Let's break down and solve **each part step-by-step**. We are given: - Shaft diameter \( d = 1.5625 \) in - Key size: \( 1'' \) long × \( .375'' \) square - Shaft speed: \( 500 \) rpm - Materials: - Key: 102 rolled steel - Shaft: 413 steel, quenched and tempered at 120°F - Factor of Safety (FoS): 3 - Motors available: in 1 hp increments We'll use standard mechanical properties (from Shigley's or similar): - 102 cold rolled steel: - Yield strength (\( S_{y,\text{key}} \)) ≈ **57 ksi** - Shear yield strength (\( S_{sy,\text{key}} \)) ≈ \( .577S_{y} \) ≈ **32.9 ksi** - 413 steel (shaft): Not needed for key design here. --- ## **Part B: Shear Failure Force (F_shear)** Key fails in shear at: \[ F_{\text{shear}} = \tau_{\text{max}} \cdot A_{\text{shear}} \] Where: - \(\tau_{\text{max}} = \frac{S_{sy}}{\text{FoS}}\) - \(A_{\text{shear}} = \) area sheared = width × length = \(.375'' \times 1'' = .375~\text{in}^2\) Plug in: \[ \tau_{\text{max}} = \frac{32,900~\text{psi}}{3} = 10,967~\text{psi} \] \[ F_{\text{shear}} = 10,967~\text{psi} \times .375~\text{in}^2 = 4,112~\text{lbf} \] **Final answer:** \[ \boxed{4,110~\text{lbf}} \] --- ## **Part C: Bearing (Crushing) Failure Force (F_bearing)** Key fails in bearing (crushing) at: \[ F_{\text{bearing}} = \sigma_{\text{max}} \cdot A_{\text{bearing}} \] Where: - \(\sigma_{\text{max}} = \frac{S_{y}}{\text{FoS}}\) - \(A_{\text{bearing}} = \) projected area = height × length = \(.5 \times .375'' \times 1'' = .5625~\text{in}^2\) - For a standard key, the height in contact is about half the key height. Plug in: \[ \sigma_{\text{max}} = \frac{57,000~\text{psi}}{3} = 19,000~\text{psi} \] \[ F_{\text{bearing}} = 19,000~\text{psi} \times .5625~\text{in}^2 = 10,688~\text{lbf} \] **Final answer:** \[ \boxed{10,700~\text{lbf}} \] --- ## **Part D: Torque Causing Key Failure** The **limiting force** is shear (from B), since it's less than the bearing limit. The torque the key can transmit: \[ T = F \cdot r \] Where: - \(F = 4,110~\text{lbf}\) (from above, shear failure) - \(r = \frac{d}{2} = \frac{1.5625}{2} = .78125~\text{in}\) \[ T = 4,110~\text{lbf} \times .78125~\text{in} = 3,211~\text{in-lbf} \] **Final answer:** \[ \boxed{3,210~\text{in-lbf}} \] --- ## **Part E: Maximum Horsepower Motor** Power (hp) is related to torque and rpm: \[ P~(\text{hp}) = \frac{T~(\text{in-lbf}) \times N~(\text{rpm})}{63025} \] \[ P = \frac{3,210 \times 500}{63,025} = 25.5~\text{hp} \] **Motors are only available in 1 hp increments**, so the maximum is **25 hp**. **Final answer:** \[ \boxed{25~\text{hp}} \] --- ## **Summary Table** | Part | Value | Units | |------|--------------|------------| | B | 4,110 | lbf | | C | 10,700 | lbf | | D | 3,210 | in-lbf | | E | 25 | hp | --- ### **Assumptions & Notes** - Key height in contact is half the key for bearing stress. - Used yield strength for bearing, shear yield for shear. - Factor of Safety applied to material strength. - Key is the limiting component, not the shaft. Let me know if you need formulas explained or a step justified!