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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.