--- title: "Limits of a 40 H7/g6 hole-shaft fit" subject: "Metrology" chapter: 9 key: "semester_3/metrology:fit-40-h7-g6-calculation" source: https://www.bctengineer.com/q/semester_3/metrology/fit-40-h7-g6-calculation site: BCT Engineering --- # Limits of a 40 H7/g6 hole-shaft fit Metrology · Chapter 9: Limits, Fits and Tolerance - Asked in: year not given, 8 marks - Video: [Mechanical Measurement and Metrology Tolerance Limits fit problem1 - MMM](https://www.youtube.com/watch?v=1BrUPNt9yMg) — Swaraj Lewis - Classes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/metrology/fit-40-h7-g6-calculation - Whole chapter: https://www.bctengineer.com/md/semester_3/metrology/chapter-9 ## Question For a 40 mm H7/g6 fit calculate the limits of the hole and the shaft, the tolerances, the maximum and minimum clearance, and state the type of fit. Use the standard tolerance unit $i = 0.45\sqrt[3]{D} + 0.001D$ ($\mu$m, $D$ in mm), IT6 $= 10i$, IT7 $= 16i$ and the fundamental deviation for shaft 'g' $= -2.5D^{0.34}$ ($\mu$m); $D$ is the geometric mean of the diameter step 30 to 50 mm. Round tolerances and deviation to the nearest standard value. ## Short answer Using D=38.73 mm, IT7=25 µm, IT6=16 µm, shaft deviation -9 µm: hole 40.000–40.025 mm, shaft 39.975–39.991 mm; maximum clearance 0.050 mm, minimum 0.009 mm; clearance fit. ## Answer #### Step 1: diameter step $$ D = \sqrt{30 \times 50} = 38.73\ \text{mm} $$ #### Step 2: tolerance unit $$ i = 0.45\,(38.73)^{1/3} + 0.001(38.73) = 0.45(3.383) + 0.039 = 1.561\ \mu\text{m} $$ - IT7 $= 16i = 24.98\ \mu$m, standard value **25 $\mu$m** - IT6 $= 10i = 15.6\ \mu$m, standard value **16 $\mu$m** #### Step 3: fundamental deviation of shaft g $$ es = -2.5\,D^{0.34} = -2.5(38.73)^{0.34} = -2.5(3.468) = -8.67\ \mu\text{m} \approx -9\ \mu\text{m} $$ #### Step 4: limits **Hole 40 H7** (lower deviation zero): - Lower limit $= 40.000$ mm - Upper limit $= 40.000 + 0.025 = 40.025$ mm - Tolerance $= 0.025$ mm **Shaft 40 g6** (upper deviation $-0.009$ mm): - Upper limit $= 40 - 0.009 = 39.991$ mm - Lower limit $= 39.991 - 0.016 = 39.975$ mm - Tolerance $= 0.016$ mm #### Step 5: clearances $$ C_{max} = \text{hole max} - \text{shaft min} = 40.025 - 39.975 = 0.050\ \text{mm} $$ $$ C_{min} = \text{hole min} - \text{shaft max} = 40.000 - 39.991 = 0.009\ \text{mm} $$ Check: $C_{max} - C_{min} = 0.041 = 0.025 + 0.016$. Both clearances are positive, so this is a **clearance fit** (close running or sliding fit). **Answer: hole 40.000 to 40.025 mm; shaft 39.975 to 39.991 mm; $C_{max}$ = 0.050 mm, $C_{min}$ = 0.009 mm; clearance fit.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What D is used in the tolerance unit formula for the 30 to 50 mm step? A. Arithmetic mean 40 mm B. Geometric mean sqrt(30 x 50) = 38.73 mm C. Difference 20 mm D. Upper limit 50 mm Answer: B. Standard tolerance formulae use the geometric mean of the diameter step limits. (Key point: Diameter step) 2. With i = 1.561 micrometre, what are the standard IT7 and IT6 values used? A. IT7 = 16 micrometre, IT6 = 25 micrometre B. IT7 = 15.6 micrometre, IT6 = 24.98 micrometre C. IT7 = 25 micrometre, IT6 = 16 micrometre D. IT7 = 10 micrometre, IT6 = 16 micrometre Answer: C. IT7 = 16i = 24.98 rounds to 25; IT6 = 10i = 15.6 rounds to 16. (Key point: Standard tolerances) 3. What are the limits of the 40 g6 shaft (es = -9 micrometre, IT6 = 16 micrometre)? A. 40.000 to 40.016 mm B. 39.991 to 40.007 mm C. 39.984 to 40.000 mm D. 39.975 to 39.991 mm Answer: D. Upper limit = 40 - 0.009 = 39.991; lower limit = 39.991 - 0.016 = 39.975. (Key point: Shaft limits) 4. What are Cmax, Cmin and the fit type? A. Cmax = 0.050 mm, Cmin = 0.009 mm; clearance fit B. Cmax = 0.041 mm, Cmin = 0.009 mm; transition fit C. Cmax = 0.050 mm, Cmin = -0.009 mm; interference fit D. Cmax = 0.025 mm, Cmin = 0.016 mm; clearance fit Answer: A. Cmax = 40.025 - 39.975 = 0.050; Cmin = 40.000 - 39.991 = 0.009. Both positive means clearance. (Key point: Clearances and fit type) ## Related in Metrology - [Basic terms of limits, fits and tolerances](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Alimits-fits-terminology) — Metrology, chapter 9 - [Classify fits from given limits of size](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Aclassify-fits-from-limits) — Metrology, chapter 9 - [Hole basis and shaft basis systems](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Ahole-basis-vs-shaft-basis) — Metrology, chapter 9 - [Clearance, transition and interference fits](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Atypes-of-fits-with-examples) — Metrology, chapter 9 - [Unilateral and bilateral tolerance conversion](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Aunilateral-bilateral-tolerance) — Metrology, chapter 9 ## Related in other subjects - [Assembly constraints and tolerance fit calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fcomputer_aided_design%3Aassembly-constraints-tolerance-fit) — Computer Aided Design, chapter 5 - [Limits, fits and tolerance calculation](https://www.bctengineer.com/question.md?key=semester_4%2Fmanufacturing_and_production_processes%3Alimits-fits-tolerance-numerical) — Manufacturing and Production Processes, chapter 3 - [Solid shaft 120 Nm: diameter from limits](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Asolid-shaft-120nm-twist-limit) — Strength of Materials, chapter 6, asked 2× - [Minimum spacing for 30 mm conductors (m = 0.95)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acorona-spacing-30mm-0-95) — High Voltage Engineering, chapter 5, asked 4×