Question:medium
A sheet metal drawing is considered feasible if the following conditions are met:
- Reduction ratio < 0.5
- Thickness to diameter ratio > 1%
A sheet metal drawing process is being planned on a sheet metal with a thickness of 2 mm with a punch of diameter 100 mm with various blanks of diameter \( D_b \).
Which ONE of the following blank diameters \( D_b \) (in mm) will result in feasible drawing?
A sheet metal drawing is considered feasible if the following conditions are met:
Which ONE of the following blank diameters \( D_b \) (in mm) will result in feasible drawing?
- Reduction ratio < 0.5
- Thickness to diameter ratio > 1%
Which ONE of the following blank diameters \( D_b \) (in mm) will result in feasible drawing?
Show Hint
Convert both feasibility conditions into inequalities for D_b using t = 2 mm and d_p = 100 mm, then check which option satisfies both.
Updated On: Aug 5, 2026
- 150
- 250
- 350
- 450
Show Solution
The Correct Option is A
Solution and Explanation
Step 1: Deciding to test each option directly:
Instead of solving the inequalities in general first, we can simply plug each candidate blank diameter into both feasibility formulas and see which one passes.
Reduction ratio is $ (D_b - 100)/D_b $ and it must stay under 0.5. Thickness to diameter ratio is $ 2/D_b $ and it must stay above 0.01.
Step 2: Testing option (A), $D_b = 150$ mm:
Reduction ratio $ = (150-100)/150 = 50/150 = 0.333 $, which is less than 0.5, so this condition passes.
Thickness ratio $ = 2/150 = 0.0133 $, which is greater than 0.01, so this condition also passes. Both conditions hold, so 150 mm works.
Step 3: Testing option (B), $D_b = 250$ mm:
Reduction ratio $ = (250-100)/250 = 150/250 = 0.6 $, which is greater than 0.5, so this already fails the first condition, no need to check further.
Step 4: Testing option (C), $D_b = 350$ mm:
Reduction ratio $ = (350-100)/350 = 250/350 = 0.714 $, which is well above 0.5, so this fails too.
Step 5: Testing option (D), $D_b = 450$ mm:
Reduction ratio $ = (450-100)/450 = 350/450 = 0.778 $, again far above 0.5, so this also fails.
Final Answer:
Direct substitution confirms that only $D_b = 150$ mm keeps both the reduction ratio and the thickness to diameter ratio within the feasible range. \[ \boxed{D_b = 150\ mm} \]
Instead of solving the inequalities in general first, we can simply plug each candidate blank diameter into both feasibility formulas and see which one passes.
Reduction ratio is $ (D_b - 100)/D_b $ and it must stay under 0.5. Thickness to diameter ratio is $ 2/D_b $ and it must stay above 0.01.
Step 2: Testing option (A), $D_b = 150$ mm:
Reduction ratio $ = (150-100)/150 = 50/150 = 0.333 $, which is less than 0.5, so this condition passes.
Thickness ratio $ = 2/150 = 0.0133 $, which is greater than 0.01, so this condition also passes. Both conditions hold, so 150 mm works.
Step 3: Testing option (B), $D_b = 250$ mm:
Reduction ratio $ = (250-100)/250 = 150/250 = 0.6 $, which is greater than 0.5, so this already fails the first condition, no need to check further.
Step 4: Testing option (C), $D_b = 350$ mm:
Reduction ratio $ = (350-100)/350 = 250/350 = 0.714 $, which is well above 0.5, so this fails too.
Step 5: Testing option (D), $D_b = 450$ mm:
Reduction ratio $ = (450-100)/450 = 350/450 = 0.778 $, again far above 0.5, so this also fails.
Final Answer:
Direct substitution confirms that only $D_b = 150$ mm keeps both the reduction ratio and the thickness to diameter ratio within the feasible range. \[ \boxed{D_b = 150\ mm} \]
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