26. A thin spherical pressure vessel of 200 mm diameter and 1 mm thickness is subjected to an internal pressure varying from 4 to 8 MPa. Assume that the yield, ultimate, and endurance strength of material are 600,800 and 400 MPa respectively. The factor of safety as per Goodman's relation is

1. 1. 2.0

   
   
   

   2. 1.6

   
   
   

   3. 1.4

   
   
   

   4. 1.2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Stress induced,
   

   σ1 = σ2 =	pr
   	2t

   
   

   σ1max =	8 × 100	= 400MPa
   	2 × 1

   
   

   σ1min =	4 × 100	= 200MPa
   	2 × 1

   
   σ_2max = 400 MPa
   σ_2min = 200 MPa
   σ_1m = 300 MPa
   σ_1a = 100 MPa
   σ_2m = 300 MPa
   σ_2a = 100 MPa
   Equivalent stress are as follows:
   σ_me = √σ_1m² + σ_2m² − σ_1mσ_2m
   = σ_me = √300² + 300² − 300 × 300
   = 300 Mpa
   Similarly, σ_ae = 100 MPa
   From Goodman equation,
   

   σae	+	σme	=	1
   Se		Sut		n

   
   

   ⇒	100	+	300	=	1
   	400		800		n

   
   = n = 1.6

   Correct Option: B

   Stress induced,
   

   σ1 = σ2 =	pr
   	2t

   
   

   σ1max =	8 × 100	= 400MPa
   	2 × 1

   
   

   σ1min =	4 × 100	= 200MPa
   	2 × 1

   
   σ_2max = 400 MPa
   σ_2min = 200 MPa
   σ_1m = 300 MPa
   σ_1a = 100 MPa
   σ_2m = 300 MPa
   σ_2a = 100 MPa
   Equivalent stress are as follows:
   σ_me = √σ_1m² + σ_2m² − σ_1mσ_2m
   = σ_me = √300² + 300² − 300 × 300
   = 300 Mpa
   Similarly, σ_ae = 100 MPa
   From Goodman equation,
   

   σae	+	σme	=	1
   Se		Sut		n

   
   

   ⇒	100	+	300	=	1
   	400		800		n

   
   = n = 1.6

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27. A carpenter glues a pair of cylindrical wooden logs by bonding their end faces at an angle of θ = 30° as shown in the figure.
    
    The glue used at the interface fails if
    Criterion 1: the maximum normal stress exceeds 2.5 MPa
    Criterion 2: the maximum shear stress exceeds 1.5 MPa
    Assume that the interface fails before the logs fail. When a uniform tensile stress of 4 MPa is applied, the interface
    

1. 1. fails only because of criterion 1

   
   
   

   2. fails only because of criterion 2

   
   
   

   3. fails because of both criteria 1 and 2

   
   
   

   4. does not fail.

   
   
   

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1. View Hint View Answer Discuss in Forum

   Normal stress on inclined plane
   σ = σ_{x cos² θ
   = 4 × cos²30 = 3 MPa}
   

   Shear stress on inclined plane τ =	σx	sin2θ
   	2

   
   = 2 × sin 60° = 1.73 MPa
   Since both the stress exceeds the given limits, answer is option (c).

   Correct Option: C

   Normal stress on inclined plane
   σ = σ_{x cos² θ
   = 4 × cos²30 = 3 MPa}
   

   Shear stress on inclined plane τ =	σx	sin2θ
   	2

   
   = 2 × sin 60° = 1.73 MPa
   Since both the stress exceeds the given limits, answer is option (c).

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28. If σ₁ and σ₃ are the algebraically largest and smallest principal stresses respectively, the value of the maximum shear stress is

1. 1. σ1 + σ3

      2

   
   
   

   2. σ1 − σ3

      2

   
   
   

   3. 

   
   
   

   4. 

   
   
   

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1. View Hint View Answer Discuss in Forum

   Maximum shear stress = Radius of Mohr’s Circle (for plane stress condition)
   

   Correct Option: B

   Maximum shear stress = Radius of Mohr’s Circle (for plane stress condition)
   

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29. The bolts in a rigid flanged coupling connecting two shafts transmitting power are subjected to
    

1. 1. shear force and bending moment
      

   
   
   

   2. axial force
      

   
   
   

   3. torsion
      

   
   
   

   4. torsion and bending moment

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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30. The bolts in a rigid flanged coupling connecting two shafts transmitting power are subjected to
    

1. 1. shear force and bending moment
      

   
   
   

   2. axial force
      

   
   
   

   3. torsion
      

   
   
   

   4. torsion and bending moment

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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