Chapter 13, Problem 13.28P

To determine

The natural frequencies.

The mode ratios.

The node locations.

Answer to Problem 13.28P

The frequencies are $1.587\text{Hz}$ and $0.549\text{Hz}$.

The mode ratios are $-0.131$ and $0.1168$.

The node locations are $0.131\text{m}$ ahead of mass center and $0.1168\text{m}$ behind the mass center.

Explanation of Solution

Given information:

The stiffness of suspension 1 is $1.95\times {10}^4\text{N}/\text{m}$, the stiffness of the suspension 2 is $2.3\times {10}^4\text{N}/\text{m}$.

Write the Equation of motion for vertical direction.

$m\ddot{x}=k_1\left(-x+L_1\theta \right)+k_2\left(-x+L_2\theta \right)$ ..... (I)

Here, Mass is $m$, displacement is $x$, angle is $\theta$, distance of suspension 1 is $L_1$ and the distance of suspension 2 is $L_2$.

Write the moment Equation.

$I_G\ddot{\theta }=-k_1\left(-x+L_1\theta \right)+k_2\left(-x-L_2\theta \right)L_2$ ..... (II)

Here, inertia is $I_G$ and the angular acceleration is $\ddot{\theta }$.

Take Laplace transform of Equation (I).

$\begin{array}{l}L\left(m\ddot{x}\right)=k_{1}L\left(-x+L_1\theta \right)+k_{2}L\left(-x+L_2\theta \right)\\ m{s}^{2}X\left(s\right)+k_{1}X\left(s\right)+k_{2}X\left(s\right)-k_1{L}_1\theta \left(s\right)+k_2{L}_2\theta \left(s\right)=0\\ \left(m{s}^2+k_1+k_2\right)X\left(s\right)+\left(k_2{L}_2-k_1{L}_1\right)\theta \left(s\right)=0\end{array}$ ..... (III)

Take laplace transform of Equation (II).

$\begin{array}{l}L\left(I_G\ddot{\theta }\right)=-k_{1}L\left(-x+L_1\theta \right)+k_{2}L\left(-x-L_2\theta \right)L_2\\ I_G{s}^2\theta \left(s\right)+k_1{L}_1^2\theta \left(s\right)+k_2{L}_2^2\theta \left(s\right)-k_{1}K{L}_{1}X\left(s\right)+k_2{L}_{2}X\left(s\right)=0\\ \left(k_2{L}_2-k_1{L}_1\right)X\left(s\right)+\left(I_G{s}^2+k_1{L}_1^2+k_2{L}_2^2\right)\theta \left(s\right)=0\end{array}$ ..... (IV)

Substitute $A_1$ for $X\left(s\right)$ and $A_2$ for $\theta \left(s\right)$ in Equation (III).

$\left(m{s}^2+k_1+k_2\right)A_1+\left(k_2{L}_2-k_1{L}_1\right)A_2=0$ ..... (V)

Substitute $A_1$ for $X\left(s\right)$ and $A_2$ for $\theta \left(s\right)$ in Equation (IV).

$\left(k_2{L}_2-k_1{L}_1\right)A_1+\left(I_G{s}^2+k_1{L}_1^2+k_2{L}_2^2\right)A_2=0$ ..... (VI)

Apply Cramer rule in Equation (V) and (VI).

$\left(\begin{array}{cc}m{s}^2+k_1+k_2& k_2{L}_2-k_1{L}_1\\ k_2{L}_2-k_1{L}_1& I_G{s}^2+k_1{L}_1^2+k_2{L}_2^2\end{array}\right)\left(\begin{array}{l}{A}_1\\ A_2\end{array}\right)=\left(\begin{array}{l}0\\ 0\end{array}\right)$ ..... (VII)

Foe a non zero solution Equation (VII) is rearranged.

$\begin{array}{l}\left(m{s}^2+k_1+k_2\right)\left(I_G{s}^2+k_1{L}_1^2+k_2{L}_2^2\right)-{\left(k_2{L}_2-k_1{L}_1\right)}^2=0\\ m{I}_G{s}^4+\left[m\left[k_1{L}_1^2-k_2{L}_2^2\right]+I_G\left(k_1+k_2\right)\right]s^2+k_1{k}_2{\left(L_1+L_2\right)}^2=0\end{array}$ ..... (VIII)

Calculation:

Substitute $1350\text{kg}{\text{.m}}^{\text{2}}$

$730\text{kg}$ for $m$

$1.5\text{m}$ for $L_1$, $1.1\text{m}$ for $L_2$, $1.95\times {10}^4\text{N}/\text{m}$ for $k_1$ and $2.3\times {10}^4\text{N}/\text{m}$ for $k_2$ in Equation (VIII).

$\begin{array}{l}\left[\begin{array}{l}\left(730\text{kg}\right)\left(1350\text{kg}{\text{.m}}^{\text{2}}\right)s^4+\\ \left[\begin{array}{l}730\text{kg}\left[\left(1.95\times {10}^4\text{N}/\text{m}\right){\left(1.5\text{m}\right)}^2-\left(2.3\times {10}^4\text{N}/\text{m}\right){\left(1.1\text{m}\right)}^2\right]+\\ \left(1350\text{kg}{\text{.m}}^{\text{2}}\right)\left(1.95\times {10}^4\text{N}/\text{m}+2.3\times {10}^4\text{N}/\text{m}\right)\end{array}\right]s^2+\\ (\left(1.95\times {10}^4\text{N}/\text{m}\right)\left(2.3\times {10}^4\text{N}/\text{m}\right){\left(1.5\text{m}+1.1\text{m}\right)}^2\end{array}\right]=0\\ 985500s^4+109719650s^2+11661\times {10}^8=0\\ s^4+111.329s^2+1.17\times {10}^4=0\end{array}$. (IX)

Substitute $u$ for $s^2$ in Equation (IX).

$u^2+111.329u+1.17\times {10}^4=0$ ..... (X)

Solve equation (X) to obtain the value of $u$ as $-99.43$ and $-11.9$.

$s^2=-99.43$ .....(XI)

$s^2=-11.9$ ..... (XII)

On solving Equation (XI and (XII).

$s=9.971j$

$s=3.45j$

Write the standard form of frequency equation.

$s=j\omega$ ..... (XIII)

Compare the values of $s$ with standard Equation

$\omega =9.971$

$\omega =3.45$

Write the Equation for frequency.

$f=\frac{\omega }{2\pi }$ ..... (XIV)

Substitute $9.971$ for $\omega$ in equation (XIV).

$\begin{array}{c}f=\frac{9.971}{2\pi }\\ =\frac{9.971}{2\left(3.1415\right)}\text{Hz}\\ =1.5869\text{Hz}\\ \approx 1.587\text{Hz}\end{array}$

Substitute $3.45$ for $\omega$ in Equation (XIV).

$\begin{array}{c}f=\frac{3.45}{2\pi }\\ =\frac{3.45}{2\left(3.1415\right)}\text{Hz}\\ =0.549\text{Hz}\end{array}$

Rearrange equation (V).

$\frac{A_1}{A_2}=\frac{k_1{L}_1-k_2{l}_2}{m{s}^2+k_1+k_2}$ ..... (XV)

Since the mode ratios are ratio of $x$ and $\theta$ substitute $X$ for $A_1$ and $\theta$ for $A_2$.

$\frac{x}{\theta }=\frac{k_1{L}_1-k_2{l}_2}{m{s}^2+k_1+k_2}$ ..... (XVI)

Substitute $730\text{kg}$ for $m$

$1.5\text{m}$ for $L_1$, $1.1\text{m}$ for $L_2$, $1.95\times {10}^4\text{N}/\text{m}$ for $k_1$ and $2.3\times {10}^4\text{N}/\text{m}$ for $k_2$ in Equation (XVI).

$\begin{array}{c}\frac{x}{\theta }=\frac{\left(1.95\times {10}^4\text{N}/\text{m}\right)\left(1.5\text{m}\right)-\left(2.3\times {10}^4\text{N}/\text{m}\right)\left(1.1\text{m}\right)}{\left(730\text{kg}\right)s^2+\left(1.95\times {10}^4\text{N}/\text{m}\right)+\left(2.3\times {10}^4\text{N}/\text{m}\right)}\\ =\frac{3.95\times {10}^3}{730s^2+4.25\times {10}^4}\\ =\frac{5.411}{s^2+58.2192}\end{array}$ ..... (XVII)

Substitute $-99.43$ for $s^2$ in Equation (XVII).

$\begin{array}{c}\frac{x}{\theta }=\frac{5.411}{-99.43+58.2192}\\ =\frac{5.411}{-41.2108}\\ =-0.131\text{m}\end{array}$

Substitute $-11.9$ for $s^2$ in Equation (XVII).

$\begin{array}{c}\frac{x}{\theta }=\frac{5.411}{-11.9+58.2192}\\ =\frac{5.411}{46.3192}\\ =0.1168\text{m}\end{array}$

Conclusion:

The frequencies are $1.587\text{Hz}$ and $0.549\text{Hz}$.

The mode ratios are $-0.131$ and $0.1168$.

The locations are $0.131\text{m}$ ahead of mass center and $0.1168\text{m}$ behind the mass center.