How much force/torque is required to turn the two axles

[Hammad]

Homework Statement

hi, I am designing a plastic bottle shredder machine, the mechanism looks like this roughly except that there are spur gears which enable the axles to rotate into each other. The axle on the left has 5 cutters which exert a total force of 1371.45 N in the given direction, this force is actually what i calculated the shear force of each cutter for a plastic bottle (PET material), and the total value of 1371.45 includes force exerted by the total weight of the blades at axle on the left. The right axle has 4 blades and total force plus weight is 1097.16 N. The curved arrow shows the direction of rotation of the respective axles.

i want to calculate the force and the torque required to move these two axles using a handle whose distance from centre of axle is 19.4 cm but the point of application of handle is 24 cm from the axle as well.

The maximum force which i have applied in the design is 400 N and a torque of 77.8 N m, it it sufficient to turn the axles for this design or not? actually i want to proof that it will require not more than 400 N to turn the handle and shred the plastic bottle.

i have not calculated the RPM, but it should be less than 60 as it is operated by a human, Horse power is also 1 or less as 1 HP is normally applied by an average human. The weight of the entire system is 40 Kg. the blades and the axle are made of stainless steel.

This is my system

 

[scottdave]

I'm not sure what you mean by "weight" of the bade where you put the arrow, Also you have the shear force pointing down but shouldn't it point up against the turning? Is that value for 1 pair of bladeaor all of them?

 

[Hammad]

I'm not sure what you mean by "weight" of the bade where you put the arrow, Also you have the shear force pointing down but shouldn't it point up against the turning? Is that value for 1 pair of bladeaor all of them?

yeah i think you are right about direction of shear force.

you are also correct about weight of the blade should be at the centre and not edge of the blade., actually i added up the weight force of the blade which is 102.25 N left and 81.8 N right so shear force left is 1269.2 N and 1015.36 N right.

the value of 1 HP is the maximum power which can be applied by a human on the handle so i have taken HP value for turning both the axles as 1 HP.

 

[CWatters]

So 1269N is the shear force required to cut the plastic.

What is the 1015N force? Is that the force required to cut the plastic on that side because the sides have a different number of teeth?

 

[Hammad]

So 1269N is the shear force required to cut the plastic.

What is the 1015N force? Is that the force required to cut the plastic on that side because the sides have a different number of teeth?

no its not number of teeth but, left axle has 5 cuttering blades, right one has four.

 

[CWatters]

Ok so to calculate the load torque add those two forces together and multiply by the radius in meters.

 

[Hammad]

Ok so to calculate the load torque add those two forces together and multiply by the radius in meters.

im actually confused, how can i add the two forces and multiple by the radius because the distances from their respective axles is different relative to the handle and they are moving counter to each other?

 

[Merlin3189]

Actually you are calculating the torque on each axle, then adding those torques because the two axles rotate at the same speed and in the compatible direction.
I use that strange description because I can't think of another term for it!.. The linking gears cause the wheels to rotate "counter" to each other, but the torques are also opposite in direction. On the left wheel the torque on the wheel is anticlockwise. On the right wheel the torque on the wheel is clockwise.

You can in such a simple case, just add the magnitudes and intuit the direction, because it is obvious.
If you want to be more abstract about it, you can assign a positive and negative direction of rotation.
Then you get a positive torque and a negative torque, but one of the torques has its direction reversed by the gears,
so you get T = TL - TR and if TL= -1269 and TR= +1015, then T= -1269 - 1015 = -2284 taking CW as positive.

You need the distance between the force and the axle to which that blade is attached. The handle is irrelevant at that stage.

Because the two wheels are linked to rotate at the same rate, you can just combine their torques (ideally with an extra torque for any frictional losses - but you'll probably neglect this for now.)

Since the handle turns at the same rate as the discs, the torque applied by the handle is this total torque, but in the opposite sense. So the resistance torque is ACW and the handle must apply a CW torque.

BTW I am assuming the wheels rotate at the same rate, because it looks like that in the pix. But you could have them rotate at different rates. Then you'd have to take the gear ratio magnitude into account, as well as the sense.

 

[CWatters]

im actually confused, how can i add the two forces and multiple by the radius because the distances from their respective axles is different relative to the handle and they are moving counter to each other?

I assumed that both forces act at the same distance from their respective axle and that both sets are the same diameter.

The length of the handle doesn't change the torque required. It just changes the force required on the handle.

Once you have calculated the force required on the handle you can work out the forces on the bearings and mounting bolts etc.