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SAIYAM KUMAR • Jan 6, 2017

# crank and crank arm calculations

hi
i want to give a motor driven crank drive to a vibratory conveyor. i have done calculations on motor rpm required. please help me out regarding crank calculations.
Jay Kosta • Jan 6, 2017
What type of crank calculations do you need?
Crank arm length?
Crank arm strength?
Crank bearing design / strength?
Motor power?
Does the motor do any other function than to drive the crank?
What is attached to the crank arm at the converyor?
If the crank produces linear motion at the conveyor, how much linear distance is needed? Does the amount of distance need to be variable, or is it constant?

Jay Kosta
Endwell NY USA
SAIYAM KUMAR • Jan 6, 2017
Yes i
Jay Kosta
What type of crank calculations do you need?
Crank arm length?
Crank arm strength?
Crank bearing design / strength?
Motor power?
Does the motor do any other function than to drive the crank?
What is attached to the crank arm at the converyor?
If the crank produces linear motion at the conveyor, how much linear distance is needed? Does the amount of distance need to be variable, or is it constant?

Jay Kosta
Endwell NY USA
SAIYAM KUMAR • Jan 6, 2017
Yes i need the calculations for crank arm length, strength bearing design. At the end of the crank arm is a 16 kg Stainless steel plate which is 1 mm thick. 8 foot long nd 1.5 foot wide. Motor is set to run at 450rpm to achieve a load speed of 1.5 inch per second on the vibratory conveyor. Motor doesnot perform any other function than to drive the crank. Linear distance is constant which must be the amount of travel of load that is 8 foot at a speed of 1.5 inch per second. Apart from it a total distributed load on the conveyor is 5 kg so that makes it approx 21 kg load on conveyor which the crank has to drive. 6 springs are there 3 on each side of conveyor which will store nd release energy during vibration.
SAIYAM KUMAR • Jan 6, 2017
Yes i need the calculations for crank arm length, strength bearing design. At the end of the crank arm is a 16 kg Stainless steel plate which is 1 mm thick. 8 foot long nd 1.5 foot wide. Motor is set to run at 450rpm to achieve a load speed of 1.5 inch per second on the vibratory conveyor. Motor doesnot perform any other function than to drive the crank. Linear distance is constant which must be the amount of travel of load that is 8 foot at a speed of 1.5 inch per second. Apart from it a total distributed load on the conveyor is 5 kg so that makes it approx 21 kg load on conveyor which the crank has to drive. 6 springs are there 3 on each side of conveyor which will store nd release energy during vibration.
Jay Kosta • Jan 6, 2017
With a motor speed of 450 rpm, the steel plate will also cycle at 450 pm.

If you want the 'total distance moved of the plate' to be 1.5 inches in 1 second, the crank arm would have to be very short. Something about (1.5/4) / 450 = tiny inches.
So the plate would vibrate at 450 pm and move only a tiny distance with each rotation of the motor.

Perhaps I do not understant the linerar motion of the plate.
Is the motor providing the 'vibration motion' to the plate?
Or is the plate supposed to move with a total 'stroke length' of 8 feet at a speed of 1.5 inch/second? (I don't think that is possible without additional gearing to reduce the rpm of the crank).
SAIYAM KUMAR • Jan 6, 2017
ok i ll clear u the concept.
i need to provide a vibratory motion to a 8 foot pan through a motor driven crank. for that i will attach the motor and a crank to a base frame which is fixed. the base frame is attached to the steel pan of mass 16kg on which i need to convey some load from one end to other linearly. the crank arm is attached to the pan. the pan is attached to the base frame by means of spring. now when the motor drives the crank the movable pan vibrates on the base frame by means of spring. while it is vibrating we put the load on one end. due to the vibration the load moves to the other end at a speed of 1.5inch/sec. the capacity of pan is 0.03TPH. and for that i need to run the motor at 450rpm which i have calculated. now what i am not able to calculate is crank arm dimensions, its strengh its design etc
Jay Kosta • Jan 6, 2017
How did you calculate that a motor speed of 450 rpm would move the load at a rate of 1.5 inch/minute WITHOUT using a crank length (vibration length) measurement?

Moving the load at 0.03 ton / hour seems like a very light weight load for the size of the pan - do I understand correctly?
What is the total weight of the load be on the pan, and what type of bearings (rollers?) support the pan? The power of the motor and strength of the crank needs to be adequate to drive that total load. It also seems that a flywheel on the crankshaft would be necessary or helpful (or do you think the springs alone are adequate)?
SAIYAM KUMAR • Jan 6, 2017
there is a formula to calculate the linear flow rate of the load. It is
TPH=WdBDs/4800
TPH is capacity in tonnes per hour=0.03
d is the width of the pan=18 inch
BD is bulk density= approx=1lb/ft3
s is the flow rate. it came out 1.5 inch/sec through this formula.
check out this link for further calculation
since my load is lumpy i assumed the amplitude 0.9mm and omega =2800 per min from the table. and velocity is coming out to be same as that of the flow rate. i took the angle of oscillation as 30 degree. rpm came out to be around 450.
yes it is a very light weight load
total weight of load on the pan is 5 kg*9.81=49.05N but the crank has also run the 16kg*9.81=156.96N of pan. so total load crank has to run is about 206N. there are no bearings which support the pan. only the springs support the pan. i dont think there is a need of flywheel as the load is very light. check out this link. it will make u clear.
Mechanical Conveyors
do take a look at the diagram.
SAIYAM KUMAR • Jan 6, 2017
correction in the formula
W is the width of the pan =18 inch
d is the depth of the material=1inch
Jay Kosta • Jan 6, 2017
SAIYAM KUMAR
...
s is the flow rate. it came out 1.5 inch/sec through this formula.
check out this link for further calculation
since my load is lumpy i assumed the amplitude 0.9mm and omega =2800 per min from the table. and velocity is coming out to be same as that of the flow rate. i took the angle of oscillation as 30 degree. rpm came out to be around 450.
yes it is a very light weight load
...
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Thank you for the link to the eBook "Mechanical Conveyors: Selection and Operation" - it helped me to better understand the device you described.

The file:///C:Users/......
So, I don't understand how your results were obtained.

Have you considered the effect of having the pan supported by springs - as is mentioned in section 5.5.3 ? especially unwanted vibrations, etc.

It seems that you already think you know the amount of eccentricity (crank stroke length) needed. And the length of the crank arm from pivot-point to pivot-point seems to only be a concern if the moving weight of the crank arm would cause problems.

I do not have any mechanical engineering training, so I cannot help with details about the minimum size and material that would be needed for the crank arm or eccentric.
But I would 'guess' that an aluminum crank arm of 1.0 inch thickness, 2.0 inch height, and 12 inch length would be strong enough, and would allow conventional roller bearings to be used at both pivot points.

Will the very low density of the material ("1lb/ft3"), react properly to the motion of a vibratory conveyor - I'm concerned about air resistance (or other air currents) not allowing the material to move as expected. Is the material some sort of foam pellets
Are you planing to have the pan inclined, or horizonal?
And will an 8 foot long pan made from 1mm SS be stiff enough - the pan does have sides, correct?

I don't think I'll be able to help you much - I don't have the knowledge ....
SAIYAM KUMAR • Jan 6, 2017
yes i know the stroke length. it is twice of amplitude so it is around 2mm.
yes the pan is horizontal and it does have sides of height 18 inches.
the pan is stiff, there is no doubt in that.
well i ll use your guess in one of the prototype.
thank you so much for your help.
can u do me favour. if you know any of your friends from mechanical background, kindly a
Jay Kosta • Jan 7, 2017
Assuming the crankarm angle is 30 degrees, the 'bottom of the stroke' will be 180 degrees from the angle of the crankarm. And at that position of the crank, I'd adjust the horizontal location of the motor so the crankarm is not exerting any pressure on the pan (the 'neutral position'). With that location of the motor, the entire stroke will keep the pan higher than its 'neutral postion' - that will keep the material flowing in the desired direction.
So yes, with a total stroke length of 2mm, the verticle 'rise' (amplitude) will raise the pan ~1 mm and move it horizontally ~1.7 mm.

For a demonstration prototype, bearings or bushings for the pivots are probably not necessary.
SAIYAM KUMAR • Jan 8, 2017
thank you so much. it would indeed help me alot. please tell me about the crank. its diameter is going to be a problem. what should i take it?
SAIYAM KUMAR • Jan 9, 2017
well i have made some calculations. to get a stroke of 2 mm i will have to offset my crank arm from the crank center by 1mm. but what about the length of the crank arm. does its length affect the geometory? how much should i take it?
SAIYAM KUMAR • Jan 9, 2017
also tell me about the belt length which has to be mounted on the crank through the motor
Jay Kosta • Jan 9, 2017
I do not understand about the belt - is it used to drive the crank (eccentric) from the motor, and also to provide motor rpm reduction to 450 rpm?
I think a simple V-belt and pulley system would work for that.
See here for a picture of a typical jackshaft / countershaft device -
Vintage Lathe Jack Shaft Assembly-Rockwell-Milwaukee Delta | eBay

I would make the crank (eccentric) like this -
1) Find the diameter of the round shaft that has 450 rpm.
2) Find a round piece of metal with diameter about 5mm larger than the shaft AND which the crank arm can fit around.
3) Drill / bore / ream a hole (the size of the 450rpm shaft) 1mm offset from the center so this crank (eccentric) piece rotates giving the 2mm crank stroke.
4) Secure the eccentric to the 450 rpm shaft with a set screw. A flat area on the shaft will probably have to be made to prevent the eccentric from slipping on the shaft.
5) The length of the crank arm doesn't affect the 'geometry', but avoid excessive length to reduce the 'moving weight'. The minimum length of the crank arm is primarily determined by how much space under the pan the motor and pulleys need for clearance.
SAIYAM KUMAR • Jan 12, 2017
thank you so much 😀
SAIYAM KUMAR • Jan 14, 2017
can you help me out with the calculations of springs. i am using a compression helical steel spring. the weight on the springs would be of the pan as well as the material on the pan. but it would also have to counter the force on it due the motor. how to calculate for the maximum weight on the spring?
Jay Kosta • Jan 14, 2017
Sorry, but I really don't know enough about springs to help.

But, earlier you said "for that i will attach the motor and a crank to a base frame which is fixed" - has that changed - and why?
I think that the motor (and pulleys and eccentric) must be solidly attached to the frame, otherwise it would induce unwanted motion and vibration to the pan.

If the goal is to have the motion of the pan be caused by the 'out of balance' movement of the motor/pullyey/etc. - that is too complicated for me.

Jay Kosta
Endwell NY USA
If only a horizontal movement is needed and the stroke is so small, why is a crank needed at all? The motor cam have an eccentric ring mounted on it with the needed traverse. This can work in a vertical slot in a plate welded at the end of the pan. When the motor runs this will move the pan back and forth equal to the eccentricity.
Jay Kosta • Jan 17, 2017
A.V.Ramani
If only a horizontal movement is needed .
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Some vertical motion is also needed - that's why the 'motion vector' is at 30 degrees above horizontal. That motion basically 'lifts and throws' the material in the desired direction.
And yes, I think an eccentric (or cam) is fine for this application.

Jay Kosta
Endwell NY USA