Engineering: Manufacturing Tolerances and Quality Control
Using specifications, measurements, and inspection data to judge product quality
Engineering: Manufacturing Tolerances and Quality Control
Using specifications, measurements, and inspection data to judge product quality
Engineering - Grade 9-12
- 1
A metal pin is specified to have a diameter of 10.00 mm with a tolerance of ±0.05 mm. What is the acceptable range of diameters?
Find the lower limit and upper limit by subtracting and adding the tolerance.
The acceptable range is 9.95 mm to 10.05 mm because 0.05 mm is subtracted from and added to the nominal diameter of 10.00 mm. - 2
A drilled hole has a specification of 6.25 mm ±0.10 mm. A technician measures one hole as 6.37 mm. Is the hole within tolerance? Explain.
The hole is not within tolerance because the acceptable range is 6.15 mm to 6.35 mm, and 6.37 mm is above the upper limit. - 3
A shaft must fit into a bearing. The shaft diameter is specified as 20.00 mm ±0.02 mm, and the bearing inner diameter is specified as 20.08 mm ±0.03 mm. What are the minimum and maximum possible clearances between the bearing and shaft?
Clearance equals bearing inner diameter minus shaft diameter. Use worst-case combinations.
The minimum clearance is 0.03 mm because the smallest bearing is 20.05 mm and the largest shaft is 20.02 mm. The maximum clearance is 0.13 mm because the largest bearing is 20.11 mm and the smallest shaft is 19.98 mm. - 4
A quality inspector measures five parts from a batch. The lengths are 49.98 mm, 50.01 mm, 50.04 mm, 49.99 mm, and 50.03 mm. The specification is 50.00 mm ±0.05 mm. How many of the parts pass inspection?
All five parts pass inspection because the acceptable range is 49.95 mm to 50.05 mm, and every measured length falls within that range. - 5
A part has a nominal mass of 125 g with a tolerance of ±2 g. A sample has a mass of 122.6 g. Does it pass? Explain your answer.
Compare the measured value with both specification limits.
The part does not pass because the acceptable range is 123 g to 127 g, and 122.6 g is below the lower limit. - 6
A manufacturer uses a go/no-go gauge to inspect a hole. The go pin is 10.00 mm and must enter the hole. The no-go pin is 10.10 mm and must not enter the hole. What does it mean if both pins enter the hole?
A no-go gauge is designed to fail when it fits.
If both pins enter the hole, the hole is too large and the part fails inspection. The go pin entering shows the hole is at least 10.00 mm, but the no-go pin entering shows the hole is larger than the allowed maximum. - 7
A process produces bolts with a target length of 40.00 mm. The specification limits are 39.90 mm and 40.10 mm. If the process average shifts from 40.00 mm to 40.08 mm, why is this a quality concern even if many parts still pass?
It is a quality concern because the process is now closer to the upper specification limit, so normal variation is more likely to produce oversized bolts. A shifted process can create more defects even before every part fails. - 8
A control chart shows that 12 points in a row are all above the centerline, but still inside the upper and lower control limits. What should a quality engineer conclude?
Control charts look for patterns as well as points outside the limits.
The engineer should conclude that the process may no longer be stable because a long run on one side of the centerline suggests a nonrandom shift. The process should be investigated even though the points are within the control limits. - 9
A part has an upper specification limit of 15.10 mm and a lower specification limit of 14.90 mm. The process standard deviation is 0.025 mm. Calculate the process capability index Cp.
Use Cp = (USL - LSL) / (6σ).
The Cp is 1.33 because Cp equals (USL - LSL) divided by 6σ, which is (15.10 - 14.90) divided by (6 × 0.025) = 0.20 divided by 0.15 = 1.33. - 10
A process has a mean of 14.98 mm, an upper specification limit of 15.10 mm, a lower specification limit of 14.90 mm, and a standard deviation of 0.025 mm. Calculate Cpk and explain what it shows.
Use the smaller of (USL - mean) / (3σ) and (mean - LSL) / (3σ).
The Cpk is 1.07 because the distance to the upper limit is 0.12 mm and the distance to the lower limit is 0.08 mm. Cpk equals the smaller distance divided by 3σ, so 0.08 divided by 0.075 = 1.07. This shows the process is capable but is not centered exactly between the limits. - 11
A factory samples 50 parts from a production run of 2,000 parts and finds 3 defective parts. Estimate the defect rate as a percent.
The estimated defect rate is 6% because 3 defective parts out of 50 sampled parts gives 3 ÷ 50 = 0.06, which is 6%. - 12
A histogram of part widths is centered near the target value, but it is very wide and extends beyond both specification limits. What does this suggest about the manufacturing process?
Think about the difference between accuracy and precision.
This suggests that the process average is near the target, but the variation is too large. The manufacturer should reduce process variation so more parts stay within the specification limits. - 13
A machine shop can either inspect every part or inspect a random sample from each batch. Give one advantage and one disadvantage of sampling inspection.
One advantage of sampling inspection is that it saves time and cost compared with inspecting every part. One disadvantage is that some defective parts may be missed because not every part is checked. - 14
A plastic clip fails inspection because its tab thickness is often below the lower specification limit. List two possible engineering actions to improve the process.
Consider both process centering and variation reduction.
Two possible actions are adjusting the mold or machine settings to increase the tab thickness and investigating sources of variation such as material shrinkage, tool wear, or temperature changes. These actions help move the process back within tolerance. - 15
A drawing shows a rectangular plate with a length of 80.0 mm ±0.2 mm and a width of 30.0 mm ±0.1 mm. A measured plate is 79.85 mm long and 30.12 mm wide. Does the plate pass inspection? Explain.
A part must meet every required dimension to pass inspection.
The plate does not pass inspection. The length passes because 79.85 mm is within the allowed range of 79.8 mm to 80.2 mm, but the width fails because 30.12 mm is above the allowed range of 29.9 mm to 30.1 mm.