Combined axial and torsion load on non-uniform cross section cylinder

[jh149]

Can someone teach me how to apply boundary condition to a hollow cylinder where both axial and torsion load is present? The cylinder is not uniform in cross section so beam element won't work.

I know that for axial load (assume z direction is the centerline) on a cylinder, I need to have boundary condition where x,y displacement is free to allow free expansion.

I also know that for torque, I split my loading surface into two equal surfaces and then apply two equal distributed line loads perpendicular to split surface dividing edges, in effect create a torque using a COUPLE of equal but opposite in direction line forces. The boundary condition on the opposite end is completely fixed.

But since the boundary condition for axial loading and torsion are different (one require free lateral expansion while the other require completely fixed end). What is a good way of apply the boundary condition?

 

[feadude]

apply a distributed axial force in z direction on your loading surface. The other end should be fixed which is what you have. Cylinder should not expand at the fixed end because it is restrained against torsion. Is this not the physical restraint? Welded or Bolted rigid? If it is it can not expand anyway.

good57morning@netzero.com

 

[jh149]

It was for a hollow drill and I want to know the stress at the drill tip. The drill will be rotating at certain RPM and it will have a torque and axial force when drill through a flexible material.

I first tried to fixed the clamping end and apply the axial force and torque at the tip, but when I apply the coupled force to simulate torque, the line force become too high so my result is not valid at the tip. And if I fixed the drill tip end and apply force and torque at the opposite end, the forces are still too high because no expansion is allowed at the tip for the axial loading.

 

[feadude]

For your torsional load on the force on edge window did you turn intensity off and then turn total force on?

good57morning@netzero.com

 

[jh149]

Yes, I did that

 

[maskingutt]

You should use cylindrical coordinates instead of cartesian for the boundary condition.

B.Maskingutt