I am a little bit confused over the concept of translational and rotational degrees of freedom (DoFs) in structures, and their relation to displacement/traction BCs.

Do displacement boundary conditions constrain translational DoFs, only, or also rotational DoFs?

For example, for a 3-D case, if I specify the 3 translational components to be zero, I am effectively constraining all 3 translational DoFs. What does this do to the rotational DoFs? Intuitively, all 3 rotational DoFs would also be constrained.

How does traction BCs play into translational and rotational DoFs?

Part of my confusion is when I look at open-source or commercial structural mechanics solvers, I only see the ability to specify displacement or traction boundary conditions, not anywhere to specify translational or rotational constraints.

Some illustrations would be very helpful in my understanding.

Mathematical expressions would be helpful as well. In particular, I am wondering how rotational and translation DoFs are related to displacements and displacement gradients mathematically.

  • $\begingroup$ If you are referring to classic elasticity in 3D, for example, there is no such a thing as a rotational degree of freedom. $\endgroup$
    – nicoguaro
    Mar 4, 2019 at 22:18
  • 1
    $\begingroup$ See en.wikipedia.org/wiki/Kirchhoff%E2%80%93Love_plate_theory for how 3D and 2D descriptions are related (for the specific case of plates). $\endgroup$ Mar 4, 2019 at 22:50
  • $\begingroup$ @nicoguaro Why does 3D elasticity have no rotational degrees of freedom? $\endgroup$ Mar 5, 2019 at 2:45
  • $\begingroup$ Mmm, I think I don't know how to answer that question properly. But, let's give it a try. Let's consider your solid as a bunch or particles, where their relative positions can change. Thus, the whole configuration of the system is described by the displacements of all the particles (measured with respect the original configuration). $\endgroup$
    – nicoguaro
    Mar 5, 2019 at 2:52
  • $\begingroup$ @nicoguaro hmm I agree based on my vague memory of molecular dynamics theory. But 2D elasticity has rotational DoFs? What about 3D inelastic materials? $\endgroup$ Mar 5, 2019 at 2:57

1 Answer 1


In classical elasticity theory, the kinematics of the body are described in terms of translational displacements. It is certainly not required that only translational displacements are used. The less-frequently-used micropolar elasticity theory also includes local rotations to describe the kinematics.

In constructing a finite element model for a solid body, it may be useful or even essential to include rotational degrees of freedom as nodal unknowns. This is most common in elements that describe bending-- i.e. beam, plate, and shell elements. In structural mechanics, the classical simplification of elasticity theory to describe this bending behavior results in higher-order derivatives of the translations. Mathematically-correct, finite element approximation of these terms frequently requires rotational degrees of freedom at the nodes. But rotational degrees of freedom are not limited to bending elements Higher-order finite element formulations for general elasticity have occasionally used rotational degrees of freedom simply to improve the approximation of the translations within the element.

Kinematic boundary conditions in classical elasticity are described in terms of translations, say, on a face of a 3D body. In a finite element model, prescribing a particular translation on a face may require also prescribing the rotations at the nodes on this face simply because the finite element approximation for this translation on this face includes those rotations.

  • $\begingroup$ Could you explain how rotational DoFs are factored into displacement and traction boundary conditions? For a Dirichlet boundary condition in a 3-D case, would prescribing the 3 components of the displacement to be zero constrain all rotational DoFs? In commercial solvers, I don't recall seeing an optional to specify a rotational BC, only traction (or force), and displacement. $\endgroup$ Mar 5, 2019 at 13:53
  • $\begingroup$ Prescribing the nodal translational displacements to be zero doesn't affect the rotational dofs. FE codes that include elements with rotational dofs often refer to them as dofs 4, 5, 6 (rotations about x, y, z axes) in the input for boundary conditions. $\endgroup$ Mar 5, 2019 at 14:33
  • $\begingroup$ Oh that makes sense for "nodal" translational displacements. If we were to specify all displacements to be zero on a face or a line, then the face and line would have no rotational DoFs right? $\endgroup$ Mar 5, 2019 at 14:36
  • $\begingroup$ I'm not sure I understand your question. If you specify both translational and rotational nodal dofs to be zero at all points on a line then displacement will be zero at all points along the line. $\endgroup$ Mar 6, 2019 at 10:40
  • $\begingroup$ So let's say you have a line aligned in the x-direction. If you specify the 3 displacement components to be zero on this line, then this line also cannot rotate about the y- or z-axis, but this line can rotate about the x axis. Therefore, by specifying all 3 displacements components on this line, we are implicitly constraining 2 rotational DoFs? For a face, if we specify all 3 displacement components to be zero, then that would implicitly constrain the face from rotating in any direction. I think this is correct, or at least this is how I visualize it in my head? $\endgroup$ Mar 6, 2019 at 15:00

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