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 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 [1] in 1892 by modifying the three-dimensional solution of Boussinesq. The stresses predicted by the Flamant solution are (in polar coordinates) σ r r = 2 C 1 cos ⁡ θ r + 2 C 3 sin ⁡ θ r σ r θ = 0 σ θ θ = 0 displaystyle begin aligned sigma _ rr &= frac 2C_ 1 cos theta r + frac 2C_ 3 sin theta r \sigma _ rtheta &=0\sigma _ theta theta &=0end aligned where C 1 , C 3 displaystyle C_ 1 ,C_ 3 are constants that are determined from the boundary conditions and the geometry of the wedge (i.e., the angles α , β displaystyle alpha ,beta ) and satisfy F 1 + 2 ∫ α β ( C 1 cos ⁡ θ + C 3 sin ⁡ θ ) cos ⁡ θ d θ = 0 F 2 + 2 ∫ α β ( C 1 cos ⁡ θ + C 3 sin ⁡ θ ) sin ⁡ θ d θ = 0 displaystyle begin aligned F_ 1 &+2int _ alpha ^ beta (C_ 1 cos theta +C_ 3 sin theta ),cos theta ,dtheta =0\F_ 2 &+2int _ alpha ^ beta (C_ 1 cos theta +C_ 3 sin theta ),sin theta ,dtheta =0end aligned where F 1 , F 2 displaystyle F_ 1 ,F_ 2 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 σ = f ( r ) g ( θ ) displaystyle sigma =f(r)g(theta ) . The stresses vary as ( 1 / r ) displaystyle (1/r) .Contents1 Forces acting on a half-plane1.1 Displacements at the surface of the half-plane2 Derivation of Flamant solution2.1 Forces acting on a half-plane2.1.1 Displacements at the surface of the half-plane3 References 4 See alsoForces acting on a half-planeElastic half-plane loaded by two point forces.For the special case where α = − π displaystyle alpha =-pi , β = 0 displaystyle beta =0 , the wedge is converted into a half-plane with a normal force and a tangential force. In that case C 1 = − F 1 π , C 3 = − F 2 π displaystyle C_ 1 =- frac F_ 1 pi ,quad C_ 3 =- frac F_ 2 pi Therefore the stresses are σ r r = − 2 π r ( F 1 cos ⁡ θ + F 2 sin ⁡ θ ) σ r θ = 0 σ θ θ = 0 displaystyle begin aligned sigma _ rr &=- frac 2 pi ,r (F_ 1 cos theta +F_ 2 sin theta )\sigma _ rtheta &=0\sigma _ theta theta &=0end aligned and the displacements are (using Michell's solution) u r = − 1 4 π μ [ F 1 ( κ − 1 ) θ sin ⁡ θ − cos ⁡ θ + ( κ + 1 ) ln ⁡ r cos ⁡ θ + F 2 ( κ − 1 ) θ cos ⁡ θ + sin ⁡ θ − ( κ + 1 ) ln ⁡ r sin ⁡ θ ] u θ = − 1 4 π μ [ F 1 ( κ − 1 ) θ cos ⁡ θ − sin ⁡ θ − ( κ + 1 ) ln ⁡ r sin ⁡ θ − F 2 ( κ − 1 ) θ sin ⁡ θ + cos ⁡ θ + ( κ + 1 ) ln ⁡ r cos ⁡ θ ] displaystyle begin aligned u_ r &=- cfrac 1 4pi mu left[F_ 1 (kappa -1)theta sin theta -cos theta +(kappa +1)ln rcos theta +right.\&qquad qquad left.F_ 2 (kappa -1)theta cos theta +sin theta -(kappa +1)ln rsin theta right]\u_ theta &=- cfrac 1 4pi mu left[F_ 1 (kappa -1)theta cos theta -sin theta -(kappa +1)ln rsin theta -right.\&qquad qquad left.F_ 2 (kappa -1)theta sin theta +cos theta +(kappa +1)ln rcos theta right]end aligned The ln ⁡ r displaystyle ln r 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 x 1 , x 2 displaystyle x_ 1 ,x_ 2 directions at the surface of the half-plane are given by u 1 = F 1 ( κ + 1 ) ln ⁡ x 1 4 π μ + F 2 ( κ + 1 ) sign ( x 1 ) 8 μ u 2 = F 2 ( κ + 1 ) ln ⁡ x 1 4 π μ + F 1 ( κ + 1 ) sign ( x 1 ) 8 μ displaystyle begin aligned u_ 1 &= frac F_ 1 (kappa +1)ln x_ 1 4pi mu + frac F_ 2 (kappa +1) text sign (x_ 1 ) 8mu \u_ 2 &= frac F_ 2 (kappa +1)ln x_ 1 4pi mu + frac F_ 1 (kappa +1) text sign (x_ 1 ) 8mu end aligned where κ = 3 − 4 ν plane strain 3 − ν 1 + ν plane stress displaystyle kappa = begin cases 3-4nu &qquad text plane strain \ cfrac 3-nu 1+nu &qquad text plane stress end cases ν displaystyle nu is the Poisson's ratio, μ displaystyle mu is the shear modulus, and sign ( x ) = + 1 x > 0 − 1 x < 0 displaystyle text sign (x)= begin cases +1&x>0\-1&x<0end cases Derivation of Flamant solution If we assume the stresses to vary as ( 1 / r ) displaystyle (1/r) , we can pick terms containing 1 / r displaystyle 1/r in the stresses from Michell's solution. Then the Airy stress function can be expressed as φ = C 1 r θ sin ⁡ θ + C 2 r ln ⁡ r cos ⁡ θ + C 3 r θ cos ⁡ θ + C 4 r ln ⁡ r sin ⁡ θ displaystyle varphi =C_ 1 rtheta sin theta +C_ 2 rln rcos theta +C_ 3 rtheta cos theta +C_ 4 rln rsin theta Therefore, from the tables in Michell's solution, we have σ r r = C 1 ( 2 cos ⁡ θ r ) + C 2 ( cos ⁡ θ r ) + C 3 ( 2 sin ⁡ θ r ) + C 4 ( sin ⁡ θ r ) σ r θ = C 2 ( sin ⁡ θ r ) + C 4 ( − cos ⁡ θ r ) σ θ θ = C 2 ( cos ⁡ θ r ) + C 4 ( sin ⁡ θ r ) displaystyle begin aligned sigma _ rr &=C_ 1 left( frac 2cos theta r right)+C_ 2 left( frac cos theta r right)+C_ 3 left( frac 2sin theta r right)+C_ 4 left( frac sin theta r right)\sigma _ rtheta &=C_ 2 left( frac sin theta r right)+C_ 4 left( frac -cos theta r right)\sigma _ theta theta &=C_ 2 left( frac cos theta r right)+C_ 4 left( frac sin theta r right)end aligned The constants C 1 , C 2 , C 3 , C 4 displaystyle C_ 1 ,C_ 2 ,C_ 3 ,C_ 4 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 becausethe unit outward normal at the vertex is undefined the forces are applied at a point (which has zero area) and hence the traction at that point is infinite.Bounded elastic wedge for equilibrium of forces and moments.To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge.[2][3] Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius a displaystyle a, . Along the arc of the circle, the unit outward normal is n = e r displaystyle mathbf n =mathbf e _ r where the basis vectors are ( e r , e θ ) displaystyle (mathbf e _ r ,mathbf e _ theta ) . The tractions on the arc are t = σ ⋅ n ⟹ t r = σ r r ,   t θ = σ r θ   . displaystyle mathbf t = boldsymbol sigma cdot mathbf n quad implies t_ r =sigma _ rr ,~t_ theta =sigma _ rtheta ~. Next, we examine the force and moment equilibrium in the bounded wedge and get ∑ f 1 = F 1 + ∫ α β [ σ r r ( a , θ )   cos ⁡ θ − σ r θ ( a , θ )   sin ⁡ θ ]   a   d θ = 0 ∑ f 2 = F 2 + ∫ α β [ σ r r ( a , θ )   sin ⁡ θ + σ r θ ( a , θ )   cos ⁡ θ ]   a   d θ = 0 ∑ m 3 = ∫ α β [ a   σ r θ ( a , θ ) ]   a   d θ = 0 displaystyle begin aligned sum f_ 1 &=F_ 1 +int _ alpha ^ beta left[sigma _ rr (a,theta )~cos theta -sigma _ rtheta (a,theta )~sin theta right]~a~dtheta =0\sum f_ 2 &=F_ 2 +int _ alpha ^ beta left[sigma _ rr (a,theta )~sin theta +sigma _ rtheta (a,theta )~cos theta right]~a~dtheta =0\sum m_ 3 &=int _ alpha ^ beta left[a~sigma _ rtheta (a,theta )right]~a~dtheta =0end aligned We require that these equations be satisfied for all values of a displaystyle a, and thereby satisfy the boundary conditions. The traction-free boundary conditions on the edges θ = α displaystyle theta =alpha and θ = β displaystyle theta =beta also imply that σ r θ = σ θ θ = 0 at     θ = α , θ = β displaystyle sigma _ rtheta =sigma _ theta theta =0qquad text at ~~theta =alpha ,theta =beta except at the point r = 0 displaystyle r=0 . If we assume that σ r θ = 0 displaystyle sigma _ rtheta =0 everywhere, then the traction-free conditions and the moment equilibrium equation are satisfied and we are left with F 1 + ∫ α β σ r r ( a , θ )   a   cos ⁡ θ   d θ = 0 F 2 + ∫ α β σ r r ( a , θ )   a   sin ⁡ θ   d θ = 0 displaystyle begin aligned F_ 1 &+int _ alpha ^ beta sigma _ rr (a,theta )~a~cos theta ~dtheta =0\F_ 2 &+int _ alpha ^ beta sigma _ rr (a,theta )~a~sin theta ~dtheta =0end aligned and σ θ θ = 0 displaystyle sigma _ theta theta =0 along θ = α , θ = β displaystyle theta =alpha ,theta =beta except at the point r = 0 displaystyle r=0 . But the field σ θ θ = 0 displaystyle sigma _ theta theta =0 everywhere also satisfies the force equilibrium equations. Hence this must be the solution. Also, the assumption σ r θ = 0 displaystyle sigma _ rtheta =0 implies that C 2 = C 4 = 0 displaystyle C_ 2 =C_ 4 =0 . Therefore, σ r r = 2 C 1 cos ⁡ θ r + 2 C 3 sin ⁡ θ r   ;     σ r θ = 0   ;     σ θ θ = 0 displaystyle sigma _ rr = frac 2C_ 1 cos theta r + frac 2C_ 3 sin theta r ~;~~sigma _ rtheta =0~;~~sigma _ theta theta =0 To find a particular solution for σ r r displaystyle sigma _ rr we have to plug in the expression for σ r r displaystyle sigma _ rr into the force equilibrium equations to get a system of two equations which have to be solved for C 1 , C 3 displaystyle C_ 1 ,C_ 3 : F 1 + 2 ∫ α β ( C 1 cos ⁡ θ + C 3 sin ⁡ θ )   cos ⁡ θ   d θ = 0 F 2 + 2 ∫ α β ( C 1 cos ⁡ θ + C 3 sin ⁡ θ )   sin ⁡ θ   d θ = 0 displaystyle begin aligned F_ 1 &+2int _ alpha ^ beta (C_ 1 cos theta +C_ 3 sin theta )~cos theta ~dtheta =0\F_ 2 &+2int _ alpha ^ beta (C_ 1 cos theta +C_ 3 sin theta )~sin theta ~dtheta =0end aligned Forces acting on a half-plane If we take α = − π displaystyle alpha =-pi and β = 0 displaystyle beta =0 , the problem is converted into one where a normal force F 2 displaystyle F_ 2 and a tangential force F 1 displaystyle F_ 1 act on a half-plane. In that case, the force equilibrium equations take the form F 1 + 2 ∫ − π 0 ( C 1 cos ⁡ θ + C 3 sin ⁡ θ )   cos ⁡ θ   d θ = 0 ⟹ F 1 + C 1 π = 0 F 2 + 2 ∫ − π 0 ( C 1 cos ⁡ θ + C 3 sin ⁡ θ )   sin ⁡ θ   d θ = 0 ⟹ F 2 + C 3 π = 0 displaystyle begin aligned F_ 1 &+2int _ -pi ^ 0 (C_ 1 cos theta +C_ 3 sin theta )~cos theta ~dtheta =0qquad implies F_ 1 +C_ 1 pi =0\F_ 2 &+2int _ -pi ^ 0 (C_ 1 cos theta +C_ 3 sin theta )~sin theta ~dtheta =0qquad implies F_ 2 +C_ 3 pi =0end aligned Therefore C 1 = − F 1 π   ;     C 3 = − F 2 π   . displaystyle C_ 1 =- cfrac F_ 1 pi ~;~~C_ 3 =- cfrac F_ 2 pi ~. The stresses for this situation are σ r r = − 2 π r ( F 1 cos ⁡ θ + F 2 sin ⁡ θ )   ;     σ r θ = 0   ;     σ θ θ = 0 displaystyle sigma _ rr =- frac 2 pi r (F_ 1 cos theta +F_ 2 sin theta )~;~~sigma _ rtheta =0~;~~sigma _ theta theta =0 Using the displacement tables from the Michell solution, the displacements for this case are given by u r = − 1 4 π μ [ F 1 ( κ − 1 ) θ sin ⁡ θ − cos ⁡ θ + ( κ + 1 ) ln ⁡ r cos ⁡ θ + F 2 ( κ − 1 ) θ cos ⁡ θ + sin ⁡ θ − ( κ + 1 ) ln ⁡ r sin ⁡ θ ] u θ = − 1 4 π μ [ F 1 ( κ − 1 ) θ cos ⁡ θ − sin ⁡ θ − ( κ + 1 ) ln ⁡ r sin ⁡ θ − F 2 ( κ − 1 ) θ sin ⁡ θ + cos ⁡ θ + ( κ + 1 ) ln ⁡ r cos ⁡ θ ] displaystyle begin aligned u_ r &=- cfrac 1 4pi mu left[F_ 1 (kappa -1)theta sin theta -cos theta +(kappa +1)ln rcos theta +right.\&qquad qquad left.F_ 2 (kappa -1)theta cos theta +sin theta -(kappa +1)ln rsin theta right]\u_ theta &=- cfrac 1 4pi mu left[F_ 1 (kappa -1)theta cos theta -sin theta -(kappa +1)ln rsin theta -right.\&qquad qquad left.F_ 2 (kappa -1)theta sin theta +cos theta +(kappa +1)ln rcos theta right]end aligned 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 x 1 displaystyle x_ 1 ( θ = 0 displaystyle theta =0 ) and negative x 1 displaystyle x_ 1 ( θ = π displaystyle theta =pi ) keeping in mind that r = x 1 displaystyle r=x_ 1 along these locations. For θ = 0 displaystyle theta =0 we have u r = u 1 = F 1 4 π μ [ 1 − ( κ + 1 ) ln ⁡ x 1 ] u θ = u 2 = F 2 4 π μ [ 1 + ( κ + 1 ) ln ⁡ x 1 ] displaystyle begin aligned u_ r =u_ 1 &= cfrac F_ 1 4pi mu left[1-(kappa +1)ln x_ 1 right]\u_ theta =u_ 2 &= cfrac F_ 2 4pi mu left[1+(kappa +1)ln x_ 1 right]end aligned For θ = π displaystyle theta =pi we have u r = − u 1 = − F 1 4 π μ [ 1 − ( κ + 1 ) ln ⁡ x 1 ] + F 2 4 μ ( κ − 1 ) u θ = − u 2 = F 1 4 μ ( κ − 1 ) − F 2 4 π μ [ 1 + ( κ + 1 ) ln ⁡ x 1 ] displaystyle begin aligned u_ r =-u_ 1 &=- cfrac F_ 1 4pi mu left[1-(kappa +1)ln x_ 1 right]+ cfrac F_ 2 4mu (kappa -1)\u_ theta =-u_ 2 &= cfrac F_ 1 4mu (kappa -1)- cfrac F_ 2 4pi mu left[1+(kappa +1)ln x_ 1 right]end aligned 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) u 1 = F 2 8 μ ( κ − 1 )   ;     u 2 = F 1 8 μ ( κ − 1 ) displaystyle u_ 1 = cfrac F_ 2 8mu (kappa -1)~;~~u_ 2 = cfrac F_ 1 8mu (kappa -1) and removing the redundant rigid body displacements u 1 = F 1 4 π μ   ;     u 2 = F 2 4 π μ   . displaystyle u_ 1 = cfrac F_ 1 4pi mu ~;~~u_ 2 = cfrac F_ 2 4pi mu ~. Then the displacements at the surface can be combined and take the form u 1 = F 1 4 π μ ( κ + 1 ) ln ⁡ x 1 + F 2 8 μ ( κ − 1 ) sign ( x 1 ) u 2 = F 2 4 π μ ( κ + 1 ) ln ⁡ x 1 + F 1 8 μ ( κ − 1 ) sign ( x 1 ) displaystyle begin aligned u_ 1 &= cfrac F_ 1 4pi mu (kappa +1)ln x_ 1 + cfrac F_ 2 8mu (kappa -1) text sign (x_ 1 )\u_ 2 &= cfrac F_ 2 4pi mu (kappa +1)ln x_ 1 + cfrac F_ 1 8mu (kappa -1) text sign (x_ 1 )end aligned where sign ( x ) = + 1 x > 0 − 1 x < 0 displaystyle text sign (x)= begin cases +1&x>0\-1&x<0end cases References^ A. Flamant. (1892). Sur la rĂ©partition des pressions dans un solide rectangulaire chargĂ© transversalement. Compte. Rendu. Acad. Sci. Paris, vol. 114, p. 1465. ^ Slaughter, W. S. (2002). The Linearized Theory of Elasticity. Birkhauser, Boston, p. 294. ^ J. R. Barber, 2002, Elasticity: 2nd Edition, Kluwer Academic Publishers.See alsoMichell solution Linear elasticit

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