Consider the bar shown in Figure 10.9 undergoing a large rotation and assumed to be elastic. Assume that E, the modulus of elasticity, me, the mass density, and A, the cross-sectional area, are all known constants relating the properties of the bar. Let us assume that the damping coefficient of the

Question

Consider the bar shown in Figure 10.9 undergoing a large rotation and assumed to be elastic. Assume that E, the modulus of elasticity, [latex]m_e[/latex], the mass density, and A, the cross-sectional area, are all known constants relating the properties of the bar. Let us assume that the damping coefficient of the bar is zero and that we have zero initial conditions. Derive the equation of motion of the bar using two one-dimensional elements assuming longitudinal deformation.

Accepted Answer

As shown in Figure 10.10, the shape function matrix [N] from the standard finiteelement method is

[latex][N]=\left[N_i N_j ight]=\left[\frac{x_j-x}{x_j-x_i} \frac{x_{-} x_i}{x_j-x_i} ight]=\left[\frac{x_j-x}{\ell} \frac{x_{-} x_i}{\ell} ight] (10.79)[/latex]

The strain-modal displacement matrix

[latex][B]=\frac{d[N]}{d x}=\left[\frac{-1}{x_j-x_i} \frac{1}{x_j-x_i} ight]=\left[-\frac{1}{\ell} \frac{1}{\ell} ight][/latex] (10.80)

The element stiffness matrix

[latex]\left[K_s ight]=\int_v[B]^T[D][B] d v[/latex] (10.81)

where D, the material property matrix, is actually the scalar E (the modulus of elasticity); hence

[latex]\left[K_s ight]=\int_{\ell j}^{\ell j}\left[\begin{array}{c}-\frac{1}{\ell} \ \frac{1}{\ell}\end{array} ight] E\left[-\frac{1}{\ell} \frac{1}{\ell} ight] A d x[/latex]

[latex]=\frac{E A}{\ell}\left[\begin{array}{cc}1 & -1 \ -1 & 1\end{array} ight][/latex] (10.82)

where A is the cross-sectional area of the element. The element mass matrix [see equation (10.68)]

[latex][M]=\int_v m_e[N]^T[N] d v[/latex] (10.68)

[latex]\begin{aligned}{\left[M^e ight] } &=\int_v m_e[N]^T[N] d v \&=\frac{m_e A}{\ell^2} \int_{\ell i}^{\ell j}\left[\begin{array}{c}x_j-x \x-x_i\end{array} ight]\left[x_j-x x-x_i ight] d x \&=\frac{m_e A}{\ell^2} \int_{\ell i}^{\ell j}\left[\begin{array}{cc}\left(x_j-x ight)^2 & \left(x_j-x ight)\left(x-x_i ight) \\left(x-x_i ight)\left(x_j-x ight) & \left(x-x_i ight)^2\end{array} ight] d x \&=\frac{1}{6} m_e A \ell\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight] (10.83)\end{aligned}[/latex]

The nodal force vector [see equation (10.71)]

[latex]\begin{aligned}\{F\}=& \int_v[N]^T\{F\} d v-\int_v m_e[N]^T\left\{\left(\left[\dot{\Omega}^{o n} ight]+\left[\Omega^{o n} ight]\left[\Omega^{o n} ight] ight)\left\{r_n ight\} ight.\&\left.+\sum_{i=1}^n\left[S^{n v} ight]\left(\left[\dot{\Omega}^{o v} ight]+\left[\Omega^{o v} ight]\left[\Omega^{o v} ight] ight)\left\{q_i ight\} ight\} d v=\left\{F_s ight\}+\left\{F_d ight\} (10.71)\end{aligned}[/latex]

[latex]\left\{F_s ight\}=\int_v[N]^T\{F\} d v[/latex]

where [latex]\{F\}=-m_e g \sin \varphi, m_e[/latex] is the mass density, and g the gravity acceleration. Then

[latex]\begin{aligned}\left\{F_s ight\} &=-\int_{\ell i}^{\ell j} m_e g \sin \varphi[N]^T d x=\frac{1}{\ell} \int_{\ell i}^{\ell j} m_e g \sin \varphi\left[\begin{array}{c}x_j-x \x-x_i\end{array} ight] d x \&=\frac{1}{2} A m_e g \ell \sin \varphi\left\{\begin{array}{l}1 \1\end{array} ight\} (10.84)\end{aligned}[/latex]

The skew matrix of the angular velocity [latex][\Omega][/latex] is defined as follows:

Since we have one body

[latex]n=1, \quad v=n-1=0[/latex]

Then

[latex]\left[\Omega^{o n} ight]=\left[\Omega^{01} ight]=\left[\dot{S}^{01} ight]\left[S^{01} ight]^{-1}=\left[\dot{S}^{01} ight]\left[S^{10} ight][/latex]

where

[latex]\left[S^{10} ight]=\left[S^{01} ight]^T=\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0 \-\sin \varphi & \cos \varphi & 0 \0 & 0 & 1\end{array} ight][/latex]

The skew matrix is then

[latex]\left[\Omega^{01} ight]=\dot{\varphi}\left[\begin{array}{ccc}0 & -1 & 0 \1 & 0 & 0 \0 & 0 & 0\end{array} ight] (10.85)[/latex]

and its product is obtained as follows

[latex]\left[\Omega^{01} ight]\left[\Omega^{01} ight]=\dot{\varphi}^2\left[\begin{array}{ccc}-1 & 0 & 0 \0 & -1 & 0 \0 & 0 & 0\end{array} ight] (10.86)[/latex]

Its derivative is simply

[latex]\left[\dot{\Omega}^{01} ight]=\ddot{\varphi}\left[\begin{array}{ccc}0 & -1 & 0 \1 & 0 & 0 \0 & 0 & 0\end{array} ight] (10.87)[/latex]

The element dynamic damping matrix [see equation (10.69)] can be expressed as:

[latex][C]=\int_v[N]^T\{\mu\}[N] d v+2 \int_v m_e[N]^T\left[\Omega^{o n} ight][N] d v=\left[C_s ight]+\left[C_d ight][/latex] (10.69)

[latex]\left[C_d ight]=2 \int_v m_e[N]^T\left[\Omega_{11}^{o n} ight][N] d v[/latex] (10.88)

Since the element is one-dimensional, we can only consider [latex]\left[\Omega_{11}^{ ext {on }} ight][/latex] in the matrix [latex]\left[\Omega^{o n} ight][/latex]. Therefore,

[latex]\left[C_d ight]=0[/latex] (10.89)

The element total damping matrix is found to be

[latex]\left[C^e ight]=\left[C_s ight]+\left[C_d ight]=0+0=0[/latex] (10.90)

The element dynamic stiffness matrix [see equation (10.70)] after substitution becomes

[latex][K]=\int_v[B]^T[D][B] d v+\int_v m_e[N]^T\left(\left[\dot{\Omega}^{o n} ight]+\left[\Omega^{ ext {on }} ight]\left[\Omega^{ ext {on }} ight] ight)[N] d v=\left[K_s ight]+\left[K_d ight][/latex] (10.70)

[latex]\begin{aligned}{\left[K_d ight] } &=\int_v m_e[N]^T\left(\left[\dot{\Omega}_{11}^{o n} ight]+\left[\Omega_{1 t}^{o n} ight]\left[\Omega_{t 1}^{o n} ight] ight)[N] d v \&=m_e \int_{\ell i}^{\ell j}\left[\begin{array}{c}x_j-x \x-x_i\end{array} ight]\left(0-\dot{\varphi}^2 ight)\left[x_j-x \quad x-x_i ight] A d x \&=-\frac{1}{6} m_e A \ell \dot{\varphi}^2\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight] (10.91)\end{aligned}[/latex]

The element total stiffness matrix is

[latex]\begin{aligned}{\left[K^e ight] } &=\left[K_s ight]+\left[K_d ight] \{\left[K^e ight] } &=\frac{E A}{\ell}\left[\begin{array}{cc}1 & -1 \-1 & 1\end{array} ight]-\frac{1}{6} m_e A \ell \dot{\varphi}^2\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight] \&=A\left[\begin{array}{cc}\frac{E}{\ell}-\frac{1}{3} m_e \ell \dot{\varphi}^2 & -\left(\frac{E}{\ell}+\frac{1}{6} m_e \ell \dot{\varphi}^2 ight) \-\left(\frac{E}{\ell}+\frac{1}{6} m_e \ell \dot{\varphi}^2 ight) & \frac{E}{\ell}-\frac{1}{3} m_e \ell \dot{\varphi}^2\end{array} ight] (10.92)\end{aligned}[/latex]

The element dynamic force vector [see equation 10.71)] is expressed as:

[latex]\begin{aligned}\left\{F_d ight\}=&-\int_v m_e[N]^T\left\{\left(\left[\dot{\Omega}_{11}^{o n} ight]+\left[\Omega_{1 t}^{o n} ight]\left[\Omega_{t 1}^{o n} ight] ight)\left\{r_n ight\} ight.\&\left.+\sum_{i=1}^n\left[S^{n v} ight]\left(\left[\dot{\Omega}^{o v} ight]+\left[\Omega^{o v} ight]\left[\Omega^{o v} ight] ight)\left\{q_i ight\} ight\} d v\end{aligned}[/latex]

where n = 1, v = 0, and [latex]\left\{r_n ight\}=x[/latex]; therefore,

[latex]\begin{aligned}\left\{F_d ight\} &=-\int_{\ell i}^{\ell j} m_e \frac{1}{\ell}\left[\begin{array}{c}x_j-x \x-x_i\end{array} ight]\left\{\left(0-\dot{\varphi}^2 ight) x+0 ight\} A d x \&=\frac{1}{6} m_e A \ell \dot{\varphi}^2\left[\begin{array}{l}3 x_j-2 \ell \3 x_i-2 \ell\end{array} ight]\end{aligned}[/latex]

or

[latex]\left\{F_d ight\}=\frac{1}{6} m_e A \ell \dot{\varphi}^2\left[\begin{array}{l}3 x_i+\ell \3 x_j-5 \ell\end{array} ight] (10.93)[/latex]

The element total force vector

[latex]\begin{aligned}\left\{F^e ight\} &=\left\{F_s ight\}+\left\{F_d ight\} \&=-\frac{1}{2} A m_e g \ell \sin \varphi\left[\begin{array}{l}1 \1\end{array} ight]+\frac{1}{6} m_e A \ell \dot{\varphi}^2\left[\begin{array}{c}3 x_j-2 \ell \3 x_i-2 \ell\end{array} ight] \&=\frac{1}{6} A m_e \ell\left[\begin{array}{l}\left(3 x_j-2 \ell ight) \dot{\varphi}^2-3 g \sin \varphi \\left(3 x_i-2 \ell ight) \dot{\varphi}^2-3 g \sin \varphi\end{array} ight] (10.94)\end{aligned}[/latex]

From equations (10.67), (10.83), (10.90), (10.92) and (10.94), we obtain the element equations of motion:

[latex][M]\{\ddot{ ho}\}+[C]\{\dot{ ho}\}+[K]\{ ho\}=\left\{F^e ight\}[/latex] (10.67)

[latex]\frac{1}{6} m_e A \ell\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight]\{\ddot{ ho}\}+A\left[\begin{array}{c}\frac{E}{\ell}-\frac{1}{3} m_e \ell \dot{\varphi}^2 \-\left(\frac{E}{\ell}+\frac{1}{6} m_e \ell \dot{\varphi}^2 ight) \-\left(\frac{E}{\ell}+\frac{1}{6} m_e \ell \dot{\varphi}^2 ight) \\frac{E}{\ell}-\frac{1}{3} m_e \ell \dot{\varphi}^2\end{array} ight][/latex]

[latex]=\frac{1}{6} A m_e \ell\left[\begin{array}{l}\left(3 x_j-2 \ell ight) \dot{\varphi}^2-3 g \sin \varphi \\left(3 x_i-2 \ell ight) \dot{\varphi}^2-3 g \sin \varphi\end{array} ight] (10.95)[/latex]

If the flexible link is divided into two elements, for element 1 we have [latex]x_j=2 \ell[/latex] and [latex]x_i=\ell[/latex]; then

[latex]\begin{aligned}&{\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight]\{\ddot{ ho}\}+\left[\begin{array}{cc}\frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) \-\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & \frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2\end{array} ight]\{ ho\}} \&=\left[\begin{array}{c}\ell \dot{\varphi}^2-3 g \sin \varphi+\frac{6 F_1}{m_e A \ell} \-2 \ell \dot{\varphi}^2-3 g \sin \varphi\end{array} ight] (10.96)\end{aligned}[/latex]

For element [latex]2, x_j=2 \ell[/latex] and [latex]x_i=\ell[/latex]; then

[latex]\begin{aligned}&{\left[\begin{array}{ll}2 & 1 \1 & 2\end{array} ight]\{\ddot{ ho}\}+\left[\begin{array}{cc}\frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) \-\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & \frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2\end{array} ight]\{ ho\}} \&=\left[\begin{array}{l}4 \ell \dot{\varphi}^2-3 g \sin \varphi \\ell \dot{\varphi}^2-3 g \sin \varphi\end{array} ight] (10.97)\end{aligned}[/latex]

Combining the two elements equations above, we obtain the global representation of the elastic body:

[latex]\left[\begin{array}{lll}2 & 1 & 0 \1 & 4 & 1 \0 & 1 & 2\end{array} ight]\left[\begin{array}{c}\ddot{ ho}_1 \\ddot{ ho}_2 \\ddot{ ho}_3\end{array} ight]+\left[\begin{array}{ccc}\frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & 0 \-\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & \frac{12 E}{m_e \ell^2}-4 \dot{\varphi}^2 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) \0 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & \frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2\end{array} ight][/latex]

[latex].\left[\begin{array}{l} ho_1 \ ho_2 \ ho_3\end{array} ight]=\left[\begin{array}{c}\ell \dot{\varphi}^2-3 g \sin \varphi+\frac{6 F_1}{m_e A \ell} \ 2 \ell \dot{\varphi}^2-6 g \sin \varphi \ \ell \dot{\varphi}^2-3 g \sin \varphi\end{array} ight][/latex] (10.98)

From the boundary conditions we have [latex] ho_1=0[/latex] and [latex]\ddot{ ho}_1=0[/latex]; then the equations above reduce to

[latex]\begin{aligned}&{\left[\begin{array}{ll}4 & 1 \1 & 2\end{array} ight]\left[\begin{array}{c}\ddot{ ho}_2 \\ddot{ ho}_3\end{array} ight]+\left[\begin{array}{cc}\frac{12 E}{m_e \ell^2}-4 \dot{\varphi}^2 & -\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) \-\left(\frac{6 E}{m_e \ell^2}+\dot{\varphi}^2 ight) & \frac{6 E}{m_e \ell^2}-2 \dot{\varphi}^2\end{array} ight]\left[\begin{array}{l} ho_2 \ ho_3\end{array} ight]} \&=\left[\begin{array}{c}2 \ell \dot{\varphi}^2-6 g \sin \varphi \\ell \dot{\varphi}^2-3 g \sin \varphi\end{array} ight] (10.99)\end{aligned}[/latex]

Let [latex]E / m_e \ell^2=a[/latex] and [latex]\{\ddot{ ho}\}=\omega^2\{ ho\}[/latex], where ω is the eigenvalue; then for the homogeneous equation (10.99), we obtain

[latex]\left[\begin{array}{c}12 a-4 \dot{\varphi}^2-4 \omega^2-\left(6 a+\dot{\varphi}^2 ight)-\omega^2 \-\left(6 a+\dot{\varphi}^2 ight)-\omega^2 6 a-2 \dot{\varphi}^2-2 \omega^2\end{array} ight]\{ ho\}=0[/latex] (10.100)

Since [latex][ ho] eq 0[/latex], the determinant of the matrix must equal zero; that is,

[latex]\left(12 a-4 \dot{\varphi}^2-4 \omega^2 ight)\left(6 a-2 \dot{\varphi}^2-2 \omega^2 ight)-\left[\left(6 a+\dot{\varphi}^2 ight)+\omega^2 ight]^2=0[/latex] (10.101)

or

[latex]7 \omega^4+\left(14 \dot{\varphi}^2-60 a ight) \omega^2+\left(7 \dot{\varphi}^4-60 a \dot{\varphi}^2+36 a^2 ight)=0[/latex] (10.102)

which yields

[latex]\omega^2=\frac{69 a-14 \dot{\varphi}^2 \pm \sqrt{\left(60 a-14 \dot{\varphi}^2 ight)^2-4 imes 7\left(7 \dot{\varphi}^4-60 a \dot{\varphi}^2+36 a^2 ight)}}{2 imes 7}[/latex]

[latex]=\frac{60 a-14 \dot{\varphi}^2 \pm \sqrt{2592} a}{14}[/latex] (10.103)

resulting in

[latex]\begin{aligned}&\omega_1^2=0.649 a-\dot{\varphi}^2 (10.104) \&\omega_2^2=7.922 a-\dot{\varphi}^2\end{aligned}[/latex]

Substituting equation (10.104) into equation (10.100), we obtain the modal shape matrix

[latex]\left[\phi_1 ight]=\left[\begin{array}{cc}1 & 1 \\sqrt{2} & \sqrt{2}\end{array} ight] (10.105)[/latex]

The normal modal shape matrix is found to be

[latex][\phi]=\left[\begin{array}{cc}\frac{1}{\sqrt{8+2 \sqrt{2}}} & \frac{1}{\sqrt{8-2 \sqrt{2}}} \\frac{\sqrt{2}}{\sqrt{8+2 \sqrt{2}}} & \frac{-\sqrt{2}}{\sqrt{8-2 \sqrt{2}}}\end{array} ight]=\left[\begin{array}{cc}0.304 & 0.440 \0.430 & -0.622\end{array} ight] (10.106)[/latex]

Let the modal coordinates be expressed as functions of the modal coordinates η, where {ρ} = {Φ}{η} and premultiply by [latex]\phi^T[/latex] on both sides of equation (10.99); we get

[latex]\{\ddot{\eta}\}+\left[\begin{array}{cc}\omega_1^2 & 0 \0 & \omega_2^2\end{array} ight]\{\eta\}=\left[\begin{array}{c}1.038 \ell \dot{\varphi}^2-3.114 g \sin \varphi \0.258 \ell \dot{\varphi}^2-0.744 g \sin \varphi\end{array} ight] (10.107)[/latex]

From the initial conditions

[latex]\{ ho(0)\}=0 \quad\{\dot{ ho}(0)\}=0[/latex] (10.108)

the modal coordinates initial conditions become

[latex]\{\eta(0)\}=0 \quad\{\dot{\eta}(0)\}=0[/latex] (10.109)

the result of which is

[latex]\eta_1=\frac{1}{\omega_1} \int_0^t\left(1.038 \ell \dot{\varphi}^2-3.114 g \sin \varphi ight) \sin \omega_1(t- au) d au[/latex] (10.110)

and

[latex]\eta_2=\frac{1}{\omega_2} \int_0^t\left(0.258 \ell \dot{\varphi}^2-0.774 g \sin \varphi ight) \sin \omega_2(t- au) d au[/latex] (10.111)

In matrix form we can express the equations above as

[latex]\begin{aligned}{\left[\begin{array}{l} ho_2 \ ho_3\end{array} ight]=} &\{ ho\}=[\phi]\{\eta\} \&=\left[\begin{array}{l}\frac{0.304}{\omega_1} \int_0^t\left(1.038 \ell \dot{\varphi}^2-3.114 g \sin \varphi ight) \sin \omega_1(t- au) d au \+\frac{0.440}{\omega_2} \int_0^t\left(0.258 \ell \dot{\varphi}^2-0.774 g \sin \varphi ight) \sin \omega_2(t- au) d au \\frac{0.430}{\omega_1} \int_0^t\left(1.038 \ell \dot{\varphi}^2-3.114 g \sin \varphi ight) \sin \omega_1(t- au) d au \-\frac{0.662}{\omega_2} \int_0^t\left(0.258 \ell \dot{\varphi}^2-0.774 g \sin \varphi ight) \sin \omega_2(t- au) d au\end{array} ight] (10.112)\end{aligned}[/latex]

where

[latex]\begin{aligned}&\omega_1=\sqrt{0.649 a-\dot{\varphi}^2} (10.113) \&\omega_2=\sqrt{7.922 a-\dot{\varphi}^2} (10.114)\end{aligned}[/latex]

and

[latex]a=\frac{E}{m_e \ell^2}[/latex]

Let [latex]\dot{\varphi}=0[/latex]; then we obtain the ordinary modal frequencies. Compared with the precise solution obtained by the partial differential equation, for [latex]\omega_1[/latex] (lowest one) the error is only 2.6 %, and for [latex]\omega_2[/latex] (second lowest) the error is 19 %. If, on the other hand, [latex]\dot{\varphi}=\omega=[/latex] constant, then

[latex]\begin{aligned} ho_2(t)=& \frac{0.316 \ell \omega^2}{\omega_1^2}\left(1-\cos \omega_1 t ight)-\frac{0.473 g}{\omega_1}\left[\frac{1}{\omega+\omega_1}\left(\sin \omega t+\sin \omega_1 t ight) ight.\&\left.-\frac{1}{\omega+\omega_1}\left(\sin \omega t-\sin \omega_1 t ight) ight]+\frac{0.114 \ell \omega^2}{\omega_2^2}\left(1-\cos \omega_2 t ight) \&-\frac{0.170 g}{\omega_2}\left[\frac{1}{\omega+\omega_2}\left(\sin \omega t+\sin \omega_2 t ight)-\frac{1}{\omega-\omega_2}\left(\sin \omega t-\sin \omega_2 t ight) ight] (10.115) \ ho_3(t)=& \frac{0.446 \ell \omega^2}{\omega_1^2}\left(1-\cos \omega_1 t ight)-\frac{0.670 g}{\omega_1}\left[\frac{1}{\omega+\omega_1}\left(\sin \omega t+\sin \omega_1 t ight) ight.\&\left.-\frac{1}{\omega-\omega_1}\left(\sin \omega t-\sin \omega_1 t ight) ight]-\frac{0.160 \ell \omega^2}{\omega_2^2}\left(1-\cos \omega_2 t ight) \&+\frac{0.241 g}{\omega_2}\left[\frac{1}{\omega+\omega_2}\left(\sin \omega t+\sin \omega_2 t ight)-\frac{1}{\omega-\omega_2}\left(\sin \omega t-\sin \omega_2 t ight) ight] (10.116)\end{aligned}[/latex]

Therefore, from equations (10.79) and (10.116) we can use the following relation

[u] = [N]{ρ} (10.117)

to find the displacement of an arbitrary point on the bar. In addition, we can evaluate the constraint force [latex]F_1[/latex] at node 1 from equation (10.98). Note that [latex]F_1[/latex] is just acting along the longitudinal direction of the bar. As shown in Figure 10.11, the total constraint force of the system is

[latex]\begin{aligned}\left\{F_c ight\} &=\left[F_1-2 m_e g A \ell \cos \varphi 0 ight]\left\{\bar{n}^1 ight\} \&=\left[F_1-3 m_e g A \ell \cos \varphi 0 ight]\left[S^{1 o} ight]\{\bar{n}\} \&=\left[F_1-2 m_e g A \ell \cos \varphi 0 ight]\left[\begin{array}{ccc}\cos \varphi & \sin \varphi & 0 \-\sin \varphi & \cos \varphi & 0 \0 & 0 & 1\end{array} ight]\left[\begin{array}{c}\bar{n}_1 \\bar{n}_2 \\bar{n}_3\end{array} ight] \&=\left(F_1 \cos \varphi+2 m_e A g \ell \cos \varphi \sin \varphi ight) \bar{n}_1+\left(F_1 \sin \varphi-2 m_e A \ell g \cos ^2 \varphi ight) \bar{n}_2\end{aligned}[/latex] (10.118)

Discussion

In this example, for simplicity in carrying out the explicit solution, we selected two one-dimensional elements and discussed the longitudinal vibration of the bar when it is undergoing a large rotation. From equations (10.113) and (10.114), we know that the modal frequencies of the system are decreased due to the angular rotation. The higher the angular velocity, the lower the modal frequencies become. If [latex]\dot{\varphi}[/latex] is time variant, the modal frequencies will vary as well. The modal shapes of the bar have nothing to do with the rotation of the system [see equation (10.106)]. The conclusion one draws is very significant. It means that the modal frequencies of an aircraft moving in space at a high speed are changeable. Therefore, modal frequencies based on laboratory experiments or calculated by material properties and geometric shapes of the system are not accurate enough to avoid resonance in the system. It is apparent from this example that coupling of rigid-body motion with flexibility is important, if the angular velocity [latex]\dot{\varphi}[/latex] takes on significant values.

Question 10.3.1: Consider the bar shown in Figure 10.9 undergoing a large rot...

Related Answered Questions

Consider the planar two-bar system shown in Figure 10.3. Find the explicit form of the velocity and acceleration of an arbitrary point p on the flexible link with the boundary conditions given by equations (10.3). ...

Verified Answer: