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Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is
[PI 2013]
- a)2.57
- b)3.29
- c)5.03
- d)6.33
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Circular blanks of 10 mm diameter are punched from an aluminum sheet o...
Punching force = πDtτ
= π × 10 × 2 × 80
= 5.023 KN
= π × 10 × 2 × 80
= 5.023 KN
Most Upvoted Answer
Circular blanks of 10 mm diameter are punched from an aluminum sheet o...
To calculate the minimum punching force required, we need to consider the shear strength of the aluminum and the geometry of the punch.
1. Calculate the shear area:
The shear area is the area of the material that will be sheared during the punch. In this case, the shear area is the circumference of the circular punch multiplied by the thickness of the sheet.
Shear Area = π * D * t
= π * 10 mm * 2 mm
= 62.83 mm²
2. Calculate the shear force:
The shear force is the force required to shear the material. It can be calculated by multiplying the shear area by the shear strength of the material.
Shear Force = Shear Area * Shear Strength
= 62.83 mm² * 80 MPa
= 5026.4 N
3. Convert the shear force to kN:
The minimum punching force required is given in kN, so we need to convert the shear force from N to kN.
Minimum Punching Force = Shear Force / 1000
= 5026.4 N / 1000
= 5.03 kN
Therefore, the minimum punching force required is 5.03 kN.
1. Calculate the shear area:
The shear area is the area of the material that will be sheared during the punch. In this case, the shear area is the circumference of the circular punch multiplied by the thickness of the sheet.
Shear Area = π * D * t
= π * 10 mm * 2 mm
= 62.83 mm²
2. Calculate the shear force:
The shear force is the force required to shear the material. It can be calculated by multiplying the shear area by the shear strength of the material.
Shear Force = Shear Area * Shear Strength
= 62.83 mm² * 80 MPa
= 5026.4 N
3. Convert the shear force to kN:
The minimum punching force required is given in kN, so we need to convert the shear force from N to kN.
Minimum Punching Force = Shear Force / 1000
= 5026.4 N / 1000
= 5.03 kN
Therefore, the minimum punching force required is 5.03 kN.
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Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer?
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Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct 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 Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct 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 Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer?.
Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct 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 Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct 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 Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer?.
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Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Circular blanks of 10 mm diameter are punched from an aluminum sheet of 2 mm thickness.The shear strength of aluminum is 80 MPa.The minimum punching force required in kN is[PI 2013]a)2.57b)3.29c)5.03d)6.33Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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