Machine Design Miscellaneous
 A ductile material having an endurance limit of 196 N/mm^{2} and the yield point at 294 N/ mm^{2} is stressed under variable load. The maximum and minimum stresses are 147 N/ mm^{2} and 49 N/mm^{2}. The fatigue stress concentration factor is 1.32. The available factor of safety for this loading is

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Consider Soderberg equation
τ_{a} + τ_{m} = 1 S_{e} S_{y} FOS τ_{a} = 147  49 = 49 MPa 2 τ_{m} = 147 + 49 = 98 MPa 2 S_{e} = 196 = 147.48 MPa 1.32 49 + 98 = 1 147.48 294 FOS
FOS = 1.5Correct Option: B
Consider Soderberg equation
τ_{a} + τ_{m} = 1 S_{e} S_{y} FOS τ_{a} = 147  49 = 49 MPa 2 τ_{m} = 147 + 49 = 98 MPa 2 S_{e} = 196 = 147.48 MPa 1.32 49 + 98 = 1 147.48 294 FOS
FOS = 1.5
 A solid shaft can resist a bending moment of 3.0 kNm and a twisting moment of 4.0 kNm together, then the maximum torque that can be applied is

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Maximum tensile stress due to bending,
f_{t} = M Z = 3 × 10^{3} {(π / 64 ) × d^{4} } d / 2 f_{t} = 30.558 kPa d^{3} Maximum shear stress due to torsion, f_{s} = TR J = 4 × 10^{3} × (d / 2) π × d^{4} 32 f_{s} = 20.7 kPa d^{3}
When both the stress acts simultaneously, maximum induced shear stress,f_{s max.} = 1 [√(f_{t})² + 4(f_{s})² ] 2
Maximum torque that can be applied,T = πd^{3} f_{s max.} 16 = πd^{3} × 1 [ √(30.558 / d^{3})² + 4(20.37 / d^{3})² ] 16 2
= 4 kNm
Correct Option: D
Maximum tensile stress due to bending,
f_{t} = M Z = 3 × 10^{3} {(π / 64 ) × d^{4} } d / 2 f_{t} = 30.558 kPa d^{3} Maximum shear stress due to torsion, f_{s} = TR J = 4 × 10^{3} × (d / 2) π × d^{4} 32 f_{s} = 20.7 kPa d^{3}
When both the stress acts simultaneously, maximum induced shear stress,f_{s max.} = 1 [√(f_{t})² + 4(f_{s})² ] 2
Maximum torque that can be applied,T = πd^{3} f_{s max.} 16 = πd^{3} × 1 [ √(30.558 / d^{3})² + 4(20.37 / d^{3})² ] 16 2
= 4 kNm
 Two shafts A and B are made of the same material. The diameter of shaft B is twice that of shaft A The ratio of power which can be transmitted by shaft A to that of shaft B is

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T_{A} = π × d^{3}τ 16 T_{B} = π × (2d)^{3}τ 16 P_{A} = T_{A} = 1 P_{B} T_{B} 8 Correct Option: C
T_{A} = π × d^{3}τ 16 T_{B} = π × (2d)^{3}τ 16 P_{A} = T_{A} = 1 P_{B} T_{B} 8
 The outside diameter of a hollow shaft that is twice its inside diameter the ratio of its torque carrying capacity to that of a solid shaft of the same material and the same outside diameter is

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T_{hollow} = π d^{4}  d^{4} 2τ ......(1) 32 16 d T_{solid} = π d^{3}τ .......(2) 16
By dividing (1) with (2)T_{hollow} = 15 T_{solid} 16 Correct Option: A
T_{hollow} = π d^{4}  d^{4} 2τ ......(1) 32 16 d T_{solid} = π d^{3}τ .......(2) 16
By dividing (1) with (2)T_{hollow} = 15 T_{solid} 16
 A large uniform plate containing a rivet hole subjected to uniform uniaxial tension of 95 MPa. The maximum stress in the plate is

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Max Stress = Avg. Force Min. Area = 95 × 10 × t = 100 MPa (10  0.5)t Correct Option: A
Max Stress = Avg. Force Min. Area = 95 × 10 × t = 100 MPa (10  0.5)t