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In direct stiffness method, the beam and column connection are assumed to be done through center-line beam and column:

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And hence we can represent them easily using frame elements ( 1D element) and do not have to model the surface contacts or using area elements.

However, in reality, beam and column may not always be connected via centerline: eg: such configuration is not uncommon

enter image description here

It is easy to see that the offset of beam shall induce additional moment on the supporting column.

How should I modify my stiffness matrix in order to accommodate the offset of beam from the column center location? The frame representation of beam and column is just too convenient to give up.

For the purpose of this question, assume that I am writing a frame analysis software from first principle ( yes, from the equations on the wiki page that I link to), and hence can't make use of any existing software package

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2 Answers 2

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Use a stiff element, as explained by AndyT, to model the offset might create numerical issues for the finite element model.

Here are some alternatives:

Alternative 1:

Build on the comment by alephzero, one can model the offset as Multi-Point Constraint (MPC). An explanation is provided here:

The basic idea of using MPC is to create a set of MPC equations that gives the relation between the DOFs of the two separated nodes. It assumes that the two corner nodes are connected by a rigid body. MPC equations are then derived using the simple kinematic relations of the DOFs on the rigid body.

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Alternative 2:

Another similar approach is to use the Direct Fabrication of offset beam stiffness ( section 20.4.3), where the stiffness matrix is explicitly modified to incorporate the offset.

enter image description here

enter image description here

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There are two main options for modelling.

Option 1 - Instead of just modelling the beam and the column, add a third member, the little beam that causes the offset.

Option 2 - Use offset geometry (if available within the software you are using). I'm not 100% sure how this works (and it may differ from software package to software package): possibly it does modify the stiffness matrix, possibly the software introduces a dummy element to connect the beam to the column, much like Option 1.

In the picture you have provided, the beam to cause the offset looks very small compared to the main beam and the column. If you went down the Option 2 route, you would not get forces and moments in that member in order to design it, and I suspect it might get overloaded. So, for the picture provided, I would recommend Option 1.

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    $\begingroup$ @Graviton - Stiffness = EI, just as it does for any other member. My Options are written from the point of view of a software user, not a software writer. Hence, the user choose to model the beam, and should know what it's E and I are. In Option 2b, if you are adding a dummy member, you should assume that the connection is rigid, for which I would take EI 6 orders of magnitude higher than the stiffest of the beam and column. Option 2a (modifying the stiffness matrix) is beyond my knowledge. $\endgroup$
    – AndyT
    Commented Jul 28, 2016 at 8:47
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    $\begingroup$ As long as there is no moment end release between the little beam and the column, then any vertical force at the main beam end of the little beam will cause a moment at the column end of the little beam, which will be transferred into the column. Also, if there is no torsion release in the main beam / moment release in the little beam, then that creates bending in the little beam, which will again be transferred into the column (assuming no moment release). This is basic statics. $\endgroup$
    – AndyT
    Commented Jul 28, 2016 at 8:55
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    $\begingroup$ @Graviton: Disagreeing with AndyT, the fictitious beam created to represent the offset is usually defined as infinitely stiff. This certifies that whatever deformations are imposed on one beam are adequately mirrored in the other. That being said, one shouldn't model infinitely stiff members as $EI=\infty$ (where $\infty$ is represented by an obscenely large number) since that can lead to numerical errors in your stiffness matrix. There are methods that adequately handle rigid members, but I'm not sure what they are. $\endgroup$
    – Wasabi
    Commented Jul 28, 2016 at 11:24
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    $\begingroup$ "There are methods that adequately handle rigid members, but I'm not sure what they are" Writing down the 6 equations that represent the translations and rotations at one end of a rigid link in terms of the other end is just high-school coordinate geometry - the 3-dimensional version of equations like $u_2 = u_1 + L \theta_1$. @Graviton, please ignore all the advice to use "short stiff beams". That is a great way to get answers that are somewhere between inaccurate and just plain wrong - and if the model results are say 20% or 50% wrong, how is the user supposed to discover that fact? $\endgroup$
    – alephzero
    Commented Jul 28, 2016 at 20:22
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    $\begingroup$ @Graviton See section 20.4 of colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.Ch20.d/…. for two different good ways to implement it. Short stiff beams are bad because (1) if they are too flexible, they introduce errors for the obvious reason, and (2) if they are too stiff, the assembled stiffness matrix will be numerically poorly conditioned. Users may not be aware of those problems, and if they are aware they don't really want to spend time running several versions of the model to check their that their "best guess" of properties for the stiff beams was OK. $\endgroup$
    – alephzero
    Commented Jul 30, 2016 at 1:01

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