Elastic instability

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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 ${\displaystyle \theta }$, the clockwise moment exerted by the force becomes ${\displaystyle M_{F}=FL\sin \theta }$. The moment equilibrium equation is given by

${\displaystyle FL\sin \theta =k_{\theta }\theta }$

where ${\displaystyle k_{\theta }}$ is the spring constant of the angular spring (Nm/radian). Assuming ${\displaystyle \theta }$ is small enough, implementing the Taylor expansion of the sine function and keeping the two first terms yields

${\displaystyle FL{\Bigg (}\theta -{\frac {1}{6}}\theta ^{3}{\Bigg )}\approx k_{\theta }\theta }$

which has three solutions, the trivial ${\displaystyle \theta =0}$, and

${\displaystyle \theta \approx \pm {\sqrt {6{\Bigg (}1-{\frac {k_{\theta }}{FL}}{\Bigg )}}}}$

which is imaginary (i.e. not physical) for ${\displaystyle FL<k_{\theta }}$ and real otherwise. This implies that for small compressive forces, the only equilibrium state is given by ${\displaystyle \theta =0}$, while if the force exceeds the value ${\displaystyle k_{\theta }/L}$ there is suddenly another mode of deformation possible.

The same result can be obtained by considering energy relations. The energy stored in the angular spring is

${\displaystyle E_{\mathrm {spring} }=\int k_{\theta }\theta \mathrm {d} \theta ={\frac {1}{2}}k_{\theta }\theta ^{2}}$

and the work done by the force is simply the force multiplied by the vertical displacement of the beam end, which is ${\displaystyle L(1-\cos \theta )}$. Thus,

${\displaystyle E_{\mathrm {force} }=\int {F\mathrm {d} x=FL(1-\cos \theta )}}$

The energy equilibrium condition ${\displaystyle E_{\mathrm {spring} }=E_{\mathrm {force} }}$ now yields ${\displaystyle F=k_{\theta }/L}$ as before (besides from the trivial ${\displaystyle \theta =0}$).

Any solution ${\displaystyle \theta }$ is stable iff a small change in the deformation angle ${\displaystyle \Delta \theta }$ results in a reaction moment trying to restore the original angle of deformation. The net clockwise moment acting on the beam is

${\displaystyle M(\theta )=FL\sin \theta -k_{\theta }\theta }$

An infinitesimal clockwise change of the deformation angle ${\displaystyle \theta }$ results in a moment

${\displaystyle M(\theta +\Delta \theta )=M+\Delta M=FL(\sin \theta +\Delta \theta \cos \theta )-k_{\theta }(\theta +\Delta \theta )}$

which can be rewritten as

${\displaystyle \Delta M=\Delta \theta (FL\cos \theta -k_{\theta })}$

since ${\displaystyle FL\sin \theta =k_{\theta }\theta }$ due to the moment equilibrium condition. Now, a solution ${\displaystyle \theta }$ is stable iff a clockwise change ${\displaystyle \Delta \theta >0}$ results in a negative change of moment ${\displaystyle \Delta M<0}$ and vice versa. Thus, the condition for stability becomes

${\displaystyle {\frac {\Delta M}{\Delta \theta }}={\frac {\mathrm {d} M}{\mathrm {d} \theta }}=FL\cos \theta -k_{\theta }<0}$

The solution ${\displaystyle \theta =0}$ is stable only for ${\displaystyle FL<k_{\theta }}$, which is expected. By expanding the cosine term in the equation, the approximate stability condition is obtained:

${\displaystyle |\theta |>{\sqrt {2{\Bigg (}1-{\frac {k_{\theta }}{FL}}{\Bigg )}}}}$

for ${\displaystyle FL>k_{\theta }}$, which the two other solutions satisfy. Hence, these solutions are stable.