Flamant solution
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 A. Flamant in 1892 by modifying the three-dimensional solution of Boussinesq. The stresses predicted by the Flamant solution are (in polar coordinates) where formula_2 are constants that are determined from the boundary conditions and the geometry of the wedge (i.e., the angles formula_3) and satisfy where formula_5 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 formula_6. The stresses vary as formula_7. Forces acting on a half-plane. For the special case where formula_8, formula_9, 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 formula_13 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. For a discussion of the issue  see http://imechanica.org/node/319. Displacements at the surface of the half-plane. The displacements in the formula_14 directions at the surface of the half-plane are given by where formula_17 is the Poisson's ratio, formula_18 is the shear modulus, and Derivation of Flamant solution. If we assume the stresses to vary as formula_7, we can pick terms containing formula_21 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 formula_24 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. Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius formula_25. Along the arc of the circle, the unit outward normal is formula_26 where the basis vectors are formula_27. 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 formula_25 and thereby satisfy the boundary conditions. The traction-free boundary conditions on the edges formula_31 and formula_32 also imply that except at the point formula_34. If we assume that formula_35 everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with and formula_37 along formula_38 except at the point formula_34. But the field formula_37 everywhere also satisfies the force equilibrium equations. Hence this must be the solution. Also, the assumption formula_35 implies that formula_42. Therefore, To find a  particular solution for formula_44 we have to plug in the expression for formula_44 into the force equilibrium equations to get a system of two equations which have to be solved for formula_2: Forces acting on a half-plane. If we take formula_8 and formula_9, the problem is converted into one where a normal force formula_50 and a tangential force formula_51 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. To find expressions for the displacements at the surface of the half plane, we first find the displacements for positive formula_56 (formula_57) and negative formula_56 (formula_59) keeping in mind that formula_60 along these locations. For formula_57 we have For formula_59 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