At the Wikipedia there are the differential formulation for Euler-Bernoulli Beam $(1)$ and Timoshenko Beam $(2)$

$$ \dfrac{d^2}{dx^2}\left(EI\dfrac{d^2w}{dx^2}\right) = q(x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$

$$ \begin{cases}\dfrac{d^2}{dx^2}\left(EI\dfrac{d\varphi}{dx}\right) = q(x) \\ \dfrac{dw}{dx} = \varphi - \dfrac{1}{\kappa AG} \cdot \dfrac{d}{dx}\left(EI\dfrac{d\varphi}{dx}\right) \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) $$

Both of formulations supposes that the undeformed beam is at $(\vec{e}_x)$ direction, the distributed charge $q$ is at $(\vec{e}_z)$ direction and $w$ is the displacement at $(\vec{e}_z)$ direction.

The deduction uses, for example that

$$ Q = \dfrac{dM}{dx} $$

But the shear force $\vec{Q}$ and the momentum $\vec{M}$ are not colinear:

$$ \vec{Q} = Q \cdot \vec{e}_{z} = \dfrac{dM}{dx} \cdot \vec{e}_z \ne \dfrac{dM}{dx} \cdot \vec{e}_{y} = \dfrac{d}{dx}\left(M \cdot \vec{e}_{y}\right) = \dfrac{d}{dx} \vec{M} $$

Question: Then, is there a formulation $(1)$ and $(2)$ using vectorial notation? Like for example

$$ \vec{Q} = \nabla \times \vec{M} $$

Cause

$$ \vec{Q} = \begin{bmatrix} 0 \\ 0 \\ Q \end{bmatrix}= \begin{bmatrix} \dfrac{-dM}{dz} \\ 0 \\ \dfrac{dM}{dx} \end{bmatrix} = \det \begin{bmatrix} \vec{e}_x & \vec{e}_y & \vec{e}_z \\ \dfrac{d}{dx} & \dfrac{d}{dy} & \dfrac{d}{dz} \\ 0 & M & 0 \end{bmatrix} = \begin{bmatrix} \dfrac{d}{dx} \\ \dfrac{d}{dy} \\ \dfrac{d}{dz} \end{bmatrix} \times \begin{bmatrix} 0 \\ M \\ 0 \end{bmatrix} $$

Motivation: I want an analytic model for a 3D beam which neutral line follows an arbitrary path $p(t) \in \mathbb{R}^{3}$. When I tried to get it, I could not use the scalar rotations cause the vectors' directions were not the same and I should use rotations.

At the Wikipedia there are the differential formulation for Euler-Bernoulli Beam \eqref{1} and Timoshenko Beam \eqref{2} $$ \begin{align} &\dfrac{d^2}{dx^2}\left(EI\dfrac{d^2w}{dx^2}\right) = q(x) \label{1}\tag{1}\\ &\begin{cases}\dfrac{d^2}{dx^2}\left(EI\dfrac{d\varphi}{dx}\right) = q(x) \\ \dfrac{dw}{dx} = \varphi - \dfrac{1}{\kappa AG} \cdot \dfrac{d}{dx}\left(EI\dfrac{d\varphi}{dx}\right) \end{cases} \label{2}\tag{2} \end{align} $$

Both of formulations supposes that the undeformed beam is at $(\vec{e}_x)$ direction, the distributed charge $q$ is at $(\vec{e}_z)$ direction and $w$ is the displacement at $(\vec{e}_z)$ direction.

The deduction uses, for example that

$$ Q = \dfrac{dM}{dx} $$

But the shear force $\vec{Q}$ and the momentum $\vec{M}$ are not colinear:

$$ \vec{Q} = Q \cdot \vec{e}_{z} = \dfrac{dM}{dx} \cdot \vec{e}_z \ne \dfrac{dM}{dx} \cdot \vec{e}_{y} = \dfrac{d}{dx}\left(M \cdot \vec{e}_{y}\right) = \dfrac{d}{dx} \vec{M} $$

Question: Then, is there a formulation $(1)$ and $(2)$ using vectorial notation? Like for example

$$ \vec{Q} = \nabla \times \vec{M} $$

Cause

$$ \vec{Q} = \begin{bmatrix} 0 \\ 0 \\ Q \end{bmatrix}= \begin{bmatrix} \dfrac{-dM}{dz} \\ 0 \\ \dfrac{dM}{dx} \end{bmatrix} = \det \begin{bmatrix} \vec{e}_x & \vec{e}_y & \vec{e}_z \\ \dfrac{d}{dx} & \dfrac{d}{dy} & \dfrac{d}{dz} \\ 0 & M & 0 \end{bmatrix} = \begin{bmatrix} \dfrac{d}{dx} \\ \dfrac{d}{dy} \\ \dfrac{d}{dz} \end{bmatrix} \times \begin{bmatrix} 0 \\ M \\ 0 \end{bmatrix} $$

Motivation: I want an analytic model for a 3D beam which neutral line follows an arbitrary path $p(t) \in \mathbb{R}^{3}$. When I tried to get it, I could not use the scalar rotations cause the vectors' directions were not the same and I should use rotations.