User:Eas4200c.f08.carbon.orear/HW5/Lecture Notes

=Lecture Notes=

Kinematic Assumptions and Stress Strain Relations (Lecture 25)
There are four zero strain components.

εxx = εyy = εzz = γxy = 0

And four zero stress components, from our stress-strain relations.

σxx = σyy = σzz = τxy = 0

And these equations become important in analyzing last weeks results in the displacement field (warping) of a non-uniform cross-section of a bar. We can see that τyz and τxz are the only stress components remaining that are non-zero. Both of these values do not depend on z! Future examination of the Prandtl Stress function will utilize these important results!

At this point we shift our focus, however. We want to look more at the stress-strain relationship and make some comments about matrix notation for our convenience.

We will examine the stress-strain relations for the case of isotropic elasticity.

What is isotropic elasticity? Isotropic Materials (from Wikipedia)

Also, here is an additional note from our Team.


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!Composite Materials and non-isotropic elasticity, from team Carbon When we think about composite materials, the properties of the fibers in the matrix are only optimal along the fiber direction. The tension strength, bending stiffness, or other beneficial properties we may desire are direction-dependant. The transverse fiber direction and the out-of-plane direction in laminar composites do not have the same poisson ratio or elastic modulus as the fiber-axial direction.

In our matrix notation, the numbers 1, 2, and 3 refer to row and column matrix position. This is not to be confused with "Material Coordinate Systems" that denote 1 as the axial fiber direction, 2 as the transverse in-plane direction, and 3 as the out-of-plane normal direction.

Isotropic materials are presumed to have properties independant of direction in the material on a macro-scale.

Contribution by carbon.orear
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We can look at either tensorial or engineering strain, as discussed in previous notes. Here we use tensorial strain for our relationship. If for some matrix shown below, where A, B , and 0 are each 3x3 sub-matrices (note that 0 is a zero matrix):

{εij} = $$\begin{bmatrix} A & 0\\0 & B \end{bmatrix}\,$$ {σij}

then it holds that, by inversion,

{εij} = $$\begin{bmatrix} A^{-1} & 0\\0 & B^{-1} \end{bmatrix}\,$$ {σij}

Note that for an invertible matrix, the forward product of a matrix and itself yields the identity matrix [ I ].

[ C ][ C -1] = [ I ]

The components of matrix A above are given in Chapter 2 of the Sun book along with the components for matrix B.

HOMEWORK SOLUTION: The homework problem given was to find a relationship for σ12. From our matrices we can see that

σ12 * {1 / G} = ε12

Equilibrium Equation for Stresses in a Non-Uniform Stress Field (Lecture 25 cont'd)
Considering the 1-dimensional case as our model:

The diagram below shows uniform axial force on the left (as in 2.3) and non-uniform on the right (as in 2.4). The model will set up our examination for analysis of a stress-field after this lecture.



Bidirectional Bending (Lecture 26)
Bidirectional Bending (see Section 4.2 and Example 4.1 in the book)

Bidirectional bending "recipe:"



$$M_y = \int_A z\sigma_{xx}dA$$

$$M_z = \int_A y\sigma_{xx}dA$$

Moment of Inertia Tensor: $$I_y, I_z, I_{yz}$$ (sometimes referred to by $$I_{yy},I_{zz},I_{yz}$$ or $$I_{22},I_{33},I_{23}$$), whose components are given by:

$$I_y = \int_A z^2dA$$

$$I_z = \int_A y^2dA$$

$$I_{yz} = \int_A yzdA$$

Remember, $$\sigma_{xx}=E\epsilon_{xx}$$

$$=>\sigma_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

$$\boldsymbol I = \begin{bmatrix} I _{11} & I _{12} & I _{13}\\ I _{21} & I _{22} & I _{23}\\ I _{31} & I _{32} & I _{33} \end{bmatrix}$$

Note: The denominator in the above equation for $$\sigma_{xx}$$ is equivalent to the determinant of the "sub-matrix" of the above matrix consisting of:

$$\begin{bmatrix} I _{22} & I _{23}\\ I _{32} & I _{33}\\ \end{bmatrix}$$

$$D=I_{22}I_{33}-(I_{23})^2$$

The Neutral Axis is where $$\sigma_{xx}=0$$

$$=>\sigma_{xx}=m_yy+m_zz=0$$

$$=>z=(-\frac{m_y}{m_z})y=(tan\beta)y$$



Equation of Equilibrium in Terms of σ (Lecture 27)
The goal is to arrive at equation 3.14:

$$\frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z} = 0$$


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!Contribution Permutation and Indicial Notation, from team Carbon
 * To convert to Indicial Notation, follow the form below (standard → indicial):
 * To convert to Indicial Notation, follow the form below (standard → indicial):
 * To convert to Indicial Notation, follow the form below (standard → indicial):

Numerator: x → 1, y → 2, z → 3 Denominator: x → x1, y → x2, z → x3

Look at the Wikipedia article below to find out more about the indicial or "Einstein" notation:

http://en.wikipedia.org/wiki/Summation_convention


 * To convert from the book notation to the class notation follow the form below (book → class):

x → y → z → x... (cyclic permutation)


 * To convert from the class notation to the book notation follow the form below (class → book):

x → z → y → x... (cyclic permutation)

Contribution by carbon.w
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Using Indicial Notation, the above equation becomes:

$$\frac{\partial \sigma _{21}}{\partial x_{2}} + \frac{\partial \sigma _{31}}{\partial x_{3}} = 0$$

In addition,

(1) $$\frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \sigma _{zy}}{\partial z} +  \frac{\partial \sigma _{xy}}{\partial x} = 0$$


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!HW Equations of Equilibrium, from team Carbon
 * Equation (1) can be deduced from the following equation:
 * Equation (1) can be deduced from the following equation:

1: $$\frac{\partial \sigma _{yy}}{\partial y} + \frac{\partial \tau _{zy}}{\partial z} +  \frac{\partial \tau _{xy}}{\partial x} = 0$$

And in indicial notation, Eqn. (1) becomes:

(1)IND $$\frac{\partial \sigma _{22}}{\partial x_{2}} + \frac{\partial \tau _{32}}{\partial x_{3}} +  \frac{\partial \tau _{12}}{\partial x_{1}} = 0$$

Similarly, the following equations (2) and (3) can be deduced from the original equations 2: and 3: and subsequently converted to indicial notation (2)IND and (3)IND:

2: $$\frac{\partial \tau _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} +  \frac{\partial \tau _{xz}}{\partial x} = 0$$

(2) $$\frac{\partial \sigma _{yz}}{\partial y} + \frac{\partial \sigma _{zz}}{\partial z} +  \frac{\partial \sigma _{xz}}{\partial x} = 0$$

(2)IND $$\frac{\partial \sigma _{23}}{\partial x_{2}} + \frac{\partial \sigma _{33}}{\partial x_{3}} +  \frac{\partial \sigma _{13}}{\partial x_{1}} = 0$$

And finally,

3: $$\frac{\partial \tau _{yx}}{\partial y} + \frac{\partial \tau _{zx}}{\partial z} +  \frac{\partial \sigma _{xx}}{\partial x} = 0$$

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

(3)IND $$\frac{\partial \sigma _{21}}{\partial x_{2}} + \frac{\partial \sigma _{31}}{\partial x_{3}} +  \frac{\partial \sigma _{11}}{\partial x_{1}} = 0$$

Contribution by carbon.w
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All of the equations above can be combined in a simple form:

$$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_{i}}} = 0$$ for j=1,2,3

This simplified form was essential to the development of Einstein's Theory of Relativity. Relativity (wikipedia article)

3D Non-Uniform Stress Field (Lecture 28)
Using the kinematic assumptions and stress strain relations,

σxx = σyy = σzz = τxy = 0

The equation, $$\sum_{i=1}^{3}{\frac{\partial \sigma_{ij}}{\partial x_{i}}} = 0$$ for j=1,2,3, can be rewriten to

$$\frac{\partial \sigma_{21}}{\delta x_2} + \frac{\partial \sigma_{31}}{\partial x_3}=0$$

$$\sum F_x=0=-\sigma(x)A + \sigma(x+dx)A+f(x)dx$$

$$\displaystyle f(x)$$ is the distributed load

$$\displaystyle 0=A[\sigma(x+dx)-\sigma(x)]+f(x)dx$$

using a taylor series expansion on the section within the square brackets to give us

$$\frac{d\sigma(x)dx}{dx}+h.o.t.$$  h.o.t. - higher order terms


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!Taylor Series Expansion, from team Carbon A Taylor Series Expansion is a way to solve a function as a infinate sum of terms. The Taylor series is represented as $$f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots\,,$$ or as $$\sum_{n=0}^{\infin} \frac{f^{(n)}(a)}{n!} (x-a)^{n}\,,$$. If the function converges then as the term number increases the term amount approaches zero and the higher order terms become so small that they make an insignifiacant different on the total and can be ignored.

For more information on Taylor Series Expansions see the Wikipedia article [Taylor Series]

Contribution by carbon.booth
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Recall: $$f(x+dx) = f(x) + \frac{df(x)dx}{dx} + \frac{1d^{2}f(x)dx}{2dx^{2}} + ...$$ and we neglect the h.o.t.

Now neglecting the higher order terms $$\frac{d\sigma}{dx}+\frac{f(x)}{A}=0$$ where A is the applied load


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!Body Force, from team Carbon Body force, $$F_{body}=\frac{F}{V} = \frac{Force}{Volume}$$, is the force on an object per unit volume. An example, as mentioned in class, is the Earth's gravitational pull on the moon. This is represented as the gravitational force of the two masses divided by the volume of the moon.

Buoyancy force is also often represented as a body force.

For more information on Body Forces see the Body Force Wikipedia article

Contribution by carbon.guidry
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Now the equilibrium equation of a 3-d non-uniform stress field will be shown, but without an applied load. Focusing on the x-direction to avoid clutter and make a more appealing figure. Each side of the cube in the figure is of length dx, dy, and dz along their respective axis. The shear forces on the front and rear faces are equal, as are the shear stresses on the top and bottom. The normal stress in the x direction can be seen on the right and left faces.

$$ \sum{F_{x}} = 0= (dydz)[-\sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)]\, $$ Face normal to X axis.

$$ + (dzdx)[\sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)]\,$$ Face normal to Y axis.

$$+ (dxdy)[\sigma_{zx}(x,y,z) + \sigma_{zx}(x,y,z+dz)]\,$$ Face normal to Z axis.

Resulting in

$$0=(dxdydz)[\frac{d\sigma_{xx}}{dx} + \frac{\sigma_{yx}}{dy} + \frac{\sigma_{zx}}{dz}]$$

$$0=[\frac{d\sigma_{xx}}{dx} + \frac{\sigma_{yx}}{dy} + \frac{\sigma_{zx}}{dz}]$$

Which is therefore, equivalent to equation (1) in Lecture 27.