Combined axial and torsion load on non-uniform cross section cylinder
Combined axial and torsion load on non-uniform cross section cylinder
(OP)
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?
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?





RE: Combined axial and torsion load on non-uniform cross section cylinder
good57morning@netzero.com
RE: Combined axial and torsion load on non-uniform cross section cylinder
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.
RE: Combined axial and torsion load on non-uniform cross section cylinder
good57morning@netzero.com
RE: Combined axial and torsion load on non-uniform cross section cylinder
RE: Combined axial and torsion load on non-uniform cross section cylinder
B.Maskingutt