformly distributed but does not have to be this is a very general equation okay but now we will try to solve it in context of the boundary conditions which we discussed which is that the plate is simply supported on all the four edges and also the other thing is that Q is not a variable bit Q is a constant so our next goal is to solve this equation so how do we solve this equation so this is a fourth order differential equation in a single variable W naught and the way we solve this equation is by making a guess so we guess an equation an expression for W naught and the guess should be such that it should satisfy all the boundary conditions and it should also satisfy the differential equation so if it satisfies all the boundary conditions and it also satisfies the differential equation then our guess is correct okay so that's how we solve it so this is the exercise we will undertake tomorrow and hopefully we will arrive at the solution which will meet all the boundary conditions and also the it will satisfy the differential differential equation exactly so that's all I wanted to discuss today and I look forward to seeing you tomorrow thank you [Music] [Music]