Elastic wedge loaded by an axial force
|
The BCs at
are
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33f03c15bc91f9dacd1912ca0e28a3b6db241464)
- 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
![{\displaystyle {\text{(31)}}\qquad {\widehat {\mathbf {n} }}{}={\widehat {\mathbf {e} }}{r}\Rightarrow \sigma _{r\theta }=t_{r}~;~~\sigma _{\theta \theta }=t_{\theta }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67af85493a8057d2dac4ae639010e07c23599b94)
For equilibrium,
.
Therefore,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3f2e7ed9890c4cbf8f0ee4ad5ae74f34c14b2872)
These constraint conditions are equivalent to the concentrated force BC.
Assume that
. This satisfies the traction
BCs on
and equation (34). Therefore,
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/655f9070ce0d746a7965c1f2bc4c2ec85bcc387e)
Hence,
![{\displaystyle {\text{(36)}}\qquad \sigma _{\theta \theta }={\frac {\partial ^{2}}{\partial \varphi \partial r}}=\zeta ^{''}(r)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ede558defc5282424c5f8d31867a1f1d5e4bdedf)
That means
is independent of
. Therefore,
in order to satisfy the BCs,
, i.e.,
![{\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 )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c561dddcc1371cabe3fa86db98d291b21cf62489)
Checking for compatibility,
, we get
![{\displaystyle {\text{(38)}}\qquad \xi ^{(IV)}(\theta )+2\xi ^{''}(\theta )+\xi (\theta )=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f96bcf4e8db1c9c35135b7ade2b7c0772177ee9a)
The general solution is
![{\displaystyle {\text{(39)}}\qquad \xi (\theta )=A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5911acbc00cd9da24af0d0b3ac65df4aa36a6691)
Therefore,
![{\displaystyle {\text{(40)}}\qquad {\varphi =r\left[A\sin \theta +B\cos \theta +C\theta \sin \theta +D\theta \cos \theta \right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36b97e7ffc3796a0b8611d7eda0751962b01c78f)
The only non-zero stress is
.
![{\displaystyle {\text{(41)}}\qquad \sigma _{rr}={\frac {1}{r}}\left[2C\cos \theta -2D\sin \theta \right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31fd1231d138ed3eb7b31d10e7787479dcc98b78)
Plugging into equation (33), we get
![{\displaystyle {\text{(42)}}\qquad -D\left[2\beta -\sin(2\beta )\right]=0\Rightarrow D=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a0854a3868d19a656908aa6b66f07d9e8ef9005)
Hence,
![{\displaystyle {\text{(43)}}\qquad \sigma _{rr}={\frac {2C}{r}}\cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/39077a038a3b8d9083f5d4a006d318343101d1b2)
Plugging into equation (32), we get
![{\displaystyle {\text{(44)}}\qquad -P=C\left[2\beta +\sin(2\beta )\right]\Rightarrow C={\frac {-P}{2\beta +\sin(2\beta )}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7e8c6f7f349559afa4a2766e51ab04bee40b288)
Therefore,
![{\displaystyle {\text{(45)}}\qquad {\varphi =Cr\theta \sin \theta ={\frac {-Pr\theta \sin \theta }{2\beta +\sin(2\beta )}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c91fd573689763ced919a6f4398d99e2f5eeb63)
The stress state is
![{\displaystyle {\text{(46)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{r[2\beta +\sin(2\beta )]}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4d80487e029788eb47a791cfa86f2451e0a5cb2)
![{\displaystyle \beta =\pi /2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89e836befcba2b8564f97f6f9a1870a7e0dc1d3c)
A concentrated point load acting on a half plane.
![{\displaystyle {\text{(47)}}\qquad {\sigma _{rr}=-{\frac {2P\cos \theta }{\pi r}}~;~~\sigma _{r\theta }=0~;~~\sigma _{\theta \theta }=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d978e5c9908c0a6391fbb7b6a2ccf410ccbe590)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/726a41d6caf467ead734138e502de68ec4917705)
where
![{\displaystyle {\begin{aligned}\nabla ^{2}{\psi }=0\\{\frac {\partial }{\partial }}{}{r}\left(r{\frac {\partial }{\partial }}{\psi }{\theta }\right)=\nabla ^{2}{\varphi }\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8db060ade76392e4d60beb217bac603595cdee78)
Plug in
,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4381058a0c623e85e5b5a65e7ca0449a3efaf009)
Plug
into
,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53037f59cd031ec3f0ff3add7a0fe1e45620461d)
Hence,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf1bf82187e9303ace479c934bdbecaf9d888470)
Solving,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5535f2dafcd9fec857a42fd9d731f8589e564fb)
Therefore,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21dd781e1d2b65c93ad595fea6cf160ff27ca866)
To fix the rigid body motion, we set
when
,
and set
when
and
.Then,
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/720d9245bd0d20da6dc9448c298b0e98f8a99652)
The displacements are singular at
and
.
At
,
![{\displaystyle {\begin{aligned}u_{r}&={\frac {\alpha C}{\mu }}\ln \left({\frac {r}{L}}\right)\\u_{\theta }&=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b0b35c95117007610dd4de1f077ac3defabca9c)
Is the small strain assumption satisfied ?