Concentrated force on a half plane
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From the Flamant Solution
![{\displaystyle {\begin{aligned}F_{1}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\cos \theta d\theta &=0\\F_{2}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\sin \theta d\theta &=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86c1dc1a16e2b407e6794e91b49efaa5727a5407)
and
![{\displaystyle \sigma _{rr}={\frac {2C_{1}\cos \theta }{r}}+{\frac {2C_{3}\sin \theta }{r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4dbde8f5b418bd68bbfcea4202e5ffec8d363e5c)
If
and
, we obtain the special case of
a concentrated force acting on a half-plane. Then,
![{\displaystyle {\begin{aligned}F_{1}+2\int _{-\pi }^{0}\left(C_{1}\cos ^{2}\theta -{\frac {C_{3}}{2}}\sin(2\theta )\right)d\theta &=0\\F_{2}+2\int _{-\pi }^{0}\left({\frac {C_{1}}{2}}\sin(2\theta )-C_{3}\sin ^{2}\theta \right)d\theta &=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac84d8e7159b4caa3d1afb38cf026a71a03c911d)
or,
![{\displaystyle {\begin{aligned}F_{1}+\pi C_{1}&=0\\F_{2}-\pi C_{3}&=0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bc651750002c6ad8d3c32b44c5decdd9ca5f8da7)
Therefore,
![{\displaystyle C_{1}=-{\frac {F_{1}}{\pi }}~;~~C_{3}={\frac {F_{2}}{\pi }}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8b477647e975551068825b9fcb8db936c094fb8)
The stresses are
![{\displaystyle \sigma _{rr}=-{\frac {2F_{1}\cos \theta }{\pi r}}-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1de3143c7920b8ca4d74fc72dc46a9dbbaac7ba6)
The stress
is obviously the superposition of the stresses
due to
and
, applied separately to the half-plane.
The tensile force
produces the stress field
![{\displaystyle \sigma _{rr}=-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/732e7e5f098d0cb8fb1dba6a9acf4f96b25cd9d9)
Stress due to concentrated force on a half plane
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The stress function is
![{\displaystyle \varphi ={\frac {F_{2}}{\pi }}r\theta \cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0363361c1d3942a5e31e1ad608a4ac642602b379)
Hence, the displacements from Michell's solution are
![{\displaystyle {\begin{aligned}2\mu u_{r}&={\frac {F_{2}}{2\pi }}\left[(\kappa -1)\theta \cos \theta +\sin \theta -(\kappa +1)\ln(r)\sin \theta \right]\\2\mu u_{\theta }&={\frac {F_{2}}{2\pi }}\left[-(\kappa -1)\theta \sin \theta -\cos \theta -(\kappa +1)\ln(r)\cos \theta \right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/33e0f21bc55b75d114b67384ab4d3f664c5c8790)
At
, (
,
),
![{\displaystyle {\begin{aligned}2\mu u_{r}=2\mu u_{1}&=0\\2\mu u_{\theta }=2\mu u_{2}&={\frac {F_{2}}{2\pi }}\left[-1-(\kappa +1)\ln(r)\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8bd12af44d96643f2d5311a8d527c8cebe80fca1)
At
, (
,
),
![{\displaystyle {\begin{aligned}2\mu u_{r}=-2\mu u_{1}&={\frac {F_{2}}{2\pi }}(\kappa -1)\\2\mu u_{\theta }=-2\mu u_{2}&={\frac {F_{2}}{2\pi }}\left[1+(\kappa +1)\ln(r)\right]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf6949aef857baf3ca0906c3579992107841d0a5)
where
![{\displaystyle {\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae4f95d69ad1803c8a4d93b89d0f0bdce92b1593)
Since we expect the solution to be symmetric about
, we superpose a
rigid body displacement
![{\displaystyle {\begin{aligned}2\mu u_{1}&={\frac {F_{2}}{4\pi }}(\kappa -1)\\2\mu u_{2}&={\frac {F_{2}}{2\pi }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7527fdf28c61291d88c15657ed7a60ac3a7da334)
The displacements are
![{\displaystyle {\begin{aligned}u_{1}&={\frac {F_{2}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6da3ec003dae5deddb0e6ded6d207c4e93830ee)
where
![{\displaystyle {\text{sign}}(x)={\begin{cases}+1&x>0\\-1&x<0\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d5c84e0ba475d609131c2b73cf071be5519ab27)
and
on
.
The tensile force
produces the stress field
![{\displaystyle \sigma _{rr}=-{\frac {2F_{2}\cos \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4b34ad5d7f9ecddb3a790caf3c340bf2b390ca4)
Stress due to concentrated force on a half plane
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The displacements are
![{\displaystyle {\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}\\u_{2}&=-{\frac {F_{1}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36ad6f8ccf29fc3cd654e6dc3077961f4a7de7b4)
Superpose the two solutions. The stresses are
![{\displaystyle \sigma _{rr}=-{\frac {2F_{1}\cos \theta }{\pi r}}-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1de3143c7920b8ca4d74fc72dc46a9dbbaac7ba6)
The displacements are
![{\displaystyle {\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}+{\frac {F_{2}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}-{\frac {F_{1}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/712535bfc78fb9331f04d5314c3c48a799581d75)