In the preceding example, let m = 1000 kg, l = 4 m, a = b = l/2 = 2 m, I = 1300 kg·m², k1 = 50 × 10³ N/m, and k2 = 70 × 10³ N/m, and c1 = c2 = 10 N·s/m. Determine the system response as a function of time if the initial conditions are given by y ( t = 0 ) = y0 = 0.03, y ( t = 0 ) = y0 = 0

Question

In the preceding example, let m = 1000 kg, l = 4 m, a = b = l/2 = 2 m, I = 1300 kg · m², [latex]k_{1} = 50 \times 10^3 N/m, k_{2} = 70 \times 10^3 N/m, \text{and} c_{1} = c_{2} = 10 N · s/m.[/latex] Determine the system response as a function of time if the initial conditions are given by

[latex]y(t = 0) = y_{0} = 0.03,\qquad \dot{y}(t = 0) = \dot{y_{0}} = 0\[0.5 cm] θ(t = 0) = θ_{0} = 0, \qquad\dot{θ}(t = 0) = \dot{θ_{0}} = 0[/latex]

Accepted Answer

The mass matrix M, the damping matrix C, and the stiffness matrix
K are given by

M = [latex]\begin{bmatrix} m & 0 \ 0 & I \end{bmatrix} = \begin{bmatrix} 1000 & 0 \ 0 & 1300 \end{bmatrix}[/latex]

C = [latex]\begin{bmatrix} c_{1}+c_{2} & c_{2}b-c_{1}a \ c_{2}b-c_{1}a & c_{1}a^{2} + c_{2}b^{2} \end{bmatrix} = \begin{bmatrix} 20 & 0 \ 0 & 80 \end{bmatrix}[/latex]

K = [latex]\begin{bmatrix} k_{1}+k_{2} & k_{2}b − k_{1}a \k_{2}b − k_{1}a & k_{1}a^{2} + k_{2}b^{2} \end{bmatrix} = 10³ \begin{bmatrix} 120 & 40 \ 40 & 480 \end{bmatrix}[/latex]

One can show that, in this example, the characteristic equation is given by

[latex]mIs^{4} + [I (c_{1} + c_{2}) + m(c_{1}a^{2} + c_{2}b^{2})]s^{3} + [m(k_{1}a^{2} + k_{2}b^{2}) + I (k_{1} + k_{2}) + (c_{1} + c_{2})(c_{1}a^{2} + c_{2}b^{2})]s^{2} + [(c_{1} + c_{2})(k_{1}a^{2} + k_{2}b^{2}) + (c_{1}a^{2} + c_{2}b^{2})(k_{1} + k_{2})]s + (k_{1} + k_{2})(k_{1}a^{2} + k_{2}b^{2}) − (k_{2}b − k_{1}a)^{2}[/latex] = 0

that is,
[latex]13 × 10^{5}s^{4} + 1.06 × 10^{5}s^{3} + 63.6 × 10^{7}s^{2} + 192 × 10^{5}s + 560 × 10^{8}[/latex] = 0

or
[latex]s^{4} + 0.08154s^{3} + 4.892 × 10^2s^{2}[/latex] + 14.769s + 4.3077 = 0

The roots of this polynomial are
[latex]s_{1} = −0.0104 + 10.7i,\qquad s_{2} = −0.0104 − 10.7i\[0.5 cm] s_{3} = −0.0304 + 19.3i,\qquad s_{4} = −0.0304 − 19.3i[/latex]

The amplitude ratios [latex]β_{i}[/latex] defined by Eq. 6.79 can be obtained using Eq. 6.64 as

[latex]β_{i} =\left(\frac{X_{1}}{X_{2}}\right)_{s=s_{i}} = \frac{X_{1i}}{X_{2i}}[/latex] (6.79)

[latex]M \ddot{x} + C \dot{x} + Kx = 0[/latex] (6.64)
[latex]β_{i} = \frac{X_{1_{i}}}{X_{2_{i}}} = −\frac{m_{12}s^{2}_{i} + c_{12}s_{i} + k_{12}}{m_{11}s^{2}_{i} + c_{11}s_{i} + k_{11}}\[0.5 cm] \qquad \qquad=− \frac{m_{22}s^{2}_{i} + c_{22}s_{i} + k_{22}}{m_{21}s^{2}_{i}+ c_{21}s_{i} + k_{21}}[/latex]

which yields
[latex]β_{1} = −8.279 − 0.01417i, \qquad β_{2} = −8.279 + 0.01417i\[0.5 cm]β_{3} = 0.106 − 4.645 \times 10^{−4}i,\qquad β_{4} = 0.106 + 4.645 \times 10^{−4}i[/latex]

The solution can then be written using Eq. 6.86 as

[latex]y(t) = e^{−0.0104t} [(β_{1}X_{21} + β_{2}X_{22}) \cos 10.7t + i(β_{1}X_{21} − β_{2}X_{22}) \sin 10.7t ]\[0.5 cm]\qquad+ e^{−0.0304t} [(β_{3}X_{23} + β_{4}X_{24}) \cos 19.3t + i(β_{3}X_{23} − β_{4}X_{24}) \sin 19.3t ]\[0.5 cm]θ(t) =e^{−0.0104t} [(X_{21} + X_{22}) \cos 10.7t + i(X_{21} − X_{22}) \sin 10.7t]\[0.5 cm]\qquad+e^{−0.0304t} [(X_{23} + X_{24}) \cos 19.3t + i(X_{23} − X_{24}) \sin 19.3t][/latex]

Note that these two equations have only four unknown amplitudes [latex]X_{21}, X_{22}, X_{23}, \text{and} X_{24}[/latex] which can be determined by using the initial conditions. To this end, we introduce the new real constants [latex]B_{1}, B_{2}, B_{3}, \text{and} B_{4}[/latex] defined as

[latex]B_{1} = X_{21} + X_{22},[/latex] [latex]B_{2} = i(X_{21} − X_{22})[/latex]
[latex]B_{3} = X_{23} + X_{24},[/latex] [latex]B_{4} = i(X_{23} − X_{24})[/latex]

Observe that since the amplitude ratios appear as complex conjugates in the form [latex]β_{i} = α_{i} ± iγ_{i}[/latex] , one has

[latex]β_{1}X_{21} + β_{2}X_{22} = α_{1}B_{1} + γ_{1}B_{2}[/latex]
[latex]i(β_{1}X_{21} − β_{2}X_{22}) = −γ_{1}B_{1} + α_{1}B_{2}[/latex]
[latex]β_{3}X_{23} + β_{4}X_{24} = α_{3}B_{3}+ γ_{3}B_{4}[/latex]
[latex]i(β_{3}X_{23} − β_{4}X_{24}) = −γ_{3}B_{3} + α_{3}B_4[/latex]

where [latex]α_{1} = α_{2}[/latex] = −8.279, [latex]γ_{1} = γ_{2}[/latex] = −0.01417, [latex]α_{3} = α_{4}[/latex] = 0.1058, and [latex]γ_{3} = γ_{4} = −4.645 × 10^{−4}[/latex]. Therefore, in terms of the new constants [latex]B_{1}, B_{2}, B_{3}, \text{and} B_{4}[/latex], the coordinates y and θ can be written as

[latex]y(t) = e^{−0.0104t} [(α_{1}B_{1} + γ_{1}B_{2}) \cos 10.7t + (−γ_{1}B_{1}+ α_{1}B_{2}) \sin 10.7t ]\[0.5 cm] \qquad+ e^{−0.0304t} [(α_{3}B_{3} + γ_{3}B_{4}) \cos 19.3t + (−γ_{3}B_{3} + α_{3}B_{4}) \sin 19.3t ]\[0.5 cm] θ(t) = e^{−0.0104t} [B_{1} \cos 10.7t + B_{2} \sin 10.7t ]\[0.5 cm] \qquad+ e^{−0.0304t} [B_{3} \cos 19.3t + B_{4} \sin 19.3t ][/latex]

Note that these two equations can be expressed in the form given by Eq. 6.87. By using the initial conditions, one can verify that the constants [latex]B_{1}, B_{2}, B_{3}[/latex], and [latex]B_{4}[/latex] are

[latex]\left.\begin{aligned} x_1(t)&=C_{11} e^{-p_1t} \sin(\Omega_t+\phi_{11})&+C_{12} e^{-p_2t} \sin(\Omega_2t+\phi_{12})\[0.5 cm] x_2(t)&=C_{21} e^{-p_1t} \sin(\Omega_t+\phi_{21})&+C_{22} e^{-p_2t} \sin(\Omega_2t+\phi_{22})\end{aligned}\right\}[/latex] (6.87)

[latex]B_{1} = −0.358 × 10^{−2},[/latex] [latex]B_{2} = −0.105 × 10^{−4}[/latex]
[latex]B^{3} = 0.358 × 10^{−2},[/latex] [latex]B_{4} = 0.915 × 10^{−5}[/latex]

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