Past exam of the elasticity and plasticity course of the University of Oxford / 2023 / 2

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.

With zero tension and zero distributed load, a free interval satisfies $w^{\prime \prime \prime \prime }=0$. Before contact the entire elastic beam is free, so $w$ is a cubic polynomial. The four end boundary conditions determine it uniquely. Reflection $x\mapsto -x$ preserves those boundary conditions, so the unique solution is an even function and its odd terms vanish. Direct substitution givesFor $\alpha \ge 0$, its minimum occurs at $x=0$ and is $-\alpha L/2$. The constraint $w\ge -H$ is therefore satisfied exactly when $\alpha \le 2H/L$. The first contact is at the midpoint:At equality the normal contact force is still zero. The uniqueness used here can also be checked by integrating $v^{\prime \prime \prime \prime }v$ for the difference of two solutions with homogeneous clamped boundary conditions, which gives ${\int }_{-L}^L(v^{\prime \prime }{)}^{2}dx=0$ and hence $v=0$.

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.

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.

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.