Consider as a simple example a rigid beam of length L, hinged in one end and free in the other, and having an angular spring attached to the hinged end. The beam is loaded in the free end by a force F acting in the compressive axial direction of the beam, see the figure to the right.
Assuming a clockwise angular deflection , the clockwise moment exerted by the force becomes . The moment equilibrium equation is given by
where is the spring constant of the angular spring (Nm/radian). Assuming is small enough, implementing the Taylor expansion of the sine function and keeping the two first terms yields
which has three solutions, the trivial , and
which is imaginary (i.e. not physical) for and real otherwise. This implies that for small compressive forces, the only equilibrium state is given by , while if the force exceeds the value there is suddenly another mode of deformation possible.
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
where and are the angles of the two beams. Linearizing by assuming these angles are small yields
The non-trivial solutions to the system is obtained by finding the roots of the determinant of the system matrix, i.e. for
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
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 bucklingmode shapes of the system.