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A shaft is subjected to fluctuating loads for which the normal torque (T) and bending moment (M) are 1000 N-m and 500 N-m respectively. If the combined shock and fatigue factor for bending is 1.5 and combined shock and fatigue factor for torsion is 2, then the equivalent twisting moment for the shaft is

A. 2000 N-m

B. 2050 N-m

C. 2100 N-m

D. 2136 N-m

Answer: Option D


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Comments ( 8 )

  1. Yerosan Yadesa
    Yerosan Yadesa :
    3 months ago

    T₁= √(1.5x500) + (2x1000)² =2136 N-m

  2. BISWAJIT
    BISWAJIT :
    3 years ago

    We know that equivalent twisting moment
    T₁= √√(KXM)² + (K,XT)² Since T = 1000 N-m; M = 500 N-m; K=1.5 and K,= 2, therefore
    T₁= √(1.5x500) + (2x1000)² =2136 N-m
    THANK YOU

  3. Kartavya Sagar
    Kartavya Sagar :
    3 years ago

    Teq = { (km * M)^2 + (kt * T)^2}^1/2

  4. Sarveshwer Chandra
    Sarveshwer Chandra :
    3 years ago

    Sqrt [(2*1000)^2 + (1.5*500)^2].

  5. Anju Prasad
    Anju Prasad :
    4 years ago

    Need full solutions

  6. Ganesh Niture
    Ganesh Niture :
    4 years ago

    How?

  7. Muhammad Usman
    Muhammad Usman :
    4 years ago

    Te=[(M*Kf)^2+(T*Kb)^2]^1/2

  8. AGALDEETI SUNIL
    AGALDEETI SUNIL :
    4 years ago

    Explain me

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