391. Wrinkling is a common defect found is

1. 1. Bent components

   
   
   

   2. Deep drawn components

   
   
   

   3. Embossed components

   
   
   

   4. Blanked component

   
   
   

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1. View Hint View Answer Discuss in Forum

   Deep drawn components having defects of wrinkles due to less blank holding force

   Correct Option: B

   Deep drawn components having defects of wrinkles due to less blank holding force

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392. A brass billet is to be extruded from its initial diameter of 100 mm to a final diameter of 50 mm. The working temperature of 700°C and the extrusion constant is 250 MPa. The force required for extrusion is

1. 1. 5.44 MN

   
   
   

   2. 2.72 MN

   
   
   

   3. 1.36 MN

   
   
   

   4. 0.36 MN

   
   
   

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1. View Hint View Answer Discuss in Forum

   R =	Ai
   	Af

   
   

   F = 250 × 106 ×	π	(0.1)2 In		0.1		2
   	4			0.05

   
   F = 2.72MN

   Correct Option: B

   R =	Ai
   	Af

   
   

   F = 250 × 106 ×	π	(0.1)2 In		0.1		2
   	4			0.05

   
   F = 2.72MN

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393. The process of hot extrusion is used to produce

1. 1. Curtain rods made of aluminium

   
   
   

   2. Steel pipes of domestic water supply

   
   
   

   3. Stainless steel tubes used in furniture

   
   
   

   4. Large size pipes used in city water mains

   
   
   

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1. View Hint View Answer Discuss in Forum

   Aluminium is ductile at higher temperature.

   Correct Option: A

   Aluminium is ductile at higher temperature.

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394. For wire drawing operation, the work material should essentially be

1. 1. ductile

   
   
   

   2. tough

   
   
   

   3. hard

   
   
   

   4. malleable

   
   
   

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1. View Hint View Answer Discuss in Forum

   For wire drawing, work material should be ductile in nature.

   Correct Option: A

   For wire drawing, work material should be ductile in nature.

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395. During open die forging process using two flat and parallel dies, a solid steel disc of initial radius (R_IN) 200 mm and initial height (H_IN) 50 mm attains a height (H_FN) of 30 mm and radius of R_FN,
     Along the die-disc interfaces,
     (i) The coefficient of friction (μ) is:
     μ = 0.35 [1 + e–R_IN/R_FN]
     (ii) In the region Rss ≤ r ≤ R_FN sllding friction prevails and P = √3 K.e^2μ(R_IN–r)/H_FN and ι = μp,
     sllding friction prevails and
     Where p and ι are the normal and the shear stresses respectively; K is the shear yield strength of steel and r is the radial distance of any point.
     (i) In the region 0 ≤ r ≤ R_IN, sticking condition prevails.
     The value of Rss (in mm), where sticking condition changes to sliding friction is

1. 1. 241.76

   
   
   

   2. 254.55

   
   
   

   3. 265.45

   
   
   

   4. 278.20

   
   
   

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1. View Hint View Answer Discuss in Forum

   At RSS,
   

   	π	d02 × h0 = πR2FN × h1
   	4

   
   R_FN
   Shear stress sticking (k) = Shear stress in sliding (μP_ss)
   

   ∴ RSS = RFN -	h	In		1
   	2μ			√3μ

   
   

   = 258.2 -	30	In		1
   	2 × 0.51			√3 × 0.51

   
   = 254.55 mm

   Correct Option: B

   At RSS,
   

   	π	d02 × h0 = πR2FN × h1
   	4

   
   R_FN
   Shear stress sticking (k) = Shear stress in sliding (μP_ss)
   

   ∴ RSS = RFN -	h	In		1
   	2μ			√3μ

   
   

   = 258.2 -	30	In		1
   	2 × 0.51			√3 × 0.51

   
   = 254.55 mm

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