User:Eas4200c.f08.radsam/HW5

 Comment: The Matlab code has been corrected and updated. Here is the HW5 original version. Eas4200c.f08.radsam.m 01:21, 11 November 2008 (UTC) =Homework #5=

=Class Notes=

October 20th: Meeting 24
 Exam Review 

The important topics that were discussed in class and will be covered in the second exam are:
 * Kinematic assumption
 * 4 Zero-strain components
 * 4 Zero-stress components
 * Expressing the $$\varepsilon = \sigma$$ relationship
 * Rewriting the $$\varepsilon = \sigma$$ relationship

$$\varepsilon = \begin{bmatrix} A & 0\\ 0 & B \end{bmatrix} \left\{\sigma \right\}$$

and $$ \sigma = \begin{bmatrix} A^{-1} & 0\\ 0 & B^{-1} \end{bmatrix} \left\{\varepsilon \right\}$$

There is a way to check the validity of this method. The product of the matrix and its inverse should yield the identity matrix.

$$\sigma _{11}=\sigma _{22}=\sigma _{33}=0\,$$

$$\sigma _{23}=2G\varepsilon _{23}$$    $$\sigma _{31}=2G\varepsilon _{31}  $$     $$\sigma _{12}=2G\varepsilon _{12}$$

 Equilibrium Equation for Stresses in a Nonuniform Stress Field  Referring back to the roadmap from several class periods ago: Part C invovles finding the equilibrium equation for stresses in a nonuniform stress field.

Here, we will consider the 1-D model as well as two different loading conditions. The first is a uniform applied stress to a beam element and the second is a nonuniform stress field.




 * Figure 1 shows a beam element under uniform stress. Here, the axial load is evenly distributed and f(x) is constant.


 * A section dx of the beam element can be analyzed in terms of the forces and stresses applied to it. Since the stress is uniform, the shear stresses are zero.  Notice also that the normal stresses are equal on both sides.


 * For the non-uniform case, Figure 3 shows that the distributed load contributes some shear stress to the differential element. This causes the normal stress on the right side of the element to be different than the axial stress on the left side.  Because of the distributed load, the normal stresses are not equal, which means that f(x) is non-constant.


 * These observations are important for our analysis of the equilibrium equations, and will be referenced in later lectures.

Contribution by radsam.s

October 22nd: Meeting 25
Exam 2 was taken during this meeting.

October 27th: Meeting 26
 Bidirectional Bending 
 * Section 4.2 in the textbook. As assigned reading read section 4.1 in the textbook to catch up, paying special attention to example 4.1 on page 124.

For the NACA Airfoil assignment for homework 5, we will have stringers at points B, E, H, F (see class notes 22-1). Note that skin and spar web do not participate in bending; only the stringers do. We need the areas of stringers $$A_{B},~A_{E},~A_{H},~A_{F}\,$$.

Slide 1:
 Bidirectional Bending Recipe 
 * Note: We refer to this as a recipe because we're simply presenting the formula without a derivation.
 * Note: This is material is covered in the applied reading of section 4.2.

$$M_y=\int_A z \sigma_{xx} dA   \,$$ (equation 4.26) Similarly, for $$M_z\,$$, see equation 4.27 in the book.
 * Note: $$dA=dydz\,$$, so $$\int_A dA = \iint dydz\,$$

Slide 2:
Moment of intertia tensor: $$I_y,~I_z,~I_{yz}\,$$
 * Note: $$I_y=I_{yy}=I_{22}\,$$ ; $$I_z=I_{zz}=I_{33}\,$$ ; $$I_{yz}=I_{23}\,$$

$$Iy= \int_A z^2 dA\,$$ (equation 4.28a)

$$Iz= \int_A y^2 dA\,$$ (equation 4.28b)

$$Iyz = \int_A yz dA\,$$(equation 4.28c)

$$\sigma_{xx}=\Epsilon \varepsilon_{xx} = \frac{I_{y}M_{z} - I_{yz}M_{y}}{I_{y}I_{z}-(I_{yz})^{2}}y + \frac{I_{z}M_{y}-I_{yz}M_{z}}{I_{y}I_{z}-I_{yz}^{2}}z\,$$ (Equation 4.29)

But surely everyone can see such a recipe is mindless learning. Our class transcends mere formulas. This means a derivation is required.

Recall: $$\mathbf{I} = \begin{bmatrix} I_{11} & I_{12} & I_{13} \\ I_{21} & I_{22} & I_{23} \\ I_{31} & I_{32} & I_{33} \\\end{bmatrix}$$ The determinant of the bottom right 4x4 matrix ($$I_{22}\,$$,$$ I_{23}\,$$,$$I_{32}\,$$,$$ I_{33}\,$$) is the denominator of the $$\sigma_{xx}\,$$ formulas seen above.

Slide 3:


I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}\,$$.
 * Note: $$D=I_y I_z - (I_{yz})^2\,$$ and can be written as $$D=I_{22} I_{33} - (I_{23})^2\,$$ which is in turn the determinant of matrix $$ \begin{bmatrix}


 * Note: $$I_{32}=I_{zy}:= \int_A zy dA\,$$, and $$zy=yz\,$$ therefore $$I_{yz}=I_{23}\,$$

Neutral Axis, where: $$\sigma_{xx}=0\,$$ From equation 4.29: $$\sigma_{xx}=m_{yy} + m_{zz} = 0\,$$ $$z=(\frac{-m_{y}}{m_{z}})y = (\tan{\beta})y\,$$ $$tan(\beta)=-tan(\alpha)\,$$

Contributed by radsam.z and radsam.e

October 29th: Meeting 27
 Roadmap Continued: Equation of Equilibrium for Stresses 

In order to continue our journey into the land of exciting airfoil computation, we must now look at the equation of equilibrium in terms of $$\sigma\,$$, which is the stress tensor with 6 components.

Our goal is:

$$(1)\frac{\partial\sigma_{yx}}{\partial y}+\frac{\partial\sigma_{zx}}{\partial z}=0\,$$ (Eqn. 3.14)

or, using (absolutely awesome) indicial notation (sadly, cyclic permutation need not be applied):

$$(2)\frac{\partial\sigma_{21}}{\partial x_{2}}+\frac{\partial\sigma_{31}}{\partial x_{3}}=0\,$$

Also, recall (as in Section 2.4 in your book):

$$(3)\frac{\partial\sigma_{xx}}{\partial\ x}+\frac{\partial\sigma_{yx}}{\partial\ y}+\frac{\partial\sigma_{zx}}{\partial\ z}=0\,$$

Similarly, see (2.22) and (2.23) --> HW5.

In indicial notation (2.21), (2.22), (2.23) can be expressed as:

$$(4)\sum_{i=1}^3 \frac{\partial \sigma_{ij}}{\partial x_i}=0\mbox{for}j=1,~2,~3\,$$

Contributed by radsam.a

October 31th: Meeting 28
For simplification purposes, we look at a a 1-dimensional model first. Equating all forces on the x-direction, we obtain:

$$\sum\ F_x = 0 = -\sigma_{(x)}\ A + \sigma_{(x + dx)}\ A + f_{(x)} dx\,$$

$$ 0 = A [ \sigma_{(x + dx)}\ - \sigma_{(x)}\ ] + f_{(x)} dx\,$$

Applying Taylor series to the term inside the brackets, it becomes:

$$\frac{\partial\sigma_{x}}{\partial\ x} dx + higher~order~terms\,$$

Recall:

$$F_{(x + dx)} = F_{(x)} + \frac{\partial\ F_{(x)}}{\partial\ dx} dx + \frac{1}{2} \frac{\partial ^2 F_{(x)}}{\partial x^2} (dx)^2 + .~.~.\,$$

Neglecting higher order terms:

$$\frac{\partial\sigma}{\partial\ x} + \frac{f_{(x)}}{A} = 0\,$$

Where $$\frac{f_{(x)}}{A}\,$$ is the applied load.

Now, we may look at a non-uniform 3-D field without applied loads, and focusing on the x-direction.

Figure 1 shows an infinitesimal element in which the stress is not uniform. However, the element must remain in equilibrium, therefore the six equations of equilibrium must be satisfied. For example, forces along the x-direction are:

$$\sum\ F_x = 0 = dy dz [ -\sigma_{xx}\ (x,y,z) + \sigma_{xx}\ (x+dx,y,z)]\,$$ Facets with normal X

$$ + dz dx [ -\sigma_{yx}\ (x,y,z) + \sigma_{yx}\ (x,y+dy,z)]\,$$ Facets with normal Y

$$ + dx dy [ -\sigma_{zx}\ (x,y,z) + \sigma_{zx}\ (x,y,z+dz)]\,$$ Facets with normal Z

$$0 = (dx dy dz) \left [ \frac{\partial\sigma_{xx}}{\partial\ x} + \frac{\partial\sigma_{yx}}{\partial\ y} + \frac{\partial\sigma_{zx}}{\partial\ z} \right ]\,$$

Contributed by radsam.d and picture by radsam.a

=Homework Problems and Solutions=

=Homework #5 Contributors= Eas4200c.f08.radsam.a. 00:22, 5 November 2008 (UTC) Eas4200c.f08.radsam.d 22:52, 6 November 2008 (UTC) Eas4200c.f08.radsam.r 00:04, 6 November 2008 (UTC) Eas4200c.f08.radsam.s 03:37, 7 November 2008 (UTC) Eas4200c.f08.radsam.e 17:55, 7 November 2008 (UTC) Eas4200c.f08.radsam.z 19:32, 7 November 2008 (UTC) Eas4200c.f08.radsam.m 22:00, 7 November 2008 (UTC)