# Energy of a 3D Timoshenko beam

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?

• Just to be abundantly clear: you are asking about bending in both directions simultaneously, right? – Wasabi Feb 1 '16 at 11:44
• 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. – Ferdi811 Feb 1 '16 at 11:55