Solving Conveyor Belt Stress for Beginners

AI Thread Summary
The discussion revolves around calculating the maximum length of a conveyor belt to prevent failure due to stress while transporting 140 t/h up a 10-degree gradient. Key factors include the belt's width, speed, ply, maximum working tension, and various load coefficients. The tension equation provided is crucial for determining the balance between belt tension and potential failure modes, such as snapping or slipping. Participants emphasize the need to analyze both tension limits to find the maximum feasible length. The complexity of the problem highlights the importance of understanding the relationship between force, velocity, and power in conveyor systems.
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Homework Statement



I am pretty confused with how to go about answering this question:
Calculate the maximum lenth of conveyor which can be used to avoid the belt failing from stress, for an installation carrying 140 t/h up a gradient of 10 degrees, if the belt width is 1.1m, the speed is 1.7 metres per second, the belt uses an 8ply belting and the max working tension is 6 kN/m ply. there's a 2 drum drivehead with an angle of wrap of 400 degrees, the coefficient of grip is 0.2.
Idler coefficients = 0.03 (when belt is empty)
0.04 (for materials)
mass of mivong parts = 65 kg/m
end pulleys etc= 35m of extra empty belt

I have looked into the belt tension equation and put a lot of effort into it, but I'm really confused,
any help in understanding it would be highly appreciated
from a confused beginner!

Homework Equations



Tb = 1.37*f*L*g*[2*mi+ (2*mb + mm)*cos (δ)] + (H*g*mm)
Where,

Tb is in Newton.

f = Coefficient of friction

L = Conveyor length in meters. Conveyor length is approximately half of the total belt length.

g = Acceleration due to gravity = 9.81 m/sec2

mi = Load due to the idlers in Kg/m.

mb = Load due to belt in Kg/m.

mm = Load due to the conveyed materials in Kg/m.

δ = Inclination angle of the conveyor in Degree.

H = vertical height of the conveyor in meters.

The Attempt at a Solution



I genuinely have no idea, i just can't understand what to do, it seems like a completely random task we've been provided

What i do know is
Tb will be somehow derived from the 7.3 kN/m ply
Also, maybe the equation F1/F1= e to the power of mu x alpha
where mu is coefficient of friction alpha is wrap angle

Any input will be highly appreciated!
 
Last edited:
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I'm just off to bed as it's late here but...

The longer the belt the higher it has to lift the material. The rate (mass per second) is fixed, so the longer the belt the more power is required. Now remembering that power is essentially force * velocity, and noting that the velocity is also fixed you can conclude the force transmitted through the belt must increase. Obviously it's more complicated but that's the idea.

Then at first glance the problem appears to provide enough info for two possible failure modes. Perhaps you have to check which is limiting with regard to belt tension?...

1) Too much tension and the belt snaps.
2) Too much tension and the belt slips on the pulley.

and then work backwards from that to the max height and length.

I've actually never done any problems to do with loads on conveyor belts but that's my take on the problem.
 
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