Calculation of the Load Coming at the Bottom of the Slow Speed Brake

Posted in: , on 16. May. 2015 - 11:48

Hello everyone,

I wish to learn to calculate the following:

1) The load coming at the bottom of a slow speed brake (Design parameters can be assumed).

2) During braking, the developed radial load / axial load.

Any book reference or guidance would be appreciated.

Regards

Rajat

Re: Calculation Of The Load Coming At The Bottom Of The Slow Sp…

Posted on 17. May. 2015 - 10:03

Hello,

In case of conveyor of positive power, brake is used to stop conveyor. Firstly designer decides (chooses) stopping time for the conveyor. Then he calculates the required torque to achieve this, taking into consideration the conveyor inbuilt resistance to motion and moving mass of the system. Such calculations are according to the well known basic laws of motion.

Your statement ‘Load at the bottom of brake’ does not have necessary clarity. I think you mean to say forces at brake unit / assembly bottom face (its foundation load). Such foundation load (force on fixing bolt-nuts) will simply be brake torque divided by arm (average span) of fixing bolt-nuts. At one end it will be uprooting and at the other end it will be pressing down.

Shoe brake does not have axial force. Caliper brake will be axially free and generally will not have axial force.

In case you have some specialised system, you can solve it accordingly.

In case of regenerative conveyor (conveyor of negative power), the situation is somewhat different and quite complex and cannot be narrated here (the braking system has to be suitable for steady speed operation whether conveyor has negative or positive power, and on command it should be able to stop conveyor, etc. The regenerative conveyor will also have operational phase of positive power when it is empty).

Ishwar G. Mulani

Author of Book: ‘Engineering Science And Application Design For Belt Conveyors’. Conveyor design basis ISO (thereby book is helpful to design conveyors as per national standards of most of the countries across world). New print Nov., 2012.

Author of Book: ‘Belt Feeder Design And Hopper Bin Silo’

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India. Tel.: 0091 (0)20 25871916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com

Deary Deary Me!

Posted on 17. May. 2015 - 11:10

Please reconsider where your brake will be mounted.

Brake torque values are 'referred' according to the square of the speed ratio and brakes are mounted accordingly.

Modern vehicles use transmission brakes and retarders whenever possible to reduce wear and tear on the wheel brakes.

Take a leaf out of their book.

Even before that 1910's development it would be unthinkable to install a brake downstream of the conveyor gearbox. My witness for the defence is the 1953 Sutcliffes Conveyor Handbook.

External shoe brakes are not self energising and therefore do not need taper relief to remove or free a siezed drum and so there is very little axial force assumed. Unless...Heat dissipation is crucial in brakes and if the thermal flux is impeded there can be considerable local overheating with consequent distortion and this can lead to fun and games. That is why any industrial brake should be foot mounted.

John Gateley johngateley@hotmail.com www.the-credible-bulk.com

Re: Calculation Of The Load Coming At The Bottom Of The Slow Sp…

Posted on 18. May. 2015 - 07:42
Quote Originally Posted by I G MulaniView Post
Hello,

In case of conveyor of positive power, brake is used to stop conveyor. Firstly designer decides (chooses) stopping time for the conveyor. Then he calculates the required torque to achieve this, taking into consideration the conveyor inbuilt resistance to motion and moving mass of the system. Such calculations are according to the well known basic laws of motion.

Your statement ‘Load at the bottom of brake’ does not have necessary clarity. I think you mean to say forces at brake unit / assembly bottom face (its foundation load). Such foundation load (force on fixing bolt-nuts) will simply be brake torque divided by arm (average span) of fixing bolt-nuts. At one end it will be uprooting and at the other end it will be pressing down.

Shoe brake does not have axial force. Caliper brake will be axially free and generally will not have axial force.

In case you have some specialised system, you can solve it accordingly.

In case of regenerative conveyor (conveyor of negative power), the situation is somewhat different and quite complex and cannot be narrated here (the braking system has to be suitable for steady speed operation whether conveyor has negative or positive power, and on command it should be able to stop conveyor, etc. The regenerative conveyor will also have operational phase of positive power when it is empty).

Ishwar G. Mulani

Author of Book: ‘Engineering Science And Application Design For Belt Conveyors’. Conveyor design basis ISO (thereby book is helpful to design conveyors as per national standards of most of the countries across world). New print Nov., 2012.

Author of Book: ‘Belt Feeder Design And Hopper Bin Silo’

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India. Tel.: 0091 (0)20 25871916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com

Dear Sir,

You are right. I wanted to mean "Forces at the bottom of a brake unit (Foundation load).

Regards

Rajat