 FAQ 002267 EN

1 December 2017

# I get internal forces which I didn't expect because they don't correspond to the release definitions. What could be the reason?

In the calculation parameters of RFEM and RSTAB, you find the option “Refer internal forces to deformed structure” for each load combination and each load case. The calculation according to the second-order analysis is required here.

We explain the relevance with a simple example of a loaded cantilever (see figure). The load of the cantilever results in a little rotation at node 3. If you perform the calculation according to the second-order analysis, you decide with this option if the internal forces on this node refer to the original or to the rotated coordinate system. If the structure is intiially calculated according to the linear static analysis, this results in the following internal forces (RO 101.6×3.6, S235):

Nx = 0
Vy = 0
Vz = 3,00 kN

Mx = 0
My = 9,00 kNm
Mz = 0

The forces and moments can be understood as vectors (Equation 1 and Equation 2). At node 3, a rotation results according to Equation 3.

Thus, the local member axes system is rotated by the angle φy. Now, the internal forces are converted to the rotated coordinate system. This is done by multiplying the vector by the so-called rotation matrix (http://en.wikipedia.org/wiki/Rotation_matrix). The rotation matrix for the rotation about the y-axis is shown in Equation 4. The conversion is carried out by Equation 5 and 6. Inserting the numbers results in Equation 7.

It becomes apparent that a smalllittle part of the shear force turns to a tension force:

Nx = 0,4326 kN
Vy = 0
Vz = 2,969 kN

The moment vector remains unchanged.

You can check the calculation for this simple case as shown in Equation 8.

It is now clarified which effect must be selected for this field in the calculation options. However, what are the “correct” internal forces? In most cases, they are the internal forces which are more accurate. A precondition for the calculation according to the second-order analysis is small rotations. Thus, the results must not differ significantly. However, if this is the case, you have to perform the large deformation analysis. Large deformations are allowed here and results always refer to the rotated coordinate system. For calculations according to the linear static analysis, internal forces always refer to the original coordinate system.

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