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2.1 HW1

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2.1.1 Problem 1

From tables we find that for cantilever beam loaded at end, the vertical deflection is $\delta =\frac{mL^{3}}{3EI}$, hence by definition $k_{b}=\frac{m}{\delta }=\frac{3EI}{L^{3}}$.

Therefore, we can model the stiffness of the system as

Therefore $$\begin{align*} k_{eq} & =k+k_{beam}+k\\ & =2k+k_{beam} \end{align*}$$

Since $k_{b}=\frac{3EI}{L^{3}}$ then the above becomes $$k_{eq}=2k+\frac{3EI}{L^{3}}$$

2.1.2 Problem 2

2.1.2.1 Part (a)

Using energy method $$\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}J_{0}\dot{\theta }^{2}=\frac{1}{2}J_{eq}\dot{\theta }_{eq}^{2}$$ But $\dot{\theta }_{eq}=\dot{\theta }$ for this part. And since $x=R\theta$ or $\dot{x}=R\dot{\theta }$, then the above becomes $$\frac{1}{2}m\left ( R\dot{\theta }\right ) ^{2}+\frac{1}{2}J_{0}\dot{\theta }^{2}=\frac{1}{2}J_{eq}\dot{\theta }^{2}$$ Simplifying gives $$J_{eq}=mR^{2}+J_{0}$$

2.1.2.2 Part (b)

Using energy method $$\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}J_{0}\dot{\theta }^{2}=\frac{1}{2}m_{eq}\dot{x}_{eq}^{2}$$ But $\dot{x}_{eq}=\dot{x}$ for this part. And since $x=R\theta$ or $\dot{x}=R\dot{\theta }$, then $\dot{\theta }=\frac{\dot{x}}{R}$ and the above becomes $$\frac{1}{2}m\dot{x}^{2}+\frac{1}{2}J_{0}\left ( \frac{\dot{x}}{R}\right ) ^{2}=\frac{1}{2}m_{eq}\dot{x}^{2}$$ Simplifying gives $$m_{eq}=m+\frac{J_{0}}{R^{2}}$$

2.1.3 Problem 3

Assuming a small deflection as shown

2.1.3.1 Mass equivalent

The kinetic energy of the system is (assuming small angles)

$$\frac{1}{2}m_{1}\left ( L_{1}\dot{\theta }\right ) ^{2}+\frac{1}{2}m_{2}\left ( L_{3}\dot{\theta }\right ) ^{2}=\frac{1}{2}I_{eq}\dot{\theta }^{2}$$

Hence

$$m_{1}L_{1}^{2}+m_{2}L_{3}^{2}=I_{eq}$$

Where $I_{eq}$ is the equivalent mass moment of inertia. The problem does not say where the equivalent mass should be located relative to the pivot point (where the torsional spring is located) so we can stop here. But assuming that distance was some $\bar{x}$, then we can write $I_{eq}=M_{eq}\bar{x}^{2}$ where equivalent  mass is used as a point mass, and simplify the above more

$$\begin{align*} m_{1}L_{1}^{2}+m_{2}L_{3}^{2} & =M_{eq}\bar{x}^{2}\\ M_{eq} & =\frac{m_{1}L_{1}^{2}+m_{2}L_{3}^{2}}{\bar{x}^{2}} \end{align*}$$

2.1.3.2 Stiffness equivalent

Using potential energy method, where energy stored by a spring due to extension or compression is $\frac{1}{2}k\Delta ^{2}$, then we see that the total energy using the above deformation is given by $$\frac{1}{2}k_{1}\left ( l_{1}\sin \theta \right ) ^{2}+\frac{1}{2}\left ( \frac{k_{3}k_{2}}{k_{3}+k_{2}}\right ) \left ( l_{2}\sin \theta \right ) ^{2}+\frac{1}{2}k_{t}\theta ^{2}=\frac{1}{2}k_{t,eq}\theta _{eq}^{2}$$

Where $\frac{k_{3}k_{2}}{k_{3}+k_{2}}$ is the equivalent stiffness of the springs $k_{2},k_{3}$ since they are in series. The above assumes small angle $\theta$, therefore we can simplify the above using $\sin \theta \approx \theta$, and obtain $$\frac{1}{2}k_{1}\left ( l_{1}\theta \right ) ^{2}+\frac{1}{2}\left ( \frac{k_{3}k_{2}}{k_{3}+k_{2}}\right ) \left ( l_{2}\theta \right ) ^{2}+\frac{1}{2}k_{t}\theta ^{2}=\frac{1}{2}k_{t,eq}\theta _{eq}^{2}$$

But here $\theta =\theta _{eq}$, therefore solving for $k_{t,eq}$ gives $$k_{t,eq}=k_{1}l_{1}^{2}+\left ( \frac{k_{3}k_{2}}{k_{3}+k_{2}}\right ) l_{2}^{2}+k_{t}$$