[ Home | News ]

PREFACE

The term "quality control" brings to mind images of control charts. A typical Chart could be developed for the following situation:

The specifications call for a dimension to be 6.000 in. ± 0.0005 in. The process selected can produce the 6.000 in mean with a standard deviation of 0.001 in. The size of the Control Chart sample would be based on the desired assurance that the chart detects a mean dimension shift that would move the process limits out of the specifications. A shift that would require reworking or scrapping parts.

The analytical solution would not only provide the numerical answers:

Sample Size

Probability of Detecting Shift of Process Limit to Specification Limit

4

84%

5

93%

6

97%

The complete solution would also display the information showing the shift of the process measurements toward a specification limit. In this case, as the process mean shifts from 6.000in.to 6.002 in. the Upper Process Limit shifts from 6.003 in. to the Upper Specification Limit, 6.005 in. A further shift would produce out of specification parts.
It would also show the shift of the sample means above the Upper Control Limit of the Control Charts being considered. For a sample size of 4 measurements, 16% of the sample means would be below the upper control Limit . 16% of the samples would not show loss of control.
For a sample size of 5 measurements, 7% of the sample means would be below the Upper Control Limit. 7% of the samples would not show loss of control. 93% would be above the Upper Control Limit and indicate trouble.
For a sample size of 6 measurements, 3% of the sample means would be below the Upper Control Limit. 97% of the samples would indicate trouble.

UNIT 1 PROBLEM SOLVING and UNITS

This chapter will cover the standard technical problem solving technique and format. In engineering we do not solve problems for our selves. In any reliable engineering organization someone technically reviews even the work of the most experienced engineer. The presentation of the solution to a problem is as important as the solution itself. This applies to neatness as well as format. This is particularly true in the classroom. A neat paper gives the impression of "correctness" until an error is found; a sloppy paper gives the assumption of error.

This chapter also covers the use of units and the conversion of units. Engineers and scientists are concerned with quantities of things. These quantities are measured in standard units such as gallons. Measurements have two parts which ANSWER the questions; "How many?" and "Of what?" To say a room is 20 feet long we ANSWER both questions. There are 20 units and the unit is feet. The number without the unit has no significance. To say that I live 50 from the school is meaningless--50 minutes, 50 feet, or 50 miles?

OBJECTIVES

After completing this section the student will be able to:

  • Select the information needed to solve a problem from the facts available.
  • Arrange the facts and the solution calculations in the standard technical problem-solving format.
  • Use English units in calculations.
  • Use S I (metric) units in calculations.
  • Make conversions within unit systems.
  • Make conversions between unit systems.

PROBLEM SOLVING TECHNIQUE

There is usually more information available than that required to solve a technical problem. The first step toward a solution is to determine the relationship of the variables pertaining to the problem. These are often in the form of an equation, often called a formula. To determine the area of a rectangular steel plate, for example, we would use the fact that the area of a rectangle is equal to the length times the width. This is shown mathematically by the equation or formula, A = LW. We would ignore other available facts such as its thickness, weight or temperature.

If we wanted to calculate the area of a rectangular metal plate with a length of 6 inches, a width of 5 inches, a thickness of 2 inches, a weight of seven pounds, and a temperature of 70 °F; we would:

First FIND the equation that is the formula for the area of a rectangle.

A = LW

Then substitute the dimensions for the letters and ignore the other information.

A = 6 in * 5 in

Then do the calculation; multiply 6 by 5 and in by in.

The solution is expressed as a complete thought (sentence) in mathematical language.

"The area equals thirty square inches." 

A = 30 in2

Note that the unit " in " was multiplied as part of the calculation. in. * in = in2 just as r * r = r2

If we were looking for the volume of the plate, we would use the equation that is the formula for the volume, V = L W H. The calculation would be:

V = L W H

V = 6 in. * 5 in. * 2 in.

V = 60 in3

TECHNICAL COMMUNICATION

There are two major purposes for technical calculations. One is to FIND the ANSWER to a problem. The second is to provide a written record of how the ANSWER was calculated. The purpose of the written record is to allow others to review the calculation. This makes the written calculation a means of communication.

Each technology has its own language, special words and symbols, to speed communication. in3 is read "cubic inches" and stands for inches times inches times inches. V = L W H, V = L * W * H, or

V = LWH mean volume equals length times width times height.

There is also a standard problem-solving format. This is a standard way of showing the steps taken to solve a technical problem.

Problem Solving Format

The presentation of a problem with its solution follows the analytical thought process. It has four parts. The GIVEN, FIND and ANSWER are always labeled. The SOLUTION section is seldom labeled.

GIVEN:

In this section you show all the data needed to solve the problem. It is often shown in a sketch or schematic diagram when this would clarify the problem and speed communication by reducing the number of words required. The wording of the problem as received is not repeated in the problem-solving format. Only the information required for the solution of the problem is shown.

 FIND:

In this section you list all the information required of the solution.

 SOLUTION:

This section will start with the formulas or principles to be used in the calculations. All calculations are shown. Every number that has units, e. g. 3 ft, will be shown with its units every time it appears.

 ANSWER:

All ANSWERs will be shown separate from the calculations. They will be labeled "ANSWER" and will be complete with title, value, and units such as:

ANSWER: AREA = 9 FT2

EXAMPLE 1

The problem as received

It is the 21st of December in Omaha. The outside temperature is 32° F. There is a box there. It has inside dimensions as follows: a base of 3 ft, a height of 4 ft and a length of 5 ft. What is the holding capacity of the box?

The problem and solution presentation.

 

Notice that the title "VOLUME" is the same in the FIND and ANSWER.

Notice also that the information shown under "GIVEN" is just that information needed to solve the problem and not the wording of the problem. The information needed to solve the problem is determined from the formula shown under "SOLUTION", V = B x H x L.

EXAMPLE 2

A 5 in. by 3 in. triangular piece of metal has a 1 in. dia. hole punched through it. What is the area of the remaining metal?

 

EXAMPLE 3

A steel box has inside dimensions as follows: a base of 3 ft, a height of 4 ft and a length of 5 ft. It has outside dimensions as follows: a base of 3.2 ft, a height of 4.1 ft, and a length of 5.3 ft. What is the volume of the steel? Note: In the solution we will calculate the volume of the outside of the box and then subtract the volume of the hole.

 

For a trapezoid, area equals the height times the average width.

A =

For H = 18 ft, W1 = 11ft, W2 = 47 ft, find the area in ft2.

EXAMPLE 4

GIVEN:

H = 18 ft

W1 = 11ft

W2 = 47 ft

FIND: A

ANSWER: A = 522ft2

A trapezoidal ditch must be dug 100 foot long. The correctional area is as in EXAMPLE 4. Earth work (outdoors its earth or soil - track it inside and it becomes dirt) is calculated in cubic yards. How much earth must be removed?

EXAMPLE 5

GIVEN:

H = 18 ft

W1 = 11ft

W2 = 47 ft

L =100 ft

FIND: V in yd3

ANSWER: V = 1933 yd3

Summary

  • The format for technical problem solutions includes the GIVEN: FIND: SOLUTION: ANSWER: sections and usually includes sketches or schematic diagrams to display information.
  • All measurements include units.

PROBLEM SET 1A

These problems provide an opportunity to demonstrate and reinforce skills in solving technical problems. Solve them using the examples as a guide. When done contact your instructor for solutions and a self-grading guide.

1. A six-inch diameter pipe will be used to distribute water at the rate of 55 gallons per minute. The measurement of six inches is the inside diameter of the pipe, (the hole). What is the area of the hole in the pipe?

2. A juice maker wants to use a can to ship his products. A can manufacturer recommends a can with inside dimensions of four inches in diameter and five inches tall. What is the volume of the can in cubic inches?

3. A triangular piece of metal has a 1.20-in. dia. hole punched through it. What is the area of the remaining metal.

4. The container shown is made of aluminum. What is its maximum holding capacity? What is the volume of aluminum?

 

[ Home | News ]

Send mail to fjg@ddps.com

with questions or comments about this web site.
Copyright © 1997 DD&G
Last modified: August 20, 1998