# Mathematical formulation of beam: get stress/strain from forces and momentum

I'm working with static beams with Euler–Bernoulli model which ODE is

$$\dfrac{d^2}{dx^2} \left(EI \cdot \dfrac{d^2w}{dx^2}\right) = q(x).$$

With a beam along the $$x$$ axis, the solution consists of finding $$\vec{U} = (u_x, \ u_y, \ u_z, \ \theta_x, \ \theta_y, \ \theta_z)$$ for every point on the beam, which I can transform into forces and momentums $$\vec{F} = (F_x, \ F_y, \ F_z, \ M_x, \ M_y, \ M_z)$$.

Question: Once I find the forces and momentums, how can I get the stress tensor $$\left(\overline{\overline{\sigma}}\right)$$ from the vector $$\vec{F}$$? I know that it must satisfy:

$$\begin{gather*} \left(F_x, \ F_y, \ F_z\right) = \int_{D} \overline{\overline{\sigma}} \cdot \vec{e}_x \ \ dy \ dz \\ \left(M_x, \ M_y, \ M_z\right) = \int_{D} \left(0, \ y, \ z\right) \times \left(\overline{\overline{\sigma}} \cdot \vec{e}_x\right) \ dy \ dz. \end{gather*}$$

I meant to get a PDE in terms of $$y$$ and $$z$$:

$$f\left(\overline{\overline{\sigma}}, \ \dfrac{\partial \overline{\overline{\sigma}}}{\partial y}, \ \dfrac{\partial \overline{\overline{\sigma}}}{\partial z}, \ \dfrac{\partial^2\overline{\overline{\sigma}}}{\partial y^2}, \ \dfrac{\partial^2 \overline{\overline{\sigma}}}{\partial y\partial z}, \ \dfrac{\partial^2 \overline{\overline{\sigma}}}{\partial z^2}, \cdots \right) = 0.$$

But I have no idea how to find this PDE. I only have the intuition it should be an elliptic PDE because it has a unique and stable solution that minimizes the total strain energy.

The boundary is that there's no stress on the external surface:

$$\overline{\overline{\sigma}} \cdot \vec{n} = \vec{0} \ \ \ \ \text{on} \ \partial D.$$

Motivation: I'm making software to compute the stress of each beam using finite elements. Solving the ODE, I can find $$\vec{U}$$ and then the effort $$\vec{F} = [K] \cdot \vec{U}$$, which I already have. Then I would like to compute the stress on $$D$$.

• if you only know the integrated stress tensor ($F$ and $M$), then I don't see how you can find the stress tensor itself without further information. Dec 10, 2022 at 20:47