Elasticity/Concentrated force on half plane

Concentrated force on a half-plane

Concentrated force on a half plane

From the Flamant Solution

${\begin{aligned}F_{1}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\cos \theta d\theta &=0\\F_{2}+2\int _{\alpha }^{\beta }\left({\frac {C_{1}\cos \theta -C_{3}\sin \theta }{a}}\right)a\sin \theta d\theta &=0\end{aligned}}$

and

$\sigma _{rr}={\frac {2C_{1}\cos \theta }{r}}+{\frac {2C_{3}\sin \theta }{r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0$

If $\alpha =-\pi \,$ and$\beta =0\,$, we obtain the special case of a concentrated force acting on a half-plane. Then,

${\begin{aligned}F_{1}+2\int _{-\pi }^{0}\left(C_{1}\cos ^{2}\theta -{\frac {C_{3}}{2}}\sin(2\theta )\right)d\theta &=0\\F_{2}+2\int _{-\pi }^{0}\left({\frac {C_{1}}{2}}\sin(2\theta )-C_{3}\sin ^{2}\theta \right)d\theta &=0\end{aligned}}$

or,

${\begin{aligned}F_{1}+\pi C_{1}&=0\\F_{2}-\pi C_{3}&=0\end{aligned}}$

Therefore,

$C_{1}=-{\frac {F_{1}}{\pi }}~;~~C_{3}={\frac {F_{2}}{\pi }}$

The stresses are

$\sigma _{rr}=-{\frac {2F_{1}\cos \theta }{\pi r}}-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0$

The stress $\sigma _{rr}\,$ is obviously the superposition of the stresses due to $F_{1}\,$ and $F_{2}\,$, applied separately to the half-plane.

Problem 1 : Stresses and displacements due to F₂

The tensile force $F_{2}\,$ produces the stress field

$\sigma _{rr}=-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0$

Stress due to concentrated force $F_{2}\,$ on a half plane

The stress function is

$\varphi ={\frac {F_{2}}{\pi }}r\theta \cos \theta$

Hence, the displacements from Michell's solution are

${\begin{aligned}2\mu u_{r}&={\frac {F_{2}}{2\pi }}\left[(\kappa -1)\theta \cos \theta +\sin \theta -(\kappa +1)\ln(r)\sin \theta \right]\\2\mu u_{\theta }&={\frac {F_{2}}{2\pi }}\left[-(\kappa -1)\theta \sin \theta -\cos \theta -(\kappa +1)\ln(r)\cos \theta \right]\end{aligned}}$

At $\theta =0$, ($x_{1}>0$, $x_{2}=0$),

${\begin{aligned}2\mu u_{r}=2\mu u_{1}&=0\\2\mu u_{\theta }=2\mu u_{2}&={\frac {F_{2}}{2\pi }}\left[-1-(\kappa +1)\ln(r)\right]\end{aligned}}$

At $\theta =-\pi$, ($x_{1}<0$, $x_{2}=0$),

${\begin{aligned}2\mu u_{r}=-2\mu u_{1}&={\frac {F_{2}}{2\pi }}(\kappa -1)\\2\mu u_{\theta }=-2\mu u_{2}&={\frac {F_{2}}{2\pi }}\left[1+(\kappa +1)\ln(r)\right]\end{aligned}}$

where

${\begin{aligned}\kappa =3-4\nu &&{\text{plane strain}}\\\kappa ={\frac {3-\nu }{1+\nu }}&&{\text{plane stress}}\end{aligned}}$

Since we expect the solution to be symmetric about $x=0\,$, we superpose a rigid body displacement

${\begin{aligned}2\mu u_{1}&={\frac {F_{2}}{4\pi }}(\kappa -1)\\2\mu u_{2}&={\frac {F_{2}}{2\pi }}\end{aligned}}$

The displacements are

${\begin{aligned}u_{1}&={\frac {F_{2}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}\end{aligned}}$

where

${\text{sign}}(x)={\begin{cases}+1&x>0\\-1&x<0\end{cases}}$

and $r=|x|\,$ on $y=0\,$.

Problem 2 : Stresses and displacements due to F₁

The tensile force $F_{1}\,$ produces the stress field

$\sigma _{rr}=-{\frac {2F_{2}\cos \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0$

Stress due to concentrated force $F_{1}\,$ on a half plane

The displacements are

${\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}\\u_{2}&=-{\frac {F_{1}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}$

Stresses and displacements due to F₁ + F₂

Superpose the two solutions. The stresses are

$\sigma _{rr}=-{\frac {2F_{1}\cos \theta }{\pi r}}-{\frac {2F_{2}\sin \theta }{\pi r}}~;~~\sigma _{r\theta }=\sigma _{\theta \theta }=0$

The displacements are

${\begin{aligned}u_{1}&=-{\frac {F_{1}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}+{\frac {F_{2}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\\u_{2}&=-{\frac {F_{2}(\kappa +1)\ln |x_{1}|}{4\pi \mu }}-{\frac {F_{1}(\kappa -1){\text{sign}}(x_{1})}{8\mu }}\end{aligned}}$

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