I am interested in the kinetic energy T and the potential energy U of a three dimensional Timoshenko beam.

For a 2D Timoshenko beam the energies read: $$ T = \dfrac{1}{2} \int\limits_o^l \left( \rho A \left( \dfrac{\partial w}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \Phi}{\partial t} \right)^2 \right) dx$$ $$ U = \dfrac{1}{2} \int\limits_o^l \left( EI \left( \dfrac{\partial \Phi}{\partial x} \right)^2 + kAG \left( \dfrac{\partial w}{\partial x} - \Phi \right)^2 \right) dx$$

But in these formulas $w$ and $\Phi$ are just defined in the x-y-Plane. How can I derive the energies for bending in both direction (x-y-Plane and y-z-Plane) and torsion (around x-axis)?

EDIT: I figured out a part of the question by myself: Because Timoshenko beams are a linear theory, one can simply add the energies of two deformations and get the energies of the superposition of the deformations. Therefore: $$ T = \dfrac{1}{2} \int\limits_o^l \left( \rho A \left( \dfrac{\partial v}{\partial t} \right)^2 + \rho A \left( \dfrac{\partial w}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \theta}{\partial t} \right)^2 + \rho I \left( \dfrac{\partial \Phi}{\partial t} \right)^2 \right) dx$$ $$ U = \dfrac{1}{2} \int\limits_o^l \left( EI \left( \dfrac{\partial \Phi}{\partial x} \right)^2 + EI \left( \dfrac{\partial \theta}{\partial x} \right)^2 + kAG \left( \dfrac{\partial v}{\partial x} - \theta \right)^2 + kAG \left( \dfrac{\partial w}{\partial x} - \Phi \right)^2 \right) dx$$ Where $w: \mathbb{R} \times \mathbb{R}^{+} \rightarrow \mathbb{R}$ is the deflection in the x-y-plane and $\Phi: \mathbb{R} \times \mathbb{R}^{+} \rightarrow \mathbb{R}$ is the slope in the x-y-plane. Respectively $v$ and $\theta$ for the y-z-plane. Can someone please confirm this?

Still the energies for torsion are missing. Can the theory of Saint-Venant applied here?

  • $\begingroup$ Just to be abundantly clear: you are asking about bending in both directions simultaneously, right? $\endgroup$ – Wasabi Feb 1 '16 at 11:44
  • $\begingroup$ Hi @Wasabi! Yes, I want to have both energies if I bend the beam both in x-y-direction and in y-z-direction simultaneously. $\endgroup$ – Ferdi811 Feb 1 '16 at 11:55

I probably do not get exactly the point, but the Timoshenko model pratically has been developed for extending the theory with the contribution of shear effects (that is usually negligible for slender beams) only.

Torsion is orthogonal to bending and consequently you are not going to consider it inside your calculation of bending deformation energy.

Just as for reference, check it out the vlasov beam theory for considering the effect of torsion-axial displacement coupling.

Other more smart models can take into account also coupling of torsion and bending and so on, but they are at least an extension of the Timoshenko model.

I hope this can be of help.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.