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{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:


 * 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{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

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}{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 $$\sigma_{33}$$ 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$$, we should have $$\sigma_{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}$$ The maximum error is

{\frac{3\nu F a}{4 b^2}} $$ The maximum tensile stress is

\sigma_{11}(a,b) = \frac{3 F a b}{2 b^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}} $$