CHAPTER 08.03: FINITE DIFF METHOD FOR ODEs: Example Part 1 of 2 

In this segment we are going to take an example of how to use Finite Difference Methods to solve ordinary differential equations. so we're going to take a simple pragmatic example here. Let's suppose somebody says that hey I have a simply supported beam and what I want to be able to do is, that I am applying some tension, T, to it let's suppose. And then I'm also applying some uniformly distributed load to it, let's suppose q. Let's suppose the length of the beam is L. Then the differential equation which governs how much this particular beam is deflecting is given by d2y/dx2 - Ty/EI = [qx(L-x)]/2EI d2y/dx2 - Ty/EI = [qx(L-x)]/2EI d2y/dx2 - Ty/EI = [qx(L-x)]/2EI That's the differential equation which governs this particular problem here. And y is the deflection, so y is the deflection which is measured like this so this is y here, this is the deflection and x is the location at a particular point along the beam. So let me just, let me just go and put some values of these variables in there, so let's suppose T is somebody gives me it, that the amount of tension which they are applying is 7200 pounds The constant load which they are applying, the uniform load which they are applying, not contant load, but uniform load which they are applying is 5400 pounds per inch. The length of the specimen is given as 75 inches. The Young's Modulus, E is the Young's Modulus, so E which you are seeing here is Young's Modulus is given as 30 msi, which is 30*10^6 psi. And let's suppose I, which is the second moment of area, please don't call it moment of inertia. Second moment of are is 120 in^4. So if somebody told me to go ahead and use Finite Difference Methods to solve this particular problem, I would say hey, what I will have to do is I will have to take this length, L, of the beam and break it up into small segments. And what I'm going to do is I'm going to take this length of the beam which is right here and I'm going to, this is 75 inches which I know which is the length which is given to me. So I'm going to call this to be node 1 let's suppose, call this to be node 2, call this to be node 3, and call this to be node 4. So what I am doing is, the problem statement is telling me that hey go ahead and choose 3 segments for doing the Finite Difference Approximation. So in that case what will happen is that this delta x which I have, which is the distance between the two nodes is going to be 75/3, is going to be 25 inches, that's what delta x is going to turn out to be. So that means that if I am able to, I already know that the value of the deflection at node 1 is zero. I already know that the deflection at node 4 is zero because it is simply supported at those two particular nodes. So I need to find out what the value of the deflection is at node 2 and node 3. So how am I going to do that? I am going to first substitute the values of the constants which I just found out and I'll d2y/dx2 - (2*10^-6)y = (7.5*10^-7)(x)(75-x) d2y/dx2 - (2*10^-6)y = (7.5*10^-7)(x)(75-x) d2y/dx2 - (2*10^-6)y = (7.5*10^-7)(x)(75-x) This particular equation I get by substituting in the values of the tension T, the uniformly distributed load q, the length L, E and I. That's what I get as my equation, so I now need to be able to set up this finite difference equations from this. So I'm going to, what I am going to do is I'm going to, I'm going to approximate my second derivative at some node i by my center divided difference approximation of that and that's what it will turn out to be. This is the center divided approximation of the second derivative of y with respect to x at any node i. What that means is that if I'm at node i, there will be a node i-1 to the left and node i+1 to the right and that will give me the approximation of the second derivative of y with respect to x. So if I do that, then I know that going back to the nodes which I have, that hey I have node 1 here, then I have node 2 here, node 3 here, and node 4 here. And let's start with x equal to zero here. This will be x equal to 25 because delta x is 25. This will be x equal to 50 and this will be x equal to 75. So when I write down this, when I substitute the second derivative approximation in my formula, this is what I am going to get, I am going to get: [y_i+1 - 2(y_i) + y_i-1]/(delta x)^2 [y_i+1 - 2(y_i) + y_i-1]/(delta x)^2 That is the approximation for the second derivative, -2*10^-6 and that will be the value of y at that particular node because now I am rewriting my ordinary differential equation for that particular node. And that will be equal to 7.5*10^-6 times the value of x at that particular node times 75 minus the value of x at that particular node x_i right there.