Introduction to Elasticity/Concentrated force on half plane

< Introduction to Elasticity

Concentrated Force on a Half-Plane

Concentrated force on a half plane

From the Flamant Solution

\begin{align} 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{align}

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{align} 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{align}

or,

\begin{align} F_1 + \pi C_1 & = 0 \\ F_2 - \pi C_3 & = 0 \end{align}

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 F_2\,


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{align} 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{align}

At \theta = 0, (x_1 > 0, x_2 = 0),

\begin{align} 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{align}

At \theta = -\pi, (x_1 < 0, x_2 = 0),

\begin{align} 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{align}

where

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

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

\begin{align} 2\mu u_1 & = \frac{F_2}{4\pi}(\kappa-1)\\ 2\mu u_2 & = \frac{F_2}{2\pi} \end{align}

The displacements are

\begin{align} 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{align}

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 F_1\,


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{align} 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{align}


Stresses and displacements due to F_1 + F_2\,


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{align} 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{align}
This article is issued from Wikiversity - version of the Wednesday, August 08, 2007. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.