Consider the system shown in Figure 9.8, where the clamped–free slender bar of a parabolic side is assumed to oscillate longitudinally. Our task is to extract the eigenvalues and the corresponding mode shapes using the Rayleigh–Ritz method discussed above.

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Accepted Answer

The mass per unit length and the stiffness along x are

[latex]m(x)=\frac{R ho}{\ell^2}\left(\ell^2-2 \ell x+x^2 ight)[/latex] (9.148)

[latex]E A(x)=\frac{E R}{\ell^2}\left(\ell^2-2 \ell x+x^2 ight)[/latex] (9.149)

To obtain the approximate first two natural frequencies and mode shapes, we use the displacement as a series of two terms.

[latex]y(x)=\sum_{i=1}^2 c_i \phi_i(x)=c_1 \phi_1(x)+c_2 \phi_2(x)[/latex] (9.150)

Let us assume that [latex]\phi_1[/latex] and [latex]\phi_2[/latex] can be approximated by sine functions; then we write

[latex]y(x)=c_1 \sin \frac{\pi x}{2 L}+c_2 \sin \frac{3 \pi x}{2 L}[/latex] (9.151)

The mode shapes will be known once the constants [latex]c_1[/latex] and [latex]c_2[/latex] are evaluated. The maximum potential energy is given by

[latex]U=\frac{1}{2} \int_0^{\ell} E A(x)\left(\frac{d y}{d x} ight)^2 d x[/latex] (9.152)

or

[latex]U=\frac{1}{2} \int_0^1 \frac{E R}{\ell^2}\left(\ell^2-2 \ell x+x^2 ight)\left(\frac{d y(x)}{d x} ight)^2 d x[/latex] (9.153)

After integration, we obtain

[latex]U=\frac{E R}{2 \ell}\left(\frac{\pi^2}{24} c_1^2+\frac{3 \pi^2}{8} c_2^2+\frac{1}{4} c_1^2+\frac{15}{8} c_1 c_2+\frac{1}{4} c_2^2 ight)[/latex] (9.154)

This final expression of U is similar to

[latex]U=\frac{1}{2} \sum_{i=1}^2 \sum_{i=1}^2 k_{i j} c_i c_j[/latex] (9.155)

which when expanded yields

[latex]U=\frac{1}{2}\left(k_{11} c_1^2+k_{12} c_1 c_2+k_{21} c_2 c_1+k_{22} c_2^2 ight)[/latex] (9.156)

Similarly, the maximum kinetic energy can be expressed as

[latex]T=\frac{\omega^2}{2} \int_0^{\ell} m(x) y^2 d x[/latex] (9.157)

Substituting for m(x) and y(x) from equations (9.148) and (9.151), we get

[latex]T=\frac{\omega^2}{2} \int_0^{\ell} \frac{R ho}{\ell^2}\left(\ell^2-2 \ell x+x^2 ight) y^2 d x[/latex] (9.158)

The above results in

[latex]T=\frac{\omega^2 R \ell ho}{2}\left(\frac{1}{6} c_1^2+\frac{1}{6} c_2^2-\frac{1}{\pi^2} c_1^2+\frac{3}{2 \pi^2} c_1 c_2-\frac{1}{9 \pi^2} c_2^2 ight)[/latex] (9.159)

which is similar to

[latex]T=\frac{\omega^2}{2} \sum_{i=1}^2 \sum_{j=1}^2 m_{i j} c_i c_j[/latex] (9.160)

Written in expanded form, we have

[latex]T=\frac{\omega^2}{2}\left(m_{11} c_1^2+m_{12} c_1 c_2+m_{21} c_2 c_1+m_{22} c_2^2 ight)[/latex] (9.161)

where ω² = λ .

Now, from the potential energy expression, the stiffness matrix can be obtained as

[latex]K=\left[\begin{array}{cc}\frac{E R}{\ell}\left(\frac{1}{4}+\frac{\pi^2}{24} ight) & \frac{15 E R}{16 \ell} \\frac{15 E R}{16 \ell} & \frac{E R}{\ell}\left(\frac{1}{4}+\frac{3 \pi^2}{8} ight)\end{array} ight] (9.162)[/latex]

and the mass matrix is found to be

[latex]M=\left[\begin{array}{cc}R \ell ho\left(\frac{1}{6}-\frac{1}{\pi^2} ight) & \frac{3 R \ell ho}{4 \pi^2} \\frac{3 R \ell ho}{4 \pi^2} & R \ell ho\left(\frac{1}{6}-\frac{1}{9 \pi^2} ight)\end{array} ight] (9.163)[/latex]

Using equation (9.146), we obtain a system of two linear equations, which can be used to solve for the eigenvalues and the constants [latex]c_1[/latex] and [latex]c_2[/latex] :

[latex]\sum_{j=1}^n\left(k_{r j}-\lambda m_{r j} ight) c_j=0[/latex] (9.146)

[K − λM]C = 0 (9.164)

where

[latex]C=\left[\begin{array}{ll}c_1 & c_2\end{array} ight]^T[/latex] (9.165)

Substituting M and K into (9.164), we obtain

[latex]\left[\begin{array}{cc}0.66123 \frac{E R}{\ell}-0.06534 \lambda R \ell ho & 0.9375 \frac{E R}{\ell}-0.0760 \lambda R \ell ho \0.9375 \frac{E R}{\ell}-0.0760 \lambda R \ell ho & 3.9511 \frac{E R}{\ell}-0.155409 \lambda R \ell ho\end{array} ight]\left[\begin{array}{l}c_1 \c_2\end{array} ight]=0[/latex] (9.166)

If we replace [latex]\lambda \ell^2 ho / E[/latex] by α, equation (9.166) reduces to

[latex]\left[\begin{array}{rr}0.66123-0.065364 \alpha & 0.9375-0.076 \alpha \0.9375-0.076 \alpha & 3.9511-0.1554 \alpha\end{array} ight]\left[\begin{array}{l}c_1 \c_2\end{array} ight]=0 (9.167)[/latex]

Setting the determinant of the system (9.167) to zero, we obtain the following characteristic equation:

1.73369 − 0.218465α + 0.0043806α² = 0 (9.168)

The roots are found to be

[latex]\alpha_1=9.90179[/latex] (9.169)

and

[latex]\alpha_2=39.9688[/latex] (9.170)

or simply

[latex]\lambda_1=9.90179 \frac{E}{ ho \ell}=\omega_1^2[/latex] (9.171)

and

[latex]\lambda_2=39.9688 \frac{E}{ ho \ell^2}=\omega_2^2[/latex] (9.172)

Thus, the first two natural frequencies and mode shapes are

[latex]\omega_1=3.1467 \sqrt{\frac{E}{ ho \ell^2}}[/latex] (9.173)

Substituting α from equation (9.169) into equation (9.167), and solving for [latex]c_1[/latex] and [latex]c_2[/latex], we obtain

[latex]y_1(x)=\sin \frac{\pi}{2 L} x-0.0707143 \sin \frac{3 \pi}{2 L} x[/latex] (9.174)

where [latex]y_1(x)[/latex] is as shown in Figure 9.9. Similarly, we find that

[latex]\omega_2=6.32208 \sqrt{\frac{E}{ ho \ell^2}}[/latex] (9.175)

and for [latex]\alpha_2[/latex] we get

[latex]y_2(x)=-\sin \frac{\pi}{2 L} x+0.92894 \sin \frac{3 \pi}{2 L} x[/latex] (9.176)

The second mode, [latex]y_2(x)[/latex], is shown in Figure 9.10.

The accuracy of the mode shapes can be enhanced if a sequence of more terms to the y(x) are used. Consider, for instance, a sequence of three terms:

[latex]y(x)=c_1 \sin \frac{\pi x}{2 \ell}+c_2 \sin \frac{3 \pi x}{2 \ell}+c_3 \sin \frac{5 \pi x}{2 \ell}[/latex] (9.177)

Once more, we need to evaluate [latex]c_1, c_2[/latex] and [latex]c_3[/latex] and the corresponding eigenvalues of the problem. Substitution of equation (9.177) into (9.153) followed by integration of the entire expression gives

[latex]\begin{aligned}U=& \frac{E R}{2 \ell}\left[\left(\frac{\pi^2}{24}+\frac{1}{4} ight) c_1^2+\left(\frac{3 \pi^2}{8}+\frac{1}{4} ight) c_2^2+\left(\frac{25 \pi^2}{24}+\frac{1}{4} ight) c_3^2 ight.\&\left.+\frac{15}{8} c_1 c_2+\frac{65}{72} c_1 c_3+\frac{255}{321} c_2 c_3 ight] (9.178)\end{aligned}[/latex]

From the potential energy expression U, we can easily deduce the values of the element of the stiffness [latex]k_{i j}[/latex] :

[latex]\begin{aligned}k_{11} &=\frac{E R}{\ell}\left(\frac{\pi^2}{24}+\frac{1}{4} ight) \k_{21} &=k_{12}=\frac{15}{16} \frac{E R}{2 \ell} k_{13}=k_{31}=\frac{65}{144} \frac{E R}{\ell} \k_{23} &=k_{32}=\frac{255}{64} \frac{E R}{2 \ell} \k_{22} &=\frac{E R}{\ell}\left(\frac{3 \pi^2}{8}+\frac{1}{4} ight) \k_{33} &=\frac{E R}{\ell}\left(\frac{25}{24} \pi^2+\frac{1}{4} ight)\end{aligned}[/latex]

Similarly, substituting equation (9.177) into (9.158) and integrating along the bar, we obtain the following closed form of the kinetic energy.

[latex]\begin{aligned}T=& \frac{\omega^2 R \ell ho}{2}\left[\left(\frac{1}{6}-\frac{1}{\pi^2} ight) c_1^2+\left(\frac{1}{6}-\frac{1}{9 \pi^2} ight) c_2^2 ight.\&\left.+\left(\frac{1}{6}-\frac{1}{25 \pi^2} ight) c_3^2+\frac{3}{2 \pi^2} c_1 c_2+\frac{15}{8 \pi^2} c_2 c_3+\frac{5}{18} c_1 c_3 ight] (9.179)\end{aligned}[/latex]

Therefore, from the expression (9.179), we also identify the values of [latex]m_{i j}[/latex] :

[latex]\begin{aligned}m_{11} &=R \ell ho\left(\frac{1}{6}-\frac{1}{\pi^2} ight), & m_{12} &=m_{21}=\frac{3}{4 \pi^2} R \ell ho \m_{22} &=R \ell ho\left(\frac{1}{6}-\frac{1}{9 \pi^2} ight), & m_{23} &=m_{32}=\frac{15}{16 \pi^2} R \ell ho \m_{33} &=R \ell ho\left(\frac{1}{6}-\frac{1}{25 \pi^2} ight), & m_{31} &=m_{13}=\frac{5}{36} R \ell ho\end{aligned}[/latex]

Thus the elements [latex]k_{i g}[/latex] and [latex]m_{i j}[/latex] of the matrices K and M, respectively, are known.

Using equation (9.164) and replacing ω² by λ and [latex]\lambda \ell^2 ho / E[/latex] by α, we obtain the following system of equations:

[latex]\left[\begin{array}{ccc}0.66123-0.065345 \alpha & 0.9375-0.076 \alpha & 0.451389-0.01407 \alpha \0.9375-0.076 \alpha & 3.9511-0.1554 \alpha & 3.9843-0.09498 \alpha \0.4513889-0.01407 \alpha & 3.9843-0.09498 \alpha & 10.5308-0.162614 \alpha\end{array} ight]\left[\begin{array}{l}c_1 \c_2 \c_3\end{array} ight]=0[/latex] (9.180)

Setting the determinant of the system (9.180) to zero, we get

10.3272 − 1.42166α + 0.041016α² − 0.0002955133α³ = 0 (9.181)

which is found to have the following roots:

[latex]\begin{aligned}&\alpha_1=9.880436 \&\alpha_2=39.55020 \&\alpha_3=89.54455\end{aligned}[/latex]

or

[latex]\begin{aligned}\lambda_1&=9.880436 \frac{E}{ ho \ell^2}=\omega_1^2\\lambda_2 &=39.55020 \frac{E}{ ho \ell^2}=\omega_2^2 \\lambda_3 &=89.54455 \frac{E}{ ho \ell^2}=\omega_3^2\end{aligned}[/latex]

and consequently, the estimated natural frequencies and natural mode shapes are

[latex]\omega_1=3.14331 \sqrt{\frac{E}{ ho \ell^2}}[/latex] (9.182)

and

[latex]y_1(x)=\sin \frac{\pi x}{2 L}-0.058134 \sin \frac{3 \pi x}{2 L}-0.015158 \sin \frac{5 \pi x}{2 L}[/latex] (9.183)

which is displayed in Figure 9.11. For

[latex]\omega_2=6.2889 \sqrt{\frac{E}{ ho \ell^2}}[/latex] (9.184)

we have

[latex]y_2(x)=-\sin \frac{\pi x}{2 L}+0.934 \sin \frac{3 \pi x}{2 L}-0.07749 \sin \frac{5 \pi x}{2 L}[/latex] (9.185)

which is shown in Figure 9.12. Finally, for

[latex]\omega_3=9.46279 \sqrt{\frac{E}{ ho \ell^2}}[/latex] (9.186)

we have

[latex]y_3(x)=\sin \frac{\pi x}{2 L}-1.01371 \sin \frac{3 \pi x}{2 L}+0.93654 \sin \frac{5 \pi x}{2 L}[/latex] (9.187)

Figure 9.13 gives the mode shape of [latex]\omega_3[/latex].

Question 9.5.1: Consider the system shown in Figure 9.8, where the clamped–f...

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