Here's a **breakdown of the information and steps to solve the problem** shown in the image: ### Given: - **Fit:** 40 mm H9g10 (hole-basis system) - **Basic size range:** 38 to 50 mm - **Find:** Upper Limit (UL) and Lower Limit (LL) of the shaft ### Tolerance Unit Formula: \[ i = 0.45 \sqrt[3]{D} + (0.001 \times D) \quad (\text{in}~\mu\text{m}) \] #### Where: - \( D \) = Basic size (here, 40 mm) ### Tolerance Values: - For IT9 = 40i - For IT10 = 64i ### Fundamental deviation for shaft (g): \[ 2.5 D^{0.34} \quad (\text{in}~\mu\text{m}) \] ### Fundamental deviation for hole: Zero --- Let's **calculate step by step**: --- #### 1. **Calculate i (Tolerance Unit):** \[ i = 0.45 \sqrt[3]{40} + (0.001 \times 40) \] \[ \sqrt[3]{40} \approx 3.419 \] \[ i = 0.45 \times 3.419 + 0.04 = 1.5386 + 0.04 = 1.5786 \ (\mu \text{m}) \] --- #### 2. **Calculate IT9 and IT10:** \[ \text{IT9} = 40i = 40 \times 1.5786 = 63.144 \ (\mu \text{m}) = 0.0631 \ (\text{mm}) \] \[ \text{IT10} = 64i = 64 \times 1.5786 = 101.0304 \ (\mu \text{m}) = 0.1010 \ (\text{mm}) \] --- #### 3. **Calculate Fundamental Deviation for Shaft (g):** \[ 2.5 D^{0.34} \] \[ D = 40, \quad D^{0.34} = 40^{0.34} \approx 3.754 \] \[ \text{Deviation (g)} = 2.5 \times 3.754 = 9.385 (\mu \text{m}) = 0.00939 \text{ mm} \] --- #### 4. **Limits for Shaft (g10):** - **Lower limit** of shaft = Basic size - (fundamental deviation + IT10) - **Upper limit** of shaft = Basic size - fundamental deviation \[ \text{Lower limit (LL)} = 40 - 0.00939 - 0.1010 = 39.8896 \text{ mm} \] \[ \text{Upper limit (UL)} = 40 - 0.00939 = 39.9906 \text{ mm} \] --- ### **Final Answers:** - **Upper Limit (UL) of shaft:** **39.9906 mm** - **Lower Limit (LL) of shaft:** **39.8896 mm** --- **Summary Table:** | Limit Type | Value (mm) | |----------------|---------------| | Shaft UL (g10) | 39.9906 | | Shaft LL (g10) | 39.8896 | Let me know if you need the limits for the hole as well!