The Euler-Bernoulli beam theory is based on the hypothesis that the beam deforms such that a straight line normal to the beam axis before deformation remains
- inextensible
- straight
- normal to the beam axis
after deformation.
I am trying to find out the consequences of these assumption number 1 and 2 for displacement field. In the following discussion, $x_1$ is the axial direction and $x_2$, and $x_3$ are in-plane directions. Similarly, $u_1$ is the displacement along axial direction, and $u_2$, and $u_3$ are in-plane displacements. The coordinate system is placed at the centroid of the cross-section of the beam.
Assumption 1: A straight line normal to the beam axis before deformation remains inextensible after deformation.
Consider a straight line $\overline{AB}$ in the cross-section of the beam as shown in Fig. 1.. The length of the beam before deformation is given by
$$l_{AB}=\sqrt{(x_{2A} - x_{2B})^2 + (x_{3A} - x_{3B})^2}$$
where, $x_{ij}$ represent the ith-coordinate of jth point. For example, $x_{3B}$ is displacement along $x_3$ direction of point B.
Similarly, the length of line $\overline{AB}$ after deformation is given by
$$l'_{AB}=\sqrt{(x_{2A} + u_{2A} - x_{2B} - u_{2B})^2 + (x_{3A} + u_{3A} - x_{3B} - u_{3B})^2}$$
Since the length of line $\overline{AB}$ cannot change, i.e., $l_{AB} = l'_{AB}$, therefore,
$$\sqrt{(x_{2A} - x_{2B})^2 + (x_{3A} - x_{3B})^2} = \sqrt{(x_{2A} + u_{2A} - x_{2B} - u_{2B})^2 + (x_{3A} + u_{3A} - x_{3B} - u_{3B})^2}$$
which is possible if and only if $u_{2A} = u_{2B}$ and $u_{3A} = u_{3B}$.
Since, points A and B are arbitrary points on the cross-section, therefore, we can conclude that the $u_2$ and $u_3$ displacement will be same for all points on the cross-section of the beam. In other words, $u_2$ and $u_3$ are functions of $x_1$ only.
$$u_2(x_1, x_2, x_3) = u_2(x_1)$$ $$u_3(x_1, x_2, x_3) = u_3(x_1)$$
Assumption 2: A straight line normal to the beam axis before deformation remains straight after deformation.
This is possible if and only if the line translate as rigid-body along axis $x_1$ and rotate as a rigid-body about axes $x_2$ and $x_3$.
Rigid-body translation: If the line moves as a rigid-body along $x_1$, its displacement along $x_1$ is given by $\overline{u_1}(x_1)$.
Rigid-body rotation about $x_2$: If the line rotates as a rigid-body about $x_2$, the displacement of any point on the line along $x_1$ is given by $x_3 \phi_2(x_1)$, where $\phi_2(x_1)$ is the angle of rotation about $x_2$.
Rigid-body rotation about $x_3$: If the line rotates as a rigid-body about $x_3$, the displacement of any point on the line along $x_1$ is given by $-x_2 \phi_3(x_1)$, where $\phi_3(x_1)$ is the angle of rotation about $x_3$.
So, the total displacement along axis $x_1$ is given by $$u(x_1, x_2, x_3) = \overline{u_1}(x_1) + x_3 \phi_2(x_1) - x_2 \phi(x_1)$$
Owing to the reasoning mentioned above for both assumptions, are the conclusions made sensible and correct?