Refrigeration & Equipment Engineering

8. FOOT FLANGER MACHINE

FOOT FLANGER MACHINE

Aim: To give flanges on both sides of the reformed can body and measure the power transmission elements of a foot flanger machine.

Apparatus:

1. Reformed can body
2. Outside caliper
3. Measuring cloth tape
4. Thread (Nylon)

Working principle: The machine is foot operated and the power from the electrical motor is transmitted to the reduction gearbox by means of pulleys and ‘V’ belt. The function of the reduction gearbox is to reduce the speed ratio. The reduced speed actuates the two punches in forward and backward direction with the help of spindle, cam and lever mechanism. Therefore by keeping the reformed can body in between the two punches on the foot flanger bed stand, the required flange can be obtained on both sides of the can body. Depending upon the length of the can body the stroke length of the punches can be adjusted on the foot flanger bed. Nearly 40 to 50 can flanges per minute can be done using the machine.

The machine essentially consists of the following parts.

1. Cast-iron body
2. Electrical motor
3. Reduction gear box
4. Spindle, cam and lever mechanism
5. Pulleys
6. Head stock with assembled punches
   

Procedure:

• The condition of the machine should be observed and if necessary greasing, oiling and cleaning should be done.
• Keep the reformed can body at the center of the bed on the stand
• Switch on the starting button
• Press the foot pedal once firmly and release
• Take the flanged can body away for the next operation
• Switch off the main switch
  

Precaution: Hold the reformed can body at the center during flanging and not at the ends of the body.

Observations: Following observations have to be recorded

Sl.No.	Parameters
1.	Horse power of the motor (HP/KW)
2.	Speed of the motor(rpm)
3.	Diameter of the driver pulley(cms)
4.	Diameter of the driven pulley(cms)
5.	Center distance between the driver and driven pulleys(cms)
6.	Speed of the driver (rpm)
7.	Speed of the driven (rpm)
8.	Type of gears used in the reduction gearbox
9.	No. of gear used (No.)
10.	No. of teeth on each gear (No.)
11.	Type of belt used
12.	No. of belts used (No.)
13.	Type of belt drive system

Calculations:
1)

N₂

Gear ratio = --------------

N₁

T₁

= -------------

T₂

Where,

N₁ = Speed of the driver gear in rpm
N₂ = Speed of the driven gear in rpm
T₁ = No of teeth on driver gear
T₂ = No of teeth on driven gear

2) Velocity ratio of belt drive

N₂

= ------------

N₁

d₁

= ------------

d₂

Where,
N₁ = Speed of the driver in rpm
N₂ = Speed of the driven in rpm
d₁ = Diameter of the driver in cm
d₂ = Diameter of the driven in cm

3) Length of the belt

(r₁ – r₂)²

L = (r₁ + r₂) + 2x + ------------

x

Where,
r₁ = radius of the driver pulley in cms
r₂ = radius of the driven pulley in cms
x = distance between the center of the pulley in cms

2 π NT

HP = ------------

4500

4) Diameter of the shaft

π

T = ------fsd³

16

Where,
Fs = Shear stress in Kg/ cm²
T = Torque in Kg-m
N = rpm
HP = Horse power
d = Diameter of driver shaft in cms