User:Eas4200c.f08.blue.a/Lecture 28

goal to determine the four zero stress components.

obtain the equations resulting from

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

which would result in

$$\frac{\delta \sigma_{11}}{\delta x_{1}}+\frac{\delta \sigma_{12}}{\delta x_{2}}+\frac{\delta \sigma_{13}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{21}}{\delta x_{1}}+\frac{\delta \sigma_{22}}{\delta x_{2}}+\frac{\delta \sigma_{23}}{\delta x_{3}}=0$$

$$\frac{\delta \sigma_{31}}{\delta x_{1}}+\frac{\delta \sigma_{32}}{\delta x_{2}}+\frac{\delta \sigma_{33}}{\delta x_{3}}=0$$

One D model problem:

$$\sum F_2=0=-\sigma_w A+\sigma_{x+dx}A+f(x)dx$$

$$0=A[\sigma_{x+dx}-\sigma_x]+f(x)dx$$

Taylor series expansion $$=\frac{d \sigma}{dx}+$$ higher order terms

now neglecting the higher order terms

$$\frac{\delta \sigma}{dx}+\frac{f(x)}{A}=0$$

So for all the forces in the X direction the sum is

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

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{xx}}{\delta_x}+\frac{\delta \sigma_{xy}}{\delta_y}+\frac{\delta \sigma_{xz}}{\delta_z})$$

So for all the forces in the Y direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{yx}(x,y,z)+\sigma_{yx}(x+dx),y,z)]+dzdx[-\sigma_{yy}(x,y,z)+\sigma_{yy}(x,y+dy,z)]+dxdy[\sigma_{yz}(x,y,z)+\sigma_{yz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{yx}}{\delta_x}+\frac{\delta \sigma_{yy}}{\delta_y}+\frac{\delta \sigma_{yz}}{\delta_z})$$

So for all the forces in the Z direction the sum is

$$\sum Fy =0= dy dz [-\sigma_{zx}(x,y,z)+\sigma_{zx}(x+dx),y,z)]+dzdx[-\sigma_{zy}(x,y,z)+\sigma_{zy}(x,y+dy,z)]+dxdy[\sigma_{zz}(x,y,z)+\sigma_{zz}(x,y,z+dz)]=0$$

which will yield

$$0=dx dy dz (\frac{\delta \sigma_{zx}}{\delta_x}+\frac{\delta \sigma_{zy}}{\delta_y}+\frac{\delta \sigma_{zz}}{\delta_z})$$