# Bevel Gear Design Problem: Step-by-Step Solution ## Problem Summary - **Type:** 20° full-depth involute bevel gears - **Shafts:** Connect at right angles (90°) - **Velocity Ratio:** 2.5:1 (Pinion : Gear) - **Materials:** - Pinion: Forged steel - Gear: Cast steel - **Power Transmitted:** 37.5 kW - **Pinion Speed (N₁):** 750 rpm - **Tasks:** 1. Design the gear drive (determine main dimensions, module, teeth, etc.) 2. Check for dynamic and wear loads. --- ## 1. **Find the Number of Teeth and Gear Ratio** ### a. **Velocity Ratio** \[ \text{Velocity Ratio (VR)} = \frac{\text{Speed of Pinion}}{\text{Speed of Gear}} = \frac{N_1}{N_2} = 2.5:1 \] \[ N_1 = 750 \text{ rpm} \implies N_2 = \frac{750}{2.5} = 300 \text{ rpm} \] ### b. **Number of Teeth** Let \( z_1 \) = Number of pinion teeth Let \( z_2 \) = Number of gear teeth \[ \frac{z_2}{z_1} = \text{Gear Ratio} = 2.5 \implies z_2 = 2.5 \, z_1 \] Select \( z_1 \) to avoid undercutting. For 20° full-depth teeth, minimum \( z_1 = 18 \) (practical, can use 20 or more). Let’s **choose \( z_1 = 20 \):** \[ z_2 = 2.5 \times 20 = 50 \] --- ## 2. **Pitch Cone Angles (Bevel Gears at 90°)** \[ \text{For shafts at 90°:} \] \[ \theta_1 = \tan^{-1} \left( \frac{z_1}{z_2} \right) = \tan^{-1} \left( \frac{20}{50} \right) = \tan^{-1} (0.4) \approx 21.8^\circ \] \[ \theta_2 = \tan^{-1} \left( \frac{z_2}{z_1} \right) = \tan^{-1} (2.5) \approx 68.2^\circ \] --- ## 3. **Calculate Pitch Diameters** \[ \text{Let } m = \text{module} \] \[ d_1 = m \cdot z_1 \] \[ d_2 = m \cdot z_2 \] --- ## 4. **Power and Tangential Force** ### a. **Torque on Pinion** \[ P = 2\pi N_1 T_1 / 60 \implies T_1 = \frac{P \times 60}{2\pi N_1} \] \[ T_1 = \frac{37,500 \times 60}{2\pi \times 750} = \frac{2,250,000}{4712.39} \approx 477.73 \text{ Nm} \] ### b. **Tangential Force at Pinion Pitch Circle** \[ F_t = \frac{2T_1}{d_1} \] But \( d_1 = m \times 20 \). So, \[ F_t = \frac{2 \times 477,730}{m \times 20} = \frac{955,460}{20m} = \frac{47,773}{m} \text{ N} \] --- ## 5. **Module Selection (Strength Equation)** For bevel gears, the **Lewis equation** (approximate for external gears): \[ F_t = \sigma_d \cdot b \cdot Y \cdot m \cdot K_v \] Where: - \( \sigma_d \) = allowable bending stress (based on weaker material) - \( b \) = face width (\( b = 10m \) recommended for bevel gears; max \( b = 1/3 \) of cone distance) - \( Y \) = Lewis form factor (for 20 teeth, 20°: \( Y \approx 0.322 \)) - \( K_v \) = velocity factor #### a. **Allowable Stress** Assume allowable bending stress (\( \sigma_{d} \)) for **cast steel** (gear) = 140 MPa (from data books) For **forged steel** (pinion) = 200 MPa Pinion is stronger. **Gear is weaker** → use gear's value. #### b. **Cone Distance (Mean Pitch Diameter)** \[ \text{Cone distance } (L) = \sqrt{(d_1/2)^2 + (d_2/2)^2} = \frac{m}{2} \sqrt{z_1^2 + z_2^2} \] \[ = \frac{m}{2} \sqrt{20^2 + 50^2} = \frac{m}{2} \sqrt{400 + 2500} = \frac{m}{2} \sqrt{2900} = \frac{m}{2} \times 53.85 = 26.93 m \] #### c. **Face Width** \[ b = 10m \qquad (\text{but } b \leq \frac{1}{3} L = \frac{1}{3} \times 26.93m = 8.98m) \] So, **limit \( b \) to \( 8m \)** (rounded). #### d. **Lewis Form Factor \( Y \)** For 20° pressure angle and \( z = 20 \): \[ Y = 0.154 - \frac{0.912}{z} = 0.154 - \frac{0.912}{20} = 0.154 - 0.0456 = 0.1084 \] #### e. **Velocity Factor \( K_v \)** Pitch line velocity, \[ v = \frac{\pi d_1 N_1}{60 \times 1000} \quad \text{(in m/s)} = \frac{\pi \cdot (m \cdot 20) \cdot 750}{60 \times 1000} = \frac{\pi \cdot m \cdot 15,000}{60,000} = \frac{\pi m \cdot 15,000}{60,000} = \frac{\pi m \cdot 15}{60} = \frac{\pi m}{4} \] Let’s try \( m = 5 \) mm first: \[ v = \frac{\pi \cdot 5}{4} \approx 3.93 \text{ m/s} \] Velocity factor (Barth’s formula): \[ K_v = \frac{3}{3 + v} \] \[ K_v = \frac{3}{3 + 3.93} = \frac{3}{6.93} = 0.433 \] #### f. **Calculate Required Module** Rewriting Lewis equation: \[ F_t = \sigma_d \cdot b \cdot Y \cdot m \cdot K_v \] \[ \frac{47,773}{m} = 140 \times 8m \times 0.1084 \times 0.433 \] \[ = 140 \times 8 \times 0.1084 \times 0.433 \times m \] \[ = 140 \times 8 = 1,120 \] \[ 1,120 \times 0.1084 = 121.408 \] \[ 121.408 \times 0.433 = 52.56 \] So, \[ \frac{47,773}{m} = 52.56 m \] \[ \Rightarrow 47,773 = 52.56 m^2 \] \[ m^2 = \frac{47,773}{52.56} = 909.76 \] \[ m = \sqrt{909.76} \approx 30.2 \text{ mm} \] This is **very large**. Let's check calculations. #### g. **Check for Calculation Error** Let’s check face width: - \( b = 8m \), so for \( m = 5 \) mm, \( b = 40 \) mm, which is less than \( 1/3 \) of cone distance (26.93m ≈ 134 mm). OK. Let’s try \( m = 8 \) mm: \[ v = \frac{\pi \cdot 8}{4} = 6.28 \text{ m/s} \] \[ K_v = \frac{3}{3 + 6.28} = \frac{3}{9.28} = 0.323 \] \[ F_t = 140 \times 64 \times 0.1084 \times 0.323 \times 8 \] But \( b = 8m = 64 \) mm. \[ \sigma_d = 140 \text{ MPa} \] \[ b = 64 \text{ mm} \] \[ Y = 0.1084 \] \[ K_v = 0.323 \] \[ m = 8 \text{ mm} \] \[ F_t = 140 \times 64 \times 0.1084 \times 0.323 = 8,960 \times 0.1084 \times 0.323 = 971.264 \times 0.323 = 313.77 \] Multiply by \( m \): \[ 313.77 \times 8 = 2,510 \text{ N} \] \[ F_t = \frac{47,773}{8} = 5,972 \text{ N} \] **Required \( F_t \) (power) is much higher than strength.** Try **larger module**: Let’s try \( m = 15 \) mm: - \( d_1 = 300 \text{ mm} \) - \( b = 8m = 120 \text{ mm} \) - \( v = \frac{\pi \cdot 15}{4} = 11.78 \text{ m/s} \) - \( K_v = \frac{3}{3 + 11.78} = 0.203 \) \[ F_t = 140 \times 120 \times 0.1084 \times 0.203 = 16,800 \times 0.1084 \times 0.203 = 1,822.12 \times 0.203 = 370 \text{ N} \] \[ 370 \times 15 = 5,550 \text{ N} \] \[ F_t = \frac{47,773}{15} = 3,185 \text{ N} \] Still not matching; perhaps our face width is too small. Let’s use the **maximum face width** (\( b = 1/3 L \)): \[ L = 26.93 m \] For \( m = 15 \) mm: \[ L = 26.93 \times 15 = 404 \text{ mm} \] \[ b = \frac{404}{3} = 134.7 \text{ mm} \] Now, \[ F_t = 140 \times 134.7 \times 0.1084 \times 0.203 \times 15 = (18,858 \times 0.1084 \times 0.203) \times 15 = (2,044.5 \times 0.203) \times 15 = (415 \times 15) = 6,225 \text{ N} \] Now, \[ F_t = \frac{47,773}{15} = 3,185 \text{ N} \] **Now, strength is sufficient.** --- ## 6. **Summary of Gear Dimensions** - **Module \( m \):** 15 mm - **Pinion Teeth \( z_1 \):** 20 - **Gear Teeth \( z_2 \):** 50 - **Pitch Diameters:** - \( d_1 = 15 \times 20 = 300 \) mm - \( d_2 = 15 \times 50 = 750 \) mm - **Face Width \( b \):** Max \( = \frac{1}{3} L = 134.7 \) mm - **Pitch Cone Angles:** - Pinion: \( 21.8^\circ \) - Gear: \( 68.2^\circ \) - **Cone Distance:** \( L = 404 \) mm --- ## 7. **Dynamic Load Check** **Dynamic load (Buckingham’s equation):** \[ F_d = F_t + \frac{21 v (F_t + b C)}{21 v + \sqrt{F_t + b C}} \] Where: - \( C \) = deformation factor (from tables, for steel, \( C \approx 118 \) N/mm for cut gears) - \( v \) = pitch line velocity at pinion \