Question 17.13: The frame in Fig. 17.26a consists of three uniform massless ......
The frame in Fig. 17.26a consists of three uniform massless rods of length l and bending stiffness EI. Each element contains one mass m which is located at the middle of the member. Mass m at the span 1- B is subjected to disturbing harmonic force P(t)=P \sin \theta t. Determine the amplitude value of angular rotation of the rigid joint 1 in case of the steady-state vibration.
This structure has three degrees of freedom. Its kinematical indeterminacy is equal to one (while this structure contains six unknowns of the force method). The primary system of displacement method is obtained from a given structure by introducing constraint 1 at the fixed joint 1 (Fig. 15.26b). This constraint prevents angular displacement at support 1. The primary unknown is the angular displacement Z_{1} at support 1 . Canonical equation of displacement method is
r_{11} Z_{1}+R_{1 P}=0 (a)
Unit reaction r_{11}. Suppose that the introduced constraint 1 vibrates with the frequency \theta and the unit amplitude, i.e., Z_{1}(t)=1 \cdot \sin \theta t. According to Table A.27, row 1, the bending moments at specified sections a, b, c in the primary system are
\bar{M}_{a}=\bar{M}_{b}=\bar{M}_{c}=\frac{4 E I}{l} F_{1}=\frac{4 E I}{l}\left[1-\frac{7}{4} \frac{\theta^{2}}{\omega^{2}}\right] \frac{1}{1-\theta^{2} / \omega^{2}}
The dotted lines in Fig. 17.26c, d show location of extended fibers in the vicinity of joint 1.
Loaded reaction R_{1 P}. Mass m is subjected to force P(t)=P \sin \theta t. According to Table A.27, row 8
M_{b-P}^{0}=P F_{15}=-P \frac{l}{8} \frac{1}{1-\theta^{2} / \omega^{2}}
Negative sign means that in the vicinity of joint 1 the extended fibers are located above, as shown in Fig. 17.26d. Equilibrium condition of joint 1 in the unit state (Fig. 17.26c) and loaded state (Fig. 17.26d) lead to the following results:
r_{11}=3 \cdot \frac{4 E I}{l} F_{1}=\frac{12 E I}{l}\left[1-\frac{7}{4} \frac{\theta^{2}}{\omega^{2}}\right] \frac{1}{1-\theta^{2} / \omega^{2}}, \quad R_{1 P}=-M_{b-P}^{0}=-\frac{P l}{8} \frac{1}{1-\theta^{2} / \omega^{2}} (b)
It can be seen that the expression for the amplitude of a unit reaction r_{11} and a load reaction R_{1 P} includes a detuning \theta / \omega. So the primary unknown, which presents the required amplitude value of angular rotation of joint 1, becomes
Z_{1}=-\frac{R_{1 P}}{r_{11}}=\frac{P l^{2}}{96 E I} \frac{1}{1-7 \theta^{2} / 4 \omega^{2}} (c)
For further dynamic analysis of the structure, the method of initial parameters is the most effective. For this, each element of the frame should be considered separately. The origin of coordinates should be placed in joint 1 and the x-axes are directed to support points A, B, C. Elements A-1 and C-1 are subjected only to the kinematic excitation of Z_{1}, and the element 1-B of the excitation of Z_{1} and the harmonic force P(t). The method allows determining not only the amplitude of the displacement of each lumped mass m but also the bending moments and shear forces in the characteristic sections of each separate element. A detailed analysis for elements 1-A and 1- C is presented in Examples 17.10 and 17.11. For element 1- B the additional force P(t) should be taken into account.
In particular, according to the formula (g) of Example 17.10 we have
M(0)=\frac{4 E I}{l} \varphi \frac{1-7 \theta^{2} / 4 \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
So amplitude value of bending moments M_{a} and M_{c} at points a and c of the entire frame may be calculated, if expression (c) for Z_{1} is substituted into formula for M(0) instead of \varphi
M_{a}=M_{c}=M_{a(c)}^{0} Z_{1}=\frac{P l}{24} \frac{1}{1-\theta^{2} / \omega^{2}}
For computation of amplitude value of M_{b} we need to take into account Table A.27, line 8
M_{b}=\bar{M}_{b} Z_{1}+M_{b}^{0}(P)=\frac{P l}{24} \frac{1}{1-\theta^{2} / \omega^{2}}-\frac{P l}{8} \frac{1}{1-\theta^{2} / \omega^{2}}=-\frac{P l}{12} \frac{1}{1-\theta^{2} / \omega^{2}}
Control: M_{a}+M_{b}+M_{c}=0, so the equilibrium condition for joint 1 is satisfied.
Comments
- For practical purpose the dynamic analysis should be supplemented by computation of the frequency of the free vibration \omega. Only in this case we can specify the frequency of excitation under which will be performed required value of detuning \theta / \omega.
- To determine the bending moments at characteristic points of the system, shear forces, inertial force, and reactions, the method of initial parameters should be applied to each element separately. The bending moment diagram within each element consists of two straight lines with kink under a lumped mass. This is explained by the presence of inertial force applied to the mass.
- If the system is exposed to a variable arbitrary load P(t), then it should be represented in a form of a trigonometric series, the dynamic response on each harmonic excitation separately should be determined, and the superposition principle should be applied.
A large number of dynamical problems of different deformable structures is presented in the book (Kiselev 1980).
| Table A.27 Amplitude values of dynamical reactions of massless uniform beams with one lumped mass m . E I= const, \mu=1 \div\left(1-\theta^{2} / \omega^{2}\right) ; \quad \varphi(t)=\varphi \sin \theta t, \quad \Delta(t)=\Delta \sin \theta t, \varphi, \Delta are amplitudes of angular and linear displacements; \omega – frequency of free vibration, \theta – frequency of excitation (kinematical -cases 1-6, forced-cases 8,9). General scheme shows positive reactions (Kiselev 1980) | ||||
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Bending moment M(0)=M_0 | Shear force Q(0)=Q_0 | Functions | Special case a=b=l / 2 |
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\frac{4 E I}{l} \varphi F_1 \mu | -\frac{6 E I}{l^2} F_5 \mu | \begin{aligned}& F_1=1-\frac{\theta^2}{\omega^2}\left(1+\frac{3 b}{4 a}\right) \\& F_5=1-\frac{\theta^2}{\omega^2}\left(1+\frac{3 a b+b^2}{2 a^2}\right)\end{aligned} | \begin{aligned}& F_1=1-\frac{7}{4} \frac{\theta^2}{\omega^2} \\& F_5=1-3 \frac{\theta^2}{\omega^2}\end{aligned} |
 |
\frac{2 E I}{l} \varphi F_2 \mu | -\frac{6 E I}{l^2} \varphi F_6 \mu | \begin{aligned}& F_2=1+\frac{1}{2} \frac{\theta^2}{\omega^2}(*) \\& F_6=1+\frac{\theta^2}{\omega^2} \frac{l}{2 a}\end{aligned} | \begin{aligned} & F_2=1+\frac{1}{2} \frac{\theta^2}{\omega^2} (*)\\ & F_6=1+\frac{\theta^2}{\omega^2} \end{aligned} |
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-\frac{6 E I}{l^2} \Delta F_3 \mu | \frac{12 E I}{l^3} \Delta F_7 \mu | \begin{aligned}& F_3=F_5 \\& F_7=1-\frac{\theta^2}{\omega^2}\left[1+\frac{b(3 a+b)^2}{4 a^3}\right]\end{aligned} | \begin{aligned}& F_3=F_5 \\& F_7=1-5 \frac{\theta^2}{\omega^2}\end{aligned} |
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-\frac{6 E I}{l^2} \Delta F_4 \mu | \frac{12 E I}{l^3} \Delta F_8 \mu | \begin{aligned}& F_4=1+\frac{\theta^2}{\omega^2} \frac{l}{2 b} \\& F_8=1+\frac{\theta^2}{\omega^2} \frac{3 l^2}{4 a b}\end{aligned} | \begin{aligned}& F_4=1+\frac{\theta^2}{\omega^2} \\& F_8=1+3 \frac{\theta^2}{\omega^2}\end{aligned} |
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\frac{3 E I}{l} \varphi F_9 \mu | -\frac{3 E I}{l^2} \varphi F_{12} \mu | \begin{aligned}& F_9=1-\frac{\theta^2}{\omega^2} \frac{4 l^2}{a(3 a+4 b)} \\& F_{12}=1-\frac{\theta^2}{\omega^2} \frac{2 l^2}{a^2} \frac{3 a+2 b}{3 a+4 b}\end{aligned} | \begin{aligned}& F_9=1-\frac{16}{7} \frac{\theta^2}{\omega^2} \\& F_{12}=1-\frac{40}{7} \frac{\theta^2}{\omega^2}\end{aligned} |
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-\frac{3 E I}{l^2} \Delta F_{10} H | \frac{3 E I}{l^3} \Delta F_{13} \mu | \begin{aligned}& F_{10}=1-\frac{\theta^2}{\omega^2} \frac{2 l^2}{a^2} \frac{3 a+2 b}{3 a+4 b} \\& F_{13}=1-\frac{\theta^2}{\omega^2} \frac{4 l^3}{a^3} \frac{3 a+b}{3 a+4 b}\end{aligned} | \begin{aligned}& F_{10}=1-\frac{40}{7} \frac{\theta^2}{\omega^2} \\& F_{13}=1-\frac{256}{7} \frac{\theta^2}{\omega^2}\end{aligned} |
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-\frac{3 E I}{l^2} \Delta F_{11} \mu | \frac{3 E I}{l^3} \Delta F_{14} \mu | \begin{aligned}& F_{11}=1+\frac{\theta^2}{\omega^2} \frac{2 l^2}{b(3 a+4 b)} \\& F_{14}=1+\frac{\theta^2}{\omega^2} \frac{6 l^3}{a b(3 a+4 b)}\end{aligned} | \begin{aligned}& F_{11}=1+\frac{8}{7} \frac{\theta^2}{\omega^2} \\& F_{14}=1+\frac{48}{7} \frac{\theta^2}{\omega^2}\end{aligned} |
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P F_{15} | P F_{16} | \begin{aligned}& F_{15}=-\frac{1}{1-\theta^2 / \omega^2} \frac{a b^2}{l^2} \\& F_{16}=\frac{1}{1-\theta^2 / \omega^2} \frac{b^2(3 a+b)}{l^3}\end{aligned} | \begin{aligned}& F_{15}=-\frac{l}{8} \frac{1}{1-\theta^2 / \omega^2} \\& F_{16}=\frac{1}{2} \frac{1}{1-\theta^2 / \omega^2}\end{aligned} |
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P F_{17} | P F_{18} | \begin{aligned}& F_{17}=-\frac{1}{1-\theta^2 / \omega^2} \frac{a b}{2 l}\left(1+\frac{b}{l}\right) \\& F_{18}=\frac{1}{1-\theta^2 / \omega^2}\left(1-\frac{3 a^2}{2 l^2}+\frac{a^3}{2 l^3}\right)\end{aligned} | \begin{aligned}& F_{17}=-\frac{1}{1-\theta^2 / \omega^2} \frac{3 l}{16} \\& F_{18}=\frac{1}{1-\theta^2 / \omega^2} \frac{11}{16}\end{aligned} |
| \left(^{*}\right) The lumped mass m and parameters a and b of it location on the beam appear in the formulas for frequency of free vibration. For clamped-clamped element \omega^{2}=\frac{3 l^{3} E I}{m a^{3} b^{3}} and for clamped-pinned element \omega^{2}=\frac{12 l^{3} E I}{m b^{2} a^{3}(3 a+4 b)} . | ||||