Let's solve step by step based on the information provided. ## **Given Data** - **Module, $m$ = 12 mm** - **Pressure angle, $\phi$ = 20°** - **Gear 2 (on shaft a)** - Rotates at $n_2 = 180$ rpm (clockwise) - Delivers $P = 150$ kW (through idler to gear 5) - **Teeth:** - $N_2 = 18$ - $N_3 = 54$ - $N_4 = 18$ - $N_5 = 54$ --- ## **1. Number of Teeth, $N$** Already provided: - Gear 2: $N_2 = 18$ - Gear 3: $N_3 = 54$ - Gear 4: $N_4 = 18$ - Gear 5: $N_5 = 54$ --- ## **2. Pitch Diameter, $d$** \[ d = m \cdot N \] - Gear 2: $d_2 = 12 \times 18 = 216$ mm - Gear 3: $d_3 = 12 \times 54 = 648$ mm - Gear 4: $d_4 = 12 \times 18 = 216$ mm - Gear 5: $d_5 = 12 \times 54 = 648$ mm --- ## **3. Outside Diameter, $d_o$** \[ d_o = m (N + 2) \] - Gear 2: $d_{o2} = 12 \times (18 + 2) = 12 \times 20 = 240$ mm - Gear 3: $d_{o3} = 12 \times (54 + 2) = 12 \times 56 = 672$ mm - Gear 4: $d_{o4} = 12 \times (18 + 2) = 240$ mm - Gear 5: $d_{o5} = 12 \times (54 + 2) = 672$ mm --- ## **4. Base Diameter, $d_b$** \[ d_b = d \cdot \cos\phi \] Where $\cos(20^\circ) \approx .9397$ - Gear 2: $d_{b2} = 216 \times .9397 = 203.98$ mm - Gear 3: $d_{b3} = 648 \times .9397 = 608.95$ mm - Gear 4: $d_{b4} = 216 \times .9397 = 203.98$ mm - Gear 5: $d_{b5} = 648 \times .9397 = 608.95$ mm --- ## **5. Inside Diameter, $d_i$** For standard gears (no internal gears), inside diameter is usually the **root diameter**: \[ d_i = m (N - 2.5) \] - Gear 2: $d_{i2} = 12 \times (18 - 2.5) = 12 \times 15.5 = 186$ mm - Gear 3: $d_{i3} = 12 \times (54 - 2.5) = 12 \times 51.5 = 618$ mm - Gear 4: $d_{i4} = 12 \times (18 - 2.5) = 186$ mm - Gear 5: $d_{i5} = 12 \times (54 - 2.5) = 618$ mm --- ## **6. Rotational Speeds, $n$** For meshing gears: \[ N_2 n_2 = N_3 n_3 \] \[ n_3 = \frac{N_2}{N_3} n_2 = \frac{18}{54} \times 180 = 600 \text{ rpm} \] Similarly, Gear 3 and Gear 4 are on the same shaft, so: \[ n_3 = n_4 = 600 \text{ rpm} \] For Gear 5: \[ N_4 n_4 = N_5 n_5 \implies n_5 = \frac{N_4}{N_5} n_4 = \frac{18}{54} \times 600 = 200 \text{ rpm} \] --- ## **7. Torque, $T$** \[ P = 2\pi n T / 60 \implies T = \frac{P \times 60}{2\pi n} \] $P = 150,000$ W - **Gear 2:** \[ T_2 = \frac{150,000 \times 60}{2\pi \times 180} = \frac{9,000,000}{11,309.73} \approx 796.3 \text{ Nm} \] - **Gear 3 (same power):** \[ T_3 = \frac{150,000 \times 60}{2\pi \times 600} = \frac{9,000,000}{3,769.91} \approx 2,387.6 \text{ Nm} \] - **Gear 4 (same shaft as 3):** \[ T_4 = T_3 = 2,387.6 \text{ Nm} \] - **Gear 5:** \[ T_5 = \frac{150,000 \times 60}{2\pi \times 200} = \frac{9,000,000}{1,256.64} \approx 7,166.9 \text{ Nm} \] --- ## **Summary Table** | | Gear 2 | Gear 3 | Gear 4 | Gear 5 | |---|---|---|---|---| | Rotational speed $n$ (rpm) | 180 | 600 | 600 | 200 | | Torque $T$ (Nm) | 796.3 | 2,387.6 | 2,387.6 | 7,166.9 | | Number of teeth $N$ | 18 | 54 | 18 | 54 | | Pitch diameter $d$ (mm) | 216 | 648 | 216 | 648 | | Outside diameter $d_o$ (mm) | 240 | 672 | 240 | 672 | | Inside diameter $d_i$ (mm) | 186 | 618 | 186 | 618 | | Base diameter $d_b$ (mm) | 204. | 609. | 204. | 609. | --- ## **Free Body Diagram (FBD) Note** - Draw forces at the points of gear contact (tangential force, radial force). - Show torques on shafts. - Indicate direction of rotation. --- ### **Final Answers Provided in the Table Above**

# Gear Analysis Solution ## **Given Data** - **Module, \( m \)**: 12 mm - **Pressure angle, \( \phi \)**: 20° - **Gear 2**: - Rotational speed, \( n_2 \): 1800 rpm (clockwise) - Power transmitted, \( P \): 150 kW - **Teeth**: - \( N_2 = 18 \) - \( N_3 = 54 \) - \( N_4 = 18 \) - \( N_5 = 54 \) --- ## **1. Number of Teeth, \( N \)** - **Gear 2**: \( N_2 = 18 \) - **Gear 3**: \( N_3 = 54 \) - **Gear 4**: \( N_4 = 18 \) - **Gear 5**: \( N_5 = 54 \) --- ## **2. Pitch Diameter, \( d \)** \[ d = m \cdot N \] - **Gear 2**: \( d_2 = 12 \times 18 = 216 \) mm - **Gear 3**: \( d_3 = 12 \times 54 = 648 \) mm - **Gear 4**: \( d_4 = 12 \times 18 = 216 \) mm - **Gear 5**: \( d_5 = 12 \times 54 = 648 \) mm --- ## **3. Outside Diameter, \( d_o \)** \[ d_o = m \cdot (N + 2) \] - **Gear 2**: \( d_{o2} = 12 \times (18 + 2) = 240 \) mm - **Gear 3**: \( d_{o3} = 12 \times (54 + 2) = 672 \) mm - **Gear 4**: \( d_{o4} = 240 \) mm - **Gear 5**: \( d_{o5} = 672 \) mm --- ## **4. Base Diameter, \( d_b \)** \[ d_b = d \cdot \cos(\phi) \] Where \( \cos(20^\circ) \approx 0.9397 \) - **Gear 2**: \( d_{b2} = 216 \times 0.9397 \approx 203.98 \) mm - **Gear 3**: \( d_{b3} = 648 \times 0.9397 \approx 608.95 \) mm - **Gear 4**: \( d_{b4} = 203.98 \) mm - **Gear 5**: \( d_{b5} = 608.95 \) mm --- ## **5. Inside Diameter, \( d_i \)** \[ d_i = m \cdot (N - 2.5) \] - **Gear 2**: \( d_{i2} = 12 \times (18 - 2.5) = 186 \) mm - **Gear 3**: \( d_{i3} = 12 \times (54 - 2.5) = 618 \) mm - **Gear 4**: \( d_{i4} = 186 \) mm - **Gear 5**: \( d_{i5} = 618 \) mm --- ## **6. Rotational Speeds, \( n \)** For meshing gears: \[ N_2 n_2 = N_3 n_3 \implies n_3 = \frac{N_2}{N_3} n_2 = \frac{18}{54} \times 1800 = 600 \text{ rpm} \] Since Gears 3 and 4 are on the same shaft: \[ n_4 = n_3 = 600 \text{ rpm} \] For Gear 5: \[ N_4 n_4 = N_5 n_5 \implies n_5 = \frac{N_4}{N_5} n_4 = \frac{18}{54} \times 600 = 200 \text{ rpm} \] --- ## **7. Torque, \( T \)** \[ P = \frac{2 \pi n T}{60} \implies T = \frac{P \times 60}{2 \pi n} \] **Power \( P = 150,000 \) W** - **Gear 2**: \[ T_2 = \frac{150,000 \times 60}{2\pi \times 1800} \approx 796.3 \text{ Nm} \] - **Gear 3**: \[ T_3 = \frac{150,000 \times 60}{2\pi \times 600} \approx 2,387.6 \text{ Nm} \] - **Gear 4**: \[ T_4 = T_3 = 2,387.6 \text{ Nm} \] - **Gear 5**: \[ T_5 = \frac{150,000 \times 60}{2\pi \times 200} \approx 7,166.9 \text{ Nm} \] --- ## **Summary Table** | | Gear 2 | Gear 3 | Gear 4 | Gear 5 | |---|---|---|---|---| | Rotational speed \( n \) (rpm) | 1800 | 600 | 600 | 200 | | Torque \( T \) (Nm) | 796.3 | 2,387.6 | 2,387.6 | 7,166.9 | | Number of teeth \( N \) | 18 | 54 | 18 | 54 | | Pitch diameter \( d \) (mm) | 216 | 648 | 216 | 648 | | Outside diameter \( d_o \) (mm) | 240 | 672 | 240 | 672 | | Inside diameter \( d_i \) (mm) | 186 | 618 | 186 | 618 | | Base diameter \( d_b \) (mm) | 203.98 | 608.95 | 203.98 | 608.95 | --- ## **Free Body Diagram (FBD) Note** - Include forces at contact points (tangential, radial). - Show torques on shafts. - Indicate direction of rotation. --- ### **Final Answers Provided in the Summary Table**