Let's solve the questions step-by-step: --- ## **Given Data** - **Material:** Steel - **Yield Strength (\( S_y \))** = 100 ksi = 100,000 psi - **Modulus of Elasticity (\( E \))** = 30 Mpsi = 30,000,000 psi - **Coefficient of friction (\( \mu \))** = .7 - **Outer diameter of cylinder (\( D_{out} \))** = 3.5 in - **Inner diameter (hole/shaft) (\( D_{in} \))** = 2. in --- ## **a) Find the max and min allowable diameters for both the hole and shaft.** ### **Medium Drive Fit Tolerances (Reference)** A "medium drive fit" typically has a **shaft slightly larger** than the hole, with an **interference fit** (shaft is .001"–.002" larger per inch of diameter). #### **Interference Calculation** - Interference per inch: **.001" – .002"** - For 2" diameter: - **Min interference:** \( .001 \times 2 = .002" \) - **Max interference:** \( .002 \times 2 = .004" \) #### **Hole and Shaft Dimensions** - **Nominal size:** 2.000" - **Hole diameter range:** 2.000" to 1.998" (lower for interference) - **Shaft diameter range:** 2.002" to 2.004" (higher for interference) ##### **Summary Table** | Part | Min Diameter (in) | Max Diameter (in) | |--------|-------------------|-------------------| | Hole | 1.998 | 2.000 | | Shaft | 2.002 | 2.004 | --- ## **b) Find the min and max torque that can be transmitted.** ### **Interference Fit and Torque Transmission** The maximum torque (\( T \)) transmitted by the interference fit (friction fit) is: \[ T = \mu \cdot F_N \cdot r \] where: - \( \mu \) = coefficient of friction = .7 - \( F_N \) = normal force (from interference) - \( r \) = shaft radius = 1 in #### **Calculate the Normal Force (\( F_N \))** - **Contact pressure (\( p \))** from interference: \[ p = \frac{\delta \cdot E}{D} \] Where \( \delta \) = interference, \( E \) = modulus, \( D \) = diameter. - **Force on interface:** \[ F_N = p \cdot A = p \cdot (\pi D L) \] Assume length \( L \) (not given, so keep as \( L \)). #### **Max Pressure (Yield Limit)** - Max allowable pressure: Don't exceed yield strength. - Set \( p_{max} = S_y = 100,000 \) psi (conservative). #### **Calculate \( F_{N,max} \):** \[ F_{N,max} = S_y \cdot \pi D L \] \[ T_{max} = \mu \cdot F_{N,max} \cdot r = .7 \cdot S_y \cdot \pi D L \cdot r \] Plug in values: \[ T_{max} = .7 \cdot 100{,}000 \cdot \pi \cdot 2 \cdot L \cdot 1 \] \[ T_{max} = .7 \cdot 100{,}000 \cdot 2\pi L \] \[ T_{max} = .7 \cdot 200{,}000 \cdot \pi L \] \[ T_{max} = 140{,}000 \pi L \quad \text{(in-lbf, as a function of length \( L \))} \] #### **Min Torque (min interference):** For minimum interference (\( \delta_{min} = .002" \)): \[ p_{min} = \frac{\delta_{min} \cdot E}{D} = \frac{.002 \times 30,000,000}{2} = 30,000 \text{ psi} \] \[ F_{N,min} = 30,000 \cdot \pi \cdot 2 \cdot L = 60,000 \pi L \] \[ T_{min} = .7 \cdot 60,000 \cdot \pi L = 42,000 \pi L \quad \text{(in-lbf)} \] ##### **Final Answers (in terms of L):** \[ \boxed{ T_{min} = 42,000 \pi L \quad \text{in-lbf} } \] \[ \boxed{ T_{max} = 140,000 \pi L \quad \text{in-lbf} } \] Where \( L \) = length of the fit (in inches). --- ## **c) If you need to incorporate these parts into a larger analysis, which dimensions would you use for each part and why?** ### **Answer/Explanation:** - **For the shaft:** Use the **minimum allowable diameter** (2.002") for strength and fit calculations, as it is the most conservative for stress and deflection analysis. - **For the hole:** Use the **maximum allowable diameter** (2.000") for assembly and tolerance stack-up, as it represents the loosest fit. - **For interference fit calculations:** Use **both extremes** (max and min) to analyze the range of possible fits, ensuring reliability for worst-case scenarios. - **For stress/strain analysis:** Use the **smaller shaft diameter** (to ensure safety against yield). **Why?** Using the extremes ensures you consider the worst-case for both assembly and performance (ensuring the parts always fit and transmit the required torque without slipping or yielding). --- ## **Summary Table** | Parameter | Value/Range | |-----------------------|-------------------------| | **Hole Diameter** | 1.998" – 2.000" | | **Shaft Diameter** | 2.002" – 2.004" | | **Min Torque** | \( 42,000 \pi L \) in-lbf| | **Max Torque** | \( 140,000 \pi L \) in-lbf| | **Use for Analysis** | Min shaft, max hole diam| --- **If you have a value for length \( L \), plug it in to get the actual torque values! Let me know if you'd like to see worked examples with a specific length.**