User:Eas4200c.f08.carbon.orear/HW4/Lecture Notes

=Lecture Notes=

Implications from MATLAB code: NACA quadrature
At this point, the MATLAB code can be examined for an interesting and important result. In our quadrature of a NACA airfoil, we examine that for a point taken outside the airfoil, the sum of Ā1 and Ā2 (shown below) gives us the Ā of the airfoil. Note that (for the given point shown) Ā1 is positive and Ā2 is negative. It doesn't matter which point we choose, because the sum with always yield an Ā for the remainder corresponding to the airfoil alone.



The cross-products we used in Homework 3 give a dA direction out of the plane of the page. The yellow area is the sum of Ā1 and Ā2. Ā1 is the sweep of both areas, and Ā2 is a negative value added to it shown by just the gray area. The resulting yellow area is the airfoil Ā.

 Ā = Ā1 + Ā2 

Considerations
We briefly want to question whether trapezoids would be more useful for the NACA airfoil quadrature. Change in curvature (as in areas noted by the arrows in the picture) cause area calculation difficulties and difficulties properly fitting trapezoids. We are used to trapezoids in a basic integration under a smooth curve, but the application is not as elegant here. This is why triangles are much more fitting. The cross-products generated are simple for calculations and the shapes can better sum to represent the airfoil.



Continuation of Single-cell airfoil
This example began in Homework 3.



$$ A = \frac{1}{2}\pi R^2 = \frac{1}{2}\pi (\frac{b}{2})^2 $$ -area for half a circle $$ A = \frac{1}{2}ba$$ -area for the triangle

$$\bar {A_i}$$ is found to be:

$$\bar {A_i} = \frac{\pi}{2}(\frac{b}{2})^2 + \frac{1}{2}ba$$

$$\bar {A} = 5.571 sq. m$$

Remember shear flow is given by: $$T = 2q\bar{A} => q = \frac{T}{2\bar{A}}$$
 * Note: T can be variable

And, the twist angle is given by: $$\theta = \frac{1}{2G\bar{A}} \sum_{j=1}^3 \frac{q_jl_j}{t_j}$$ where $$ j = 1,2,3 $$ index for segment number.

For this example, $$\theta = \frac{q}{2G\bar{A}}(1239.912677)$$ which is a function of q, since G will be given and Ā is known.

We assume that the allowable shear stress equals the max shear stress.

τmax = {q} / {min(t1,t2,t3)}

In this case, the minimum of these is t1. Since $$q = \frac{T}{2\bar{A}}$$, it follows that

Tallow = 2Āτallow(min{t1,t2,t3})

Which becomes, given that τallow = 100 GPa Tallow = 8.913 GNm

Multicell Airfoil
To consider an example of an airfoil separated by spars (creating multicells), one must first consider the simplified case using right angles. In the figure below:

t1 = .3cm, t2 = .5cm, t3 = .4cm



Looking back to the Roadmap Section K, and using the equation $$\theta_i = \frac{1}{2G_i\bar{A_i}} \oint \frac{q_i}{t_i} ds$$ we can find θ as a function of T and J (torsional constant).

(1) $$T = T_{1} + T_{2} = 2q_{1}\bar{A_{1}} + 2q_{2}\bar{A_{2}}$$, where $$\bar{A_{1}} = ac$$ and $$\bar{A_{2}} = bc$$

(3) $$\theta_1 = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$

(4) $$\theta_2 = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

Because the two cells are connected the angle of each cell must rotate together. In other words, C1 and C2 have the same rate of twist angle, where q1 and q2 are two unknowns.

(2) $$\theta_1 = \theta_2$$ Equations (1) and (2) can be used to solve for an expression of q1 and q2 in terms of the resultant of T.


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!HW Torsion of Multicell Airfoil, from team Carbon
 * Use $$T = T_{1} + T_{2}$$ to find the first equation:
 * Use $$T = T_{1} + T_{2}$$ to find the first equation:

$$\bar{A}_{1} = a*c = (.3m)(.4m) = .12m^{2}$$ and $$\bar{A}_{2} = b*c = (.6m)(.4m) = .24m^{2}$$

→ $$T = 2q_{1}\bar{A}_{1} + 2q_{2}\bar{A}_{2} = .24q_{1} + .48q_{2}$$ (1)

Now solve for $$\theta _{1}$$ and $$\theta _{2}$$:


 * $$\theta _{1} = \frac{1}{2G\bar{A_1}}[\frac{2q_{1}a}{t_{1}} + \frac{q_{1}c}{t_{1}} + \frac{(q_{1} - q_{2})c}{t_{12}}]$$, where $$G_{1} = G_{2} = G$$

$$\theta _{1} = \frac{1}{.24G}[200q_{1} + 133.33q_{1} + 100(q_{1} - q_{2})]$$

$$\theta _{1} = \frac{1}{.24G}(433.33q_{1} - 100q_{2})$$


 * $$\theta _{2} = \frac{1}{2G\bar{A_2}}[\frac{2q_{2}a}{t_{2}} + \frac{q_{2}c}{t_{2}} + \frac{(q_{2} - q_{1})c}{t_{12}}]$$

$$\theta _{2} = \frac{1}{.48G}[240q_{2} + 80q_{2} + 100(q_{2} - q_{1})]$$

$$\theta _{2} = \frac{1}{.48G}(420q_{2} - 100q_{1})$$


 * $$\theta _{1} = \theta _{2}$$

$$\frac{1}{.24}(433.33q_{1} - 100q_{2}) = \frac{1}{.48}(420q_{2} - 100q_{1})$$

$$2013.875q_{1} = 1291.67q_{2}$$, where G cancels. (2)


 * Solving the system of equations of (1) and (2):

from (1): $$q_{1} = 4.167T - 2q_{2}$$

substitute into (2): $$(2013.87)(4.167T - 2q_{2}) = 1291.67q_{2}$$

→ $$q_{2} = .000783T$$


 * Substitute into (2):

→ $$q_{1} = .0005024T$$

Since $$\theta _{1} = \theta _{2} = \theta $$, we can plug into either $$\theta _{1}$$ or $$\theta _{2}$$

$$\theta _{2} = \theta = \frac{1}{.48G}(420q_{2} - 100q_{1})$$

$$\theta = \frac{.58}{G}T$$


 * Because $$\theta = \frac{T}{2GJ}$$, therefore $$\frac{.58}{G}T = \frac{T}{2GJ}$$

→ → $$J = .862m^{4}$$

Contribution by carbon.w
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Lecture 20
Recall that $$\displaystyle T = GJ \theta$$

Once $$\displaystyle \theta$$ is found as a function of T, use $$\theta = \frac{T}{GJ}$$ or $$J = \frac{T}{G\theta}$$ to find J.

The theory behind this method can be outlined as follows:

$$\displaystyle T = GJ\theta$$ (which has been previously derived)

$$T = 2q\bar{A}$$ (which has been previously derived)

The engineering, or "ad-hoc" derivation of $$\theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$

Uniform Bar with a Non-circular Cross Section Subjected to Twist



Displacement PP' due to $$\displaystyle \alpha$$:

$$\frac{PP'}{OP} = tan \alpha \cong\alpha$$ for small $$\displaystyle \alpha$$

Project the displacement PP' on the direction perpendicular to OP':

$$\displaystyle PP'' = PP'cos\alpha$$

$$\displaystyle => PP = (OPtan\alpha)cos\alpha = (OPcos\alpha)tan\alpha$$, where $$\displaystyle OP = OPcos\alpha$$

Recall that $$\displaystyle OP=r$$ (the radial coordinate) and $$\displaystyle OP'' = \rho$$

$$\displaystyle PP''=(rcos\alpha)tan\alpha$$

where:

PP'' is the displacement of P in the firection "tangent" to the lateral surface of the bar

$$\displaystyle rcos\alpha$$ is $$\displaystyle \rho$$

$$tan\alpha \cong \alpha$$

Strain: $$\gamma = \frac{PP''}{dx} = \frac{\rho\alpha}{dx} = \rho\theta$$, with $$\theta = \frac{\alpha}{dx}$$ (the rate of twist).


 * Recall that for $$\displaystyle \alpha$$ very small, it can be denoted by $$\displaystyle d\alpha$$.

Formal Justification by Elasticity Theory
Hooke's Law:

$$\displaystyle \tau=G\gamma = G\rho\theta $$

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

Integrate the contur of the shape, C:

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

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

$$\oint_{C}^{}{\rho(s) ds}=2\bar{A}$$

This is how we get the earlier expresson $$\theta = \frac{1}{2G\bar{A}}\oint\frac{q}{t}ds$$

What is the ad hoc about the above derivation of expression for $$\theta$$ that we found above and the derivation of $$T=2q\bar{A}$$?

1) Strain $$\gamma$$ must be abtained using the displacement of P in the direction tangent to C at P, but PP" is not nessecarily tangent to C. It is however actually very close to being tangent to C. 2)$$\tau=\frac{q}{t}$$ was obtained from the ad hoc assumtion that $$\tau$$ was uniform across the wall thickness, t.

Now formal justification (derivation) by elasticity theory:

Road Map (from HW#3)

A. Kinematic Assumption

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

Where $$\displaystyle \theta$$ is considered to be constant with respect to $$\displaystyle x$$. $$\displaystyle u_{y}(x,z)=-\theta xz$$

$$\displaystyle u_{z}(x,y)=\theta xy$$

To transform the equation in the book by Sun[2006] to our uniform notation, use cyclic permation

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

$$\epsilon _{xx}=\frac{\partial u_{x}(y,z)}{\partial x}=0$$

$$\epsilon _{yy}=\frac{\partial u_{y}(x,z)}{\partial y}=0$$

$$\epsilon _{zz}=\frac{\partial u_{z}(x,y)}{\partial z}=0$$

$$\gamma _{yz}=\frac{\partial u_{y}(x,z)}{\partial z}+\frac{\partial u_{z}(x,y)}{\partial y}$$

Where:

$$\frac{\partial u_{y}(x,z)}{\partial z}=-\theta x$$

$$\frac{\partial u_{z}(x,y)}{\partial y}=+\theta x$$

Lecture 22
Engineering derivation continued:

Point 3) The inconsistency in the assumption of size of $$\alpha$$: To get line PP', assume $$\alpha$$ is small; to get PP", assume $$\alpha$$ is finite (cos$$\alpha$$), and $$\rho = rcos\alpha$$; reintroduce small $$\alpha$$ after that.

Back to the formal derivation (from P. 21-3). How many strain components in 3-D? The answer is there a six unique components, although there are nine coefficients as seen below:

$$\boldsymbol{\varepsilon } = \begin{bmatrix} \varepsilon _{xx} & \varepsilon _{xy} & \varepsilon _{xz}\\ \varepsilon _{yx} & \varepsilon _{yy} & \varepsilon _{yz}\\ \varepsilon _{zx} & \varepsilon _{zy} & \varepsilon _{zz} \end{bmatrix}$$


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!Material Coordinates, from team Carbon
 * In the following matrix we use matrix row and column location to denote each component. This is NOT to be confused with "material coordinate systems."  In the material coordinate systems of composite layups, for example, the "1" direction represents the axial fiber direction, the "2" direction represents the normal in-plane transverse direction, and "3" denotes out-of-plane normal transverse direction.  carbon.orear
 * }
 * }

$$\boldsymbol{\varepsilon } = \begin{bmatrix} \varepsilon _{11} & \varepsilon _{12} & \varepsilon _{13}\\ \varepsilon _{21} & \varepsilon _{22} & \varepsilon _{23}\\ \varepsilon _{31} & \varepsilon _{32} & \varepsilon _{33} \end{bmatrix} = [\varepsilon _{ij}]$$ (tensorial notation), where x → 1, y → 2, z → 3, i = row, j = column, and i,j = 1,2,3

By using the tensorial notation, it allows you to only need to memorize one equation for $$\boldsymbol{\varepsilon }$$ rather than three:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}}  \frac{\partial u_{j}}{\partial x_{i}})$$

where x → x1, y → x2, z → x3

$$\varepsilon _{11} = \varepsilon _{xx} = \frac{1}{2}(\frac{\partial u_{1}}{\partial x_{1}}  \frac{\partial u_{1}}{\partial x_{1}}) = \frac{\partial u_{1}}{\partial x_{1}} = \frac{\partial u_{x}}{\partial x}$$


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!HW Strain Symmetry, from team Carbon
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:
 * By symmetry, $$\varepsilon _{ij} = \varepsilon _{ji}$$. If we make this assumption, then:

$$\varepsilon _{ij} = \frac{1}{2}(\frac{\partial u_{i}}{\partial x_{j}} + \frac{\partial u_{j}}{\partial x_{i}}) = \varepsilon _{ji} = \frac{1}{2}(\frac{\partial u_{j}}{\partial x_{i}} +  \frac{\partial u_{i}}{\partial x_{j}})$$ which holds.

Contribution by carbon.w
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Due to this symmetry, there are only 6 independent components of $$\displaystyle \epsilon$$.

Also there are only 6 independent componets of $$\displaystyle \sigma$$ in 3-D

Is the symmetry of $$\epsilon $$ related to isotrophy of materials?

No Isotropic elasticity is related to the material behavior (a $$\sigma$$ - $$\epsilon$$ relation)

$$\displaystyle \epsilon _{xx}=\epsilon _{yy}=\epsilon _{zz}=\gamma _{yz}=0$$ Fallowing this equation we get

$$\displaystyle \sigma _{xx}=\sigma _{yy}=\sigma _{zz}=\tau _{yz}=0$$ due to the stress - strain ($$\displaystyle \sigma - \epsilon$$) relation

Lecture 23
Poisson's ratio, $$\nu $$, (see external link ) is the ratio of the distance a material is stretched or compressed to it's cross-sectional area. Poisson's ratio can be used to relate normal strains to normal stresses by

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


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!HW Showing $$\varepsilon _{yy}$$, Team Carbon $$\varepsilon _{yy} = \frac{\sigma _{yy}}{E} - \frac{\nu \sigma _{xx} }{E} - \frac{\nu \sigma _{zz} }{E}$$

Contribution by carbon.guidry
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!HW Showing $$\varepsilon _{zz}$$, Team Carbon $$\varepsilon _{zz} = \frac{\sigma _{zz}}{E} - \frac{\nu \sigma _{xx} }{E} - \frac{\nu \sigma _{yy} }{E}$$

Contribution by carbon.guidry
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$$\gamma _{xy} = 2\sigma _{xy} = \frac{\tau _{xz}}{G}$$


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!HW Showing $$\gamma _{xz}$$, Team Carbon $$\gamma _{xz} = 2\sigma _{xz} = \frac{\tau _{xz}}{G}$$

Contribution by carbon.guidry
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!HW Showing $$\gamma _{yz}$$, Team Carbon $$\gamma _{yz} = 2\sigma _{yz} = \frac{\tau _{yz}}{G}$$

Contribution by carbon.guidry
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Voigt Notation:

Due to the symmetry of $$\varepsilon = \begin{bmatrix} \varepsilon _{ij} \end{bmatrix}$$ 3X3 and $$\sigma = \begin{bmatrix} \sigma _{ij} \end{bmatrix}$$3X3, $$\sigma _{ij}$$ and $$\varepsilon _{ij}$$ can be arranged in a column matrix.

$$ \begin{Bmatrix} \varepsilon _{ij} \end{Bmatrix}=

\begin{Bmatrix} \varepsilon _{11} \\ \varepsilon _{22} \\ \varepsilon _{33} \\ \varepsilon _{23} \\ \varepsilon _{31} \\ \varepsilon _{12} \end{Bmatrix}

\begin{Bmatrix} \sigma _{ij} \end{Bmatrix}=

\begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \sigma _{33} \\ \sigma _{23} \\ \sigma _{31} \\ \sigma _{12} \end{Bmatrix} $$

Hooke's Law for isotropic elasticity

$$ \begin{Bmatrix} \varepsilon _{11} \\ \varepsilon _{22} \\ \varepsilon _{33} \\ \sigma _{yz} \\ \sigma _{xz} \\ \sigma _{xy} \end{Bmatrix}=

\begin{vmatrix} \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}{G} & 0 & 0\\ 0 & 0 & 0 & 0 & \frac{1}{G} & 0\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{G} \end{vmatrix}

\begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \sigma _{33} \\ \sigma _{yz} \\ \sigma _{xz} \\ \sigma _{xy} \end{Bmatrix} $$


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!HW Hooke's Law Matrix Rewritten, Team Carbon
 * using $$\varepsilon _{xy}=\frac{\gamma _{xy}}{2}$$
 * using $$\varepsilon _{xy}=\frac{\gamma _{xy}}{2}$$

$$ \begin{Bmatrix} \varepsilon _{11} \\ \varepsilon _{22} \\ \varepsilon _{33} \\ \varepsilon _{23} \\ \varepsilon _{31} \\ \varepsilon _{12} \end{Bmatrix}=

\begin{vmatrix} \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{vmatrix}

\begin{Bmatrix} \sigma _{11} \\ \sigma _{22} \\ \sigma _{33} \\ \sigma _{23} \\ \sigma _{31} \\ \sigma _{12} \end{Bmatrix} $$

Contribution by carbon.guidry
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The value of Poisson's Ratio if dependent on the properties of the specific material. For regular materials, Poisson's ratio has a value between 0 and .5. Any artificial or otherwise not natural material would have a negative ratio. Other Poisson's Ratios for common materials include steel, .3, rubber, .5, and a unique case in cork, which has a ratio of 0. Cork is special because as it is stretched or compressed, the cross-sectional area does not change.