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The term "quality control" brings to mind images of control
charts. A typical 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:
UNIT 1 PROBLEM SOLVING and UNITS After completing this section the student will be able to:
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. 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.
EXAMPLE 4 GIVEN:
FIND: A ANSWER: A = 522ft2
FIND: V in yd3 ANSWER: V = 1933 yd3
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?
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Last modified: August 20, 1998