Question 17.11: Uniform clamped-clamped beam AB with lumped mass m is subjec......
Uniform clamped-clamped beam A B with lumped mass m is subjected to unit harmonic rotation \varphi(t)=\varphi \sin \theta t, \varphi=1 of the left support A (Fig. 17.24a). Determine the reactions at the right support B, inertial force and bending moment under the lumped mass. Assume a=b=l / 2. Use formulas ( \mathrm{g}, \mathrm{h}) from previous example.
In the case of a static rotation of a clamped support A, displacement in the middle of the beam according to Example 17.10, (or Table A.9, row 1) equals y=\frac{l}{8} \varphi. Therefore the equation of motion of the mass m becomes
y(t)=\frac{l}{8} \varphi \sin \theta t \frac{1}{1-\theta^{2} / \omega^{2}}
Since the frequency of the free vibration \omega=\sqrt{\frac{1}{m \delta_{11}}}=\sqrt{\frac{192 E I}{m l^{3}}}, then the mass m may be presented in terms of frequency vibration m=\frac{192 E I}{\omega^{2} l^{3}}. Therefore the expression for inertial force of mass m becomes
J^{*}(t)=m \theta^{2} y(t)=\frac{192 E I}{\omega^{2} l^{3}} \cdot \theta^{2} \cdot \frac{l}{8} \varphi \sin \theta t \frac{1}{1-\theta^{2} / \omega^{2}}=\frac{24 E I}{l^{2}} \varphi \frac{\theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
The bending moment and shear force at the left clamped support are
\begin{aligned}& M(0)=\frac{4 E I}{l} \varphi \sin \theta t-\frac{J^{*} l}{8}=\frac{4 E I}{l} \varphi \frac{1-7 \theta^{2} / 4 \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t \\& Q(0)=\frac{6 E I}{l^{2}} \varphi \sin \theta t+\frac{J^{*}}{2}=-\frac{6 E I}{l^{2}} \varphi \frac{1-3 \theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t .\end{aligned}
The expressions for M(0) and Q(0) coincide with (g), which had been obtained by initial parameters method. Bending moment under lumped mass m
M\left(\frac{l}{2}\right)=M(0)+Q(0) \frac{l}{2}=\frac{4 E I}{l} \varphi \sin \theta t-\frac{J^{*} l}{8}=\frac{E I}{l} \varphi \frac{1+2 \theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
Bending moment at the right clamped support B is
M(l)=-\frac{2 E I}{l} \varphi \sin \theta t-\frac{J^{*} l}{8}=-\frac{2 E I}{l} \varphi \frac{1+\theta^{2} / 2 \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
Reaction at the right clamped support B may be determined from equation
\sum Y=-R_{A 1}-J^{*}(t)+R_{B 1}=0 \rightarrow R_{B 1}=\frac{6 E I}{l^{2}} \varphi \frac{1+\theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
The scheme of forces for beam under given excitation (first state) is presented in Fig. 17.24b.
According to Table A.27, row 1, reactions at the right clamped support B are unknown; this is shown in Fig. 17.24a. For computation of these reactions we will apply the reciprocal reactions theorem. For this purpose we need to create additional state 2 for beam with same boundary condition (Fig. 17.24c). For this state there are tabulated reactions at the left support A and unknown reactions at the right support B.
According to the reciprocal reactions theorem we have M_{B 1}=M_{A 2} ; \quad R_{B 1}=R_{A 2} ; \quad M_{B 2}=M_{A 1} ; \quad R_{B 2}=R_{A 1}. Reactions which correspond these relationships are shown in the blocks of the same type.
Verification: bending moment under mass m taking into account the right forces
M_{m}=-M_{B 1}+R_{B 1} \frac{l}{2}=\left[-\frac{2 E I}{l}\left(1+\frac{1}{2} \frac{\theta^{2}}{\omega^{2}}\right)+\frac{6 E I}{l^{2}}\left(1+\frac{\theta^{2}}{\omega^{2}}\right) \frac{l}{2}\right] \varphi \frac{\sin \theta t}{1-\theta^{2} / \omega^{2}}=\frac{E I}{l} \varphi \frac{1+2 \theta^{2} / \omega^{2}}{1-\theta^{2} / \omega^{2}} \sin \theta t
Now we will demonstrate application of the reciprocal reactions and displacements theorem. Let us consider a beam shown in Fig. 17.24a subjected to harmonic kinematical excitation of the left clamped support. We need to determine vertical dynamical displacement y(l / 2) where the mass m is located. For this purpose we need to create additional state, which presents same clamped-clamped massless beam with one lumped mass m attached at the middle of the span. Corresponding scheme is shown in Table A.27, row 8. According to the reciprocal theorem, the displacement y(l / 2) caused by harmonic angular displacement of the left support is equal to the bending moment at the left clamped support caused by harmonic force which is applied at the point where displacement should be determined, i.e.,
\left.y\left(\frac{l}{2}\right)\right|_{\text {row } 1}=-\left.r(0)\right|_{\text {row } 8}=\frac{l}{8} \frac{\sin \theta}{1-\theta^{2} / \omega^{2}} .
For both design diagrams, the amplitudes of the kinematical and forced excitations are equal to unity.
The initial parameters method in form (17.39) allows us to create a library of design diagrams for weightless homogeneous beams with a lumped mass; these beams are subjected to different harmonic excitations, such as kinematical and force excitations. This set of different design diagrams of separate beams together with the displacement method is very effective for dynamic analysis of steady-state vibration of various rod systems, especially for continuous beams, and frames.
\begin{aligned}\varphi(x) & =\varphi_{0}+\frac{1}{E I}\left[M_{0} x+Q_{0} \frac{x^{2}}{2}+P \frac{\left(x-a_{P}\right)^{2}}{2}+R \frac{\left(x-a_{R}\right)^{2}}{2}+M\left(x-a_{M}\right)\right]+\frac{\theta^{2}}{E I}\left[m y_{m} \frac{\left(x-a_{m}\right)^{2}}{2}+J \varphi_{m}\left(x-a_{m}\right)\right] \\M(x) & =M_{0}+Q_{0} x+\left[M+P\left(x-a_{P}\right)+R\left(x-a_{R}\right)\right]+\theta^{2}\left[m y_{m}\left(x-a_{m}\right)+J \varphi_{m}\right] \\Q(x) & =Q_{0}+(P+R)+\theta^{2} m y_{m}\end{aligned} ( 17.39)
| Table A.9 Elastic curves of one span uniform beams caused by the unit angular and linear displacements (\varphi=1 ; \quad \Delta=1) | |||
| # |  |
y\left(x=\frac{l}{2}\right) | Expression for elastic curve |
| 1 |  |
\frac{1}{8} l | f_1=x-2 \frac{x^2}{l}+\frac{x^3}{l^2} |
| 2 |  |
\frac{-1}{8} l | f_2=-\frac{x^2}{l}+\frac{x^3}{l^2} |
| 3 |  |
-\frac{1}{2} | f_3=-\left(1-3 \frac{x^2}{l^2}+2 \frac{x^3}{l^3}\right) |
| 4 |  |
\frac{1}{2} | f_4=3 \frac{x^2}{l^2}-2 \frac{x^3}{l^3} |
| 5 |  |
\frac{3}{16} l | f_5=x-3 \frac{x^2}{l}+\frac{x^3}{2 l^2} |
| 6 |  |
\frac{3}{16} l | f_6=-\left(1-3 \frac{x^2}{l^2}+\frac{x^3}{2 l^3}\right) |
| 7 |  |
\frac{5}{16} | f_7=3 \frac{x^2}{2 l^2}-\frac{x^3}{2 l^3} |
| 8 |  |
-\frac{1}{2} | f_8(x)=-\left(1-\frac{x}{l}\right) |
| 9 |  |
\frac{1}{2} | f_9(x)=\frac{x}{l} |
| 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) | ||||
 |
Bending moment M(0)=M_0 | Shear force Q(0)=Q_0 | Functions | Special case a=b=l / 2 |
 |
\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} |
 |
-\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} |
 |
-\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} |
 |
\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} |
 |
-\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} |
 |
-\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} |
 |
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} |
 |
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)} . | ||||
