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10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm.The blanking force during the operation will be
[ME 2004]
- a)22.6 kN
- b)37.7 kN
- c)61.6 kN
- d)94.3 kN
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
10 mm diameter holes are to be punched in a steel sheet of 3 mm thickn...
Given, Shear strength,
τ = 400 N/mm2, Sheet thickness,
t = 3 mm
Diameter of hole,
d = 10 mm,
Shear on punch,
t1 = 2 mm
Penetration of punch on fraction,
2p = 0.4
Blanking force when shear is applied on the punch
= 22.61 kN
τ = 400 N/mm2, Sheet thickness,
t = 3 mm
Diameter of hole,
d = 10 mm,
Shear on punch,
t1 = 2 mm
Penetration of punch on fraction,
2p = 0.4
Blanking force when shear is applied on the punch
= 22.61 kN
Most Upvoted Answer
10 mm diameter holes are to be punched in a steel sheet of 3 mm thickn...
Given, Shear strength,
τ = 400 N/mm2, Sheet thickness,
t = 3 mm
Diameter of hole,
d = 10 mm,
Shear on punch,
t1 = 2 mm
Penetration of punch on fraction,
2p = 0.4
Blanking force when shear is applied on the punch
= 22.61 kN
τ = 400 N/mm2, Sheet thickness,
t = 3 mm
Diameter of hole,
d = 10 mm,
Shear on punch,
t1 = 2 mm
Penetration of punch on fraction,
2p = 0.4
Blanking force when shear is applied on the punch
= 22.61 kN
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Community Answer
10 mm diameter holes are to be punched in a steel sheet of 3 mm thickn...
Calculation of Blanking Force
To calculate the blanking force required to punch holes in a steel sheet, we use the following formula:
Blanking Force (F) = Shear Area × Shear Strength
Given Data
- Diameter of hole (d) = 10 mm
- Thickness of sheet (t) = 3 mm
- Shear strength (τ) = 400 N/mm²
- Penetration (p) = 40% of thickness = 0.4 × 3 mm = 1.2 mm
- Shear provided on the punch (s) = 2 mm
Shear Area Calculation
1. Shear Area (A) = Perimeter of the hole × Effective thickness
- Perimeter of the hole = π × Diameter = π × 10 mm ≈ 31.42 mm
- Effective thickness = Thickness - Penetration = 3 mm - 1.2 mm = 1.8 mm
- Shear Area (A) = 31.42 mm × 1.8 mm ≈ 56.556 mm²
Blanking Force Calculation
2. Blanking Force (F) = Shear Area × Shear Strength
- F = 56.556 mm² × 400 N/mm² = 22622.4 N
- Converting to kN: F ≈ 22.62 kN
Conclusion
The total blanking force required to punch the holes in the steel sheet is approximately 22.6 kN, which corresponds to option A.
To calculate the blanking force required to punch holes in a steel sheet, we use the following formula:
Blanking Force (F) = Shear Area × Shear Strength
Given Data
- Diameter of hole (d) = 10 mm
- Thickness of sheet (t) = 3 mm
- Shear strength (τ) = 400 N/mm²
- Penetration (p) = 40% of thickness = 0.4 × 3 mm = 1.2 mm
- Shear provided on the punch (s) = 2 mm
Shear Area Calculation
1. Shear Area (A) = Perimeter of the hole × Effective thickness
- Perimeter of the hole = π × Diameter = π × 10 mm ≈ 31.42 mm
- Effective thickness = Thickness - Penetration = 3 mm - 1.2 mm = 1.8 mm
- Shear Area (A) = 31.42 mm × 1.8 mm ≈ 56.556 mm²
Blanking Force Calculation
2. Blanking Force (F) = Shear Area × Shear Strength
- F = 56.556 mm² × 400 N/mm² = 22622.4 N
- Converting to kN: F ≈ 22.62 kN
Conclusion
The total blanking force required to punch the holes in the steel sheet is approximately 22.6 kN, which corresponds to option A.
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10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm.The blanking force during the operation will be[ME 2004]a)22.6 kNb)37.7 kNc)61.6 kNd)94.3 kNCorrect answer is option 'A'. Can you explain this answer?
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10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm.The blanking force during the operation will be[ME 2004]a)22.6 kNb)37.7 kNc)61.6 kNd)94.3 kNCorrect answer is option 'A'. 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 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm.The blanking force during the operation will be[ME 2004]a)22.6 kNb)37.7 kNc)61.6 kNd)94.3 kNCorrect answer is option 'A'. 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 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm2 and penetration is 40%. Shear provided on the punch is 2 mm.The blanking force during the operation will be[ME 2004]a)22.6 kNb)37.7 kNc)61.6 kNd)94.3 kNCorrect answer is option 'A'. Can you explain this answer?.
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