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The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is 't' and diameter of the blanked part is 'd'. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is
[ME 2007]
- a)3.0
- b)4.5
- c)6.7
- d)8.0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The force requirement in a blanking operation of low carbon steel shee...
Let t be the shear stress
F = t × d × t = 5.0 kN
F1 = r × 1.5 d × 4 × t = .6 × 5 = 3 kN
F = t × d × t = 5.0 kN
F1 = r × 1.5 d × 4 × t = .6 × 5 = 3 kN
Most Upvoted Answer
The force requirement in a blanking operation of low carbon steel shee...
Given:
Force requirement in blanking operation of low carbon steel sheet = 5.0 kN
Thickness of the sheet = t
Diameter of the blanked part = d
To find:
New blanking force when diameter is increased to 1.5d and thickness is reduced to 0.4t
Solution:
We know that the force required for blanking operation depends on the area of the blanked part, thickness of the sheet, and the shear strength of the material.
The formula for calculating blanking force is:
F = (T x A x S) / 1000
Where,
F = blanking force in kN
T = sheet thickness in mm
A = area of the blanked part in mm²
S = shear strength of the material in N/mm²
Let's assume the shear strength of the low carbon steel sheet to be constant.
Now,
Area of the blanked part = π/4 x d²
New diameter of the blanked part = 1.5d
New area of the blanked part = π/4 x (1.5d)² = 2.25 x π/4 x d²
New thickness of the sheet = 0.4t
Using the formula,
New blanking force F' = (T' x A' x S) / 1000
Where,
T' = new thickness of the sheet
A' = new area of the blanked part
Substituting the values,
F' = (0.4t x 2.25 x π/4 x d² x S) / 1000
F' = 0.5625 x t x d² x S / 1000
We can see that the new blanking force F' is proportional to the thickness of the sheet and the square of the blanked part diameter.
To find the new blanking force, we need to find the ratio of the new force to the old force:
F' / F = (0.5625 x t x d² x S / 1000) / (t x d² x S / 1000)
F' / F = 0.5625
F' = 0.5625 x F
F' = 0.5625 x 5.0 = 2.81 kN
Rounding off to one decimal place, we get F' = 3.0 kN
Therefore, the new blanking force is 3.0 kN.
Answer: (a) 3.0 kN.
Force requirement in blanking operation of low carbon steel sheet = 5.0 kN
Thickness of the sheet = t
Diameter of the blanked part = d
To find:
New blanking force when diameter is increased to 1.5d and thickness is reduced to 0.4t
Solution:
We know that the force required for blanking operation depends on the area of the blanked part, thickness of the sheet, and the shear strength of the material.
The formula for calculating blanking force is:
F = (T x A x S) / 1000
Where,
F = blanking force in kN
T = sheet thickness in mm
A = area of the blanked part in mm²
S = shear strength of the material in N/mm²
Let's assume the shear strength of the low carbon steel sheet to be constant.
Now,
Area of the blanked part = π/4 x d²
New diameter of the blanked part = 1.5d
New area of the blanked part = π/4 x (1.5d)² = 2.25 x π/4 x d²
New thickness of the sheet = 0.4t
Using the formula,
New blanking force F' = (T' x A' x S) / 1000
Where,
T' = new thickness of the sheet
A' = new area of the blanked part
Substituting the values,
F' = (0.4t x 2.25 x π/4 x d² x S) / 1000
F' = 0.5625 x t x d² x S / 1000
We can see that the new blanking force F' is proportional to the thickness of the sheet and the square of the blanked part diameter.
To find the new blanking force, we need to find the ratio of the new force to the old force:
F' / F = (0.5625 x t x d² x S / 1000) / (t x d² x S / 1000)
F' / F = 0.5625
F' = 0.5625 x F
F' = 0.5625 x 5.0 = 2.81 kN
Rounding off to one decimal place, we get F' = 3.0 kN
Therefore, the new blanking force is 3.0 kN.
Answer: (a) 3.0 kN.
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The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct answer is option 'A'. Can you explain this answer?
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The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct 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 The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct 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 The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct answer is option 'A'. Can you explain this answer?.
The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct 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 The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct 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 The force requirement in a blanking operation of low carbon steel sheet is 5.0 kN. The thickness of the sheet is t and diameter of the blanked part is d. For the same work material, if the diameter of the blanked part is increased to 1.5d and thickness is reduced to 0.4t, the new blanking force in kN is[ME 2007]a)3.0b)4.5c)6.7d)8.0Correct answer is option 'A'. Can you explain this answer?.
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