Question 16.14: Determine the frequency of free vibration for frame shown in......
Determine the frequency of free vibration for frame shown in Fig. 16.23a. Mass per unit length for all members is m.
Parameter of the corrected functions \psi for vertical bar is equal u=\sqrt[4]{\frac{m \omega^{2} l^{4}}{E I}}, while for crossbar the parameter becomes
\sqrt[4]{\frac{m \omega^{2} l^{4}}{4 E I}}=\sqrt[4]{\frac{m \omega^{2} l^{4}}{E I}} \sqrt[4]{\frac{1}{4}}=0.707 u
Primary system is shown in Fig. 16.23b. Bending moment diagrams caused by unit primary displacements Z_{1}=1 and Z_{2}=1 are shown in Fig. 16.23c. For calculation of unit reaction the Smirnov’s functions can be used. Unit reactions according to Table A.28 are
r_{11}=4 i \psi_{2}(u)+3 \times 4 i \psi_{1}(0.707 u)=4 i \psi_{2}(u)+12 i \psi_{1}(0.707 u), \quad r_{12}=r_{21}=-\frac{6 i}{l} \psi_{5}(u)
When calculating the unit reaction r_{22}, one should take into account not only the shear forces in the vertical bar but also the inertial forces of the distributed masses of the crossbar.
If introduced constraint 2 executes a harmonic linear displacement with a frequency of free vibration \omega, then inertial force of the distributed masses of the cross-rod 1-2 is equal
J_{1-2}=m l \omega^{2}=m l \frac{u^{4} E I}{m l^{4}}=\frac{i}{l^{2}} u
Therefore, the total reaction r_{22} is equal to
r_{22}=\frac{12 i}{l^{2}} \psi_{10}(u)+\frac{3 i}{l^{2}} \psi_{12}(u)-\frac{i}{l^{2}} u
The frequency equation
D=\left|\begin{array}{ll}r_{11} & r_{12} \\r_{21} & r_{22}\end{array}\right|=r_{11} r_{22}-r_{12}^{2}=0
In expanded form, the frequency equation becomes
D=\left[4 i \psi_{2}(u)+12 i \psi_{1}(0.707 u)\right]\left[\frac{12 i}{l^{2}} \psi_{10}(u)+\frac{3 i}{l^{2}} \psi_{12}(u)-\frac{i}{l^{2}} u\right]-\left[\frac{6 i}{l} \psi_{5}(u)\right]^{2}=0
After division on i^{2} / l^{2} we get
D=\left[4 \psi_{2}(u)+12 \psi_{1}(0.707 u)\right]\left[12 \psi_{10}(u)+3 \psi_{12}(u)-u\right]-\left[6 \psi_{5}(u)\right]^{2}=0
For solution of this equation the analytical expressions of Smirnov’s function (Table A.28) may be used, or tables of numerical data of these functions (Karnovsky and Lebed 2004a). The smallest root of this equation is u_{\min }=1.68. Fundamental frequency of free vibration
\omega=\frac{u^{2}}{l^{2}} \sqrt{\frac{E I}{m}}=\frac{2.82}{l^{2}} \sqrt{\frac{E I}{m}}
It is useful to discuss some aspects of the analysis due to the change in the design diagram shown in Fig. 16.23a.
(a) Modified design diagram contains additional massless element 1-A of length l_{1}, bending stiffness E I_{1}, and with roller support A (Fig.16.24a). All supports of the structure allow horizontal displacement of the crossbar. Primary system contains two introduced constraints 1 and 2 (Fig. 16.24b). The bending stiffness per unit length of element 1-A is E I_{1} / l_{1}=i_{1}.
Let the stiffness i_{1} of additional member 1- A and stiffness i of basic vertical element be related as i_{1}=k i, where \kappa is some positive number. Since element 1-A is massless, then in case of displacement Z_{1}=1 of introduced constraint 1 the bending moment diagram within portion 1-A is linear (Fig. 16.24c). Therefore, the unit reaction r_{11} takes an additional term 3ki (Table A.3, row 1) and becomes r_{11}=4 i \psi_{2}(u)+12 i \psi_{1}(0.707 u)+3 k i. The remaining coefficients of canonical equations do not change.
(b) Let support A of modified structure be adopted as pinned one. In this case the structure does not have a side displacement; therefore the number of unknowns of displacement method is equal to one. The primary system contains only one introduced constraint 1 , so the canonical equation of displacement method becomes r_{11} Z_{1}=0. The frequency equation is r_{11}=0, or r_{11}=4 \psi_{2}(u)+12 \psi_{1}(0.707 u)+3 k=0. In this case the parameters of the right vertical bar (E I, m, l) are not included in the frequency equation.
Table A.3 Reactions of fixed-pinned beams |
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| No | Loading conditions | Reactions and bending moment diagrams | Expressions for bending moments and reactions | ||||||||||||||||
| 1 |  |
 |
\begin{aligned} & M_A=\frac{3 E I}{l}=3 i, \quad i=\frac{E I}{l} \\ & R_A=R_B=\frac{3 E I}{l^2}=\frac{3 i}{l} \end{aligned} | ||||||||||||||||
| 2 |  |
 |
\begin{aligned} & M_A=\frac{3 E I}{l^2}=\frac{3 i}{l} \\ & R_A=R_B=\frac{3 E I}{l^3}=\frac{3 i}{l^2} \end{aligned} | ||||||||||||||||
| 3 |  |
 |
\begin{aligned} & M_A=\frac{P l}{2} \boldsymbol{\boldsymbol{\boldsymbol{v}}}\left(1-\boldsymbol{\boldsymbol{\boldsymbol{v}}}^2\right) \\ & M_C=\frac{P l}{2} u^2 \boldsymbol{\boldsymbol{\boldsymbol{v}}}(3-u) \\ & R_A=\frac{P \boldsymbol{\boldsymbol{\boldsymbol{v}}}}{2}\left(3-\boldsymbol{\boldsymbol{\boldsymbol{v}}}^2\right) \\ & R_B=\frac{P u^2}{2}(3-u) \\ & \underline{u=\boldsymbol{\boldsymbol{\boldsymbol{v}}}=0.5} \\ & R_A=\frac{11}{16} P ; \quad R_B=\frac{5}{16} P \\ & M_A=\frac{3}{16} P l ; \quad M_C=\frac{5}{32} P l \end{aligned} | ||||||||||||||||
| 4 |  |
 |
\begin{aligned} & M_A=\frac{M}{2}\left(1-3 \boldsymbol{v}^2\right) \\ & M_{C B}=\frac{3 M}{2} \boldsymbol{v}\left(1-\boldsymbol{v}^2\right) \\ & R_A=R_B=\frac{3 M}{2 l}\left(1-\boldsymbol{v}^2\right) \\ & \underline{u=\boldsymbol{v}=0.5:} \quad M_A=\frac{M}{8} \\ & R_A=R_A=\frac{9}{8} \frac{M}{l} \\ & M_{C A}=\frac{7}{16} M ; \quad M_{C B}=\frac{9}{16} M \\ & \underline{\boldsymbol{v}=0}: \quad M_A=M / 2 \\ & R_A=R_B=\frac{3 M}{2 l} \end{aligned} | ||||||||||||||||
| 5 |  |
 |
\begin{aligned} & M_A=\frac{q l^2}{8}, \quad M\left(\frac{l}{2}\right)=\frac{q l^2}{16} \\ & R_A=\frac{5}{8} q l ; \quad R_B=\frac{3}{8} q l \end{aligned} | ||||||||||||||||
| 6 |  |
 |
\begin{aligned} & M_A=\frac{q l^2}{15} \\ & R_A=\frac{2}{5} q l, R_B=\frac{1}{10} q l \end{aligned} | ||||||||||||||||
| 7 |  |
 |
\begin{aligned} & M_A=\frac{7 q l^2}{120}, \\ & R_A=\frac{9}{40} q l, R_B=\frac{11}{40} q l \end{aligned} | ||||||||||||||||
| 8 |  |
 |
\begin{aligned} & M_A=\frac{3 E I \alpha\left(t_1-t_2\right)}{2 h} \\ & R_A=R_B=\frac{3 E I \alpha\left(t_1-t_2\right)}{2 h l} \end{aligned} | ||||||||||||||||
| 9 |  |
 |
\begin{aligned}& M_A=\frac{q l^2}{8} u^2(2-u)^2=q l^2 k_1 \\& R_B=\frac{q a}{2} u-q l k_1, R_A=q a-R_B\end{aligned}
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| Table A.28 The exact dynamic reactions of uniform beams with distributed mass due by kinematic harmonic excitation of the ends supports \varphi(t)=1 \cdot \sin \theta t, \quad \Delta(t)=1 \cdot \sin \theta t , and forced excitation P \sin \theta t . The length of a beam is l , distributed mass per unit length m_0 , the bending stiffness is E I=c o n s t , the frequency excitation \theta, i=E I / l , dimensional parameter u=\sqrt[4]{m_0 \theta^2 l^4 / E I} (Darkov 1989) | ||
| Design diagram | Amplitude values of moments | Amplitude values of shear |
 |
\begin{aligned}& M_A=4 i \psi_2(u) \\& M_B=2 i \psi_3(u)_{,}\end{aligned} | \begin{aligned}& Q_A=\frac{6 i}{l} \psi_5(u) \\& Q_B=\frac{6 i}{l} \psi_6(u)\end{aligned} |
 |
\begin{aligned}& M_A=\frac{6 i}{l} \psi_5(u) \\& M_B=\frac{6 i}{l} \psi_6(u)\end{aligned} | \begin{aligned}& Q_A=\frac{12 i}{l^2} \psi_{10}(u) \\& Q_B=\frac{12 i}{l^2} \psi_{11}(u)\end{aligned} |
 |
M_A=3 i \psi_1(u) | \begin{aligned}Q_A & =\frac{3 i}{l} \psi_4(u) \\Q_B & =\frac{3 i}{l} \psi_7(u)\end{aligned} |
 |
M_A=\frac{3 i}{l} \psi_7(u) | \begin{aligned}& Q_A=\frac{3 i}{l^2} \psi_9(u), \\& Q_B=\frac{3 i}{l^2} \psi_{12}(u)\end{aligned} |
 |
M_A=\frac{6 i}{l} \psi_4(u) | \begin{aligned}& Q_A=\frac{3 i}{l^2} \psi_8(u), \\& Q_B=\frac{3 i}{l^2} \psi_9(u)\end{aligned} |
 |
M_A=M_B=\frac{P l}{8} \psi_{1 P}(u) \text {, } | Q_A=Q_B=\frac{P}{8} \psi_{2 P}(u) |
