We have been asked to redesign the conveyor belt system on a Cold Milling Machine (CMM) by PaviNgZ Inc.

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Introduction

We have been asked to redesign the conveyor belt system on a Cold Milling Machine (CMM) by PaviNgZ Inc. This is because PaviNgZ are extending their range of CMM by making one with an extended conveyor belt. We are to outline the details of the extended belt, the power transmission and the requirements.

The conveyor has to

  • Transport material 2m high
  • Transport material 6m across
  • Material will enter at 30-120 ton/hr
  • Belt speed can’t exceed 2m/s
  • Conveyor system has to be built around the frame provided
  • Only minor modifications can be made to the frame
  • These alterations have to be made from standard steel sections of the same grade
  • Shafts to be manufactured from carbon steel
  • Conveyor has to last for 5 years, running 200 days a year and at an average of 4 hours per day. This comes to 4000 hours
  • The belt must be powered by a 3 phase AC motor
  • The design must consider its operating environment
  • The design must be safe
  • The belt must have a take up of 100mm
  • The max angle of surcharge of material is 25 degrees
  • The average density of material is 800kg/m^3
  • The design must be practical and efficient

Our approach to this project

  • Compile a list of specifications
  • Calculate the tensions in the belt based on idler spacing and belt velocity that we specify
  • Select power transmission components that will deliver the required power – geared motor, chain drive
  • Selection of other components – bearings, take up units
  • Design mounts for power transmission system
  • Design cover for any hazardous areas

Performance

The conveyor uses 860 W of power and has the ability to transport 120 tonnes of bulk material an hour.

Component list


Weights of Motor mount, take up units, pillow stock and pulley cover are only approximations.

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Total weight of the conveyor comes to 591.2kg


Calculations

1. Determining Force on tail bearing

Tt = 2368.6 N

Summing forces in Y direction:

Tt*sin25 ° - 0.5mg = 912.7 N

 Where m is mass of pulley

Forces in x direction Tt*in25 = 2146.7 N

Magnitude of force on the bearing = √ (912.7^2 + 2146.7 ^2) = 2332.7 N

Total force on each bearing on the tail pulley is 2332.7 N

The life of this bearing is given by the equation

L = (C/P) ^k * 10 ^6

Where k ...

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