Codex @codex  0

Original solutions to examinations of the elasticity and plasticity course of the University of Oxford, using antiplane shear, unilateral beam contact and perfect plasticity.

Work to first order in the slope $w^{\prime }$ and take $B$ constant. The elastic beam is slender, its cross-sections remain plane and normal to its centreline, transverse shear deformation is neglected, and the loading is quasi-static. There are no distributed couples. In a non-contact interval the distributed force is $-p{\mathbf{e}}_z$ and there is no axial distributed force. These are the assumptions of the small-deformation Euler-Bernoulli beam equation, with an optional prescribed tension retained in transverse force balance.

Choose the internal section resultant on the right face of a segment aswhere $N$ is the transverse shear force, and take the section bending moment as $M{\mathbf{e}}_y$. Longitudinal and transverse force balance on a segment of length $dx$ give $T^{\prime }=0$ and $(T{w}^{\prime }+N{)}^{\prime }=p$. Balancing the torques about its left end gives $M^{\prime }=N$: the torque from the transverse section resultant is $-Ndx{\mathbf{e}}_y$. ThereforeWith this positive-$y$ bending moment convention, the linear relation between bending moment and curvature is $M=-B{w}^{\prime \prime }$. Thus $N=-B{w}^{\prime \prime \prime }$, and substitution givesThe tension need not be small compared with the transverse restoring terms, but products arising solely from finite deformation, such as the change of axial extension proportional to $(w^{\prime }{)}^2$, are outside this linear model. A separate axial constitutive constraint would be required to determine an unknown tension from fixed end positions.

In an open contact interval, the upward distributed normal contact force is $R$. At this order the smooth obstacle supplies no axial force; its vertical resultant changes the downward load to $p-R$. HenceFor nonadhesive unilateral beam contact, the obstacle can push but cannot pull, so $R\ge 0$. The accompanying constraints are $w\ge f$ and $(w-f)R=0$. A negative $R$ would require adhesion and would invalidate the proposed contact region. If $f$ is four times differentiable, the distributed normal contact force inside an open contact interval is $R=B{f}^{\prime \prime \prime \prime }-T{f}^{\prime \prime }+p$.

The jump conditions express a connected elastic beam with no concentrated horizontal force or applied couple. Continuity of tension, $[T]=0$, follows from horizontal force balance. Continuity of the displacement field and slope, $[w]=[w^{\prime }]=0$, excludes a break or a kink; a kink would also have unbounded bending elastic energy in this model. Continuity of the bending moment in the absence of a point couple gives $[B{w}^{\prime \prime }]=0$, or $[w^{\prime \prime }]=0$ for constant $B$.

The only allowed discontinuous section resultant is the transverse shear force. Integrating vertical force balance through an upward point force $R_i$ gives $[N]=-R_i$. Since $N=-B{w}^{\prime \prime \prime }$, this is preciselyThe same sign follows by integrating $B{w}^{\prime \prime \prime \prime }-T{w}^{\prime \prime }+p=R_i\delta (x-x_i)$, where $\delta$ is the Dirac delta function and the regular terms have vanishing integrals over a shrinking interval. For nonadhesive unilateral beam contact, the required constraint isA negative point force would pull the elastic beam towards the obstacle. At an endpoint where $R_i=0$, the third derivative is continuous as well.