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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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