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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.

Let the contact interval be $[-s,s]$, where $w=-H$. Its slope and curvature vanish. Continuity of the displacement field, slope and bending moment at $x=s$ givesOn the free interval $s<x<L$, the cubic polynomial satisfying these three conditions is $w=-H+C(x-s{)}^3$. Put $d=L-s$. The end boundary conditions becomeConsequentlyThis requires $\alpha \ge 3H/L$, with $s=0$ at the transition. The full displacement field isIt satisfies nonpenetration. In the open contact interval the distributed normal contact force is $R=B{w}^{\prime \prime \prime \prime }=0$, since the obstacle is flat and $p=T=0$. The normal contact force is concentrated at the two endpoints. At $x=s$, $w^{\prime \prime \prime }$ jumps from zero to $6H/d^3$; at $x=-s$, it jumps from $-6H/d^3$ to zero. Both upward point forces are thereforeAt $s=0$ they merge into the single midpoint point force of the preceding regime. These formulae belong to the small-slope model; interpreting arbitrarily large $\alpha$ as a finite physical rotation would exceed its assumptions.

Seek an even function with isolated contact at $x=0$. On $0\le x\le L$, the contact and reflection conditions impose $w(0)=-H$ and $w^{\prime }(0)=0$, while the free interval still has $w^{\prime \prime \prime \prime }=0$. WriteThe two end boundary conditions giveand therefore the full displacement field isThe jump condition at zero gives the upward point forceIt is nonnegative precisely when $\alpha \ge 2H/L$. In that range $A_3\ge 0$. Nonpenetration near zero requires $A_2\ge 0$, equivalently $\alpha \le 3H/L$. Conversely, when both inequalities hold,with strict inequality away from zero. Thus this is an admissible isolated-contact solution throughoutAt the upper endpoint the contact curvature $w^{\prime \prime }(0)=2A_2$ vanishes, and the profile remains above the obstacle for $x\ne 0$ because $A_3>0$. Increasing $\alpha$ further would make $A_2<0$, forcing penetration near zero. An interval of contact must then develop.

The admissible solution is unique within this linear model: the bending elastic energy $B\int (w^{\prime \prime }{)}^2/2$ is strictly convex on the convex set of admissible profiles with the prescribed end data. Integration by parts shows that a solution with nonnegative contact point forces satisfies the corresponding minimum condition, since every admissible variation lifts a point already on the obstacle. Reflection therefore justifies the assumed even function also after contact.

The net upward force is the sum of the contact point forces, since the distributed normal contact force vanishes in every regime here. Before contact it is zero. During isolated contact it is $R_0=12B(\alpha L-2H)/L^3$. For an interval of contact it is $R_{-s}+R_s=12BH/d^3$, with $d=3H/\alpha$. ThereforeThe formulae agree at both transitions. At $\alpha =3H/L$ they give $12BH/L^3$, and their first derivatives also agree there. As a check on the sign and normalization, integrating the Euler-Bernoulli beam equation over the whole elastic beam gives $F_{\mathrm{up}}=B(w^{\prime \prime \prime }(L^{-})-w^{\prime \prime \prime }(-L^{+}))$, which yields the same values.

An elastic beam above a smooth obstacle cannot penetrate it, and a nonadhesive obstacle can only push. The displayed constraints express this when $R$ is an upward distributed normal contact force; concentrated point forces must also be nonnegative. In the small-slope Euler-Bernoulli beam equation,Integrating across a contact point with no applied couple gives jump conditions $[w]=[w^{\prime }]=[B{w}^{\prime \prime }]=0$ and $[B{w}^{\prime \prime \prime }]=R_i$, provided the tension is continuous. For a flat obstacle, no distributed load and zero tension, an open contact interval has $R=0$; its entire normal contact force can therefore be concentrated at its endpoints.