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A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.
What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)
Correct answer is '62.7'. Can you explain this answer?
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A cold rolled steel cup with an inside radius of 30 mm and a thicknes...
Given:
Inside radius of the cup (r) = 30 mm
Thickness of the cup (t) = 3 mm
Diameter of the plank (D) = 40 mm
Shear yield stress (τy) = 210 N/mm^2
Maximum allowable stress (σmax) = 600 N/mm^2
Coefficient of friction (μ) = 0.1
β = 0.05
Assumptions:
1. The material is homogeneous and isotropic.
2. The deformation is elastic-plastic.
3. The drawing process is axisymmetric.
4. The cup is assumed to be perfectly cylindrical.
Approach:
The drawing force required for the cup can be determined using the maximum drawing force formula:
F = (σmax + τy) × (π/4) × (D^2 - d^2) × β
Where:
F = Drawing force
σmax = Maximum allowable stress
τy = Shear yield stress
D = Diameter of the plank
d = Diameter of the cup after drawing (d = D - 2t)
β = Friction factor
Calculation:
1. Calculate the diameter of the cup after drawing:
d = D - 2t
= 40 mm - 2 × 3 mm
= 34 mm
2. Calculate the drawing force:
F = (σmax + τy) × (π/4) × (D^2 - d^2) × β
= (600 N/mm^2 + 210 N/mm^2) × (π/4) × (40 mm^2 - 34 mm^2) × 0.05
= 810 N/mm^2 × (π/4) × (1600 mm^2 - 1156 mm^2) × 0.05
= 810 N/mm^2 × (π/4) × 444 mm^2 × 0.05
= 810 N/mm^2 × 3.1416/4 × 22.2 mm^2 × 0.05
= 235.61904 N/mm^2 × 22.2 mm^2 × 0.05
= 261.865216 N × mm/mm^2 × mm^2 × 0.05
= 261.865216 N × 0.05
= 13.0932608 N
3. Convert the drawing force to kN:
F = 13.0932608 N / 1000
= 0.0130932608 kN
Therefore, the drawing force required for the cup is approximately 0.0131 kN, which when rounded to one decimal place is 0.01.3 kN.
Inside radius of the cup (r) = 30 mm
Thickness of the cup (t) = 3 mm
Diameter of the plank (D) = 40 mm
Shear yield stress (τy) = 210 N/mm^2
Maximum allowable stress (σmax) = 600 N/mm^2
Coefficient of friction (μ) = 0.1
β = 0.05
Assumptions:
1. The material is homogeneous and isotropic.
2. The deformation is elastic-plastic.
3. The drawing process is axisymmetric.
4. The cup is assumed to be perfectly cylindrical.
Approach:
The drawing force required for the cup can be determined using the maximum drawing force formula:
F = (σmax + τy) × (π/4) × (D^2 - d^2) × β
Where:
F = Drawing force
σmax = Maximum allowable stress
τy = Shear yield stress
D = Diameter of the plank
d = Diameter of the cup after drawing (d = D - 2t)
β = Friction factor
Calculation:
1. Calculate the diameter of the cup after drawing:
d = D - 2t
= 40 mm - 2 × 3 mm
= 34 mm
2. Calculate the drawing force:
F = (σmax + τy) × (π/4) × (D^2 - d^2) × β
= (600 N/mm^2 + 210 N/mm^2) × (π/4) × (40 mm^2 - 34 mm^2) × 0.05
= 810 N/mm^2 × (π/4) × (1600 mm^2 - 1156 mm^2) × 0.05
= 810 N/mm^2 × (π/4) × 444 mm^2 × 0.05
= 810 N/mm^2 × 3.1416/4 × 22.2 mm^2 × 0.05
= 235.61904 N/mm^2 × 22.2 mm^2 × 0.05
= 261.865216 N × mm/mm^2 × mm^2 × 0.05
= 261.865216 N × 0.05
= 13.0932608 N
3. Convert the drawing force to kN:
F = 13.0932608 N / 1000
= 0.0130932608 kN
Therefore, the drawing force required for the cup is approximately 0.0131 kN, which when rounded to one decimal place is 0.01.3 kN.
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Community Answer
A cold rolled steel cup with an inside radius of 30 mm and a thicknes...
Formulas to be used
We first calculate the blank holding force Fh from the given data as,
Fh = 0.05 x π x 402 x 210 N = 52778 N
⇒ Fh = 52778 N
Next, we find the value of σr at r = rd by using Eq. (1).
Thus,
Now using Equation (2), we get
σz = 94.8 x e0.1 x π/2 N/mm2
⇒ σz = 110.9 N/mm2
It should be noted that this is much less than the fracture strength though rj – rp = 10 mm = 3.33t i.e., very close to the limit set by the condition of plastic buckling. From Equation (3), the drawing force is found is given as,
F = 2π x 30 x 3 x 110.9 N = 62680 N = 62.7 kN
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A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer?
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A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. 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 A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. 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 A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer?.
A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. 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 A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. 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 A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer?.
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A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer?, a detailed solution for A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer? has been provided alongside types of A cold rolled steel cup with an inside radius of 30 mm and a thickness of 3 mm is to be drawn from a plank of diameter 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm2 and 600 N/mm2, respectively.What will be the drawing force (in kN) if the coefficient of friction μ = 0.1 and β = 0.05? (Answer up to one decimal place)Correct answer is '62.7'. Can you explain this answer? theory, EduRev gives you an
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