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

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

Which one is correct ?

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

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

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

Please give the procedure