Changing orientation of Mx, My, and Mxy

[paulofficer]

I am using Staad Pro to carry out a FE design of an irregular floor slab.

The problem is this: From the output, I can get Mx, My, and Mxy in the element local axes, but only Mx and My in Global axes. Mxy is not accounted for in this global moment (i.e. Wood-Armer) and is not given to allow me to calculate the design moment myself.

I think it is possible to convert Mx and My from local to global coords by multiplying by the Cosine of the angle. This is based on the the fact that if a direction of a force can be modified by cos(angle) then a moment can be converted in the same way as it is just a force at an eccentricity. For Mxy though, I'm not sure if this argument still holds true.

Can anyone shed some light on this?

Thanks, Paul

 

[idly123]

Mx My and Mxy are not forces they are the stress resultants(moment resultants). A change in orientation of the axis would affect the stress state . A transformation in stress state would require and then its desired to compute the moment resultants by integrating over thickness.

may be to get Mxy values u can check the output options in STAAD pro. I think it is possible to get it.
Hope it helps
regards
raj Raj

 

[Arydanrekan]

hey paulo..
Try read Basic of FEA for Frame Element,which Mxy is result from torsion.

You will get Normal,Torsion,Moment local Forces by multiplying matrix BNE with local displacement. By the way Normal,torsion, and moment global is found from multipyed Matrix BNE x MAtrix Transformation (containing Cosinus and Sinus Transformation) x Global Displacement.

When you get the STAAD Pro you get Global Displacement. So you must have your own Matrix BNE ( residual matrix for calculating inside forces),and transformation matrix
FEM S2 Student: ABQ,NAS,COS,ANS,MAR,SAP,
STaaD,ETABS,GTSTRUDL, Civil Eng.
LOOKING FOR A JOB ....

 

[vonlueke]

paulofficer: You are correct...almost. Moments are vectors, just like forces, except they have a double arrowhead. Therefore they handle, add, and transform vectorially, just like force vectors.

In 2-D, you convert from local to global by using not just cosine but also sine of the angle between the two coordinate systems.

But your stated question is a 3-D problem. Therefore, you first construct the 3-D, local-to-global rotational transformation matrix R, as follows. The global coordinates of a unit vector lying on (or parallel to) a local coordinate system axis are the vector's direction cosines, and are laid into the local-to-global rotational transformation matrix column-wise. (I.e., the global coordinates of a local x axis unit vector are laid into column 1 of the 3 by 3 matrix R; the global coordinates of a local y axis unit vector are laid into column 2; and the global coordinates of a local z axis unit vector are laid into column 3. Tip: If any two of these are measured or known, the third can be computed using cross product.) If the resultant matrix is a valid rotational transformation matrix, its inverse will equal its transpose (i.e., it will be orthogonal).

Once R is constructed, as described above, proceed as follows, where multiplication means dot product. If 3 by 1 vector X is the global (x,y,z) coordinates of a point in space, then X = R X' and X' = transpose(R) X, where X is the coordinates of the point in the global CS and X' is the coordinates of the point in the local CS.

Note in the above discussion, in your case, X' = (Mx,My,Mxy). Your local Mx, My, and Mxy are the three local CS components of your one moment vector M. Good luck.