Flamant solution

In continuum mechanics, the Flamant solution provides expressions for the stresses and displacements in a linear elastic wedge loaded by point forces at its sharp end.

This solution was developed by Alfred-Aimé Flamant [fr] in 1892[1] by modifying the three dimensional solutions for linear elasticity of Joseph Valentin Boussinesq.

The stresses predicted by the Flamant solution are (in polar coordinates) where

are constants that are determined from the boundary conditions and the geometry of the wedge (i.e., the angles

are the applied forces.

The wedge problem is self-similar and has no inherent length scale.

Also, all quantities can be expressed in the separated-variable form

, the wedge is converted into a half-plane with a normal force and a tangential force.

In that case Therefore, the stresses are and the displacements are (using Michell's solution) The

dependence of the displacements implies that the displacement grows the further one moves from the point of application of the force (and is unbounded at infinity).

This feature of the Flamant solution is confusing and appears unphysical.

directions at the surface of the half-plane are given by where

is the shear modulus, and If we assume the stresses to vary as

in the stresses from Michell's solution.

Then the Airy stress function can be expressed as Therefore, from the tables in Michell's solution, we have The constants

can then, in principle, be determined from the wedge geometry and the applied boundary conditions.

However, the concentrated loads at the vertex are difficult to express in terms of traction boundary conditions because To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge.

[3][4] Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius

Along the arc of the circle, the unit outward normal is

The tractions on the arc are Next, we examine the force and moment equilibrium in the bounded wedge and get We require that these equations be satisfied for all values of

and thereby satisfy the boundary conditions.

The traction-free boundary conditions on the edges

also imply that except at the point

everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with and

everywhere also satisfies the force equilibrium equations.

into the force equilibrium equations to get a system of two equations which have to be solved for

, the problem is converted into one where a normal force

In that case, the force equilibrium equations take the form Therefore The stresses for this situation are Using the displacement tables from the Michell solution, the displacements for this case are given by To find expressions for the displacements at the surface of the half plane, we first find the displacements for positive

θ = π

we have We can make the displacements symmetric around the point of application of the force by adding rigid body displacements (which does not affect the stresses) and removing the redundant rigid body displacements Then the displacements at the surface can be combined and take the form where