Question 17.14: Design diagram of the uniform continuous massless two-span b......
Design diagram of the uniform continuous massless two-span beam A-1-B with equal spans of length l and two equal lumped masses m_{1}=m_{2}=m is shown in Fig. 17.27a. Bending stiffness of each span is EI. This structure is subjected to disturbing harmonic force P(t)=P \sin \theta t. Provide total dynamic analysis of the steady-state vibration.
This structure has two degrees of freedom, while its kinematical indeterminacy is the first degree. The primary unknown is the angular displacement Z_{1} at support 1 . The primary system are obtained from a given structure by introducing constraint 1 at middle support 1 (Fig. 17.27b); this constraint prevents angular displacement at support 1. Canonical equation of displacement method is
r_{11} Z_{1}+R_{1 P}=0 (a)
The bending moment diagram in primary system due to angular harmonic rotation Z(t)=1 \cdot \sin \theta t of introduced constraint 1 is shown in Fig. 17.27c. The bending moment diagram in primary system due to given force P(t)=P \sin \theta t is shown in Fig. 17.27d. Notation M_{1 A} and Q_{1 A} means bending moment and shear infinitely close left to the point 1 (i.e., this point belongs to the element 1 A ), while M_{1 B} and Q_{1 B} infinitely close right to the point 1 (element 1 B ); (here superscript ” 0 ” for bending moments and shear forces in primary system is omitted).
Table 17.4 contains expressions for bending moments, shear forces, and inertial forces at the characteristic points of the primary system caused by harmonic rotation of introduced constraint 1 with unit amplitude Z(t)=1 \cdot \sin \theta t. This table also contains the bending moments at points A, 1-A, and 1-B caused by the given load P(t)=P \cdot \sin \theta t.
All expressions are composed on the basis of Table A.27, and in some cases, in combination with the reciprocity reaction theorem. Numerical values are calculated under the following assumption: the rods are uniform, detuning \theta / \omega=0.5 ( \omega is a frequency of free vibration of the entire structure), and each mass is located in the middle of the span.
Unit reaction (Fig. 17.27e) and loaded term (Fig. 17.27f) are
\begin{aligned}r_{11} & =M_{1 A}^{0}+M_{1 B}^{0}=3 \frac{E I}{l}+1.7142 \frac{E I}{l}=4.7142 \frac{E I}{l}, \\R_{1 P} & =-M_{1 B-P}^{0}=-0.25 P l\end{aligned} (b)
where M_{1 B-P}^{0} means bending moment at section 1, which belongs to span 1 B, caused by load P in the primary system. Finally, the primary unknown becomes
Z_{1}=-\frac{R_{1 P}}{r_{11}}=\frac{0.25 P l}{4.7142 E I / l}=0.05303 \frac{P l^{2}}{E I}(\mathrm{rad}) (c)
Amplitude Values of the Inertia Forces
For computation of the dynamic characteristics of the system, it is necessary to take into account the inertial forces that arise during vibration of the system. Their calculation for each span is presented in Table 17.5. General expression for inertia forces is J=m \theta^{2} y. For each element of the primary system, it is easy to write the expression for the frequency of free vibrations. This expression allows representing the mass m through the square of the frequency of free vibration \omega^{2}. The substitution m=m \left(\omega^{2}\right) in the expression for J leads to the expression which will contain the square of the detuning \theta^{2} / \omega^{2}. The frequency of free vibration of each span is determined by the formula \omega=\sqrt{1 / m \delta_{11}}, where \delta_{11} is the unit displacement. The parameter y is a dynamic displacement of a mass.
Computation of final dynamic bending moments and shear forces at specified points are presented in Tables 17.6 and 17.7, respectively
Final amplitude bending moment and shear diagrams, reactions R and inertial forces J are shown in Fig. 17.27g, h.
Control For total verification of obtained results we will use the different approaches, i.e., considering each span separately and the structure in a whole:
Left span A-1. Shear forces within portion A-m and 1-m are
Q_{A m}=\frac{d M}{d x}=\frac{-0.106-0.159}{l / 2} P l=-0.530 P, \quad Q_{1 m}=\frac{-0.159-(-0.106)}{l / 2} P l=-0.106 P
Inertial force of mass m_{1} is
J_{1}=Q_{\text {right }}-Q_{\text {left }}=Q_{1 m}-Q_{A m}=-0.106 P-(-0.530 P)=0.424 P(\uparrow)
Right span 1-B. Shear forces within portion 1- m and m-B are
Q_{1 m}=\frac{0.159-(-0.2538)}{l / 2} P l=0.8256 P, \quad Q_{B m}=-0.5076 P
Total force at mass m_{2} includes the given and inertial forces; they are P and J
P+J_{2}=Q_{B m}-Q_{1 m}=-0.5076 P-0.8256 P=-1.3332 P, \quad J_{2}=-0.3332 P(\downarrow)
Amplitude values of reactions for entire structure and the equilibrium condition for entire structure
\begin{aligned}& R_A=Q_A=-0.5303 P(\downarrow) ; \\& R_1=Q_1^{\text {right }}-Q_1^{\text {left }}=0.8258-(-0.106)=0.9318 P(\uparrow), \\& R_B=-Q_B=0.5076 P(\uparrow)\end{aligned}
\sum Y=-0.5303 P+0.4242 P+0.9318 P-P-0.3332+0.5076 P=-1.8635 P+1.8634 P
Bending moment at support 1, using the left and right forces, respectively
M_1^{l e f t}=M_A-R_A l-J_1 \frac{l}{2}=\left(0.159-0.530 \cdot 1+0.424 \frac{1}{2}\right) P l=(0.371-0.530) P l=-0.159 P l
M_1^{r i g h t}=R_B l-P \frac{l}{2}-J_2 \frac{l}{2}=\left(0.5076-1.3332 \frac{1}{2}\right) P l=-0.159 P l
Assume the beam is loaded by the static force P at the middle of the span 1- B. In this case, the bending moment at support 1 is M_{1}\left(P_{\text {stat }}\right)=-\frac{2}{7} P l v\left(1-v^{2}\right). If v=0.5, then M_{1}\left(P_{\text {stat }}\right)=-\frac{3}{28} P l=0.107 P l, M_{A}\left(P_{\text {stat }}\right)=-\frac{3}{54} P l (see problem 10.4). Dynamic coefficient becomes \mu=\frac{0.159 P l}{0.107 P l}=1.49.
Comments
In this example, the relative frequency of excitation, i.e., a certain detuning \theta / \omega, had been accepted, while the frequency of free vibration \omega and the frequency of excitation \theta separately are not computed. Let us show procedure for computation of absolute value of frequency excitation \theta.
In order to form the frequency equation of free vibration we need to put R_{1 P}=0 in Eq. (a). Since Z_{1} \neq 0, then the frequency equation is r_{11}=0. According to Fig. 17.27e, the frequency equation is r_{11}=M_{1 A}+M_{1 B}=0. According to Table 17.4 (the first row for both spans), the frequency equation in extended form after simplification becomes
\underbrace{4\left[1-\frac{7}{4} \frac{\theta^{2}}{\omega_{f-f}^{2}}\right]}_{\text {member } A 1}+\underbrace{3\left[1-\frac{16}{7} \frac{\theta^{2}}{\omega_{f-p}^{2}}\right]}_{\text {member } 1 B}=0 (c)
The peculiarity of the equation is that it contains frequencies \omega_{f-f}, \omega_{f-p} related to elements of various types, such as fixedfixed (f-f) and fixed-pinned (f-p). For these elements \omega_{f-f}^{2}=\frac{192 E I}{m l^{3}} and \omega_{f-p}^{2}=\frac{768 E I}{7 m l^{3}}. As the base element we take fixed-fixed beam; thus, \frac{\omega_{f-p}^{2}}{\omega_{f-f}^{2}}=\frac{4}{7} \rightarrow \omega_{f-p}^{2}=\frac{4}{7} \omega_{f-f}^{2}.
Substituting this ratio into (c) and replacing \theta=\omega_{1}, we get
\underbrace{4\left[1-\frac{7}{4} \frac{\omega_{1}^{2}}{\omega_{f-f}^{2}}\right]}_{\text {member }}+\underbrace{3\left[1-\frac{16}{7} \frac{\omega_{1}^{2}}{\frac{4}{7} \omega_{f-f}^{2}}\right]}_{\text {member } 1 B}=0
This equation leads to the following equation for \omega_{1}^{2} in terms of frequency \omega_{f-f}^{2} of fixed-fixed beam
7-19 \frac{\omega_{1}^{2}}{\omega_{f-f}^{2}}=0,
so the required square of frequency of free vibration of original structure becomes
\omega_{1}^{2}=\frac{7}{19} \omega_{f-f}^{2}=\frac{7}{19} \frac{192 E I}{m l^{3}}=70.7368 \frac{E I}{m l^{3}}
Verification. As the base element we take fixed-pinned member. In this case \omega_{f-f}^{2}=\frac{7}{4} \omega_{f-p}^{2}. Similar procedure leads to the following equation for \omega_{1}^{2} in terms of frequency \omega_{f-p}^{2} of fixed-pinned beam, and required square of frequency
7-\frac{76}{7} \frac{\omega_{1}^{2}}{\omega_{f-p}^{2}}=0 \rightarrow \omega_{1}^{2}=\frac{49}{76} \omega_{f-p}^{2}=\frac{49}{76} \frac{768 E I}{7 m l^{3}}=70.7368 \frac{E I}{\mathrm{ml}^{3}}
Since \theta / \omega=0.5, then the absolute value of the frequency excitation becomes
\theta=0.5 \times 8.4105 \sqrt{\frac{E I}{m l^{3}}}=4.205 \sqrt{\frac{E I}{m l^{3}}}
| Table 17.4 Bending moments and shear at characteristic points of the beam; \theta / \omega=0.5 | |
| \text { Left } \operatorname{span} A-1 | \text { Right span } 1-B |
| \begin{aligned} & M_{1 A}=\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}=3 \frac{E I}{l} \\\\ & M_A=\frac{2 E I}{l} F_2=\frac{2 E I}{l}\left[1+\frac{\theta^2}{\omega^2} \frac{1}{2}\right] \frac{1}{1-\theta^2 / \omega^2}=\frac{3 E I}{l} \\\\ & Q_A=-\frac{6 E I}{l^2} F_6=-\frac{6 E I}{l^2}\left[1+\frac{\theta^2}{\omega^2}\right] \frac{1}{1-\theta^2 / \omega^2}=-\frac{10 E I}{l^2} \\\\ & Q_{1 A}=-\frac{6 E I}{l^2} F_5=\frac{6 E I}{l^2}\left[1-3 \frac{\theta^2}{\omega^2}\right] \frac{1}{1-\theta^2 / \omega^2}=-\frac{2 E I}{l^2} \end{aligned}
(reciprocal reactions theorem, Table A.27, rows 1 and 2 ) R_A=Q_A=-\frac{10 E I}{l^2}(\downarrow), \quad R_{1 A}=Q_A=-\frac{2 E I}{l^2}(\uparrow), |
\begin{aligned}& M_{1 B}=\frac{3 E I}{l} F_9=\frac{3 E I}{l}\left[1-\frac{16}{7} \frac{\theta^2}{\omega^2}\right] \frac{1}{1-\theta^2 / \omega^2}=1.7142 \frac{E I}{l} \\\\& Q_{1 B}=-\frac{3 E I}{l^2} F_{12}=-\frac{3 E I}{l^2}\left[1-\frac{40}{7} \frac{\theta^2}{\omega^2}\right] \frac{1}{1-\theta^2 / \omega^2}=1.7142 \frac{E I}{l^2} \\\\& Q_B=-\frac{3 E I}{l^2} F_{11}=-\frac{3 E I}{l^2}\left[1+\frac{8}{7} \frac{\theta^2}{\omega^2}\right] \frac{1}{1-\theta^2 / \omega^2}=-5.1428 \frac{E I}{l^2}\end{aligned}
(reciprocal reactions theorem, Table A27, rows 5 and 7 ) \begin{aligned}& R_{1 B}=1.7143 \frac{E I}{l^2}(\uparrow), \quad R_B=5.1428 \frac{E I}{l^2}(\uparrow) \\\\& \text { Control: } \\\\& J_{m 2}=Q_B-Q_{1 B}=-5.1428 \frac{E I}{l^2}-\frac{1.7143 E I}{l^2}=-6.8571 \frac{E I}{l^2}(\downarrow) \\\\& M_{1 B-P}^0=P F_{17}=-P \frac{3 l}{16} \cdot \frac{1}{1-\theta^2 / \omega^2}=-0.25 P l\end{aligned} |
| M_{m 1}=M_A-Q_A \frac{l}{2}=3.0 \frac{E I}{l}-10.0 \frac{E I}{l^2} \cdot \frac{l}{2}=-2.0 \frac{E I}{l} | M_{m 2}=M_{1 B}+Q_{1 B} \frac{l}{2}=1.7142 \frac{E I}{l}+1.7142 \frac{E I}{l^2} \cdot \frac{l}{2}=2.5714 \frac{E I}{l} |
| Table 17.6 Amplitudes of bending moments at the specified points due to harmonic disturbing force P(t)=P \sin \theta t \left(Z=0.05303 P l^2 / E I \quad r a d\right) | |||||
| span | Point | M_{Z=1} | M_{Z=1} Z | M_P{ }^0 | M=M_{Z=1} Z+M_P^0 |
| Left A-1 | \begin{gathered}A \\m \\1^{\text {left }}\end{gathered} | 3.0\\ -2.0\\ -3.0 | \begin{gathered}0.1591 \\-0.1061 \\-0.1591\end{gathered} | \begin{aligned}& 0 \\& 0 \\& 0\end{aligned} | \begin{gathered}0.1591 \\-0.1061 \\-0.1591\end{gathered} |
| Right 1-B | 1^{\text {right }}\\ m | \begin{aligned}& 1.7142 \\& 0.8571\end{aligned} | \begin{gathered}0.0909 \\0.04545\end{gathered} | -0.25 \\ 0.2083 | \begin{array}{r}-0.1591 \\0.2538\end{array} |
| Factor | EI/l | P l | P l | P l | |
| Table 17.5 Inertial forces due to unit Z_1=1 and actual primary nknown Z_1=0.05303 \mathrm{Pl}^2 / E I(\theta / \omega=0.5) | ||||
| Span of primary system | Frequency of free vibration \omega^2=1 / m \delta_{11} | Displacement y of the mass due by displacement Z_1=1 \cdot \sin \theta t | Amplitude inertial force \bar{J}=m \theta^2 y due by Z_1=1(*) | Amplitude inertial force J=m \theta^2 y due by \mathrm{Z}_1=0.05303 P l^2 / E I |
| Left (Fixed-fixed beam) | \frac{192 E I}{m^3} | \frac{1}{8} | \frac{24 E I}{l^2} \frac{\theta^2}{\omega^2} \cdot \frac{1}{1-\theta^2 / \omega^2}=8.00(\uparrow) | 0.4242 |
| Right (Fixed-pinned beam) | \frac{768 E I}{7 m l^3} | \frac{3 l}{16} | \frac{144 E I}{7 l^2} \frac{\theta^2}{\omega^2} \cdot \frac{1}{1-\theta^2 / \omega^2}=6.8571(\downarrow) | 0.3636^{(* *)} |
| Factor | 1 \cdot \sin \theta t /\left(1-\theta^2 / \omega^2\right) | E I / l^2 | P | |
| Table 17.7 Amplitudes of bending moments at the specified points due to harmonic disturbing force P(t)=P \sin \theta t \left(Z=0.05303 P l^2 / E I \quad r a d\right) | |||||
| span | Point | Q_{Z=1} | Q_{Z=1}Z | Q_(P)^0 | Q=Q_{Z=1}Z+Q_P^0 |
| Left A-1 | A \\ 1^{left} | -10.0 \\ -2.0 | \begin{gathered}-0.5303 \\-0.1061\end{gathered} | 0 \\ 0 | \begin{aligned}& -0.5303 \\& -0.1061\end{aligned} |
| Right 1-B | 1 ^{right} \\ B | \begin{aligned}& -1.7142 \\& -1.7142\end{aligned} | \begin{gathered}-0.0909 \\-0.0909\end{gathered} | \begin{gathered}0.9167 \\-0.4167\end{gathered} | \begin{gathered}0.8258 \\-0.5076\end{gathered} |
| Factor | EI/l^2 | P | P | P | |
| 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} |
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\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)} . | ||||