Elasticity/Plane strain example 1

Example 1

Given:

The plane strain solution for the stresses in a rectangular block with $0<x_1<a$, $-b<x_2<b$, and $-c<x_3<c$ with a given loading is

${\sigma }_{11}=\frac{3F{x}_1{x}_2}{2b^3};{\sigma }_{12}=\frac{3F(b^2-x_2^2)}{4b^3};{\sigma }_{22}=0;{\sigma }_{33}=-\frac{3\nu F{x}_1{x}_2}{2b^3}.$

Find:

1. Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
2. We wish to use this solution to solve the corresponding problem in which the surfaces $x_3=\pm c$ are traction-free. Determine an approximate corrective solution for this problem by offloading the unwanted force and moment results using elementary bending theory.
3. Find the maximum error in the stress ${\sigma }_{33}$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

Solution

The tractions acting on the block are:

$\begin{array}{cc}\text{at}{x}_1& =0,\widehat{𝐧}\equiv (-1,0,0),t_i=n_1{\sigma }_{1i}\equiv (-{\sigma }_{11},-{\sigma }_{12},-{\sigma }_{13})\\ \text{at}{x}_1& =a,\widehat{𝐧}\equiv (1,0,0),t_i=n_1{\sigma }_{1i}\equiv ({\sigma }_{11},{\sigma }_{12},{\sigma }_{13})\\ \text{at}{x}_2& =-b,\widehat{𝐧}\equiv (0,-1,0),t_i=n_2{\sigma }_{2i}\equiv (-{\sigma }_{21},-{\sigma }_{22},-{\sigma }_{23})\\ \text{at}{x}_2& =b,\widehat{𝐧}\equiv (0,1,0),t_i=n_2{\sigma }_{2i}\equiv ({\sigma }_{21},{\sigma }_{22},{\sigma }_{23})\\ \text{at}{x}_3& =-c,\widehat{𝐧}\equiv (0,0,-1),t_i=n_3{\sigma }_{3i}\equiv (-{\sigma }_{31},-{\sigma }_{32},-{\sigma }_{33})\\ \text{at}{x}_3& =c,\widehat{𝐧}\equiv (0,0,1),t_i=n_3{\sigma }_{3i}\equiv ({\sigma }_{31},{\sigma }_{32},{\sigma }_{33})\end{array}$

Plugging in the expressions for stress,

$\begin{array}{cc}\text{at}{x}_1& =0,t_i\equiv (0,-\frac{3F(b^2-x_2^2)}{4b^3},0)\\ \text{at}{x}_1& =a,t_i\equiv (\frac{3Fa{x}_2}{2b^3},\frac{3F(b^2-x_2^2)}{4b^3},0)\\ \text{at}{x}_2& =-b,t_i\equiv (0,0,0)\\ \text{at}{x}_2& =b,t_i\equiv (0,0,0)\\ \text{at}{x}_3& =-c,t_i\equiv (0,0,\frac{3\nu F{x}_1{x}_2}{2b^3})\\ \text{at}{x}_3& =c,t_i\equiv (0,0,-\frac{3\nu F{x}_1{x}_2}{2b^3})\end{array}$

These tractions are illustrated in the following figure

Tractions on the beam

To unload the tractions on the faces $x_3=\pm c$, we have to superpose the solution to a problem with equal and opposite tractions and moments. In order to use elementary bending theory, we have set up a problem with simple boundary conditions.

Let us first consider the force distribution required for the superposed problem. Since the loading is antisymmetric at $x_3=\pm c$, there is no net force is the $x_3$ direction. Similarly, there is no net moment about the $x_2$ axis.

However, there is a net moment about the $x_1$ axis. Hence, the problem to be superposed should have a bending stress distribution ${\sigma }_{33}=C{x}_2$, where $C$ is a constant that is chosen so as to make the total bending moment (original problem + superposed problem) equal to zero. (Note: Think of a beam in the $x_3-x_2$ plane subjected to bending moments at the ends.)

The total stress for the corrected problem is

${\sigma }_{33}=-\frac{3\nu F{x}_1{x}_2}{2b^3}+C{x}_2$

The bending moment for a cross-section of the beam in the $x_3-x_2$ plane about the $x_1$ axis is $M=-{\sigma }_{33}I/x_2$, where $I={\int }_0^a{\int }_{-b}^b{x}_2^{2}d{x}_{2}d{x}_1$.

Since ${\sigma }_{33}$ varies with $x_1$, the total bending moment for the beam is given by

$\begin{array}{cc}M& ={\int }_0^a{\int }_{-b}^b\left(\frac{-{\sigma }_{33}}{x_2}\right)x_2^{2}d{x}_{2}d{x}_1\\ & ={\int }_0^a{\int }_{-b}^b(\frac{3\nu F{x}_1{x}_2}{2b^3}-C{x}_2)x_{2}d{x}_{2}d{x}_1\\ & ={\int }_0^a{[\frac{3\nu F{x}_1{x}_2^3}{6b^3}-\frac{C{x}_2^3}{3}]}_{-b}^{b}d{x}_1\\ & ={\int }_0^a[\frac{\nu F{x}_1{b}^3}{b^3}-\frac{2C{b}^3}{3}]d{x}_1\\ & ={[\frac{\nu F{x}_1^2{b}^3}{2b^3}-\frac{2C{x}_1{b}^3}{3}]}_0^a\\ & =\frac{\nu F{a}^2{b}^3}{2b^3}-\frac{2Ca{b}^3}{3}\end{array}$

Setting the bending moment to zero, we have

$C=\frac{3\nu Fa}{4b^3}$

Therefore, the corrected solution is

${\sigma }_{33}=-\frac{3\nu F{x}_1{x}_2}{2b^3}+\frac{3\nu Fa{x}_2}{4b^3}$

Ideally, for a problem with zero tractions on $x_3=\pm c$, we should have ${\sigma }_{33}=0$. Therefore, the error in our solution is

${\sigma }_{33}^{\text{err}}=\text{abs}(\frac{3\nu F{x}_1{x}_2}{2b^3}-\frac{3\nu Fa{x}_2}{4b^3})$

The error is maximum at $(0,-b)$, $(0,b)$, $(a,-b)$, and $(a,b)$. Thus,

$\begin{array}{cc}{{\sigma }_{33}^{\text{err}}|}_{(0,-b)}& =\text{abs}\left(\frac{3\nu Fab}{4b^3}\right)=\text{abs}\left(\frac{3\nu Fa}{4b^2}\right)\\ {{\sigma }_{33}^{\text{err}}|}_{(0,b)}& =\text{abs}\left(\frac{3\nu Fab}{4b^3}\right)=\text{abs}\left(\frac{3\nu Fa}{4b^2}\right)\\ {{\sigma }_{33}^{\text{err}}|}_{(a,-b)}& =\text{abs}\left(\frac{3\nu Fab}{4b^3}\right)=\text{abs}\left(\frac{3\nu Fa}{4b^2}\right)\\ {{\sigma }_{33}^{\text{err}}|}_{(a,b)}& =\text{abs}\left(\frac{3\nu Fab}{4b^3}\right)=\text{abs}\left(\frac{3\nu Fa}{4b^2}\right)\end{array}$

The maximum error is

$\frac{3\nu Fa}{4b^2}$

The maximum tensile stress is

${\sigma }_{11}(a,b)=\frac{3Fab}{2b^3}=\frac{3Fa}{2b^2}$

Therefore, the ratio of the maximum error in ${\sigma }_{33}$ to the maximum tensile stress is

$\text{Ratio}=\frac{\nu }{2}$

Elasticity