Dear Mr Rao

The torque (T), transmitted from a drum to a belt is given by the formule :

T= (Force in pulling belt part - Force in non pulling belt part) x Diam/2

The force ratio between the pulling part and the non pulling part of the belt is :

F(pulling) / F(non pulling) = e^(2.f.alpha)

in which :

f = friction factor between drum and belt

alpha = covered angle of belt over drum.

The torque can now be written as :

T = F(pulling) x (1 - e^-(2.f.alpha)) x D/2

T = F(non pulling) x (1 + e^(2.f.alpha)) x D/2

From this equation can be read :

Torque transmittal is higher with higher pre-stress

Torque transmittal is higher with higher friction factor

Torque transmittal is higher with larger covered angle

Torque transmittal is higher with larger diameter of drum

All common sense.

This explains your friends statement

best regards ■

Dear Shri Rao,

You have to first decide the pulley diameter which is suitable for the belt, in accordance with its carcass, as per DIN / ISO. Also, decide the torque value, which you want to transmit. This torque value is difference of belt tensions T1 and T2 multiplied by the pulley radius. Now, you have to maintain the ratio of T1 and T2 so that there is adequate frictional grip between pulley and belt. This is given by very well know formula ratio T1/T2^mu*wrap angle. This is the limiting ratio and tension T1 and T2 should be selected such that their actual ratio is less than or equal to the above selected ratio, in all working situations. The pulley diameter as selected above will rarely be too small to transmit the torque.

There is also one consideration that the pressure between belt and pulley periphery should not exceed certain value, but one rarely reaches to this level for the kind of application you are referring.

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916 ■

dear Mr Mulani,

The well known formula ratio should be:

T1/T2 = e^(2*mu*wrap angle).

May be you typed too fast

The 2 comes from the vector triangle, representing the pulling force in the belt and the normal force (N) to generate the friction.

Action = Reaction leads to the differential equation :

mu * 2 * pulling force * d(wrap angle) = d(pulling force)

Integration between wrap angle = 0 and complete wrap angle gives :

T1/T2 = e^(2*mu*wrap angle).

Mathematics must be handled with maximum accuracy.

All for now ■

Referring to my earlier reply, I had missed to mention e in the formula. Please read the formula as ratio T1/T2 should be less equal to e raised to (mu x wrap angle in radian).

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916 ■

Dear Mr Mulani,

I still miss the "2" in the exponent of the formula, or am I wrong that it should be there.

I made the derivation of the pormula again and came up with the "2" again.

If you wish, I can e-mail it to you

best regards ■

Dear Mr. Teus,

This is a very common formula. You can find the derivation of this formula in many of the mechanical engineering books pertaining to Design. There is no scope for error in this formula. Kindly check the engineering literature with which you are familiar. If you have got pertaining literature from companies like ContiTech Germany publication page-45, Good Year handbook page 2-4, DIN 22101 page 11, etc., you will find this formula.

Please do inform whether you found the matching references and satisfied with this answer.

Regards,

Ishwar G Mulani.

Author of Book : Engineering Science and Application Design for Belt Conveyors.

Author of Book : Belt Feeder Design and Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Email : parimul@pn2.vsnl.net.in

Tel.: 0091 (0)20 25882916 ■

Mr Mulani has given the correct formula.

It is still imprinted on my brain from my college training 30 years ago. ■

Dear Mr Mulani and designer,

As I said, "Mathematics must be handled with maximum accuracy"

You are right.

I checked my derivation again and found that I made an error in the force triangle. I took mistakenly d(belt angle) as 2*the d(wrap angle) and that was wrong.

Problem solved.

Thank you for your attention.

I will be more carefull next time.

best regards ■

Mr Rao,

The preceding mathematical treatise does confirm, even though it has not admitted it, that there is no correlation twixt pulley diameter & tramsmissable torque. Torque limits depend on the maximum allowable tension that the belt can sustain. ■

Dear Mr. Rao,

I suggest that to get good idea you please go through Mr. Mulani 's book- a very correct approach for design of pulley, shaft etc..in relation to torque.

Regards.

A.Banerjee ■

True John. However, there is a direct correlation between the belt strength and pulley diameter. By example, a typical drive pulley diameter should be sized to meet or exceed the two main factors:

1. Rubber shear strength under the steel cord should not excced 15 kg/sq cm if memory serves me correctly. Rubber shear strength is proportional to Fn=T/R where Fn denotes the radial force acting under the cable, T denotes the maximum unit belt tension on one cable, and R denotes the pulley radius

2. Pulley diameter should exceed 150 times (CDI criteria) the steel cord diameter (depends on weave) for a safety factor of SF= 6.7 :1. Thus, a 5 mm cord (ST-1500 N/mm) should have a high tension pulley diameter = 750 mm plus lagging or about 790 mm.

The ST-1500 N/mm belt, SF=6.7:1 can carry 769 kW for 1000 mm wide belt, at a belt speed = 5 m/s, and drive loss eff = 97%. Add or take away some belt width and hit the desired kW. At 200 kW, the belt will be pretty narrow and we still did not hit the 800mm pulley diameter. This was factored at a material density of 1500 kg/m^3.

Dear Graham, this is a theorectical exercise to illustrate the thread starters request. ■