Past exam of the elasticity and plasticity course of the University of Oxford / 2011 / 3 / b / Solution

Put $P=-F>0$. With $G=0$, the Euler elastica equation is $EI{\theta }^{\prime \prime }+P\sin \theta =0$. Linearization about the straight configuration gives $EI{\theta }^{\prime \prime }+P\theta =0$. Under the boundary conditions explicitly stated, $\theta (0)=\theta (L)=0$, a nonzero solution requiresThus with only the stated boundary conditions, the first critical compressive force is $\boxed{F_c=-{\pi }^{2}EI/L^2}$. The corresponding linear transverse displacement field, with $y(0)=0$, isand $x(s)=s$ to first order. These are the successive small-amplitude buckling modes. The nonlinear branches retain the simple pendulum equation; if the angular amplitude is ${\theta }_m$ and $m=\sin ({\theta }_m/2)$, fitting $n$ half-periods into the rod gives $P=4EI{n}^{2}K(m{)}^2/L^2$, which tends to $P_n$ as $m\downarrow 0$. Here $K$ is the complete elliptic integral of the first kind, with $m$ its modulus, not its squared parameter.

Endpoint position compatibility for the Euler elastica adds a condition when fully fixed clamps have equal transverse height. If the word “clamped” also imposes $y(L)=y(0)$, then ${\int }_0^L{\theta }_{n}ds=0$. With $G=0$, this retains only even $n=2j$, yieldingThe same exclusion is exact on the nonlinear branches: integrating the elastica equation gives $P\int \sin \theta ds=-EI[{\theta }^{\prime }(L)-{\theta }^{\prime }(0)]$, and an odd number of half-periods reverses the nonzero endpoint slope. Hence odd $n$ changes the endpoint height. The conventional fixed-clamp first Euler buckling load is the second boxed value, whereas the source's explicitly listed boundary conditions alone give the first. Additional clamped modes with a nonzero transverse reaction require releasing the imposed $G=0$. Load-controlled buckling also permits axial shortening; an exactly inextensible rod whose endpoint separation is fixed equal to its full length has no room to buckle.