1. According to Von-Mises' distortion energy theory, the distortion energy under three dimensional stress state is represented by

1. 1. 1	[σ²1 + σ²2 + σ²3 - 2m(σ1σ2 + σ3σ2 + σ1σ3)]
      εE

   
   
   

   2. 1 - 2μ	[σ²1 + σ²2 + σ²3 - 2m(σ1σ2 + σ3σ2 + σ1σ3)]
      6E

   
   
   

   3. 1 + μ	[σ²1 + σ²2 + σ²3 - 2m(σ1σ2 + σ3σ2 + σ1σ3)]
      3E

   
   
   

   4. 1	[σ²1 + σ²2 + σ²3 - 2m(σ1σ2 + σ3σ2 + σ1σ3)]
      3E

   
   
   

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1. View Hint View Answer Discuss in Forum

   1 + μ	[σ²1 + σ²2 + σ²3 - (σ1σ2 + σ2σ3 + σ3σ1)]
   3E

   Correct Option: C

   1 + μ	[σ²1 + σ²2 + σ²3 - (σ1σ2 + σ2σ3 + σ3σ1)]
   3E

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2. A bimetallic cylindrical bar of cross sectional area 1 m² is made by bonding Steel (Young's modulus = 210 GPa) and Aluminium (Young's modulus = 70 GPa) as shown in the figure. To maintain tensile axial strain of magnitude 10^–6 in steel bar and compressive axial strain of magnitude 10^–6 in Aluminum bar, the magnitude of the required force P(in kN) along the indicated direction is
   

1. 1. 70

   
   
   

   2. 140

   
   
   

   3. 210

   
   
   

   4. 280

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   Let R_A & R_B be the reation at the supports A & B . For the equilibrium of the bar these reaction must act towards left. so that; R_A + R_B = P
   σ_A = ∊Es       σ_B = ∊E_AL
   R_A = A∊Es        σR_B = A∊E_AL
   R_A = 210KN       R_B = 70 KN
   ∴ P = R_A +R_B = 70+210uN
   P = 280 KN.

   Correct Option: D

   
   Let R_A & R_B be the reation at the supports A & B . For the equilibrium of the bar these reaction must act towards left. so that; R_A + R_B = P
   σ_A = ∊Es       σ_B = ∊E_AL
   R_A = A∊Es        σR_B = A∊E_AL
   R_A = 210KN       R_B = 70 KN
   ∴ P = R_A +R_B = 70+210uN
   P = 280 KN.

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3. A steel rod of length L and diameter D, fixed at both ends, is uniformly heated to a temperature rise of ∆T. The Young's modulus is E and the coefficient of linear expansion is α. The thermal stress in the rod is

1. 1. 0

   
   
   

   2. α∆T

   
   
   

   3. Eα∆T

   
   
   

   4. Eα∆TL

   
   
   

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1. View Hint View Answer Discuss in Forum

   Since rod is free to expand, therefore ΔL = elongation = LαΔt
   

   ∴	ΔL	= strain = αΔt
   	L

   
   Thermal stress = EαΔt
   Since rod is fixed at both ends, so thermal strain will be zero but there will be thermal stresses.

   Correct Option: C

   Since rod is free to expand, therefore ΔL = elongation = LαΔt
   

   ∴	ΔL	= strain = αΔt
   	L

   
   Thermal stress = EαΔt
   Since rod is fixed at both ends, so thermal strain will be zero but there will be thermal stresses.

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4. A bar having a cross-sectional area of 700 mm² is subjected to axial loads at the positions indicated.
   The value of stress in the segment QR is
   

1. 1. 40MPa

   
   
   

   2. 50 MPa

   
   
   

   3. 70 MPa

   
   
   

   4. 120 MPa

   
   
   

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1. View Hint View Answer Discuss in Forum

   
   

   Stress =	Force
   	Area

   
   

   =	28 × 10³	N/m²
   	700 × 10-6

   
   = 4 × 10^-5 N/m² = 40 MPa

   Correct Option: A

   
   

   Stress =	Force
   	Area

   
   

   =	28 × 10³	N/m²
   	700 × 10-6

   
   = 4 × 10^-5 N/m² = 40 MPa

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5. Match the following criteria of material failure, under biaxial stresses σ₁ and σ₂ and yield stress σ_y, with their corresponding graphic representations:
   

1. 1. P–M, Q–L, R–N

   
   
   

   2. P–M, Q–N, R–L

   
   
   

   3. P–N, Q–M, R–L

   
   
   

   4. Q–N, Q–L, R–M

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: C

   NA

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