1. For blacking and piercing operations, clearance is provided on the _______ and the _______ respectively.
(1 Mark, 1987)

Ans: punch, die
2. Wrinkling is a common defect found in
(a) Bent components
(b) Deep drawn components
(c) Embossed component
(d) Blanked component
(1 Mark, 1990)

Ans: b
3. The thickness of the blank needed to produce, by power spinning a missile cone of thickness 1.5 mm and half cone angle 30°, is:
(a) 3.0 mm
(b) 2.5 mm
(c) 2.0 mm
(d) 1.5 mm
(1 Mark, 1992)

Ans: a
t = t_{o}.Sinα
t_{o} = Thickness of blank,
t = Thickness of missile conet = 1.5 \times sin 30 = 3 mm
4. In deep drawing of sheets, the values of limiting drawing ratio depends on
(a) Percentage elongation of sheet metal
(b) Yield strength of sheet metal
(c) Type of press used
(d) Thickness of sheet
(1 Mark, 1994)

Ans: b
In deep drawing process, the limiting drawing ratio depends on the characteristics of the material, die and punch design and friction condition.
5. A 50 mm diameter disc is to be punched out from a carbon steel sheet 1.0 mm thick. The diameter of the die should be
(a) 49.925 mm
(b) 50.00 mm
(c) 50.075 mm
(d) None of the above
(2 Mark, 1996)

Ans: b
It is a blanking operation.
For blanking operation:
Die size = Size of the product
Punch size = Die size – 2\times clearance
6. Identify the stress–strain in the flange portion of a partially drawn cylindrical cup when deep drawing without a blank holder
(a) Tensile in all three directions
(b) No stress in the flange at all, because there is no blank – holder
(c) Tensile stress in one direction and compressive in the other direction
(d) Compressive in two directions and tensile in the third direction
(2 Mark, 1999)

Ans: b
7. The cutting force in punching and blanking operations mainly depends on
(a) The modulus of elasticity of metal
(b) The shear strength of metal
(c) The bulk modulus of metal
(d) The yield strength of metal
(1 Mark, 2001)

Ans: b
8. In a blanking operation, the clearance is provided on
(a) The die
(b) The punch
(c) Both the die and the punch equally
(d) Neither the punch nor the die
(1 Mark, 2002)

Ans: b
For blanking operation:
Die size = Size of the product
Punch size = Die size – 2\times clearance
9. A shell of 100 mm diameter and 100 mm height with the corner radius of 0.4 mm is to be produced by cup drawing. The required blank diameter is
(a) 118 mm
(b) 161 mm
(c) 224 mm
(d) 312 mm
(2 Mark, 2003)

Ans: c
Required punch diameter (D)
= \sqrt { { d }^{ 2 }+4dh } when d\ge 20r
= \sqrt { { 100 }^{ 2 }+4\times 100\times 100 }
\simeq 224 mm
10. A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the die clearance is 3%. The required punch diameter is
(a) 19.88 mm
(b) 19.94 mm
(c) 20.06 mm
(d) 20.12 mm
(2 Mark, 2003)

Ans: a
For blanking operation:
Die size = Size of the product
Punch size = Die size – 2\times clearance
Clearance = 2%\times 3 = 0.02\times 3 = 0.06
Punch diameter = 20 – 2\times 0.06 = 19.88 mm
11. 10 mm diameter holes are to be punched in a steel sheet of 3 mm thickness. Shear strength of the material is 400 N/mm^{2} and penetration is 40%. Shear provided on the punch is 2 mm. The blanking force during the operation will be
(a) 22.6 kN
(b) 37.7 kN
(c) 61.6 kN
(d) 94.3 kN
(2 Mark, 2004)

Ans: a
Blanking force (F) = \frac { W }{ S }
=> F = \frac { \pi dt\tau \times \left( pt \right) }{ S }
=> F = \frac { \pi \times 10\times 3\times 400\times \left( 0.4\times 3 \right) }{ 2 }
=> F = 22.6 kN
12. A 2 mm thick metal sheet is to be bent at an angle of one radian with a bend radius of 100 mm. If the stretch factor is 0.5, the bend allowance is
(a) 99 mm
(b) 100 mm
(c) 101 mm
(d) 102 mm
(2 Mark, 2005)

Ans: c
Bend allowance, B = α(R + Kt)
α = Bend angle, radians,
R = Inside radius of the bend,
t = Sheet thickness, mm
K = Location of neutral axis
= 0.33 when R< 2t
= 0.50 when R> 2tB = 1\left( 100+0.5\times 2 \right) = 101 mm
13. Match the items in columns I and II.
(a) P – 6, Q – 3, R – 1, S – 2
(b) P – 4, Q – 5, R – 6, S – 1
(c) P – 2, Q – 5, R – 3, S – 4
(d) P – 4, Q – 3, R – 1, S – 2
(2 Mark, 2006)

Ans: d
14. Match the correct combination for following metal working processes.
(a) P – 2, Q – 1, R – 3, S – 4
(b) P – 3, Q – 4, R – 1, S – 5
(c) P – 5, Q – 4, R – 3, S – 1
(d) P – 3, Q – 1, R – 2, S – 4
(2 Mark, 2007)

Ans: d
15. The force requirement in a blanking operation of low carbon steel sheet is 5 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
(a) 3.0
(b) 4.5
(c) 5.0
(d) 8.0
(2 Mark, 2007)

Ans: a
Blank force (F) = \tau dt
=> \frac { { F }_{ 2 } }{ { F }_{ 1 } } =\frac { { d }_{ 2 }{ t }_{ 2 } }{ { d }_{ 1 }{ t }_{ 1 } }
=> \frac { { F }_{ 2 } }{ 5 } =\frac { 1.5d\times 0.4t }{ d\times t }
=> { F }_{ 2 } = 3 kN
16. In the deep drawing of cups, blanks show a tendency to wrinkle up around the periphery (flange). The most likely cause and remedy of the phenomenon are, respectively,
(a) Buckling due to circumferential compression; Increase blank holder pressure.
(b) High blank holder pressure and high friction; Reduce blank holder pressure and apply lubricant.
(c) High temperature causing increase in circumferential length; Apply coolant to blank.
(d) Buckling due to circumferential compression; decrease blank holder pressure.
(2 Mark, 2008)

Ans: a
17. The shear strength of a sheet metal is 300MPa. The blanking force required to produce a blank of 100mm diameter from a 1.5 mm thick sheet is close to
(a) 45kN
(b) 70kN
(c) 141kN
(d) 3500kN
(2 Mark, 2011)

Ans: c
Blacking force = \pi dt\tau
= \pi \times 100\times 1.5\times 300
= 1413716 N
\simeq 141 kN
Statement for Linked Answer Questions: 18 & 19
In a shear cutting operation, a sheet of 5mm thickness is cut along a length of 200mm. The cutting blade is 400mm long and zeroshear (S=0) is provided on the edge. The ultimate shear strength of the sheet is 100MPa and penetration to thickness ratio is 0.2. Neglect friction.
18. Assuming force vs displacement curve to be rectangular, the work done (in J) is
(a) 100
(b) 200
(c) 250
(d) 300
(2 Mark, 2010)

Ans: a
Length to cut (L) = 200 mm
Don’t take 400 mm. (It is the length of cutter)
penetration to thickness ratio (p) = 0.2
Work done = punching force\times Punch travel
= Lt\tau \times pt
= 200\times (5\times { 10 }^{ 3 })\times (100\times { 10 }^{ 3 })\times 0.2\times (5\times { 10 }^{ 3 })
= 100 J
19. A shear of 20mm (S = 20mm) is now provided on the blade. Assuming force vs displacement curve to be trapezoidal, the maximum force (in kN) exerted is
(a) 5
(b) 10
(c) 20
(d) 40
(2 Mark, 2010)

Ans: b
The effective shear is 10 mm (half of 20 mm). It is because the length of workpiece to be cut is half of the length of the cutter.
Maximum force when shear is applied
= \frac { W }{ S }
= \frac { 100 }{ 10\times { 10 }^{ 3 } } = 10 kN
20. Calculate the punch size in mm, for a circular blanking operation for which details are given below.
Size of the blank = 25 mm,
Thickness of the sheet = 2 mm,
Radial clearance between punch and die = 0.06 mm,
Die allowance = 0.05 mm
(a) 24.83
(b) 24.89
(c) 25.01
(d) 25.17
(2 Mark, 2012)

Ans: a
Punch diameter = Blank diameter – 2\times radial clearance – Die allowance
= 25 – 2\times 0.06 – 0.05
= 24.83 mm
21. A rectangular hole of size 100 mm × 50 mm is to be made on a 5 mm thick sheet of steel having ultimate tensile strength and shear strength of 500 MPa and 300 MPa, respectively. The hole is made by punching process. Neglecting the effect of clearance, the punching force (in kN) is
(a) 300
(b) 450
(c) 600
(d) 750
(2 Mark, 2014[2])

Ans: b
Punching force (F) = \tau \times Pt
= 300\times 2\left( 100+50 \right) \times 5
= 450,000 N
= 450 kN
22. In a sheet metal of 2 mm thickness a hole of 10 mm diameter needs to be punched. The yield strength in tension of the sheet material is 100 MPa and its ultimate shear strength is 80 MPa. The force required to punch the hole (in kN) is ____.
(2 Mark, 2016[3])

Ans: 5.02
The force required to punch (F) = \pi dt\tau
= \pi \times 10\times 2\times 80
= 5026 N = 5.026 kN
23. A 10 mm deep cylindrical cup with diameter of 15mm is drawn from a circular blank. Neglecting the variation in the sheet thickness, the diameter (upto 2 decimal points accuracy) of the blank is ____mm.
(2 Mark, 2017[1])

Ans: 28.72
Diameter of the blank (D) = \sqrt { { d }^{ 2 }+4dh }
= \sqrt { { 15 }^{ 2 }+4\times 15\times 10 } = 28.72 mm
24. The percentage scrap in a sheet metal blanking operation of a continuous strip of sheet metal as shown in the figure is _______ (correct to two decimal places).
(2 Mark, 2018[1])

Ans: 53.25
Area of hole (a) = \frac { \pi }{ 4 } { D }^{ 2 }
Area of Sheet metal required for a single blanking operation (A) = \left( \frac { 2D }{ 5 } +D \right) \times \left( \frac { D }{ 5 } +D \right)
The percentage of scrap in sheet metal:
= \left( 1\frac { a }{ A } \right) \times 100
= \left( 1\frac { \frac { \pi }{ 4 } { D }^{ 2 } }{ \left( \frac { 2D }{ 5 } +D \right) \times \left( \frac { D }{ 5 } +D \right) } \right) \times 100
= 53.25%