What is the Torque of My Machine's Shaft?

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The discussion centers on calculating the torque required for a pharmaceutical machine's rotating drum, which has four baffles and holds approximately 50 kg of medicines. The formula for torque, T = F x r x sin(θ), is acknowledged, with a crude estimate of 375 Nm provided for maximum torque at a constant speed of 15 rpm. Participants clarify that while the drum's weight does not affect torque at constant speed, it should be considered when calculating starting torque due to its contribution to rotational inertia. Additionally, friction in the bearings may require some torque, though it is generally minimal. Understanding these principles is essential for accurate torque calculations in mechanical design.
Saurav Anand
hello everyone,
I am a mechanical design engineer in pharmaceutical machine maker company. my question is about the torque of the shaft in my machine. In picture you can see the pan, in which the medicines are placed for drying and colour. the details is placed on picture. please tell me how to find torque in detail. Thank you in Advance.
 

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Welcome to the PF.

Since you are an ME, can you show us your work so far? Can you Upload your sketches on this?
 
The machine appears to show a rotating drum of some sort. The torque required to rotate it will depend to some extent on what's in the drum and how it moves. eg Will there be baffles that lift the contents and allow it to fall back down?

As Berkman says... As an ME we would expect you to have some ideas already.
 
Thank you berkeman and CWatters for replying me.
Yes there are four baffles in pan. the raw medicines put in the pan and rotates that drum on max. speed of 15 rpm.
The medicines weight is 50 kgs approximately. That machine is already made by my company hundreds times but no one knows the principles behind it.
I have just 1.5 years experience in this field so now i eager to know the calculations behind it.
I know that torque = F x r x sinθ. and all that but i don't know how to implement it.
please guys help me out.
 
Saurav Anand said:
The medicines weight is 50 kgs approximately... snip... I know that torque = F x r x sinθ. and all that but i don't know how to implement it.

Try making a sketch drawing of a section through the drum. Perhaps assume all 50kg of the medicines have been lifted up on one baffle. Mark on the sketch F, r and θ.
 
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CWatters please can you explain it with equation. you can make any assumption. like r should be 500 mm or 1000 mm whatever. i just need to know which parameters are used there and how to use it. thank you for your reply.
 
Assumptions:
1) All 50kg of pills have been lifted up on a baffle until it reaches the horizontal position (θ = 90, Sinθ = 1)
2) The average distance of the pills from the centre is R = 750mm

Drum.jpg


Then using..
T = F*R*Sin(θ)

F = 50*g ≈ 500N
RSin(θ) = 0.75 * 1 = 0.75

gives

T = 500 * 0.75
= 375Nm

That's a crude estimate of the max torque required to turn the drum at a constant speed.
 
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CWatters,
Thank you very much for your reply.
now i realize that its easy to calculate the torque.
my next question is what's the role of axis in torque ? in this matter we have a perpendicular axis. what about the linear axis ?
2 ) you assume the pills weight of 50 kg but what about the drum`s weight ? it has minimum 150 kg weight so should we not considered it ?
 
I don't understand your question about the axis.

Regarding the drum...

If the drum is symmetrical then no torque is required to turn it at a constant speed. However a torque is required to accelerate the drum. Eg from rest up to its normal operating speed.

Torque = moment of inertia * angular acceleration

When starting from rest the pills will also contribute to the moment of inertia.

Some torque might also be needed to overcome friction in the bearings but that is likely to be small.
 
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  • #10
CWatters said:
I don't understand your question about the axis.

Regarding the drum...

If the drum is symmetrical then no torque is required to turn it at a constant speed. However a torque is required to accelerate the drum. Eg from rest up to its normal operating speed.

Torque = moment of inertia * angular acceleration

When starting from rest the pills will also contribute to the moment of inertia.

Some torque might also be needed to overcome friction in the bearings but that is likely to be small.
 
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sir, should we add the weight of drum ?

Thank you for your reply. please help me out.
 
  • #12
Saurav Anand said:
sir, should we add the weight of drum ?

No. If the weight of the drum is uniformly distributed then you do not need to add the weight to calculate the torque needed to rotate it at a constant speed.

You might need to account for the weight of the drum if you want to calculate the torque required to change the speed (eg the starting torque). However to understand this you need to know about "rotational inertia" and "Moments of Inertia".
 
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  • #13
Sir,
Thank you very much to share your knowledge with me. its very helpful to me in future.
 
  • #14
CWatters said:
Some torque might also be needed to overcome friction in the bearings but that is likely to be small.
"Small" though is relative to the weight of the drum and neediness of the drive system. It *may* be big enough to be relevant.
 
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