User:Eas4200c.f08.vqcrew.f/hw5

Equation of Equilibrium Stresses

We now want to look at a non uniform stress field. To accomplish this we must first start with the basics and consider a 1-D case as a model. We will have to consider both a Uniform and Nonuniform case. The uniform case can be seen in Figure ##, while the nonuniform case can be seen in Fig #. The uniform stress includes a single force acting on the end of a cantilever beam, while the nonuniform case has a varying shear stress acting horizontally across the top face of the beam along with the force on the end of the beam.

In the nonuniform case, the shear stress f(x) changes along the beam, thus f(x) is not a constant.



Bidirectional Bending



The recipe for Bidirectional Bending can be seen in Figure ##. The figure shows nonuniform stresses acting on a nonuniform cylinder. The equations below show the Moments about the y and z axis from $$ \displaystyle \sigma_{xx}$$. The Area over which the following integrals will be integrated are $$ dA = dy dz\;$$.


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


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

The next step is to look at the Moment of Inertia Tensors. First it is helpful to look at how the tensors can be transformed into both uniform and indicial notations.


 * $$ I_y = I_{yy} = I_{22}\;$$


 * $$ I_z = I_{zz} = I_{33}\;$$


 * $$ I_{yz} = I_{23}\;$$

Now from Figure ##, the Moment of Inertia Tensors can be determined.


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


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


 * $$ I_{yz} = \int_A yz \,dA$$

Now we will examine Hooke's Law:


 * $$\displaystyle \sigma_{xx} = E \epsilon_{xx}$$

Another way of looking of looking at Hooke's Law:


 * $$\displaystyle \sigma_{xx} = M_yy + M_zz$$


 * $$\displaystyle \sigma_{xx} = \frac{I_y M_z - I_{yz} M_z}{I_y I_z - I_{yz}^2}y + \frac{I_z M_z - I_{yz} M_{z}}{I_y I_z - I_{yz}^2}z$$

Note that the denominator is the determinant of the moment of inertia matrix.

Note that the nuetral axis where $$\displaystyle \sigma_{xx} = 0$$. Thus Hooke's Law can be transformend into the following:


 * \displaystyle \sigma_{xx} = M_yy + M_zz = 0


 * $$ z = \frac{-M_y}{M_z}y$$

By looking at the simple geometry of Fig ##.


 * $$tan \displaystyle \beta = \frac{M_y}{M_z}$$

Thus the equation can be simplified:


 * $$ z = (tan\displaystyle \beta)y$$