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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

Answer & Solution
Correct Answer: Option B
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6 Comments
Abdullah
Abdullah 5 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
Sumeet Sadawarti
Sumeet Sadawarti 5 years ago
Which one is correct ?
Viddesh Gharat
Viddesh Gharat 6 years ago
F. O. S = yield stress / bending stress
= 280/80
= 7/2
= 3.5
Aruth Mari
Aruth Mari 6 years ago
stress 1 = √ 80^2 + (4*30^2) = 100
max shear stress = 100-0/2 = 50
50 = 280/ 2*Fos
2.8
Saurabh Kumar
Saurabh Kumar 6 years ago
I think ans should be (d) 3.5
(80-0)/2 = 280/(2*F.S)
Managobinda Sahoo
Managobinda Sahoo 7 years ago
Please give the procedure