Introduction to Elasticity/Williams asymptotic solution

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[edit] Williams' Asymptotic Solution

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, \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 rα as we approach the point r = 0, the strain energy is given by
\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. Hence, singular stress fields are acceptable only if the exponent on the stress components exceeds − 1.

[edit] Stresses near the notch corner

  • 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 in equation (3) is an integer. Let n = λ − 1.
\begin{align} \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{align}

The stresses are

\begin{align} \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{align}

The BCs are σrθ = σθθ = 0 at θ = α.Hence,

\begin{align} 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{align}

The BCs are σrθ = σθθ = 0 at θ = − α.Hence,

\begin{align} 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{align}

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


[edit] Eigenvalues of λ

Adding equations (9) and (10),

\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),

\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),

\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),

\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

\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

\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

\text{(19)} \qquad \lambda\sin(2\alpha) + \sin(2\lambda\alpha) = 0

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

\text{(20)} \qquad \lambda\sin(2\alpha) - \sin(2\lambda\alpha) = 0
  • From equation (4), acceptable singular stress fields must have λ > 0.Hence, λ = 0 is not acceptable.
  • The term with the smallest eigenvalue of λ dominates the solution. Hence, this eigenvalue is what we seek.
  • λ = 1 leads to \varphi = a_4 \sin(0). Unacceptable.
  • We can find the eiegnvalues for general wedge angles using graphical methods.


[edit] Special case : \alpha = \pi = 180^o \,

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

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

The lowest eigenvalue is 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

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

\begin{align} \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_{r\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_{\theta\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{align}

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

\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.

\begin{align} \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_{r\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_{\theta\theta} & = \frac{K_{II}}{\sqrt{2\pi r}} \left[\frac{1}{4}\cos\left(\frac{\theta}{2}\right) + \frac{3}{4}\sin\left(\frac{3\theta}{2}\right)\right] \text{(29)} \qquad \end{align}
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