$
\varepsilon_{x}=\frac{\sigma_{x}}{E}
$
$
\varepsilon_{y}=\varepsilon_{z}=-\frac{\upsilon\sigma_{x}}{E}
$
$
\gamma_{xy}=\frac{\tau_{xy}}{G}=\frac{\pi}{2}-\theta=\theta_{1}+\theta_{2}
$
$
G=\frac{E}{2(1+\upsilon)}
$
# Stress-strain Cube
![[Pasted image 20260907145936.png|center|400]]
$
\sigma_{x}=\lim_{ \Delta A \to 0 } \frac{\Delta F_{x}}{\Delta A}\quad\tau_{xy}=\lim_{ \Delta A \to 0 } \frac{\Delta F_{y}}{\Delta A}\quad\tau_{xz}=\lim_{ \Delta A \to 0 } \frac{\Delta F_{z}}{\Delta A}
$
- **Naming convention**
- $\sigma_{i}$ is the normal stress on face "$i
quot;
- $\tau_{ij}$ is the shear stress in the "$jquot; direction on face "$iquot;
- **Sign convention**
- A normal stress $\sigma_{i}$ is positive if it points outward on face "$iquot; of the cube, for $i=x,y,z.$ Note that a positive normal stress corresponds to tension
- A shear stress $\tau_{ij}$ is positive if it points in the positive $j$-direction on the positive $i$-face of the stress cube. Otherwise, the shear stress is negative
- Components on the positive faces, along with those on the negative faces, are shown on the stress cube, where a "primed" quantity ($\sigma_{x}',\tau_{xy}'$) are those components on the negative face
- 6 components of normal stress: $(\sigma_{x},\sigma_{y},\sigma_{z},\sigma_{x}',\sigma_{y}',\sigma_{z}')$
- 12 components of shear stress: $\left(\tau_{xy}, \tau_{xz}, \tau_{yx}, \tau_{yz}, \tau_{zx}, \tau_{zy}, \tau_{xy}', \tau_{xz}', \tau_{yx}', \tau_{yz}', \tau_{zx}', \tau_{zy}'\right)$
$
\sigma_{x}'=\sigma_{x}\qquad\sigma_{y}'=\sigma_{y}\qquad\sigma_{z}'=\sigma_{z}
$
$
\tau_{xy}'=\tau_{xy}=\tau_{yx}'=\tau_{yx}\qquad\tau_{yz}'=\tau_{yz}=\tau_{zy}'=\tau_{zy}\qquad \tau_{zx}'=\tau_{zx}=\tau_{xz}'=\tau_{xz}
$
# Thermal Strain
Thermal strain is the change in size or shape of a material caused by a change in temperature. Since it is uniform and proportional to $\Delta T,$ we can write it as
$
\varepsilon_{x,T}=\varepsilon_{y,T}=\varepsilon_{z,T}=\alpha\Delta T
$
where $\alpha$ is the coefficient of thermal expansion.
Temperature changes produce only extensional strains, no shear strains
# Generalized Hooke's Law
Strains due to mechanical loading in the x-direction:
$
\varepsilon_{x}=\frac{\sigma_{x}}{E}
$
$
\varepsilon_{y}=-\upsilon\varepsilon_{x}=-\frac{\upsilon\sigma_{x}}{E}
$
$
\varepsilon_{z}=-\upsilon\varepsilon_{x}=-\frac{\upsilon\sigma_{x}}{E}
$
Strains due to mechanical loading in the y-direction:
$
\varepsilon_{x}=-\upsilon\varepsilon_{y}=-\frac{\upsilon\sigma_{y}}{E}
$
$
\varepsilon_{y}=\frac{\sigma_{y}}{E}
$
$
\varepsilon_{z}=-\upsilon \varepsilon_{y}=-\frac{\upsilon\sigma_{y}}{E}
$
And so on for the z-direction...
Generalized formulas:
$
\varepsilon_{x}=\frac{1}{E}[\sigma_{x}-\upsilon(\sigma_{y}+\sigma_{z})]+\alpha\Delta T
$
^7e92b9
$
\varepsilon_{y}=\frac{1}{E}[\sigma_{y}-\upsilon(\sigma_{x}+\sigma_{z})]+\alpha\Delta T
$
^cfb52e
$
\varepsilon_{z}=\frac{1}{E}[\sigma_{z}-\upsilon(\sigma_{x}+\sigma_{y})]+\alpha\Delta T
$
^5e2215
$
\gamma_{xy}=\frac{\tau_{xy}}{G}\quad\gamma_{xz}=\frac{\tau_{xz}}{G}\quad\gamma_{yz}=\frac{\tau_{yz}}{G}
$