Calculating Torque of a Crank Mechanism

[ham_revilo]

Hi,
Im currently designing a small machine which uses a simple crank mechanism but my crank mechanism is not a typical one. My application:

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My application uses the crank wheel to drive the block (10kg). Hence, I need to calculate the torque require to drive the crank wheel. So I can select a suitable motor to do so.

The problem is my crank wheel and block is offset. Hence, it’s different from a typical crank mechanism where it’s all in a straight line.

Therefore, my question is:
1. In my application, is there a top/bottom dead centre where torque is 0Nm since its offset?

2. If my case doesn’t have a top/bottom dead centre, how would I calculate the torque required?

3. I came out with a theory where the Torque=F*12.5. Is it correct?

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4. If I’m wrong, Can I use the normal crank mechanism formula (Torque=F*r*sin α) to calculate the torque required?

5. How is the force reaction like so I can resolve it and calculate the torque?

Thanks in advance :)

 

[Simon Bridge]

1. In my application, is there a top/bottom dead centre where torque is 0Nm since its offset?

Yes - you have drawn one of them. But I think you have your thinking backwards - you are driving the block from the wheel not the other way around - the block moves along a line so torque may not help much here.

2. If my case doesn’t have a top/bottom dead centre, how would I calculate the torque required?

The torque you need to apply to the shaft to drive the block will vary with time - depending on how you want the block (or the wheel) to move.

eg. You should be able to plot the position of the black against the angular position of the wheel. If you intend to crank the wheel at a constant angular speed, you can use that to find a position-time graph for the block.

3. I came out with a theory where the Torque=F*12.5. Is it correct?

Who knows - how did you calculate this figure? What were the conditions?

4. If I’m wrong, Can I use the normal crank mechanism formula (Torque=F*r*sin α) to calculate the torque required?

Not easily.

5. How is the force reaction like so I can resolve it and calculate the torque?

Your last diagram appears to show an applied force on the block, but your description says the wheel is driving the block - which is it? If you push on the block like that diagram shows, the force is transmitted along the shaft. So the direction shown for force at the pivot between the wheel and the shaft is incorrect.

There is not enough information: for instance, the torque will be different if you want the block to have a constant speed for part of it's motion, a constant acceleration, or perhaps you want to drive the wheel at a constant speed - or with a constant torque (and you want the minimum torque for this).

 

[ham_revilo]

Yes - you have drawn one of them. But I think you have your thinking backwards - you are driving the block from the wheel not the other way around - the block moves along a line so torque may not help much here.

yup, my application requires the wheel to drive the block. but then, i need to select a suitable motor to drive the crank/crankwheel, then the linkage and then the block. Therefore, I need the torque require to calculate my power needed right? if no, then do you have suggestion how much power/torque to drink the crankwheel in order to make the block move?

The torque you need to apply to the shaft to drive the block will vary with time - depending on how you want the block (or the wheel) to move.

eg. You should be able to plot the position of the black against the angular position of the wheel. If you intend to crank the wheel at a constant angular speed, you can use that to find a position-time graph for the block.

yes, it will vary in time. just to let you know, the crankwheel is going to continuously run/rotate 360deg. I already set/came out with the position of the block i need. also, i want to plot a torque of the crankwheel against time/angular position of the block.

Who knows - how did you calculate this figure? What were the conditions?

if you look at my 3rd picture, the 12.5mm is the distance and F is the force required to move the block. this is just my theory. it might be wrong or right. I am not too sure. hence, i need more suggestion & guideline :)

Not easily.

do you have any suggestion/guideline what formula to use or how to resolve the force to come out with the torque require to drive the crankwheel?

Your last diagram appears to show an applied force on the block, but your description says the wheel is driving the block - which is it? If you push on the block like that diagram shows, the force is transmitted along the shaft. So the direction shown for force at the pivot between the wheel and the shaft is incorrect.

There is not enough information: for instance, the torque will be different if you want the block to have a constant speed for part of it's motion, a constant acceleration, or perhaps you want to drive the wheel at a constant speed - or with a constant torque (and you want the minimum torque for this).

just to clarify, my wheel is driving the block. the reason why i calculate backward from the block to the wheel is because, the only information i have is the weight of the block. therefore, i calculate the force acting from the block and resolve it to the crankwheel in order to find the torque required to drive the block.

also, i understand that the torque will be different during the motion of the crankwheel. assume my crankwheel will rotate at constant angular velocity, how do i calculate or plot the torque-time graph or torque-angular graph.

if i can come out with the torque graph, i could know the maximum torque in the crank cycle. hence, i am able to choose the right motor :)

lastly, thanks a lot for the reply. appreciate it :)

 

[Simon Bridge]

Well your first problem is that the top of the wheel is higher than the shaft can go. (The wheel diameter is 50, same as the length of the shaft, but the bottom of the wheel is well above the track the mass runs on.)

At and angle of 30 degrees, the mass is 50+25=75 away from the axle, so the height of the axle above the mass is 75sin(30)=62.5 > 50The applied torque will pull on the end of the shaft - <sigh> there must be a better way to write that - the shaft pulls diagonally on the mass - part lifting and part sliding it. Use free body diagrams to work out how the torque turns into ma.

Hint: F=\tau/r at the end - with the angle +90deg.

So it is going to get stuck.

 

[ham_revilo]

Well your first problem is that the top of the wheel is higher than the shaft can go. (The wheel diameter is 50, same as the length of the shaft, but the bottom of the wheel is well above the track the mass runs on.)

At and angle of 30 degrees, the mass is 50+25=75 away from the axle, so the height of the axle above the mass is 75sin(30)=62.5 > 50


The applied torque will pull on the end of the shaft - <sigh> there must be a better way to write that - the shaft pulls diagonally on the mass - part lifting and part sliding it. Use free body diagrams to work out how the torque turns into ma.

Hint: F=\tau/r at the end - with the angle +90deg.

So it is going to get stuck.

my bad. this is a simplify diagram. the actual dwg, the locus, shaft length, crank wheel diameter is perfect where it can rotate and it won't get stuck. sorry for the confusion on the diagram. the actual machine looks complicated, so i came out with a simplified one just to describe the problem. so sorry