Past exam of the elasticity and plasticity course of the University of Oxford / 2008 / 3

For an admissible variation $\eta$ with $\eta (0)=\eta (L)=0$,Stationarity for every variation therefore gives the Euler-Lagrange equationThe first energy term is bending energy: the curvature of an inextensible planar beam is ${\theta }_s$. The shortening relative to the straight reference is $\int (1-\cos \theta )ds$. A prescribed compressive dead load does positive work on that shortening, so its potential-energy contribution is minus $P$ times that integral. A minimizer must satisfy the stationary equation, but a stationary solution need not be a minimizer.

Set $\boxed{\xi =s/L,\lambda =P{L}^2/({\pi }^{2}EI)}$. The endpoint interval becomes $[0,1]$, and multiplication of the dimensional equation by $L^2/EI$ givesThe energy in these variables is $U=(EI/L){\int }_0^1[{\theta }_{\xi }^2/2-{\pi }^2\lambda (1-\cos \theta )]d\xi$. Thus $\lambda$ measures load relative to the first linearized eigenvalue for the angular endpoint conditions actually imposed.

Linearization gives ${\theta }^{\prime \prime }+{\pi }^2\lambda \theta =0$. The two homogeneous Dirichlet conditions select $\boxed{\lambda =n^2,\theta =A\sin (n\pi \xi ),n=1,2,\dots }$. For any small trial amplitude in this mode, ${\int }_0^1{\sin }^2(n\pi \xi )={\int }_0^1{\cos }^2(n\pi \xi )=1/2$, soFor $\lambda <1$ every nonzero quadratic variation is positive by the sharp Poincaré inequality. For $\lambda >1$, the first mode gives negative energy at sufficiently small amplitude, so zero cannot be a local minimum.

The printed strict “only if” conclusion omits the critical case. At $\lambda =1$ the quadratic test is degenerate, not negative. In fact zero is still the unique minimum of the full displayed functional: use $1-\cos \theta \le {\theta }^2/2$ and $\int {\theta }^{\prime 2}\ge {\pi }^2\int {\theta }^2$ to get $U\ge 0$, with strict inequality for a nonzero continuous function because $1-\cos t<t^2/2$ when $t\ne 0$. In particular on the critical first mode,Thus strict positivity of the second variation holds for $\lambda <1$, while local minimality of the full energy holds also at $\lambda =1$. This is the distinction captured by critical-load minimality of the angular Euler strut.

Near the first eigenvalue, load detuning supplies a term $ϵ{\lambda }_1\theta$, while the first nonlinearity in $\sin \theta =\theta -{\theta }^3/6+\cdots$ is cubic. Balancing them gives ${\theta }^2=O(ϵ)$, so $\boxed{\theta =O(\sqrt{ϵ})}$.

Put $\theta =\sqrt{ϵ}[A_0\sin \pi \xi +ϵ{\Theta }_1+\cdots ]$. At order ${ϵ}^{3/2}$,The operator has kernel $\sin \pi \xi$, so Fredholm solvability requires its forcing to be orthogonal to that function. Since $\int {\sin }^2\pi \xi =1/2$ and $\int {\sin }^4\pi \xi =3/8$,When this holds the nonresonant forcing is $-{\pi }^2{A}_0^3\sin 3\pi \xi /24$; for instance ${\Theta }_1=A_0^3\sin 3\pi \xi /192$ is a solution after absorbing its arbitrary first-mode component into the amplitude convention.

The branches are $\boxed{A_0=0\text{ for all }{\lambda }_1,A_0=\pm \sqrt{8{\lambda }_1}\text{ for }{\lambda }_1\ge 0}$. This is a supercritical pitchfork bifurcation. The leading reduced energy is proportional to $A_0^4-16{\lambda }_1{A}_0^2$: the nonzero branches are minima for ${\lambda }_1>0$, while zero loses minimality on that side. At the origin the quartic term still stabilizes the critical state.