16. A 60 mm long and 6 mm thick fillet weld carries a steady load of 15 kN along the weld. The shear strength of the weld material is equal to 200 MPa. The factor of safety is

1. 1. 2.4

   
   
   

   2. 3.4

   
   
   

   3. 4.8

   
   
   

   4. 6.8

   
   
   

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1. View Hint View Answer Discuss in Forum

   P = .707 t × L_e × τ_s.
   

   	15 × 103	= τs
   	.707 × 6 × 60

   
   ∴ τ_s = 58.94 MPa
   Given shear strength of material = 200 MPa
   

   FOS =	200	= 3.4
   	58.94

   
   

   Correct Option: B

   P = .707 t × L_e × τ_s.
   

   	15 × 103	= τs
   	.707 × 6 × 60

   
   ∴ τ_s = 58.94 MPa
   Given shear strength of material = 200 MPa
   

   FOS =	200	= 3.4
   	58.94

   
   

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17. Endurance limit of a beam subjected to pure bending decreases with

1. 1. incr ease i n t he sur face r oughness and increase in the size of the beam

   
   
   

   2. decr ease in t he sur face r oughness and increase in the size of the beam

   
   
   

   3. decr ease in t he sur face r oughness and decrease in the size of the beam

   
   
   

   4. incr ease i n t he sur face r oughness and decrease in the size of the beam

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

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18. Fatigue life of a material for a fully reversed loading condition is estimated from σ_a = 1100 N^–0.15 where σ_a is the stress amplitude in MPa and N is the failure life in cycles. The maximum allowable stress amplitude (in MPa) for a life of 1 × 10⁵ cycles under the same loading condition is_______(correction to two decimal places).

1. 1. 195.61 N

   
   
   

   2. 194.61 N

   
   
   

   3. 193.61 N

   
   
   

   4. 192.61 N

   
   
   

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1. View Hint View Answer Discuss in Forum

   	σmax - σmin	= 1100N-0.15
   	2

   
   

   	σmax - (- σmin)	= 1100N-0.15
   	2

   
   

   	2σmax	= 1100N-0.15(N = 105)
   	2

   
   ∴ We get, σ_max = 195.61 N

   Correct Option: A

   	σmax - σmin	= 1100N-0.15
   	2

   
   

   	σmax - (- σmin)	= 1100N-0.15
   	2

   
   

   	2σmax	= 1100N-0.15(N = 105)
   	2

   
   ∴ We get, σ_max = 195.61 N

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19. A machine element has an ultimate strength (σ_u) of 600 N/mm², and endurance limit (σ_en) of 250 N/mm². The fatigue curve for the element on a log-log plot is shown below. If the element is to be designed for a finite life of 10000 cycles, the maximum amplitude of a completely reversed operating stress is ______N/mm².
    

1. 1. 386.19 MPa

   
   
   

   2. 376.19 MPa

   
   
   

   3. 366.19 MPa

   
   
   

   4. 385.19 MPa

   
   
   

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1. View Hint View Answer Discuss in Forum

   		log Sf - log (0.8 × 600)	=	4 - 3
   		log 250 - log (0.6 × 600)		6 - 3

   
   

   ⇒		log Sf - log (480)	=	1
   		log 250 - log (360)		3

   
   S_f = 386.19 MPa

   Correct Option: A

   		log Sf - log (0.8 × 600)	=	4 - 3
   		log 250 - log (0.6 × 600)		6 - 3

   
   

   ⇒		log Sf - log (480)	=	1
   		log 250 - log (360)		3

   
   S_f = 386.19 MPa

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20. At a critical point in a component, the state of stress is given as σ_xx = 100 MPa, σ_yy = 220 MPa, σ_xy = σ_yx = 80 M Pa and al l ot her st r ess components are zero. The yield strength of the material is 468 MPa. The factor of safety on the basis of maximum shear stress theory is_____ (round off to one decimal place).

1. 1. 1.8

   
   
   

   2. 1.4

   
   
   

   3. 2.8

   
   
   

   4. 2.4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Given, σ_x = 100 MPa
   σ_y = 220 MPa
   σ_xy = 80 MPa
   
   = 160 ± √55² + 80²
   σ_{1, 2} = 160 ± 97.0824
   ⇒ σ₁ = 257.08 Mpa, σ₂ = 62.9 Mpa
   As per maximum shear stress theory
   
   

   		Syt	=	σ1	=	257.08
   		2N		2		2

   
   

   N =	468	= 1.8
   	257.08

   
   

   Correct Option: A

   Given, σ_x = 100 MPa
   σ_y = 220 MPa
   σ_xy = 80 MPa
   
   = 160 ± √55² + 80²
   σ_{1, 2} = 160 ± 97.0824
   ⇒ σ₁ = 257.08 Mpa, σ₂ = 62.9 Mpa
   As per maximum shear stress theory
   
   

   		Syt	=	σ1	=	257.08
   		2N		2		2

   
   

   N =	468	= 1.8
   	257.08

   
   

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