
#1
Nov3011, 10:28 PM

P: 5

Hi,
Im currently designing a small machine which uses a simple crank mechanism but my crank mechanism is not a typical one. My application: [/URL] 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? [/URL] 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 :) 



#2
Dec111, 06:06 AM

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PF Gold
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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 positiontime graph for the block. 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). 



#3
Dec111, 06:54 PM

P: 5

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 torquetime graph or torqueangular 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 alot for the reply. appreciate it :) 



#4
Dec111, 09:55 PM

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Calculating Torque of a Crank Mechanism
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: [itex]F=\tau/r[/itex] at the end  with the angle +90deg. So it is going to get stuck. 



#5
Dec111, 10:01 PM

P: 5




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