axial stress for an unsymmetrical section loaded by Mx, My and Mz.

[FJCV]

Dear fellow engineers,
I am looking for a closed form solution for the axial stress Sx for an unsymmetrical cross section loaded by Mx, My and Mz. I expect it will be a more generic equation than the equation for the special case for when Mx=0 as shown in the attached figure. Thank you for your help.

Reference: Equation for when Mx=0

http://files.engineering.com/getfil...-8f34-f52c602570db&file=equation_for_Mx=0.jpg

 

[klaus.]

Is Mz Torsion ?? in your Question


best regards
Klaus

 

[Denial]

I think Mx is the torsion.[ ] The formula attached to the original post looks a bit like the formula for biaxial bending of a cross-section lacking any axis of symmetry, but I've never seen it in that form before (and I haven't checked it).

FJCV.[ ] The x-direction (axial) stresses resulting from an applied torsion do not lend themselves to being expressed in a simple explicit formula.[ ] Calculating them for a general cross-section is a very complicated exercise.[ ] Their only saving grace is that once you have managed to calculate them they can simply be added to the axial stresses that result from the biaxial bending.

 

[FJCV]

Denial,Klaus,
Mx is indeed torsion.
I attached a link for the derivation of the shown equation for completeness. Thx.


reference:

http://www.public.iastate.edu/~e_m.424/unsymm bending.pdf

 

[rb1957]

it looks like it's working in principal axes.

dumm the problem down ... apply My to an asymmetric section. With Iy, Iz and Iyz, you calculate the principal axes. You then transform the applied moment into those axes, and standard bending stress calcs. Then apply Mz, and so the general solution with My and Mz.

another day in paradise, or is paradise one day closer ?

 

[rb1957]

looking at it again, I notice "y" and "z" as the co-ords for the point in question. Maybe they're using the unrotated co-ord sys (the non-principal axes if you will) to define the point ? maybe this "simplifies" the equation ? so I'd ...
1) calc the principal axes, y' and z';
2) apply the moment in these axes (ie take components), My' and Mz' (a simple applied moment (My) would be easier to start with);
3) calc bending stresses in the principal axes, My'*z'/Iy'y'
4) now reverse engineer the equation, substituting the expressions from the original axes ... My' = My*cos(theta) - Mz*sin(theta), etc and see how it falls out.

Your inclusion of Mx (and the "odd" axes system, x and y are more normally inplane axes) in the thread title only confuses things.

another day in paradise, or is paradise one day closer ?