Michell solution explained

The Michell solution is a general solution to the elasticity equations in polar coordinates (

r,\theta

) developed by J. H. Michell. The solution is such that the stress components are in the form of a Fourier series in

\theta

.

Michell[1] showed that the general solution can be expressed in terms of an Airy stress function of the form\begin \varphi(r,\theta) &= A_0 r^2 + B_0 r^2 \ln(r) + C_0 \ln(r) \\ & + \left(I_0 r^2 + I_1 r^2 \ln(r) + I_2 \ln(r) + I_3 \right) \theta \\ & + \left(A_1 r + B_1 r^ + B_1' r \theta + C_1 r^3 + D_1 r \ln(r)\right) \cos\theta \\ & + \left(E_1 r + F_1 r^ + F_1' r \theta + G_1 r^3 + H_1 r \ln(r)\right) \sin\theta \\ & + \sum_^ \left(A_n r^n + B_n r^ + C_n r^ + D_n r^\right) \cos(n\theta) \\ & + \sum_^ \left(E_n r^n + F_n r^ + G_n r^ + H_n r^\right) \sin(n\theta) \end The terms

A1r\cos\theta

and

E1r\sin\theta

define a trivial null state of stress and are ignored.

Stress components

The stress components can be obtained by substituting the Michell solution into the equations for stress in terms of the Airy stress function (in cylindrical coordinates). A table of stress components is shown below.[2]

\varphi

\sigmarr

\sigmar\theta

\sigma\theta\theta

r2

2

0

2

r2~lnr

2~lnr+1

0

2~lnr+3

lnr

r-2

0

-r-2

\theta

0

r-2

0

r3~\cos\theta

2~r~\cos\theta

2~r~\sin\theta

6~r~\cos\theta

r\theta~\cos\theta

-2~r-1~\sin\theta

0

0

r~lnr~\cos\theta

r-1~\cos\theta

r-1~\sin\theta

r-1~\cos\theta

r-1~\cos\theta

-2~r-3~\cos\theta

-2~r-3~\sin\theta

2~r-3~\cos\theta

r3~\sin\theta

2~r~\sin\theta

-2~r~\cos\theta

6~r~\sin\theta

r\theta~\sin\theta

2~r-1~\cos\theta

0

0

r~lnr~\sin\theta

r-1~\sin\theta

-r-1~\cos\theta

r-1~\sin\theta

r-1~\sin\theta

-2~r-3~\sin\theta

2~r-3~\cos\theta

2~r-3~\sin\theta

rn+2~\cos(n\theta)

-(n+1)(n-2)~rn~\cos(n\theta)

n(n+1)~rn~\sin(n\theta)

(n+1)(n+2)~rn~\cos(n\theta)

r-n+2~\cos(n\theta)

-(n+2)(n-1)~r-n~\cos(n\theta)

-n(n-1)~r-n~\sin(n\theta)

(n-1)(n-2)~r-n~\cos(n\theta)

rn~\cos(n\theta)

-n(n-1)~rn-2~\cos(n\theta)

n(n-1)~rn-2~\sin(n\theta)

n(n-1)~rn-2~\cos(n\theta)

r-n~\cos(n\theta)

-n(n+1)~r-n-2~\cos(n\theta)

-n(n+1)~r-n-2~\sin(n\theta)

n(n+1)~r-n-2~\cos(n\theta)

rn+2~\sin(n\theta)

-(n+1)(n-2)~rn~\sin(n\theta)

-n(n+1)~rn~\cos(n\theta)

(n+1)(n+2)~rn~\sin(n\theta)

r-n+2~\sin(n\theta)

-(n+2)(n-1)~r-n~\sin(n\theta)

n(n-1)~r-n~\cos(n\theta)

(n-1)(n-2)~r-n~\sin(n\theta)

rn~\sin(n\theta)

-n(n-1)~rn-2~\sin(n\theta)

-n(n-1)~rn-2~\cos(n\theta)

n(n-1)~rn-2~\sin(n\theta)

r-n~\sin(n\theta)

-n(n+1)~r-n-2~\sin(n\theta)

n(n+1)~r-n-2~\cos(n\theta)

n(n+1)~r-n-2~\sin(n\theta)

Displacement components

(ur,u\theta)

can be obtained from the Michell solution by using the stress-strain and strain-displacement relations. A table of displacement components corresponding the terms in the Airy stress function for the Michell solution is given below. In this table

\kappa=\begin{cases} 3-4~\nu&\rm{for~plane~strain}\\ \cfrac{3-\nu}{1+\nu}&\rm{for~plane~stress}\\ \end{cases}

where

\nu

is the Poisson's ratio, and

\mu

is the shear modulus.

\varphi

2~\mu~ur

2~\mu~u\theta

r2

(\kappa-1)~r

0

r2~lnr

(\kappa-1)~r~lnr-r

(\kappa+1)~r~\theta

lnr

-r-1

0

\theta

0

-r-1

r3~\cos\theta

(\kappa-2)~r2~\cos\theta

(\kappa+2)~r2~\sin\theta

r\theta~\cos\theta

1
2

[(\kappa-1)\theta~\cos\theta+\{1-(\kappa+1)lnr\}~\sin\theta]

-1
2

[(\kappa-1)\theta~\sin\theta+\{1+(\kappa+1)lnr\}~\cos\theta]

r~lnr~\cos\theta

1
2

[(\kappa+1)\theta~\sin\theta-\{1-(\kappa-1)lnr\}~\cos\theta]

1
2

[(\kappa+1)\theta~\cos\theta-\{1+(\kappa-1)lnr\}~\sin\theta]

r-1~\cos\theta

r-2~\cos\theta

r-2~\sin\theta

r3~\sin\theta

(\kappa-2)~r2~\sin\theta

-(\kappa+2)~r2~\cos\theta

r\theta~\sin\theta

1
2

[(\kappa-1)\theta~\sin\theta-\{1-(\kappa+1)lnr\}~\cos\theta]

1
2

[(\kappa-1)\theta~\cos\theta-\{1+(\kappa+1)lnr\}~\sin\theta]

r~lnr~\sin\theta

-1
2

[(\kappa+1)\theta~\cos\theta+\{1-(\kappa-1)lnr\}~\sin\theta]

1
2

[(\kappa+1)\theta~\sin\theta+\{1+(\kappa-1)lnr\}~\cos\theta]

r-1~\sin\theta

r-2~\sin\theta

-r-2~\cos\theta

rn+2~\cos(n\theta)

(\kappa-n-1)~rn+1~\cos(n\theta)

(\kappa+n+1)~rn+1~\sin(n\theta)

r-n+2~\cos(n\theta)

(\kappa+n-1)~r-n+1~\cos(n\theta)

-(\kappa-n+1)~r-n+1~\sin(n\theta)

rn~\cos(n\theta)

-n~rn-1~\cos(n\theta)

n~rn-1~\sin(n\theta)

r-n~\cos(n\theta)

n~r-n-1~\cos(n\theta)

n(~r-n-1~\sin(n\theta)

rn+2~\sin(n\theta)

(\kappa-n-1)~rn+1~\sin(n\theta)

-(\kappa+n+1)~rn+1~\cos(n\theta)

r-n+2~\sin(n\theta)

(\kappa+n-1)~r-n+1~\sin(n\theta)

(\kappa-n+1)~r-n+1~\cos(n\theta)

rn~\sin(n\theta)

-n~rn-1~\sin(n\theta)

-n~rn-1~\cos(n\theta)

r-n~\sin(n\theta)

n~r-n-1~\sin(n\theta)

-n~r-n-1~\cos(n\theta)

Note that a rigid body displacement can be superposed on the Michell solution of the form

\begin{align} ur&=A~\cos\theta+B~\sin\theta\\ u\theta&=-A~\sin\theta+B~\cos\theta+C~r\\ \end{align}

to obtain an admissible displacement field.

See also

Notes and References

  1. On the direct determination of stress in an elastic solid, with application to the theory of plates. Proc. London Math. Soc.. 1899-04-01. J. H.. Michell. 31. 1. 100–124. 10.1112/plms/s1-31.1.100.
  2. J. R. Barber, 2002, Elasticity: 2nd Edition, Kluwer Academic Publishers.