Question 16.13: Determine the frequencies of free vibration for the frame sh......
Determine the frequencies of free vibration for the frame shown in Fig. 16.22a. Each element of the frame has uniformly distributed mass m, the length l, and the bending stiffness E I= const.
The primary system of the displacement method and the bending moment diagram due to the unit angular displacement Z_1=1 are given in Fig. 16.22b. Since the distributed inertial forces are taken into account, then the bending moment diagram within each element is curvilinear.
Reaction of the introduced constraint 1 is presented in Krylov-Duncan form. According to Table A.25 line 1 we have
M_{1 A}=M_{1 B}=\frac{4 E I}{l} f_3(\lambda), \quad f_3=\frac{\lambda}{4} \frac{T U-S V}{U^2-T V},
so the unit reaction is r_{11}=2 \cdot \frac{4 E I}{l} f_3(\lambda). Canonical equation is r_{11} Z_1=0. Since r_{11} \neq 0, then nontrivial solution occurs if r_{11}=0. Thus, the frequency equation isr_{11}=T U-S V=0. In the equivalent form, taking into account (16.26), and after simplest transformation, the frequency equation becomes
\tan k l=\tanh k l (a)
\begin{aligned} & S T-U V=\frac{1}{2}(\cosh k x \sin k x+\sinh k x \cos k x) \\ & T U-S V=\frac{1}{2}(\cosh k x \sin k x-\sinh k x \cos k x) \\ & S^2-U^2=\cosh k x \cos k x ; \quad T^2-V^2=2\left(S U-V^2\right)=\sinh k x \sin k x \\ & U^2-T V=\frac{1}{2}(1-\cosh k x \cos k x) ; \quad S^2-T V=\frac{1}{2}(1+\cosh k x \cos k x) \\ & T^2-S U=S U-V^2=\frac{1}{2} \sinh k x \sin k x ; \quad 2 S U=T^2+V^2 \end{aligned} (6.26)
The roots of this equation are \lambda=k l=3.926,7.0685, \ldots The exact frequencies are
\omega_1=\frac{3.926^2}{l^2} \sqrt{\frac{E I}{m_0}}, \quad \omega_2=\frac{7.0685^2}{l^2} \sqrt{\frac{E I}{m_0}}, \ldots
It is worth mentioning that the Smirnov’s functions are very effective for analyzing the free vibrations of continuous beams and frame systems using the displacement method. These functions refer to uniform single-span beams. Table A.28 contains beam reactions due to force and kinematic harmonic excitation. A feature of the functions is that they are composed taking into account the distributed mass of the beam. These reactions are presented in an analytical form through trigonometric functions, and tabulated.
| 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 |
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\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} |
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\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} |
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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} |
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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} |
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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} |
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M_A=M_B=\frac{P l}{8} \psi_{1 P}(u) \text {, } | Q_A=Q_B=\frac{P}{8} \psi_{2 P}(u) |
| Table A.25 Amplitude values of dynamical reactions of uniform beams with distributed mass m_0 ; \lambda=k l , \varphi(t)=\varphi \sin \theta t ; \Delta(t)=\Delta \sin \theta t ; \varphi,\Delta are amplitudes of angular and linear displacements, \varphi=1, \Delta=1(\mathrm{Kiselev} 1980) | ||||
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Bending moment M(0)=M_0 | Bending moment M(l) | Shear force Q(0) | Shear Force Q(l) |
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\begin{gathered}\frac{4 E I}{l} f_3(\lambda) \\f_3=\frac{\lambda}{4} \frac{T U-S V}{U^2-T V}\end{gathered} | \begin{gathered}\frac{2 E I}{l} f_5(\lambda) \\f_5=\frac{\lambda}{2} \frac{V}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_4(\lambda) \\f_4=\frac{\lambda}{4} \frac{S U-T^2}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_6(\lambda) \\f_6=\frac{\lambda^2}{6} \frac{U}{U^2-T V}\end{gathered} |
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\begin{gathered}-\frac{6 E I}{l^2} f_4(\lambda) \\f_4=\frac{\lambda}{4} \frac{S U-T^2}{U^2-T V}\end{gathered} | \begin{gathered}-\frac{6 E I}{l^2} f_8(\lambda) \\f_8=\frac{\lambda^2}{6} \frac{U}{U^2-T V}\end{gathered} | \begin{gathered}\frac{12 E I}{l^3} f_7(\lambda) \\f_7=\frac{\lambda^3}{12} \frac{S T-U V}{U^2-T V}\end{gathered} | \begin{gathered}\frac{12 E I}{l^3} f_9(\lambda) \\f_9=\frac{\lambda^3}{12} \frac{U}{U^2-T V}\end{gathered} |
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\begin{gathered}\frac{3 E I}{l} f_{10}(\lambda) \\f_{10}=\frac{\lambda}{3} \frac{T^2-V^2}{T U-S V}\end{gathered} | 0 | \begin{gathered}-\frac{3 E I}{l^2} f_{11}(\lambda) \\f_{11}=\frac{\lambda^2}{3} \frac{U V-S T}{T U-S V}\end{gathered} | \begin{gathered}-\frac{3 E I}{l^2} f_{12}(\lambda) \\f_{12}=\frac{\lambda^2}{3} \frac{T}{T U-S V}\end{gathered} |
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\begin{gathered}-\frac{3 E I}{l^2} f_{11}(\lambda) \\f_{11}=\frac{\lambda^2}{3} \frac{U V-S T}{T U-S V}\end{gathered} | 0 | \begin{gathered}-\frac{3 E I}{l^3} f_{13}(\lambda) \\f_{13}=\frac{\lambda^2}{3} \frac{U^2-S^2}{T U-S V}\end{gathered} | \begin{gathered}-\frac{3 E I}{l^3} f_{14}(\lambda) \\f_{14}=-\frac{\lambda^2}{3} \frac{S}{T U-S V}\end{gathered} |
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\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | -\frac{P}{k} \frac{U_{k l} V_{k a}-V_{k l} U_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | P \frac{U_{k l} U_{k a}-T_{k l} V_{k a}}{U_{k l}^2-T_{k l} V_{k l}} | P \frac{U_{k l} U_{k a}-T_{k l} V_{k a}}{U_{k l}^2-T_{k l} V_{k l}} |