User:Eas4200c.f08.wiki.b/HW5

=Mtg. 26 - Monday 10/27=

Bidirectional Bending Recipe
The following is a general recipe of equations used for solving a bidirectional bending problem.

The moment in the y-direction is

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

Similarly, the moment in the z-direction is

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

Where $$\displaystyle dA = dydz $$

The moment of intertia in the y-direction is

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

Similarly, the moment of intertia in the z-direction is

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

 $$\displaystyle I_{yz} = \int_A yz dA $$

The stress in the x-direction is

 $$\displaystyle \sigma_{xx} = E \epsilon_{xx} = {I_y M_z - I_{yz} M_y \over I_y I_z - (I_{yz})^2} y + {I_z M_y - I_{yz} M_z \over I_y I_z - (I_{yz})^2} z $$

Recall that

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

Note that the determinat is defined as

 $$\displaystyle D := I_y I_z - (I_{yz})^2 = I_{22} I_{33} - (I_{23})^2 $$

for $$\displaystyle \begin{bmatrix} I_{22} & I_{23} \\ I_{32} & I_{33} \end{bmatrix}$$

 $$\displaystyle I_{32} = I_{zy} = \int_A zy dA = I_{yz} = I_{23} $$

At the neutral axis, or where $$\displaystyle \sigma_{xx} = 0 $$

 $$\displaystyle \sigma_{xx} = m_y y + m_z z $$

Thus,

 $$\displaystyle z = (- {m_y \over m_z})y = (\tan \beta) y $$



=Mtg. 28 - Friday 10/31=

Taking the previous equation

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

The equation can be rewritten as

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

Notice that

 $$\displaystyle [ \sigma(x+dx) - \sigma(x) ] = {d \sigma (x) \over dx} $$

Also recall that

 $$\displaystyle f(x+dx) = f(x) + {df(x) \over dx} dx + {1 \over 2} {d^2 f(x) \over dx^2} (x)(dx)^2 + ... $$

If the higher orter terms are neglected, we obtain

 $$\displaystyle {d \sigma (x) \over dx} + {f(x) \over A} = 0 $$

where $$\displaystyle {f(x) \over A} $$ is the applied load or body force, which is equal to the force divided by the volume.



As seen in the above image, the sum of the forces in the x-direction can be expressed as

<p style="text-align:center;"> $$\displaystyle \sum F_x = 0 = dydz [- \sigma_{xx}(x,y,z) + \sigma_{xx}(x+dx,y,z)] + dzdx [- \sigma_{yx}(x,y,z) + \sigma_{yx}(x,y+dy,z)] $$ <p style="text-align:center;"> $$\displaystyle + dxdy [- \sigma_{zx}(x,y,z) + \sigma_{zx}(x,y,z+dz)] $$

Simplifying yields

<p style="text-align:center;"> $$\displaystyle 0 = dxdydz [ {\partial \sigma_{xx} \over \partial x} + {\partial \sigma_{yx} \over \partial y} + {\partial \sigma_{zx} \over \partial z} ] $$