Stresses in wedges

Wedge with Boundary Tractions

edit
 
Elastic wedge with normal and shear surface tractions

Suppose

  • The tractions on the boundary vary as  .
  • No body forces.

Then

 

To find   plug into  .

 

If   and  ,

 

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


Williams' Asymptotic Solution

edit

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,  .
  • 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   as we approach the point  , the strain energy is given by
 

This integral is bounded only if  . Hence, singular stress fields are acceptable only if the exponent on the stress components exceeds  .

Stresses near the notch corner

edit
  • 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   in equation (3) is an integer. Let  .
 

The stresses are

 

The BCs are   at  .Hence,

 

The BCs are   at  .Hence,

 

The above equations will have non-trivial solutions only for certain eigenvalues of  , one of which is  . Using the symmetries of the equations, we can partition the coefficient matrix.


Eigenvalues of λ

edit

Adding equations (9) and (10),

 

Subtracting equation (10) from (9),

 

Adding equations (11) and (12),

 

Subtracting equation (12) from (11),

 

Therefore, the two independent sets of equations are

 

and

 

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

 

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

 
  • From equation (4), acceptable singular stress fields must have  .Hence,   is not acceptable.
  • The term with the smallest eigenvalue of   dominates the solution. Hence, this eigenvalue is what we seek.
  •   leads to  . Unacceptable.
  • We can find the eiegnvalues for general wedge angles using graphical methods.


Special case : α = π = 180°

edit

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

 

The lowest eigenvalue is  . 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

 

where   is a constant. The singular stress field at the crack tip is then

 

where,   is the { Mode I Stress Intensity Factor.}

 

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

 

Axially Loaded Wedge

edit
 
Elastic wedge loaded by an axial force

The BCs at   are

 

What about the concentrated force BC?

edit
  • What is   at the vertex ?
  • The traction is infinite since the force is applied on zero area. Consider equilibrium of a portion of the wedge.

At  , the BCs are

 

For equilibrium,  . Therefore,

 

These constraint conditions are equivalent to the concentrated force BC.

Solution Procedure

edit

Assume that  . This satisfies the traction BCs on   and equation (34). Therefore,

 

Hence,

 

That means   is independent of  . Therefore, in order to satisfy the BCs,  , i.e.,

 

Checking for compatibility,  , we get

 

The general solution is

 

Therefore,

 

The only non-zero stress is  .

 

Plugging into equation (33), we get

 

Hence,

 

Plugging into equation (32), we get

 

Therefore,

 

The stress state is

 

Special Case : β = π/2

edit

A concentrated point load acting on a half plane.

 

Displacements

edit
 

where

 

Plug in  ,

 

Plug   into  ,

 

Hence,

 

Solving,

 

Therefore,

 

To fix the rigid body motion, we set   when  , and set   when   and  .Then,

 

The displacements are singular at   and  . At  ,

 

Is the small strain assumption satisfied ?


The Flamant Solution

edit
 
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  .
  • The stresses vary as   (the area of action of the force decreases with increasing  ). How about a conical wedge ?

From Michell's solution, pick terms containing   in the stresses. Then,

 

Therefore, from Tables,

 

From traction BCs,  . From equilibrium,

 

After algebra,

 

Special Case : α = -π, β = 0

edit
 

The displacements are

 

where

 

and

 
edit

Introduction to Elasticity