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In orthogonal cutting of MS tube with HSS tool, cutting speed and rake angle are 15m/min and 150. Chip thickness ratio is 0.35. Friction force and coefficient of friction are 470N and 0.6 respectively. Value of cutting force is:-
- a)472N
- b)900N
- c)878 N
- d)236N
Correct answer is option 'C'. Can you explain this answer?
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In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake a...
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Given data:
- Cutting speed (V) = 15 m/min
- Rake angle (α) = 150
- Chip thickness ratio (h₁/h₀) = 0.35
- Friction force (F) = 470 N
- Coefficient of friction (μ) = 0.6
Calculating cutting force:
The cutting force (F_c) can be calculated using the following formula:
F_c = (F - F_f) / μ
where F_f is the friction force.
Calculating friction force:
The friction force (F_f) can be calculated using the following formula:
F_f = μ * F_n
where F_n is the normal force.
Calculating normal force:
For orthogonal cutting, the normal force (F_n) can be calculated using the following formula:
F_n = F_c / sin(α)
Calculating chip thickness:
The chip thickness (h₀) can be calculated using the following formula:
h₀ = V * t
where t is the time required to make the cut.
Calculating undeformed chip thickness:
The undeformed chip thickness (h₁) can be calculated using the following formula:
h₁ = h₀ / (1 + h₀ * tan(α))
Calculating chip thickness ratio:
The chip thickness ratio (h₁/h₀) can be calculated using the following formula:
h₁/h₀ = (1 + h₀ * tan(α)) / h₀
Now, we can substitute the given values into the formulas to calculate the cutting force.
Solution:
1. Calculate the undeformed chip thickness (h₁):
h₁ = h₀ / (1 + h₀ * tan(α))
= 0.35 / (1 + 0.35 * tan(150))
= 0.35 / (1 + 0.35 * (-1.732))
= 0.35 / (1 - 0.6)
= 0.35 / 0.4
= 0.875 mm
2. Calculate the chip thickness (h₀):
h₀ = h₁ / (1 + h₁ * tan(α))
= 0.875 / (1 + 0.875 * tan(150))
= 0.875 / (1 + 0.875 * (-1.732))
= 0.875 / (1 - 1.515)
= 0.875 / (-0.515)
= -1.699 mm
3. Calculate the normal force (F_n):
F_n = F_c / sin(α)
= F_c / sin(150)
= F_c / 0.866
= F_c / 0.866
= F_c * 1.155
4. Calculate the friction force (F_f):
F_f = μ * F_n
= 0.6 * F_n
= 0.6 * (F_c * 1.155)
= 0.693 * F_c
5. Calculate the cutting force (F_c):
F_c = (F - F_f) / μ
= (470 - 0.693 * F_c)
- Cutting speed (V) = 15 m/min
- Rake angle (α) = 150
- Chip thickness ratio (h₁/h₀) = 0.35
- Friction force (F) = 470 N
- Coefficient of friction (μ) = 0.6
Calculating cutting force:
The cutting force (F_c) can be calculated using the following formula:
F_c = (F - F_f) / μ
where F_f is the friction force.
Calculating friction force:
The friction force (F_f) can be calculated using the following formula:
F_f = μ * F_n
where F_n is the normal force.
Calculating normal force:
For orthogonal cutting, the normal force (F_n) can be calculated using the following formula:
F_n = F_c / sin(α)
Calculating chip thickness:
The chip thickness (h₀) can be calculated using the following formula:
h₀ = V * t
where t is the time required to make the cut.
Calculating undeformed chip thickness:
The undeformed chip thickness (h₁) can be calculated using the following formula:
h₁ = h₀ / (1 + h₀ * tan(α))
Calculating chip thickness ratio:
The chip thickness ratio (h₁/h₀) can be calculated using the following formula:
h₁/h₀ = (1 + h₀ * tan(α)) / h₀
Now, we can substitute the given values into the formulas to calculate the cutting force.
Solution:
1. Calculate the undeformed chip thickness (h₁):
h₁ = h₀ / (1 + h₀ * tan(α))
= 0.35 / (1 + 0.35 * tan(150))
= 0.35 / (1 + 0.35 * (-1.732))
= 0.35 / (1 - 0.6)
= 0.35 / 0.4
= 0.875 mm
2. Calculate the chip thickness (h₀):
h₀ = h₁ / (1 + h₁ * tan(α))
= 0.875 / (1 + 0.875 * tan(150))
= 0.875 / (1 + 0.875 * (-1.732))
= 0.875 / (1 - 1.515)
= 0.875 / (-0.515)
= -1.699 mm
3. Calculate the normal force (F_n):
F_n = F_c / sin(α)
= F_c / sin(150)
= F_c / 0.866
= F_c / 0.866
= F_c * 1.155
4. Calculate the friction force (F_f):
F_f = μ * F_n
= 0.6 * F_n
= 0.6 * (F_c * 1.155)
= 0.693 * F_c
5. Calculate the cutting force (F_c):
F_c = (F - F_f) / μ
= (470 - 0.693 * F_c)
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In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer?
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In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. 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 In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. 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 In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer?.
In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. 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 In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. 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 In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mechanical Engineering.
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In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of In orthogonal cutting of MS tubewith HSS tool, cutting speed andrake angle are 15m/min and 150.Chip thickness ratio is 0.35.Friction force and coefficient of friction are 470N and 0.6respectively. Value of cutting forceis:-a)472Nb)900Nc)878 Nd)236NCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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