To find the bearing force on the beam, the vertical force that the mass exerts on the left edge of the beam is first found. This requires finding the acceleration of the mass \(m\) and from that \(\mathbf{F}=m\mathbf{a}\) is used to find the force. Therefore, the first step is to find the absolute acceleration vector \(\mathbf{a}\) of the mass \(m\) treated as a particle.
The direction of the angular acceleration vector \(N\) is fixed in space. Hence the body fixed coordinates system will have its origin at left edge of the shaft, and its \(y\) axis in the same direction as \(Y\) axis of the reference frame (inertial frame in this case). The position vector of \(m\) in body fixed coordinates c.s. is\[ \boldsymbol{\rho }=-l\cos \alpha \mathbf{k}+l\sin \alpha \mathbf{j}\] Its relative velocity is\[ \boldsymbol{\dot{\rho }}_{r}=0 \] Since the mass does not move relative to the c.s. It follows also that \[ \boldsymbol{\ddot{\rho }}_{r}=0 \] Now, the angular acceleration of the body fixed c.s. is\[ \boldsymbol{\omega }=\omega \mathbf{K}+N\mathbf{j}\] Since \(\mathbf{K}\) is aligned with \(\mathbf{k}\) all the time, the above can be written using c.s. basis vectors\[ \boldsymbol{\omega }=\omega \mathbf{k}+N\mathbf{j}\] This is valid for all time. Now \(\boldsymbol{\dot{\omega }}\) is found. The only angular velocity vector which changes direction is \(\omega \mathbf{k}\). The angular velocity vector \(N\mathbf{j}\) does not change direction. Therefore\[ \boldsymbol{\dot{\omega }}=\left \{ \dot{\omega }\mathbf{k}+\left ( N\mathbf{j\times }\omega \mathbf{k}\right ) \right \} +\left \{ \dot{N}\mathbf{j+0}\right \} \] Since all angular velocities are zero then \(\dot{\omega }\mathbf{k}=0\) and \(\dot{N}\mathbf{j}=\mathbf{0}\). The above becomes\begin{align*} \boldsymbol{\dot{\omega }} & =N\mathbf{j\times }\omega \mathbf{k}\\ & =N\omega \mathbf{i} \end{align*}
Now all the terms needed have been found, the absolute acceleration vector is determined\begin{align*} \mathbf{a} & =\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \\ & =\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \\ & =\left ( N\omega \mathbf{i}\times \left ( -l\cos \alpha \mathbf{k}+l\sin \alpha \mathbf{j}\right ) \right ) +\left ( \omega \mathbf{k}+N\mathbf{j}\right ) \times \left ( \left ( \omega \mathbf{k}+N\mathbf{j}\right ) \times \left ( -l\cos \alpha \mathbf{k}+l\sin \alpha \mathbf{j}\right ) \right ) \\ & =\left ( N\omega l\cos \alpha \mathbf{j}+N\omega l\sin \alpha \mathbf{k}\right ) +\left ( \omega \mathbf{k}+N\mathbf{j}\right ) \times \left ( -\omega l\sin \alpha \mathbf{i}-Nl\cos \alpha \mathbf{i}\right ) \\ & =\mathbf{j}\left ( \omega ^{2}l\sin \alpha -\omega Nl\cos \alpha +N\omega l\cos \alpha \right ) +\mathbf{k}\left ( -N\omega l\sin \alpha +N^{2}l\cos \alpha +N\omega l\sin \alpha \right ) \\ & =\omega ^{2}l\sin \alpha \mathbf{j}+N^{2}l\cos \alpha \mathbf{k} \end{align*}
Therefore, the downward vertical force on the beam is \begin{align*} \mathbf{f}_{y} & =m\mathbf{a}_{y}\\ & =m\omega ^{2}l\sin \alpha \mathbf{j} \end{align*}
And\begin{align*} \mathbf{f}_{z} & =m\mathbf{a}_{z}\\ & =mN^{2}l\cos \alpha \mathbf{k} \end{align*}
Drawing a free body diagram of the beam, the reactions can be found
Taking moments around point \(A\) gives\begin{align*} \left \vert \mathbf{f}_{y}\right \vert b+V_{B}L & =0\\ V_{B} & =-m\omega ^{2}l\sin \alpha \frac{b}{L} \end{align*}
And taking moments around point \(B\) gives\begin{align*} \left \vert \mathbf{f}_{y}\right \vert \left ( b+L\right ) -V_{A}L & =0\\ V_{A} & =m\omega ^{2}l\sin \alpha \frac{\left ( b+L\right ) }{L} \end{align*}
Now that \(V_{A}\) and \(V_{B}\) (the reactions) are found and the load on the end is also known, the bending moment and shear diagrams can also be found if needed. Internal stress at any section can also be found.
The first step is to find the absolute acceleration \(\mathbf{a}\) of a unit mass of tube. A body fixed coordinates system is setup which has its origin where the tube is attached to the vertical shaft and attached to the vertical shaft as shown in this diagram
The analysis starts by assuming the oil tube is rigid. Once the forces are found, then the tube is assumed to be elastic in order to find the end deflection. The position vector \(\boldsymbol{\rho }\) of unit mass \(dm\) of length \(d\rho \) is shown above in gray area is\[ \boldsymbol{\rho }=\rho \sin \theta \mathbf{j}+\rho \cos \theta \mathbf{k}\] And \(\boldsymbol{\dot{\rho }}_{r}=\boldsymbol{\ddot{\rho }}_{r}=0\). The angular velocity of the body fixed c.s. is\[ \boldsymbol{\omega }=\omega \mathbf{K}=\omega \mathbf{k}\] Since the angular acceleration \(\boldsymbol{\omega }\) is constant, then\[ \boldsymbol{\dot{\omega }}=\dot{\omega }\mathbf{K}=\dot{\omega }\mathbf{k}=\mathbf{0}\] The absolute acceleration of \(dm\) is given by\[ \mathbf{a}=\mathbf{\ddot{R}}+\boldsymbol{\ddot{\rho }}_{r}+2\left ( \boldsymbol{\omega }\times \boldsymbol{\dot{\rho }}_{r}\right ) +\left ( \boldsymbol{\dot{\omega }}\times \boldsymbol{\rho }\right ) +\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \] Since \(\mathbf{\ddot{R}}=\mathbf{0}\) and \(\boldsymbol{\dot{\omega }}=\boldsymbol{0}\) the above simplifies to\begin{equation} \mathbf{a}=\boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) \tag{1} \end{equation} Hence\begin{align*} \boldsymbol{\omega }\times \boldsymbol{\rho } & =\omega \mathbf{k}\times \left ( \rho \sin \theta \mathbf{j}+\rho \cos \theta \mathbf{k}\right ) \\ & =-\omega \rho \sin \theta \mathbf{i} \end{align*}
Therefore\begin{align*} \boldsymbol{\omega }\times \left ( \boldsymbol{\omega }\times \boldsymbol{\rho }\right ) & =\omega \mathbf{k}\times \left ( -\omega \rho \sin \theta \mathbf{i}\right ) \\ & =-\omega ^{2}\rho \sin \theta \mathbf{j} \end{align*}
Eq. (1) becomes\[ \mathbf{a}=-\omega ^{2}\rho \sin \theta \mathbf{j}\] Since \[ dm=\frac{m}{L}d\rho \] Then the force acting on \(dm\) due the above acceleration is\begin{align*} d\mathbf{F} & \mathbf{=a}dm\\ & =-\omega ^{2}\rho \sin \theta \frac{m}{L}d\rho \mathbf{j} \end{align*}
The force up to some point \(\varsigma \) in the tube is found by integration\begin{align*} \mathbf{F}\left ( \varsigma \right ) & =-\int _{0}^{\varsigma }\omega ^{2}\rho \sin \theta \frac{m}{L}d\rho \mathbf{j}\\ & =-\omega ^{2}\frac{\varsigma ^{2}}{2}\sin \theta \frac{m}{L}\mathbf{j} \end{align*}
The total force is\[ \mathbf{F}\left ( L\right ) =-\omega ^{2}\frac{L}{2}\sin \theta m\mathbf{j}\] At a section distance \(\varsigma \) the forces are shown below
Now that the force vector at a distance along the tube is found, the bending moment at a section distance \(\varsigma \) is calculated.
The weight of the tube is \(\frac{m}{L}\) per unit length, which can be modeled as uniform distributed load. A free body diagram of the oil tube is given below. The force in the \(y\) direction is resolved as axial force and as perpendicular force to the tube.
Resolving \(\omega ^{2}\frac{L}{2}\sin \theta m\mathbf{j}\) along the tube length, and perpendicular to the tube length gives an axial force of \(\omega ^{2}\frac{\varsigma ^{2}}{2}\sin ^{2}\theta \frac{m}{L}\) and perpendicular force \(\omega ^{2}\frac{\varsigma ^{2}}{2}\sin \theta \frac{m}{L}\cos \theta \) as shown in this diagram. The axial force does not produce bending moment. The weight of tube is \(\frac{m}{L}g\) per unit length and acts in the \(z\) direction. The weight is also resolved so that it acts perpendicular to the tube as well giving \(\frac{m}{L}g\sin \theta \) pre unit length. Therefore for distance \(\varsigma \) from the origin, the total weight is \(\frac{m}{L}g\varsigma \sin \theta \)
Therefore, the bending moment at section distance \(\varsigma \) is\begin{align*} M\left ( \varsigma \right ) & =\left ( \omega ^{2}\frac{\varsigma ^{2}}{2}\sin \theta \frac{m}{L}\cos \theta \right ) \varsigma -\left ( \frac{m}{L}g\sin \theta \varsigma \right ) \frac{\varsigma }{2}\\ & =\omega ^{2}\frac{\varsigma ^{3}}{2}\sin \theta \frac{m}{L}\cos \theta -\frac{m}{2L}g\sin \theta \varsigma ^{2} \end{align*}
Unit check: Moment is force times distance. Hence units is \(\frac{ML^{2}}{T^{2}}\). Checking units of each term in the RHS above it agrees.
To find end point deflection, the tube is treated as elastic and viewed as follows
For purpose of finding end point deflection at steady state, only forces acting in the transverse direction to the tube as shown need to be considered .The end force is found by letting \(\varsigma =L\) in the above which gives the force at the free end as \[ P=\omega ^{2}m\frac{L}{2}\sin \theta \] let \(\beta \) be the weight per unit length. Using cantilever beam end deflection formula the end deflection is given by\[ \eta =\frac{PL^{3}}{3EI}-\frac{\beta L^{4}}{8EI}\] A positive sign is given to deflection to due to \(P\) since it acts up, and the weight acts down. Hence end point deflection is\begin{align*} \eta & =\frac{\omega ^{2}m\frac{L}{2}\sin \theta L^{3}}{3EI}-\frac{\frac{m}{L}g\sin \theta L^{4}}{8EI}\\ & =\frac{\omega ^{2}mL^{4}\sin \theta }{6EI}-\frac{mg\sin \theta L^{3}}{8EI}\\ & =\frac{4\omega ^{2}mL^{4}\sin \theta -3mg\sin \theta L^{3}}{24EI}\\ & =\frac{mL^{3}\sin \theta \left ( 4\omega ^{2}L-3g\right ) }{24EI} \end{align*}