User:EAS4200C.F08.WIKI.A/Homework6

=Equation of Equilibrium Continued=

Recall the 1-D case from earlier: dimensional analysis.

$$\left. [f]=\frac{F}{L} \right \} \;\left [ \frac{f}{A} \right] = \frac{f}{L^3}$$

Notes :


 * The square brackets mean "dimension of"


 * F = Force


 * L = Length

$$[A]=L^2$$

$$[\sigma] = \frac{F}{L^2} \Rightarrow \left [ \frac{d \sigma}{dx} \right] = \frac{[d \sigma]}{[dx]} = \frac{\frac{F}{L^2}}{L} = \frac{F}{L^3}$$
 * F/L3 = Force/Volume or Body Force

$$[dx] = L$$

$$\left. \begin{matrix} \varepsilon = \frac{du}{dx}\\ E = \frac{\Delta L}{L} \end{matrix}  \right \} [\varepsilon] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

$$\nu =\frac{\varepsilon_{yy}}{\varepsilon_{xx}} \Rightarrow [\nu] = \frac{[\varepsilon_{yy}]}{[\varepsilon_{xx}]} = 1$$

=Torsional Analysis Continued=

The torsional analysis previously started is continued by looking at the dimensions of the derivative of stress:

$$[\frac{\partial \sigma _{ij}}{\partial x_{i}}] = \frac{F}{L^3} = \frac{force}{volume}$$

Recalling - D. Prandtl stress function - from the road map previously constructed in class we see that:

$$\sigma _{yx} = \frac{\partial \phi }{\partial z}$$

and

 $$ \sigma _{zx} = -\frac{\partial \phi }{\partial y}$$

where $$\phi$$ plays the role of a potential function and $$(\sigma _{yx}, \sigma _{zx}$$ are composed of the "gradient" of $$\phi$$ with respect to (y,z).

As a review of the gradient operators, recall that for scalar functions f(x,y,z):  $$\bigtriangledown f(x,y,z) = \frac{\partial f}{\partial x}\vec{i} +\frac{\partial f}{\partial y}\vec{j} +\frac{\partial f}{\partial z}\vec{k}$$

Where f is the potential function.

Now, substituting the values for $$\sigma _{yx}$$ and $$\sigma _{zx}$$ into the previously given Laplace's Equation we get:

$$\frac{\partial }{\partial y}(\frac{\partial \phi}{\partial z}) + \frac{\partial }{\partial z}(-\frac{\partial \phi}{\partial y}) = 0$$

Which can be rearranged to show that:  $$\frac{\partial ^2 \phi}{\partial y \partial x} = \frac{\partial ^2 \phi}{\partial z \partial y}$$

Due to $$\phi$$ being a continuous and smooth function, the second derivatives are interchangeable in the order they are performed.

Likewise:

$$\frac{\partial ^2 \phi}{\partial y^2} + \frac{\partial ^2 \phi}{\partial z^2} = -2G\theta $$

The lefthand side can also be written as $$\bigtriangledown ^2 \phi$$ which is the Laplacian of $$\phi$$.

$$[t]_{3x1} = [\sigma]_{3x3}[n]_{3x1}$$

Where t is the component of the traction force t, $$\sigma$$ is the component of the stress tensor, and n is the component of the normal vector $$\vec{n}$$

A graphical depiction of the 2-D case is shown below:



2-D Case Continued


Here we will find the sum of the forces in the y-direction.
$$\displaystyle \begin{Vmatrix} \bar{n} \end{Vmatrix}= 1$$ $$\displaystyle \sum{F_y} = -\sigma_{yy}\cdot (dz\cdot 1) - \sigma(dy\cdot1) + t_y(ds\cdot1) = 0 = \begin{cases}dz =ds\cos \theta \\ dy = ds\sin\theta \end{cases}$$

$$\displaystyle 0 = -\sigma_{yy}n_y - \sigma_yzn_z + t_y ds$$

$$\displaystyle n_y = \cos\theta, n_z = \sin\theta $$ is the unit normal vector unit depth along x-direction.

This is equation one which will be combined with two later equations to give us the generalized 3-D case:

<p style="text-align:center;">$$t_y=\sigma_{yy}n_y+\sigma_{yz}n_z\;$$

Note: $$\left[t_y \right]=\frac{F}{L^2}$$

$$\vec{t}\;$$:traction vector (distribution surface force)

<p style="text-align:center;">$$\left[t_y \right]=\left[\sigma \right]$$

This is equation 2:

Taking the sum of the forces in the z-direction
<p style="text-align:center;">$$t_z=\sigma _{yz}\cdot n_y+\sigma _{zz}n_z$$

This is equations one and two combined into matrix format: <p style="text-align:center;"> $$\begin{Bmatrix} t_y\\ t_z \end{Bmatrix}=\begin{bmatrix} \sigma_{yy} &\sigma_{yz} \\ \sigma_{zy}& \sigma_{zz} \end{bmatrix}\begin{Bmatrix} n_y\\ n_z \end{Bmatrix}$$

Generalized to 3-D case

<p style="text-align:center;">$$\begin{Bmatrix} t_1\\ t_2\\ t_3 \end{Bmatrix}=\begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} &\sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix} \begin{Bmatrix} n_1\\ n_2\\ n_3 \end{Bmatrix}$$

<p style="text-align:center;"> $$t_i=\sum_{j=1}^{3}{\sigma_{ij}n_j}$$ $$i=1,2,3$$

<p style="text-align:center;"> $$\left\{ti \right\}_{3x1}=\left[\sigma_{ij} \right]_{3x3}\left\{n_j \right\}_{3x1}$$

Two Cases of Torsional Analysis
There are two specific cases that we will concern ourselves with. The first being the closed thin walled cross-section and the second being the solid cross section. the two are depicted below.

Case 1: Closed Thin Walled Cross-Section (compare equations)
This case can be compared to the NACA Airfoil that is continuously being analyzed. It is important to note that the Prandtl stress function $$\displaystyle \Phi$$ is constant on the outer surface $$\displaystyle S_o$$ and the inner surface $$\displaystyle S_i$$. $$\displaystyle S_i$$ can also be represented by $$\displaystyle S_1$$ as well. The image below represents this case. It is comparable to Figure 3.11 in the text (Sun 2006).

<p style="text-align:center;">

Case 2: Solid Cross-Section
The one thing that is important to note about this case is that the Prandtl stress function $$\displaystyle \Phi$$ is zero on the outter surface $$\displaystyle S_o$$. This case is depicted below.

<p style="text-align:center;">

Uniform Bar With Solid Circular Cross-Section
This section coincides with section 3.3 in the text (Sun 2006). The following figure depicts the cross section of a solid circular bar and we will be determining the expression for the torque and polar moment of inertia in terms of the the Prandtl stress function.

<p style="text-align:center;">

Torque In Terms of $$\displaystyle \phi$$
The axis of the bar coincides with the origin of the coordinate system. The boundary of the bar can be expressed as the following equation;

<p style="text-align:center;">$$\displaystyle x^2 +y^2 = a^2$$ where $$\displaystyle a^2$$ is the radius of the cross section

It is important to note that this is equivalent to the equation of a circle. We will assume that the stress function equals the following;

<p style="text-align:center;"> $$\displaystyle \phi = C \left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right)$$

With this information we are able to compute the torque. This is done below.

<p style="text-align:center;">$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy} =2 C \int \int_{A}^{}{\left(\frac{x^2}{a^2} + \frac{y^2}{a^2} -1 \right) dxdy} = 2C\int \int_{A}^{}{\left(\frac{r^2}{a^2} -1 \right) dA} = 2C \left(\frac{{\pi} a^2}{2} - ({\pi}a^2) \right) = -C \pi a^2$$

We must now solve for the variable $$\displaystyle C$$. We know that the Prandtl stress function is equal to zero on the outer surface. We also that the gradient of the Prandtl stress function is equal to the following;

<p style="text-align:center;">$$\displaystyle \nabla^2 \phi = -2G\theta = \frac{\partial^2 \phi }{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} = C\frac{2}{a^2} + C\frac{2}{a^2} = C\frac{4}{a^2}$$

Solving for $$\displaystyle C$$ yields;

<p style="text-align:center;"> $$\displaystyle C = -\frac{a^2G\theta}{2}$$

Plugging this back into the equation for torque we can obtain the final expression.

<p style="text-align:center;">$$\displaystyle T = \frac {\theta G \pi a^4}{2}$$

Polar Moment of Inertia In Terms of $$\displaystyle \phi$$
We know that the polar moment of inertia $$\displaystyle J$$ equals the following;

<p style="text-align:center;"> $$\displaystyle J = \frac {T}{G\theta}$$

where

<p style="text-align:center;">$$\displaystyle T = 2\int \int_{A}^{}{\phi dxdy}= -C\pi a^2$$ <br \> <p style="text-align:center;">$$\displaystyle G \theta = -\frac{\nabla^2 \phi}{2} = \frac{2C}{a^2}$$

The polar moment of inertia then equals the following;

<p style="text-align:center;">$$\displaystyle J = \frac {T}{G\theta} = --\frac{C\pi a^2}{\frac{2C}{a^2}} = \frac{\pi a^4}{2}$$

This is the exact equation that has been derived multiple time before using other methods.

{| class="toccolours collapsible collapsed" style="width:100%" ! Relationship Between K and k_c
 * The function K can be expressed in terms of k_c as the following;
 * The function K can be expressed in terms of k_c as the following;

<p style="text-align:center;">$$\displaystyle K = k_c \frac{\pi^2}{12}$$

How do we come to this equation? First we have to recognize that the critical buckling load is given as follows;

<p style="text-align:center;">$$\displaystyle (P_{x})_{cr} = k_c\frac{\pi^2D}{b}$$ where b is the width of the plate

The variable $$\displaystyle k_c = \left(\frac {mb}{a}+\frac{a}{mb} \right)^2$$. $$\displaystyle m $$ and $$\displaystyle n $$ are the number of half wave lengths and take on the value of whole numbers only. We next have to note that the critical buckling stress is given by the follwing equation;

<p style="text-align:center;">$$\displaystyle (\sigma_{xx})_{cr} = \frac{(P_x)_{cr}}{bh} = k_c \frac{\pi^2D}{b^2h} $$ D is known as the plate bending stiffness <br \> <p style="text-align:center;">$$\displaystyle D = \frac{Eh^3}{12(1-v^2)}$$

For the case of a simply supported rectangular plate under a distributed compressive load on both sides, the critical buckling stress can be represented as the following;

<p style="text-align:center;">$$\displaystyle (\sigma_{xx})_{cr} = K\frac{E}{1-v^2}\left(\frac{t}{b} \right)^2$$ where E is the elastic modulus, v is poissons ratio, and t is the thickness of the plate h.

Equating the two expressions for the critical buckling stress and solving for K, and making sure to plug in for the plate bending stiffness D, the following results can be obtained. As shown most of the variables cancel and

<p style="text-align:center;">$$\displaystyle K\frac{E}{1-v^2}\left(\frac{h}{b} \right)^2 = k_c \frac{\pi^2Eh^3}{12b^2h(1-v^2)}$$ <br \><p style="text-align:center;"> $$\displaystyle K = k_c\frac{\pi^2}{12}$$

Verification
We will verify $$\displaystyle K $$ for the following values: <p style="text-align:center;"> $$\displaystyle \frac{a}{b} = 1, m=1$$

The value of $$\displaystyle k_c $$ would then equal the following;

<p style="text-align:center;">$$\displaystyle k_c = \left(\frac {mb}{a}+\frac{a}{mb} \right)^2 = \left(\frac {(1)(1)}{1}+\frac{1}{(1)(1)} \right)^2 = 4$$

Plugging this value into the equation for $$\displaystyle K $$ we can obtain the final answer.

<p style="text-align:center;">$$\displaystyle K = k_c\frac{\pi^2}{12} = 4\frac{\pi^2}{12} = 1.0472$$
 * }

<p style="text-align:center;">

Torque can be expressed as

<p style="text-align:center;"> $$\displaystyle T = 2 \int_A \phi dA $$

From the theory of elasticity, the torque becomes

<p style="text-align:center;"> $$\displaystyle T = 2C ( {J \over a^2} - A) $$

Where the area, $$\displaystyle A $$, since the radius is $$\displaystyle a $$, is simply

<p style="text-align:center;"> $$\displaystyle A = \pi a^2 $$

And where $$\displaystyle J $$ is

<p style="text-align:center;"> $$\displaystyle J = \int_A r^2 dA = {1 \over 2} \pi a^4 $$

Using the ad-hoc method

<p style="text-align:center;"> $$\displaystyle T = GJ \theta $$

Recall that the Prandtl Stress function of $$\displaystyle \phi $$ states that the stress in the x-direction normal to the y-axis is

<p style="text-align:center;"> $$\displaystyle \sigma_{yx} = {\partial \phi \over \partial z} $$

Taking this partial derivative, you obtain

<p style="text-align:center;"> $$\displaystyle \sigma_{yx} = 2C {z \over a^2} $$

Where substitution for $$\displaystyle C $$ yields

<p style="text-align:center;"> $$\displaystyle \sigma_{yx} = - G \theta z $$  (1.1)

Similarly, the stress in the x-direction normal to the z-axis is

<p style="text-align:center;"> $$\displaystyle \sigma_{zx} = - {\partial \phi \over \partial y} $$

Again, taking this partial derivative, you obtain

<p style="text-align:center;"> $$\displaystyle \sigma_{zx} = -2C {y \over a^2} $$

Where, again, substitution for $$\displaystyle C $$ yields

<p style="text-align:center;"> $$\displaystyle \sigma_{zx} = G \theta y $$  (1.2)

Also note that shear stress, $$\displaystyle \tau $$, is

<p style="text-align:center;"> $$\displaystyle \tau = {Tr \over J} $$

Proof of No Warping
Now, using the above equations in addition to the stress strain relation, we can prove that no warping takes place.

First, recall that

<p style="text-align:center;"> $$\displaystyle \gamma = {\sigma \over G} $$

Or more specifically in the context of this problem

<p style="text-align:center;"> $$\displaystyle \gamma_{yx} = {\sigma_{yx} \over G} $$

<p style="text-align:center;"> $$\displaystyle \gamma_{zx} = {\sigma_{zx} \over G} $$

Also recall that

<p style="text-align:center;"> $$\displaystyle \gamma_{yx} = {\partial u_x \over \partial y} - \theta z $$

<p style="text-align:center;"> $$\displaystyle \gamma_{zx} = {\partial u_x \over \partial z} + \theta y $$

Thus,

<p style="text-align:center;"> $$\displaystyle {\sigma_{yx} \over G} = {\partial u_x \over \partial y} - \theta z $$  (2.1)

<p style="text-align:center;"> $$\displaystyle {\sigma_{zx} \over G} = {\partial u_x \over \partial z} + \theta y $$  (2.2)

Now, substituting equation (1.1) into equation (2.1), and similarly substituting equation (1.2) into equation (2.2) the result is

<p style="text-align:center;"> $$\displaystyle {-G \theta z \over G} = {\partial u_x \over \partial y} - \theta z $$

<p style="text-align:center;"> $$\displaystyle {G \theta y \over G} = {\partial u_x \over \partial z} + \theta y $$

In both equations, each $$\displaystyle G $$ cancels out. Also, $$\displaystyle \theta x $$ and $$\displaystyle \theta y $$ cancel out when subtracted from each side of the respective equation, thus leaving

<p style="text-align:center;"> $$\displaystyle {\partial u_x \over \partial y} = 0 $$

<p style="text-align:center;"> $$\displaystyle {\partial u_x \over \partial z} = 0 $$

Thus, we can conclude that there is no warping since

<p style="text-align:center;"> $$\displaystyle u_x (y,z) = 0 $$

Flexural Shear Flow in Thin-Walled Cross Sections
<p style="text-align:center;">

Unsymmetrical thin walled cross sections
There will be a comparison drawn between cross sections with symmetry and cross sections of solids without symmetrical cross sections.

This is the generalized formula for a unsymmetrical cross section. (Equation 5.1)

<p style="text-align:center;">$$ \int_{A_i}^{}{\frac{d\sigma _{xx}}{dx} dA}=-q_A$$

<p style="text-align:center;">

Case 1: Symmetrical about the Y-axis
The Stress factor is given under symmetrical cross sectional conditions as follows:

<p style="text-align:center;">$$\sigma _{xx}=\frac{M_yZ}{I_y} $$

The shear flow factor $$q\;$$ is:

<p style="text-align:center;">$$q(s)=\frac{-V_zQ_y}{I_y}$$

$$Q_y=\int_{A_S}^{}{ZdA}=Z_cA_s$$

Case 2:Unsymmetrical about the Y-axis
The Stress factor is given under unsymmetrical cross sectional conditions as follows:

<p style="text-align:center;">$$\sigma _{xx}=(K_yM_z-K_{yz}M_y)y+(K_yM_y-K_{yz}M_z)z \;$$

$$K_y=\frac{I_y}{D}$$

$$K_{yz}=\frac{I_yz}{D}$$

$$K_z=\frac{I_Z}{D}$$

The stress in the x-direction can be solved for in matrix format.

<p style="text-align:center;">$$\sigma _{xx}=\begin{vmatrix} y & z \end{vmatrix}_{1x2}\begin{vmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{vmatrix}\begin{Bmatrix} M_z\\M_y

\end{Bmatrix}\Leftrightarrow \sigma _{xx}=\begin{vmatrix} y & z \end{vmatrix}_{1x2}\begin{vmatrix} K_yM_Z & -K_{yz}M_y\\ -K_{yz}M_z & K_zM_y \end{vmatrix}$$

$$\sigma _{xx}=\begin{vmatrix} z & y \end{vmatrix}_{1x2}\begin{vmatrix} K_y & -K_{yz}\\ -K_{yz} & K_z \end{vmatrix}\begin{Bmatrix} M_y\\M_z

\end{Bmatrix}$$ It is more convenient for convention issues to use this version of the formula for $$\sigma _{xx}$$

Setting the moment about the z-axis to zero we are particularizing the to cross section of a symmetrical case

<p style="text-align:center;">$$M_z=0\;$$

$$I_{yz}=0\Rightarrow D=I_yI_z$$

D being previously stated, $$K_y$$ and $$K_z$$ are as follows: $$K_y=\frac{1}{I_z}$$

<p style="text-align:center;">$$K_y=0\;$$

$$K_z=\frac{1}{I_y}$$

The Shear flow for the particularize case is:

<p style="text-align:center;">$$q(s)=-(K_yV_y-K_{yz}V_z)Q_z-(K_zV_z-K_{zy}V_y)Q_y\;$$

<p style="text-align:center;"> $$Q_Z=\int_{A_s}^{}{ydA}\; \; \; \; \;

Q_y=\int_{A_s}^{}{zdA} $$

= MATLAB =

Airfoil


Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the top skin. Due to the graph being concave up, it can be seen that the load is compressive on the top skin of the wing section.



Above is a graph of $$\sigma_{xx}$$ at all y positions in the middle wing section on the bottom skin. Due to the graph being concave down, it can be seen that the load is tensile on the top skin of the wing section.

The average compressive stress was found to be 144.2 MPa.

Simply-Supported Boundary Conditions
A simply-supported rectangular plate under an in-plane distributed compressive load is to be evaluated:

Buckling mode shapes can be determined by the following equation: $$\displaystyle \psi (x,y) = c_{mn}\sin\left(\frac{m \pi x}{a}\right)\sin\left(\frac{n \pi y}{b}\right)\, \ {\rm for} \ m,n = 1, 2, 3, \ldots$$ 1

Perspective views of the buckling shape of a simply-supported rectangular plate for various cases of $$\displaystyle\ m$$ and $$\displaystyle\ n$$



Now considering the factor with $$\displaystyle\ x$$ as the variable in the expression for $$\displaystyle\ \psi_{mn}(x,y)$$, the period, $$\displaystyle\ T$$, is calculated as follows:

$$ \sin ( \frac{m \pi (x + T)}{a} ) = \sin ( \frac{m \pi x}{a} )$$

$$\frac{m\pi (x+T)}{a}=\frac{m\pi (x)}{a}+2\pi n$$ where $$\displaystyle\ n=1,2,3,...$$

Evaluating at $$\displaystyle\ n=1$$

$$\frac{m\pi (x+T)}{a}=\frac{m\pi (x)}{a}+2\pi$$

$$ x+T = x + \frac{2\pi a}{m\pi}$$

$$ T = \frac{2a}{m}$$

The values of $$\displaystyle\ T$$ along with plots of the function are calculated and displayed below for values of $$\displaystyle\ m=1,...,5$$ to show that $$\displaystyle\ m$$ is indeed the number of half wave-lengths generated.

The critical buckling load $$\displaystyle\ (P_x)_{cr}$$ is calculated as follows:

$$\displaystyle (P_x)_{cr}=k_c\frac{\pi ^2 D}{b}$$ 1

where $$\displaystyle k_c(m, a/b):=\left(\frac{m b}{a}+\frac{a}{m b}\right)^2$$

and $$\displaystyle D:=\frac{E h^3}{12 (1 - \nu^2)}$$

A graph of $$\displaystyle\ k_c$$ versus aspect ratio, $$\displaystyle\ a/b$$, for various values of $$\displaystyle\ m$$ is shown below

Going one step further, the critical buckling stress can be calculated by

$$\displaystyle (\sigma_{xx})_{cr}=\frac{(P_x)_{cr}}{bh}=k_c\frac{\pi ^2 D}{b^2 h}$$ 1

Clamped Boundary Conditions
Next a rectangular plate with clamped boundary conditions is to be examined

The critcal buckling stress under an in-plane compressive load can be calculated as follows:

$$\displaystyle\ (\sigma_{xx})_{cr}=K\frac{E}{1-\nu ^{2}}(\frac{t}{b})^2$$

where $$\displaystyle K$$ is a function different from $$\displaystyle k_c$$, and $$\displaystyle t \equiv h$$ the plate thickness.

The buckling shape is now calculated using

$$\displaystyle\psi (x,y)=\frac{c}{4}\left(1 - \cos \frac{2 \pi x}{a}\right)\left(1 - \cos \frac{2 \pi y}{b}\right)$$



Conclusions
The figure below shows the critical buckling stress as a function of aspect ratio for the top panel (under compressive load from prescribed conditions) of the NACA 2415 Airfoil under both simply-supported and clamped boundary conditions. Also shown in green is the average compressive stress in the panel. Since stringers support the panel, it is neither exactly simply-supported or clamped; therefore, the actual critical stress must fall within the envelope prescribed by the simply-supported and clamped cases. It can be seen that the average compressive stress of the panel does encounter the critical stress for the clamped case but not the simply-supported one; consequently, depending on how close the panel relates to the two cases, buckling is a possibility if the panel behaves closely to the clamped boundary condition.



= Team Contribution =

Christopher Pergola EAS4200C.F08.WIKI.A 19:27, 21 November 2008 (UTC)

Chris FontanaEas4200c.f08.WIKI.E 19:11, 17 November 2008 (UTC)

Jeff Huenink Eas4200c.f08.wiki.f 17:30, 18 November 2008 (UTC)

Braden Barnes Eas4200c.f08.wiki.b 19:51, 20 November 2008 (UTC)

Eric Viale Eas4200c.f08.wiki.c 04:07, 21 November 2008 (UTC)

Michael Rolle Eas4200c.f08.wiki.d 20:49, 21 November 2008 (UTC)