A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 1650 rev/min,

Question

A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 1650 rev/min, this pinion is expected to carry a steady load of 1.2 kW. Determine the bending stress.

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Hưng Khoa 6 months 2021-08-02T15:56:13+00:00 1 Answers 5 views 0

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    2021-08-02T15:57:32+00:00

    Answer:

    The value of bending stress on the pinion 35.38 M pa

    Explanation:

    Given data

    m = 2 mm

    Pressure angle \phi = 20°

    No. of teeth T = 17

    Face width (b) = 20 mm

    Speed N = 1650 rpm

    Power = 1200 W

    Diameter of the pinion gear

    D = m T

    D = 2 × 17

    D = 34 mm

    Velocity of the pinion gear

    V =\pi D( \frac{N}{60} )

    V = 3.14 (0.034) \frac{(1650)}{60}

    V = 2.93 \frac{m}{s}

    Form factor for the pinion gear is

    Y = 0.303

    Now

    K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

    Force on gear tooth

    F = \frac{P}{V}

    F = \frac{1200}{2.93}

    F = 408.73 N

    Now the bending stress is given by the formula

    \sigma = \frac{K_{v} F}{m b y}

    \sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

    \sigma = 35.38 M pa

    This is the value of bending stress on the pinion

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