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A shaft is subjected to a maximum bending stress of 80 N/mm2 and maximum shearing stress equal to 30 N/mm2 at a particular section. If the yield point in tension of the material is 280 N/mm2 and the maximum shear stress theory of failure is used, then the factor of safety obtained will be

A. 2.5

B. 2.8

C. 3.0

D. 3.5

Answer: Option B


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Comments ( 6 )

  1. Abdullah
    Abdullah :
    2 years ago

    For maximum shear stress = √(sigma b)² + 4(tou)²
    = √ 80² + 4(30)² = 100
    So Factor Of safety = Yield stress/working stress
    =280N/mm²/100N/mm²
    =2.8

  2. Sumeet Sadawarti
    Sumeet Sadawarti :
    3 years ago

    Which one is correct ?

  3. Viddesh Gharat
    Viddesh Gharat :
    3 years ago

    F. O. S = yield stress / bending stress
    = 280/80
    = 7/2
    = 3.5

  4. Aruth Mari
    Aruth Mari :
    3 years ago

    stress 1 = √ 80^2 + (4*30^2) = 100
    max shear stress = 100-0/2 = 50
    50 = 280/ 2*Fos
    2.8

  5. Saurabh Kumar
    Saurabh Kumar :
    3 years ago

    I think ans should be (d) 3.5
    (80-0)/2 = 280/(2*F.S)

  6. Managobinda Sahoo
    Managobinda Sahoo :
    4 years ago

    Please give the procedure

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