From Wikiversity

[edit] 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

$\sigma_{11} = \frac{3 F x_1 x_2}{2 b^3}~;~~ \sigma_{12} = \frac{3 F (b^2 - x_2^2)}{4 b^3}~;~~ \sigma_{22} = 0~;~~ \sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3}.$

Find:

Find the tractions on the surfaces of the block and illustrate the results on a sketch of the block.
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.
Find the maximum error in the stress $σ 33$ in the corrected solution and compare it with the maximum tensile stress in the plane strain solution.

[edit] Solution

The tractions acting on the block are:

$\begin{align} \text{at}~ x_1 &= 0~,~~\widehat{\mathbf{n}}{} \equiv (-1,0,0)~,~~t_i = n_1 \sigma_{1i} \equiv (-\sigma_{11},-\sigma_{12},-\sigma_{13}) \\ \text{at}~ x_1 &= a~,~~\widehat{\mathbf{n}}{} \equiv (1,0,0)~,~~t_i = n_1 \sigma_{1i} \equiv (\sigma_{11},\sigma_{12},\sigma_{13}) \\ \text{at}~ x_2 &= -b~,~~\widehat{\mathbf{n}}{} \equiv (0,-1,0)~,~~t_i = n_2 \sigma_{2i} \equiv (-\sigma_{21},-\sigma_{22},-\sigma_{23}) \\ \text{at}~ x_2 &= b~,~~\widehat{\mathbf{n}}{} \equiv (0,1,0)~,~~t_i = n_2 \sigma_{2i} \equiv (\sigma_{21},\sigma_{22},\sigma_{23}) \\ \text{at}~ x_3 &= -c~,~~\widehat{\mathbf{n}}{} \equiv (0,0,-1)~,~~t_i = n_3 \sigma_{3i} \equiv (-\sigma_{31},-\sigma_{32},-\sigma_{33}) \\ \text{at}~ x_3 &= c~,~~\widehat{\mathbf{n}}{} \equiv (0,0,1)~,~~t_i = n_3 \sigma_{3i} \equiv (\sigma_{31},\sigma_{32},\sigma_{33}) \end{align}$

Plugging in the expressions for stress,

$\begin{align} \text{at}~ x_1 &= 0~,~~t_i \equiv (0,-\frac{3 F (b^2 - x_2^2)}{4 b^3},0) \\ \text{at}~ x_1 &= a~,~~t_i \equiv (\frac{3 F a x_2}{2 b^3},\frac{3 F (b^2 - x_2^2)}{4 b^3},0)\\ \text{at}~ x_2 &= -b~,~~t_i \equiv (0,0,0) \\ \text{at}~ x_2 &= b~,~~t_i \equiv (0,0,0) \\ \text{at}~ x_3 &= -c~,~~t_i \equiv (0,0,\frac{3 \nu F x_1 x_2}{2 b^3}) \\ \text{at}~ x_3 &= c~,~~t_i \equiv (0,0,-\frac{3 \nu F x_1 x_2}{2 b^3}) \end{align}$

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 $σ 33 = C x 2$ , 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

$\sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3} + C x_2$

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 $I = \int_0^a \int_{-b}^b x_2^2 dx_2 dx_1$ .

Since $σ 33$ varies with $x 1$ , the total bending moment for the beam is given by

$\begin{align} M & = \int_0^a \int_{-b}^b \left(\frac{-\sigma_{33}}{x_2}\right) x_2^2 dx_2 dx_1 \\ & = \int_0^a \int_{-b}^b \left(\frac{3 \nu F x_1 x_2}{2 b^3} - C x_2 \right) x_2 dx_2 dx_1 \\ & = \int_0^a \left[\frac{3\nu F x_1 x_2^3}{6 b^3} - \frac{C x_2^3}{3} \right]_{-b}^b dx_1 \\ & = \int_0^a \left[\frac{\nu F x_1 b^3}{b^3} - \frac{2C b^3}{3} \right] dx_1 \\ & = \left[\frac{\nu F x_1^2 b^3}{2b^3} - \frac{2C x_1 b^3}{3} \right]_0^a \\ & = \frac{\nu F a^2 b^3}{2b^3} - \frac{2C a b^3}{3} \end{align}$

Setting the bending moment to zero, we have

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

Therefore, the corrected solution is

${ \sigma_{33} = -\frac{3 \nu F x_1 x_2}{2 b^3} + \frac{3\nu F a x_2}{4 b^3}}$

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

$\sigma_{33}^{\text{err}} = \text{abs}\left(\frac{3 \nu F x_1 x_2}{2 b^3} - \frac{3\nu F a x_2}{4 b^3}\right)$

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

$\begin{align} \left.\sigma_{33}^{\text{err}}\right|_{(0,-b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(0,b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(a,-b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \\ \left.\sigma_{33}^{\text{err}}\right|_{(a,b)} & = \text{abs}\left(\frac{3\nu F a b}{4 b^3}\right) = \text{abs}\left(\frac{3\nu F a}{4 b^2}\right) \end{align}$