noc18-me58 Lec 49-Anticlastic curvature

Welcome to advanced composites, today is the start of the 9th week of this course and what we have done till so far is that we have finished treatment of finite plates. And we will now proceed into solving some more examples of finite plates, where the methods which we have discussed earlier may not be necessarily applicable. Earlier we had said that when we want to solve the problem of finite plates, we have to select functions for u, v and w such that they meet all the boundary conditions associated with the plate. And they also should satisfy the governing equations, but usually it is not possible to quickly arrive at such functions. So, when it is not possible to arrive at such functions, what do we do? In that context, we will be introducing the notion of virtual work and we will see that how the principle of virtual work can be used to solve the problem of finite plates, when the displacement functions for u, v and w are not easily available in the sense that these functions exactly satisfy all the boundary conditions as well as the governing differential equation. Before we embark on this particular journey, we will have a small diversion and we will discuss what is quite often known as anticlastic curvature. And we will discuss this effect or the phenomena of anticlastic curvature in context of the semi-infinite beam case, specifically case a which we had studied. And we will see what this anticlastic curvature is. The same concept though we are going to explain in context of semi-infinite beam, the same concept is also applicable to other plates also which may not be necessarily semi-infinite in geometry, but they may be finite plate. So, we will first revisit case a for the semi-infinite beam in context of anticlastic curvature. So, let us first very quickly recap the case of the semi-infinite beam. So, this is case a and here the beam lamination sequence was 0, 90 symmetric and the boundary conditions at the pin end were W0 is equal to 0, U0 is equal to 0, V0 is equal to 0 and Mx0 equals 0. And at the roller end, the boundary conditions were W0 equals 0, Mx equals 0 and then because it is free to move because of roller support Nx plus is equal to 0 and Nxy plus equals 0. And also I will just write this lamination sequence below. So, the lamination sequence is 0, 90 symmetric. The plate was also uniformly loaded with a load intensity of Q or actually Q0. It is a uniformly distributed load. For such a problem, we had generated a solution for this problem and the solution was that U0 equals 0, V0 equals 0 and W0 was found to be Q a to the power of 4 over 384 d 11 into 16 x over a 4. So, this entire length is a and our coordinate system is in the middle. So, 16 x over a whole 4 minus 24 x over a square plus 5. So, this was the expression for W and now that we have U, V and W, we also calculated Nx was found to be 0, Ny was found to be 0, Nxy was found to be 0, Mxy was found to be 0 and Mx was determined to be Q a square over 2 1 minus x over a whole square and then My equals d 12 over d 11 Mx. And if we have to calculate curvatures, Kx is equal to del 2 W0 over del x square. So, this is not equal to 0, Ky because W is not a function of y and Ky is defined at del 2 over del y square. So, this is equal to 0 and Kxy is also equal to 0. So, this is the overall solution. Now and finally, we will also write an expression for Qx effective. So, we had said that Qx effective is defined as dMx over dx or actually partial derivative plus 2 del Mxy over del y and because Mxy is 0, this term is 0 and Mx is Qa square times 1 minus x by a whole square. So, if I differentiate this in x, I get minus Qa square over 2 into minus twice of x over a twice of x over into 1 over a. So, what I get is minus Qx. So, that is why actually this is. So, this negative should not be there and I get minus Q of x. So, this is the overall thing. Now, let us look at the curvatures of this plate. So, we have said that the curvature of the plate in the y direction is 0. This is what the exact solution is telling us. In the x direction, it is not 0 and the twisting curvature Kxy is not 0. So, the plate if I look at it, let us say this is the mid plane of the plate in the undeformed position and when the plate deforms because the curvature is only in the x direction and it is not in the y direction, the plate will bend something like this. So, this green line is the reflects the curvature of the plate. So, this is my x axis, this is my y axis and vertical is the z axis. So, the plate bends only x direction, no other curvature. Now, we will look at. So, this is how the plate is going to bend. This is what we should expect and this is what actually happens as observed in physical systems. Now, let us look at the moments on the plate. So, if I take the plate mid plane, we have seen that Mx is not equal to 0, My is not equal to 0 and Mxy is equal to 0. This is what we have computed, Mx is equal to qA square times 1 minus x over A whole square and My is d12 divided by d11 times Mx. So, Mx is not equal to 0. So, if this is the mid plane of the plate, the mid plane of the plate will experience Mx and My. So, this is how the Mx will get, this is the positive Mx on a small infinitesimal small mid plane surface area of the plate. So, this is Mx and also there is My and of course, Mx and My, they vary with respect to position, but they are not 0 along the entire length. So, this is My. So, let us mark these points ABCD. So, we will make some comments. So, the first thing is ABCD is portion of mid plane, then only three forces are acting. So, we should not call forces and literally force, but only three things are acting on the plate, three forces. So, what are those three things? Mx because we computed it to be nonzero, My we computed to be nonzero and q, which is the vertical because the plate is transversely loaded. So, that is q, q is there. Then we make some more comments, we say that since k y equals 0. So, what is k y? k y is the curvature in the y direction. So, this is x, this is y. So, k y is equal to 0 and if k y is equal to 0, it means the curvature along the line AB and CD that curvature is 0 and if that is going to be 0 lines AB and CD will remain straight even after load q is applied, more observations. So, k y is 0 and that implies that this line, this line will remain straight and also this line will remain straight. If it was straight to begin with it will also remain straight because the curvature in the y direction is 0. The other thing is there is a vertical force q acting on the plate, the force is acting like this vertical force and what balances that, that force if there was nothing to balance it the plate will just drop. So, what balances it? There is a moment like this Mx, moment like this Mx and My on these two edges. So, Mx and My are responsible for balancing the force against external load q. So, let us write that down vertical force q x y, actually I should not call force, it is load and it is a distributed load is balanced by Mx and My, but My equals d 12 over d 11 Mx. And typically for the composite system which we are discussing d 12 over d 11 is approximately equal to 0.022. So, what that means is that My is equal to 0.022 Mx. So, My is very small compared to Mx. So, most of the balancing is being done by actually Mx does. So, for balancing My plays a little role. So, My it does balance, but its role is not significant, but My still is important and that we are going to see in a moment. If My was equal to 0. So, what will happen if My was equal to 0, this plate. So, suppose this is x axis, so the x axis along the length of this paper is x and this is y axis, if My was equal to 0. So, what is My, My is acting like this, no My is acting like this and Mx is acting like this, Mx is acting like this, My is acting like this. If My was not 0, then what would happen is that the plate will develop a curvature like this, it will develop a curvature in the y direction, it will develop a curvature like this. So, if My then ABCD gets curved as shown. So, let us draw that. So, this is A, this is B. So, this is the deformed position A prime and B prime and this is C prime, this is D prime. So, this is how the plate bends. So, here A B prime curves upwards and not downwards, it does not do this because A B prime has to expand on upper side, remember because the load is acting and we have seen that it has to expand on the upper side and contract underneath. So, this would happen if, so this is what will happen if My was equal to 0. So, this curvature, so this is anticlastic curvature. So, this curvature will develop in absence of My. Now, what does My do? My acts in a way that it opposes this curvature. So, My acts to oppose this curvature and keeps A B and C D straight, it ensures that A B and C D are straight. So, this is the role of My. If My was not there, then the plate would bend as per anticlastic curvature. Now, what is My? My equals D12 over D11 times Mx and what is D12? D12 is directly proportional to Poisson ratio of the plate. So, My has a role due to Poisson effect. Thus, it is the nature of material that ensures del over del y equals 0 because curvature is what second derivative of W with respect to y. So, it is the nature of material that ensures that del over del y is 0 and because it is related to Poisson effect. So, this is what is all about anticlastic curvature and you will see the same effects in other plates, but it is comes out very clearly because in finite plates things become very complicated. So, you cannot separate out different effects, but here because things are simple you can exactly see what is happening and you see that in absence of My, the plate would develop a curvature in y direction and it is My which ensures that plate does not develop a curvature and the curvature which it opposes is known as anticlastic curvature. So, that is what I wanted to discuss today. Starting tomorrow, we will start talking about the principle of virtual work and how it can be used to solve again some more and a different type of different class of finite plate problems. So, that is all for today and I look forward to seeing you tomorrow. Thank you.