The bar A B shown in Fig. 6.5, which represents a simplified model for the chassis of a vehicle, has length l, mass m, and mass moment of inertia I about its mass center C. The bar is supported by two linear springs which have constants k1 and k2. Determine the differential equations of motion

Question

The bar A B shown in Fig. 6.5, which represents a simplified model for the chassis of a vehicle, has length l, mass m, and mass moment of inertia I about its mass center C. The bar is supported by two linear springs which have constants [latex]k_{1}[/latex] and [latex]k_{2}[/latex]. Determine the differential equations of motion assuming that the motion of the bar in the horizontal direction is small and can be neglected.

Accepted Answer

Since the displacement of the bar in the x direction is neglected, the system configuration can be identified using the two variables y and θ, where y is the vertical displacement of the center of mass and θ is the angular rotation of the bar. It is left to the reader as an exercise to show that the weight of the bar cancels with the deflection of the springs at the static equilibrium position. Therefore, the differential equations of motion are given by

m[latex]\ddot{y} = −k_{1}\left( y − \frac{l}{2}θ\right) − k_2\left(y + \frac{l}{2}θ\right)[/latex]

I [latex]\ddot{θ} = k_1 (y − \frac{l}{2}θ) \frac{l}{2}\ cos θ − k_2 (y + \frac{l}{2}θ) \frac{l}{2}[/latex] cos θ

For small angular oscillations, cos θ ≈ 1, and the differential equations reduce to

m[latex]\ddot{y} + (k_{1} + k_{2})y + (k_{2} − k_{1}) \frac{l}{2}[/latex] θ = 0

I [latex]\ddot{θ} + (k_{2} + k_{1}) \frac{l^2}{4} θ + (k_{2} − k_{1}) \frac{l}{2}y[/latex] = 0

which can be written in matrix form as

[latex]\begin{bmatrix} m & 0 \ 0 & I \end{bmatrix}[/latex] [latex]\begin{bmatrix} \ddot{y} \ \ddot{θ} \end{bmatrix}+ \begin{bmatrix} k_1 + k_2 & (k_2 − k_1) \frac{l}{2} \ (k_2 − k_1) \frac{l}{2} & (k_2 + k_1) \frac{l^2}{4} \end{bmatrix}[/latex] [latex]\begin{bmatrix} y \ θ \end{bmatrix}=[/latex] [latex]\begin{bmatrix} 0 \ 0 \end{bmatrix}[/latex]

which can be written in a more compact form as
M[latex]\ddot{x}[/latex] + Kx = 0

where M and K are the symmetric mass and stiffness matrices given by

M = [latex]\begin{bmatrix} m & 0 \ 0 & I \end{bmatrix}[/latex], K = [latex]\begin{bmatrix} k_1 + k_2 & (k_2 − k_1) \frac{l}{2} \ (k_2 − k_1) \frac{l}{2} & (k_2 + k_1) \frac{l^2}{4} \end{bmatrix}[/latex]

and the vectors x and [latex]\ddot{x}[/latex] are

[latex]x=\begin{bmatrix} y \ θ \end{bmatrix},\qquad \ddot{x}=\begin{bmatrix} \ddot{y} \ \ddot{θ} \end{bmatrix}[/latex]

One can then assume a solution in the form x = Xsin(ωt + Φ), where X is the vector of amplitudes defined as

X= [latex]\begin{bmatrix} Y\\Theta \end{bmatrix}[/latex]

The acceleration vector [latex]\ddot{x}[/latex]is given by [latex]\ddot{x}= −ω^2X\sin(ωt + \phi)[/latex] . Substituting x and [latex]\ddot{x}[/latex] into the differential equation one obtains
[K − ω²M]X = 0
For a nontrivial solution, the determinant of the coefficient matrix must be equal to zero, and hence
|K − ω²M| = 0
which can be written in a more explicit form as

[latex]\left|\begin{matrix}k_{11} − ω^{2}m&k_{12}\k_{21}&k_{22} − ω^{2}I\end{matrix}\right|[/latex] = 0

where [latex]k_{11} = k_{1} +k_{2}, k_{12} = k_{21} = (k_{2} −k_{1})(l/2), [/latex] and [latex] k_{22} = (k_{1} +k_{2}[/latex])(l²/4).
The characteristic equation is then given by
([latex]k_{11} − ω^{2}m)(k_{22} − ω^{2}I) − k^{2}_{12}[/latex] = 0
or
[latex]ω^{4}Im − ω^{2}[mk_{22} + Ik_{11}] + k_{11}k_{22} − k^{2}_{12}[/latex] = 0

which yields the following equation:

[latex]\omega^4Im-\omega^2(k_1+k_2)\left[m\frac{l^2}{4}+I\right]+4k_1k_2\frac{l^2}{4}=0[/latex]

The roots of this equation are given by

[latex]ω^{2}_{1}= \frac{ −b +\sqrt{b^2 − 4ac}}{2a}[/latex] [latex] ω^{2}_{2}=\frac{ −b -\sqrt{b^2 − 4ac}}{2a}[/latex]

where
[latex]a = ml, b = −(k_1 + k_2) \left(m\frac{l^2}{4}+ I\right) , c = 4k_1k_2 \frac{l^2}{4}[/latex]

The amplitude ratios [latex]β_1[/latex] and [latex]β_2[/latex] are given by

[latex]β_1 = \frac{X_{11}}{X_{21}} = -\frac{(k_2 − k_1)\frac{l}{2}}{k_1 + k_2 − ω^2_1 m} =- \frac{(k_1 + k_2) \frac{l^2}{4}− ω^2_1I}{(k_2 − k_1) \frac{l}{2}}\[0.5 cm] β_2 =\frac{X_{12}}{X_{22}} = -\frac{(k_2 − k_1) \frac{l}{2}}{k_1 + k_2 − ω_2^2m}= -\frac{(k_1 + k_2) \frac{l^2}{4} −ω^2_2I}{(k_2 − k_1) \frac{l}{2}}[/latex]

The solution is then given by
[latex]x_1 = β_1X_{21} \sin(ω_1t + \phi_1) + β_2X_{22} \sin(ω_2t + \phi_2)\[0.5 cm] x_2 = X_{21} \sin(ω_1t + \phi_1) + X_{22} \sin(ω_2t + \phi_2)[/latex]

Question 6.4: The bar A B shown in Fig. 6.5, which represents a simplified...

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