Take $C=0.69, σ_yt = 450N/mm^2$

Mumbai University > Mechanical Engineering > Sem 5 > Production process 3

Marks: 8 M

Year: May 2015

Given data:

d = 50 mm

r = 1.6 mm

t = 1.2 mm

h = 50 mm

1.Size of the blank

The ratio of d/r = 50/1.6 = 31.25

The ratio is more than 20,

$D = \sqrt{d2+4dh} = + \sqrt{50X50±4X50X50}$

D = 111.8 mm

Blank diameter is theoretical in order to get a smooth edge we add 3.2 mm on each side

D = 111.8 + 3.2 × 2 = 118.2 mm

2.Percentage reduction

Percentage reduction is given by

= (1 – d/D) X 100 + (1-50/118.2) X 100 = 57.7 %

%reduction permissible for the first draw = 45 to 50%

Hence cup cannot be drawn in one draw

3.Number of draws

Ratio of h/D = 50/50 =1

Therefore, No of draws = 2

Standard Table for number of draws

4.Punch and die radii (R)

The punch and die radii is considered as 4 times the die thickness

R = 4t = 4 X 1.2 = 4.8 mm

R = 4.8 mm

5.Die clearance (C)

The die clearance is considered as 1.25 times the blank thickness

C = 1.25 t = 1.25 X 1.2 = 1.5 mm

C = 1.5 mm

6.Drawing pressure (F)

F = πdtσt(D/d -C)

= π X 50 X 1.2 X 450 X (118.2/50 – 0.69)

F = 142 X 10^3 N = 142kN