Subjects machine design

Bevel Gear Forces

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Bevel Gear Forces


1. **Problem Statement:** We have a bevel gearset with shaft AB rotating at 600 rev/min transmitting 10 hp, and gears with a 20° pressure angle. We need to find: (a) The bevel angle $\gamma$ for the gear on shaft AB. (b) The pitch-line velocity. (c) The tangential, radial, and axial forces acting on the pinion. 2. **Given Data:** - Speed of shaft AB, $N_1 = 600$ rev/min - Power transmitted, $P = 10$ hp - Pressure angle, $\phi = 20^\circ$ - Dimensions from the figure (in inches): - Shaft D length segments: 6.50 in total, with 2.50 in and 3.63 in distances to gear center E - Shaft AB length segments: 3.88 in and 3 in - Diameters: 1.13 in (shaft D bottom), 0.88 in (shaft AB) 3. **Step (a): Determine bevel angle $\gamma$ for gear on shaft AB** - The bevel angle $\gamma$ is the angle between the gear face and the shaft axis. - For bevel gears, $\tan \gamma = \frac{r_2}{r_1}$ where $r_1$ and $r_2$ are pitch radii of the two gears. - From the drawing, the shafts are at right angles, so the bevel angle for shaft AB gear is the angle between shaft AB and the gear face. - Using the given distances, the bevel angle $\gamma$ can be found by geometry: $$\gamma = \tan^{-1}\left(\frac{3.63}{3.88 + 3}\right) = \tan^{-1}\left(\frac{3.63}{6.88}\right)$$ Calculate: $$\gamma = \tan^{-1}(0.5279) \approx 27.8^\circ$$ 4. **Step (b): Determine pitch-line velocity $v$** - Power $P$ in hp can be converted to watts: $P = 10 \times 745.7 = 7457$ W - Speed $N_1 = 600$ rev/min - Torque $T$ on shaft AB: $$T = \frac{P \times 60}{2 \pi N_1} = \frac{7457 \times 60}{2 \pi \times 600}$$ Calculate: $$T = \frac{447420}{3769.91} \approx 118.7 \text{ Nm}$$ - Pitch diameter $d_1$ of pinion (shaft AB gear) is approximately the diameter at gear center, say $d_1 = 2 \times r_1$. - From figure, approximate pitch radius $r_1 = 3.88 + 3 = 6.88$ in = $6.88 \times 0.0254 = 0.1747$ m - So pitch diameter $d_1 = 2 \times 0.1747 = 0.3494$ m - Pitch line velocity: $$v = \pi d_1 N_1 / 60 = \pi \times 0.3494 \times 600 / 60 = \pi \times 0.3494 \times 10 = 10.97 \text{ m/s}$$ 5. **Step (c): Determine tangential, radial, and axial forces on pinion** - Tangential force $F_t$: $$F_t = \frac{2T}{d_1} = \frac{2 \times 118.7}{0.3494} = 679.8 \text{ N}$$ - Radial force $F_r$: $$F_r = F_t \tan \phi = 679.8 \times \tan 20^\circ = 679.8 \times 0.3640 = 247.4 \text{ N}$$ - Axial force $F_a$: $$F_a = F_t \tan \gamma = 679.8 \times \tan 27.8^\circ = 679.8 \times 0.5279 = 358.6 \text{ N}$$ **Final answers:** - (a) Bevel angle $\gamma \approx 27.8^\circ$ - (b) Pitch-line velocity $v \approx 10.97$ m/s - (c) Forces on pinion: - Tangential force $F_t \approx 680$ N - Radial force $F_r \approx 247$ N - Axial force $F_a \approx 359$ N These calculations assume approximate pitch radius from given dimensions and standard unit conversions.