User:Eas4200c.f08.aero.e/Week 8

=Monday=



Hooke's Law: $$\tau (s)=G\gamma =G\rho \theta $$

$$\tau (s)=G\rho (s)\theta (x)$$

$$\tau =\frac{q(s)}{t(s)}$$

If we now integrate along the contour $$\varsigma$$:

$$\oint_{\varsigma }^{}{}\tau (s)ds=G\theta (x)\oint_{\varsigma }^{}{}\rho (s)dA$$

$$\oint_{\varsigma }^{}{}\rho (s)dA=2\bar{A}$$

$$\theta =\frac{d\alpha }{dx}$$

What is Ad hoc about the above expression $$(\theta )$$ and the derivative for $$T=2q\bar{A}$$ ?

1) Strain $$(\gamma)$$ must be obtained using the displacement of $$(\rho)$$ in the direction tangent to $$ (\varsigma)$$ at (roe),          but PP’’ is not necessarily tangent to $$(\varsigma)$$.  (However, it is close)

2) $$\tau=q/t$$ obtained from the ad-hoc assumption that $$\tau$$ was uniform across the wall thickness.

Now lets visit a formal justification by the Elasticity Theory:

A. Kinematic assumption.

$$U_{x}(y,z) = \theta \psi (y,z)$$

$$U_{z}(x,y) = \theta x Y$$

$$U_{y}(x,z) = \theta x Z$$

It is important to mention that there is a difference in notation with respect to the text Sun (2006). In order to Rectify the difference we change to the unified notation using cyclic permutation.



$$\varepsilon _{xx}=\varepsilon _{yy}=\varepsilon _{zz}=\gamma _{yz}=0$$

$$\varepsilon _{xx}=\frac{dU_{x}}{dx}=0$$

$$\gamma _{yz}=\frac{dU_{y}}{dz} + \frac{dU_{z}}{dy} = 0$$

$$\frac{dU_{y}}{dz} = -\theta x $$

$$\frac{dU_{z}}{dy} = \theta x $$

Homework assignment for this lecture was to perform the above operation for $$\varepsilon _{yy}$$ &  $$\varepsilon _{zz}$$

$$\varepsilon _{yy}=\frac{dU_{y}}{dy}=0$$

$$\varepsilon _{zz}=\frac{dU_{z}}{dz}=0$$

=Wednesday=

As a continuation of the previous lecture here is a third Ad Hoc assumption for the engineering derivative.

3) There is an inconsistancy in the assumption on the size of $$\alpha$$.               *To get PP' assume a small $$\alpha$$               *To get PP" assume $$\alpha$$ is infinite and $$\rho=r\cos\alpha$$                *Then reintroduce a small $$\alpha$$ again

The following question was posed in class: How many strain components exist in three dimensions?

answer = 6

$$\varepsilon = \begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz}\\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz}\\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix} = \begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix}$$

The matrix transformation shown above was changed to indicial notation where x=1, y=2 and z=3.

$$\varepsilon = \left[\varepsilon _{ij} \right]$$ i,j = 1,2,3

* i = Row * j = Colum

tensorial rotation;

Symmetry of $$\varepsilon$$ : $$\varepsilon _{ij} = \frac{1}{2}(\frac{du_{i}}{dx_{j}} + \frac{du_{j}}{dx_{i}})$$

Coord.

$$x = x_{1}$$

$$y = x_{2}$$

$$z = x_{3}$$

$$\varepsilon _{11} = \varepsilon _{yx} = \frac{1}{2}(\frac{du_{1}}{dx_{1}} + \frac{du_{1}}{dx_{1}}) = \frac{du_{1}}{dx_{1}} = \frac{du_{x}}{dx} $$

$$\varepsilon _{12}=\varepsilon _{21} \Leftrightarrow \varepsilon _{xy}=\varepsilon _{yx}$$ *Only six independent components of $$\varepsilon$$ *Similarly for the stress tensor $$\sigma$$

$$\sigma =\left[\sigma _{ij} \right]_{3x3}$$ *Also only six independent components

$$\sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\tau _{yz}=0$$ *Due to the stress strain relation *See page 70 in Sun(2006) *Isotropic elastic material

MatLab Guide



1) Set $$P_{o}=D$$, then the sweep area from B to L and then from L to E. (Area $$A_{1}$$)

1) Set $$P_{o}=E$$, then the sweep area from F to B. (Area $$A_{21}$$)

1) Set $$P_{o}=F$$, then the sweep area from E to H. (Area $$A_{22}$$)

1) Set $$P_{o}=G$$, then the sweep area from H to T and then from T to F. (Area $$A_{3}$$)

$$\bar{A}=\bar{A_{1}}+\bar{A_{21}}+\bar{A_{22}}+\bar{A_{3}}$$

=Friday=

Normal Strains:

$$ \epsilon _{xx}=\frac{\sigma _{xx}}{E}-\frac{\nu \sigma _{yy}}{E}-\frac{\nu \sigma _{zz}}{E}$$

$$ \epsilon _{yy}=\frac{\sigma _{yy}}{E}-\frac{\nu \sigma _{xx}}{E}-\frac{\nu \sigma _{zz}}{E}$$

$$\epsilon _{zz}=\frac{\sigma _{zz}}{E}-\frac{\nu \sigma _{yy}}{E}-\frac{\nu \sigma _{xx}}{E}$$

 Shear Strains:

$$\gamma_{xy}=2\epsilon_{xy}=\frac{\tau_{xy}} {G}$$

$$\gamma_{yz}=2\epsilon_{yz}=\frac{\tau_{yz}} {G}$$

$$\gamma_{xz}=2\epsilon_{xz}=\frac{\tau_{xz}} {G}$$

Due to symmetry of strain tensor and stress tensor, we can write the relations in column matrix form:
 * $$\underline{\underline{\epsilon}} = \begin{bmatrix}

\epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12}\end{bmatrix}$$


 * $$\underline{\underline{\sigma}} = \begin{bmatrix}

\sigma_{11}\\ \sigma_{22}\\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12}\end{bmatrix}$$

Hooke's Law for isotropic elasticity is as follows:


 * $$\underline{\underline{\epsilon}} = \begin{bmatrix}

\epsilon_{11}\\ \epsilon_{22}\\ \epsilon_{33} \\ \epsilon_{23} \\ \epsilon_{31} \\ \epsilon_{12}\end{bmatrix}=\begin{bmatrix} \frac{1}{E} & \frac{-\nu}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E} & \frac{1}{E} & \frac{-\nu}{E} & 0 & 0 & 0\\ \frac{-\nu}{E} & \frac{-\nu}{E} & \frac{1}{E} & 0 & 0 & 0\\ 0 & 0 & 0 & \frac{1}{2G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{2G} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{2G}\end{bmatrix}\begin{bmatrix} \sigma_{11}\\ \sigma_{22}\\ \sigma_{33} \\ \sigma_{23} \\ \sigma_{31} \\ \sigma_{12}\end{bmatrix}$$

Note: The Poisson ratio is determined by the type of material. The only material with a virtually zero Poisson ratio is cork, used for closing wine bottles. As the material is stretched or compressed in on direction, the other directions do not experience a strain. Most materials have Poisson ratios between 0 and 0.5. Steel has a Poisson ratio of 0.3, while rubber has a 0.5 Poisson ratio.