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 in 1892 by modifying the three dimensional solutions for linear elasticity of Joseph Valentin Boussinesq.

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

) and satisfy

are the applied forces.

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

Forces acting on a half-plane

, 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)

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.

Displacements at the surface of the half-plane

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

where

is the Poisson's ratio is the shear modulus, and

Derivation

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

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

1. the unit outward normal at the vertex is undefined

2. the forces are applied at a point (which has zero area) and hence the traction at that point is infinite.

To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge.. The tractions on the arc are

Next, we examine the force and moment equilibrium in the bounded wedge and get

and thereby satisfy the boundary conditions.

The traction-free boundary conditions also imply that

.

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

.

Therefore,

:

Forces acting on a half-plane

act on a half-plane. 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

Displacements at the surface of the half-plane

along these locations.

we have

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

References