506. In orthogonal turning of medium carbon steel, the specific machining energy is 2.0 J/mm³. The cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev and 2 mm respectively. The main cutting force in N is

1. 1. 40

   
   
   

   2. 80

   
   
   

   3. 400

   
   
   

   4. 800

   
   
   

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   F_C = Cutting force;
   V = Cutting velocity
   Specific machining energy = 2.0 J/mm³
   ∴  F_C × V =2 × MRR
   ⇒  F_C = 2 × .2 × 10⁻³ × 2 × 10³ = 800 N

   Correct Option: D

   F_C = Cutting force;
   V = Cutting velocity
   Specific machining energy = 2.0 J/mm³
   ∴  F_C × V =2 × MRR
   ⇒  F_C = 2 × .2 × 10⁻³ × 2 × 10³ = 800 N

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507. In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90°, the shear angle in degree is

1. 1. 20.56

   
   
   

   2. 26.56

   
   
   

   3. 30.56

   
   
   

   4. 36.56

   
   
   

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1. View Hint View Answer Discuss in Forum

   Chip thickness,
   

   r =	feed	=	0.24	= 0.5
   	chip thickness		0.48

   
   rake angle = 0
   

   tanφ =	rcosα	= .5
   	1 − sinα

   
   ∴  φ = 26.56

   Correct Option: B

   Chip thickness,
   

   r =	feed	=	0.24	= 0.5
   	chip thickness		0.48

   
   rake angle = 0
   

   tanφ =	rcosα	= .5
   	1 − sinα

   
   ∴  φ = 26.56

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508. The percentage of total energy dissipated due to friction at the tool-chip interface is

1. 1. 30%

   
   
   

   2. 42%

   
   
   

   3. 58%

   
   
   

   4. 70%

   
   
   

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1. View Hint View Answer Discuss in Forum

   Total heat generated, Q₁ = F_c × V = 24000 J/min
   Heat dissipated due to friction, Q₂ = F × r × v
   Frictional force, F = F_c sin α + F_f cos α = 503.76 N
   Q₂ =7193.69 J/min.
   

   ∴	Q2	= 0.299; 30%
   	Q2

   Correct Option: A

   Total heat generated, Q₁ = F_c × V = 24000 J/min
   Heat dissipated due to friction, Q₂ = F × r × v
   Frictional force, F = F_c sin α + F_f cos α = 503.76 N
   Q₂ =7193.69 J/min.
   

   ∴	Q2	= 0.299; 30%
   	Q2

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509. The values of shear angle and shear strain, respectively, are

1. 1. 30.3° and 1.98

   
   
   

   2. 30.3° and 4.23

   
   
   

   3. 40.2° and 2.97

   
   
   

   4. 40.2° and 1.65

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here, r = cutting ratio,
   

   n =	t1	=	0.5	= 0.714
   	t2		0.7

   
   

   ∴  tanφ =	rcosα
   	1 − rsinα

   
   where, φ = shear angle
   and α = rake angle
   ∴  φ = 40.2º
   Shear strain = cot φ + tan (φ – α) = 1.65

   Correct Option: D

   Here, r = cutting ratio,
   

   n =	t1	=	0.5	= 0.714
   	t2		0.7

   
   

   ∴  tanφ =	rcosα
   	1 − rsinα

   
   where, φ = shear angle
   and α = rake angle
   ∴  φ = 40.2º
   Shear strain = cot φ + tan (φ – α) = 1.65

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510. Two tools P and Q have signatures 5°-50-60-6°- 8°-30°-0 and 5°-5°-7°-7°-8°-15°-0 (both ASA) respectively. They are used to turn components under the same machining conditions. If hp and hQ denote the peak-to-valley heights of surfaces produced by the tools P and Q, the ratio h_p /h_Q will be

1. 1. tan 8º + cot15º

      tan8º + cot30º

   
   
   

   2. tan15º + cot8º

      tan30º + cot8º

   
   
   

   3. tan15º + cot7º

      tan30º + cot7º

   
   
   

   4. tan7º + cot15º

      tan7º + cot30º

   
   
   

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1. View Hint View Answer Discuss in Forum

   Hmax =	f
   	tan(ψ) + cot(ψ)

   
   ψ = SCEA
   ψ₁ = ECEA
   

   hP	=	tan15° + Cot8°
   hQ		tan30° + cot8°

   Correct Option: B

   Hmax =	f
   	tan(ψ) + cot(ψ)

   
   ψ = SCEA
   ψ₁ = ECEA
   

   hP	=	tan15° + Cot8°
   hQ		tan30° + cot8°

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