User:Eas4200c.f08.vqcrew.c/Homework 6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems. Eml4500.f08 13:27, 23 November 2008 (UTC)

In addition to fixing the errors placed in the comment boxes in the report below, I have also made a change to the overall MATLAB code as a result of a discussion with a group member of Team Carbon. The end result of this change is the new average compressive stress calculated is roughly 3 MPa, which is much lower than the previously calculated value of 59 MPa. This error was also present in the code from HW5, where the maximum buckling stress was found to be 81 MPa. I changed the code for HW5, and found the max buckling stress to be on the order of 4 MPa. Thus, the ratio between the values remains constant, but the overall magnitude is now lower- I am not sure of the correctness of this solution, but I am certain the moment of inertia components are now being calculated correctly. I cannot find any error in my stress calculations, so I am presenting these results as our final solution. Eas4200c.f08.vqcrew.c 03:47, 28 November 2008 (UTC)

Please add a table similar to that at the end of your HW5 report to list the newly corrected values of the moment of inertia components together other corrected results so to document the differences (e.g., angle for neutral axis, etc.) between what you obtained now and what you obtained before the correction of the bug. Eml4500.f08 21:54, 29 November 2008 (UTC)

Equation of Equilibrium
Recall the 1-dimensional case and apply dimensional analysis:


 * $$[f]=\frac{F}{L}$$

where F is force units and L is length units. Additionally, square brackets "[]" represent the "dimension of" the variable. Furthermore, dimensions of other parameters can be assessed, as shown:


 * $$[A]=L^2\;$$
 * $$[\frac{f}{A}]=\frac{F}{L^3}$$
 * $$[\sigma]=\frac{F}{L^2}$$
 * $$[\frac{d\sigma}{dx}]=\frac{[d\sigma]}{[dx]}=\frac{\frac{F}{L^2}}{L}=\frac{F}{L^3}=\frac{force}{volume}=body\;force$$
 * $$\epsilon=\frac{du}{dx}=\frac{\Delta L}{L}\therefore[\epsilon]=\frac{[du]}{[dx]}=\frac{L}{L}=1\;nondimensional$$
 * $$\nu=-\frac{\epsilon_{yy}}{\epsilon_{xx}}\therefore[\nu]=\frac{[\epsilon_{yy}]}{[\epsilon_{xx}]}=\frac{1}{1}=1\;nondimensional$$

Step D: Prandtl Stress Function
For a little clarification from the previous lectures, we know that the change in stresses with respect to the change in location is analogous to the total force applied with respect to the volume...


 * $$ \begin{bmatrix} \frac {d \sigma_{ij}}{dx_i} \end{bmatrix} = \frac {F}{L^3} = \frac {Force}{Volume} \, $$

But now, on to the next step of our Road Map! Next up: Step D, the Prandtl Stress Function! The equations that governs this relationship are (3.15), shown as:


 * $$ \sigma_{yx} = \frac {d\phi}{dz} \, $$


 * $$ \sigma_{zx} = \frac {-d\phi}{dy} \, $$

In this case $$\phi_\,$$ plays the role of a potential function, where as $$ \sigma_{yx} \, $$ and $$ \sigma_{zx} \, $$ serve as components of the "gradient" of $$\phi_ \,$$ with respect to (y,z). Now recalling the scalar function for the f(x,y,z) gradient...


 * $$ \frac{df}{dx}\vec{i} + \frac{df}{dy}\vec{j} + \frac{df}{dz}\vec{k} \, $$

which is of course a vector, thus implying that F is a potential. Now, if we analyze equation (3.15) stated above utilizing equation (1), from lecture notes 27-1, we derive the following relation:


 * $$ \frac {d}{dy}(\frac{d\phi}{dz}) + \frac {d}{dz}(\frac{-d\phi}{dy}) = \frac {d^2\phi}{dydz} - \frac {d^2\phi}{dzdy} = 0 \, $$

The reason this relation works is that $$ \phi_ \, $$ is continuous and smooth, such that its second mixed derivative is interchangeable, ie:


 * $$ \frac {d^2\phi}{dydz} = \frac {d^2\phi}{dzdy} \, $$

The next step is to take the above relation and put it in Laplacian terms. These manipulations give us the following result for the Laplacian of $$ \phi_ \, $$.


 * $$ \frac {d^2\phi}{dydz} + \frac {d^2\phi}{dzdy} = -2G\theta \, $$


 * $$ \nabla^2\phi = -2G\theta \, $$

Traction Force
The traction force is defined as the product of the components of the stress tensor matrix, $$ \begin{bmatrix}\sigma\end{bmatrix} \;$$, and the components of the normal vector, $$  \begin{Bmatrix}n\end{Bmatrix} \;$$.

This relation can be derived using Figures 1-3. The normal vector, $$ \begin{Bmatrix}n\end{Bmatrix} \;$$, is a unit vector with magnitude equal to one.


 * $$ \begin{Bmatrix}t\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma\end{bmatrix}_{3x3} \begin{Bmatrix}n\end{Bmatrix}_{3x1} \;$$, where $$\begin{Vmatrix}n\end{Vmatrix} = 1\;$$

The traction force can then broken into its components in the y and z directions.


 * $$\vec{t} = t_y\vec{j} + t_z\vec{k}\;$$


 * $$\vec{n} = n_y\vec{j} + n_z\vec{k}\;$$

The differential lengths of the section and the components of the normal vector can be found using the surface length, ds, and the angle between the normal vector and the y-axis, $$\theta\;$$.


 * $$dz = ds(cos\theta)\;$$


 * $$dy = ds(sin\theta)\;$$


 * $$n_y = cos\theta\;$$


 * $$n_z = sin\theta\;$$

The sum of the forces in the y and z directions must be zero, so an expression relating the traction force, normal force and stresses can be derived. The depth of the section is set to be equal to one. Below is the development for the component of the traction force in the y-direction. The development of the component in the z-direction is reserved for the homework section.


 * $$\sum F_y = 0 = -\sigma_{yy}(dz)(1) - \sigma_{yz}(dy)(1) + t_y(ds)(1)\;$$


 * $$ 0 = -\sigma_{yy}(ds)(n_y) - \sigma_{yz}(ds)(n_z) + t_y(ds)\;$$


 * $$t_y = \sigma_{yy}(n_y) + \sigma_{yz}(n_z)\;$$

The components of the traction force can be written in matrix form.


 * $$\begin{Bmatrix}t_y\\t_z\end{Bmatrix} = \begin{bmatrix}\sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\end{bmatrix}\begin{Bmatrix}n_y\\ n_z \end{Bmatrix}\;$$

The three dimensional case can be expanded from the two dimesional case.


 * $$\begin{Bmatrix}t_1\\t_1\\t_3\end{Bmatrix} = \begin{bmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{11} & \sigma_{12} & \sigma_{13}\end{bmatrix}\begin{Bmatrix}n_1\\ n_2\\n_3 \end{Bmatrix}\;$$

This result then can be compressed into a compacet notation. This brings the derivation back to the origin point with the original definition of the traction force in a general notation.


 * $$t_i = \sum_{j=1}^3\sigma_{ij}n_j$$ where i = 1, 2, 3


 * $$\begin{Bmatrix}t_i\end{Bmatrix}_{3x1} = \begin{bmatrix}\sigma_{ij}\end{bmatrix}_{3x3} \begin{Bmatrix}n_{j}\end{Bmatrix}_{3x1}\;$$

Boundary Conditions for Phi
For solid cross sections, the constant value of $$\phi \;$$ is arbitrary, therefore it can be set to zero. Thus, we have the boundary condition $$\phi =0 \;$$ on the lateral surface of a solid bar. This boundary condition is depicted in Figure 4, where $$\phi \;$$ is equal to zero on $$S_0$$.

In the case of a thin-walled cross section, $$\phi \;$$ has a different constant value on each surface. This is depicted in Figure 5, where $$\phi =C_0\;$$ on $$S_0$$ and $$\phi =C_1\;$$ on $$S_1$$.

For a solid bar with circular cross section, as shown in Figure 6, the Prandtl stress function is assumed to be
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right)$$
 * {| class="collapsible collapsed"

!HW Problem - Proof that this equation satisfies the boundary condition We can prove that this satisfies the surface boundary condition of $$\phi =0\;$$ as follows:
 * $$\phi = C\left(\frac{y^2}{a^2} + \frac{z^2}{a^2}-1 \right) = C\left(\frac{y^2+z^2}{a^2}-1 \right)$$

Substituting in the Pythagorean theorem, $$a^2=y^2+z^2$$, we get:
 * $$\phi = C\left(\frac{a^2}{a^2}-1 \right)=C(1-1)=0$$

To find C, we can plug this function into the compatibility equation:
 * }
 * $$\frac{\partial^2 \phi }{\partial y^2} + \frac{\partial^2 \phi }{\partial z^2}=-2G \theta$$

which results in:
 * $$\frac{2C}{a^2} + \frac{2C}{a^2} = -2G \theta \Rightarrow C=-\frac{1}{2}a^2G\theta$$

This can also be expressed in terms of torque $$T$$ and polar moment of inertia $$J$$:
 * $$C=-\frac{1}{2}a^2TJ$$

Relating $$\theta$$ to $$\phi$$
It can be shown that torque and $$\phi$$ can be related by the relation


 * $$T = 2\ \int_A \phi\ dA$$

where $$\phi$$ is the Prandtl stress function. It can be defined that


 * $$T = G\ J\ \theta$$

Relating these together it can be deduced that


 * $$2\ \int_A \phi dA = G\ J\ \theta$$

It can also be defined that the Prandtl stress function


 * $$\tau_{yx} = \frac{\partial \phi}{\partial z}$$
 * $$\tau_{zx} = - \frac {\partial \phi}{\partial y}$$

It can be further shown that the shear strain can be related as


 * $$\tau_{yx} = -G\ \theta\ z$$
 * $$\tau_{zx} = G\ \theta\ y$$

Therefore it is shown that


 * $$\frac{\partial \phi}{\partial z} = -G\ \theta\ z$$
 * $$\frac{\partial \phi}{\partial y} = G\ \theta\ y$$


 * {| class="collapsible collapsed"

!HW Problem - Proof of Zero Warping for Solid Circular Bar Consider the following two relations for the shear strains $$\gamma_{yx}\;$$ and $$\gamma_{zx}\;$$:


 * $$\gamma_{yx} = \frac{\sigma_{yx}}{G} = \frac{\partial u_x(x,y)}{\partial y} - \theta_z$$


 * $$\gamma_{zx} = \frac{\sigma_{zx}}{G} = \frac{\partial u_x(x,y)}{\partial z} + \theta_y$$

Note also the following two relations for the shear stresses $$\sigma_{yx}\;$$ and $$\sigma_{zx}\;$$:


 * $$\sigma_{yx} = -G \theta_z\;$$


 * $$\sigma_{zx} = G \theta_y\;$$

Substituting the shear stress equations into the shear strain equations yields the following relationships:


 * $$\frac{-G \theta_z}{G} = -\theta_z = \frac{\partial u_x(x,y)}{\partial y} - \theta_z$$


 * $$\frac{G \theta_y}{G} = \theta_y = \frac{\partial u_x(x,y)}{\partial z} + \theta_y$$

By moving the $$\theta\;$$ from the left hand side to the right hand side of the above relations, the following conditions are immediately apparent:


 * $$\frac{\partial u_x(x,y)}{\partial y} = 0$$


 * $$\frac{\partial u_x(x,y)}{\partial z} = 0$$

The combination of these two conditions results in the overall condition $$u_x(y,z) = c\;$$, where $$c\;$$ is a constant that is not necessarily 0. Given that the bar is under "clamped" conditions at $$x\;=\;0$$, and that this in turn means that $$u_x(y,z) = 0\;$$ at $$x\;=\;0$$, noting that the displacement in the x plane is not dependent on $$\;y$$ or $$\;z$$ leads directly to the condition $$c\;=\;0$$; thus, there is no warping for a uniform bar with a solid circular cross-section.
 * }

 Error: You need to use the fact that the bar is "clamped" at $$\displaystyle x=0$$, i.e., $$\displaystyle u_x (y,z) = 0$$ at $$\displaystyle x=0$$, and since $$\displaystyle u_x (y,z)$$ is independent of $$\displaystyle x$$, and hence $$\displaystyle c = 0$$. Eml4500.f08 13:21, 23 November 2008 (UTC)

This error has been fixed. Eas4200c.f08.vqcrew.c 22:46, 27 November 2008 (UTC)

Flexural Shear Flow in Thin Walled Sections




In dealing with Aerospace Structures it is sometimes easier to examine idealized cases to have an easier look at the forces acting on the structure. For this case we would like to examine an aircraft wing and idealize the wing as a cantilever beam. In Figure 7 a nonuniform stress field can be seen acting on the beam.

The cantilever beam can then be broken up even more into two parts to examine the internal shear force and bending moments acting on a specific spot on the beam a distance x from the end. This image can be seen in Figure 8 The beam has been cut at a distance x, and separated for a view of the internal forces.

Asymmetrical Thin-walled Cross-Section
Now that we have examined a simple cantilever beam, we can now broaden our scopes to an unsymmetrical case. Figure 9 shows a sample unsymmetrical case over a small area, A_s, of the cross-section. The cross-section shown displays that a variable thickness and area along the path s. Examining the small section, s, of the cross section gives us the following relationship for the general case for unsymmetrical cross-sections:

$$\int_{A_s} \frac {d\sigma_{xx}}{dx} \,dA = -q(s)$$



Now that we have a brief introduction to unsymmetrical cross sections, it will help to examine the similarities and differences between symmetrical cross-sections about the z-axis and unsymmetrical cross-sections.

For Symmetrical cross-sections about the z-axis the relations come straight forward. The stress field can be written as the following:


 * $$\displaystyle \sigma_{xx} = \frac{M_y z}{I_y}$$

The shear flow can be represented as the following:


 * $$q(s) = - \frac{V_z Q_y}{I_y}$$

$$Q\;$$ is known as the first moment of the, which can be represented as the following:


 * $$Q_y = \int_{A_s} z \,dA = Z_c A_s$$

Now that we have the equations for a symmetrical case, lets examine the unsymmetrical case and see the differences:


 * $$\displaystyle \sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z$$

Where the coefficients for k determined by the following equations:


 * $$k_y = \frac{I_y}{I_y I_z - (I_{yz})^2}\;$$


 * $$k_z = \frac{I_z}{I_y I_z - (I_{yz})^2}\;$$


 * $$k_{yz} = \frac{I_{yz}}{I_y I_z - (I_{yz})^2}\;$$

The equation for the stress can also be looked at in matrix form.


 * $$\left[\displaystyle \sigma_{xx} \right]_{1x1} = \left[ z y \right] \begin{bmatrix} k_y & -k_{yz} \\ -k_{yz} & k_z \end{bmatrix} \begin{Bmatrix} M_y \\ M_z \end{Bmatrix}$$

This can then be simplified to a 1x2 matrix and a 2x1 matrix forming the correct 1x1 matrix from $$\sigma_{xx}$$.


 * $$\left[\displaystyle \sigma_{xx} \right]_{1x1} = \left[ z y \right]_{1x2} \begin{bmatrix} k_z M_y - k_{yz} M_z \\ -kyz M_y + k_y M_z \end{bmatrix}_{2x1}$$

Now look at the shear flow:


 * $$q(s) = -(k_y V_y - k_{yz} V_z)Q_z - (k_z V_z - k_{yz} V_y) Q_y\;$$

Where:


 * $$ Q_z = \int_{A_s} y \,dA$$


 * $$ Q_y = \int_{A_s} z \,dA$$

Read and Report - Plate Buckling Page
The plate buckling analysis page is split into two main sections: one for simply-supported boundary conditions and one for clamped boundary conditions.

The first portion of the simply-supported section considers the buckling mode shape relation $$\psi(x,y)\;$$, followed by a a relation for the out-of-plane displacement due to buckling. These relations describe the physical response of a "flat plate" to buckling stresses. The section further describes the critical buckling load $$(P_x)_{cr}\;$$, followed by a discussion of the dimensions of the quantities involved. This information can be used to determine the critical buckling stress $$(\sigma_{xx})_{cr}\;$$, which can be used to characterize the plate's resistance to buckling loads.

The second portion consists of essentially the salient equations from the simply-supported section, but adjusted for the different boundary conditions.

MATLAB Coding Portion

 * {| class="collapsible collapsed"

!Full MATLAB Code - Buckling Analysis


 * }


 * {| class="collapsible collapsed"

!Command Window Output for MATLAB code


 * }

The above tables contain the full MATLAB code for buckling analysis as well as the command window output for a run of the code using the parameters listed in | Homework 5. The first part of the question requested a graph of of normal bending stress in the panels BF and EH (as described in Figure 10), with another curve showing the average compressive stress in the appropriate panel. This graph is shown in Figure 11. The red curve represents the stress in panel BF, the blue curve panel EH, and the average stress by the green curve. It is readily apparent from this plot that the compressive stress is experienced by panel BF.

Next, the problem requests a replotting of a figure from the MIT OpenCourseWare notes relating to the loading and support conditions on a flat plat; this is provided in Figure 12. The buckling mode shapes for the cases of $$m\;=\;1\;,\;n\;=\;1$$ and $$m\;=\;2\;,\;n\;=\;1$$ are plotted in Figures 13 and 14 respectively. The next portion of the question refers to determining the period of a portion the buckling mode shape function; that analysis is provided below.

 The figures you plotted were not the perspective view of the buckling mode shapes; see Team Aero. Eml4500.f08 15:42, 23 November 2008 (UTC)

This error has been fixed. Eas4200c.f08.vqcrew.c 03:42, 28 November 2008 (UTC)

Determining Period of Buckling Mode Shape Function
Consider the term of the $$\psi (x,y)\;$$ function represented as $$\;\sin ( \frac{m \pi x}{a} )$$. In order to find the period of this function, the following relationship is useful:


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

Rudimentary algebra makes it immediately obvious that the following is also true:


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

Noting the trigonometric relationship that $$\sin(\theta + 2\pi) = \sin(\theta) \;$$, the following can be written:


 * $$\frac{m\pi T}{a} = 2\pi\;$$

Thus, an equation for the period of this function is immediately apparent:


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

For $$m\;=\;1\;$$, the period of this function is $$2\;a$$, indicating one half-wavelength present in the plate. For $$m\;=\;2\;$$, the period is $$a\;$$, indicating the presence of two half-wavelengths, or one full wavelength. Finally, for $$m\;=\;3\;$$, the period is $$\frac{2a}{3}$$, indicating the presence of 3 half-wavelengths, or 1.5 full wavelengths. Thus, it is apparent that $$\;m$$ is the number of half-wavelengths.

MATLAB Coding Portion Continued
The problem further requests the replotting of another figure from the MIT notes; this time, the $$k_c\;$$ vs. $$a/b\;$$ curves for various values of m. This plot is provided in Figure 15. Each curve refers to a different value of m; $$m\;=\;1$$ is plotted as a solid blue line, $$m\;=\;2$$ as a red dotted line, $$m\;=\;3$$ as a green dash-dotted line, and $$m\;=\;4$$ as a magenta dashed line.

The final portion of the question referred to an analysis of the airfoil skin panel as a flat plate, such as the type shown in Figure 12. The height of this idealized plate will be equivalent to the airfoil's skin thickness, and the length $$b\;$$ equivalent to the length of the airfoil skin panel. The skin material was considered to be Aluminum 2024-T3, as enumerated in a separate email about the problem. Using this idealized plate, the critical buckling stress $$(\sigma_{xx})_{cr}\;$$ for both simply supported and clamped boundary conditions was calculated for a range of aspect ratio such that $$(a/b) \in [.5,2]\;$$, and plotted in Figure 16. The red curve is for the clamped conditions, and the blue curve for the simply supported. Also, the buckling mode shape for the clamped case was also determined, and is provided in Figure 17.  Error: In Fig.16, the critical buckling stress for the clamped plate is lower than that for the simply-supported plate; there must be something wrong. Eml4500.f08 22:01, 29 November 2008 (UTC)

One important note about the determination of critical buckling stress for the simply-supported conditions relates to the determination of $$m\;$$ for the calculations. There was no condition enumerated in any communication on the problem about this issue; thus, the assumption of using the $$m\;$$ corresponding to the minimum $$k_c\;$$ at a given aspect ratio would be used. Thus, for aspect ratios less than $$1.414\;$$, $$m\;=\;1$$ was used; otherwise, $$m\;=\;2$$ was used.

When comparing the previously calculated average compressive stress to these two critical buckling stress curves, when noting that the actual support condition is somewhere "inbetween" the two conditions considered, it is apparent that the airfoil skin is not particularly suited for bending resistance. The average compressive stress was approximately $$3x10^6\;$$ Pascals. From the critical stress graph, for any aspect ratio greater than approximately .75, this would be a higher stress than could be handled by the simply supported condition, which indicates the airfoil skin would experience buckling failure under those conditions. Only very small aspect ratio panels would be able to handle that kind of loading. Thus, it is apparent that the airfoil skin is not useful from the standpoint of bending resistance.

Read and Report: Section 3.2 Torsion of Uniform Bars
When carrying out structural analyses on aircraft, torque is a major factor that must be considered. Torque is a moment, with units of N*M applied about the longitudinal axis of a beam. Due to Saint-Venant's principle, a torque symbol can represent infinitely many different statically equivalent load distributions. It must be noted, however, that for shafts made up of thin-walled cross-sections, Saint-Venant's principle may not apply.

In an elementary analysis of torsion, two significant assumptions are made:
 * 1. Upon torque application, deformations do not cause plane sections of the shaft to become non-circular or non-planar.
 * 2. All diameters within the plane section continue to be straight upon torque application and subsequent deformation.

These assumptions, however, are not valid in the current analysis due to the fact that the particular shafts being considered are of non-circular cross-section. In order to solve these types of structural analysis problems, two approaches can be utilized:
 * 1. The Prandtl stress function method
 * 2. The Saint-Venant warping function method

The latter will first be derived. Consider a constant cross-section straight shaft upon which equal and opposite torques T are applied to each side. The center of twist (COT) of the cross-section is chosen as the origin of the coordinate system. $$\alpha$$ signifies the twist angle (or angle of rotation) at z relative to the end at z = 0. The twist angle per unit length at z can be expressed as:


 * $$\theta=\frac{\alpha}{z}$$

Here it is assumed that plane sections warp during the application of torsional deformation, but projections of those plane sections rotate on the x-y plane as a rigid body. Consider P, an arbitrary point at z on the cross-section that, upon application of torque, moves to P' through a small angle $$\alpha$$. Note that the cross-section at z = 0 does not move. The displacement components at this arbitrary point P can be expressed as follows:


 * $$u=-r\alpha sin(\beta)=-\alpha y=-\theta zy\;$$
 * $$v=r\alpha cos(\beta)=\alpha x=\theta zx\;$$

where r is defined as the distance between the origin and point P. w, the displacement in the z-direction, is given by:


 * $$w(x,y)=\theta \psi (x,y)\;$$

where w and $$\theta$$, the rate of twist, are independent of z. In the previous equation, $$\psi(x,y)$$ is the warping function. The displacement given by the previous three equations yields:


 * $$\epsilon_{xx}=\epsilon_{yy}=\epsilon_{zz}=\gamma_{xy}=0\;\therefore\;\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{xy}=0$$

Therefore, $$\tau_{yz}$$ and $$\tau_{xz}$$ are the only two remaining components of stress. Next, the equations of equilibrium become:


 * $$\frac{\delta \tau_{xz}}{\delta x}+\frac{\delta \tau_{yz}}{\delta y}=0$$

A stress function $$\phi(x,y)$$ is subsequently introduced:


 * $$\tau_{xz}=\frac{\delta \phi}{\delta y}\;\;\;\;\;\tau_{xz}=-\frac{\delta \phi}{\delta x}$$

These stress components clearly conform to the equations of equilibrium. From the strain-displacement relations, the following is derived:


 * $$\gamma_{xz}=\frac{\delta w}{\delta x}-\theta y$$
 * $$\gamma_{yz}=\frac{\delta w}{\delta y}+\theta x$$
 * $$\therefore\;\frac{\delta \gamma_{yz}}{\delta x}-\frac{\delta \gamma_{xz}}{\delta y}=2\theta$$

Utilizing the stress-strain relations and the newly defined Prandtl stress function:


 * $$\frac{\delta^2 \phi}{\delta x^2}+\frac{\delta^2 \phi}{\delta y^2}=-2G\theta$$

Taking into account that no loads are applied on the lateral surface of the bar, the stress function can be found using traction t:


 * $$t_x=0\;\;\;\;\;t_y=0$$
 * $$t_z=\tau_{xz}n_x+\tau_{yz}n_y=\frac{\delta \phi}{\delta y} n_x-\frac{\delta \phi}{\delta x} n_y$$

where $$n_x=\frac{dy}{ds}\;and\;n_y=-\frac{dx}{ds}$$

Plugging these derivations into the previous equations yields:


 * $$t_z=\frac{\delta \phi}{\delta y}\frac{dy}{ds}+\frac{\delta \phi}{\delta x}\frac{dx}{ds}=\frac{d\phi}{ds}$$

The boundary condition for traction $$t_z=0$$ now corresponds to a boundary condition of:


 * $$\frac{d \phi}{ds}=0\;\therefore\;\phi=constant$$ on the lateral surface. This constant can be set to zero when analyzing solid sections with a single contour boundary.  The total resultant torque is subsequently obtained with the integration of dT over all of the cross-section:


 * $$T=-\int\int(x\frac{\delta \phi}{\delta x}+y\frac{\delta \phi}{\delta y})dxdy=2\int\int\phi dxdy-\int[x\phi]_{x_1}^{x_2}dy-\int[y\phi]_{y_1}^{y_2}dx$$
 * $$\therefore T=2\int\int\phi dxdy$$

Applying these results to the equation of the torsion constant J yields:


 * $$J=-\frac{4}{\nabla^2 \phi}\int\int\phi dxdy$$

Read and Report: Section 3.3 Bars with Circular Cross-Sections
Now, a uniform bar of circular cross-section is considered. For this case, the origin of the coordinate system lies on the center of the cross-section. Let a be the radius of the circular boundary. Then the stress function is:


 * $$\phi=C(\frac{x^2}{a^2}+\frac{y^2}{a^2}-1)$$

Analyzing the compatibility equation with this new phi considered yields:


 * $$C=-\frac{1}{2}a^2G\theta$$

The torque is then:


 * $$T=2C(\frac{I_p}{a^2}-A)\;where\;I_p=\int\int r^2 dA=\frac{1}{2}\pi a^4$$ denotes the polar moment of inertia of the cross-section.

Substituting $$a^2 A=2I_p\;$$ into the torque equation yields:


 * $$T=-\frac{2CI_p}{a^2}=\theta G I_p$$

The shear stresses then become:


 * $$\tau_{xz}=\frac{\delta \phi}{\delta y} = 2C\frac{y}{a^2}=-G\theta y$$
 * $$\tau_{yz}=-\frac{\delta \phi}{\delta x}=-2C\frac{x}{a^2}=G\theta x$$

Using relations for traction it can be shown that:


 * $$t_z=-G\theta\frac{xy}{r}+G\theta\frac{xy}{r}=0$$

Subsequently,


 * $$\tau=-t_z=G\theta r=\frac{Tr}{J}$$

From all of the previous equation and the stress-strain relations it can be shown that:


 * $$w=0\;$$

Which indicates that there is no warping in a circular cross-section under torsion.

Contributing Team Members
Adam Edstrand - Eas4200c.f08.vqcrew.a 20:56, 21 November 2008 (UTC)

Javier Stober - Eas4200c.f08.vqcrew.b 21:09, 21 November 2008 (UTC)

John Saxon - Eas4200c.f08.vqcrew.c 21:27, 21 November 2008 (UTC)

Darin Toscano - Eas4200c.f08.vqcrew.d 20:46, 21 November 2008 (UTC)

Kevin Klauk - Eas4200c.f08.vqcrew.E 02:17, 21 November 2008 (UTC)

Steven Hepsworth - Eas4200c.f08.vqcrew.f 18:23, 19 November 2008 (UTC)

Casey Barnard - Eas4200c.f08.vqcrew.g 20:03, 19 November 2008 (UTC)