Stresses in wedges

## Contents

* 1 Wedge with Boundary Tractions
* 2 Williams' Asymptotic Solution
* 3 Stresses near the notch corner
* 4 Eigenvalues of λ
* 4.1 Special case : α = π = 180°
* 5 Axially Loaded Wedge
* 5.1 What about the concentrated force BC?
* 6 Solution Procedure
* 6.1 Special Case : β = π/2
* 7 Displacements
* 8 The Flamant Solution
* 8.1 Special Case : α = -π, β = 0
* 9 Related Content

## Wedge with Boundary Tractions[edit | edit source]

Elastic wedge with normal and shear surface tractions

Suppose

* The tractions on the boundary vary as  r n {\displaystyle r^{n}} .
* No body forces.

Then

(1) φ = r n + 2 f ( θ ) {\displaystyle {\text{(1)}}\qquad \varphi =r^{n+2}f(\theta )}

To find  f ( θ ) {\displaystyle f(\theta )} plug into  ∇ 4 φ = 0 {\displaystyle \nabla ^{4}{\varphi }=0} .

(2) ( d 2 d θ 2 + ( n + 2 ) 2 ) ( d 2 d θ 2 + n 2 ) f ( θ ) = 0 {\displaystyle {\text{(2)}}\qquad \left({\frac {d^{2}}{d\theta ^{2}}}+(n+2)^{2}\right)\left({\frac {d^{2}}{d\theta ^{2}}}+n^{2}\right)f(\theta )=0}

If  n ≠ 0 {\displaystyle n\neq 0} and  n ≠ − 2 {\displaystyle n\neq -2} ,

(3) φ = r n + 2 [ a 1 cos ⁡ { ( n + 2 ) θ } + a 2 cos ⁡ ( n θ ) + a 3 sin ⁡ { ( n + 2 ) θ } + a 4 sin ⁡ ( n θ ) ] {\displaystyle {\text{(3)}}\qquad \varphi =r^{n+2}\left[a_{1}\cos\\{(n+2)\theta \\}+a_{2}\cos(n\theta )+a_{3}\sin\\{(n+2)\theta \\}+a_{4}\sin(n\theta )\right]}

The corresponding stresses and displacements can be found from Michell's solution. We have to take special care for the case where  n = 0 {\displaystyle n=0} , i.e., the traction on the surface is constant.

## Williams' Asymptotic Solution[edit | edit source]

Ref: M.L. Williams, ASME J. Appl. Mech., v. 19 (1952), 526-528.

The Williams' solution

* Stress concentration at the notch.
* Singularity at the sharp corner, i.e,  σ i j -> ∞ {\displaystyle \sigma _{ij}\rightarrow \infty } .
* William's solution involves defining the origin at the corner and expanding the stress field as an asymptotic series in powers of r.
* If the stresses (and strains) vary with  r α {\displaystyle r^{\alpha }} as we approach the point  r = 0 {\displaystyle r=0} , the strain energy is given by

(4) U = 1 2 ∫ 0 2 π ∫ 0 r σ i j ε i j r d r d θ = C ∫ 0 r r 2 a + 1 d r {\displaystyle {\text{(4)}}\qquad U={\frac {1}{2}}\int _{0}^{2\pi }\int _{0}^{r}\sigma _{ij}\varepsilon _{ij}~r~dr~d\theta =C\int _{0}^{r}r^{2a+1}dr}

This integral is bounded only if  a > − 1 {\displaystyle a>-1} . Hence, singular stress fields are acceptable only if the exponent on the stress components exceeds  − 1 {\displaystyle -1} .

## Stresses near the notch corner[edit | edit source]

* Use a separated-variable series as in equation (3).
* Each of the terms satisfies the traction-free BCs on the surface of the notch.
* Relax the requirement that  n {\displaystyle n} in equation (3) is an integer. Let  n = λ − 1 {\displaystyle n=\lambda -1} .

(5) φ = r λ + 1 [ a 1 cos ⁡ { ( λ + 1 ) θ } + a 2 cos ⁡ ( λ − 1 ) θ + a 3 sin ⁡ { ( λ + 1 ) θ } + a 4 sin ⁡ ( λ − 1 ) θ ] {\displaystyle {\begin{aligned}{\text{(5)}}\qquad \varphi =r^{\lambda +1}\left[\right.&a_{1}\cos\\{(\lambda +1)\theta \\}+a_{2}\cos {(\lambda -1)\theta }+\\\&a_{3}\sin\\{(\lambda +1)\theta \\}+a_{4}\sin {(\lambda -1)\theta }\left.\right]\end{aligned}}}

The stresses are

σ r r = r λ − 1 [ − a 1 λ ( λ + 1 ) cos ⁡ { ( λ + 1 ) θ } − a 2 λ ( λ − 3 ) cos ⁡ { ( λ − 1 ) θ } − a 3 λ ( λ + 1 ) sin ⁡ { ( λ + 1 ) θ } − a 4 λ ( λ − 3 ) sin ⁡ { ( λ − 1 ) θ } ] (6) σ r θ = r λ − 1 [ + a 1 λ ( λ + 1 ) sin ⁡ { ( λ + 1 ) θ } + a 2 λ ( λ − 1 ) sin ⁡ { ( λ − 1 ) θ } − a 3 λ ( λ + 1 ) cos ⁡ { ( λ + 1 ) θ } − a 4 λ ( λ − 1 ) cos ⁡ { ( λ − 1 ) θ } ] (7) σ θ θ = r λ − 1 [ + a 1 λ ( λ + 1 ) cos ⁡ { ( λ + 1 ) θ } + a 2 λ ( λ + 1 ) cos ⁡ { ( λ − 1 ) θ } + a 3 λ ( λ + 1 ) sin ⁡ { ( λ + 1 ) θ } + a 4 λ ( λ + 1 ) sin ⁡ { ( λ − 1 ) θ } ] (8) {\displaystyle {\begin{aligned}\sigma _{rr}=r^{\lambda -1}\left[\right.&-a_{1}\lambda (\lambda +1)\cos\\{(\lambda +1)\theta \\}-a_{2}\lambda (\lambda -3)\cos\\{(\lambda -1)\theta \\}\\\&-a_{3}\lambda (\lambda +1)\sin\\{(\lambda +1)\theta \\}-a_{4}\lambda (\lambda -3)\sin\\{(\lambda -1)\theta \\}\left.\right]{\text{(6)}}\qquad \\\\\sigma _{r\theta }=r^{\lambda -1}\left[\right.&+a_{1}\lambda (\lambda +1)\sin\\{(\lambda +1)\theta \\}+a_{2}\lambda (\lambda -1)\sin\\{(\lambda -1)\theta \\}\\\&-a_{3}\lambda (\lambda +1)\cos\\{(\lambda +1)\theta \\}-a_{4}\lambda (\lambda -1)\cos\\{(\lambda -1)\theta \\}\left.\right]{\text{(7)}}\qquad \\\\\sigma _{\theta \theta }=r^{\lambda -1}\left[\right.&+a_{1}\lambda (\lambda +1)\cos\\{(\lambda +1)\theta \\}+a_{2}\lambda (\lambda +1)\cos\\{(\lambda -1)\theta \\}\\\&+a_{3}\lambda (\lambda +1)\sin\\{(\lambda +1)\theta \\}+a_{4}\lambda (\lambda +1)\sin\\{(\lambda -1)\theta \\}\left.\right]{\text{(8)}}\qquad \end{aligned}}}

The BCs are  σ r θ = σ θ θ = 0 {\displaystyle \sigma _{r\theta }=\sigma _{\theta \theta }=0} at  θ = α {\displaystyle \theta =\alpha } .Hence,

0 = r λ − 1 λ [ + a 1 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } + a 2 ( λ − 1 ) sin ⁡ { ( λ − 1 ) α } − a 3 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } − a 4 ( λ − 1 ) cos ⁡ { ( λ − 1 ) α } ] (9) 0 = r λ − 1 λ [ − a 1 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } − a 2 ( λ − 1 ) sin ⁡ { ( λ − 1 ) α } − a 3 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } − a 4 ( λ − 1 ) cos ⁡ { ( λ − 1 ) α } ] (10) {\displaystyle {\begin{aligned}0=r^{\lambda -1}\lambda \left[\right.&+a_{1}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}+a_{2}(\lambda -1)\sin\\{(\lambda -1)\alpha \\}\\\&-a_{3}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}-a_{4}(\lambda -1)\cos\\{(\lambda -1)\alpha \\}\left.\right]{\text{(9)}}\qquad \\\0=r^{\lambda -1}\lambda \left[\right.&-a_{1}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}-a_{2}(\lambda -1)\sin\\{(\lambda -1)\alpha \\}\\\&-a_{3}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}-a_{4}(\lambda -1)\cos\\{(\lambda -1)\alpha \\}\left.\right]{\text{(10)}}\qquad \end{aligned}}}

The BCs are  σ r θ = σ θ θ = 0 {\displaystyle \sigma _{r\theta }=\sigma _{\theta \theta }=0} at  θ = − α {\displaystyle \theta =-\alpha } .Hence,

0 = r λ − 1 λ [ + a 1 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } + a 2 ( λ + 1 ) cos ⁡ { ( λ − 1 ) α } + a 3 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } + a 4 ( λ + 1 ) sin ⁡ { ( λ − 1 ) α } ] (11) 0 = r λ − 1 λ [ + a 1 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } + a 2 ( λ + 1 ) cos ⁡ { ( λ − 1 ) α } − a 3 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } − a 4 ( λ + 1 ) sin ⁡ { ( λ − 1 ) α } ] (12) {\displaystyle {\begin{aligned}0=r^{\lambda -1}\lambda \left[\right.&+a_{1}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}+a_{2}(\lambda +1)\cos\\{(\lambda -1)\alpha \\}\\\&+a_{3}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}+a_{4}(\lambda +1)\sin\\{(\lambda -1)\alpha \\}\left.\right]{\text{(11)}}\qquad \\\0=r^{\lambda -1}\lambda \left[\right.&+a_{1}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}+a_{2}(\lambda +1)\cos\\{(\lambda -1)\alpha \\}\\\&-a_{3}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}-a_{4}(\lambda +1)\sin\\{(\lambda -1)\alpha \\}\left.\right]{\text{(12)}}\qquad \end{aligned}}}

The above equations will have non-trivial solutions only for certain eigenvalues of  λ {\displaystyle \lambda } , one of which is  λ = 0 {\displaystyle \lambda =0} . Using the symmetries of the equations, we can partition the coefficient matrix.

## Eigenvalues of λ[edit | edit source]

Adding equations (9) and (10),

(13) a 3 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } + a 4 ( λ − 1 ) cos ⁡ { ( λ − 1 ) α } = 0 {\displaystyle {\text{(13)}}\qquad a_{3}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}+a_{4}(\lambda -1)\cos\\{(\lambda -1)\alpha \\}=0}

Subtracting equation (10) from (9),

(14) a 1 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } + a 2 ( λ − 1 ) sin ⁡ { ( λ − 1 ) α } = 0 {\displaystyle {\text{(14)}}\qquad a_{1}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}+a_{2}(\lambda -1)\sin\\{(\lambda -1)\alpha \\}=0}

Adding equations (11) and (12),

(15) a 1 ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } + a 2 ( λ + 1 ) cos ⁡ { ( λ − 1 ) α } = 0 {\displaystyle {\text{(15)}}\qquad a_{1}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}+a_{2}(\lambda +1)\cos\\{(\lambda -1)\alpha \\}=0}

Subtracting equation (12) from (11),

(16) a 3 ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } + a 4 ( λ + 1 ) sin ⁡ { ( λ − 1 ) α } = 0 {\displaystyle {\text{(16)}}\qquad a_{3}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}+a_{4}(\lambda +1)\sin\\{(\lambda -1)\alpha \\}=0}

Therefore, the two independent sets of equations are

(17) [ ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } ( λ − 1 ) sin ⁡ { ( λ − 1 ) α } ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } ( λ + 1 ) cos ⁡ { ( λ − 1 ) α } ] [ a 1 a 2 ] = [ 0 0 ] {\displaystyle {\text{(17)}}\qquad {\begin{bmatrix}(\lambda +1)\sin\\{(\lambda +1)\alpha \\}&(\lambda -1)\sin\\{(\lambda -1)\alpha \\}\\\\(\lambda +1)\cos\\{(\lambda +1)\alpha \\}&(\lambda +1)\cos\\{(\lambda -1)\alpha \\}\end{bmatrix}}{\begin{bmatrix}a_{1}\\\a_{2}\end{bmatrix}}={\begin{bmatrix}0\\\0\end{bmatrix}}}

and

(18) [ ( λ + 1 ) cos ⁡ { ( λ + 1 ) α } ( λ − 1 ) cos ⁡ { ( λ − 1 ) α } ( λ + 1 ) sin ⁡ { ( λ + 1 ) α } ( λ + 1 ) sin ⁡ { ( λ − 1 ) α } ] [ a 3 a 4 ] = [ 0 0 ] {\displaystyle {\text{(18)}}\qquad {\begin{bmatrix}(\lambda +1)\cos\\{(\lambda +1)\alpha \\}&(\lambda -1)\cos\\{(\lambda -1)\alpha \\}\\\\(\lambda +1)\sin\\{(\lambda +1)\alpha \\}&(\lambda +1)\sin\\{(\lambda -1)\alpha \\}\end{bmatrix}}{\begin{bmatrix}a_{3}\\\a_{4}\end{bmatrix}}={\begin{bmatrix}0\\\0\end{bmatrix}}}

Equations (17) have a non-trivial solution only if

(19) λ sin ⁡ ( 2 α ) + sin ⁡ ( 2 λ α ) = 0 {\displaystyle {\text{(19)}}\qquad \lambda \sin(2\alpha )+\sin(2\lambda \alpha )=0}

Equations (18) have a non-trivial solution only if

(20) λ sin ⁡ ( 2 α ) − sin ⁡ ( 2 λ α ) = 0 {\displaystyle {\text{(20)}}\qquad \lambda \sin(2\alpha )-\sin(2\lambda \alpha )=0}

* From equation (4), acceptable singular stress fields must have  λ > 0 {\displaystyle \lambda >0} .Hence,  λ = 0 {\displaystyle \lambda =0} is not acceptable.
* The term with the smallest eigenvalue of  λ {\displaystyle \lambda } dominates the solution. Hence, this eigenvalue is what we seek.
* λ = 1 {\displaystyle \lambda =1} leads to  φ = a 4 sin ⁡ ( 0 ) {\displaystyle \varphi =a_{4}\sin(0)} . Unacceptable.
* We can find the eiegnvalues for general wedge angles using graphical methods.

### Special case : α = π = 180°[edit | edit source]

In this case, the wedge becomes a crack.In this case,

(21) λ = 1 2 , 1 , 3 2 , {\displaystyle {\text{(21)}}\qquad \lambda ={\frac {1}{2}},1,{\frac {3}{2}},}

The lowest eigenvalue is  1 / 2 {\displaystyle 1/2} . If we use, this value in equation (17), then the two equations will not be linearly independent and we can express them as one equation with the substitutions

(22) a 1 = A 2 sin ⁡ ( α 2 ) ; a 2 = − 3 A 2 sin ⁡ ( 3 α 2 ) {\displaystyle {\text{(22)}}\qquad a_{1}={\frac {A}{2}}\sin \left({\frac {\alpha }{2}}\right)~;~~a_{2}=-{\frac {3A}{2}}\sin \left({\frac {3\alpha }{2}}\right)}

where  A {\displaystyle A} is a constant. The singular stress field at the crack tip is then

σ r r = K I 2 π r [ 5 4 cos ⁡ ( θ 2 ) − 1 4 cos ⁡ ( 3 θ 2 ) ] (23) σ θ θ = K I 2 π r [ 3 4 cos ⁡ ( θ 2 ) + 1 4 cos ⁡ ( 3 θ 2 ) ] (24) σ r θ = K I 2 π r [ 1 4 sin ⁡ ( θ 2 ) + 1 4 sin ⁡ ( 3 θ 2 ) ] (25) {\displaystyle {\begin{aligned}\sigma _{rr}&={\frac {K_{I}}{\sqrt {2\pi r}}}\left[{\frac {5}{4}}\cos \left({\frac {\theta }{2}}\right)-{\frac {1}{4}}\cos \left({\frac {3\theta }{2}}\right)\right]{\text{(23)}}\qquad \\\\\sigma _{\theta \theta }&={\frac {K_{I}}{\sqrt {2\pi r}}}\left[{\frac {3}{4}}\cos \left({\frac {\theta }{2}}\right)+{\frac {1}{4}}\cos \left({\frac {3\theta }{2}}\right)\right]{\text{(24)}}\qquad \\\\\sigma _{r\theta }&={\frac {K_{I}}{\sqrt {2\pi r}}}\left[{\frac {1}{4}}\sin \left({\frac {\theta }{2}}\right)+{\frac {1}{4}}\sin \left({\frac {3\theta }{2}}\right)\right]{\text{(25)}}\qquad \end{aligned}}}

where,  K I {\displaystyle K_{I}} is the { Mode I Stress Intensity Factor.}

(26) K I = 3 A π 2 {\displaystyle {\text{(26)}}\qquad K_{I}=3A{\sqrt {\frac {\pi }{2}}}}

If we use equations (18) we can get the stresses due to a mode II loading.

σ r r = K I I 2 π r [ − 5 4 sin ⁡ ( θ 2 ) + 3 4 sin ⁡ ( 3 θ 2 ) ] (27) σ θ θ = K I I 2 π r [ − 3 4 sin ⁡ ( θ 2 ) − 3 4 sin ⁡ ( 3 θ 2 ) ] (28) σ r θ = K I I 2 π r [ 1 4 cos ⁡ ( θ 2 ) + 3 4 cos ⁡ ( 3 θ 2 ) ] (29) {\displaystyle {\begin{aligned}\sigma _{rr}&={\frac {K_{II}}{\sqrt {2\pi r}}}\left[-{\frac {5}{4}}\sin \left({\frac {\theta }{2}}\right)+{\frac {3}{4}}\sin \left({\frac {3\theta }{2}}\right)\right]{\text{(27)}}\qquad \\\\\sigma _{\theta \theta }&={\frac {K_{II}}{\sqrt {2\pi r}}}\left[-{\frac {3}{4}}\sin \left({\frac {\theta }{2}}\right)-{\frac {3}{4}}\sin \left({\frac {3\theta }{2}}\right)\right]{\text{(28)}}\qquad \\\\\sigma _{r\theta }&={\frac {K_{II}}{\sqrt {2\pi r}}}\left[{\frac {1}{4}}\cos \left({\frac {\theta }{2}}\right)+{\frac {3}{4}}\cos \left({\frac {3\theta }{2}}\right)\right]{\text{(29)}}\qquad \end{aligned}}}

## Axially Loaded Wedge[edit | edit source]

Elastic wedge loaded by an axial force

The BCs at  θ = ± β {\displaystyle \theta =\pm \beta } are

(30) t r = t θ = 0 ; n ^ = ± e ^ θ ⇒ σ r θ = σ θ θ = 0 {\displaystyle {\text{(30)}}\qquad t_{r}=t_{\theta }=0~;~~{\widehat {\mathbf {n} }}{}=\pm {\widehat {\mathbf {e} }}{\theta }\Rightarrow \sigma _{r\theta }=\sigma _{\theta \theta }=0}

### What about the concentrated force BC?[edit | edit source]

* What is  n ^ {\displaystyle {\widehat {\mathbf {n} }}{}} at the vertex ?
* The traction is infinite since the force is applied on zero area. Consider equilibrium of a portion of the wedge.

At  r = a {\displaystyle r=a} , the BCs are

(31) n ^ = e ^ r ⇒ σ r θ = t r ; σ θ θ = t θ {\displaystyle {\text{(31)}}\qquad {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{r}\Rightarrow \sigma _{r\theta }=t_{r}~;~~\sigma _{\theta \theta }=t_{\theta }}

For equilibrium,  ∑ F 1 = ∑ F 2 = ∑ M 3 = 0 {\displaystyle \sum F_{1}=\sum F_{2}=\sum M_{3}=0} . Therefore,

P 1 + ∫ − β β [ σ r r ( a , θ ) cos ⁡ θ − σ r θ ( a , θ ) sin ⁡ θ ] a d θ = 0 (32) ∫ − β β [ σ r r ( a , θ ) sin ⁡ θ + σ r θ ( a , θ ) cos ⁡ θ ] a d θ = 0 (33) ∫ − β β [ a σ r θ ( a , θ ) ] a d θ = 0 (34) {\displaystyle {\begin{aligned}P_{1}+\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\cos \theta -\sigma _{r\theta }(a,\theta )\sin \theta \right]a~d\theta =0{\text{(32)}}\qquad \\\\\int _{-\beta }^{\beta }\left[\sigma _{rr}(a,\theta )\sin \theta +\sigma _{r\theta }(a,\theta )\cos \theta \right]a~d\theta =0{\text{(33)}}\qquad \\\\\int _{-\beta }^{\beta }\left[a\sigma _{r\theta }(a,\theta )\right]a~d\theta =0{\text{(34)}}\qquad \end{aligned}}}

These constraint conditions are equivalent to the concentrated force BC.

## Solution Procedure[edit | edit source]

Assume that  σ r θ ( r , θ ) = 0 {\displaystyle \sigma _{r\theta }(r,\theta )=0} . This satisfies the traction BCs on  θ = ± β {\displaystyle \theta =\pm \beta } and equation (34). Therefore,

(35) σ r θ = ∂ ∂ r ( 1 r ∂ ∂ φ θ ) = 0 ⇒ φ = r η ( θ ) + ζ ( r ) {\displaystyle {\text{(35)}}\qquad \sigma _{r\theta }={\frac {\partial }{\partial }}{}{r}\left({\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }\right)=0\Rightarrow \varphi =r\eta (\theta )+\zeta (r)}

Hence,

(36) σ θ θ = ∂ 2 ∂ φ ∂ r = ζ ″ ( r ) {\displaystyle {\text{(36)}}\qquad \sigma _{\theta \theta }={\frac {\partial ^{2}}{\partial \varphi \partial r}}=\zeta ^{''}(r)}

That means  σ θ θ {\displaystyle \sigma _{\theta \theta }} is independent of  θ {\displaystyle \theta } . Therefore, in order to satisfy the BCs,  σ θ θ = 0 {\displaystyle \sigma _{\theta \theta }=0} , i.e.,

(37) ζ ( r ) = C 1 r + C 2 ⇒ φ = r η ( θ ) + C 1 r = r [ η ( θ ) + C 1 ] = r ξ ( θ ) {\displaystyle {\text{(37)}}\qquad \zeta (r)=C_{1}r+C_{2}\Rightarrow \varphi =r\eta (\theta )+C_{1}r=r[\eta (\theta )+C_{1}]=r\xi (\theta )}

Checking for compatibility,  ∇ 4 φ = 0 {\displaystyle \nabla ^{4}{\varphi }=0} , we get

(38) ξ ( I V ) ( θ ) + 2 ξ ″ ( θ ) + ξ ( θ ) = 0 {\displaystyle {\text{(38)}}\qquad \xi ^{(IV)}(\theta )+2\xi ^{''}(\theta )+\xi (\theta )=0}

The general solution is

(39) ξ ( θ ) = A sin ⁡ θ + B cos ⁡ θ + C θ sin ⁡ θ + D θ cos ⁡ θ {\displaystyle {\text{(39)}}\qquad \xi (\theta )=A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta }

Therefore,

(40) φ = r [ A sin ⁡ θ + B cos ⁡ θ + C θ sin ⁡ θ + D θ cos ⁡ θ ] {\displaystyle {\text{(40)}}\qquad {\varphi =r\left[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta \right]}}

The only non-zero stress is  σ r r {\displaystyle \sigma _{rr}} .

(41) σ r r = 1 r [ 2 C cos ⁡ θ − 2 D sin ⁡ θ ] {\displaystyle {\text{(41)}}\qquad \sigma _{rr}={\frac {1}{r}}\left[2C\cos \theta -2D\sin \theta \right]}

Plugging into equation (33), we get

(42) − D [ 2 β − sin ⁡ ( 2 β ) ] = 0 ⇒ D = 0 {\displaystyle {\text{(42)}}\qquad -D\left[2\beta -\sin(2\beta )\right]=0\Rightarrow D=0}

Hence,

(43) σ r r = 2 C r cos ⁡ θ {\displaystyle {\text{(43)}}\qquad \sigma _{rr}={\frac {2C}{r}}\cos \theta }

Plugging into equation (32), we get

(44) − P = C [ 2 β + sin ⁡ ( 2 β ) ] ⇒ C = − P 2 β + sin ⁡ ( 2 β ) {\displaystyle {\text{(44)}}\qquad -P=C\left[2\beta +\sin(2\beta )\right]\Rightarrow C={\frac {-P}{2\beta +\sin(2\beta )}}}

Therefore,

(45) φ = C r θ sin ⁡ θ = − P r θ sin ⁡ θ 2 β + sin ⁡ ( 2 β ) {\displaystyle {\text{(45)}}\qquad {\varphi =Cr\theta \sin \theta ={\frac {-Pr\theta \sin \theta }{2\beta +\sin(2\beta )}}}}

The stress state is

(46) σ r r = − 2 P cos ⁡ θ r [ 2 β + sin ⁡ ( 2 β ) ] ; σ r θ = 0 ; σ θ θ = 0 {\displaystyle {\text{(46)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{r[2\beta +\sin(2\beta )]}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}}

### Special Case : β = π/2[edit | edit source]

A concentrated point load acting on a half plane.

(47) σ r r = − 2 P cos ⁡ θ π r ; σ r θ = 0 ; σ θ θ = 0 {\displaystyle {\text{(47)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{\pi r}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}}

## Displacements[edit | edit source]

2 μ u r = − ∂ ∂ φ r + α r ∂ ∂ ψ θ 2 μ u θ = − 1 r ∂ ∂ φ θ + α r 2 ∂ ∂ ψ r {\displaystyle {\begin{aligned}2\mu u_{r}&=-{\frac {\partial }{\partial }}{\varphi }{r}+\alpha r{\frac {\partial }{\partial }}{\psi }{\theta }\\\2\mu u_{\theta }&=-{\frac {1}{r}}{\frac {\partial }{\partial }}{\varphi }{\theta }+\alpha r^{2}{\frac {\partial }{\partial }}{\psi }{r}\end{aligned}}}

where

∇ 2 ψ = 0 ∂ ∂ r ( r ∂ ∂ ψ θ ) = ∇ 2 φ {\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0\\\\{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\end{aligned}}}

Plug in  φ = C r θ sin ⁡ θ {\displaystyle \varphi =Cr\theta \sin \theta } ,

∂ ∂ r ( r ∂ ∂ ψ θ ) = ∇ 2 φ ⇒ ∂ ∂ r ( r ∂ ∂ ψ θ ) = 2 C r cos ⁡ θ ⇒ r ∂ ∂ ψ θ = 2 C ln ⁡ r cos ⁡ θ + A ( θ ) ⇒ ∂ ∂ ψ θ = 2 C ln ⁡ r r cos ⁡ θ + A ( θ ) r ⇒ ψ = 2 C ln ⁡ r r sin ⁡ θ + η ( θ ) r + ξ r {\displaystyle {\begin{aligned}&{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\\\\\Rightarrow &{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)={\frac {2C}{r}}\cos \theta \\\\\Rightarrow &r{\frac {\partial }{\partial }}{\psi }{\theta }=2C\ln r\cos \theta +A(\theta )\\\\\Rightarrow &{\frac {\partial }{\partial }}{\psi }{\theta }=2C{\frac {\ln r}{r}}\cos \theta +{\frac {A(\theta )}{r}}\\\\\Rightarrow &\psi =2C{\frac {\ln r}{r}}\sin \theta +{\frac {\eta (\theta )}{r}}+\xi {r}\end{aligned}}}

Plug  ψ {\displaystyle \psi } into  ∇ 2 ψ = 0 {\displaystyle \nabla ^{2}{\psi }=0} ,

1 r 3 η ″ ( θ ) + 1 r 3 η ( θ ) + ξ ″ ( r ) + 1 r ξ ′ ( r ) − 4 C r 3 sin ⁡ θ = 0 ⇒ η ″ ( θ ) + η ( θ ) + r 3 ξ ″ ( r ) + r 2 ξ ′ ( r ) − 4 C sin ⁡ θ = 0 {\displaystyle {\begin{aligned}&{\frac {1}{r^{3}}}\eta ^{''}(\theta )+{\frac {1}{r^{3}}}\eta (\theta )+\xi ^{''}(r)+{\frac {1}{r}}\xi ^{'}(r)-{\frac {4C}{r^{3}}}\sin \theta =0\\\\\Rightarrow &\eta ^{''}(\theta )+\eta (\theta )+r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)-4C\sin \theta =0\end{aligned}}}

Hence,

η ″ ( θ ) + η ( θ ) − 4 C sin ⁡ θ = b r 3 ξ ″ ( r ) + r 2 ξ ′ ( r ) = − b r 3 {\displaystyle {\begin{aligned}\eta ^{''}(\theta )+\eta (\theta )-4C\sin \theta &=b\\\r^{3}\xi ^{''}(r)+r^{2}\xi ^{'}(r)&=-{\frac {b}{r^{3}}}\end{aligned}}}

Solving,

η ( θ ) = − 2 C θ cos ⁡ θ + d cos ⁡ θ + e sin ⁡ θ + b ξ ′ ( r ) = f r + b r 2 {\displaystyle {\begin{aligned}\eta (\theta )&=-2C\theta \cos \theta +d\cos \theta +e\sin \theta +b\\\\\xi ^{'}(r)&={\frac {f}{r}}+{\frac {b}{r^{2}}}\end{aligned}}}

Therefore,

2 μ u r = 2 α C ln ⁡ r cos ⁡ θ + ( 2 α − 1 ) C θ sin ⁡ θ + α ( e − 2 C ) cos ⁡ θ − α d sin ⁡ θ 2 μ u θ = − 2 α C ln ⁡ r sin ⁡ θ + ( 2 α − 1 ) C sin ⁡ θ + ( 2 α − 1 ) C θ cos ⁡ θ − α d cos ⁡ θ − α e sin ⁡ θ + α f r {\displaystyle {\begin{aligned}2\mu u_{r}&=2\alpha C\ln r\cos \theta +(2\alpha -1)C\theta \sin \theta +\alpha (e-2C)\cos \theta -\alpha d\sin \theta \\\2\mu u_{\theta }&=-2\alpha C\ln r\sin \theta +(2\alpha -1)C\sin \theta +(2\alpha -1)C\theta \cos \theta -\alpha d\cos \theta -\alpha e\sin \theta +\alpha fr\end{aligned}}}

To fix the rigid body motion, we set  u θ = 0 {\displaystyle u_{\theta }=0} when  θ = 0 {\displaystyle \theta =0} , and set  u r = 0 {\displaystyle u_{r}=0} when  θ = 0 {\displaystyle \theta =0} and  r = L {\displaystyle r=L} .Then,

u r = α C μ ln ⁡ ( r L ) cos ⁡ θ + ( 2 α − 1 ) C 2 μ θ sin ⁡ θ u θ = − α C μ ln ⁡ ( r L ) sin ⁡ θ + ( 2 α − 1 ) C 2 μ θ cos ⁡ θ − C 2 μ sin ⁡ θ {\displaystyle {\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\cos \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \sin \theta \\\u_{\theta }&=-{\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\sin \theta +{\frac {(2\alpha -1)C}{2\mu }}\theta \cos \theta -{\frac {C}{2\mu }}\sin \theta \end{aligned}}}

The displacements are singular at  r = 0 {\displaystyle r=0} and  r = ∞ {\displaystyle r=\infty } . At  θ = 0 {\displaystyle \theta =0} ,

u r = α C μ ln ⁡ ( r L ) u θ = 0 {\displaystyle {\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\\\u_{\theta }&=0\end{aligned}}}

Is the small strain assumption satisfied ?

## The Flamant Solution[edit | edit source]

Elastic wedge loaded by two forces at the tip

* This problem is also self-similar (no inherent length scale).
* 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)} (the area of action of the force decreases with increasing  r {\displaystyle r} ). How about a conical wedge ?

From Michell's solution, pick terms containing  1 / r {\displaystyle 1/r} in the stresses. Then,

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}r\theta \sin \theta +C_{2}r\ln r\cos \theta +C_{3}r\theta \cos \theta +C_{4}r\ln r\sin \theta }

Therefore, from Tables,

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 {2\cos \theta }{r}}\right)+C_{2}\left({\frac {\cos \theta }{r}}\right)+C_{3}\left({\frac {2\sin \theta }{r}}\right)+C_{4}\left({\frac {\sin \theta }{r}}\right)\\\\\sigma _{r\theta }&=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}}}

From traction BCs,  C 2 = C 4 = 0 {\displaystyle C_{2}=C_{4}=0} . From equilibrium,

F 1 + 2 ∫ α β ( C 1 cos ⁡ θ − C 3 sin ⁡ θ a ) a cos ⁡ θ d θ = 0 F 2 + 2 ∫ α β ( C 1 cos ⁡ θ − C 3 sin ⁡ θ a ) a sin ⁡ θ d θ = 0 {\displaystyle {\begin{aligned}F_{1}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\cos \theta d\theta &=0\\\F_{2}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\sin \theta d\theta &=0\end{aligned}}}

After algebra,

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 _{r\theta }=0~;~~\sigma {\theta \theta }=0}

### Special Case : α = -π, β = 0[edit | edit source]

C 1 = − F 1 π ; C 2 = F 2 π {\displaystyle C_{1}=-{\frac {F_{1}}{\pi }}~;~~C_{2}={\frac {F_{2}}{\pi }}}

The displacements are

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}|}{4\pi \mu }}+{\frac {F_{2}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}-{\frac {F_{1}(\kappa +1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}}

where

κ = 3 − 4 ν plane strain κ = 3 − ν 1 + ν plane stress {\displaystyle {\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\\\\\end{aligned}}}

and

sign ( x ) = { + 1 x > 0 − 1 x < 0 {\displaystyle {\text{sign}}(x)={\begin{cases}+1&x>0\\\\-1&x<0\end{cases}}}

## Related Content[edit | edit source]

Introduction to Elasticity