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Chapter 31: Problem 52

When parts are machined, it is important that they are created with enough precision so that they can be assembled with other parts. No machine can hold dimensions exactly, so it is important to have an agreed upon level of variation. A company that creates nuts and bolts makes its parts with specific tolerances that follow rules established by an international standard. The nut (or hole) has a slightly larger tolerance than the bolt (or shaft) so that the nuts and bolts will work together. This company uses process control, with samples taken five times during each hour to ensure that the processes are stable and running on target. For the nuts, the process is running with \(x=10.004 \mathrm{~mm}\) and a sigma estimate of all measurements \(s=0.002 \mathrm{~mm}\). For the bolts, \(x=10.000 \mathrm{~mm}\) with a sigma estimate of all measurements \(s=0.001 \mathrm{~mm}\). Compute the natural tolerances for both the nuts and bolts. What issue do you see with where the process is currently running?

Short Answer

Expert verified
Natural tolerances for nuts: (9.998, 10.010) mm; bolts: (9.997, 10.003) mm. Overlap is limited, which can cause assembly issues.

Step by step solution

01

Understand the Problem Statement

The problem involves calculating the natural tolerances of nuts and bolts with given mean measurements and standard deviations. For nuts, we have a mean \( x = 10.004 \, \mathrm{mm} \) and standard deviation \( s = 0.002 \, \mathrm{mm} \); for bolts, the mean is \( x = 10.000 \, \mathrm{mm} \) and the standard deviation is \( s = 0.001 \, \mathrm{mm} \). We need to find the range within which the means for nuts and bolts vary naturally.
02

Determine the Natural Tolerance Calculation

The natural tolerances of a process are typically calculated as \( \bar{x} \pm 3s \), where \( \bar{x} \) is the process mean and \( s \) is the standard deviation of the process.
03

Calculate Natural Tolerances for Nuts

Using the formula \( \bar{x} \pm 3s \) for the nuts:\[ 10.004 \, \mathrm{mm} \pm 3 \times 0.002 \, \mathrm{mm} = (10.004 - 0.006, 10.004 + 0.006) \]Calculating the interval gives:\( 10.004 \, \mathrm{mm} \pm 0.006 \, \mathrm{mm} = (9.998, 10.010) \, \mathrm{mm} \).
04

Calculate Natural Tolerances for Bolts

Using the formula \( \bar{x} \pm 3s \) for the bolts:\[ 10.000 \, \mathrm{mm} \pm 3 \times 0.001 \, \mathrm{mm} = (10.000 - 0.003, 10.000 + 0.003) \]This gives us the interval:\( 10.000 \, \mathrm{mm} \pm 0.003 \, \mathrm{mm} = (9.997, 10.003) \, \mathrm{mm} \).
05

Identify Issues with Current Process

In comparing the natural tolerance ranges, the nuts have a tolerance range of \((9.998, 10.010)\) while the bolts have a tolerance range of \((9.997, 10.003)\). The issue is that the overlap between these ranges is only from \(9.998\) to \(10.003\), suggesting potential assembly issues since the nut is positioned more towards its higher tolerance range.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Process Control
Process control is an essential aspect of manufacturing and production. It involves monitoring and controlling a process to ensure that it operates at its optimal performance level. This not only means keeping the process efficient but also ensuring that the resulting products meet certain quality standards.
The company that manufactures nuts and bolts employs process control strategies, such as taking measurements periodically鈥攆ive times an hour, in this instance鈥攖o maintain consistency and accuracy.
Here are some advantages of using process control:
  • Ensures product quality and consistency.
  • Reduces waste and increases production efficiency.
  • Allows for quick response to process deviations, minimizing defects.
Effective process control minimizes the variability in the production process, ensuring that the dimensions of the manufactured parts remain within a defined tolerance range. This is crucial for pieces like nuts and bolts, which must fit together precisely.
In the case of this exercise, the company measured the average dimensions and the standard deviation of these parts over time, providing data to calculate natural tolerances and assess the process's effectiveness.
Statistical Quality Control
Statistical Quality Control (SQC) plays a vital role in enhancing the quality of the manufacturing process. This method uses statistical methods to monitor and control the quality of a product or process.
A key component of SQC is the use of control charts and process capability indices to ensure that a process remains in control and performs according to the design specifications.
By employing techniques such as:
  • Control charts: graphical tools used to determine if a manufacturing or business process is in a state of statistical control.
  • Process capability analysis: compares the output of a process to its specification limits.
  • Sampling: gathering and assessing data to predict overall performance.
In our example, the SQC approach was used to calculate the natural tolerance, giving insight into the process's variability and guiding necessary adjustments.
The calculation of natural tolerances helps detect any deviations from the desired standards, allowing corrective actions to be implemented before significant defects arise. However, in this case, the narrow overlap in tolerance ranges raises concerns about the possibility of assembly issues.
Standard Deviation
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much individual measurements of a process deviate from the average or mean.
Understanding standard deviation is crucial for assessing the precision and reliability of a manufacturing process.
  • A smaller standard deviation indicates that the values are closely clustered around the mean, signifying a more consistent process.
  • A larger standard deviation suggests a wider spread of values, indicating more variation and possibly pointing towards quality issues.
For the nuts and bolts example, the standard deviation values were used to calculate the natural tolerance of the manufacturing process.
Natural tolerance is derived by adding and subtracting three times the standard deviation from the process mean, symbolizing the range where most of the production will lie. In this exercise, estimating this tolerance helped identify if the parts can fit together within the prescribed limits, crucial for ensuring smooth assembly.
If the standard deviation figures were larger, it might indicate potential problems in maintaining the required product specifications.

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Most popular questions from this chapter

The Boston Marathon has been run each year since 1897. Winning times for men were highly variable in the early years, but control improved as the best runners became more professional. A clear downward trend continued until the 1980 s. Rick plans to make a control chart for the winning times from 1950 to the present. The first few times are \(153,148,152,139,141\), and 138 minutes. Calculation from the winning times from 1950 to 2019 gives $$ x=133.615 \text { minutes and } s=6.210 \text { minutes } $$ Rick draws a center line at \(x\) and control limits at \(x \pm 3 s\) for a plot of individual winning times. Explain carefully why these control limits are too wide to effectively signal unusually fast or slow times.

A luxury sports car dealership offers its clients a complimentary shuttle service to and from the dealership when their cars are serviced. Currently, the dealership's driver shuttles clients to and from locations. However, the dealership has only one driver, and clients sometimes have to wait for an extended period. In hopes of improving service and pleasing clients, the dealership decides to change from an in-house shuttle service to using a ride- sharing service. The dealership wants to monitor the impact of this change to see if the percentage of clients who take advantage of the transportation service changes. First, the dealership gathers historical data to determine the percentage of clients who have been using the shuttle service. It looks at records for the past six months. The average number of clients who visit the dealership each month is 190 , with relatively little month-to-month variation. During the past six months, a total of 354 clients have requested rides. a. What is the estimated total number of clients requesting rides during these six months? What is \(p\) ? b. Give the center line and control limits for a \(p\) chart on which to plot the future monthly proportions of clients requesting rides.

A manufacturer of consumer electronic equipment makes full use not only of statistical process control but also of automated testing equipment that efficiently tests all completed products. Data from the testing equipment show that finished products have only \(3.0\) defects per million opportunities. a. What is \(p\) for the manufacturing process? If the process turns out 4000 pieces per day, how many defects do you expect to see per day? In a typical month of 24 working days, how many defects do you expect to see? b. What are the center line and control limits for a \(p\) chart for plotting daily defect proportions? c. Explain why a \(p\) chart is of no use at such high levels of quality.

The inside diameter of automobile engine piston rings is important to the proper functioning of the engine. The manufacturer checks the control of the piston ring forging process by measuring a sample of five consecutive items during each hour's production. The target diameter for a ring is \(\mu=74.000 \mathrm{~mm}\). The process has been operating in control, with the center close to the target and \(\sigma=0.015 \mathrm{~mm}\). a. What center line and control limits should be drawn on the \(s\) chart? On the \(x\) chart? b. A different manufacturer creates the pistons in which the rings will be fitted. This manufacturer has a target value of \(73.945 \mathrm{~mm}\) for the piston diameter. The manufacturer checks control of the piston diameter four times each hour. Recently, the process has been running high with \(\mu=74.000 \mathrm{~mm}\) and \(\sigma=0.005 \mathrm{~mm}\). Do you see any issues that might arise for the manufacturer of the engine when the two parts from the different manufacturers are assembled?

A manufacturer of ultrasonic parking sensors samples four sensors during each production shift. The expectation is that a sensor will initially alarm when an object comes within 60 inches of the sensor. The sensors are put on a rack, and an object is moved toward each sensor, one at a time, at a 90 -degree angle until the sensor alarms. The distance from the object to the sensor at that point is recorded. This results in four measurements, one for each sensor on the rack, and the mean of these four measurements is recorded. The process mean should be \(\mu=60\) inches. Past experience indicates that the response varies with \(\sigma=1.0\) inches. The mean response distance is plotted on an \(x\) control chart. The center line for this chart is a. \(1.0\) inches. b. 4 inches. c. 60 inches.
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