Steps for Solving an Engineering Problem

In this segment, we'll talk about steps of solving an engineering problem the reason why it is important to talk about this in a numerical methods course is because the engineering problem solution is obtained by solving some kind of a mathematical model and many times that mathematical model requires numerical method.

So, how do we go about solving engineering problems? So, what we basically do is we describe the problem, then we develop a mathematical model for it, and this might be coming from analysis, this might be coming from experimental data, this might be coming from numerical data; so, it can be all sorts of things or a combination of them. And then what we want to do is we want to take that mathematical model and to solve it, and then of course the big part is also to take that solution which you just found for the mathematical model and be able to apply it to the solution of the problem.

So, what we're going to do is we're going to take an example and go through these four steps. So, the first step is problem description. So here what you are seeing is a drawbridge which you find along the waterways throughout the world. It simply is a first class, and it has a fulcrum right there, so it basically opens and closes just like a seesaw. In fact, the bascule bridge name comes from the French word for see-saw. And this is a close review of it when it's being constructed and what I would like you to pay attention to is the fulcrum right here and it is not entirely visible there, but the function is right there up this bascule.

So, this is a close-up this is a close-up of the vascular bridge fulcrum here and consists of three components basically that you have the trunnion, which is simply a hollow or a solid cylinder, and then what you have is that it is put in this uh item called a hub. And then the trunnion hub is put into the girder of the bridge and that's how this first-class lever is made for a for a drawbridge.

So, how does the trunnion hub girder assembly procedure work? So, here are the three components which are shown here: so, this is the trunnion, this is the hub, and this is the girder. So, what you do first is that you take the trunnion this one right here and you immerse it in ice & alcohol mixture. Why do you need to do that? Because it won't otherwise go into the hub so because the outer diameter of the trunnion is bigger than the inner diameter of the hub. So, we immersed it in ice and alcohol mixture, it contracts the diameter, we shrink fit it into the in-into the into the hub and then it simply uh is left to warm up. But it's not finished yet, what you got to do now is that you got to take this tuning hub assembly and put it in ice & alcohol mixture so that the outer diameter of the hub starts to decrease so that it can go into the hole in the girder and that's what the eventual assembly turns out to be.

So, what was the issue in in doing this trunnion-hub-girder assembly? Was that in one of the bridges what happened was that when the trunnion was cooled down, the trunnion is supposed to go all the way up to this point right here into the hub. So, this point here has to match uh this uh cross section of the of the hub, but it got stuck halfway through. So, we want to figure out why did that happen.

So, the magnitude of the contraction which is required in the trunnion is uh was supposed to be 0.015 meters, so you wanted the magnitude of the delta d to be greater than or equal to 0.015 inches, so it had to contract more than 0.015 inches. How did we get that number? That number is obtained by knowing what kind of fit was required for the specifications, in this case it was an fn2 fit and that required the contraction to be at least 0.015 inches.

So, what we got to figure out is hey did you contract enough? So, in order to do that we have to do mathematical modeling, that is the second step of solving an engineering problem. So, this problem was given to a consultant and what the consultant did was simply said “hey i know my physics 101 so delta d which is the change in diameter will be the diameter multiplied by thermal expansion coefficient of the material [which was cast steel] and multiplied by delta t which is the change in temperature,” because that's what comes from the definition of the thermal expansion coefficient. The quotient of linear thermal expansion is the change in length per unit length per unit temperature. So, using that definition right here we have the diameter, we get our alpha from a handbook (this is the room temperature alpha), the change in temperature is 80 degrees - let's suppose if we assume it to be the room temperature - 108 is the temperature of the ice & alcohol mixture so the change in temperature is - 188 degrees Fahrenheit. So, if we apply this formula right here, d is 12.363, the alpha is given to a 6.47x10^(-6) and the temperature change is – 188. Multiply the three numbers and you get this number right here. So, this, if you look at the magnitude of the contraction is 0.01504 inches, and it is more than what we wanted. Right? The specifications required that the temperature that the change in the diameter should be more than 0.015 inches, which it turns out the case.

So, what's the issue here? The issue here is that, is this the correct formula to use in the first place? Delta d is equal to d times alpha times delta t - when you can recognize from literature that alpha, the thermal expansion coefficient, is a function of temperature. And you can see that as the temperature decreases, the thermal expansion coefficient also decreases. So, you're not getting the same amount of contraction at every temperature. In fact, as it gets cooler the amount of contraction which you're getting, per unit temperature, starts becoming smaller. So, we have to recognize that: is this really the correct formula to use?

The correct model would-would be to account for the varying thermal expansion coefficient. So, what that implies is that we'll have to use this formula. Is the delta d is equal to d times alpha times delta t - this might work for temperature not varying too much, so, it's an okay formula to use if alpha is not varying a whole lot between the temperatures, it will be subjected to, but it is not the right formula to use when thermal expansion coefficient is changing. In fact, you'll find out that this formula does reduce to this formula right here if alpha is a constant.

So, the next step is a solution of mathematical model. Let's go and see how do we go about solving this mathematical model. So, the mathematical model which we have is that hey the change in the diameter is not given by d times alpha times delta t, but it's given by this integral right here. And now what you’re seeing here is that you got data which is not given as a continuous function of temperature - so what that means is that hey I cannot simply use my integral calculus knowledge of exact integration to be able to find out what the change in diameter is, because I got to do this integral first and then multiply by the diameter. So, how do I go about doing that? So, let's go and see can we roughly estimate the contraction, so what we can do is we can say hey I was at 80 degrees (that's the alpha 80 degrees) at -108 it is this, so, I'm going to simply somehow say hey, let me draw a trapezoid here. And this is 80, this is -108, if I’m just doing it rough estimation on the first day of class. So, let me see what the area under the curve is uh although the the curve is given only a discrete data points, can I find out the area under the curve? So, this area under the curve, which you're finding out here, is a trapezoid – so, I'm going to say hey you know this one looks about 5(or so)x10^(-6). That is the height of that parallel side, and the height of this parallel side is about 6.5x10^(-6) I divided by 2 so that gives me the average length of the parallel sides. Multiplied by the perpendicular distance between the two parallel sides, which is -188. So, that is the area of a trapezoid right, average length of the parallel sides multiplied by the perpendicular distance between the two parallel sides and what I get here is -1081x10^(-6). So, if I find my delta d, delta d will be 12.363 times -1081x10^(-6), and this number here, turns out to be 0.013364 inches. And this quantity right here, the magnitude of this number is not bigger than 0.015 inches. So right there we can see that hey in fact actually we did not get enough contraction.

Another question which people ask is that “hey how did I get this minus from?” although the area under the curve right here is above the x-axis, in this case is temperature, but look at this - we are integrating from higher temperature to lower temperature - we integrate from right to left and that's why this integral quantity right here, the area in the curve, is a negative number right here. But what we have shown is that hey delta d magnitude is turning out to be less than 0.015 inches, in this case.

Now in the previous slide we we talked about how to estimate the contraction with a very rough estimate. Can we do it a little bit more accurately? Yes, what we can do is we can approximate this data, which is given to us, by a second-order polynomial regression model. So, these terms which I’m just talking about polynomial regression model may not be familiar with you, but later on in the course when we talk about regression uh you will you'll be able to uh see that “hey where do we get this from?” So, at this point what you got to think about is that hey if somebody gives me several data points uh relating one variable with another, I can get an approximate formula (maybe a second-order polynomial, first order polynomial) to approximate the data. So, this is what we get to approximate the data, a second-order polynomial, but the great thing about is that we can just simply uh include it in here, in in this form - in this integral here, and since everybody knows how to integrate a second-order polynomial exactly and we already know what the value of d is, then we can find out what delta d is. So, we're getting very close to what we got previously. Even with the rough estimate, we got -0.0133 and here we’re getting -0.0137 inches, and that's still, this this magnitude delta d, is still less than 0.015 inches, even if we do a more accurate estimate of the change in diameter.

So, the question is that hey although we have been able to answer the question that hey why it got stuck, but that's not what real life is looking for. Real life is looking for a solution hey okay you got a solution to this mathematical model; can you use this to maybe come up with the solution which whether it will not get stuck. So here is uh one way of doing.

So, one of the ways of doing that is that hey that we don't put it into dry ice and alcohol mixture, but what we do is we put in liquid nitrogen. Liquid nitrogen temperature is boiling temperatures, -321 degrees Fahrenheit, so by conducting the same exact calculations as we have a for dry ice & alcohol mixture by putting alpha right here and putting the lower limit & upper limit of integration. This the diameter of the trunnion, we get a delta d of -0.0244 inches. And this delta d right here, is greater than 0.015 inches. So, what we have done is that we have first we have used the correct model to show that hey delta d (the change in diameter) is not enough for it to get uh, for it to have what's it called, a proper shrink fit. So, what we need to do is we need to put it into a cooler medium and that cooler medium would be let's suppose liquid nitrogen.

So, let's go and revisit the steps to solve an engineering problem. We just solved an engineering problem, let's go and revisit the steps.

What is the problem description? The-the problem description was trunnion got stuck - now this is uh too reductive in nature, but problem description in real life could run into pages, but here I’m just recalling that hey trunnion got stuck, it should have gone into the hub, but it didn't go.

What was the mathematical model? We used a mathematical model which was this, but we recognize that this is the wrong model to use, in this case, because thermal expansion coefficient is changing over temperature. So, the correct model to use is-is this.

And then the solution of the mathematical model, when we did it by using the rough method, we just said hey! Let's do trapezoid the rule - where we just approximated the area of the curve by a trapezoidal rule. The more accurate method was to regress to a second order polynomial (right) plus integrate. That's what we did, we said hey! Let's go ahead and find a second-order polynomial approximation of the thermal expansion coefficient as a function of temperature, then we can plug it into the integral which we have right here, and we can find out what delta d is.

Then how do we use the solution? We use the solution by saying that hey, let's put it in liquid nitrogen rather than putting in ice and alcohol mixture, which did not give us enough contraction, let's go ahead and put in liquid nitrogen where we get enough contraction.