Angle of Twist and Power to Shaft Questions

If you and everyone else in the class is lost on this problem - just asking - why aren't any of you asking the teacher to explain?

sounds like its due tomorrow....

It's not, but thanks for the input.

I kinda want to avoid the teacher's input for now because he might just make the question more difficult if he sees we've grasped the concept. I know what I'm doing I'm just not sure how he's arrived at the numbers he has. I guess I wasn't clear, I'm not having difficulty with concepts I'm having trouble arriving at the same calculations.

First problem : make your units coherent (lbs : rads : s : in) then you will easily solve the problem. The equations are straight forward.

Second problem : you might have noticed the trick question: hollow shaft outer diameter 80mm with thickness 40mm ? Looks quite solid to me.
The torques seem to add up to 0 which is nice.
Your bar is 8m long I take it ?
So take one end or the other as the fixed point of your bar and apply the equations to the 3 remaining torques. It shouldn't matter which end you take as your starting point, in fact you can do it both ways so that your result is checked.

Hope this helps

I actually don't like that the torques add up to zero because it gives me "0" for an answer and apparently that's not the answer.

If the torques didn't add up to 0, then you would have a spinning bar. Hence the reason why it is good that they are 0. You can treat one of them as the reaction torque which has no effect on the twist of the bar as it is at the origin. It is your starting point.

Then apply the 3 other torques at the lengths you have been given. The length of bar from the origin to the applied torque contributes to the twist. Maybe you haven't taken that into account ? Angle of twist = TL/GJ. Obviously the L is different for each torque.

It is simple superposition of the different torques and the twists. Note that for every individual torque applied, it has an effect between the torque and the origin but has no further twist effect after it because it is assumed that plane sections remain plane (not exactly true but it is the simplified assumption in this type of analysis and is true for a circular section due to the radial symmetry).

You could calculate the overall torque on each particular length of bar and do the calc for each section. It gives the same result.

You know what forget I asked. I wish you could delete posts here.

Don't give up. You are almost there. I am at work and don't have the time to work it out, but you seem to be on the right track.

Start the problem again on a white board or new paper, begin with the free body diagram and work from there. Make sure that you have visualized the problem clearly and the answer is just a little algebra away.

Drew K

Don't give up just yet! As a start, tell us what you got for the 1st part, power from a 6" shaft. If you got that right I (and maybe others) might try to help with the next question.

this why your teacher is making the big union bucks. consult them prior to a forum of strangers.