Punching or blanking is a process in which the punch removes a portion of material from the larger piece or a strip of sheet metal. If the small removed piece is discarded, the operation is called punching, whereas if the small removed piece is the useful part and the rest is scrap, the operation is called blanking.

Hence, In punching and blanking operations the shear is provided respectively on punch face and die face respectively because of the cutting action will occur accordingly.

Note:

Blanking operation: Die size = blank size

Punch size = blank size – 2 x clearance

Piercing operation:

Punch size = blank size

Explanation:

Given:

D = 70 mm

Calculation:

For first draw, reduction ratio is 40 %

Thus,

\(\frac{{40}}{{100}} = 1 - \frac{{{d_1}}}{D}\)

\(\frac{{{d_1}}}{{70}} = 0.6\)

∴ d₁ = 42 mm

For second draw, reduction ratio is 30 %, thus

\(\frac{{30}}{{100}} = 1 - \frac{{{d_2}}}{{{d_1}}}\)

\(\frac{{{d_2}}}{{42}} = 0.7\)

∴ d₂ = 29.4 mm

Explanation:

Given:

D = 60 mm, C = 0.2 mm, Die allowance = 0.08 mm

Calculation:

For blanking operation,

Punch size = D – 2C – die allowance

Punch size = 60 – (2 × 0.2) – (0.08)

∴ Punch size = 59.52 mm

Concept: Work done during blanking operation = Maximum Force x Punch travel = F_max × (p × t)

Where F_max is the maximum Punching Force or Blanking Force, p is the percentage penetration required for rupture, t is the plate thickness.

Calculation:

Given, F_max = 2 × 10⁵ N, t = 4mm, p = 0.25 

Workdone in this blanking operation = Fmax × (p × t) = 2 × 10⁵ × 0.25 × 4 = 200 J

Explanation:

Given:

t = 4 mm, d = 40 mm, τ = 450 MPa

Calculation:

This is a punching operation.

Radial clearance = 0.0032 t × √τ

Thus, diametrical clearance = 2 × radial clearance

c = 0.0064 t√τ

\(c = \left( {0.0064} \right)\left( 4 \right)\sqrt {450} \) 

c = 0.543 mm

Now,

Punching force is given by

F = (πdt)τ

F = (π × 40 × 4)(450)

F = 226.19 ≃ 226.2 kN

Option 2) is correct.

Calculation:

Given:

D = 100 mm, t = 0.6 mm, d = 40 mm

\({\rm{Diameter\;of\;blank\;}}\left( D \right) = \sqrt {{d^2} + 4dh}\)

\(100 = \sqrt {{{40}^2} + 4\left( {40} \right)h}\)

∴ h = 52.5 mm

Now,

For a reduction ratio of 35 %,

After first draw,

\(\frac{{35}}{{100}} = 1 - \frac{{{d_1}}}{D}\)

\(\frac{{{d_1}}}{{100}} = 0.65\)

∴ d₁ = 65 mm

After second draw,

\(\frac{{35}}{{100}} = 1 - \frac{{{d_2}}}{{{d_1}}}\)

\(\frac{{{d_2}}}{{65}} = 0.65\)

∴ d₂ = 42.25 mm

Explanation:

Given:

D = 60 mm, t = 5 mm, τ = 450 MPa

P = 0.3

Calculation:

Given case is of punching,

Thus, diameter of punch = diameter of hole

d_punch = 60 mm

diameter of die = D + 2C

Also,

\(C = 0.0032t\sqrt \tau\) 

\(C = 0.0032 \times 5 \times \sqrt {450} = 0.339\) 

D_die = 60 + 2(0.339) = 60.678 mm

Now,

Work done = F × (pt)

Work done = τ (πdt) (pt)

Work done = (450) (π × 60 × 5) (0.3 × 5)

∴ Work done = 636.17 J

Hence, correct options (b) and (c) 

Explanation:

Given:

Punch size (d) = 85.2 mm = 8.52 cm, Yield strength (τ) = 2300 kg/cm²

Thickness of sheet metal (t) = 2.4 mm, Blank diameter (D) = 136 mm

Now,

Die opening size = Punch size + 2 (thickness of sheet metal)

∴ Die opening size= 85.2 mm + 2(2.4)

∴ Die opening size = 90 mm

Now,

Drawing force (F) \(= \pi \times d \times t \times \tau \left[ {\frac{D}{d} - C} \right] \)

Drawing force (F) \(= \pi \times 9 \times 0.24 \times 2300 \times \left[ {\frac{{136}}{{90}} - 0.6} \right]\)

Explanation:

Given:

D = 75 mm, t = 3.2 mm

C = 5 % of t

Calculation:

For blanking operation,

Diameter of die = Diameter of blank to be produced

Thus,

D_die = 75 mm

The clearance will be given on punch,

Thus, Diameter of punch = D – 2C

\({\rm{C}} = {\rm{\;}}5{\rm{\;\% \;of\;}}t = \frac{5}{{100}} \times 3.2\)

∴ C = 0.16 mm

Now,

D_punch = 75 – (2 × 0.16)

D_punch = 75 – 0.32 = 74.68 mm

∴ Only (d) is the correct answer.

Explanation:

Given:

k = 0.33, R = 5 mm, t = 3 mm

Calculation:

For semi-circle, initial length

L₁ = πR = 5π mm, L₂ = 20 mm

And, for quarter-circle, initial length

\({L_3} = \frac{{2\pi R}}{4} = \frac{{5\pi }}{2}mm\)

Since,

Bend allowance = θ (R + kt)

For semi-circle, bend allowance = π (5 + 0.33 × 3)

∴ Bend allowance = 18.818 mm

Now,

For straight part, bend allowance = 0

For quarter-circle,

bend allowance \({\rm{}} = \frac{\pi }{2}\left( {5 + 0.33 \times 3} \right)\)

∴ Bend allowance = 9.409 mm

Thus,

Total bend allowance = 18.818 + 0 + 9.409

Total bend allowance = 28.227 mm