Question 17.17: The simply supported beam of length l, bending stiffness EI,......

Solution For dynamic analysis of this beam we will apply the initial parameter method. The origin is placed at point A of the beam. Due to symmetry, consider the equivalent left part of the beam (Fig. 17.31b). At the point on the axis of symmetry the sliding support is located. This boundary condition provides same elastic curve as the original design diagram. In the section on the axis of symmetry the angle of rotation \varphi(a)=0 and shear force Q(a)=P / 2. Therefore, from the system of equations of the initial parameters method we use two equations, namely for \varphi(x) and Q(x).

The general expressions for these functions are

\begin{aligned}& \varphi(x)=y_{0} V_{x} k+\varphi_{0} S_{x}+M_{0} T_{x} \frac{1}{E I k}+Q_{0} U_{x} \frac{1}{E I k^{2}}, \\& Q(x)=y_{0} T_{x} E I k^{3}+\varphi_{0} U_{x} E I k^{2}+M_{0} V_{x} k+Q_{0} S_{x}, \quad k=\sqrt[4]{\frac{m_{0} \theta^{2}}{E I}}\end{aligned}     (a)

Taking into account initial parameters y_{0}=0, and M_{0}=0, at section x=a we have

\begin{aligned}& \varphi(a)=\varphi_{0} S_{a}+Q_{0} U_{a} \frac{1}{E I k^{2}}=0 \\& Q(a)=\varphi_{0} U_{a} E I k^{2}+Q_{0} S_{a}=\frac{P}{2}\end{aligned}            (b)

Here and further notation S_{a}, T_{a}, etc., means S(k a), T(k a), etc.

Homogeneous set of Eq. (b) allows compiling the equation of the frequencies of free vibration

D=\left|\begin{array}{cc}S_{a} & U_{a} \frac{1}{E I k^{2}} \\U_{a} E I k^{2} &  S_{a}\end{array}\right|=0

In expanded form, the frequency equation and its solution

S_{a}^{2}-U_{a}^{2}=\cosh k a \cos k a=0 \rightarrow \cos k a=\cos k \frac{l}{2}=0 \rightarrow k \frac{l}{2}=\frac{\pi}{2} \rightarrow k=\frac{\pi}{l}

Thus, the frequency of free vibration becomes \omega=k^{2} \sqrt{\frac{E I}{m_{0}}}=\frac{\pi^{2}}{l^{2}} \sqrt{\frac{E I}{m_{0}}}

Solution of nonhomogeneous set of Eq. (b) is

\begin{aligned}\varphi(0) & =-\frac{P}{2 E I k^{2}} \frac{U_{a}}{S_{a}^{2}-U_{a}^{2}} \\Q(0) & =\frac{P}{2} \frac{S_{a}}{S_{a}^{2}-U_{a}^{2}} \end{aligned}          (c)

The bending moment and displacement at section under harmonic force are

\begin{aligned}& M(a)=\varphi_{0} V_{a} k E I+Q_{0} T_{a} \frac{1}{k}=\frac{P}{2 k} \frac{S_{a} T_{a}-U_{a} V_{a}}{S_{a}^{2}-U_{a}^{2}}, \\& y(a)=\varphi_{0} T_{a} \frac{1}{k}+Q_{0} V_{a} \frac{1}{E I k^{3}}=\frac{P}{2 k^{3} E I} \frac{S_{a} V_{a}-T_{a} U_{a}}{S_{a}^{2}-U_{a}^{2}} \end{aligned}           (d)

Formulas (c, d) present the amplitude values of corresponding response functions. Dynamic response involves function \sin \theta t. According to (16.26), formulas (c, d) can be easily presented in terms of hyperbolic-trigonometric functions.

\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}        (16.26)

Dynamic coefficient for any parameter is determined as ratio of its amplitude dynamic value and static one; for example, dynamic coefficient for displacement at y(a=l / 2) becomes

\mu_{d y n}^{y(a)}=\frac{y(a)}{y(a)_{\text {stat }}}=\frac{y(a)}{P l^{3} / 48 E I}=\frac{24}{k^{3} l^{3}} \frac{S_{a} V_{a}-T_{a} U_{a}}{S_{a}^{2}-U_{a}^{2}} .           (e)

Since the amplitude dynamic expression and static one for any function (displacement, reaction, etc.) linearly depend on P, the dynamic coefficient does not depend on P. Obviously, expression (e) is approximate, since the static displacement is defined taking into account only the force P. Static displacement taking into account P and uniformly distributed load q_{0}=m_{0} g\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right) should be taken in form y(a)_{\text {stat }}=P l^{3} / 48 E I+5 q_{0} 4^{4} / 384 E I.

Let us find out the effect of excitation frequency \theta on dynamic coefficients. Two versions of computation may be recommended.

Version 1. Calculate the parameter k in terms of the absolute frequency of the excitation \theta

k=\sqrt[4]{\frac{q_{0} \theta^{2}}{g E I}}=\sqrt[4]{\frac{q_{0}}{g E I}} \sqrt{\theta}

Let q_{0}=263 \mathrm{~N} / \mathrm{m}, \quad E=21.5 \cdot 10^{10} \mathrm{~N} / \mathrm{m}^{2}, \quad I=2.14 \cdot 10^{-5} \mathrm{~m}^{4}, l=4.75 \mathrm{~m}. In this case parameter k=k(\theta) is

 k=\sqrt[4]{\frac{263}{9.81 \times 21.5 \times 10^{10} \times 2.14 \times 10^{-5}}} \sqrt{\theta}=0.04913 \sqrt{\theta}\left(\mathrm{m}^{-1}\right) .

If \theta=144 \mathrm{~s}^{-1}, then parameter k becomes k=0.04913 \sqrt{144}=0.5896\left(\mathrm{~m}^{-1}\right). Argument of Krylov functions k a=k l / 2=0.5896 \times 4.75 / 2=1.4. Krylov-Duncan functions of this argument are

S_{a}=1.16043, \quad T_{a}=1.44487, \quad U_{a}=0.99046, \quad V_{a}=0.45942

For dynamic coefficient we get the following value:

\begin{aligned}& \mu_{\mathrm{dyn}}^{y(a)}=\frac{24}{k^{3} l^{3}} \frac{S_{a} V_{a}-T_{a} U_{a}}{S_{a}^{2}-U_{a}^{2}}= \\& \frac{24}{0.5896^{3} \times 4.75^{3}} \frac{1.16043 \times 0.45942-1.44487 \times 0.99046}{0.36558}=2.685\end{aligned}

Dynamic coefficients for Q(0) and M(0) are 3.174 and 2.386, respectively. Note that the dynamic coefficients related to the various characteristics of the system state (reaction, bending moments, etc.) are different.

Version 2. Calculate the parameter k in terms of the relative frequency of the excitation i.e., with respect to ratio \theta / \omega.

The frequency of free vibration of entire beam is \omega=\frac{\pi^{2}}{l^{2}} \sqrt{\frac{E I}{m_{0}}} \rightarrow m_{0}=\frac{\pi^{4}}{l^{4}} \frac{E I}{\omega^{2}}, so parameters k and k a may be presented as follows:

k=\sqrt[4]{\frac{m_{0} \theta^{2}}{E I}}=\frac{\pi}{l} \sqrt[4]{\frac{\theta^{2}}{\omega^{2}}}=\frac{\pi}{l} \sqrt{\frac{\theta}{\omega}} \rightarrow k a=k \frac{l}{2}=\frac{\pi}{2} \sqrt{\frac{\theta}{\omega}}

Finally the relative excitation frequency (detuning \theta / \omega ) and the factor in the expression (e) may be presented in the form

\frac{\theta}{\omega}=\left(\frac{2 k a}{\pi}\right)^{2}, \quad \frac{24}{k^{3} l^{3}}=\frac{24}{k^{3}(2 a)^{3}}=\frac{3}{(k a)^{3}} 

The dynamic coefficient computation for displacement in the middle point of the beam \mu_{\mathrm{dyn}}^{y(a)} as a function of the k a parameter are presented in Table 17.10. In this table, the second and third column contains the relative excitation frequency \theta / \omega and coefficient 3 /(\mathrm{ka})^{3}, respectively. Columns with values of Krylov-Duncan function omitted. Detailed tables of Krylov-Duncan functions may be found in (Karnovsky and Lebed 2001, 2004a, b).

Dynamic coefficients for all types of response, in particular, formula (e) for displacement in the middle of the beam, contains expression S_{a}^{2}-U_{a}^{2} in the denominator. The equation S_{a}^{2}-U_{a}^{2}=S^{2}(k a)-U^{2}(k a)=0, according to tables of Krylov functions, is satisfied for the following parameters k a: \pi / 2,3 \pi / 2, \ldots Graph of dynamic coefficient is shown in Fig. 17.32.

The dynamic coefficient is plotted as a function of dimensionless parameter k a. To each value of k a there is corresponding dimensionless detuning \theta / \omega. The axis of parameters k a is divided into intervals 0-\pi / 2-3 \pi / 2-. The boundaries of these intervals is the detuning \theta / \omega at which resonance occurs; they are k a=\pi / 2,3 \pi / 2. The dimensionless interval k a=[0-\pi / 2] corresponds to the first frequency interval \theta / \omega_1=[0.0-1.0]; parameter k a=1.4 in the example above corresponds to the first frequency interval (row 2 of Table 17.10).

When the detuning is equal to zero, the dynamic coefficient is equal to one (static case). With increasing detuning \theta / \omega, the dynamic coefficient increases, and if the frequency excitation \theta and free vibration \omega coincide, the dynamic coefficient becomes infinity. Parameter k a=\pi / 2, in which \theta=\omega correspond to resonance. On the frequency interval \theta / \omega_1=[0-1.0], the symmetric vibration with one half wave is observed (curve 1 is shown by the dotted line).

Passing through point \pi / 2 a change in the nature of the vibration occurs, which in the interval [\pi / 2-3 \pi / 2] becomes antisymmetric with two half waves with one nodal point (curve 2). When the detuning is equal to \pi, an antiresonance phenomenon is observed, which consists in a noticeable weakening of the dynamic coefficient. If detuning \theta / \omega passing through parameter k a=3 \pi / 2, a resonance phenomenon and a transition to a symmetrical vibration with two nodal points (curve 3 ) are observed.

Comments

1. Original equation (17.43) relates to the Bernoulli-Euler’s classical theory of transversal vibration of thin rods. It is assumed that the height of such rods is small compared to its length, h<<l. Passing from symmetric form of vibration with one halfwave to antisymmetric form with two half-waves, the original “height-length” ratio of the beam is violated. It happens because a nodal point appears in the middle of the beam (curve 2), i.e., the length of each half-wave of the beam is halved. Thus, if for the initial beam the ratio h / l was taken at the upper permissible boundary [h / l], then when passing to the antisymmetric form, the value h \div(l / 2) exceeds the maximum permissible [h / l]. This means that when analyzing higherorder vibrations, modified bending theories should be applied.

Various theories of transverse vibration of beams (Rayleigh, Bresse, Volterra, etc.), their assumptions, features, and dispersion curves are discussed in the books (Karnovsky and Lebed 2001, 2004b)

2.  Table 17.11 contains the amplitudes of the dynamic reactions of typical homogeneous beams loaded with harmonic concentrated force; the position u l of the force is arbitrary. The formulas are obtained by the initial parameters method.

Obviously, these expressions can be considered as equations of the dynamic influence lines of the bending moment and the shear force in the section x=0. Each ordinate of the corresponding function represents the amplitude of the dynamic reaction \left(M_0, Q_0\right) caused by the harmonic force of the unit amplitude, which is located above of this ordinate. It is easy to verify that if the force is located at the left support (u=0), then M_0=0, Q_0=P \sin \theta t.