# CHAPTER 18 - Forming and Shaping Plastics and Composite Materials¶

## EXAMPLE 18.1 - PG NO 484¶

In [1]:
#example 18.1
#page no. 491
# Given that
W=400.#in mm Lateral(width) Dimension of a plastic shopping bag

# Sample Problem on page no. 484

print("\n # Blown Film # \n")

#part(a)

P=2.*W#in mm Perimeter of bag
D=P/3.14#in mm blown diameter calculated from Permeter=3.14*diameter
#Given in this process, a tube is expanded to form 1.5 to 2.5 in times the extrusion die diameter, so take maximum value 2.5
Dd=D/2.5#Extrusion die diameter
print'%s %d %s' %("\n\n Extrusion Die Diameter =",Dd,"mm")

#part(b) is theoritical

 # Blown Film #

Extrusion Die Diameter = 101 mm


## EXAMPLE 18.2 - PG NO 488¶

In [2]:
#example 18.2
#page no. 488
# Given that
W=250.#in ton Weight of injection moulding machine
d=4.5#in inch diameter of spur gear
t=0.5#in inch thickness of spur gear
#Gears have a fine tooth profile

# Sample Problem on page no. 488

print("\n # Injection Molding of Parts # \n")

#because of fine tooth profile pressure required in the mould cavity is assumed to be of the order 100MPa or 15Ksi

p=15#inKsi
A=(3.14*(d**2))/4#in inch^2 area of the gear
F=A*15*1000
n=(W*2000)/F #weight is converted into lb by multiplying it by 2000
print'%s %d' %("\n\n Number of gears that can be injected =",n)

#print'%s %d %s' %("\n\n Force required is = ",A/10000,"MN" )

# Second part of this question is theoritical

 # Injection Molding of Parts #

Number of gears that can be injected = 2