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Elasticity/Prandtl stress function

Prandtl Stress Function φ
The traction free BC is obviously difficult to satisfy if the cross-section is not a circle or an ellipse.

To simplify matters, we define the Prandtl stress function $$\phi(x_1,x_2)\,$$ using

{ \sigma_{13} = \phi_{,2} ; \sigma_{23} = -\phi_{,1} } $$ You can easily check that this definition satisfies equilibrium.

It can easily be shown that the traction-free BCs are satisfied if

{ \frac{d\phi}{ds} = 0 \forall~(x_1,x_2) \in \partial\text{S} } $$ where $$s$$ is a coordinate system that is tangent to the boundary.

If the cross section is simply connected, then the BCs are even simpler:

{ \phi = 0 \forall~(x_1,x_2) \in \partial\text{S} } $$

From the compatibility condition, we get a restriction on $$\phi$$

{ \nabla^2{\phi} = C \forall~(x_1,x_2) \in \text{S} } $$ where $$C$$ is a constant.

Using relations for stress in terms of the warping function $$\psi$$, we get

{ \nabla^2{\phi} = -2\mu\alpha \forall~(x_1,x_2) \in \text{S} } $$

Therefore, the twist per unit length is

{ \alpha = -\frac{1}{2\mu} \nabla^2{\phi} } $$

The applied torque is given by

{ T = -\int_{S} (x_1 \phi_{,1} + x_2 \phi_{,2}) dA \, } $$ For a simply connected cylinder,

{ T =2 \int_{S} \phi dA \, } $$

The projected shear traction is given by

{\tau = \sqrt{(\phi_{,1})^2+ (\phi_{,2})^2}} $$

The projected shear traction at any point on the cross-section is tangent to the contour of constant $$\phi\,$$ at that point.

The relation between the warping function $$\psi\,$$ and the Prandtl stress function $$\phi\,$$ is

{ \psi_{,1} = \frac{1}{\mu\alpha} \phi_{,2} + x2 ~; \psi_{,2} = -\frac{1}{\mu\alpha} \phi_{,1} - x1 } $$

Membrane Analogy
The equations

\nabla^2{\phi} = -2\mu\alpha \forall~(x_1,x_2) \in \text{S}; \phi = 0 \forall~(x_1,x_2) \in \partial\text{S} $$ are similar to the equations that govern the displacement of a membrane that is stretched between the boundaries of the cross-sectional curve and loaded by an uniform normal pressure.

This analogy can be useful in estimating the location of the maximum shear stress and the torsional rigidity of a bar.

 * The stress function is proportional to the displacement of the membrane from the plane of the cross-section.
 * The stiffest cross-sections are those that allow the maximum volume to be developed between the deformed membrane and the plane of the   cross-section for a given pressure.
 * The shear stress is proportional to the slope of the membrane.

title: Elasticity/Prandtl stress function


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Metaphysical philosophy

Welcome Beloved I AM!
This page was created for those of us who travel inward while thinking deeply; only to come to the realization that the only thing real in this universe is that which is eternal. To digress us from this truth is legion and his lies, which have infiltrated us at every level. The time has come whereby the true manifestations of the highest living God beloved I AM THAT I AM, rise up as one Beloved I AM, to bring light to all of those dark places exposing all lies while seeking only real truth. I and my father are one!

The door that lies before us is that which is eternal, and the path that lies ahead of us involves expanding the mind to think both finitely and infinitely. Please feel free to join in and help to grow this department.

With that I leave you with this Decree!
In the name of my constant, loving I AM Presence, I AM THAT I AM and my holy Christ self. I call to Beloved I AM, Archangel Michael and to Beloved I AM, Saint Germain to keep intact through every passing moment the shimmering shower of Gods beautiful light manifesting as divinely charged energy radiating from this page transcending with the ever expanding violet fire of Freedom's forgiving, transmuting flame! Cause the ever expanding energy of this flame projected downward into the force field of my human energies to completely change every negative condition into the positive polarity of my own Great God Self! I decree this in memory of the blessed dawning day when all discord—cause, effect, record, and memory—is forever changed into the Victory of Light and the peace of the ascended Jesus Christ. I AM now constantly accepting the full power and manifestation of this fiat of Light and calling it into instantaneous action by my own God-given free will and the power to accelerate without limit this sacred release of assistance from God's own heart until all men are ascended and God-free in the Light that never, never, never fails!

And for you non-believers, I dare ask the question: Can nothing create something?

VagyokC4 21:05, 5 August 2007 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Metaphysical philosophy


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Elasticity/Solution strategy for Prandtl stress function

Solution strategy using the Prandtl stress function
The equation $$\nabla^2{\phi} = -2\mu\alpha$$ is a Poisson equation.

Since the equation is inhomogeneous, the solution can be written as

\phi = \phi_p + \phi_h \, $$ where $$\phi_p\,$$ is a particular solution and $$\phi_h\,$$ is the solution of the homogeneous equation.

Examples of particular solutions are, in rectangular coordinates,

\phi_p = -\mu\alpha x_1^2 ; \phi_p = -\mu\alpha x_2^2 \, $$ and, in cylindrical co-ordinates,

\phi_p = -\frac{\mu\alpha r^2}{2} $$

The homogeneous equation is the Laplace equation $$\nabla^2{\phi}=0$$, which is satisfied by both the real and the imaginary parts of any analytic function $$f(z)\,$$ of the complex variable

z = x_1 + i x_2 = r e^{i\theta} \, $$ Thus,

\phi_h = \text{Re}(f(z)) \text{or} \phi_h = \text{Im}(f(z)) $$

Suppose $$f(z) = z^n\,$$. Then, examples of $$\phi_h\,$$ are

\phi_h = C_1 r^n\cos(n\theta) ; \phi_h = C_2 r^n\sin(n\theta) ;   \phi_h = C_3 r^{-n}\cos(n\theta) ;   \phi_h = C_4 r^{-n}\sin(n\theta) $$ where $$C_1\,$$, $$C_2\,$$, $$C_3\,$$, $$C_4\,$$ are constants.

Each of the above can be expressed as polynomial expansions in the $$x_1\,$$ and $$x_2\,$$ coordinates.

Approximate solutions of the torsion problem for a particular cross-section can be obtained by combining the particular and homogeneous solutions and adjusting the constants so as to match the required shape.

Only a few shapes allow closed-form solutions. Examples are
 * Circular cross-section.
 * Elliptical cross-section.
 * Circle with semicircular groove.
 * Equilateral triangle.


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Elasticity/Torsion of triangular cylinder

Example: Equilateral triangle
The equations of the three sides are

 * $$\begin{align}

\text{side}~\partial S^{(1)} ~: & f_1(x_1,x_2) = x_1 - \sqrt{3} x_2 + 2a = 0 \\ \text{side}~\partial S^{(2)} ~: & f_2(x_1,x_2) = x_1 + \sqrt{3} x_2 + 2a = 0\\ \text{side}~\partial S^{(3)} ~: & f_3(x_1,x_2) = x_1 - a = 0 \end{align}$$

Let the Prandtl stress function be

\phi = C f_1 f_2 f_3 \, $$

Clearly, $$\phi = 0\,$$ at the boundary of the cross-section (which is what we need for solid cross sections).

Since, the traction-free boundary conditions are satisfied by $$\phi\,$$, all we have to do is satisfy the compatibility condition to get the value of $$C\,$$. If we can get a closed for solution for $$C\,$$, then the stresses derived from $$\phi\,$$ will satisfy equilibrium.

Expanding $$\phi\,$$ out,

\phi = C (x_1 - \sqrt{3} x_2 + 2a)(x_1 + \sqrt{3} x_2 + 2a)(x_1 - a) $$

Plugging into the compatibility condition

\nabla^2{\phi} = 12 C a = -2\mu\alpha $$ Therefore,

C = -\frac{\mu\alpha}{6a} $$ and the Prandtl stress function can be written as

\phi = -\frac{\mu\alpha}{6a} (x_1^3+3ax_1^2+3ax_2^2-3x_1x_2^2-4a^3) $$

The torque is given by

T = 2\int_S \phi dA = 2\int_{-2a}^{a} \int_{-(x_1+2a)/\sqrt{3}}^{(x_1+2a)/\sqrt{3}} \phi dx_2 dx_1 = \frac{27}{5\sqrt{3}} \mu\alpha a^4 $$

Therefore, the torsion constant is

\tilde{J} = \frac{27 a^4}{5\sqrt{3}} $$

The non-zero components of stress are
 * $$\begin{align}

\sigma_{13} = \phi_{,2} & = \frac{\mu\alpha}{a}(x_1-a)x_2 \\ \sigma_{23} = -\phi_{,1} & = \frac{\mu\alpha}{2a}(x_1^2+2ax_1-x_2^2) \end{align}$$

The projected shear stress

\tau = \sqrt{\sigma_{13}^2+ \sigma_{23}^2} $$ is plotted below The maximum value occurs at the middle of the sides. For example, at $$(a,0)$$,

\tau_{\text{max}} = \frac{3\mu\alpha a}{2} $$

The out-of-plane displacements can be obtained by solving for the warping function $$\psi$$. For the equilateral triangle, after some algebra, we get

u_3 = \frac{\alpha x_2}{6a} (3x_1^2 - x_2^2) $$ The displacement field is plotted below


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Adobe Flash

Short Introduction
Adobe Flash is a 2D animation program, which uses vector graphics. It is considered as one of the best private animation tools, and one of the few which is not based to be used within animation studios. It was originally developed by Macromedia, but was recently bought by Adobe.

Tools
Before explaining animation, we will explain the tools:

Selection Tool:

This is the selection tool. It is used for selecting objects, or portions of one. When selecting a brushed area it will only select the same color of brushed area which is connected to the original selected area, lines will also confine a selected area. You can also move things with this tool.

Subselection Tool:

This is the subselection tool, it is used for selecting points in the vector graphic. When you mark something with this tool, you will highlight possible selection points. By selecting these points you can drag them to another position, when you do this the lines between the two nearest points on each side follow. However unlike the selection tool this shows all outer lines of an object instead of the filling.

Free Transform Tool:

This is the free transform tool. It is basically the same as the selection tool, however it can stretch, enlarge/shrink, and change the center of the selected object.

Gradient Transform Tool:

This is the gradient transform tool, with this you can change the angles and length of gradient filling. You can also move selected filling with it.

Line Tool:

The line tool is used to draw straight marks from one point to another. Using the Selection Tool, you can bend lines after they are created, to get a curve effect.

Lasso Tool:

Like the Selection tool, the Lasso tool allows you to select certain objects. Rather than drawing a square marqueé over the objects, the Lasso tool lets you draw an outline of what you want to select by dragging your mouse around the object, creating whatever shape is drawn. Whatever is inside the shape will be selected.

Text Tool:

The text tool will let you put text directly into the Stage. The color can be changed, and whatever fonts you have on your computer will be available to use.

Oval Tool:

The oval tool can be used to draw an oval or circle. Like the rectangle tool, hɛolding SHIFT while drawing will make the oval a perfect circle.

Rectangle Tool/Polystar Tool:

The rectangle tool is used to create simple four sided polygons. Additionally, one can use the line tool and the fill tool to draw the shape, but it is not recommended. The Polystar tool (click and hold the rectangle tool button to receive a drop-down menu, where you can select Polystar) can be used to create polygons (the number of sides can be adjusted in Options in the Properties tab) and stars (the number of points can be adjusted as well in Options in the Properties tab).

title: Adobe Flash


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Elasticity/Torsion of thin walled open sections

Torsion of thin-walled open sections
Examples are I-beams, channel sections and turbine blades.

We assume that the length $$b\,$$ is much larger than the thickness $$t\,$$, and that $$t\,$$ does not vary rapidly with change along the length axis $$\xi\,$$.

Using the membrane analogy, we can neglect the curvature of the membrane in the $$\xi\,$$ direction, and the Poisson equation reduces to

\frac{d^\phi}{d\eta^2} = -2\mu\alpha $$ which has the solution

\phi = \mu\alpha\left(\frac{t^2}{4}-\eta^2\right) $$ where $$\eta$$ is the coordinate along the thickness direction.

The stress field is

\sigma_{3\xi} = \frac{\partial }{\partial} {\phi}{\eta} = -e\mu\beta\eta ; \sigma_{3\eta} = 0 $$ Thus, the maximum shear stress is

\tau_{\text{max}} = \mu\beta t_{\text{max}}\, $$


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Elasticity/Torsion of thin walled closed sections

Torsion of thin-walled closed sections
The Prandtl stress function $$\phi\,$$ can be approximated as a linear function between $$\phi_1\,$$ and $$0\,$$ on the two adjacent boundaries.

The local shear stress is, therefore,

\sigma_{3s} = \frac{\phi_1}{t} $$ where $$s\,$$ is the parameterizing coordinate of the boundary curve of the cross-section and $$t\,$$ is the local wall thickness.

The value of $$\phi_1\,$$ can determined using

\phi_1 = \frac{2\mu\alpha A}{\oint_S \frac{dS}{t}} $$ where $$A\,$$ is the area enclosed by the mean line between the inner and outer boundary.

The torque is approximately

T = 2A\phi_1 \, $$


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Elasticity/Concentrated force on 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 F2
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 $$

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 F1
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 $$ The displacements are
 * $$\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 F1 + F2
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}$$


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Elasticity/Distributed force on half plane

Distributed force on a half-plane

 * Applied load is $$p(\xi)\,$$ per unit length in the $$x_2\,$$ direction.
 * We already know the stresses and displacements due to a  concentrated force. The stresses and displacements due to the   distributed load can be found by superposition.
 * The Flamant solution is used as a Green's function, i.e., the distributed load is taken as the limit of a set of point   loads of magnitude $$p(\xi)\delta\xi\,$$.

At the point $$P\,$$

u_2 = - \frac{(\kappa+1)}{4\pi\mu} \int_A p(\xi)\ln|x - \xi|~d\xi $$ As $$x \rightarrow \infty\,$$, $$u_2\,$$ is unbounded. However, if we are interested in regions far from $$A\,$$, we can apply the distributed force as a statically equivalent concentrated force and get displacements using the concentrated force solution.

The avoid the above issue, contact problems are often formulated in terms of the displacement gradient

\frac{du_2}{dx_1} = - \frac{(\kappa+1)}{4\pi\mu} \int_A \frac{p(\xi)}{x - \xi}~d\xi $$ If the point $$P\,$$ is inside $$A\,$$, then the integral is taken to be the sum of the integrals to the left and right of $$P\,$$.


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Elasticity/Flat punch indentation

Indentation due to a frictionless rigid flat punch

 * Start with uneven surface profile $$u_0(x_1)\,$$.
 * Unsymmetric load $$F\,$$, but sufficient for complete contact over the area $$A\,$$.

Displacement in $$x_2\,$$ direction is

u_2 = - u_0(x_1) + C_1 x_1 + C_0\, $$ where $$C_0\,$$ is a rigid body translation and $$C_1 x_1\,$$ is a rigid body rotation.

Rigid body motions can be determined using a statically equivalent set of forces and moments
 * $$\begin{align}

\int_A p(\xi)~d\xi & = -F \\ \int_A p(\xi)\xi~d\xi & = -F~d \end{align}$$

The displacement gradient is

-\frac{du_0}{dx_1} +C_1 = - \frac{(\kappa+1)}{4\pi\mu} \int_{-a}^a \frac{p(\xi)}{x - \xi}~d\xi ~; -a < x < a $$

Integral is a Cauchy Singular Integral that appears often and very naturally when the problem is solved using complex variable methods.

Note that the only thing we are interested in is the distribution of contact forces $$p(\xi)\,$$.If we change the variables so that

$$x = a \cos\phi\,$$ and $$\xi = a \cos\theta\,$$, then

\frac{1}{a\sin\phi}\frac{du_0}{d\phi}+C_1 = - \frac{(\kappa+1)}{4\pi\mu} \int_0^{\pi} \frac{p(\theta)\sin\theta}{\cos\phi - \cos\theta}~d\theta ~; 0 < \phi < \pi $$

If we write $$p(\theta)\,$$ and $$du_0/d\phi\,$$ as
 * $$\begin{align}

p(\theta) & = \sum_0^{\infty} \frac{p_n \cos(n\theta)}{\sin\theta} \\ \frac{du_0}{d\phi} & = \sum_1^{\infty} u_n \sin(n\phi) \end{align}$$ and do some algebra, we get
 * $$\begin{align}

p_0 & = -\frac{F}{\pi a} \\ p_1 & = -\frac{F d}{\pi a^2} \\ p_n & = -\frac{4\mu u_n}{(\kappa+1)a} ~; n > 1 \end{align}$$

Flat punch with symmetric load

 * $$u_0 = C\,$$

In this case,

\frac{du_0}{d\phi} = 0 \Rightarrow u_n = 0 ~; n = 1 {\infty} $$ Also, $$d = 0\,$$ (origin at the center of $$A\,$$), hence $$p_1 = 0\,$$. Therefore,

p(x) = \frac{p_0}{\sin\phi} = -\frac{F}{\pi \sqrt{a^2 - x^2}} $$ At $$x = \pm a$$, the load is infinite, i.e. there is a singularity.

title: Elasticity/Flat punch indentation


text:
Elasticity/Hertz contact

The Hertz problem : rigid cylindrical punch

 * The contact length $$a$$ depends on the load $$F\,$$.
 * There is no singularity at $$x = \pm a$$.
 * The radius of the cylinder ($$R\,$$) is large.

We have,

\frac{d^2 u_0}{dx^2} = -\frac{1}{R} $$

Hence,

u_0 = C_0 - \frac{x^2}{2R} = C_0 - \frac{a^2\cos(2\phi)}{4R}-\frac{a^2}{4R} $$ and

\frac{d u_0}{d\phi} = -\frac{a^2\sin(2\phi)}{2R} $$

Therefore,

u_1 = 0 ~; u_2 = \frac{a^2}{2R} ~; u_n = 0 ~(n > 2) $$ and

p_0 = -\frac{F}{\pi a} ~; p_1 = 0 ~; p_2 = \frac{2\mu a}{R(\kappa+1)} ~; p_n = 0 ~(n > 2) $$

Plug back into the expression for $$p(\theta)$$ to get

p(\theta) = \left(-\frac{F}{\pi a} + \frac{2\mu a}{R(\kappa+1)}\cos(2\theta) \right)/\sin\theta $$

This expression is singular at $$\theta=0$$ and $$\theta=\pi$$, unless we choose

\frac{F}{\pi a} = \frac{2\mu a}{R(\kappa+1)} \Rightarrow a = \sqrt{\frac{F(\kappa+1)R}{2\pi\mu}} $$

Plugging $$a$$ into the equation for $$p(\theta)$$,

p(\theta) = -\frac{2F\sin\theta}{\pi a} \Rightarrow p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2} $$

Two deformable cylinders
If instead of the half-plane we have an cylinder; and instead of the rigid cylinder we have a deformable cylinder, then a similar approach can be used to obtain the contact length $$a\,$$

a = \sqrt{\frac{FR_1R_2}{2\pi(R_1+R_2)}\left( \frac{\kappa_1+1}{\mu_1} + \frac{\kappa_2+1}{\mu_2}\right)} $$ and the force distribution $$p$$

p(x) = -\frac{2F\sqrt{a^2-x^2}}{\pi a^2} $$

title: Elasticity/Hertz contact


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Elasticity/Transversely loaded wedge

Wedge loaded transversely by a concentrated load
Given:

A wedge of infinite length with a concentrated load $$\mathbf{P} = P~\widehat{\mathbf{e}}_{2}$$ per unit wedge thickness at the vertex. Plane stress/strain.

Find:

The stress field in the wedge.

Solution
From the Flamant solution, we know that the stress field in the wedge is
 * $$\begin{align}

\sigma_{rr} & = \frac{2}{r}\left(C_1\cos\theta+C_2\sin\theta\right) \\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}$$ The constants $$C_1\,$$ and $$C_2\,$$ can be found by using the equilibrium conditions
 * $$\begin{align}

2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right)\cos\theta ~d\theta & = 0 \\ P + 2\int_{-\beta}^{\beta} \left(C_1\cos\theta - C_2\sin\theta\right) \sin\theta ~d\theta & = 0 \end{align}$$ or,
 * $$\begin{align}

C_1 \left[2\beta + \sin(2\beta)\right] & = 0 \\ P + C_2\left[\sin(2\beta) - 2 \beta\right] & = 0 \end{align}$$ Therefore,

C_1 = 0 ~; C_2 = \frac{P}{2\beta - \sin(2\beta)} $$ Hence, the stresses are
 * $$\begin{align}

\sigma_{rr} & = \frac{2P\sin\theta}{r\left[2\beta-\sin(2\beta)\right]}\\ \sigma_{r\theta} & = 0 \\ \sigma_{\theta\theta} & = 0 \end{align}$$

title: Elasticity/Transversely loaded wedge


text: File:Wedge with vertical force.png

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Elasticity/Torsion of rectangular sections

Torsion of a bar with a rectangular cross section
Given:

A cylinder with a rectangular cross sections $$a\times b$$ under torison.

Show:

Why a function of the form

\varphi = m\left(\frac{x_1^2}{a^2}-1\right)\left(\frac{x_2^2}{b^2}-1\right) $$ cannot be used as a Prandtl stress function for this cross section.

Solution
For compatibility, we need

\nabla^2{\varphi} = C $$ However, for the given $$\varphi$$,

\nabla^2{\varphi} = -\frac{2m}{a^2b^2}\left[a^2+b^2-(x_1^2+x_2^2)\right] $$ which is not constant. Hence, the given function cannot be used as a Prandtl stress function.


text: File:Torsion rectangle.png

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Elasticity/Stress example 1

Example 1
For a unit cube that is in statical equilibrium and has a homogeneous state of stress, show that there are only six possible components of stress.

Solution
Since the stress in the cube is homogeneous, the forces on opposite faces of the cube should be equal and opposite. Hence, the possible components of stress are

\sigma_{xx},~\sigma_{yy},~\sigma_{zz},~\tau_{xy},~\tau_{yx},~ \tau_{yz},~\tau_{zy},~\tau_{zx},~\tau_{xz}. $$ The homogeneity of the stress also means that the three components of force on any face must pass through the center of the face.

Since the cube is in static equilibrium, the total moment about an axis passing through the center of the cube and parallel to the $$x\,$$ axis should be zero. (Recall that the moment of a force about an axis measures the tendency to rotate a rigid body about the axis. Also recall that the sign of a moment is positive or negative depending on the clockwise or anticlockwise direction of rotation due to the applied force).

Hence, $$\tau_{yz} = \tau_{zy}\,$$. Similarly, $$\tau_{zx} = \tau_{xz}\,$$, and $$\tau_{xy} = \tau_{yx}\,$$.

Therefore, the six possible components of stress are

\sigma_{xx},~\sigma_{yy},~\sigma_{zz},~\tau_{xy},~\tau_{yz},~\tau_{zx}. $$

title: Elasticity/Stress example 1


text:
Elasticity/Kinematics example 1

Example 1
Take a unit cube of material. Rotate it 90 degrees in the clockwise direction around the z-axis. Calculate the strains. Discuss your results - their accuracy and the reasons for your conclusions.

Solution
The strains are related to displacements by

\epsilon_{xx} = \frac{\partial u}{\partial x};~ \epsilon_{yy} = \frac{\partial v}{\partial y};~ \epsilon_{zz} = \frac{\partial w}{\partial z};~ \gamma_{xy} = \frac{\partial u}{\partial y} + \frac{\partial v}{\partial x};~ \gamma_{yz} = \frac{\partial v}{\partial z} + \frac{\partial w}{\partial y};~ \gamma_{zx} = \frac{\partial w}{\partial x} + \frac{\partial u}{\partial z} $$

Let us consider rotation about the center of the cube. Since the problem concerns a pure rotation, a cylindrical co-ordinate system is appropriate. This problem also provides us a easy way of trying out Maple. Here are the steps that you can follow to find the strains at a point in the cube.

r := \sqrt{x^{2} + y^{2}} $$

\theta := arctan(\frac{y}{x}) $$

x1 := \frac{\sqrt{x^2 + y^2}}{\sqrt{1 + \frac{y^2}{x^2}}} $$

y1 := \frac{\sqrt{x^2 + y^2}\,y}{x\,\sqrt{1 + \frac{y^2}{x^2}}} $$

x2 := -\frac{\sqrt{x^2 + y^2}\,y}{x\,\sqrt{1 + \frac{y^2}{x^2}}} $$

y2 := \frac{\sqrt{x^2 + y^2}}{\sqrt{1 + \frac {y^2}{x^2}}} $$

u := -\frac{\sqrt{x^2 + y^2}\,y}{x\,\sqrt{1 + \frac {y^2}{x^2}}} -\frac{\sqrt{x^2 + y^2}}{\sqrt{1 + \frac {y^2}{x^2}}} $$

v := \frac{\sqrt{x^2 + y^2}}{\sqrt{1 + \frac {y^2}{x^2} }} - \frac{\sqrt{x^2 + y^2}\,y}{x\,\sqrt{1 + \frac {y^2}{x^2}}} $$

epsx := -\frac{\sqrt{x^2 + y^2}}{x\,\sqrt{\frac{x^2 + y^2}{x^2}}} $$

epsy := -\frac{\sqrt{x^2 + y^2}}{x\,\sqrt{\frac{x^2 + y^2}{x^2}}} $$

gamxy := 0 $$ From the above Maple calculation, and noting that there is no motion in the $$z$$ direction, the strains in the cube are

\epsilon_{xx} = -1;~\epsilon_{yy} = -1; \epsilon_{zz} = 0; \gamma_{xy} = 0; \gamma_{yz} = 0; \gamma_{zx} = 0 $$ A pure rigid body rotation should not result in any non-zero strains.

Therefore, the measure of strain we have used is not appropriate for large rigid body motions.

title: Elasticity/Kinematics example 1


text:
Elasticity/Constitutive example 1

Example 1
Take a uniform bar of length "56", cross-section area "6", density "4", Young's modulus "8", and hang it from a rigid, hypothetical ceiling. Calculate how much the bar increases in length due to its own weight. Assume that the acceleration due to gravity is "6" and the bar is fixed rigidly to the ceiling.

Solution
The tensile stress at a point P, at a distance $$x$$ from the ceiling is

\sigma_{2x} = \rho~g~(L - x) $$ From Hooke's law

\epsilon_{xx} = \frac{\sigma_{xx}}{E} = \frac{\rho~g~(L-x)}{E} $$ Now,

\frac{\partial u}{\partial x} = \epsilon_{xx} $$ Integrating,

u(x) = \frac{\rho~g~(2Lx - x^2)}{2E} + A $$ Applying the boundary conditions, $$u(x) = 0$$ at $$x = 0$$, we get $$A = 0$$. Therefore, the increase in length is

u(L) = \frac{\rho~g~L^2}{2E} $$

title: Elasticity/Constitutive example 1


text:
Elasticity/Transformation example 1

Example 1
Derive the transformation rule for second order tensors ($$T^{'}_{ij} = l_{ip} l_{jq} T_{pq}$$). Express this relation in matrix notation.

Solution
A second-order tensor $$\mathbf{T}$$ transforms a vector $$\mathbf{u}$$ into another vector $$\mathbf{v}$$. Thus,

\mathbf{v} = \mathbf{T}\mathbf{u} = \mathbf{T}\bullet\mathbf{u} $$ In index and matrix notation,
 * $$\text{(1)} \qquad

v_i = T_{ij} u_i \leftrightarrow v_p = T_{pq} u_q ~\text{or,}~ \left[v\right] = \left[T\right] \left[u\right] $$ Let us determine the change in the components of $$\mathbf{T}$$ with change the basis from ($$\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$$) to ($$\mathbf{e}_1^{'},\mathbf{e}_2^{'},\mathbf{e}_3^{'}$$). The vectors $$\mathbf{u}$$ and $$\mathbf{v}$$, and the tensor $$\mathbf{T}$$ remain the same. What changes are the components with respect to a given basis. Therefore, we can write
 * $$\text{(2)} \qquad

v^{'}_i = T^{'}_{ij} u^{'}_i ~\text{or,}~ \left[v\right]^{'} = \left[T\right]^{'} \left[u\right]^{'} $$ Now, using the vector transformation rule,
 * $$\begin{align}\text{(3)} \qquad

v^{'}_i & = l_{ip} v_p ~;~ u^{'}_i = l_{ip} u_p ~\text{or,}~ \left[v\right]^{'} = \left[L\right] \left[v\right] ~; \left[u\right]^{'} = \left[L\right] \left[u\right] \\ v_q & = l_{iq} v^{'}_i ~;~ u_q = l_{iq} u^{'}_i ~\text{or,}~ \left[v\right] = \left[L\right]^{T} \left[v\right]^{'} ~; \left[u\right] = \left[L\right]^{T} \left[u\right]^{'} \end{align}$$ Plugging the first of equation (3) into equation (2) we get,
 * $$\text{(4)} \qquad

l_{ip} v_p = T^{'}_{ij} u^{'}_i ~\text{or,}~ \left[L\right] \left[v\right] = \left[T\right]^{'} \left[u\right]^{'} $$ Substituting for $$v_p$$ in equation~(4) using equation~(1),
 * $$\text{(5)} \qquad

l_{ip} T_{pq} u_q = T^{'}_{ij} u^{'}_i ~\text{or,}~ \left[L\right] \left[T\right] \left[u\right] = \left[T\right]^{'} \left[u\right]^{'} $$ Substituting for $$u_q$$ in equation (5) using equation (3),
 * $$\text{(6)} \qquad

l_{ip} T_{pq} l_{iq} u^{'}_i = T^{'}_{ij} u^{'}_i ~\text{or,}~ \left[L\right] \left[T\right] \left[L\right]^{T} \left[u\right]^{'} = \left[T\right]^{'} \left[u\right]^{'} $$ Therefore, if $$\mathbf{u} \equiv \left[u\right]$$ is an arbitrary vector,

l_{ip} T_{pq} l_{iq} = T^{'}_{ij} \Rightarrow T^{'}_{ij} = l_{ip} l_{jq} T_{pq} ~\text{or,}~ \left[T\right]^{'} = \left[L\right] \left[T\right] \left[L\right]^{T} $$ which is the transformation rule for second order tensors.

Therefore, in matrix notation, the transformation rule can be written as

\left[T\right]^{'} = \left[L\right] \left[T\right] \left[L\right]^{T} $$

title: Elasticity/Transformation example 1


text:
Elasticity/Constitutive example 2

Example 1
Convert the stress-strain relation for isotropic materials (in matrix form) into an equation in index notation. Show all the steps in the process.

Solution
The stress-strain relation is

\begin{bmatrix} \varepsilon_{11}\\\varepsilon_{22}\\\varepsilon_{33}\\\varepsilon_{23}\\\varepsilon_{31}\\\varepsilon_{12} \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 1 & -\nu & -\nu & 0 & 0 & 0 \\ -\nu & 1 & -\nu & 0 & 0 & 0 \\ -\nu & -\nu & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1+\nu & 0 & 0\\ 0 & 0 & 0 & 0 & 1+\nu & 0\\ 0 & 0 & 0 & 0 & 0 & 1+\nu \\ \end{bmatrix} \begin{bmatrix} \sigma_{11}\\\sigma_{22}\\\sigma_{33}\\\sigma_{23}\\\sigma_{31}\\\sigma_{12} \end{bmatrix} $$ Let us expand out the terms and put all of them in a similar form. Thus,
 * $$\begin{align}

E \varepsilon_{11} & = (1+\nu)\sigma_{11} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{22} & = (1+\nu)\sigma_{22} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{33} & = (1+\nu)\sigma_{33} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(1)\\ E \varepsilon_{23} & = (1+\nu)\sigma_{23} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0)\\ E \varepsilon_{31} & = (1+\nu)\sigma_{31} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0)\\ E \varepsilon_{12} & = (1+\nu)\sigma_{12} - \nu(\sigma_{11}+\sigma_{22}+\sigma_{33})(0) \end{align}$$ We know that $$\sigma_{11}+\sigma_{22}+\sigma_{33} = \sigma_{kk}$$. Also a quantity that is $$1$$ when $$i=j$$ and $$0$$ when $$i \ne j$$ can be represented by the Kronecker $$\delta$$. Therefore, we can write the above equations as

E \varepsilon_{ij} = (1+\nu)\sigma_{ij} - \nu\sigma_{kk}\delta_{ij} $$ or,

\varepsilon_{ij} = \frac{(1+\nu)}{E}\sigma_{ij} - \frac{\nu}{E}\sigma_{kk}\delta_{ij} $$

title: Elasticity/Constitutive example 2


text:
Elasticity/Equilibrium example 1

Example 1
 Given:

Euler's second law for the conservation of angular momentum
 * $$ \text{(1)} \qquad

\int_{\partial B} e_{ijk}~x_j~n_l~\sigma_{lk}~dS + \int_B \rho~e_{ijk}~x_j~b_k~dV = \frac{d}{dt}\left(\int_B \rho~e_{ijk}~x_j~v_k~dV \right) $$ The divergence theorem
 * $$ \text{(2)} \qquad

\int_{\partial B} n_i~\sigma_{ij}~dS = \int_B \frac{\partial \sigma_{ij}}{\partial x_i}~dV $$ The equilibrium equation (Cauchy's first law)
 * $$ \text{(3)} \qquad

\frac{\partial \sigma_{ij}}{\partial x_i} + \rho~b_j = \frac{d}{dt}\left(\rho~v_j\right) $$

 Show:

 * $$ \text{(4)} \qquad

\sigma_{ij} = \sigma_{ji} $$

Solution
Let us first look at the first term of equation~(1) and apply the divergence theorem (2). Thus,
 * $$\begin{align}

\int_{\partial B} e_{ijk}~x_j~n_l~\sigma_{lk}~dS & = \int_{\partial B} n_l~(e_{ijk}~x_j~\sigma_{lk})~dS \\ & = \int_{B} \frac{\partial{(e_{ijk}~x_j~\sigma_{lk})}}{\partial x_l}~dV \\ & = \int_{B} \left(e_{ijk}~\frac{\partial x_j}{\partial x_l}~\sigma_{lk} +              e_{ijk}~x_j~\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV\\ & = \int_{B} \left(e_{ijk}~\delta_{jl}~\sigma_{lk} +              e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV\\ & = \int_{B} \left(e_{ilk}~\sigma_{lk} +              e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV \end{align}$$ Plugging this back into equation~(1) gives

\int_{B} \left(e_{ilk}~\sigma_{lk} + e_{ijk}~x_j\frac{\partial\sigma_{lk}}{\partial x_l}\right)~dV + \int_B \rho~e_{ijk}~x_j~b_k~dV = \frac{d}{dt}\left(\int_B \rho~e_{ijk}~x_j~v_k~dV \right) $$ Therefore, bringing all terms to the left hand side,
 * $$\text{(5)} \qquad

\int_{B} \left[e_{ilk}~\sigma_{lk} + e_{ijk}~x_j\left(\frac{\partial\sigma_{lk}}{\partial x_l}  + \rho~b_k  - \frac{d}{dt}\left(\rho~v_k\right)\right)\right]~dV = 0 $$ Using the equilibrium equations~(3), equation~(5) reduces to
 * $$\text{(6)} \qquad

\int_{B} e_{ilk}~\sigma_{lk}~dV = 0 $$ Since this holds for any $$B$$, we have
 * $$\text{(7)} \qquad

e_{ilk}~\sigma_{lk} = 0 $$ If you work this expression out, you will see that $$\sigma_{ij} = \sigma_{ji}$$. Hence, the stress tensor is symmetric.

title: Elasticity/Equilibrium example 1


text:
BCP/Centaurea cyanus

Centaurea cyanus

title: BCP/Centaurea cyanus


text:
BCP/Centaurea cyanus/Logs

This page contains recent bloom logs for the plant Centaurea cyanus (Bachelor's Button).

2007 Logs

 * Hazel 16:40, 3 August 2007 (UTC)
 * --Luai lashire 14:10, 9 September 2007 (UTC)

2008 Logs

 * Ngravagna 16:46, 22 May 2008 (UTC)
 * Ngravagna 17:04, 7 June 2008 (UTC)
 * Ngravagna 20:03, 15 June 2008 (UTC)
 * Ngravagna 17:19, 21 June 2008 (UTC)

2010 Logs
(If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).

Recent Logs

 * Anna reg 10:50, 9 May 2011 (UTC)

title: BCP/Centaurea cyanus/Logs


text:
Elasticity/Stress example 2

Example 2
 Given: A homogeneous stress field with components in the basis $$(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)\,$$ given by

\left[\boldsymbol{\sigma}\right] = \begin{bmatrix} 3 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} \text{(MPa)} $$

 Find:

 * 1) The traction ($$\mathbf{t}\,$$) acting on a surface with unit normal $$\widehat{\mathbf{n}} = (\widehat{\mathbf{e}}_2+\widehat{\mathbf{e}}_3)/\sqrt{2}$$.
 * 2) The normal traction ($$\mathbf{t}_n\,$$) acting on a surface with unit normal $$\widehat{\mathbf{n}} = (\widehat{\mathbf{e}}_2+\widehat{\mathbf{e}}_3)/\sqrt{2}$$.
 * 3) The projected shear traction ($$\mathbf{t}_s$$) acting on a surface with unit normal $$\widehat{\mathbf{n}} = (\widehat{\mathbf{e}}_2+\widehat{\mathbf{e}}_3)/\sqrt{2}$$.
 * 4) The principal stresses.
 * 5) The principal directions of stress.

Solution
Here's how you can solve this problem using Maple.

\sigma :=  \begin{bmatrix} 3 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} $$

\mathit{e2} := \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$

\mathit{e3} := \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $$

n := \begin{bmatrix} 0 \\ { \frac {\sqrt{2}}{2}} \\ [2ex] { \frac {\sqrt{2}}{2}} \end{bmatrix} $$

\mathit{sigmaT} := \begin{bmatrix} 3 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{bmatrix} $$

\mathbf{t} := \begin{bmatrix} \sqrt{2} \\ \sqrt{2} \\ \sqrt{2} \end{bmatrix} ~ \text{Solution for Part 1} $$

\mathit{tT} := \begin{bmatrix} \sqrt{2} & \sqrt{2} & \sqrt{2} \end{bmatrix} $$

N := \begin{bmatrix} 2 \end{bmatrix} ~ \text{Solution for Part 2} $$

\mathit{tdott} := \begin{bmatrix} 6 \end{bmatrix} $$

S := \sqrt{2} ~ \text{Solution for Part 3} $$

\mathit{sigPrin} := 1, \,-2, \,4 ~ \text{Solution for Part 4} $$

\mathit{dirPrin} := [1, \,1, \,\{[-1, \,1, \,1]\}], \,[-2, \,1, \,\{[0, \,-1, \,1]\}], \,[4, \,1, \,\{[2, \,1, \,1]\}] $$

[1, \,1, \,\{[-1, \,1, \,1]\}] ~ \text{Solution for Part 5} $$

[-2, \,1, \,\{[0, \,-1, \,1]\}] ~ \text{Solution for Part 5} $$

[4, \,1, \,\{[2, \,1, \,1]\}] ~ \text{Solution for Part 5} $$

title: Elasticity/Stress example 2


text:
Diploma Web Development Qualification in Australia

Qualification name: ICA50605 Diploma of Information Technology (Website Development)

From the official document:
 * "This qualification provides the skills and knowledge for an individual to be competent as a senior ICT professional with responsibilities spanning design, development, site performance, database integration through to implementation and acceptance testing."

Core Units

 * BSBPM505A Manage project quality [[Image:00%.png]]
 * ICAA5054A Validate quality and completeness of system design specifications [[Image:00%.png]]
 * ICAA5056A Prepare disaster recovery and contingency planns[[Image:00%.png]]
 * ICAA5141A Design dynamic websites to meet technical requirements[[Image:00%.png]]
 * ICAA5146A Develop website information architecture[[Image:00%.png]]
 * ICAA5151A Gather data to identify business requirements[[Image:00%.png]]
 * ICAA5158A Translate business needs into technical requirements[[Image:00%.png]]
 * ICAB5156A Create dynamic pages[[Image:00%.png]]
 * ICAB5180A Integrate database with a website[[Image:00%.png]]
 * ICAP5039A Match IT needs with the strategic direction of the enterprise[[Image:00%.png]]
 * ICAS5102A Establish and maintain client user liason[[Image:00%.png]]
 * ICAS5202A Ensure privacy for users[[Image:00%.png]]
 * ICAT5081A Perform systems test[[Image:00%.png]]
 * ICAT5083A Develop and conduct client acceptance test[[Image:00%.png]]
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 * PSPPM502B Manage complex projects[[Image:00%.png]]

Electives
This qualification allows for a large variety of electives. Here are some suggested electives:
 * ICAA5153A Model data objects[[Image:75%.png]]
 * ICAA5154A Model data processes[[Image:00%.png]]
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 * ICAI3101A Install and manage network protocols[[Image:00%.png]]
 * ICAI4097A Install and configure a network[[Image:00%.png]]
 * ICAS5192A Configure an internet gateway[[Image:00%.png]]


text:
Elasticity/Stress example 3

Example 3
Given:

A stress field whose components in the basis $$(\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3)\,$$ are given by the matrix

[\boldsymbol{\sigma}] = \begin{bmatrix} 6~x_1~x_3^2 & 0 & -2~x_3^3 \\ 0 & 1 & 2 \\   -2~x_3^3 & 2 & 3~x_1^2 \end{bmatrix} $$

 Find:

 * 1) Assuming negligible body forces, show whether this field satisfies equilibrium.
 * 2) Find the traction acting at a point P whose position vector is $$ \mathbf{x} = 2~\mathbf{e}_1 + 3~\mathbf{e}_2 + 2~\mathbf{e}_3 \, $$ acting on a plane $$2~x_1 + x_2 - x_3 = 5\,$$.
 * 3) Determine the normal traction at this point on the plane.
 * 4) Determine the projected shear traction at this point on the plane.
 * 5) Determine the principal stresses at point P.
 * 6) Determine the principal directions of stress at point P.

Solution
This problem is very similar to example 2.

The stress is a function of $$x_1\,$$ and $$x_3\,$$. We first plug the stress into the equilibrium equations (Cauchy's first law) and find that the sum is zero. Hence the stress satifies equilibrium.

Next, we find the stress at the given point byplugging in the values of $$x_1\,$$ and $$x_3\,$$ into the given stress.

Then we find the normal to the given plane $$a_i~x_i = b\,$$ using the relation

\widehat{\mathbf{n}} = \pm \cfrac{a_i~\mathbf{e}_i}{\sqrt{a_j~a_j}} $$ and find the traction vector and thenormal and projected shear tractions.

We find the eigenvalues (principal stresses) by first forming the cubic characteristic equation and then solving for it. Then we plug the values of the principal stresses into the eigenfunction and solve for the direction of the principal strains.

Since the stress tensor is symmetric, we need an extra relation between $$n_1\,$$, $$n_2\,$$, and $$n_3\,$$ to reduce the system down to two equations and two unknowns. This relation is $$n_1^2 + n_2^2 + n_3^2 = 1\,$$, which means that each eigenvector is a unit vector.

The Maple output is shown below. > with(linalg): > sigma := > linalg[matrix](3,3,[6*x1*x3^2,0,-2*x3^3,0,1,2,-2*x3^3,2,3*x1^2]);

[      2              3]                           [6 x1 x3     0    -2 x3 ] [                      ]                  sigma := [   0        1      2   ] [                      ]                           [      3              2 ]                           [ -2 x3      2    3 x1  ]

> e1 := linalg[matrix](3,1,[1,0,0]): > e2 := linalg[matrix](3,1,[0,1,0]): > e3 := linalg[matrix](3,1,[0,0,1]): > sigi1i := diff(sigma[1,1],x1)+diff(sigma[2,1],x2)+diff(sigma[3,1],x3);

sigi1i := 0

> sigi2i := diff(sigma[1,2],x1)+diff(sigma[2,2],x2)+diff(sigma[3,2],x3);

sigi2i := 0

> sigi3i := diff(sigma[1,3],x1)+diff(sigma[2,3],x2)+diff(sigma[3,3],x3);

sigi3i := 0

> X := evalm(2*e1 + 3*e2 + 2*e3): > sig := linalg[matrix](3,3): > for i from 1 to 3 do > for j from 1 to 3 do >   sig[i,j] := eval(eval(eval(sigma[i,j], x1 = X[1,1]), x2 =X[2,1]), x3 > = X[3,1]); > end do; > end do; > evalm(sig);

[ 48   0    -16]                          [               ]                          [  0    1      2]                          [               ]                          [-16    2     12]

> a1 := 2: a2 := 1: a3 := -1: ajaj := sqrt(a1*a1 + a2*a2 + a3*a3): > n := evalm(a1*e1/ajaj + a2*e2/ajaj + a3*e3/ajaj);

[ 1/2 ]                                 [ 6    ]                                 [  ]                                 [  3   ]                                 [      ]                                 [  1/2 ]                            n := [ 6    ] [ ]                                 [  6   ]                                 [      ]                                 [   1/2]                                 [  6   ]                                 [- ]                                 [   6  ]

> sigT := transpose(sig): > t := evalm(sigT&*n);

[   1/2]                                 [56 6   ]                                 [---]                                 [   3   ]                                 [       ]                            t := [   1/2 ] [ 6    ]                                 [-  ]                                 [   6   ]                                 [       ]                                 [    1/2]                                 [-7 6   ]

> tT := transpose(t): > N := evalm(tT&*n): Nt := evalf(N[1,1]);

Nt := 44.16666667

> tdott := evalm(tT&*t): > St := evalf(sqrt(tdott[1,1] - N[1,1]^2));

St := 20.83599984

> I1 := sig[1,1]+sig[2,2]+sig[3,3];

I1 := 61

> I2 := sig[1,1]*sig[2,2] + sig[2,2]*sig[3,3] + sig[3,3]*sig[1,1] - > sig[1,2]*sig[2,1] - sig[2,3]*sig[3,2] - sig[3,1]*sig[1,3];

I2 := 376

> I3 := det(sig);

I3 := 128

> prinSig := solve(x^3 - I1*x^2 + I2*x - I3 = 0, x): > s1 := evalf(prinSig[1]); s2 := evalf(prinSig[2]); s3 := > evalf(prinSig[3]);

-8                   s1 := 54.09271724 - 0.2 10   I

-8              s2 := 0.361501126 - 0.8160254040 10   I

-8              s3 := 6.545781610 + 0.9160254040 10   I

> sig1 := 54.09271724; sig2 := 6.545781610; sig3 := .361501126;

sig1 := 54.09271724

sig2 := 6.545781610

sig3 := 0.361501126

> with(LinearAlgebra): Id := IdentityMatrix(3): > Left1 := evalm(sig - sig1*Id): > n3 := sqrt(1 - n1^2 - n2^2): > n := linalg[matrix](3,1,[n1,n2,n3]): > Ax1 := evalm(Left1 &* n): > sols1 := solve({Ax1[1,1] = 0, Ax1[2,1] = 0});

sols1 := {n2 = 0.01340427809, n1 = -0.9344527487}

> N1_1 := sols1[2]; N1_2 := sols1[1]; N1_3 := sqrt(1 - > (.1340427809e-1)^2 - (-.9344527487)^2); >

N1_1 := n1 = -0.9344527487

N1_2 := n2 = 0.01340427809

N1_3 := 0.3558347730

> Left2 := evalm(sig - sig2*Id): > Ax2 := evalm(Left2 &* n): > sols2 := solve({Ax2[1,1] = 0, Ax2[2,1] = 0});

sols2 := {n1 = 0.3412802400, n2 = 0.3188798847}

> N2_1 := sols2[1]; N2_2 := sols1[2]; N2_3 := sqrt(1 - (.3412802400)^2 - > (.3188798847)^2);

N2_1 := n1 = 0.3412802400

N2_2 := n1 = -0.9344527487

N2_3 := 0.8842191001

> Left3 := evalm(sig - sig3*Id): > Ax3 := evalm(Left3 &* n): > sols3 := solve({Ax3[1,1] = 0, Ax3[2,1] = 0});

sols3 := {n1 = 0.1016162335, n2 = -0.9477003447}

> N3_1 := sols3[1]; N3_2 := sols3[2]; N3_3 := sqrt(1 - (.1016162335)^2 - > (-.9477003447)^2);

N3_1 := n1 = 0.1016162335

N3_2 := n2 = -0.9477003447

N3_3 := 0.3025528017

title: Elasticity/Stress example 3


text:
Elasticity/Stress example 4

Example 4
Given:

The octahedral plane is the plane that is equally inclined to the directions of the three principal stresses. For any given stress of state there are eight such planes.

Show:

 * 1) The normal traction on an octahedral plane is given by
 * $$ \mathbf{t}_n^{\text{oct}} = \cfrac{1}{3}\left(\sigma_1 + \sigma_2 + \sigma_3\right) = \cfrac{1}{3}~I_{\boldsymbol{\sigma}} $$.

 * 1) The projected shear traction on an octahedral plane is given by
 * $$ \mathbf{t}_s^{\text{oct}} = \cfrac{1}{3}\sqrt{\left(\sigma_1 - \sigma_2\right)^2 + \left(\sigma_2 - \sigma_3\right)^2 + \left(\sigma_3 - \sigma_2\right)^2} = \cfrac{1}{3}\sqrt{2~I_{\boldsymbol{\sigma}}^2 - 6~II_{\boldsymbol{\sigma}}}.$$

Here $$(\sigma_1, \sigma_2, \sigma_3)\,$$ are the principal stresses and $$(I_{\boldsymbol{\sigma}}, II_{\boldsymbol{\sigma}})\,$$ are the first two invariants of the stress tensor ($$\boldsymbol{\sigma}\,$$).

Solution
Let us take the basis as the directions of the principal stresses $$\widehat{\mathbf{n}}{1}$$, $$\widehat{\mathbf{n}}{2}$$, $$\widehat{\mathbf{n}}{3}$$. Then the stress tensor is given by

\left[\boldsymbol{\sigma}\right] = \begin{bmatrix} \sigma_1 & 0 & 0 \\ 0 & \sigma_2 & 0 \\ 0 & 0 & \sigma_3 \end{bmatrix} $$ If $$\widehat{\mathbf{n}}_{o}$$ is the direction of the normal to an octahedral plane, then the components of this normal with respect to the principal basis are $$n_{o1}\,$$, $$n_{o2}\,$$, and $$n_{o3}\,$$. The normal is oriented in such a manner that it makes equal angles with the principal directions. Therefore, $$n_{o1} = n_{o2} = n_{o3} = n_o\,$$. Since $$n_{o1}^2 + n_{o2}^2 + n_{o3}^2 = 1\,$$, we have $$n_o = 1/\sqrt(3)$$.

The traction vector on an octahedral plane is given by

\mathbf{t}_o = \widehat{\mathbf{n}}_{o} \bullet \left[\boldsymbol{\sigma}\right] = {n_o \sigma_1, n_o \sigma_2, n_o \sigma_3} $$ The normal traction is,

N = \mathbf{t}_o \bullet \widehat{\mathbf{n}}_{o} = n_o^2 \sigma_1 + n_o^2 \sigma_2 + n_o^2 \sigma_3 $$ Now, $$I_\sigma = (\sigma_1 + \sigma_2 + \sigma_3)\,$$. Therefore,

N = (1/3)(\sigma_1 + \sigma_2 + \sigma_3) = (1/3)I_\sigma\, $$ The projected shear traction is given by

S = \sqrt{\mathbf{t}_o\bullet\mathbf{t}_o - N^2} $$ Therefore,

S = \sqrt{n_o^2\sigma_1^2 + n_o^2\sigma_2^2 + n_o^2\sigma_3^2 - (1/9)(\sigma_1 + \sigma_2 + \sigma_3)^2} $$ Also,

II_{\sigma} = \sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_3\sigma_1\, $$ If you do the algebra for S, you will get the required relations.

title: Elasticity/Stress example 4


text:
Australian Vocational Education and Training/ICAA5153A Model data objects

From the official document:
 * "This unit defines the competency required to understand business operations, identify entities and data, diagrammatically represent their relationships and prepare a data model."

Basically, that means you'll be able to design and prepare the data models that your client application will need. As a practical example, if you were to create a simple blog application, your models would include a Blog Post (with the fields Title, Date posted, Body) and a Comment model (with the fields PostID relating back to your blog post, name, website, comment, date posted).

This unit is also an excellent chance to learn a web-framework that provides Object-Relational Mapping - where you simply define your object models and their relationships, then the framework allows you to use them in your application (looking after the database for you)!

You can read more about this unit on the National Training Information Systems Official document for ICABA5153A.

This unit forms part of the following qualifications:
 * Diploma Web Development Qualification in Australia

How this unit will help you
After learning and applying the skills required by this unit you'll be able to demonstrate your skills in:

 * Analyse a businesses processes to understand:
 * what data is being used (the type of data)
 * how that data can be grouped into real-world objects (such as a Blog-post, Book review or product detail)
 * how the different data objects relate to one another (for example, a book may have many reviews, a product may have many transactions)
 * Document the data objects and their relationships using an entity-relationship diagram (or if you are familiar with a language that allows you to prototype quickly, then a working prototype that demonstrates the models and their relationships is even better.)
 * After analysing and understanding the business process and it's data, you may be able to review the process to suggest improvements as well as review the way the data is stored (making sure you don't have unwanted duplicates or inefficient storage of data), all the while determining the impact of these improvements on the process.
 * Documenting and communicating your models and relationships to your client for signoff. Again, the best way to communicate the models is with a demonstrable prototype - if you are familiar (or keen to learn) a tool that allows quick prototyping.

Ideas for learning the required skills
An excellent task to start learning how to model data is to create your own simple blogging application (with the help of your facilitator!) You'll need to:
 * Define the data models (start with the suggestions in the example above, but you'll most likely find you'll need to add more data/relationships).
 * Document/prototype the models and communicate them back to your facilitator.

You may then want to progress to a real client web application requiring multiple data objects, such as an ordering system or a product catelogue.

It is recommended that you begin learning a web-framework that will help you to model your data, such as the Django web framework (Python-based web application framework), CakePHP (a PHP-based web application framework) or Ruby On Rails. Undertaking to learn a web-framework can be a daunting task, but the benefits are huge!

Ideas for demonstrating this unit
The best way to demonstrate these skills is through the web applications that you create. These might include activities you've done while learning, or a real client web application. Normally part of your demonstration will include answering questions about your sites and making small modifications to verify that it is your work and you understand everything included in the pages.


text:
Elasticity/Kinematics example 2

Example 2
 Given: A body occupies the unit cube $$X_i \in [0,1]$$ in the reference configuration. The mapping between the current and the reference configuration given by $$x_1 = X_1 + \kappa~X_2$$, $$x_2 = X_2\,$$, $$x_3 = X_3\,$$.

 Find:

 * 1) Sketch current configuration.
 * 2) Show that motion is isochoric.
 * 3) Find stretches in the directions $$\widehat{\mathbf{e}}_{1}$$, $$\widehat{\mathbf{e}}_{2}$$, $$(1/\sqrt{2})(\widehat{\mathbf{e}}_{1}+\widehat{\mathbf{e}}_{2})$$, and $$(1/\sqrt{2})(\widehat{\mathbf{e}}_{1}-\widehat{\mathbf{e}}_{2})$$.
 * 4) Find the orthogonal shear strain between the reference material directions $$\widehat{\mathbf{e}}_{1}$$ and $$\widehat{\mathbf{e}}_{2}$$. Also between directions    $$(1/\sqrt{2})(\widehat{\mathbf{e}}_{1}+\widehat{\mathbf{e}}_{2})$$ and $$(1/\sqrt{2})(\widehat{\mathbf{e}}_{1}-\widehat{\mathbf{e}}_{2})$$.
 * 5) Find principal stretches and principal directions of stretch ($$\kappa = 0.4\,$$).

Solution:

 * All parallelograms that have the equal heights and the same base have equal areas. Hence, there is no volume change in this deformation.  Hence isochoric.

 * Stretches in the material direction $$\widehat{\mathbf{G}}$$ are given by

\lambda(\widehat{\mathbf{G}}) = \sqrt{\widehat{\mathbf{G}} : \boldsymbol{C} : \widehat{\mathbf{G}}} $$ where $$\boldsymbol{C}\,$$ is the Cauchy-Green deformation tensor

\boldsymbol{C} = \boldsymbol{F}^T \bullet \boldsymbol{F} $$ We will use Maple to calculate the stretches in the four directions. with(linalg): x := array(1..3): X := array(1..3): x[1] := X[1] + k*X[2]: x[2] := X[2]: x[3] := X[3]: F := linalg[matrix](3,3): for i from 1 to 3 do       for j from 1 to 3 do          F[i,j] := diff(x[i],X[j]); end do; end do; evalm(F); C := evalm(transpose(F)&*F); e1 := linalg[matrix](1,3,[1,0,0]): e2 := linalg[matrix](1,3,[0,1,0]): e1pe2 := evalm((e1 + e2)/sqrt(2)): e1me2 := evalm((e1-e2)/sqrt(2)):'' lambda[1] := sqrt(evalm(evalm(e1&*C)&*transpose(e1))[1,1]); lambda[2] := sqrt(evalm(evalm(e2&*C)&*transpose(e2))[1,1]); lambda[3] := simplify(sqrt(evalm(evalm(e1pe2&*C)&*transpose(e1pe2))[1,1])); lambda[4] := simplify(sqrt(evalm(evalm(e1me2&*C)&*transpose(e1me2))[1,1]));

F := \begin{bmatrix} 1 & k & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

C := \begin{bmatrix} 1 & k & 0 \\ k & k^{2} + 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$
 * $$ \lambda_{1} := 1 \,$$
 * $$ \lambda_{2} := \sqrt{k^{2} + 1} $$
 * $$ \lambda_{3} := \frac{\sqrt{4 + 4k + 2k^2}}{2} $$
 * $$ \lambda_{4} := \frac{\sqrt{4 - 4k + 2k^2}}{2} $$

The following figure shows that the calculated stretches are correct.

 * The orthogonal shear strain between two orthogonal units vectors $$\widehat{\mathbf{G}}_{1}$$ and $$\widehat{\mathbf{G}}_{2}$$ in the reference material co-ordinate system is given by

\gamma(\widehat{\mathbf{G}}_{1},\widehat{\mathbf{G}}_{2}) = \sin^{-1}\left(\frac{\widehat{\mathbf{G}}_{1}:\mathbf{C}:\widehat{\mathbf{G}}_{2}}        {\lambda(\widehat{\mathbf{G}}_{1})\lambda(\widehat{\mathbf{G}}_{2})}\right) $$ Once again, we will use Maple to calculate these strains. The steps are the same upto the calculation of the stretches. numer1 := evalm(evalm(e1&*C)&*transpose(e2))[1,1]: denom1 := lambda[1]*lambda[2]: ratio1 := numer1/denom1: gam1 := arcsin(ratio1); numer2 := simplify(evalm(evalm(e1pe2&*C)&*transpose(e1me2))[1,1]): denom2 := lambda[3]*lambda[4]: ratio2 := numer2/denom2: gam2 := arcsin(ratio2);

\gamma_1 := \sin^{-1}(\frac {k}{\sqrt{k^2 + 1}}) $$

\gamma_2 := - \sin^{-1}(\frac{2k^2}{\sqrt{4+4k+2k^2}\sqrt{4-4k+2k^2}}) $$ The following figure shows that the calculated orthogonal shear strains are correct.

The value of $$\gamma_{1}\,$$ can easily be verified using the definition of $$\sin{\gamma}\,$$. For the verification of the value of $$\gamma_{2}\,$$, the cosine law

\cos(A) = \frac{b^2+c^2-a^2}{2bc}\, $$ has to be used. Note that the actual length of the sides $$b\,$$ and $$c\,$$ of the triangle is obtained after multiplying the values shown in the figure by the length of the diagonal ($$\sqrt{2}$$). Hence, the calculated values are correct. b := sqrt(2)/2*lambda[4]: c := sqrt(2)/2*lambda[3]: a := 1: A := (b^2 + c^2 - a^2)/(2*b*c);

A := \frac{2k^2}{\sqrt{4+4k+2k^2}\sqrt{4-4k+2k^2}} $$

 * The principal stretches are given by the square roots of the eigenvalues of $$\boldsymbol{C}$$. The principal directions are the eigenvectors of $$\boldsymbol{C}$$.

Once again, we use Maple for our calculations. CC := linalg[matrix](3,3): for i from 1 to 3 do       for j from 1 to 3 do          CC[i,j] := eval(C[i,j], k=0.4); end do; end do; evalm(CC); eigvals := eigenvals(CC); PrinStretch[1] := sqrt(eigvals[1]); PrinStretch[2] := sqrt(eigvals[2]); PrinStretch[3] := sqrt(eigvals[3]); PrinDir := eigenvects(CC);

CC :=\begin{bmatrix} 1 & 0.4 & 0 \\ 0.4 & 1.16 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$

\text{eigvals} := 0.67, 1.0, 1.49 $$

\text{PrinStretch}[1] := 0.82 ~, \text{PrinStretch}[2] := 1.0 ~, \text{PrinStretch}[3] := 1.22 $$

\text{PrinDir}[1] := [0.77, -0.63, 0.] ~; \text{PrinDir}[2] := [0, 0, 1] ~; \text{PrinDir}[3] := [0.63, 0.77, 0.] $$

title: Elasticity/Kinematics example 2


text:
Elasticity/Kinematics example 3

Example 3
 Given:

Unit square $$(X_1,X_2) \in [0,1]$$ with displacement fields :

 * 1) $$\mathbf{u} = \kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.
 * 2) $$\mathbf{u} = -\kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.
 * 3) $$\mathbf{u} = \kappa X_1^2 \widehat{\mathbf{e}}_{2}$$.

 Sketch: Deformed configuration in $$x_1,x_2$$ plane.

Solution
The displacement $$\mathbf{u} = \mathbf{x} - \mathbf{X}$$. Hence, $$\mathbf{x} = \mathbf{u} + \mathbf{X}$$. In the reference configuration, $$\mathbf{u} = 0$$ and $$\mathbf{x} = \mathbf{X}$$. Hence, in the $$(x_1, x_2)$$ plane, the initial square is the same shape as the unit square in the $$(X_1, X_2)$$ plane. We can use Maple to find out the values of $$x_1$$ and $$x_2$$ after the deformation $$\mathbf{u}$$.

with(linalg): X := array(1..3): x := array(1..3): u = array(1..3): e1 := array(1..3,[1,0,0]): e2 := array(1..3,[0,1,0]): e3 = array(1..3,[0,0,1]): ua := evalm(k*X[2]*e1 + k*X[1]*e2): ub := evalm(-k*X[2]*e1 + k*X[1]*e2); uc := evalm(k*X[1]^2*e2);

\mathit{ua} := \left[   k{X_{2}}, k{X_{1}}, 0  \right] $$

\mathit{ub} := \left[    - k{X_{2}}, k{X_{1}}, 0   \right] $$

\mathit{uc} := \left[   0, k{X_{1}}^{2}, 0   \right] $$ xa := evalm(ua + X); xb := evalm(ub + X); xc := evalm(uc + X);

\mathit{xa} := \left[   k{X_{2}} + {X_{1}}, k{X_{1}} + {X_{2}}, {X_{3}}   \right] $$

\mathit{xb} := \left[    - k{X_{2}} + {X_{1}}, k{X_{1}} + {X_{2}}, {X_{3}}   \right] $$

\mathit{xc} := \left[   {X_{1}}, k{X_{1}}^{2} + {X_{2}}, {X_{3}}   \right] $$

Plots of the deformed body are shown below

title: Elasticity/Kinematics example 3


text:
Elasticity/Kinematics example 4

Example 4
 Given:

Displacement field $$\mathbf{u} = \kappa X_2 \widehat{\mathbf{e}}_{1} + \kappa X_1 \widehat{\mathbf{e}}_{2}$$.

 Find:

 * 1) The Lagrangian Green strain tensor $$\boldsymbol{E}\,$$.
 * 2) The infinitesimal strain tensor $$\boldsymbol{\varepsilon}\,$$.
 * 3) The infintesimal rotation tensor $$\boldsymbol{\omega}\,$$.
 * 4) The infinitesimal rotation vector $$\boldsymbol{\theta}\,$$.
 * 5) The exact longitudinal strain in the reference material direction $$\mathbf{e}_1\,$$.
 * 6) The approximate longitudinal strain in the direction $$\mathbf{e}_1\,$$ based on the infinitesimal strain tensor $$\boldsymbol{\varepsilon}\,$$.

Solution
The Maple output of the computations are shown below:

with(linalg): with(LinearAlgebra): X := array(1..3): x := array(1..3): e1 := array(1..3,[1,0,0]): e2 := array(1..3,[0,1,0]): e3 := array(1..3,[0,0,1]): u := evalm(k*X[2]*e1 + k*X[1]*e2);

u := \left[  \! k\,{X_{2}}, \,k\,{X_{1}}, \,0 \! \right] $$

x := evalm(u + X);

x := \left[  \! k\,{X_{2}} + {X_{1}}, \,k\,{X_{1}} + {X_{2}}, \, {X_{3}} \! \right] $$ F := linalg[matrix](3,3): for i from 1 to 3 do   for j from 1 to 3 do      F[i,j] := diff(x[i],X[j]); end do; end do; evalm(F);

F := \begin{bmatrix} 1 & k & 0 \\ k & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Id := IdentityMatrix(3): C := evalm(transpose(F)&*F); E := evalm((1/2)*(C - Id));

C := \begin{bmatrix} 1 + k^{2} & 2\,k & 0 \\ 2\,k & 1 + k^{2} & 0 \\ 0 & 0 & 1   \end{bmatrix} $$

E := \begin{bmatrix} { \frac {k^{2}}{2}} & k & 0 \\ [2ex] k & { \frac {k^{2}}{2}} & 0 \\ [2ex] 0 & 0 & 0  \end{bmatrix} $$ gradu := linalg[matrix](3,3): for i from 1 to 3 do   for j from 1 to 3 do      gradu[i,j] := diff(u[i],X[j]); end do; end do; evalm(gradu);

gradu := \begin{bmatrix} 0 & k & 0 \\ k & 0 & 0 \\ 0 & 0 & 0    \end{bmatrix} $$ epsilon := evalm((1/2)*(gradu + transpose(gradu)));

\varepsilon := \begin{bmatrix} 0 & k & 0 \\ k & 0 & 0 \\ 0 & 0 & 0    \end{bmatrix} $$ omega := evalm((1/2)*(gradu - transpose(gradu)));

\omega := \begin{bmatrix} 0 & 0 & 0 \\      0 & 0 & 0 \\       0 & 0 & 0    \end{bmatrix} $$ stretch1 :=  sqrt(evalm(evalm(e1&*C)&*transpose(e1))[1,1]): longStrain1 := stretch1 - 1;

\mathit{stretch1} := \sqrt{1 + k^{2}} $$

\mathit{longStrain1} := \sqrt{1 + k^{2}} - 1 $$ approxLongStrain1 := evalm(evalm(e1&*epsilon)&*transpose(e1))[1,1];

\mathit{approxLongStrain1} := 0 $$

The geometrical difference between the large strain and small strain cases can be observed by looking at the figures from the previous examples.

title: Elasticity/Kinematics example 4


text:
Joe Perry

Back to Requirements Analysis

Background
Joe was born in the United Kingdom but spent most of his childhood in Vancouver, Canada. At university he studied English literature and decided to spend a year teaching English as a foreign language, and did a 4 week training course straight after university.

Now 28 Joe has spent more than 5 years teaching in various parts of Europe, including Sicily, Turkey, Georgia and currently Prague.

Joe has become fed up with teaching and has decided that he needs a "proper job".

Interests
Joe reads a lot. His teaching job gives him plenty of time for reading as his classes are spread around the city.

title: Joe Perry


text:
User:Mysha~enwikiversity

 * 1) REDIRECT

title: User:Mysha~enwikiversity


text:
User:Vakho~enwikiversity

I want to in wikiversity,i want to speak with many people in all over the world,and speak with them about everythin in all over the world.

title: User:Vakho~enwikiversity


text:
Technical writing/Technical writing Persona Joe Perry

Back to Requirements Analysis

Background
Joe was born in the United Kingdom but spent most of his childhood in Vancouver, Canada. At university he studied English literature and decided to spend a year teaching English as a foreign language, and did a 4 week training course straight after university.

Now 28 Joe has spent more than 5 years teaching in various parts of Europe, including Sicily, Turkey, Georgia and currently Prague.

Joe has become fed up with teaching and has decided that he needs a "proper job".

Interests
Joe reads a lot. His teaching job gives him plenty of time for reading as his classes are spread around the city.


text:
BCP/Salvia splendens

Salvia splendens

title: BCP/Salvia splendens


text:
BCP/Salvia splendens/Logs

This page contains recent bloom logs for the plant Salvia splendens (Scarlet Sage).

Recent Logs

 * --SB_Johnny | talk 07:06, 1 October 2007 (UTC)
 * SB Johnny-NHAMP 09:18, 9 October 2007 (UTC)
 * --SB_Johnny | talk 10:12, 2 November 2007 (UTC)
 * --SB_Johnny | talk 08:09, 19 June 2008 (UTC)
 * --SB_Johnny | talk 10:28, 14 July 2008 (UTC)

=====Archived Logs===== (If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * --SB_Johnny | PA! 19:28, 28 July 2007 (UTC)
 * --SB_Johnny | PA! 10:29, 6 August 2007 (UTC)
 * SB Johnny-LM 09:42, 8 September 2007 (UTC)

title: BCP/Salvia splendens/Logs


text:
BCP/Impatiens pallida

Impatiens pallida

title: BCP/Impatiens pallida


text:
BCP/Impatiens pallida/Logs

This page contains recent bloom logs for the plant Impatiens pallida (Pale Jewelweed).

Recent Logs

 * --SB_Johnny | PA! 12:04, 4 August 2007 (UTC)
 * --Luai lashire 18:45, 25 August 2007 (UTC)
 * --SB_Johnny | PA! 09:30, 26 August 2007 (UTC)

=====Archived Logs=====

(If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * (no archived logs yet)

title: BCP/Impatiens pallida/Logs


text:
User:Sniperworld

Hi, This page was created to share my knowledge in computers and basics in internet. I Also want to share more ideas on HR Practices and strategies.

title: User:Sniperworld


text:
User:Potrivitu George

My pages here, on Wikiversity:
 * go to: Department of Astrochemistry
 * go to: Department of Tropical Diseases-help me with this!It's very important and it's a big topic!

title: User:Potrivitu George


text:
User:KrakowJohn

Hi

My name's John and I'm an EFL teacher and writer living in Krakow. I've just discovered Wikiversity. I hope to gain and share learning and information about writing (especially drama), language learning and linguistics.

title: User:KrakowJohn


text:
User talk:Potrivitu George

Hello Potrivitu George, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
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 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 13:52, 6 August 2007 (UTC)

Astrochemistry
Topic:Astrochemistry is a content development project where everyone can collaborate to plan and develop learning resources about Astrochemistry. Individual learning resources for Astrochemistry should be placed on their own pages in the main namespace. --JWSchmidt 14:10, 6 August 2007 (UTC)
 * "i wrote something, what i knew about this!this is wrong?" <-- nothing is "wrong", just be sure to create a new page as soon as you have identified a specific learning resource that is needed. Pages in the topic namespace are for planning and should not contain the actual learning resources. For example, Topic:Astrochemistry could link to Introduction to Astrochemistry and Advanced Astrochemistry. --JWSchmidt 14:22, 6 August 2007 (UTC)

tropical diseases
A "Department of tropical diseases" is a good idea. I have previous experience in virology and microbiology, so I hope I have a chance to help you with such topics. --JWSchmidt 15:53, 7 August 2007 (UTC)

 * "i want to help me someone to show me what i must to do" This is a good place to start --> "i'll write also general thinks about tropical diseases". We are trying to develop some learning resources about how to use Wikiversity and participate. See Introduction to Wikiversity. --JWSchmidt 13:46, 8 August 2007 (UTC)

Tropical Diseases
Greetings George-Cristian Potrivitu!

Your Servant will be pleased to do what he can for the School of Biology Department of Tropical Diseases.

On the School of Biology Listing of Divisions and Departments, Your Servant  has taken the Liberty of Moving the Department into Alphabetical Order and Editing the Link.

What sort of assistance would you wish with reference to Tropical Diseases?

Best regards,

(s) Dionysios  (talk), Date: 2007-08-08 (August 8, 2007) Time:  1751 UTC

title: User talk:Potrivitu George


text:
User talk:KrakowJohn

Hello KrakowJohn, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 14:03, 6 August 2007 (UTC)

title: User talk:KrakowJohn


text:
User talk:Jana~enwikiversity

Hello Jana, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 14:05, 6 August 2007 (UTC)

Image copyright problems

 * Your currently affected uploads are listed below. For each, please provide a license template and source. Thanks. – Mike.lifeguard | @en.wb 23:45, 13 February 2008 (UTC)

 * 2007-08-22 09:04, Jana: Jana S.jpg: untagged! ; 0 (0)
 * 2007-08-23 10:43, Jana: Jana S1.jpg: untagged! ; 0 (0)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Jana. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Jana~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:39, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Jana~enwikiversity


text:
User talk:Pfthom01

Hello Pfthom01, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --Pfthom01 15:22, 6 August 2007 (UTC)

title: User talk:Pfthom01


text:
Astrochemistry

Astrochemistry is a relatively new domain of chemistry but is very important because it helps us to know how the Universe was made.

We know that all the Universe is made of material and chemical elements. All the elements have an age. For example Lead was born from the transformation of Uranium 4.6 milliard years ago. Let's talk about stars! Our Sun is also a "chemistry object". We know that the stars are made of Hydrogen which is permanently transformed into Helium so that the stars became big "plasma balls" -- Plasma being the 4th state of aggregation after solid, liquid, and gas.

A recent article, "Milky Way Fortified with Calcium" by Sara Goudarzi, February, 2007 discloses that our Galaxy, the Milky Way, "is fortified with Calcium".

This course is a cooperative effort from the school of chemistry and the department of radiation astronomy of the school of physics and astronomy.

Lectures

 * 1) Astrochemistry/Lecture
 * 2) Atmospheric astronomy
 * 3) Why planets lose their atmospheres
 * 4) Gaseous-object astronomy
 * 5) Intergalactic medium
 * 6) Interplanetary medium
 * 7) Interstellar medium
 * 8) Liquid-object astronomy
 * 9) Meteorites
 * 10) Minerals
 * 11) Mineral astronomy
 * 12) Molecule astronomy
 * 13) Nebulas
 * 14) Radiation chemistry
 * 15) Rocky-object astronomy and
 * 16) Rocks

Laboratories

 * 1) Astrognosy/Laboratory
 * 2) Chemical analysis of the spectrum of Vega
 * 3) Rocks/Meteorites/Laboratory

Quizzes

 * 1) Atmospheric astronomy/Quiz
 * 2) Chemicals/Quiz
 * 3) Comets/Quiz
 * 4) Gaseous-object astronomy/Quiz
 * 5) Intergalactic medium/Quiz
 * 6) Interplanetary medium/Quiz
 * 7) Interstellar medium/Quiz
 * 8) Liquid-object astronomy/Quiz
 * 9) Meteorites/quiz
 * 10) Minerals astronomy/Quiz
 * 11) Minerals/Quiz
 * 12) Nebulas/Quiz
 * 13) Radiation astrochemistry/Quiz
 * 14) Rocks/Quiz
 * 15) Rocky-object astronomy/Quiz

title: Astrochemistry


text: File:Kinematics ex2 fig1.png

title: File:Kinematics ex2 fig1.png


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title: File:Kinematics ex2 fig2.png


text: File:Kinematics ex2 fig3.png

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text: File:Kinematics ex3 fig.png

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text:
Elasticity/Constitutive example 3

Example 3
 Given: The strain energy density for a material undergoing small strain
 * $$ \text{(1)} \qquad

U(\boldsymbol{\varepsilon}) = \int_0^{\boldsymbol{\varepsilon}} \boldsymbol{\sigma} : d\boldsymbol{\varepsilon}~. $$

 Show: For linear elastic deformations and small strains,
 * $$ \text{(2)} \qquad

U(\boldsymbol{\varepsilon}) = \frac{1}{2} \boldsymbol{\sigma} : \boldsymbol{\varepsilon}~. $$

Solution
If the strain energy density is given by equation (1), then (for linear elastic materials) the stress and strain can be related using
 * $$ \text{(3)} \qquad

\sigma_{ij} = \frac{\partial U(\boldsymbol{\varepsilon})}{\partial \varepsilon_{ij}} $$

We will show that equation (2) is equivalent to equation (3). We start off with equation (2) and work backward.
 * $$ \text{(4)} \qquad

U(\boldsymbol{\varepsilon}) = \frac{1}{2} \sigma_{ij} \varepsilon_{ij} $$ For linear elastic materials,
 * $$ \text{(5)} \qquad

\sigma_{ij} = C_{ijkl} \varepsilon_{kl} $$ Substituting equation (5) into equation (4), we get,
 * $$ \text{(6)} \qquad

U(\boldsymbol{\varepsilon}) = \frac{1}{2} C_{ijkl} \varepsilon_{kl} \varepsilon_{ij} $$ Recall that, for a second order tensor $$\boldsymbol{A}\,$$,

\frac{\partial A_{ij}}{\partial A_{kl}} = \delta_{ik} \delta_{jl} $$ and that for a fourth order rensor $$\mathsf{C}\,$$ (substitution rule),

C_{ijkl} \delta_{ir} = C_{rjkl} \, $$ Differentiating equation (6) with respect to $$\varepsilon_{rs}\,$$, we have,
 * $$\begin{align}

\frac{\partial U(\boldsymbol{\varepsilon})}{\partial \varepsilon_{rs}} & = \frac{1}{2} C_{ijkl} \varepsilon_{kl} \delta_{ir} \delta_{js} + \frac{1}{2} C_{ijkl} \varepsilon_{ij} \delta_{kr} \delta_{ls} \\ & = \frac{1}{2} C_{rskl} \varepsilon_{kl} + \frac{1}{2} C_{ijrs} \varepsilon_{ij} \end{align}$$ Using the symmetry of the stiffness tensor, we have,
 * $$\begin{align}

\frac{\partial U(\boldsymbol{\varepsilon})}{\partial \varepsilon_{rs}} & = \frac{1}{2} C_{rskl} \varepsilon_{kl} + \frac{1}{2} C_{rsij} \varepsilon_{ij} \\ & = \frac{1}{2} \sigma_{rs} + \frac{1}{2} \sigma_{rs} \\ & = \sigma_{rs} \end{align}$$ Therefore,

\sigma_{ij} = \frac{\partial U(\boldsymbol{\varepsilon})}{\partial \varepsilon_{ij}} $$ which is the same as equation (3). Hence shown.

title: Elasticity/Constitutive example 3


text:
Elasticity/Constitutive example 4

Example 4
 Given:

A monoclinic crystal has inversion symmetric about the $$\widehat{\mathbf{e}}_{1}$$-$$\widehat{\mathbf{e}}_{2}$$ plane. Therefore, the material properties do not change for a mirror-reflection through this plane. The stress-strain relations must therefore remain unchanged under this transformation. The transformation matrix $$\left[L\right]\,$$ for this for the mirror inversion is given by
 * $$ \left[L\right] = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1

\end{bmatrix} $$

 Show:

If we apply this transformation to the stress and strain tensors, then the stiffness matrix of the material (in Voigt notation) is

\left[C\right] = \begin{bmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & C_{16} \\ C_{21} & C_{22} & C_{23} & 0 & 0 & C_{26} \\ C_{31} & C_{32} & C_{33} & 0 & 0 & C_{36} \\ 0     & 0      & 0      & C_{44} & C_{45} & 0 \\ 0     & 0      & 0      & C_{54} & C_{55} & 0 \\ C_{61} & C_{62} & C_{63} & 0 & 0 & C_{66} \\ \end{bmatrix} $$

Solution
In 3$$\times$$3 matrix form, the strain tensor is given by

\boldsymbol{\varepsilon} = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & \varepsilon_{13} \\ \varepsilon_{21} & \varepsilon_{22} & \varepsilon_{23} \\ \varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33} \end{bmatrix} $$ The transformation rule for a second order tensor $$\boldsymbol{A}\,$$ is

\left[A\right]^{'} = \left[L\right] \left[A\right] \left[L\right]^{T} $$ Applying this transformation to the strain tensor, we have
 * $$\begin{align}

\begin{bmatrix} \varepsilon_{11}^{'} & \varepsilon_{12}^{'} & \varepsilon_{13}^{'} \\ \varepsilon_{21}^{'} & \varepsilon_{22}^{'} & \varepsilon_{23}^{'} \\ \varepsilon_{31}^{'} & \varepsilon_{32}^{'} & \varepsilon_{33}^{'} \end{bmatrix} & =  \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & \varepsilon_{13} \\ \varepsilon_{21} & \varepsilon_{22} & \varepsilon_{23} \\ \varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33} \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \\ & =  \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & -\varepsilon_{13} \\ \varepsilon_{21} & \varepsilon_{22} & -\varepsilon_{23} \\ \varepsilon_{31} & \varepsilon_{32} & -\varepsilon_{33} \end{bmatrix} \\ & =  \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & -\varepsilon_{13} \\ \varepsilon_{21} & \varepsilon_{22} & -\varepsilon_{23} \\ -\varepsilon_{31} & -\varepsilon_{32} & \varepsilon_{33} \end{bmatrix} \end{align}$$ In engineering notation (Voigt notation),
 * $$\begin{align}

\left[\boldsymbol{\varepsilon}\right] & = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{22} & \varepsilon_{33} & 2\varepsilon_{23} & 2\varepsilon_{31} & 2\varepsilon_{12} \end{bmatrix}^{T} \\ & = \begin{bmatrix} \varepsilon_{1} & \varepsilon_{2} & \varepsilon_{3} & \varepsilon_{4} & \varepsilon_{5} & \varepsilon_{6} \end{bmatrix}^{T} \end{align}$$ Therefore, the transformed strain tensor can be written as
 * $$\begin{align}

\left[\boldsymbol{\varepsilon}\right]^{'} & = \begin{bmatrix} \varepsilon_{1}^{'} & \varepsilon_{2}^{'} & \varepsilon_{3}^{'} & \varepsilon_{4}^{'} & \varepsilon_{5}^{'} & \varepsilon_{6}^{'} \end{bmatrix}^{T} \\ & = \begin{bmatrix} \varepsilon_{1} & \varepsilon_{2} & \varepsilon_{3} & -\varepsilon_{4} & -\varepsilon_{5} & \varepsilon_{6} \end{bmatrix}^{T} \\ \end{align}$$ The expression for the strain energy density of a linear elastic material imposes a constraint on the components of the stiffness tensor in the presence of planes of material symmetry. This constraint is

C_{ij} (\varepsilon_i \varepsilon_j - \varepsilon_i^{'} \varepsilon_j^{'}) = 0 ~(i,j = 1\dots 6) $$ where $$C_{ij}$$ are the components of the 6$$\times$$6 matrix that contains the independent components of the stiffness tensor.

Therefore,
 * $$\begin{align}

C_{11} (\varepsilon_1 \varepsilon_1 - \varepsilon_1^{'} \varepsilon_1^{'}) + C_{12} (\varepsilon_1 \varepsilon_2 - \varepsilon_1^{'} \varepsilon_2^{'}) + C_{13} (\varepsilon_1 \varepsilon_3 - \varepsilon_1^{'} \varepsilon_3^{'}) & + \\ C_{14} (\varepsilon_1 \varepsilon_4 - \varepsilon_1^{'} \varepsilon_4^{'}) + C_{15} (\varepsilon_1 \varepsilon_5 - \varepsilon_1^{'} \varepsilon_5^{'}) + C_{16} (\varepsilon_1 \varepsilon_6 - \varepsilon_1^{'} \varepsilon_6^{'}) & + \\ C_{21} (\varepsilon_2 \varepsilon_1 - \varepsilon_2^{'} \varepsilon_1^{'}) + C_{22} (\varepsilon_2 \varepsilon_2 - \varepsilon_2^{'} \varepsilon_2^{'}) + C_{23} (\varepsilon_2 \varepsilon_3 - \varepsilon_2^{'} \varepsilon_3^{'}) & + \\ C_{24} (\varepsilon_2 \varepsilon_4 - \varepsilon_2^{'} \varepsilon_4^{'}) + C_{25} (\varepsilon_2 \varepsilon_5 - \varepsilon_2^{'} \varepsilon_5^{'}) + C_{26} (\varepsilon_2 \varepsilon_6 - \varepsilon_2^{'} \varepsilon_6^{'}) & + \\ C_{31} (\varepsilon_3 \varepsilon_1 - \varepsilon_3^{'} \varepsilon_1^{'}) + C_{32} (\varepsilon_3 \varepsilon_2 - \varepsilon_3^{'} \varepsilon_2^{'}) + C_{33} (\varepsilon_3 \varepsilon_3 - \varepsilon_3^{'} \varepsilon_3^{'}) & + \\ C_{34} (\varepsilon_3 \varepsilon_4 - \varepsilon_3^{'} \varepsilon_4^{'}) + C_{35} (\varepsilon_3 \varepsilon_5 - \varepsilon_3^{'} \varepsilon_5^{'}) + C_{36} (\varepsilon_3 \varepsilon_6 - \varepsilon_3^{'} \varepsilon_6^{'}) & + \\ C_{41} (\varepsilon_4 \varepsilon_1 - \varepsilon_4^{'} \varepsilon_1^{'}) + C_{42} (\varepsilon_4 \varepsilon_2 - \varepsilon_4^{'} \varepsilon_2^{'}) + C_{43} (\varepsilon_4 \varepsilon_3 - \varepsilon_4^{'} \varepsilon_3^{'}) & + \\ C_{44} (\varepsilon_4 \varepsilon_4 - \varepsilon_4^{'} \varepsilon_4^{'}) + C_{45} (\varepsilon_4 \varepsilon_5 - \varepsilon_4^{'} \varepsilon_5^{'}) + C_{46} (\varepsilon_4 \varepsilon_6 - \varepsilon_4^{'} \varepsilon_6^{'}) & + \\ C_{51} (\varepsilon_5 \varepsilon_1 - \varepsilon_5^{'} \varepsilon_1^{'}) + C_{52} (\varepsilon_5 \varepsilon_2 - \varepsilon_5^{'} \varepsilon_2^{'}) + C_{53} (\varepsilon_5 \varepsilon_3 - \varepsilon_5^{'} \varepsilon_3^{'}) & + \\ C_{54} (\varepsilon_5 \varepsilon_4 - \varepsilon_5^{'} \varepsilon_4^{'}) + C_{55} (\varepsilon_5 \varepsilon_5 - \varepsilon_5^{'} \varepsilon_5^{'}) + C_{56} (\varepsilon_5 \varepsilon_6 - \varepsilon_5^{'} \varepsilon_6^{'}) & + \\ C_{61} (\varepsilon_6 \varepsilon_1 - \varepsilon_6^{'} \varepsilon_1^{'}) + C_{62} (\varepsilon_6 \varepsilon_2 - \varepsilon_6^{'} \varepsilon_2^{'}) + C_{63} (\varepsilon_6 \varepsilon_3 - \varepsilon_6^{'} \varepsilon_3^{'}) & + \\ C_{64} (\varepsilon_6 \varepsilon_4 - \varepsilon_6^{'} \varepsilon_4^{'}) + C_{65} (\varepsilon_6 \varepsilon_5 - \varepsilon_6^{'} \varepsilon_5^{'}) + C_{66} (\varepsilon_6 \varepsilon_6 - \varepsilon_6^{'} \varepsilon_6^{'}) & = 0 \end{align}$$ For a monoclinic material, replacing the transformed strain components by the equivalent original strain components, we get
 * $$\begin{align}

C_{14} (\varepsilon_1 \varepsilon_4 + \varepsilon_1 \varepsilon_4) + C_{15} (\varepsilon_1 \varepsilon_5 + \varepsilon_1 \varepsilon_5) + C_{24} (\varepsilon_2 \varepsilon_4 + \varepsilon_2 \varepsilon_4) & + \\ C_{25} (\varepsilon_2 \varepsilon_5 + \varepsilon_2 \varepsilon_5) + C_{34} (\varepsilon_3 \varepsilon_4 + \varepsilon_3 \varepsilon_4) + C_{35} (\varepsilon_3 \varepsilon_5 + \varepsilon_3 \varepsilon_5) & + \\ C_{41} (\varepsilon_4 \varepsilon_1 + \varepsilon_4 \varepsilon_1) + C_{42} (\varepsilon_4 \varepsilon_2 + \varepsilon_4 \varepsilon_2) + C_{43} (\varepsilon_4 \varepsilon_3 + \varepsilon_4 \varepsilon_3) & + \\ C_{46} (\varepsilon_4 \varepsilon_6 + \varepsilon_4 \varepsilon_6) + C_{51} (\varepsilon_5 \varepsilon_1 + \varepsilon_5 \varepsilon_1) + C_{52} (\varepsilon_5 \varepsilon_2 + \varepsilon_5 \varepsilon_2) & + \\ C_{53} (\varepsilon_5 \varepsilon_3 + \varepsilon_5 \varepsilon_3) + C_{56} (\varepsilon_5 \varepsilon_6 + \varepsilon_5 \varepsilon_6) + C_{64} (\varepsilon_6 \varepsilon_4 + \varepsilon_6 \varepsilon_4) & + \\ C_{65} (\varepsilon_6 \varepsilon_5 + \varepsilon_6 \varepsilon_5) & = 0 \end{align}$$ or,
 * $$\begin{align}

2~C_{14} \varepsilon_1 \varepsilon_4 + 2~C_{15} \varepsilon_1 \varepsilon_5 + 2~C_{24} \varepsilon_2 \varepsilon_4 + 2~C_{25} \varepsilon_2 \varepsilon_5 + 2~C_{34} \varepsilon_3 \varepsilon_4 + 2~C_{35} \varepsilon_3 \varepsilon_5 & + \\ 2~C_{41} \varepsilon_4 \varepsilon_1 + 2~C_{42} \varepsilon_4 \varepsilon_2 + 2~C_{43} \varepsilon_4 \varepsilon_3 + 2~C_{46} \varepsilon_4 \varepsilon_6 + 2~C_{51} \varepsilon_5 \varepsilon_1 + 2~C_{52} \varepsilon_5 \varepsilon_2 & + \\ 2~C_{53} \varepsilon_5 \varepsilon_3 + 2~C_{56} \varepsilon_5 \varepsilon_6 + 2~C_{64} \varepsilon_6 \varepsilon_4 + 2~C_{65} \varepsilon_6 \varepsilon_5 & = 0 \end{align}$$ Using the symmetry of the stiffness matrix, we have
 * $$\begin{align}

4~C_{14} \varepsilon_1 \varepsilon_4 + 4~C_{15} \varepsilon_1 \varepsilon_5 + 4~C_{24} \varepsilon_2 \varepsilon_4 + 4~C_{25} \varepsilon_2 \varepsilon_5 + 4~C_{34} \varepsilon_3 \varepsilon_4 + 4~C_{35} \varepsilon_3 \varepsilon_5 &+ \\ 4~C_{46} \varepsilon_4 \varepsilon_6 + 4~C_{56} \varepsilon_5 \varepsilon_6 & = 0 \end{align}$$ Since the strains can be arbitrary, the above condition is satisfied only if

C_{14} = C_{15} = C_{24} = C_{25} = C_{34} = C_{35} = C_{46} = C_{56} = 0 $$ Therefore, the stiffness matrix is given by

\left[C\right] = \begin{bmatrix} C_{11} & C_{12} & C_{13} & 0 & 0 & C_{16} \\ C_{21} & C_{22} & C_{23} & 0 & 0 & C_{26} \\ C_{31} & C_{32} & C_{33} & 0 & 0 & C_{36} \\ 0     & 0      & 0      & C_{44} & C_{45} & 0 \\ 0     & 0      & 0      & C_{54} & C_{55} & 0 \\ C_{61} & C_{62} & C_{63} & 0 & 0 & C_{66} \\ \end{bmatrix} $$ Hence shown.

title: Elasticity/Constitutive example 4


text:
Elasticity/Constitutive example 5

Example 5
 Given:

An isotropic material with Young's modulus $$E$$ and Poisson's ration $$\nu$$.

 Find:

The compliance matrix of the material in terms of the Young's modulus and Poisson's ratio.

Solution
The strain is related to the stress via the compliance matrix by the equation

\begin{bmatrix} \varepsilon_1 \\ \varepsilon_2 \\ \varepsilon_3 \\ \varepsilon_4 \\ \varepsilon_5 \\ \varepsilon_6 \end{bmatrix} =   \begin{bmatrix} S_{11} & S_{12} & S_{13} & S_{14} & S_{15} & S_{16} \\ S_{21} & S_{22} & S_{23} & S_{24} & S_{25} & S_{26} \\ S_{31} & S_{32} & S_{33} & S_{34} & S_{35} & S_{36} \\ S_{41} & S_{42} & S_{43} & S_{44} & S_{45} & S_{46} \\ S_{51} & S_{52} & S_{53} & S_{54} & S_{55} & S_{56} \\ S_{61} & S_{62} & S_{63} & S_{64} & S_{65} & S_{66} \end{bmatrix} \begin{bmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \\ \sigma_4 \\ \sigma_5 \\ \sigma_6 \end{bmatrix} $$ For an isotropic material

\varepsilon_{ij} = \frac{1}{E} \left[(1+\nu)\sigma_{ij} - \nu\sigma_{kk}\delta_{ij}\right] $$ Therefore,
 * $$\begin{align}

\varepsilon_{11} & = \frac{1}{E} \left[\sigma_{11} - \nu\sigma_{22} - \nu\sigma_{33}\right] \\ \varepsilon_{22} & = \frac{1}{E} \left[\sigma_{22} - \nu\sigma_{11} - \nu\sigma_{33}\right] \\ \varepsilon_{33} & = \frac{1}{E} \left[\sigma_{33} - \nu\sigma_{11} - \nu\sigma_{22}\right] \\ \varepsilon_{23} & = \frac{1}{E} \left[(1+\nu)\sigma_{23}\right] \\ \varepsilon_{31} & = \frac{1}{E} \left[(1+\nu)\sigma_{31}\right] \\ \varepsilon_{12} & = \frac{1}{E} \left[(1+\nu)\sigma_{12}\right] \end{align}$$ In engineering notation,
 * $$\begin{align}

\varepsilon_{1} & = \frac{1}{E} \left[\sigma_{1} - \nu\sigma_{2} - \nu\sigma_{3}\right] \\ \varepsilon_{2} & = \frac{1}{E} \left[\sigma_{2} - \nu\sigma_{1} - \nu\sigma_{3}\right] \\ \varepsilon_{3} & = \frac{1}{E} \left[\sigma_{3} - \nu\sigma_{1} - \nu\sigma_{2}\right] \\ \varepsilon_{4} & = \frac{1}{E} \left[2(1+\nu)\sigma_{4}\right] \\ \varepsilon_{5} & = \frac{1}{E} \left[2(1+\nu)\sigma_{5}\right] \\ \varepsilon_{6} & = \frac{1}{E} \left[2(1+\nu)\sigma_{6}\right] \end{align}$$ Converting into matrix notation,

\begin{bmatrix} \varepsilon_1 \\ \varepsilon_2 \\ \varepsilon_3 \\ \varepsilon_4 \\ \varepsilon_5 \\ \varepsilon_6 \end{bmatrix} = \frac{1}{E} \begin{bmatrix} 1 & -\nu & -\nu & 0 & 0 & 0 \\ -\nu & 1 & -\nu & 0 & 0 & 0 \\ -\nu & -\nu & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2(1+\nu) & 0 & 0 \\ 0 & 0 & 0 & 0 & 2(1+\nu) & 0 \\ 0 & 0 & 0 & 0 & 0 & 2(1+\nu) \end{bmatrix} \begin{bmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \\ \sigma_4 \\ \sigma_5 \\ \sigma_6 \end{bmatrix} $$ We may also write the above equation as

\begin{bmatrix} \varepsilon_1 \\ \varepsilon_2 \\ \varepsilon_3 \\ \varepsilon_4 \\ \varepsilon_5 \\ \varepsilon_6 \end{bmatrix} =   \begin{bmatrix} 1/E & -\nu/E & -\nu/E & 0 & 0 & 0 \\ -\nu/E & 1/E & -\nu/E & 0 & 0 & 0 \\ -\nu/E & -\nu/E & 1/E & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/\mu & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/\mu & 0 \\ 0 & 0 & 0 & 0 & 0 & 1/\mu \end{bmatrix} \begin{bmatrix} \sigma_1 \\ \sigma_2 \\ \sigma_3 \\ \sigma_4 \\ \sigma_5 \\ \sigma_6 \end{bmatrix} $$ where $$\mu$$ is the shear modulus.

title: Elasticity/Constitutive example 5


text:
Elasticity/Constitutive example 6

Example 6
 Given:

For an isotropic material

K = \lambda + \frac{2}{3} \mu~, E = \frac{\mu(3\lambda+2\mu)}{\lambda+\mu}~, \nu = \frac{\lambda}{2(\lambda+\mu)} $$

 Verify:

 * 1) $$\mu = \frac{E - 3\lambda + r}{4} $$
 * 2) $$K = \frac{E + 3\lambda + r}{6} $$

where $$r = \sqrt{E^2 + 9\lambda^2 + 2E\lambda}$$.

Solution
From the second equation that has been given

E\lambda + E\mu = 3\mu\lambda + 2\mu^2 $$ or,

2\mu^2 - (E - 3\lambda)\mu - E\lambda = 0 $$ Therefore,

\mu = \frac{(E-3\lambda) \pm \sqrt{(E-3\lambda)^2 + 8E\lambda}}{4} $$ or,

\mu = \frac{(E-3\lambda) \pm \sqrt{E^2 + 9\lambda^2 + 2E\lambda}}{4} $$ or,

\mu = \frac{E-3\lambda \pm r}{4} $$ To find out whether the plus or the minus sign should be placed before $$r$$ in the above equation, we put everything in terms of $$\nu$$ and $$E$$. Thus,
 * $$\begin{align}

\mu & = \frac{E}{2(1+\nu)} \\ \lambda & = \frac{\nu E}{(1+\nu)(1-2\nu)} \\ E - 3\lambda & = \frac{E(1 - 2\nu^2 - 4\nu)}{(1+\nu)(1-2\nu)} \\ 8E\lambda & = \frac{8\nu E^2}{(1+\nu)(1-2\nu)} \end{align}$$ Plugging these into the equation for $$\mu$$, multiplying both sides by $$(1+\nu)(1-2\nu)$$ and dividing by $$E$$, we get

2(1-2\nu) = (1-2\nu^2-4\nu) \pm \sqrt{(1-2\nu^2-4\nu)^2 + 8\nu(1+\nu)(1-2\nu)} $$ The limiting value of $$\nu$$ is 0.5. Plugging this value into the above equation, we get,

0 = -1.5 \pm 1.5 $$ The above can be true only if the sign is positive. Therefore, the correct relation is

\mu = \frac{E-3\lambda + r}{4} $$ Plugging this relation into the first of the given equations, we have,
 * $$\begin{align}

K & = \lambda + \cfrac{2}{3}\left(\cfrac{E-3\lambda+r}{4}\right) \\ & = \cfrac{6\lambda + E -3\lambda+r}{6} \\ K & = \cfrac{E +3\lambda+r}{6} \end{align}$$

title: Elasticity/Constitutive example 6


text:
Elasticity/Compatibility example 1

Example 1
Given:

The compatibility equations in terms of the strains are (in index notation)

e_{ikr}~e_{jls}~\varepsilon_{ij,kl} = 0 $$ The stress-strain relations for a linear elastic material are

\varepsilon_{ij} = \frac{1}{E}\left[(1+\nu)~\sigma_{ij} - \nu~\sigma_{mm}~\delta_{ij}\right] $$

Show:

Substituting the stress-strain relations into the compatibility equations, show that the compatibility equation of stress can be expressed as

\sigma_{ii,jj}~ \delta_{rs} - \sigma_{ii,rs} - (1+\nu)(\sigma_{ij,ij}~\delta_{rs} + \sigma_{rs,ii}           -\sigma_{is,ir} - \sigma_{ir,is}) = 0 $$

Solution
Substituting the stress-strain relations into the left hand side of the compatibility equations and multiplying both sides by $$E$$, we get,
 * $$\begin{align}

E~e_{ikr}~e_{jls}~\varepsilon_{ij,kl} & = e_{ikr}~e_{jls}~\left[(1+\nu)~\sigma_{ij,kl} - \nu~\sigma_{mm,kl}~\delta_{ij}\right] \\ & = (1+\nu)~e_{ikr}~e_{jls}~\sigma_{ij,kl} - \nu~e_{ikr}~e_{jls}~\delta_{ij}~\sigma_{mm,kl} \\ & = (1+\nu)~e_{ikr}~e_{jls}~\sigma_{ij,kl} - \nu~e_{nkr}~e_{nls}~\sigma_{mm,kl} \\ & = (1+\nu)~e_{ikr}~e_{jls}~\sigma_{ij,kl} - \nu~e_{krn}~e_{lsn}~\sigma_{mm,kl} \end{align}$$ Now, the $$\delta-e$$ rule states that

e_{ijk}~e_{pqk} = \delta_{ip}\delta_{jq} - \delta_{iq}\delta_{jp} $$ Therefore,
 * $$\begin{align}

E~e_{ikr}~e_{jls}~\varepsilon_{ij,kl} & = (1+\nu)~e_{ikr}~e_{jls}~\sigma_{ij,kl} - \nu~\left(\delta_{kl}\delta_{rs} - \delta_{ks}\delta_{rl}\right) ~\sigma_{mm,kl} \\ & = (1+\nu)~e_{ikr}~e_{jls}~\sigma_{ij,kl} - \nu~\left(\sigma_{mm,nn}\delta_{rs} - \sigma_{mm,sr}\right) \end{align}$$ Recall that,
 * $$\begin{align}

e_{ikr}~e_{jls} & = \text{det} \begin{bmatrix} \delta_{ij} & \delta_{il} & \delta_{is} \\ \delta_{kj} & \delta_{kl} & \delta_{ks} \\ \delta_{rj} & \delta_{rl} & \delta_{rs} \end{bmatrix} \\ & = \delta_{ij}\delta_{kl}\delta_{rs} - \delta_{ij}\delta_{ks}\delta_{rl} - \delta_{il}\delta_{kj}\delta_{rs} + \delta_{il}\delta_{ks}\delta_{rj} +\delta_{is}\delta_{kj}\delta_{rl} - \delta_{is}\delta_{kl}\delta_{rj} \end{align}$$ Therefore,
 * $$\begin{align}

\delta_{ij}\delta_{kl}\delta_{rs}\sigma_{ij,kl} &= \sigma_{ii,jj}\delta_{rs} \\ \delta_{ij}\delta_{ks}\delta_{rl}\sigma_{ij,kl} &= \sigma_{ii,rs} \\ \delta_{il}\delta_{kj}\delta_{rs}\sigma_{ij,kl} &= \sigma_{ij,ij}\delta_{rs} \\ \delta_{il}\delta_{ks}\delta_{rj}\sigma_{ij,kl} &= \sigma_{ir,is} \\ \delta_{is}\delta_{kj}\delta_{rl}\sigma_{ij,kl} &= \sigma_{is,ir} \\ \delta_{is}\delta_{kl}\delta_{rj}\sigma_{ij,kl} &= \sigma_{sr,jj} \end{align}$$ Hence,
 * $$\begin{align}

E~e_{ikr}~e_{jls}~\varepsilon_{ij,kl} & = (1+\nu)\left(\sigma_{ii,jj}\delta_{rs} - \sigma_{ii,rs} -             \sigma_{ij,ij}\delta_{rs} + \sigma_{ir,is} +              \sigma_{is,ir} - \sigma_{sr,jj}\right) - \nu~\left(\sigma_{ii,jj}\delta_{rs} - \sigma_{ii,rs}\right) \\ & = \sigma_{ii,jj}\delta_{rs} - \sigma_{ii,rs} - (1+\nu)\left(          \sigma_{ij,ij}\delta_{rs} + \sigma_{rs,ii}            -\sigma_{is,ir} - \sigma_{ir,is}\right) = 0 \end{align}$$ Hence shown.

title: Elasticity/Compatibility example 1


text:
Elasticity/Airy example 1

Example 1 - Beltrami solution
Given:

Beltrami's solution for the equations of equilibrium states that if

\boldsymbol{\sigma} = \boldsymbol{\nabla} \times \boldsymbol{\nabla} \times \mathbf{A} $$ where $$\mathbf{A}$$ is a stress function, then

\boldsymbol{\nabla} \bullet \boldsymbol{\sigma} = 0~;\boldsymbol{\sigma} = \boldsymbol{\sigma}^{T} $$ Airy's stress function is a special form of $$\mathbf{A}$$, given by (in 3$$\times$$3 matrix notation)

\left[A\right] = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & \varphi \end{bmatrix} $$

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Verify that the stresses when expressed in terms of Airy's stress function satisfy equilibrium.

Solution
In index notation, Beltrami's solution can be written as

\sigma_{ij} = e_{imn}~e_{jpq}~A_{mp,~nq} $$ For the Airy's stress function, the only non-zero terms of $$A_{mp,~nq}\,$$ are $$A_{33,~nq} = \varphi_{,nq}\,$$ which can have nine values. Therefore,
 * $$\begin{align}

\sigma_{11} & = e_{13n}~e_{13q}~\varphi_{,~nq} \\ \sigma_{22} & = e_{23n}~e_{23q}~\varphi_{,~nq} \\ \sigma_{33} & = e_{33n}~e_{33q}~\varphi_{,~nq} \\ \sigma_{23} & = e_{23n}~e_{33q}~\varphi_{,~nq} \\ \sigma_{31} & = e_{33n}~e_{13q}~\varphi_{,~nq} \\ \sigma_{12} & = e_{13n}~e_{23q}~\varphi_{,~nq} \end{align}$$ Since $$e_{33k} = 0\,$$ for $$k=1,2,3$$, the above set of equations reduces to
 * $$\begin{align}

\sigma_{11} & = e_{13n}~e_{13q}~\varphi_{,~nq} \\ \sigma_{22} & = e_{23n}~e_{23q}~\varphi_{,~nq} \\ \sigma_{33} & = 0 \\ \sigma_{23} & = 0 \\ \sigma_{31} & = 0 \\ \sigma_{12} & = e_{13n}~e_{23q}~\varphi_{,~nq} \end{align}$$ Now, $$e_{13k}\,$$ is non-zero only if $$k = 2$$, and $$e_{23k}\,$$ is non-zero only if $$k=1$$. Therefore, the above equations further reduce to
 * $$\begin{align}

\sigma_{11} & = e_{132}~e_{132}~\varphi_{,~22} \\ \sigma_{22} & = e_{231}~e_{231}~\varphi_{,~11} \\ \sigma_{33} & = 0 \\ \sigma_{23} & = 0 \\ \sigma_{31} & = 0 \\ \sigma_{12} & = e_{132}~e_{231}~\varphi_{,~21} \end{align}$$ Therefore, (using the values of $$e_{132}\,$$, $$e_{231}\,$$ and the fact that the order of differentiation does not change the final result), we get
 * $$\begin{align}

\sigma_{11} & = \varphi_{,~22} \\ \sigma_{22} & = \varphi_{,~11} \\ \sigma_{33} & = 0 \\ \sigma_{23} & = 0 \\ \sigma_{31} & = 0 \\ \sigma_{12} & = -\varphi_{,~12} \end{align}$$ The equations of equilibrium (in the absence of body forces) are given by

\sigma_{ji,j} = 0 $$ or,
 * $$\begin{align}

\sigma_{11,1} + \sigma_{21,2} + \sigma_{31,3} & = 0 \\ \sigma_{12,1} + \sigma_{22,2} + \sigma_{32,3} & = 0 \\ \sigma_{13,1} + \sigma_{23,2} + \sigma_{33,3} & = 0 \end{align}$$ Plugging the stresses in terms of $$\varphi$$ into the above equations gives,
 * $$\begin{align}

\varphi_{,221} - \varphi_{,122} + 0 & = 0 \\ -\varphi_{,121} + \varphi_{,112} + 0 & = 0 \\ 0 + 0 + 0 & = 0 \end{align}$$ Noting that the order of differentiation is irrelevant, we see that equilibrium is satisfied by the Airy stress function.

title: Elasticity/Airy example 1


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