In continuum mechanics, the Flamant solution provides expressions for the stresses and displacements in a linear elastic wedge loaded by point forces at its sharp end. This solution was developed by in 1892[1] by modifying the three dimensional solutions for linear elasticity of Joseph Valentin Boussinesq.
The stresses predicted by the Flamant solution are (in polar coordinates)
\begin{align} \sigmarr&=
2C1\cos\theta | |
r |
+
2C3\sin\theta | |
r |
\\ \sigmar\theta&=0\\ \sigma\theta\theta&=0 \end{align}
C1,C3
\alpha,\beta
\begin{align} F1&+
\beta | |
2\int | |
\alpha |
(C1\cos\theta+C3\sin\theta)\cos\thetad\theta=0\\ F2&+
\beta | |
2\int | |
\alpha |
(C1\cos\theta+C3\sin\theta)\sin\thetad\theta=0\end{align}
F1,F2
\sigma=f(r)g(\theta)
(1/r)
For the special case where
\alpha=-\pi
\beta=0
C1=-
F1 | |
\pi |
, C3=-
F2 | |
\pi |
\begin{align} \sigmarr&=-
2 | |
\pir |
(F1\cos\theta+F2\sin\theta)\\ \sigmar\theta&=0\\ \sigma\theta\theta&=0 \end{align}
\begin{align} ur&=-\cfrac{1}{4\pi\mu}\left[F1\{(\kappa-1)\theta\sin\theta-\cos\theta+(\kappa+1)lnr\cos\theta\}+\right.\\ & \left.F2\{(\kappa-1)\theta\cos\theta+\sin\theta-(\kappa+1)lnr\sin\theta\}\right]\\ u\theta&=-\cfrac{1}{4\pi\mu}\left[F1\{(\kappa-1)\theta\cos\theta-\sin\theta-(\kappa+1)lnr\sin\theta\}-\right.\\ & \left.F2\{(\kappa-1)\theta\sin\theta+\cos\theta+(\kappa+1)lnr\cos\theta\}\right] \end{align}
lnr
The displacements in the
x1,x2
\begin{align} u1&=
F1(\kappa+1)ln|x1| | + | |
4\pi\mu |
F2(\kappa+1)sign(x1) | |
8\mu |
\\ u2&=
F2(\kappa+1)ln|x1| | + | |
4\pi\mu |
F1(\kappa+1)sign(x1) | |
8\mu |
\end{align}
\kappa=\begin{cases}3-4\nu& planestrain\\ \cfrac{3-\nu}{1+\nu}& planestress\end{cases}
\nu
\mu
sign(x)=\begin{cases}+1&x>0\\ -1&x<0\end{cases}
If we assume the stresses to vary as
(1/r)
1/r
\varphi=C1r\theta\sin\theta+C2rlnr\cos\theta+C3r\theta\cos\theta+C4rlnr\sin\theta
\begin{align} \sigmarr&=
C | ||||
|
\right)+
C | ||||
|
\right)+
C | ||||
|
\right)+
C | ||||
|
\right)\\ \sigmar\theta&=
C | ||||
|
\right)+
C | ||||
|
\right)\\ \sigma\theta\theta&=
C | ||||
|
\right)+
C | ||||
|
\right)\end{align}
C1,C2,C3,C4
However, the concentrated loads at the vertex are difficult to express in terms of traction boundary conditions because
To get around this problem, we consider a bounded region of the wedge and consider equilibrium of the bounded wedge.[3] [4] Let the bounded wedge have two traction free surfaces and a third surface in the form of an arc of a circle with radius
a
n=er
(er,e\theta)
t=\boldsymbol{\sigma} ⋅ n \impliestr=\sigmarr,~t\theta=\sigmar\theta~.
Next, we examine the force and moment equilibrium in the bounded wedge and get
\begin{align} \sumf1&=F1+
\beta | |
\int | |
\alpha |
\left[\sigmarr(a,\theta)~\cos\theta -\sigmar\theta(a,\theta)~\sin\theta\right]~a~d\theta=0\\ \sumf2&=F2+
\beta | |
\int | |
\alpha |
\left[\sigmarr(a,\theta)~\sin\theta +\sigmar\theta(a,\theta)~\cos\theta\right]~a~d\theta=0\\ \summ3&=
\beta | |
\int | |
\alpha |
\left[a~\sigmar\theta(a,\theta)\right]~a~d\theta=0 \end{align}
a
The traction-free boundary conditions on the edges
\theta=\alpha
\theta=\beta
\sigmar\theta=\sigma\theta\theta=0 at~~\theta=\alpha,\theta=\beta
r=0
If we assume that
\sigmar\theta=0
\begin{align} F1&+
\beta | |
\int | |
\alpha |
\sigmarr(a,\theta)~a~\cos\theta ~d\theta=0\\ F2&+
\beta | |
\int | |
\alpha |
\sigmarr(a,\theta)~a~\sin\theta ~d\theta=0\end{align}
\sigma\theta\theta=0
\theta=\alpha,\theta=\beta
r=0
\sigma\theta\theta=0
\sigmar\theta=0
C2=C4=0
Therefore,
\sigmarr=
2C1\cos\theta | |
r |
+
2C3\sin\theta | |
r |
~;~~ \sigmar\theta=0~;~~\sigma\theta\theta=0
To find a particular solution for
\sigmarr
\sigmarr
C1,C3
\begin{align} F1&+
\beta | |
2\int | |
\alpha |
(C1\cos\theta+C3\sin\theta)~\cos\theta~d\theta=0\\ F2&+
\beta | |
2\int | |
\alpha |
(C1\cos\theta+C3\sin\theta)~\sin\theta~d\theta=0\end{align}
If we take
\alpha=-\pi
\beta=0
F2
F1
\begin{align} F1&+
0 | |
2\int | |
-\pi |
(C1\cos\theta+C3\sin\theta)~\cos\theta~d\theta=0 \impliesF1+C1\pi=0\\ F2&+
0 | |
2\int | |
-\pi |
(C1\cos\theta+C3\sin\theta)~\sin\theta~d\theta=0 \impliesF2+C3\pi=0 \end{align}
C1=-\cfrac{F1}{\pi}~;~~C3=-\cfrac{F2}{\pi}~.
\sigmarr=-
2 | |
\pir |
(F1\cos\theta+F2\sin\theta)~;~~ \sigmar\theta=0~;~~\sigma\theta\theta=0
\begin{align} ur&=-\cfrac{1}{4\pi\mu}\left[F1\{(\kappa-1)\theta\sin\theta-\cos\theta+(\kappa+1)lnr\cos\theta\}+\right.\\ & \left.F2\{(\kappa-1)\theta\cos\theta+\sin\theta-(\kappa+1)lnr\sin\theta\}\right]\\ u\theta&=-\cfrac{1}{4\pi\mu}\left[F1\{(\kappa-1)\theta\cos\theta-\sin\theta-(\kappa+1)lnr\sin\theta\}-\right.\\ & \left.F2\{(\kappa-1)\theta\sin\theta+\cos\theta+(\kappa+1)lnr\cos\theta\}\right] \end{align}
To find expressions for the displacements at the surface of the half plane, we first find the displacements for positive
x1
\theta=0
x1
\theta=\pi
r=|x1|
For
\theta=0
\begin{align} ur=u1&=\cfrac{F1}{4\pi\mu}\left[1-(\kappa+1)ln|x1|\right]\\ u\theta=u2&=\cfrac{F2}{4\pi\mu}\left[1+(\kappa+1)ln|x1|\right] \end{align}
\theta=\pi
\begin{align} ur=-u1&=-\cfrac{F1}{4\pi\mu}\left[1-(\kappa+1)ln|x1|\right]+\cfrac{F2}{4\mu}(\kappa-1)\\ u\theta=-u2&=\cfrac{F1}{4\mu}(\kappa-1)- \cfrac{F2}{4\pi\mu}\left[1+(\kappa+1)ln|x1|\right] \end{align}
u1=\cfrac{F2}{8\mu}(\kappa-1)~;~~u2=\cfrac{F1}{8\mu}(\kappa-1)
u1=\cfrac{F1}{4\pi\mu}~;~~u2=\cfrac{F2}{4\pi\mu}~.
\begin{align} u1&=\cfrac{F1}{4\pi\mu}(\kappa+1)ln|x1|+\cfrac{F2}{8\mu}(\kappa-1)sign(x1)\\ u2&=\cfrac{F2}{4\pi\mu}(\kappa+1)ln|x1|+\cfrac{F1}{8\mu}(\kappa-1)sign(x1) \end{align}
sign(x)=\begin{cases}+1&x>0\\ -1&x<0\end{cases}