View source for Elasticity/Plate with hole in shear

== Circular hole in a shear field ==
{|align="center"
 |[[image:elastic_plate_with_hole_shear.png|thumb|400px|Elastic plate with circular hole under shear]] 
 |}

'''Given:'''
*Large plate in pure shear.
*Stress state perturbed by a small hole.

The BCs are
: at <math> r = a </math>
: <math>
    \text{(103)} \qquad 
    t_r = t_{\theta} = 0 ~;~~ \widehat{\mathbf{n}} = -\widehat{\mathbf{e}}~r
    \Rightarrow \sigma_{rr} = \sigma_{r\theta} = 0
  </math>

: at <math> r \rightarrow \infty </math>
: <math>
  \text{(104)} \qquad
  \sigma_{12} \rightarrow S ~;~~
         \sigma_{11} \rightarrow 0 ~;~~ \sigma_{22} \rightarrow 0
  </math>

We will solve this problem by superposing a perturbation due to the hole on the unperturbed solution.  The effect of the perturbation will decrease with increasing distance from the hole, i.e. the effect will be proportional to <math>r^{-n} \,</math>.

=== Unperturbed Solution===
:<math> \text{(105)} \qquad
    \sigma_{11} = \sigma_{22} = 0 ~;~~ \sigma_{12} = S
  </math>
Therefore,
:<math>\text{(106)} \qquad
    \sigma_{12} = -\varphi_{,12} = S
  </math>
Integrating,
:<math>\text{(107)} \qquad
    \varphi_{,1} = -Sx_2 + f(x_1) \Rightarrow
    \varphi = -Sx_1x_2 + \int f(x_1) dx_1
  </math>
Since <math>\varphi</math> is a potential, we can neglect the integration constants (these do not affect the stresses - which are what we are interested in).  Hence,
:<math> \text{(108)} \qquad
    \varphi = -Sx_1x_2 = -S (r\cos\theta) (r\sin\theta) =
              -\cfrac{Sr^2}{2}\sin(2\theta)
  </math>
or,
:<math> \text{(109)} \qquad
    \varphi = -\cfrac{Sr^2}{2}\sin(2\theta) 
  </math>
Note that we have arranged the expression so that it has a form similar to the Fourier series of the previous section.

=== Perturbed Solution===

For this we have to add terms to <math>\varphi</math> in such a way that

*The unperturbed solution continues to be true as <math>r\rightarrow\infty \,</math>.
*The terms have the same form as the unperturbed solution,i.e., <math>sin(2\theta) \,</math> terms.
*The new <math>\varphi</math> leads to stresses that are proportional to <math>r^{-n}\,</math>.

Recall,
:<math>
    \varphi = \sum^{\infty}_{n=0} f_n(r) \cos(n\theta) +
              \sum^{\infty}_{n=0} g_n(r) \sin(n\theta)
  </math>
where,
:<math>\begin{align}
    f_0(r) & = A_0 r^2 + B_0 r^2 \ln r + C_0 + D_0 \ln r \\
    f_1(r) & = A_1 r^3 + B_1 r  + C_1 r \ln r + D_1 r^{-1} \\
    f_n(r) & = A_n r^{n+2} + B_n r^n  + C_n r^{-n+2} + D_n r^{-n} ~,~~n > 1
  \end{align}
  </math>
So the appropriate stress function for the perturbation is
:<math>\text{(110)}\qquad
    \varphi = g_2(r) \sin(2\theta) = \left(C_2 r^{-2+2} + D_2 r^{-2}\right)
        \sin(2\theta)
  </math>
or,
:<math>\text{(111)} \qquad
    \varphi = \left(C_2  + D_2 r^{-2}\right)\sin(2\theta)
  </math>
Hence, the stress function appropriate for the superposed solution is
:<math>\text{(112)}
    \varphi = -\cfrac{Sr^2}{2}\sin(2\theta) +
              \left(C_2  + D_2 r^{-2}\right)\sin(2\theta)
  </math>
We determine <math>C_2</math> and <math>D_2</math> using the boundary conditions at <math>r=a</math>.

The stresses are
:<math>\begin{align}
    \text{(113)}\qquad\sigma_{rr} & = \cfrac{1}{r}\cfrac{\partial\varphi}{\partial r} +
       \cfrac{1}{r^2}\cfrac{\partial^2\varphi}{\partial \theta^2}
       = \left(S - 4C_2r^{-2} - 6D_2r^{-4}\right)\sin(2\theta) \\
    \text{(114)}\qquad\sigma_{\theta\theta} & = \cfrac{\partial^2\varphi}{\partial r^2}
       = \left(-S + 6D_2r^{-4}\right)\sin(2\theta) \\
    \text{(115)}\qquad\sigma_{r\theta} & = -\cfrac{\partial}{\partial r}
       \left(\cfrac{1}{r}\cfrac{\partial\varphi}{\partial \theta}\right)
       = \left(S + 6D_2r^{-4}\right)\cos(2\theta)
  \end{align}</math>
Hence,
:<math>\begin{align}
    \text{(116)}\qquad\left.\sigma_{rr}\right|_{r=a} & = 0
       = \left(S - 4C_2a^{-2} - 6D_2a^{-4}\right)\sin(2\theta) \\
    \text{(117)}\qquad\left.\sigma_{r\theta}\right|_{r=a} & = 0
       = \left(S + 2C_2a^{-2} + 6D_2a^{-4}\right)\cos(2\theta)
  \end{align}</math>
or,
:<math>\begin{align}
    \text{(118)}\qquad 4C_2a^{-2} + 6D_2a^{-4} & = S \\
    \text{(119)}\qquad 2C_2a^{-2} + 6D_2a^{-4} & = -S 
  \end{align}</math>
Solving,
:<math>\text{(120)}\qquad
    C_2 = Sa^2 ~;~~ D_2 = -\cfrac{Sa^4}{2}
  </math>
Back substituting,
:<math>\begin{align}
    \text{(121)}\qquad \sigma_{rr} & =
       S\left(1 - 4\cfrac{a^2}{r^2} + 3\cfrac{a^4}{r^4}\right)\sin(2\theta)
       \\
    \text{(122)}\qquad \sigma_{\theta\theta} & =
       S\left(-1 - 3\cfrac{a^4}{r^4}\right)\sin(2\theta)
       \\
    \text{(123)}\qquad \sigma_{r\theta} & =
       S\left(1 + 2\cfrac{a^2}{r^2} - 3\cfrac{a^4}{r^4}\right)\cos(2\theta)
  \end{align}</math>

== Example homework problem ==
Consider the elastic plate with a hole subject to pure shear.
{|align="center" 
|[[image:elastic_plate_with_hole_shear.png|thumb|400px|Elastic plate with a circular hole under pure shear]]
|}
The stresses close to the hole are given by
:<math>\begin{align}
   \text{(29)} \qquad   \sigma_{rr} & = 
       S\left(1 - 4\frac{a^2}{r^2} + 3\frac{a^4}{r^4}\right)\sin(2\theta)
       \\
   \text{(30)} \qquad  \sigma_{\theta\theta} & = 
       S\left(-1 - 3\frac{a^4}{r^4}\right)\sin(2\theta)
        \\
   \text{(31)} \qquad   \sigma_{r\theta} & = 
       S\left(1 + 2\frac{a^2}{r^2} - 3\frac{a^4}{r^4}\right)\cos(2\theta)   
\end{align}</math>
    
*Show that the normal and shear traction boundary conditions far from the hole are satisfied by these stresses.    
*Calculate the stress concentration factors at the hole, i.e., (<math>\tau_{\text{max}}/S</math>) (shear) and (<math>\sigma_{\text{max}}/\sigma_0</math>) (normal).   
*Calculate the displacement field corresponding to this stress field (for plane stress).  Plot the deformed shape of the hole.  


=== Solution ===

Far from the hole, <math>r = \infty</math>.  Therefore,
:<math>\begin{align}
   \text{(32)} \qquad   \sigma_{rr} & = S\sin(2\theta) \\
   \text{(33)} \qquad  \sigma_{\theta\theta} & = -S\sin(2\theta)  \\
    \text{(34)} \qquad   \sigma_{r\theta} & = S\cos(2\theta)
  \end{align}</math>
To rotate the stresses back to the <math>(x_1,x_2)</math> coordinate system, we use the tensor transformation rule
:<math>\text{(35)} \qquad 
  \begin{bmatrix}
    \sigma_{11} & \sigma_{12} & \sigma_{13} \\
    \sigma_{21} & \sigma_{22} & \sigma_{23} \\
    \sigma_{31} & \sigma_{32} & \sigma_{33} 
  \end{bmatrix} = 
  \begin{bmatrix}
    \cos\theta & -\sin\theta & 0 \\
    \sin\theta & \cos\theta & 0 \\
    0 & 0 & 1
  \end{bmatrix}
  \begin{bmatrix}
    \sigma_{rr} & \sigma_{r\theta} & \sigma_{rz} \\
    \sigma_{r\theta} & \sigma_{\theta\theta} & \sigma_{\theta z} \\
    \sigma_{rz} & \sigma_{\theta z} & \sigma_{zz} 
  \end{bmatrix} 
  \begin{bmatrix}
    \cos\theta & \sin\theta & 0 \\
    -\sin\theta & \cos\theta & 0 \\
    0 & 0 & 1
  \end{bmatrix}
</math>
Setting <math>\sigma_{rz} = 0</math> and <math>\sigma_{\theta z} = 0</math>, we get the simplified set of equations
:<math>\begin{align}
  \text{(36)} \qquad \sigma_{11} &= \sigma_{rr}\cos^2\theta + \sigma_{\theta\theta}\sin^2\theta
                 - \sigma_{r\theta}\sin(2\theta)  \\
  \text{(37)} \qquad \sigma_{22} &= \sigma_{rr}\sin^2\theta + \sigma_{\theta\theta}\cos^2\theta
                 + \sigma_{r\theta}\sin(2\theta)  \\
  \text{(38)} \qquad \sigma_{12} &= \frac{\sigma_{rr}-\sigma_{\theta\theta}}{2}\sin(2\theta)
                 + \sigma_{r\theta}\cos(2\theta)  
\end{align}</math>
Plugging in equations (32-34) in the above, we have
:<math>\begin{align}
  \text{(39)} \qquad \sigma_{11} &= -S\left[\sin(2\theta)\cos(2\theta)-\sin(2\theta)\cos(2\theta)
                   \right] = 0  \\
  \text{(40)} \qquad \sigma_{22} &= S\left[\sin(2\theta)\cos(2\theta)-\sin(2\theta)\cos(2\theta)
                   \right] = 0  \\
  \text{(41)} \qquad \sigma_{12} &= S\left[\sin(2\theta)\sin(2\theta)+\cos(2\theta)\cos(2\theta)
                   \right] = S 
\end{align}</math>
Hence, ''the far field stress BCs are satisfied.''


The stresses at the hole (<math>r = a</math>) are
:<math>\begin{align}
    \text{(42)} \qquad \sigma_{rr} & = 
       S\left(1 - 4 + 3\right)\sin(2\theta) = 0
        \\
    \text{(43)} \qquad \sigma_{\theta\theta} & = 
       S\left(-1 - 3\right)\sin(2\theta) = -4S\sin(2\theta)
        \\
    \text{(44)} \qquad \sigma_{r\theta} & = 
       S\left(1 + 2 - 3\right)\cos(2\theta) = 0
        
  \end{align}</math>

The maximum (or minimum) hoop stress at the hole is at the locations
where <math>d\sigma_{\theta\theta}/d\theta = -8S\cos(2\theta) = 0</math>. These locations are <math>\theta = \pi/4</math> and <math>\theta = 3\pi/4</math>.  The value of the hoop stress is 
:<math>\begin{align}
  \text{(45)} \qquad \text{at}~ \theta = \frac{\pi}{4} & & \sigma_{\theta\theta} = -4S
    \\
  \text{(46)} \qquad \text{at}~ \theta = \frac{3\pi}{4} &  &\sigma_{\theta\theta} = 4S
\end{align}</math>
The maximum shear stress is given by
:<math> \text{(47)} \qquad 
  \tau_{\text{max}} = \frac{1}{2}\left|\sigma_{rr}-\sigma_{\theta\theta}\right|
     = 2S
</math>
Therefore, the stress concentration factors are
:<math>\text{(48)} \qquad 
  \frac{\sigma_{\text{max}}}{S} = 4 ~;~~ 
  \frac{\tau_{\text{max}}}{S} = 2 
</math>

The stress function used to derive the above results was
:<math>\text{(49)} \qquad 
  \varphi = -\frac{S}{2}r^2\sin(2\theta) +
              Sa^2 \sin(2\theta) - \frac{Sa^4}{2}r^{-2}\sin(2\theta) 
</math>
From [[Michell solution|Michell's solution]], the displacements corresponding
to the above stress function are given by
:<math>\begin{align}
  \text{(50)} \qquad 2\mu u_r & = -\frac{S}{2}\left[-2r\sin(2\theta)\right] 
               +Sa^2\left[(\kappa+1)r^{-1}\sin(2\theta)\right] 
               -\frac{Sa^4}{2}\left[2r^{-3}\sin(2\theta)\right]
                \\ 
  \text{(51)} \qquad 2\mu u_{\theta} & = -\frac{S}{2}\left[-2r\cos(2\theta)\right] 
               +Sa^2\left[(\kappa-1)r^{-1}\cos(2\theta)\right] 
               -\frac{Sa^4}{2}\left[-2r^{-3}\cos(2\theta)\right]
               
\end{align}</math>
or,
:<math>\begin{align}
  \text{(52)} \qquad u_r & = \frac{Sr\sin(2\theta)}{2\mu}\left[
               1 + (\kappa+1)\frac{a^2}{r^2} - \frac{a^4}{r^4}\right]
                \\ 
  \text{(53)} \qquad u_{\theta} & = \frac{Sr\cos(2\theta)}{2\mu}\left[
               1 + (\kappa-1)\frac{a^2}{r^2} + \frac{a^4}{r^4}\right]
               
\end{align}</math>
For plane stress, <math>\kappa = (3-\nu)/(1+\nu)</math>.  Hence, 
:<math>\begin{align}
  \text{(54)} \qquad u_r & = \frac{Sr\sin(2\theta)}{2\mu}\left[
               1 + \left(\frac{4}{1+\nu}\right)\frac{a^2}{r^2} 
               - \frac{a^4}{r^4}\right]
                \\ 
  \text{(55)} \qquad u_{\theta} & = \frac{Sr\cos(2\theta)}{2\mu}\left[
               1 + 2\left(\frac{1-\nu}{1+\nu}\right)\frac{a^2}{r^2} 
               + \frac{a^4}{r^4}\right]
               
\end{align}</math>
At <math>r = a</math>,
:<math>\begin{align}
  \text{(56)} \qquad u_r & = \frac{Sa\sin(2\theta)}{\mu}\left(\frac{2}{1+\nu}\right)
                \\ 
  \text{(57)} \qquad u_{\theta} & = \frac{Sa\cos(2\theta)}{\mu}\left(\frac{2}{1+\nu}\right)
               
\end{align}</math>
Now <math>\mu = E/2(1+\nu)</math>.  Hence, we have
:<math>\begin{align}
  \text{(58)} \qquad  u_r & = \frac{4Sa\sin(2\theta)}{E} \\ 
  \text{(59)} \qquad u_{\theta} & = \frac{4Sa\cos(2\theta)}{E} 
\end{align}</math>
The deformed shape is shown below
{|align="center" 
|[[image:shear_hole_deform.png|thumb|400px|Displacement field near a hole in plate under pure shear]]
  |}


{{subpage navbar}}

[[Category:Elasticity]]