Tensions in Multiple V-belt Drive System

You should find the formulas here....

https://books.google.com/books?id=c_ZLBgAAQBAJ&pg=PA215&lpg=PA215&dq=dynamic+loads+3+pulley+drive+system+shaft+bearing+loads&source=bl&ots=ZGxgRq4K99&sig=KuyPyfXr8wIYkIVkH0WDDOob0XM&hl=en&sa=X&ei=vykcVaXpCYyZgwTL-ICABg&ved=0CD8Q6AEwCA#v=onepage&q=dynamic%20loads%203%20pulley%20drive%20system%20shaft%20bearing%20loads&f=false

Solar Eagle,

Thanks but nowhere in the link you have sent solves tensions for a multiple V-drive system. So it doesn't help.

I should point out that I know absolutely nothing about this sort of thing, but:-

Try here

https://books.google.co.uk/books?id=E-_gSWmjkzAC&pg=PA38&dq=tension+in+V+belt&hl=en&sa=X&ei=mDocVeycHo-p7AaBwYHIBA&ved=0CDQQ6AEwAA#v=onepage&q=tension%20in%20V%20belt&f=false

If you move down to page 47 it shows an idler tensioner applied to the outside of the belt rather than the inside. This enables you to use a smaller tension ratio because the ratio is dependent on the arc of contact.

It also shows the slack side tension equal at both sides of the idler.

If you can implement the design in that way I think things will be much simpler.

SP section = metric V-belt I assume?

There are links on this page http://www.engineersedge.com/vee_flat_belt_menu.shtml from our friends at Engineer's Edge which should help. Once you determine the minimum tensions, you can factor that up, and with a little vector geometrytrig math, you should be able to calculate the shaft loads. Or am I missing something?

There is probably design software available, but I didn't see any in my quick search.

Thanks guys for the contributions. I guess there are some clues to the puzzle but it still requires the tedious hand calculations. I'll keep looking for a software.

An obvious thought experiment comes into picture. Suppose one brought in a third pulley C to take up the slack in a 2 pulley drive system, and then kept moving the pulley in the X direction away from A and B, you'd expect the tensions in strands 3 and 1&2 to gradually increase. The Wallace Erickson book link shared earlier points out that the slack side tensions in both strands 1&2 should be equal. Therefore, there are only 2 unique tensions to the problem, T_tight in strand 3 and T_slack in strands 1&2 (equal). Would you all agree with this?

Another perhaps childishly curious question I have is about power flow in the system. Suppose it was pulley B at which I needed 90 kW and pulley A was connected to the auxiliary side of an internal combustion engine with enough juice. What power value does the engine settle at in providing to pulley A? I believe it has to be some value greater than 90kW , from which pulley C takes a bit and passes on the rest to B. For example, if the engine supplied 120kW to A, then B would take 90kW while C takes the remaining 30kW. Is my intuition correct on this point?

I repeat my disclaimer: I know nothing about this sort of thing, but:-

For example, if the engine supplied 120kW to A, then B would take 90kW while C takes the remaining 30kW. Is my intuition correct on this point?

I think you'll find that most of the "lost" 30KW is between A and the belt; the second greatest loss is between B and the belt, and, the smallest loss is between C and the belt.

The loss due to C can be reduced more by reducing the length of belt in contact with C: does it really need to turn through a right angle as shown? I think that's what Tornado is saying is "goofy" below.

Thanks Randall.

Why is most of the lost power between A and the belt ?(I consider V-belts to be 90-95% efficient) Perhaps you could point out to me the loss percentages for all three points?

I know this drive does look a little goofy but both B and C are attached to the frame of an air cooled heat exchanger and there's little I can do to shift the c-c distances drastically. The idler diameter could be increased to 1.2~1.4 times the size of A but I don't know if that'll make it any less goofier.

I know this drive does look a little goofy but both B and C are attached to the frame of an air cooled heat exchanger and there's little I can do to shift the c-c distances drastically.

Can you move the motor and pulley A?

I think all the calculations are in the book I linked to in post 3.

Hi Randall,

Unfortunately I cant do that, the centers of A and C have to be vertically inline. Eitherway, how is the second arrangement superior? Is it because the arc of contact is lesser at the idler, therefore the losses are lesser?

Not as important as increased arc of contact with driver and driven pulleys.

Most important is that the idler does not have a V groove: it's pressing against the flat back of the belt; that's why the two "slack" legs have the same tension.

If you can get through it, this Optibelt manual addresses drives with idlers:
http://www.optibelt.com/fileadmin/power_transmission/service/dokumentation/TH_KR_GB.pdf

Unsolvable without knowing how hard the idler is forced into the "loose" portions of the belt. (The layout is goofy, anyway.)

Without knowing the length of the belt and its coefficient of elasticity, you cannot solve it. It's simple to solve for the differences of the tensions once you know the torque supplied by 2 of the 3 pulleys, but the sum of the three tensions depends on the belt.(If it gets tighter, all three tensions will increase.)

Missed the idler part. You need the spring force on the idler to solve.

Surely the only relevant figure is the drive side tension? The other sides will be less and thus irrelevant.

Del

Damn, I've gone invisible again...

Del