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An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15o and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________ (correct to two
decimal places)
    Correct answer is '(0.4408)'. Can you explain this answer?
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    An orthogonal cutting operations is being carried out in which uncut t...

    Where F represents frictional force
    Ratio of frictional energy to total energy


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    An orthogonal cutting operations is being carried out in which uncut t...
    Calculation of Friction Energy and Total Energy:

    Uncut thickness (t1) = 0.010 mm
    Cutting speed (V) = 130 m/min
    Width of cut (b) = 6 mm
    Rake angle (α) = 15o
    Chip thickness (t2) = 0.015 mm
    Cutting force (Fc) = 60 N
    Thrust force (Ft) = 25 N

    1. Calculation of Shear Angle (φ):
    tan(φ) = t1/t2
    tan(φ) = 0.010/0.015
    φ = tan^-1(0.667) = 33.75o

    2. Calculation of Shear Strain (γ):
    γ = tan(φ) - tan(α)
    γ = tan(33.75o) - tan(15o)
    γ = 0.406

    3. Calculation of Specific Cutting Energy (Ec):
    Ec = Fc/(b*t2*V)
    Ec = 60/(6*0.015*130) = 0.641 J/mm^3

    4. Calculation of Total Energy (Et):
    Et = Ec*t2*b*V
    Et = 0.641*0.015*6*130 = 0.748 J/mm

    5. Calculation of Friction Energy (Ef):
    Ef = Ft*tan(φ)*t2*b*V
    Ef = 25*tan(33.75o)*0.015*6*130 = 0.308 J/mm

    6. Calculation of Ratio of Friction Energy to Total Energy:
    Ratio of Friction Energy to Total Energy = Ef/Et
    Ratio of Friction Energy to Total Energy = 0.308/0.748 = 0.4408

    Therefore, the ratio of friction energy to total energy is 0.4408 (correct to two decimal places).
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    An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15o and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________ (correct to twodecimal places)Correct answer is '(0.4408)'. Can you explain this answer?
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    An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15o and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________ (correct to twodecimal places)Correct answer is '(0.4408)'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15o and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________ (correct to twodecimal places)Correct answer is '(0.4408)'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15o and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________ (correct to twodecimal places)Correct answer is '(0.4408)'. Can you explain this answer?.
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