Torque applied to both ends of a rod, resulting shear?

[LordBerkley]

Hi,

Can anyone explain what would be the maximum shear force experienced by a rod if equal and opposite torques (T) were applied at each end?

In comparison to a rod that was fixed at one end and had a torque T applied at the other end would the maximum shear experienced by the rod be double or the same?


Any help much appreciated

LB

 

[SteamKing]

We know from beam theory that if V is the shear and M is the bending moment in a beam, dM/dx = V.

Since both beams have constant moments applied, dM/dx = dc/dx = 0 = V.

Both beams are undergoing pure bending and consequently experience no shear.

 

[PhanthomJay]

Ask yourself what is the internal torque at any point along the rod for each case.

 

[LordBerkley]

Well, I was thinking that at the centre of the rod the net torque would be zero. This is where I became a little confused. Does the shear double at this point or is there none at all?

My instinct would say that the shear is doubled. However, because there is effectively no torque at this point there can't be any shear!

 

[PhanthomJay]

I think first you need to clarify the applied torque direction...is it a twisting torque causing torsional shear stresses or a bending torque causing bending stresses? I assume it is a twisting torque. But in any case, when you draw a free body diagram at a cut section of the rod, why do you say the moment is zero at the center?

 

[LordBerkley]

It is a twisting torque. At one end of there is a torque, T, acting clockwise. At the other end there is a torque, T, acting anti clockwise. I assumed that the opposing torques would cancel out at the centre of the rod, but thinking about it with a FBD it doesn't work like that.

 

[PhanthomJay]

It is a twisting torque. At one end of there is a torque, T, acting clockwise. At the other end there is a torque, T, acting anti clockwise. I assumed that the opposing torques would cancel out at the centre of the rod, but thinking about it with a FBD it doesn't work like that.

That is right, it does not. So how does it work? What it is the torque at any section for case 1? For case 2?

 

[LordBerkley]

For case 1 the torque in any section is equal to 2T? - still not sure :/

For case 2 the torque in any section is equal to T

 

[PhanthomJay]

For case 2, with one end fixed and with a torque T applied at the opposite end, the torque at the fixed end is T and the internal torque at any section is also T. This follows from a FBD of say the right part of the rod from the cut section to the end where the torque is applied. For equilibrium, the torque at the cut section must be equal to the applied torque, equal and opposite.

Now for case 1, where you have an applied torque T at each free end, draw a similar FBD per above and now what is the internal torque at any cut section? It is not 2T.

This may not be a good analogy, but suppose you attached a rope to a wall and pulled on it horizonatlly with a force P. The tension in the rope would be P, at the wall and throughout. Now suppose instead of attaching the rope to the wall, instead you pulled on one end with a force P and your buddy pulled on the other end with a force P, what would be the tension in the rope, P, 2P or 0?

Free Body Diagrams reveal the answer.

 

[LordBerkley]

Yes, it makes sense now, it has to be T. Effectively if you make a cut into the case 1 rod it is apparent that at the cut, like case 2, the torque can only ever be T in the opposite direction. Thank you for your help, much appreciated!