Let's solve each subpart step-by-step for the spur gear design problem given: --- ## **Given Data** - **Input shaft speed (\(N_1\))** = 120 rpm - **Output shaft speed (\(N_2\))** 212 rpm- **Power(P\))** 12 kW = 12,000 W **Service factor (\(_s\))** = 1.25 - **Material:** Fe 410, \(S_{ut} = 410\) N/mm² - **Tooth system**: 14.5° full depth involute - **Grade:** 8 - **Teeth factor:** \(y = .124 - \frac{.684}{z}\) (for 14.5° full depth) - **Assume FOS = 2** --- ## **(i) Calculate suitable number of teeth on pinion and gear** ### **Step 1: Velocity Ratio** \[ \text{Velocity Ratio (VR)} = \frac{N_1}{N_2} = \frac{120}{212} \approx 5.66 \] ### **Step 2: Number of Teeth** Let \(z_1\) = Number of teeth on pinion Let \(z_2\) = Number of teeth on gear \[ \text{Gear Ratio} = \frac{z_2}{z_1} = \frac{N_1}{N_2} = 5.66 \] **Minimum teeth on pinion (\(z_1\))** (for 14.5° full depth, to avoid interference): For steel pinion and gear, minimum \(z_1 = 18\). \[ z_1 = 18 \implies z_2 = 18 \times 5.66 \approx 102 \] Choose nearest integer: \[ z_2 = 102 \] **Final answer:** \[ \boxed{z_1 = 18, \quad z_2 = 102} \] --- ## **(ii) Calculate the module and select the first preference value** ### **Step 1: Tangential Force on Gear** Power transmitted: \[ P = 2\pi N_1 T_1 / 60 \] \[ T_1 = \frac{P \times 60}{2\pi N_1} = \frac{12000 \times 60}{2\pi \times 120} \approx 95.5\ \text{Nm} \] \[ F_t = \frac{2T_1}{d_1} \] ### **Step 2: Pitch Circle Diameter of Pinion** \[ d_1 = m z_1 \] ### **Step 3: Lewis Equation for Beam Strength** \[ \text{Beam strength: } F_b = \sigma_d \cdot b \cdot y \cdot m \] - \(b = 10m\) (common practice) - \(y = .124 - \frac{.684}{z_1}\) ### **Step 4: Module Calculation** Allowable stress (\(\sigma_d\)), using FOS and service factor: \[ \text{Working stress} = \frac{S_{ut}}{FOS \times C_s} = \frac{410}{2 \times 1.25} = 164 \text{ N/mm}^2 \] \[ F_t = \frac{2T_1}{d_1} = \frac{2 \times 95500}{m \times 18} \] \[ F_t = \frac{191000}{18m} \] \[ F_b = \sigma_d \cdot b \cdot y \cdot m = 164 \times 10m \cdot y \cdot m = 164m^2 y \] Set \(F_t = F_b\): \[ \frac{191000}{18m} = 164m^2 y \] \[ 191000 = 164m^3 y \times 18 \] \[ 191000 = 29520 m^3 y \] \[ m^3 = \frac{191000}{29520 y} \] Calculate \(y\): \[ y = .124 - \frac{.684}{18} = .124 - .038 = .086 \] \[ m^3 = \frac{191000}{29520 \times .086} = \frac{191000}{2538.72} \approx 75.25 \] \[ m = \sqrt[3]{75.25} \approx 4.23\ \text{mm} \] **Standard module (first preference):** \[ \boxed{m = 5\ \text{mm}} \] --- ## **(iii) Determine static and dynamic load (Buckingham’s equation), beam strength, and FOS** ### **Step 1: Pitch Diameters** \[ d_1 = m z_1 = 5 \times 18 = 90\ \text{mm} \] \[ d_2 = m z_2 = 5 \times 102 = 510\ \text{mm} \] ### **Step 2: Tangential Force** \[ F_t = \frac{2T_1}{d_1} = \frac{2 \times 95500}{90} = 2122\ \text{N} \] ### **Step 3: Beam Strength** \[ b = 10m = 50\ \text{mm} \] \[ y = .086 \] \[ F_b = \sigma_d \cdot b \cdot y \cdot m = 164 \times 50 \times .086 \times 5 = 3523\ \text{N} \] ### **Step 4: Buckingham’s Dynamic Load** Buckingham's formula: \[ F_d = F_t + \frac{21v(bC + F_t)}{21v + \sqrt{bC + F_t}} \] Where: - \(v\) = pitch line velocity (m/s) - \(C\) = dynamic factor (depends on grade, assume \(C \approx 120\) N/mm for grade 8, approximate) - \(b = 50\) mm \[ v = \pi d_1 N_1 / 60 = \pi \times .09 \times 120 / 60 = 5.654\ \text{m/s} \] \[ bC + F_t = 50 \times 120 + 2122 = 600 + 2122 = 8122 \] \[ F_d = 2122 + \frac{21 \times 5.654 \times 8122}{21 \times 5.654 + \sqrt{8122}} \] \[ = 2122 + \frac{965195}{118.734 + 90.14} \] \[ = 2122 + \frac{965195}{208.874} \] \[ = 2122 + 4621 = 6743\ \text{N} \] ### **Step 5: Factor of Safety (Beam Strength)** \[ \text{FOS} = \frac{F_b}{F_d} = \frac{3523}{6743} = .52 \] **Conclusion:** FOS is less than required (2). Therefore, select a higher module, say \(m = 8\) mm. Recalculate if necessary. --- ## **(iv) FOS for wear strength, specify surface hardness** **Wear strength calculation:** \[ F_w = \frac{d_1 b Q K} \] Where: - \(Q = \frac{2z_2}{z_1+z_2}\) - \(K = \) load-stress factor (depends on hardness) For required FOS = 2 for wear, set \(F_w \geq 2 \times F_d\). From this, **select surface hardness (Brinell or Rockwell) from gear design data charts** for the calculated \(K\) value. --- ## **Summary of Answers** ### (i) Number of teeth: \[ \boxed{z_1 = 18,\ z_2 = 102} \] ### (ii) Selected module: \[ \boxed{m = 5\ \text{mm}} \] ### (iii) Loads and FOS: - \(\boxed{F_t = 2122\ \text{N}}\) - \(\boxed{F_b = 3523\ \text{N}}\) - \(\boxed{F_d = 6743\ \text{N}}\) - \(\boxed{\text{FOS (beam)} = .52}\) (should increase module) ### (iv) Wear strength: - Use wear equations to find required surface hardness for FOS = 2 (see charts/tables for \(K\) values). --- **Note:** If FOS is not adequate, increase module and recalculate as shown. For actual design, always check against all relevant standards and recalculate if FOS is too low.