106. A ductile material having an endurance limit of 196 N/mm² and the yield point at 294 N/ mm² is stressed under variable load. The maximum and minimum stresses are 147 N/ mm² and 49 N/mm². The fatigue stress concentration factor is 1.32. The available factor of safety for this loading is
     

1. 1. 3.0
      

   
   
   

   2. 1.5
      

   
   
   

   3. 1.33
      

   
   
   

   4. 4.0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Consider Soderberg equation
   

   	τa	+	τm	=	1
   	Se		Sy		FOS

   
   

   τa =	147 - 49	= 49 MPa
   	2

   
   

   τm =	147 + 49	= 98 MPa
   	2

   
   

   Se =	196	= 147.48 MPa
   	1.32

   
   

   	49	+	98	=	1
   	147.48		294		FOS

   
   FOS = 1.5

   Correct Option: B

   Consider Soderberg equation
   

   	τa	+	τm	=	1
   	Se		Sy		FOS

   
   

   τa =	147 - 49	= 49 MPa
   	2

   
   

   τm =	147 + 49	= 98 MPa
   	2

   
   

   Se =	196	= 147.48 MPa
   	1.32

   
   

   	49	+	98	=	1
   	147.48		294		FOS

   
   FOS = 1.5

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107. 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
     

1. 1. 7.0 kNm
      

   
   
   

   2. 3.5 kNm
      

   
   
   

   3. 4.5 kNm
      

   
   
   

   4. 5.0 kNm

   
   
   

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1. View Hint View Answer Discuss in Forum

   Maximum tensile stress due to bending,
   

   ft =	M
   	Z

   
   

   =	3 × 103
   	{(π / 64 ) × d4 }
   	d / 2

   
   

   ft =	30.558	kPa
   	d3

   
   

   Maximum shear stress due to torsion, fs =	TR
   	J

   
   

   =	4 × 103 × (d / 2)
   	π	× d4
   	32

   
   

   fs =	20.7	kPa
   	d3

   
   When both the stress acts simultaneously, maximum induced shear stress,
   

   fs max. =	1	[√(ft)² + 4(fs)² ]
   	2

   
   Maximum torque that can be applied,
   

   T =	πd3	fs max.
   	16

   
   

   =	πd3	×	1	[ √(30.558 / d3)² + 4(20.37 / d3)² ]
   	16		2

   
   = 4 kNm
   

   Correct Option: D

   Maximum tensile stress due to bending,
   

   ft =	M
   	Z

   
   

   =	3 × 103
   	{(π / 64 ) × d4 }
   	d / 2

   
   

   ft =	30.558	kPa
   	d3

   
   

   Maximum shear stress due to torsion, fs =	TR
   	J

   
   

   =	4 × 103 × (d / 2)
   	π	× d4
   	32

   
   

   fs =	20.7	kPa
   	d3

   
   When both the stress acts simultaneously, maximum induced shear stress,
   

   fs max. =	1	[√(ft)² + 4(fs)² ]
   	2

   
   Maximum torque that can be applied,
   

   T =	πd3	fs max.
   	16

   
   

   =	πd3	×	1	[ √(30.558 / d3)² + 4(20.37 / d3)² ]
   	16		2

   
   = 4 kNm
   

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108. 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
     

1. 1. 1

      2

      
      

   
   
   

   2. 1

      4

      
      

   
   
   

   3. 1

      8

      
      

   
   
   

   4. 1

      16

   
   
   

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1. View Hint View Answer Discuss in Forum

   TA =	π	× d3τ
   	16

   
   

   TB =	π	× (2d)3τ
   	16

   
   

   	PA	=	TA	=	1
   	PB		TB		8

   Correct Option: C

   TA =	π	× d3τ
   	16

   
   

   TB =	π	× (2d)3τ
   	16

   
   

   	PA	=	TA	=	1
   	PB		TB		8

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109. 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
     

1. 1. 15

      16

      
      

   
   
   

   2. 3

      4

      
      

   
   
   

   3. 1

      2

      
      

   
   
   

   4. 1

      16

   
   
   

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1. View Hint View Answer Discuss in Forum

   Thollow =	π		d4 -	d4		2τ	......(1)
   	32			16		d

   
   

   Tsolid =	π	d3τ .......(2)
   	16

   
   By dividing (1) with (2)
   

   	Thollow	=	15
   	Tsolid		16

   Correct Option: A

   Thollow =	π		d4 -	d4		2τ	......(1)
   	32			16		d

   
   

   Tsolid =	π	d3τ .......(2)
   	16

   
   By dividing (1) with (2)
   

   	Thollow	=	15
   	Tsolid		16

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110. A large uniform plate containing a rivet hole subjected to uniform uniaxial tension of 95 MPa. The maximum stress in the plate is
     
     

1. 1. 100 MPa
      

   
   
   

   2. 285 MPa
      

   
   
   

   3. 190 MPa
      

   
   
   

   4. Indeterminate

   
   
   

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1. View Hint View Answer Discuss in Forum

   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

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