Introduction To Numerical Method Pdf

[Blanche Bunnell]

Inthe truss problem, one can solve \(1000\) simultaneous linear equations for \(1000\) unknowns without using a calculator. One can use fractions, long divisions, and long multiplications to get the exact answer. But just the thought of such a task is laborious. The task may seem less laborious if we are allowed to use a calculator, but it would still fall under the category of an intractable, if not an impossible, problem. So, we need to find a numerical technique and convert it into a computer program that solves a set of \(n\) equations and \(n\) unknowns.

Again, what are numerical methods? They are techniques to solve a mathematical problem approximately. As we go through the course, you will see that numerical methods let us find solutions close to the exact one, and we can quantify the approximate error associated with the answer. After all, what good is an approximation without quantifying how good the approximation is?

Numerical methods are used by engineers and scientists to solve problems. However, numerical methods are just one step in solving an engineering problem. There are four steps for solving an engineering problem, as shown in Figure \(\PageIndex2.1\).

The first step is to describe the problem. The description would involve writing the background of the problem and the need for its solution. The second step is developing a mathematical model for the problem, and this could include the use of experiments or/and theory. The third step involves solving the mathematical model. The solution may consist of analytical or/and numerical means. The fourth step is implementing the solution to see if the problem is solved.

To make the fulcrum (Figure \(\PageIndex2.2\)) of a bascule bridge, a long hollow steel shaft called the trunnion is shrunk-fit into a steel hub. The resulting steel trunnion-hub assembly is then shrunk-fit into the girder of the bridge.

The shrink-fitting is done by first immersing the trunnion in a cold medium such as a dry-ice/alcohol mixture. After the trunnion reaches the steady-state temperature, that is, the temperature of the cold medium, the outer diameter of the trunnion contracts. The trunnion is taken out of the medium and slid through the hole of the hub (Figure \(\PageIndex2.3\)).

When the trunnion heats up, it expands and creates an interference fit with the hub. In 1995, on one of the bridges in Florida, this assembly procedure did not work as designed. Before the trunnion could be inserted fully into the hub, the trunnion got stuck. Luckily, the trunnion was taken out before it got stuck permanently. Otherwise, a new trunnion and hub would need to be ordered at the cost of \(\$50,000\). Coupled with construction delays, the total loss could have been more than a hundred thousand dollars.

A hollow trunnion of an outside diameter \(12.363^\prime\prime\) is to be fitted in a hub of inner diameter \(12.358^\prime\prime\). The trunnion was put in a dry ice/alcohol mixture (temperature of the fluid - dry-ice/alcohol mixture is \(- 108^\circ\textF\)) to contract the trunnion so that it can be slid through the hole of the hub. To slide the trunnion without sticking, a diametrical clearance of at least \(0.01^\prime\prime\) is required between the trunnion and the hub. Assuming the room temperature is \(80^\circ\textF\), is immersing the trunnion in a dry-ice/alcohol mixture a correct decision?

To calculate the contraction in the diameter of the trunnion, the thermal expansion coefficient at room temperature is used. In that case, the reduction \(\Delta D\) in the outer diameter of the trunnion is

So, according to this calculation, immersing the steel trunnion in dry-ice/alcohol mixture gives the desired contraction of greater than \(0.015^\prime\prime\) as the predicted contraction is \(0.01504^\prime\prime\). But, when the steel trunnion was put in the hub, it got stuck. Why did this happen? Was our mathematical model adequate for this problem, or did we create a mathematical error?

So, one needs to curve fit the data to find the coefficient of thermal expansion as a function of temperature. This curve is found by regression where we best fit a function to the data given in Table 1. In this case, we may fit a second-order polynomial

The values of the coefficients in the above Equation \((\PageIndex2.3)\) will be found by polynomial regression (we will learn how to do this later in the chapter on Nonlinear Regression). At this point, we are just going to give you these values, and they are

\[\displaystyle \beginsplit \alpha &= a_0 + a_1T + a_2T^2\\ &= 6.0150 \times 10^- 6 + 6.1946 \times 10^- 9T - 1.2278 \times 10^- 11T^2 \endsplit \;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex2.4) \nonumber\]

\[\displaystyle \beginsplit \Delta D &= 12.363\beginbmatrix 6.0150 \times 10^- 6 \times ( - 108 - 80) \\ + 6.1946 \times 10^- 9\displaystyle \frac\left( ( - 108)^2 - (80)^2 \right)2 \\ - 1.2278 \times 10^- 11\displaystyle \frac(( - 108)^3 - (80)^3)3 \\ \endbmatrix\\ &= - 0.013689^\prime\prime\endsplit \nonumber\]

Although we were able to find out why the trunnion got stuck in the hub, we still need to find and implement a solution. What if the trunnion were immersed in a medium that was cooler than the dry-ice/alcohol mixture of \(- 108^\circF\) , say liquid nitrogen, which has a boiling temperature of \(- 321^\circF\)? Will that be enough for the specified contraction in the trunnion?

\[\displaystyle \beginsplit \Delta D &= 12.363\beginbmatrix 6.0150 \times 10^- 6 \times ( - 321 - 80) \\ + 6.1946 \times 10^- 9 \displaystyle\frac\left( ( - 321)^2 - (80)^2 \right)2 \\ \ - 1.2278 \times 10^- 11 \displaystyle\frac(( - 321)^3 - (80)^3)3 \\ \endbmatrix\\ & \\ &= - 0.024420^\prime\prime \endsplit \nonumber\]

2) Rather than regressing the thermal expansion coefficient data to a second-order polynomial so that one can find the contraction in the trunnion OD, how would you use the trapezoidal rule of integration for unequal segments?

Numerical methods are techniques to approximate mathematical processes. This introductory numerical methods course will develop and apply numerical techniques for the following mathematical processes:

of finding the roots of a quadratic equation \(\displaystyle ax^2 + bx + c = 0\) goes back to the ancient world. But in the real world, we get equations that are not just the quadratic ones. They can be polynomial equations of a higher order and transcendental equations.

Assume that the ball has a density of \(600\ \textkg/\textm^3\) and has a radius of \(0.055\ \textm\). On applying the Newtons laws of motion and hence equating the weight of the ball to the buoyancy force, one finds that the depth, \(x\) in meters, to which the ball is underwater and is given by

A cubic equation can be solved exactly by radicals, but it is a tedious process. The same is true but even more complicated for a general fourth-order polynomial equation as well. However, there is no closed-form solution available for a general polynomial equation of fifth-order or more. So, one has to resort to numerical techniques to solve polynomial and other transcendental nonlinear equations (e.g., finding the nonzero roots of \(\tan x = x\)).

Solving a few simultaneous linear equations such as the above set can be done without the knowledge of numerical techniques. However, imagine that instead of three given points, you were given 10 data points. Now the setting up as well as solving the set of 10 simultaneous linear equations without numerical techniques becomes laborious, if not impossible.

Interpolation involves that given a function as a set of data points. How does one find the value of the function at points that are not given?. For this, we choose a function, called an interpolant, and make it pass through all the points involved.

You may think that you have already used interpolation in courses such as Thermodynamics and Statistics. After all, it was just taking two points from a table at the back of the textbook or online and finding the value of the function at a point in between by using a straight line.

Although this is possibly enough for courses such as Thermodynamics and Statistics, there are two questions to ask. Is the value calculated accurately, and how accurate is it? To know that, one needs to calculate at least more than one value. In the above example of a rocket velocity vs. time, one can instead use a second-order polynomial interpolant and set up the three equations and three unknowns to find the unknown coefficients, \(a\), \(b\), and \(c\) as given in the previous section.

When we talked about curve fitting by interpolation, the chosen interpolant needs to go through all the points considered. What happens when we are given many data points, and we instead want a simplified formula to explain the relationship between two variables. See, for example, in Figure \(\PageIndex3.4a\), we are given the coefficient of linear thermal expansion data for cast steel as a function of temperature. Looking at the data, one may proclaim that a straight line could explain the data, and that is drawn in Figure \(\PageIndex3.4b\). How we draw this straight line is what is called regression. It would be based on minimizing some form of the residuals between what is observed (given data points) and what is predicted (straight line). It does not mean that every time you have data given to you, you draw a straight line. It is possible that a second-order polynomial or a transcendental function other than the first-order polynomial will be a better representation of this particular data. So these are the questions that we will answer when we discuss regression. We will also discuss the adequacy of linear regression models.

In the previous lesson, we looked at the example of contracting the diameter of a trunnion for a bascule bridge fulcrum assembly by dipping it in a mixture of dry ice and alcohol. The contraction is given by

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