Potential Energy Functional - Timoshenko Shear Locking

[bugatti79]

Folks,

The total potential energy functional for an isolated finite element timoshenko beam is given as

$\displaystyle \Pi_e(w, \Psi)=\int_{x_e}^{x_{e+1}} \left[ \frac{EI}{2} \left (\frac{d \Psi}{dx}\right )^2 + \frac{ G A K_s}{2} \left ( \frac {dw}{dx} + \Psi \right )^2 +...\right]dx +...$

Where the first term in the integral is the bending energy of the element. The author states that a constant state of $\Psi(x)$ is not admissible because the bending energy of the element would be zero leading to the numerical problem of shear locking.

Not sure I understand this concept. It is just a term that will go to 0 on the first derivative but the rest of the integral can still be evaluated. Why is it not admissible?

 

[Achy47]

it is important to understand the concepts and principles behind the equations and models we use in our research. In this case, the concept of admissibility plays a crucial role in understanding why a constant state of $\Psi(x)$ is not acceptable in the total potential energy functional for an isolated finite element Timoshenko beam.

Admissibility refers to the requirement that the displacement and rotation functions used to describe the behavior of a beam must satisfy certain conditions in order to accurately represent the physical behavior of the beam. In this case, the total potential energy functional is used to calculate the potential energy of the beam, which is a measure of the energy stored in the beam due to its deformation.

When considering the bending energy term in the integral, we can see that it is dependent on the second derivative of $\Psi(x)$, which represents the rotation of the beam. If we were to assume a constant state of $\Psi(x)$, then its first derivative would be zero, and the second derivative would also be zero, resulting in a zero bending energy term. This means that the beam would have no potential energy due to bending, which is not physically realistic.

Furthermore, in a Timoshenko beam, the shear deformation is also taken into account, which is represented by the second term in the integral. Again, assuming a constant state of $\Psi(x)$ would result in a zero shear deformation term, leading to a numerical problem known as shear locking. This means that the beam would be unable to deform in the shear direction, which is not accurate.

In summary, a constant state of $\Psi(x)$ is not admissible because it does not accurately represent the physical behavior of the beam, leading to numerical problems and inaccurate results. It is important to carefully consider the concepts of admissibility and physical plausibility when using equations and models in our research.