User:Eas4200c.f08.radsam/HW6

 Comment: This page is identical to the previous one, except that the "Read and Report" section was added, which was accidentally omitted. Eas4200c.f08.radsam.a. 22:49, 6 December 2008 (UTC) =Homework #6=

=Class Notes=

November 3th - Plate Buckling
Plate buckling occurs when the applied axial load exceeds a critical value. Consider plate of dimensions:

a = dimension along the x-axis b = dimension along the y-axis

First, we look at a 1-D case:

$$[f] = \frac{F}{L}[A] = L^2\,\,$$

Therefore:

$$\left [ \frac{f}{A} \right ] = \frac{F}{L^3}$$ $$[ \sigma ] = \frac{F}{L^2} \Rightarrow \left [ \frac{\partial\sigma}{\partial x} \right ] = \frac{\frac{F}{L^2}}{L}\,$$ $$\epsilon = \frac{du}{dx}and\epsilon = \frac{\triangle L}{L}\Rightarrow [\epsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1\,$$

$$\nu = -\frac{\epsilon_{yy}}{\epsilon_{xx}}\Rightarrow [\nu] = 1\,$$

Eas4200c.f08.radsam.d 21:39, 21 November 2008 (UTC)

November 5 - Torsional Analysis
 Read and Report: Eqn 3.14 

$$\frac{\partial \tau _{xz}}{\partial x} - \frac{\partial \tau _{yz}}{\partial y}=0$$  (Equation 3.14)

This equation shows that $$\tau _{yz}$$ and $$\tau _{xz}$$ are the only two nonvanishing stress components. Both are independent of z and are a reduction of the equations of equilibrium.

Looking back at the roadmap from earlier in the semester, recall the D-prandtl stress function:

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

Notice that phi plays the role of a potential function. Thus, $$\sigma _{yx}$$ and $$\sigma _{zx}$$ are components of the "gradient" of phi with respect to y and z.

Also recall the scalar function f(x,y,z):

grad(f(x,y,z)) = $$\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}$$    where f is the potential function

By plugging this into our equations of equilibrium, we see that the equation is satisfied:

$$\frac{\partial }{\partial x}(\frac{\partial \phi }{\partial z})+\frac{\partial }{\partial x}(-\frac{\partial \phi }{\partial y}) = \frac{\partial^{2} \phi }{\partial y \partial z} - \frac{\partial^{2} \phi }{\partial z \partial y} = 0$$

This is true because phi is continuous and smooth, so the second derivative is interchangable. Thus, they are equal and subtracting them yields zero.

 Read and Report: Eqn 3.15 - 3.19 

The same procedure can be followed for strain-displacement relations to put in terms of the Prandtl stress function:

$$\gamma _{xz} = \frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}$$ and $$\gamma _{yz} = \frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}$$

Thus, we obtain:

$$\gamma _{xz} = \frac{\partial w}{\partial x}-\theta y$$ and $$\gamma _{yz} = \frac{\partial w}{\partial y}+\theta x$$

We can now easily derive the following:

$$\frac{\partial \gamma _{yz}}{\partial x}-\frac{\partial \gamma _{xz}}{\partial y} = 2\theta$$

This is called the compatibility equation. Recalling what we proved earlier for the stress-strain relations:

$$\frac{\partial \tau _{yz}}{\partial x}-\frac{\partial \tau _{xz}}{\partial y} = 2G\theta$$

Finally, in terms of the Prandtl stress function, we end up with:

$$\frac{\partial^{2} \phi }{\partial y^{2}} + \frac{\partial^{2} \phi }{\partial z^{2}} = -2G\theta $$ where the left hand side of the equation is the LAPLACIAN of phi.

 Read and Report: Through end of Section 3.2 

The following images show how the traction force and normal force vectors are defined.


 * The first image refers to a general 2D case (Figure 2)


 * The second image is an example of the traction force applied to an aircraft (Figure 3)

The vector $$t$$ vanishes because no loads are applied on the lateral surface: $${t} = [\sigma] {n}$$

this way, the stress vector may be evaluated on the lateral surface, knowing that $$n_z = 0$$, thus: $$\begin{Bmatrix} t_x\\ t_y\\ t_z \end{Bmatrix} = \begin{bmatrix} 0 &0 & \tau _{xz}\\ 0 &0 &\tau _{yz} \\ \tau _{xz} &\tau_{yz} &0 \end{bmatrix} \begin{Bmatrix} n_x\\ n_y\\ 0 \end{Bmatrix}\, $$

Therefore, we have: $$t_x = 0t_y = 0$$ $$t_z = \tau_{xz} n_x + \tau_{yz} n_y = \frac{\partial \phi }{\partial y}n_x - \frac{\partial \phi}{\partial y}n_y$$

If we look at Figure 4, we may easily derive: $$n_x = \sin \eta = \frac{dy}{ds}\,$$ $$n_y = \cos \eta = \frac{dx}{ds}$$

So we may express $$t_z\,$$ as:

$$t_z = \frac{\partial \phi}{\partial y}\frac{dy}{ds} + \frac{\partial \phi}{\partial x}\frac{dx}{ds} = \frac{d\phi}{ds}\,$$ So, the free boundary condition $$t_z = 0\,$$ is given by: $$\frac{d\phi}{ds} = 0 or  \phi = constant$$ on the lateral surface.

Note: For solid sections with a single contour boundary, this constant may be approximated to zero.

We are interested in the shear stresses $$\tau_{xz}~and~\tau_{yz}\,$$ and the resultant torque. Considering a small area $$dA = dx dy\,$$, the torque is:

$$dT = x \tau_{yz} dA - y \tau_{xz} dA\,$$ $$dT = \begin{pmatrix} -x\frac{\partial \phi}{\partial x} -y\frac{\partial \phi}{\partial y} \end{pmatrix} dA\,$$

Taking the integral:

$$T = -\int \int_{A}^{}{}(x\frac{\partial \phi }{\partial x}+y\frac{\partial \phi }{\partial y})dxdy$$

$$ = 2\int \int_{A}^{}{\phi dxdy}-\int \left[x\phi  \right]_{x1}^{x2}dy-\int \left[y\phi  \right]_{y1}^{y2}dx$$

Recall, phi vanishes on the boundary contour, so the last two terms in the equation vanish. Thus:

$$ = 2\int \int_{A}^{}{\phi dxdy} $$

Using our equations from before, combined with the warping equations developed in previous homework presenations, we can use the integration to find the torsion constant J:

$$J=\frac{T}{G\theta }$$

Eas4200c.f08.radsam.s 20:54, 21 November 2008 (UTC) Eas4200c.f08.radsam.d 21:40, 21 November 2008 (UTC)

November 10th - Torsional Analysis (Cont')
Homework 6: Find relationship between K and K_c. Verify K for a/b=1, m=1.

Torsional Analysis Continuation
 Case 1 : Thin-walled cross-section (closed) e.g. NACA airfoil.



$$\displaystyle \phi~=~C_o~on~S_o$$    (3.40)   (Fig 3.11)

$$\displaystyle \phi~=~C_1~on~S_1$$ (3.41)

 Case 2 :



$$\displaystyle \phi~=~0~on~S_o$$    (3.246)

Uniform bar with solid, circular cross-section (sec 3.3)

Look at page 74

Expression for T & J in terms of $$\displaystyle \phi$$

$$\displaystyle \phi(y,z)~=~C(\frac{y^2}{a^2}~+~\frac{z^2}{b^2}~-~1)$$

$$\displaystyle b~=~a$$(circle) $$\displaystyle \phi~=~0~on~S_o$$ $$\displaystyle \delta^2 \phi = -2G\Theta$$ $$C~=~-~\frac{1}{2}a^2G\Theta$$

November 12th


If we were to look at a cross section of a homogeneous circular bar, and place the origin of a coordinate system in its center...

$$x^2+y^2=a^2\,$$

Where a is the radius of the circular boundary.

Now, if we were to study the lateral surface of the bar, we would observe that the stress vector (traction) would be equal to zero. This means that the traction free boundary condition states that that the change of the stress function over the lateral surface is zero.

$$\frac{d\phi}{ds}=0\,$$

or

$$\phi=constant\,$$

This would mean that for single contour boundary-solid sections, the constant is arbitrary and thus can be zero. Hence the boundary condition can be defined on the lateral surface of the bar as:

$$\phi=0\,$$

Now that we have defined the boundary condition (stated above), we write the stress function in a form that satisfies this condition:

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

Next we take the Prandtl stress function:

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

Having obtained the compatibility equation above, we plug into it the stress function, which yields:

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

The stress function and the equation for C above, indicate the fact that the center of the circular cross section is the center of twist.

Recalling the definition of torque:

$$T= 2\iint_A \phi dx dy\,$$

We have the torque as:

$$T=2C\iint_A (\frac{x^2}{a^2} + \frac{y^2}{a^2}-1)dxdy\,$$

$$T=2C\iint_A (\frac{r^2}{a^2}-1)dA\,$$

$$T=2C(\frac{I_p}{a^2}-A)\,$$

Where:

$$I_p=\iint_A r^2dA= \frac{1}{2}\pi a^4\,$$

Where the above eq. corresponds to the polar moment of inertia and also:

$$A=\pi a^2\,$$ since that is the cross sectional are of the circular bar.

Since

$$a^2A=2I_p\,$$

Therefore:

$$T=-\frac{2CI_p}{a^2} = \theta GI_p\,$$

and

$$J = I_p\,$$

The shear stresses are:

$$\tau_{xz}=\frac{\partial \phi}{\partial y} = 2C\frac{y}{a^2}=-G\theta y\,$$

$$\tau_{yz}=-\frac{\partial \phi}{\partial x} = -2C\frac{x}{a^2} = G\theta x\,$$

It is also worth noting that the shear stress can be expressed as:

$$\tau=\frac{Tr}{J}\,$$

We can also show that for bars with circular cross-sections under torsion, no warping occurs:

$$\gamma_{xz}=\frac{\sigma_{xz}}{G}=\frac{\partial w_{x}}{\partial z} - \theta_{y}\,$$

$$\gamma_{yz}=\frac{\sigma_{yz}}{G}=\frac{\partial w_x}{\partial y} + \theta_{x}\,$$

Now recalling the shear stresses derived above:

$$\sigma_{xz}=-G\theta y\,$$

$$\sigma{yz}=G\theta x\,$$

We can see that the warping is zero:

$$w=0\,$$

Flexural Shear Flow in thin-walled sections

(Chapter 5) --> 5.1, 5.1.1, 5.1.2, 5.1.3, 5.3, 5.3.1, 5.3.2, 5.4

HW #5: Effects of M(x) on NACA airfoil

HW #6: Plate buckling, compressive in-plane load NACA airfoil

HW #7: Effects of V(x) on NACA airfoil, Plate Buckling, Shear Load.

Contribution by radsam.z

November 14th
 Flexural Shear Flow in Open, Asymmetric, Thin-Walled Sections 

Sections 5.1, 5.1.1, 5.1.2

As $$\Delta x\,$$ approaches 0, the following equation can be used for either symmetric or asymmetric cross sections:

$$\int_{A_s}^{} \frac{d\sigma_{xx}}{dx} dA = -q_s$$

$$I_{yz}=0\,$$ due to cross section symmetry about either the y or z axis.

For symmetric cross sections, we have the following equations:

$$\sigma_{xx}= \frac{M_yz}{I_y}$$

$$q(s)=-\frac{V_zQ}{I_y}$$

$$Q=\int_{A_s}^{}z dA = z_c A_s$$

For asymmetric cross sections, we have:

$$\sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z(5.3)\,$$

$$k_y= \frac {I_y}{I_yI_z-I_{yz}^2}\,$$

$$k_z = \frac {I_z}{I_yI_z-I_{yz}^2}\,$$

$$k_{yz}= \frac {I_{yz}}{I_yI_z-I_{yz}^2}\,$$

$$\sigma_{xx_{(1x1)}}=\begin{bmatrix}z & y\end{bmatrix}_{1x2}\begin{bmatrix}k_z & -k_{yz}\\ -k_{yz} & k_y\end{bmatrix}\begin{Bmatrix}M_y \\ M_z\end{Bmatrix}\,$$

Where the last two terms of the above product are: $$ \begin{Bmatrix} k_z M_y - k_{yz} M_z \\ -k_{yz} M_y + k_y M_z \end{Bmatrix}_{2x1} \,$$

We now particularize to a symmetric cross-section.

$$I_{yz}=0\,$$. [Why? See HW6 contribution below.] Consider $$M_z=0\,$$.

$$I_{yz}=0 => D=I_{y}I_{z}\,$$

$$k_y=\frac{1}{I_z}, k_{yz}=0, k_z=\frac{1}{I_y}\,$$

$$=>\sigma_{xx}=\frac{z M_{y}}{I_{y}}\,$$

For the asymmetric case:

$$q(s)=-(k_{y}V_{y}-k_{yz}V_{z})Q_{z}-(k_{z}V_{z}-k_{yz}V_{y})Q_{y}(5.5)\,$$

$$Q_{z}=\int_{A_{s}}ydA\,$$

$$Q_{y}=\int_{A_{s}}zdA\,$$

Stringer-web sections: Thickness of skin and spar webs is very small so neglect in components $$I_y\,$$, $$I_z\,$$, $$I_{yz}\,$$, $$Q_y\,$$, and $$Q_z\,$$. (Use only stringer areas.)

Symmetric with respect to y-axis:

Ex. 5.2, page 154 in book:

$$A_3=A_2\,$$

$$A_4=A_1\,$$

Asymmetric:

$$A_1=A\,$$

$$A_2=2A\,$$

$$A_3=3A\,$$

$$A_4=4A\,$$

Contribution by radsam.a

=Read and Report=

Chapter 3, Section 2 - Torsion of Uniform Bars
Torque is commonly encountered in an aircraft and is a very important stress component to consider. A torque is a applied about the longitudinal axis of a shaft or beam. A torque symbol (represented by a curved arrow) can contain multiple statically equivalent load distributions.

Two assumptions are made when deriving the equations for shaft analysis:
 * 1 - Cross sections remain planar upon the application of torque--thus, there is not out of plane deformation.
 * 2 - Diameters do not change upon deformation.

For non-circular cross sections, however, a different approach must be taken. There are two methods, this first being the Prandtl stress function method, and the second the Saint-Venant warping function method. Saint-Venant's warping function method will be derived as follows. Consider a straight shaft with a constant cross-section where equal and opposite torques, $$T\,$$, are applied at each end. The center of twist (COT) is defined as the location in which no in-plane displacements occur. The origin of the coordinate system is to be chosen at this location.

The twist angle per unit length at $$z\,$$ is given by:

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

Where $$\alpha\,$$ is the twist or rotation angle at $$z\,$$ relative to the end at of the beam, at $$z=0\,$$.

It can be assumed that cross sections warp, but the cross section projections rotate on the x-y plane in a rigid manner. We now look at the point, $$P\,$$, on the edge of the cross-section located at $$z\,$$. Given a torque, $$P\,$$ moves to $$P'\,$$ over the angle, $$\alpha\,$$, which is considered to be small. Assuming that the cross-section at the base of the shaft doesn't move, the displacement components at $$P\,$$ are given by:

$$u=-r\alpha sin\beta=-\alpha y=-\theta zy\;$$

$$v=r\alpha cos\beta=\alpha x=\theta zx\;$$

where $$r\,$$ is the distance from the origin to point $$P\,$$.

The displacement in the $$z\,$$-direction, $$w\,$$, can be written:

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

Which is the warping function. It should be noted that $$w\,$$ and $$\theta\,$$ do not depend on $$z\,$$. The given equations yield the following:

$$\epsilon_{xx}=\epsilon_{yy}=\epsilon_{zz}=\gamma_{xy}=0\,$$

Thus:

$$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}=\tau_{xy}=0\,$$

So $$\tau_{yz}\,$$ and $$\tau_{xz}\,$$ are the two remaining stress components.

The equations of equilibrium can now be reduced to:

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

From Prandtl's stress function, $$\phi(x,y)\,$$, we have:

$$\tau_{xz}=\frac{\delta\phi}{\delta y}\,$$

$$\tau_{xz}=-\frac{\delta\phi}{\delta x}\,$$

In which both stress components automatically satisfy the equilibrium equations.

We then gather the following from the strain-displacement relations:

$$\gamma_{xz}=\frac{\delta w}{\delta x}+\frac{\delta u}{\delta z}\,$$

$$\gamma_{yz}=\frac{\delta w}{\delta y}+\frac{\delta v}{\delta z}\,$$

Which can be written as:

$$\gamma_{xz}=\frac{\delta w}{\delta x}-\theta y\,$$

$$\gamma_{yz}=\frac{\delta w}{\delta y}+\theta x\,$$

So the following torsion compatibility equation can be derived:

$$\frac{\delta \gamma_{yz}}{\delta x}-\frac{\delta \gamma_{xz}}{\delta y}=2\theta\,$$

This expressed in terms of the Prandtl stress function gives:

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

Because of no loading on the lateral bar surface, the stress vector, $$t\,$$, can be evaluated as:

$$t_x=0\,$$

$$t_y=0\,$$

$$t_z=\tau_{xz}n_x+\tau_{yz}n_y=\frac{\delta \phi}{\delta y} n_x-\frac{\delta \phi}{\delta x} n_y$$

Where:

$$n_x=\frac{dy}{ds}\,$$ $$n_y=-\frac{dx}{ds}\,$$

Substituting into the stress vector equations gives:

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

$$t_z=0\,$$ is now expressed as:

$$\frac{d \phi}{ds}=0\mbox{or}\phi=constant$$

On the lateral surface. Choosing $$\phi=0\,$$ is allowable in that we are analyzing solid sections with a single contour boundary.

By integrating dT across the entire cross-section, the total resultant torque is given by:

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

So:

$$T=2\int\int_A\phi dxdy$$

Finally, from:

$$J=\frac{T}{G\theta}\,$$

We get:

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

Contribution by radsam.a

Chapter 3, Section 3 - Bars with Circular Cross-Sections
Looking at a uniform cross-section, the contour can be expressed as:

$$x^2+y^2=a^2\,$$

Where $$a\,$$ is the contour radius.

Then the stress function is then given by:

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

Substituting into the compatibility equations results in:

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

So the center of the circular cross-section is also the center of twist!

The torque is given by:

$$T=2C(\frac{I_p}{a^2}-A)\,$$

Where:

$$I_p=\int\int_A r^2dA=\frac{1}{2}\pi a\,$$

gives the cross-section polar moment of inertia.

Also:

$$a^2 A=2I_p\;$$

So, substituting this into the equation for torque gives:

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

And, $$J = I_p\,$$.

The shear stresses then given by:

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

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

And traction gives:

$$t_x=t_y=0\,$$

$$t_z=\tau_{xz}n_x+\tau_{yz}n_y\,$$

While:

$$n_x=cos\beta=\frac{x}{r}\,$$

$$n_y=sin\beta=\frac{y}{r}\,$$

So we end up with:

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

Also:

$$n_z=0\,$$

And:

$$t_z=-G\theta{r}\,$$

So a little manipulation gives:

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

We result in finding out that:

$$w=0\;$$

And we have confirmed that there is, indeed, no warping for circular cross-sections.

=Homework Problems and Solutions=

=Homework #5 Contributors= Eas4200c.f08.radsam.s 20:49, 21 November 2008 (UTC) Eas4200c.f08.radsam.d 19:42, 21 November 2008 (UTC) Eas4200c.f08.radsam.a. 21:20, 21 November 2008 (UTC) Eas4200c.f08.radsam.z 20:31, 20 November 2008 (UTC) Eas4200c.f08.radsam.r 14:09, 20 November 2008 (UTC) Eas4200c.f08.radsam.m 21:09, 21 November 2008 (UTC)