AE4630 - Aerospace Structural Design:
Lecture 6
Development of the Stress Equilibrium Equations
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Equations of motion: equilibrium and stress
2D conservation of linear momentum (translational equilibrium equations)
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Equilibrium parallelepiped - a differential volume element
![]()
\[\begin{align}
\sum F_x &= \; m a_x \\
\end{align}\]
\[\begin{align}
\delta y \left({{\sigma_{xx}}(x+\delta x,y)}-{{\sigma_{xx}}(x,y)}\right)
&\\
+ \delta x \left({\tau_{yx} (x,y+\delta y)}-{{\tau_{yx}}(x,y)}\right)
&\\
+ \delta x \delta y \; b_x &= \delta x \delta y \; \rho \, a_x\\
\end{align}\]
\[\begin{align}
\frac{{{\sigma_{xx}}(x+\delta x,y)}-{{\sigma_{xx}}(x,y)}}{\delta x}
+ \frac{{\tau_{yx} (x,y+\delta y)}-{{\tau_{yx}}(x,y)}}{\delta y} + b_x &= \rho \, a_x\\
\end{align}\]
\[\begin{align}
{\frac{\partial {{\sigma_{xx}}(x,y)}}{\partial x}}+ {\frac{\partial {{\tau_{yx}}(x,y)}}{\partial y}}+b_x &= \rho \, a_x\\
\end{align}\]
Similarly after summing the forces in \(y\):
\[\begin{align}
{\frac{\partial {{\sigma_{yy}}(x,y)}}{\partial y}}+ {\frac{\partial {{\tau_{xy}}(x,y)}}{\partial x}}+ b_y &= \rho \, a_y\\
\end{align}\]
Generalizing: the stress equilibrium equations are:
\[\begin{align}
{{\sigma_{xx}} {}_{,x}} + {{\sigma_{yx}} {}_{,y}} + {{\sigma_{zx}} {}_{,z}} + b_x = \rho \, a_x\\
{{\sigma_{xy}} {}_{,x}} + {{\sigma_{yy}} {}_{,y}} + {{\sigma_{zy}} {}_{,z}} + b_y = \rho \, a_y\\
{{\sigma_{xz}} {}_{,x}} + {{\sigma_{yz}} {}_{,y}} + {{\sigma_{zz}} {}_{,z}} + b_z = \rho \, a_z\\
\end{align}\]
The form of the equations of motion depend on the coordinate system
used.
2D moment equilibrium equations (conservation of angular momentum)
For an infinitesimal body:
![]()
\[\sum M_{C z} = I_M \ddot{\theta}\]
- (neglect inertia for brevity)
\[\begin{align}
\frac{\delta x}{2} \delta y \left[{\tau_{xy} (x+\delta x,y)}+ {{\tau_{xy}}(x,y)}\right] &\\
- \frac{\delta y}{2} \delta x \left[{\tau_{yx} (x,y+\delta y)}+ {{\tau_{yx}}(x,y)}\right] &=
0 \\
\end{align}\]
\[\begin{align}
\left[{\tau_{xy} (x+\delta x,y)}+ {{\tau_{xy}}(x,y)}\right] & \\
- \left[{\tau_{yx} (x,y+\delta y)}+{{\tau_{yx}}(x,y)}\right] &= 0 \\
\end{align}\]
\[\begin{align}
\left[{\tau_{xy} (x+\delta x,y)}+ {{\tau_{xy}}(x,y)}\right] &= \left[{\tau_{yx} (x,y+\delta y)}+{{\tau_{yx}}(x,y)}\right] \\
\end{align}\]
\[\begin{align}
\lim_{\delta x, \delta y \rightarrow 0} & \\
\end{align}\]
\[\begin{align}
2 \, {\tau_{xy}}&= 2 \, {\tau_{yx}}\\
\end{align}\]
Conclusion: the stress tensor is symmetric
\[\begin{align}
{\tau_{xy}}&= {\tau_{yx}}\\
{\tau_{xz}}&= {\tau_{zx}}\\
{\tau_{yz}}&= {\tau_{zy}}\\
\end{align}\]
\[\left[\sigma\right] =
\left[
\begin{array}{ccc}
{\sigma_{xx}}& {\sigma_{xy}}& {\sigma_{xz}}\\
{\sigma_{xy}}& {\sigma_{yy}}& {\sigma_{yz}}\\
{\sigma_{xz}}& {\sigma_{yz}}& {\sigma_{zz}}\\
\end{array}
\right]\]
There are only 6 unique components in a stress tensor.
Conclusion
Close examination of these equations illustrates that the normal
stress and shear stress are coupled together.
![Lap joint. Shear and normal stress are coupled. As the axial stress decreases along one leg of the joint, the load is passed into the adhesive via shear load transfer. It is subsequently transferred across the adhesive and is transferred via shear again to become axial stress in the other leg of the joint. A further observation is that the changing shear stress also requires a changing normal stress in the perpendicular direction. Hence a high concentration of peel stress exists where axial stress is changing.]()
Additional discussion about the coupling:
Equilibrium in cylindrical coordinates
By similar arguments, equilibrium in cylindrical coordinates
(\(r, \theta, z\)) requires:
\[\begin{align}
{\frac{\partial {\sigma_{rr}}}{\partial r}} + \frac{1}{r}{\frac{\partial {\sigma_{r\theta}}}{\partial \theta}} + {\frac{\partial {\sigma_{rz}}}{\partial z}} + \frac{{\sigma_{rr}}-{\sigma_{\theta\theta}}}{r} + b_r =& \; \rho \; a_r\\
{\frac{\partial {\sigma_{r\theta}}}{\partial r}} + \frac{1}{r}{\frac{\partial {\sigma_{\theta\theta}}}{\partial \theta}} + {\frac{\partial {\sigma_{\theta z}}}{\partial z}} + \frac{2 \, {\sigma_{r\theta}}}{r} + b_\theta =& \; \rho \; a_\theta\\
{\frac{\partial {\sigma_{rz}}}{\partial r}} + \frac{1}{r}{\frac{\partial {\sigma_{\theta z}}}{\partial \theta}} + {\frac{\partial {\sigma_{zz}}}{\partial z}} + \frac{{\sigma_{rz}}}{r} + b_z =& \; \rho \; a_z\\
\end{align}\]
Equilibrium in spherical coordinates
In spherical coordinates (\(r, \theta, \phi\)), equilibrium requires:
\[\begin{align}
{\frac{\partial {\sigma_{rr}}}{\partial r}}
+ \frac{1}{r}
{\frac{\partial {\sigma_{\theta r}}}{\partial \theta}}
+ \frac{1}{r \sin \theta}
{\frac{\partial {\sigma_{\phi r}}}{\partial \phi}}
+ \frac{1}{r}
\left[
2 \, {\sigma_{rr}}- {\sigma_{\phi\phi}}- {\sigma_{\theta\theta}}- {\sigma_{r\theta}}\cot \theta
\right]
+ b_r
=& \; \rho \; a_r\\
{\frac{\partial {\sigma_{r\theta}}}{\partial r}}
+ \frac{1}{r}
{\frac{\partial {\sigma_{\theta\theta}}}{\partial \theta}}
+ \frac{1}{r \sin \theta}
{\frac{\partial {\sigma_{\phi \theta}}}{\partial \phi}}
+ \frac{1}{r}
\left[
({\sigma_{\theta\theta}}-{\sigma_{\phi\phi}}) \cot \theta + 3 {\sigma_{r\theta}}
\right]
+ b_{\theta}
=& \; \rho \; a_{\theta}\\
{\frac{\partial {\sigma_{r\phi}}}{\partial r}}
+ \frac{1}{r}
{\frac{\partial {\sigma_{\theta \phi}}}{\partial \theta}}
+ \frac{1}{r \sin \theta}
{\frac{\partial {\sigma_{\phi\phi}}}{\partial \phi}}
+ \frac{1}{r}
\left[
3 {\sigma_{r\phi}}+ 2 \, {\sigma_{\theta \phi}}\cot \theta
\right]
+ b_{\phi}
=& \; \rho \; a_{\phi}\\
\end{align}\]
