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UDL on pulley from rope tension 1

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shorion

Mechanical
Nov 5, 2013
29
Untitled_tlr0ad.png


A lot of posts, ifs, ahhhhs and opinions but no posts that seem to be able to point to an equation and go 'heres the equation'.

Theres the good ole 2T/Dd
Throwing some figures at it
(2 x 10,000) / (0.5x0.006) = 6.67 MPa
as a udl on say 135 deg angle of wrap and a 6mm rope squash width
F = PA = 6.67*(2*pi*0.5)*(135/360)*0.006 = 47 kN
as a UDL
47 / (2*pi*0.5)*(135/360) = 39.9 kN/m?

sound about right?... Thoughts?
 
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but a browse of the responses is ...

I cannot directly assist you ...
I must admit I do not have extensive knowledge in this area....
i propose you seek out some sources on belt conveyor re belt pulley design for deeper insight....

The internet sigh... lol
 
You used the value of the radius of the pulley rather than the diameter, so divide your answer by 2.
 
Compositepro

Cheers - mega thanks. Got a wishlist or something similar I can buy something for you, wine, spirits etc? :D

Test result on the disk came out the same. Had load cells in spokes to see if the approximation would meet the calculation. Came out real nice, (very close) and I got a pat on the back.
 
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