Solved Examples - Determination of Natural Frequencies and Mode Shapes

Example 1 - Estimate the fundamental natural freuqency of a simply supported beam carrying three indetical equally spaced measses as shown in Figure 1.
Figure 1 - Simply supported beam with three bodies of masses $m_1$, $m_2$, and $m_3$. Solution - The flexibility influence coefficient required for the application of Dunkerley's formula are given by: $$\begin{eqnarray} a_{11} &=& a_{33} =\frac{3}{256}\frac{l^2}{EI}\nonumber\\ a_{22} &=& \frac{1}{48}\frac{l^3}{EI}\nonumber \end{eqnarray}$$ The masses $m_1, m_2,$ and $m_3$ are equal. $$\begin{eqnarray} \frac{1}{\omega_1^2} &=& \left(\frac{3}{256} + \frac{1}{48} + \frac{3}{256}\right)\frac{ml^3}{EI} = 0.04427\frac{ml^3}{EI}\nonumber\\ &=& 4.75375\sqrt{\frac{EI}{ml^3}}\nonumber \end{eqnarray}$$
Example 2 Estimate the fundamental frequency of vibration of the system shown in Figure 2. Assume that $m_1=m_2=m_3 = m$, and $k_1 = k_2 = k_3 = k$, and the mode shape is $\vec{X} = \begin{Bmatrix}1\\2\\3 \end{Bmatrix}$.
Figure 2 - Three degree of freedom system which consist of three springs with stiffness $k_1 = k_2 = k_3 = k$ and three bodies with masses $m_1 = m_2 = m_3 = m$. Solution: The stiffness and mass matrices of the system are: $$\begin{eqnarray} \mathrm{\mathbf{K}} &=& k\begin{bmatrix}2&-1&0\\ -1&2&-1\\ 0&-1&1\end{bmatrix}\nonumber\\ \mathrm{\mathbf{M}} &=& m\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}\nonumber \end{eqnarray}$$ Substituting the assumed mode shape in the expression for Rayleigh's quotient, the following is obtained: $$\begin{eqnarray} R(\vec{X}) &=& \omega^2 = \frac{\vec{X}^T\mathrm{\mathbf{K}}\vec{X}}{\vec{X}^T\mathrm{\mathbf{M}}\vec{X}} = \frac{\begin{pmatrix}1&2&3\end{pmatrix}k\begin{bmatrix}2&-1&0\\ -1&2&-1\\ 0&-1&1\end{bmatrix} \begin{Bmatrix}1\\2\\3 \end{Bmatrix}}{\begin{pmatrix}1&2&3\end{pmatrix}m\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}\begin{Bmatrix}1\\2\\3 \end{Bmatrix}}\nonumber\\ R(\vec{X}) &=& 0.2143 \frac{k}{m}\nonumber\\ \omega_1 &=& 0.4629 \sqrt{\frac{k}{m}}\nonumber \end{eqnarray}$$ Example 3 Estimate the fundamental frequency of the lateral vibration of a shaft carrying three rotors (masses), as shown in Figure 3 with $m_1 = 20 kg$, $m_2 = 50 kg$, $m_3 = 40 kg$, $l_1 = 1 m$, $l_2 = 3m$, $l_3 = 4 m$, and $l_4 = 2 m$. The shaft is made of steel with a solid circular cross section of diameter 10 cm.
Figure 3 - Shaft with three different bodies with masses $m_1$, $m_2$, and $m_3$.
In Figure 4 the position of load P that is causing deflection, is shown. Figure 4 - Beam under static load P. Solution: The deflection of the beam shown in Figure 4 due to the static load P can be written as: $$\begin{eqnarray} w(x) &=& \begin{cases} \frac{Pbx}{6EIl}(l^2 - b^2 -x^2) \quad 0 \leq x \leq a\\ -\frac{Pa(l-x)}{6EIl}(a^2 + x^2 -2lx) \quad a\leq x\leq l\end{cases}\nonumber \end{eqnarray}$$ Before calculating natural freuqency of the system it is neceassary to obtain the total deflections caused by the bodies with massess $m_1$, $m_2$, and $m_3$. To obtain the total deflection caused by these bodies it is necessary to investigate and calculate the deflections caused by each body and in each analysis neglected the remaining two bodies. Deflection due to the weight of $m_1$ at the location of mass $m_1$: $$\begin{eqnarray} w_1' &=& \frac{20\cdot 9.81 \cdot 9 \cdot 1}{6EI\cdot 10} (100-81.1) = \frac{529.74}{EI}\nonumber \end{eqnarray}$$ A the location $m_2$: $$\begin{eqnarray} w_2' &=& -\frac{20\cdot 9.81 \cdot 1 \cdot 6}{6EI\cdot 10} ( 1+ 16-2\cdot 10 \cdot 4) = \frac{1236.06}{EI} \end{eqnarray}$$ At the location of $m_3$: $$\begin{eqnarray} w_3' &=& -\frac{20\cdot 9.81 \cdot 1 \cdot 2}{6EI\cdot 10} (1+64-2\cdot 10 \cdot 8) = \frac{621.3}{EI}\nonumber \end{eqnarray}$$ The deflection caused by the weight $m_2$. At the location of $m_1$: $$\begin{eqnarray} w_1'' &=& \frac{Pbx}{6EIl}(l^2 - b^2 - x^2) \nonumber\\ &=& \frac{5\cdot 9.81\cdot 6 \cdot 1}{6\cdot 10 EI}(100-36-1) \nonumber\\ &=& \frac{3090.15}{EI}.\nonumber \end{eqnarray}$$ At the location of $m_2$ using the first equation of $w(x)$: $$\begin{eqnarray} w_2'' &=& \frac{Pbx}{6EIl}(l^2 - b^2 - x^2)\nonumber\\ &=& \frac{50\cdot 9.831 \cdot 6 \cdot 4}{6EI\cdot 10}(100-36-16)\nonumber\\ &=& \frac{9417.6}{EI}.\nonumber \end{eqnarray}$$ At the location of $m_3$ using the second equation of $w(x)$: $$\begin{eqnarray} w_3''&=& -\frac{Pa(l-x)}{6EIl}(a^2 + x^2 - 2lx)\nonumber\\ &=& -\frac{50\cdot 9.81 \cdot 4 \cdot 2}{6\cdot 10 \cdot EI}(16+64-2\cdot 10\cdot 8)\nonumber \\ &=& \frac{5232.0}{EI}.\nonumber \end{eqnarray}$$ Deflection caused by the weight of $m_3$ at the location of $m_1$ with a use of the first equation of $w_(x)$: $$\begin{eqnarray} w_1''' &=& \frac{Pbx}{6EIl}(l^2 - b^2 - x^2)\nonumber\\ &=& \frac{40\cdot 9.81\cdot 2 \cdot 1}{6\cdot 10 EI}(100-4-1)\nonumber\\ &=& \frac{1242.6}{EI}.\nonumber \end{eqnarray}$$ At the location of $m_2$ using first equation of $w(x)$ the $w_2'''$ is equal to: $$\begin{eqnarray} w_2'''&=&\frac{Pbx}{6EIl}(l^2 - b^2 - x^2)\nonumber\\ &=& \frac{40\cdot 9.81\cdot 2 \cdot 4}{6\cdot 10 EI}(100-4-16)\nonumber\\ &=& \frac{4185.6}{EI}.\nonumber \end{eqnarray}$$ A the location the $m_3$ using first equation of $w(x)$ the $w_3'''$ is equal to: $$\begin{eqnarray} w_3'''&=& \frac{Pbx}{6EIl}(l^2 - b^2 - x^2)\nonumber\\ &=& \frac{40\cdot 9.81\cdot 2 \cdot 8}{6\cdot 10 EI}(100-4-64)\nonumber\\ &=& \frac{3348.48}{EI}.\nonumber \end{eqnarray}$$ The total deflections of the bodies with mass $m_1,m_2,$ and $m_3$ are equal to: $$\begin{eqnarray} w_1 &=& w_1' + w_1'' + w_1''' = \frac{4962.49}{EI},\nonumber\\ w_2 &=& w_2' + w_2'' + w_2''' = \frac{14839.26}{EI},\nonumber\\ w_3 &=& w_3' + w_3'' + w_3''' = \frac{9201.78}{EI}.\nonumber\\ \end{eqnarray}$$ The general form of natural frequency equation for specific case can be wirtten as: $$\begin{eqnarray} \omega &=& \sqrt{\frac{g(m_1 w_1 + m_2 w_2 + m_3w_3)}{m_1w_1^2 + m_2w_2^2 + m_3w_3^2}}.\nonumber \end{eqnarray}$$ By substituting masses and deflections into previous equation the natural frequency can be written as: $$\begin{eqnarray} \omega &=& \sqrt{\frac{9.81\cdot(20\cdot 4862.49 + 50\cdot 14839.26 + 40\cdot 9201.78)EI}{20\cdot (4862.49)^2 + 50\cdot (14839.26)^2 + 40\cdot 9201.78^2 }}\nonumber \\ \omega &=& 0.028222\sqrt{EI}\nonumber \end{eqnarray}$$