Holding torque in epicyclic gear train

 

This topic contains 1 reply, has 2 voices, and was last updated by  Surjeet Sankararaj 2 years, 9 months ago.

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  • #3238

    nikhilmkhairnar
    Participant
    Points: 26

    In both questions torques are opposite in direction. Still in first que. they have added them directly (2000+100)
    and in second question they have subtracted them from each other..(10-1=9). Both solutions are contradicting with each other please help…
    What is actual procedure to find holding torque in epicyclic gear train…?

    And my heartily thanks to mechteacher.com and surjeet sir for solving my previous doubts….

    #3243

    Surjeet Sankararaj
    Keymaster
    Points: 627

    According to “Theory of Machines” book by R.S. Khurmi and J.K. Gupta, no matter whatever the conditions are, use this equation:

    T1ω1+T2ω2+T3ω3 = 0

    As ω3 = 0, this equation can be reduced to
    T1ω1+T2ω2 = 0

    In the above equation, you will get T2 with negative sign. Substitute T2 as it is with its negative sign in the following equation to get holding torque (T3)

    T3 = -(T1+T2)

    The second image you posted contains the right method.

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