Question 17.20: Dynamic analysis of the frame shown in Fig. 17.34a, had been......

The primary system of displacement method is shown in Fig. 17.35b. Configuration of bending moment diagrams due to unit displacement of introduced constraints will be as in Fig. 17.34c,d; however the values of bending moment and reactions are taken from Table A.29 and shown in Fig. 17.35. For element 1-A, 2-B the parameter i=E I / l, while for element 1-2 i_{12}=4 E I / l=4 i; parameter k=m_{0} \omega^{2} l^{3}, where \omega is a required frequency of free vibration.

Unit reactions are

\begin{aligned}& r_{11}=M_{1 A}^{(1)}+M_{12}^{(1)}=\underbrace{\left(4 i-\frac{k}{105}\right)}_{\text {case } 2}+\underbrace{\left(3 i_{1-2}-\frac{2}{105} k\right)}_{\text {case } 4}=4 i-\frac{k}{105}+3 \times 4 i-\frac{2}{105} k=16 i-\frac{3}{105} k ; \\& r_{12}=-M_{1 A}^{(2)}=\underbrace{-\left(\frac{6 i}{l}-\frac{11}{210} \frac{k}{l}\right)}_{\text {case } 1} ; \quad r_{21}=-Q_{1 A}^{(1)}=-\underbrace{\left(\frac{6 i}{l}-\frac{11}{210} \frac{k}{l}\right)}_{\text {case } 2}\end{aligned}            (a)

Parameter k=m l^{3} \omega^{2}, \omega^{2}=k / m l^{3}, so inertial force of crossbar is J_{1-2}=m l \omega^{2}=\frac{k}{l^{2}}, and unit reaction r_{22} becomes

r_{22}=Q_{1 A}^{(2)}+Q_{2 B}^{(2)}-J_{1-2}=\underbrace{\left(\frac{12 i}{l^{2}}-\frac{13}{35} \frac{k}{l^{2}}\right)}_{\text {case } 1}+\underbrace{\left(\frac{3 i}{l^{2}}-\frac{33}{140} \frac{k}{l^{2}}\right)}_{\text {case } 5}-\frac{k}{l^{2}}=\frac{15 i}{l^{2}}-1.607 \frac{k}{l^{2}}          (b)

The frequency equation is r_{11} r_{22}-r_{12}^{2}=0. In our case we have

 \left(16 i-\frac{3}{105} k\right)\left(\frac{15 i}{l^{2}}-1.607 \frac{k}{l^{2}}\right)-\left[-\left(\frac{6 i}{l}-\frac{11}{210} \frac{k}{l}\right)\right]^{2}=0      (c)

After elementary simplification this equation may be presented in the form

k^{2}-524.506 \frac{E I}{l} k+4194.079\left(\frac{E I}{l}\right)^{2}=0 

The smallest root of this equation is k_{1}=8.122 E I / l. Corresponding approximate frequency of free vibration becomes

\omega=\sqrt{\frac{k}{m_{0} l^{3}}}=\sqrt{\frac{8.122}{m_{0} l^{3}} \frac{E I}{l}}=\frac{2.85}{l^{2}} \sqrt{\frac{E I}{m_{0}}}      (d)

The exact value equals \omega=\frac{2.82}{l^{2}} \sqrt{\frac{E I}{m_{0}}} (Darkov 1989).

It can be seen that in the problems of determining the frequencies of free vibration, the Bolotin method leads to significant simplifications of computational procedures and, at the same time, to a small error in comparison with accurate analysis. This may be explained by the fact that the character of a static elastic curve of a structure is similar to a dynamic elastic line in case of free vibration. It should be expected that in case of the forced vibration the errors will be more noticeable. Let us consider this question in more detail.

Suppose that the frequency of steady-state vibration

\theta=0.8 \omega=0.8 \frac{2.85}{l^{2}} \sqrt{\frac{E I}{m_{0}}}=\frac{2.256}{l^{2}} \sqrt{\frac{E I}{m_{0}}}

Corresponding parameter k^{*} is

k^{*}=m_{0}(0.8 \omega)^{2} l^{3}=0.64 m_{0} \omega^{2} l^{3}=0.64 \times 8.122 i=5.198 i        (e)

Free terms of canonical equation in case of approximate method of solution remain the same. Therefore, the canonical equations of the displacement method that correspond to the steady-state vibration of a system with an excitation frequency \theta=0.8 \omega take the form:

\begin{aligned}& \left(16 i-\frac{3}{105} k^{*}\right) Z_{1}-\left(\frac{6 i}{l}-\frac{11}{210} \frac{k^{*}}{l}\right) Z_{2}=0 \\& -\left(\frac{6 i}{l}-\frac{11}{210} \frac{k^{*}}{l}\right) Z_{1}+\left(\frac{15 i}{l^{2}}-1.607 \frac{k^{*}}{l^{2}}\right) Z_{2}=P \end{aligned}     (f)

Solution of these equation is

Z_{1}=0.0789 \frac{P l^{2}}{E I} \quad\left(0.0759 \frac{P l^{2}}{E I}\right), \quad Z_{2}=-0.218 \frac{P l^{3}}{E I} \quad\left(-0.210 \frac{P l^{3}}{E I}\right)       (g)

In brackets are the exact values of the primary unknowns. The error is 3.95 \% and 3.81 \%, respectively. These results can be used to evaluate the ordinates of the bending moment diagram. For example, the bending moment at point 1 of the crossbar equals

M_{12}^{(1)}=\left(3 i_{1-2}-\frac{2}{105} k^{*}\right) Z_{1}=\left(3 \times 4 i-\frac{2}{105} \times 5.198 i\right) 0.0789 \frac{P l^{2}}{E I}=0.939 P l    (h)

The precise value of the bending moment is 0.903 \mathrm{Pl}, error is 3.99 \%. In the presence of a harmonic force applied in the span of the beam (frame), the magnitude of the difference between the approximate and exact solutions will depend significantly on the position of the force.

Summary

Determination of the frequencies of free vibration by the exact method has considerable computational difficulties, since it is necessary to determine the roots of the transcendental equation. It is much easier to perform this step using the Bolotin’s functions. Calculation of the primary unknowns Z is not difficult, since conversion of the coefficients of the canonical equations turns out to be elementary.