User:Eas4200c.f08.aero.e/Week 10

=Monday=

Birectional Bending: See sections 4.2 & 4.1 and Example 4.1 on Page 124 of the text book.



Homework 5. Stringers @ points B,E,H,F on lecture 22-1

*Skin and span web do not contribute in beding; only stringers

*Need areas of stringers $$A_{B}, A_{E}, A_{H}, A_{F}$$

Bidirectional bending recipe: (no symmetry)

$$M_{y}=\int_{A}^{}{}Z\sigma _{xx}dA$$ (sec. 4.26)

Moment of inertia tensor: $$I_{y}=I_{yy} I_{x}=I_{xx}  I_{z}=I_{zz}$$

$$I_{y}=\int_{A}^{}{}z^2dA$$ (sec. 4.28a)

$$I_{Z}$$ (sec. 4.28b)

$$I_{yz}=\int_{A}^{}{}yzdA$$ (sec.4.28c)

$$\sigma_{xx}=Eepsilon_{xx}=$$

=Wednesday= HW.5 continued from pg (24-3)

Equation of Equilibrium (Stresses)

Goal: $$\frac{\delta \sigma_{yx}}{\delta y} + \frac{\delta \sigma_{zx}}{\delta z}=0$$ (compare with 3.14 in book)

This equation derived in a section detailing the governing equations of applying torque to uniform bars. Essentially, cross-sections of a uniform body undergoing an angular displacement can be treated as rigid bodies to which a rotation has been applied. Because the properties associated with each cross-section are independent of z, the strains and all but the above shear stress values equal zero.

Rewriting the equilibrium equation in indicial notation:

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

Recall: (Sec 2.4)

$$\frac{\delta \sigma_{zx}}{\delta x} + \frac{\delta \sigma_{yz}}{\delta y} + \frac{\delta \sigma_{zx}}{\delta z}=0$$  (Eq. 2-21)

Similarly, see (2.22 and 2.23)

In indicial notation, (2.21, 2.22, and 2.23):

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

=Friday=

Derivation of equation:

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

We will do the 1-D case first for simplification.

And we will be referring to the following figure:



$$\sum{F_{x}} = 0 = -\sigma (x)A + \sigma (x + dx)A + F(x)dx$$

$$0 = A\left[ \sigma (x+dx)-\sigma (x)+F(x)dx\right] $$

Now, if we use Taylor series expansion the above expression can be converted to another useful expression as follows:

$$ 0 = \left[ \sigma (x + dx) - \sigma (x)\right]A + F(x)dx $$ = $$\frac{\partial  \sigma (x)dx}{\partial x} + H.O.T = 0$$

Where H.O.T = Higher Order Terms

Recall: $$f(x+dx) = f(x) + \frac{df(x)dx}{dx} + \frac{1d^{2}f(x)dx}{2dx^{2}} + ...$$ and we neglect H.O.T

Then, we get:

$$\frac{d\sigma }{dx} + \frac{f(x)}{A} = 0$$ = Applied load

Now, non-uniform stress field in 3-D, but without applied load and focusing on the x-direction only (i.e: without the other stress components to avoid cluttering of the figure):



Where we will use the following convention: σij, where i indicates that is normal to the facet and j indicates the direction of stress.

Then we obtain:

$$\sum{}F_{x} = 0 = dy dz[-\sigma _{xx}(x,y,z)+\sigma _{xx}(x+dx,y,z)]+ dzdx[-\sigma _{yz}(x,y,z)+\sigma _{yx}(x,y+dy,z)]+ dxdy[-\sigma _{zx}(x,y,z)+\sigma _{zx}(x,y,z+dz)]$$

$$(dx)(dy)(dz)[\frac{\partial \sigma _{xx}}{\partial x}+\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{xz}}{\partial z}]= 0$$