Statics: 2 shafts interconnected by a universal joint

[SoylentBlue]

Homework Statement

I got part A of the problem right. I am stuck on part B. As you can see, they ask for the reactions at B, D and E.[/B]

Homework Equations

The Attempt at a Solution

OK, here's the book's the answers: at D, 350N k. At E, -100Nk, and at B -250N k.
So, we have a 50Nm torque applied at the shaft along axis x. At the edge of the yoke, the force applied would be 50/d, where d is an unknown quantity (1/2 the width of the yoke).

Wouldn't we want to find the reactions at D, E and B by starting from the edge of the yoke?

Then, for the reaction at D, we'd have an equation like this:
-(50/d)j x (-dj + .08i)
The d values cancel conveniently, yielding a value of 4Nk, which disagrees with the book.
Where am I making a wrong turn?
All help appreciated...thank you :^)
[/B]

 

[haruspex]

Please post your answer for part a).
What plane must the torque vector on the crosspiece from shaft CF lie in?
What plane must the torque vector on the crosspiece from the other shaft lie in?

 

[SoylentBlue]

The book's answer for part a is 43.3Nm. This happens to be (cos 30)(50Nm). So I looked at it this way: the 30-degree angle of shaft BC costs some rotational energy, and it does so as a function of the angle.

So if the 50Nm is the output of an engine, then the torque that actually makes it to the wheels (at point A) is only 43.3Nm; the rest is wasted through the U-joint.

For part B, I have thought about this some more.
To answer your question, when CF is vertical, the 50Nm (at the top of the yoke) is pointing into the paper. So we have a force of -k times a distance d (1/2 of the width of the yoke) in the j direction, giving us a torque vector in the i direction. So its torque is purely along the x-axis and is therefore in the yz plane.
The BC yoke is horizontal. But it is at a 30-degree angle. So its torque is in the yz plane that is tilted 30 degrees; so it is also trying to impart a torque along the xy plane, right? (That is the wasted energy.)

The yz component of the torque from BC would have to cancel the 50Nm yz rotation. So wouldn't that only leave the xy component from BC as the only torque that could impact B, D and E?

I noticed that the book's answer has the force at D and E in opposite directions, which has further confused me.

Thank you for any insights...I am trying to work through statics on my own, so I have no prof or other classmates to bounce questions off of.

 

[haruspex]

the 30-degree angle of shaft BC costs some rotational energy

No, you must assume there is no loss of energy. That would only arise in consequence of frictional torque. It's to do with mechanical advantage. As the crosspiece rotates, the mechanical advantage changes. In one orientation it is maximal in favour one shaft, while a quarter turn later it is maximally in favour of the other.
In the diagram, note the plane in which the crosspiece lies. It is normal to the BC shaft, but not to the CE shaft. I'm struggling to find a clear way to prove it, but it is intuitively obvious to me that this means the counter-torque from the BC shaft only needs to match that component of the CE torque which aligns with the BC axis, i.e. 50 cos(30). I'll see if I can explain that better.

when CF is vertical, the 50Nm (at the top of the yoke) is pointing into the paper

Not sure what you mean. Do you mean when the arm of the x-piece attached to the CE shaft is vertical? That is in the next question. In the first question, that arm is horizontal. The arm attached to the BC shaft is at 30 degrees to the vertical.

Having established the torque along each shaft, what is their net torque on the x-piece?