User:Eas4200c.f08.vqcrew.d/hw5

Equation of Equilibrium
The goal of this section is to prove that the equation below must be equal to zero in order for the body to be in equilibrium. This equation is thus called the equilibrium equation. The equilibrium equation in the x direction is:
 * $$\frac{\partial \sigma _{xx}}{\partial x}+ \frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}=0$$

This equation can be expressed in indicial notation with $$x=1$$, $$y=2$$, and $$z=3$$ and compressed as follows:
 * $$\sum_{i=1}^{3}{\frac{\partial \sigma _{ij}}{\partial x_i}}=0$$ for $$j=1, 2, 3$$

We can also note that in the case of pure torsion, the normal stress component $${\sigma _{xx}}\;$$ is equal to zero, so the equilibrium equation simplifies to:
 * $$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{zx}}{\partial z}=0$$

Before moving into the derivation of the equilibrium equation for the 3-D case, The 1-D case shown in Figure ## should first be examined. In order for this stress element to be in equilibrium, the sum of the forces must equal zero.
 * $$\sum{F_x}=0=-\sigma (x)A+\sigma (x+dx)A+f(x)dx$$
 * $$0= A \left[ \sigma (x+dx)-\sigma (x)\right]+f(x)dx$$

Now, recall the Taylor Series expansion:
 * $$f(x+dx)=f(x) + \frac{df(x)}{dx}(dx) + \frac{1}{2}\frac{d^2f(x)}{dx^2}(dx)^2 + ...$$



We can replace the term in square brackets with the first order Taylor Series expansion to simplify the equation as follows:
 * $$\frac{d \sigma }{dx}+\frac{f(x)}{A}=0$$

In this equation, $$f(x)\;$$ is in units of force per length, and the term $$\frac{f(x)}{A}$$ is called the applied load. We will now extend this derivation into the three-dimensional case. We will consider a stress element in a nonuniform stress field, without an applied load, and focus only on the stress components in the x-direction to avoid over-complicating the problem. The derivation will yield the same results if we consider the stresses acting in the y- and z-directions. The 3-D problem is depicted in Figure ##. Note that the first subscript in the stress terms defines the direction normal to the plane of the stress, and the second subscript defines the direction of the stress. The origin of the coordinate system is at the location $$(x,y,x)\;$$.

The sum of the forces is expressed as:


 * $$\sum{F_x}=0=dydz\left[-\sigma _{xx}(x,y,z) + \sigma _{xx}(x+dx,y,z)\right] + dxdz\left[-\sigma _{yx}(x,y,z) + \sigma _{yx}(x,y+dy,z) \right] + dxdy\left[-\sigma _{zx}(x,y,z) + \sigma _{zx}(x,y,z+dz) \right]$$

Simplifying the equation using the Taylor Series expansion, as in the 1-D case, yields the following:
 * $$0=(dxdydz)\left[\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}\right]$$

The volume term cancels out, and we are left with the equilibrium equation for stresses in the x direction for the 3-D problem:
 * $$0=\frac{\partial \sigma _{xx}}{\partial x} + \frac{\partial \sigma _{yx}}{\partial y} + \frac{\partial \sigma _{zx}}{\partial z}$$