Question 17.21: (Smirnov 1947) The design diagram of a non-regular frame is ......

This structure contains two specific parts. One of them presents absolutely rigid element C D with one degree of freedom, and left part of structure consists of three deformable elements with uniformly distributed masses. Designate one of the elements as the base one and express the parameters of all other elements through the base one. Let element B C be the base one, then the parameters i=E I / l; and k=m \omega^{2} l^{3} for other elements are presented in Table 17.12.

The primary system is shown in Fig. 17.36b: introduced constraint 1 prevents angular displacement of joint B and constraint 2 prevents the linear vertical displacement of hinge C. Bending moment diagrams M_{1} and M_{2} in Fig. 17.36c,d are caused by amplitude values of harmonic displacements Z_{1}=1 \times \sin \omega t, \quad Z_{2}=1 \times \sin \omega t. Equilibrium equations for joint B in the first and second states (bending moment diagrams M_{1} and M_{2} ) lead to the following values of unit reactions

\begin{aligned}& r_{11}=\underbrace{\left(3 \frac{i}{2}-\frac{2}{105} \cdot 8 k\right)}_{\text {rod } A B, \text { case } 4}+\underbrace{\left(4 i-\frac{k}{105}\right)}_{\text {rod } B E \text {, case } 2}+\underbrace{\left(3 i-\frac{2 k}{105}\right)}_{\text {rod } B C, \text { case } 4}=8.5 i-0.181 k \\& r_{22}=\underbrace{\left(\frac{3 i}{l^{2}}-\frac{33 k}{140 l^{2}}\right)}_{\text {rod } B E, \text { case } 5}-\underbrace{\frac{k}{3 l^{2}}}_{\text {rod } C D, \text { case } 6}=\frac{3 i}{l^{2}}-0.57 \frac{k}{l^{2}} \\& r_{12}=r_{21}=-\underbrace{\left(\frac{3 i}{l}+\frac{11 k}{280 l}\right)}_{\text {rod } B E, \text { case } 5}=-\left(\frac{3 i}{l}+0.04 \frac{k}{l}\right)\end{aligned}

Canonical equation of the free vibration of the displacement method are

\left[\begin{array}{ll}r_{11} & r_{12} \\r_{21} & r_{22}\end{array}\right]\left[\begin{array}{l}Z_{1} \\Z_{2}\end{array}\right]=0 

In the extended form

\begin{aligned}& (8.5-0.181 \mu) Z_{1}-(3+0.04 \mu) \frac{Z_{2}}{l}=0 \\& -(3+0.04 \mu) Z_{1}+(3-0.57 \mu) \frac{Z_{2}}{l}=0\end{aligned}        (a)

where \mu=\frac{k}{i}=\frac{m \omega^{2} l^{4}}{E I}, \quad \omega^{2}=\frac{\mu E I}{m l^{4}}.

The frequency equation

\begin{aligned}& r_{11} r_{22}-r_{12} r_{21}=(8.5-0.181 \mu)(3-0.57 \mu)-(3+0.04 \mu)^{2}=0 \\& \text { or } \quad 0.101 \mu^{2}-5.63 \mu+16.5=0\end{aligned}

The roots of the frequency equation and corresponding the frequencies of free vibration are

\mu_{1}=3.1, \quad \mu_{2}=52.5, \quad \omega_{1}=\sqrt{\frac{3.1 E I}{m l^{4}}}, \quad \omega_{1}=\sqrt{\frac{52.5 E I}{m l^{4}}}

From the first equation (a) we get

Z_{1}=\frac{3+0.04 \mu}{8.5-0.181 \mu} \frac{Z_{2}}{l}

Amplitude distribution coefficients according to the first and second frequency of the free vibrations

\begin{aligned}& Z_1\left(\mu_1\right)=\frac{3+0.04 \times 3.1}{8.5-0.181 \times 3.1} \frac{Z_2}{l}=0.394 \frac{Z_2}{l} \\& Z_1\left(\mu_2\right)=\frac{3+0.04 \times 52.5}{8.5-0.181 \times 52.5} \frac{Z_2}{l}=-5.087 \frac{Z_2}{l}\end{aligned}

Verification: from the second equation (a) we get

Z_1\left(\mu_1\right)=\frac{(3-0.57 \mu)}{(3+0.04 \mu)} \frac{Z_2}{l}=\frac{(3-0.57 \times 3.1)}{(3+0.04 \times 3.1)} \frac{Z_2}{l}=0.395 \frac{Z_2}{l}

Two mode shapes of vibrations are presented in Fig. 17.36e,f.