Elastic Instability - Multiple Degrees of Freedom-systems

Multiple Degrees of Freedom-systems

By attaching another rigid beam to the original system by means of an angular spring a two degrees of freedom-system is obtained. Assume for simplicity that the beam lengths and angular springs are equal. The equilibrium conditions become

$F L ( \sin \theta_1 + \sin \theta_2 ) = k_\theta \theta_1$

$F L \sin \theta_2 = k_\theta ( \theta_2 - \theta_1 )$

where and are the angles of the two beams. Linearizing by assuming these angles are small yields

$\begin{pmatrix} F L - k_\theta & F L \\ k_\theta & F L - k_\theta \end{pmatrix} \begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

The non-trivial solutions to the system is obtained by finding the roots of the determinant of the system matrix, i.e. for

$\frac{F L}{k_\theta} = \frac{3}{2} \mp \frac{\sqrt{5}}{2} \approx \left\{\begin{matrix} 0.382\\2.618 \end{matrix}\right.$

Thus, for the two degrees of freedom-system there are two critical values for the applied force F. These correspond to two different modes of deformation which can be computed from the nullspace of the system matrix. Dividing the equations by yields

$\frac{\theta_2}{\theta_1} \Big|_{\theta_1 \ne 0} = \frac{k_\theta}{F L} - 1 \approx \left\{\begin{matrix} 1.618 & \text{for } F L/k_\theta \approx 0.382\\ -0.618 & \text{for } F L/k_\theta \approx 2.618 \end{matrix}\right.$

For the lower critical force the ratio is positive and the two beams deflect in the same direction while for the higher force they form a "banana" shape. These two states of deformation represent the buckling mode shapes of the system.

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