\begin{equation} \sum \vec{ \textbf{F}} = 0 \qquad (\textnormal{Condition 1 for equilibrium}) \tag{12.1} \end{equation} \begin{equation} \sum F_x = 0 \qquad \sum F_y = 0 \qquad \sum F_z = 0 \end{equation} \begin{equation} \sum \vec{ \boldsymbol{\tau}} = 0 \qquad (\textnormal{Condition 2 for equilibrium}) \tag{12.2} \end{equation} \begin{equation} \sum \tau_x = 0 \qquad \sum \tau_y = 0 \qquad \sum \tau_z = 0 \end{equation} \begin{equation} \vec{\textbf{r}}_{cm} = \frac{m_1 \, \vec{\textbf{r}}_1 \, + m_2 \, \vec{\textbf{r}}_2 \, + \, m_3 \, \vec{\textbf{r}}_3 \, + \, \, ...}{m_1 + m_2 + m_3 \, + \, ...} = \frac{\sum_i m_i \, \vec{\textbf{r}}_i}{\sum_i m_i} \qquad (\textnormal{Center of mass}) \tag{12.3} \end{equation} \begin{equation} \frac{\textnormal{stress}}{\textnormal{strain}} = \textnormal{Elastic modulus} \qquad (\textnormal{Hooke's law}) \tag{12.4} \end{equation} \begin{equation} Y = \frac{\textnormal{tensile stress}}{\textnormal{tensile strain}} = \frac{F_{\perp}/A}{\Delta l / l_0} = \frac{F_{\perp} \, l_0}{A \, \Delta l} \qquad (\textnormal{Young's modulus}) \tag{12.4} \end{equation} \begin{equation} p = \frac{F_{\perp}}{A} \qquad (\textnormal{pressure in a fluid}) \tag{12.5} \end{equation} \begin{equation} B = \frac{\textnormal{Bulk stress}}{ \textnormal{Bulk strain}} = - \frac{\Delta p}{\Delta V / V_0} \qquad (\textnormal{Bulk modulus}) \tag{12.6} \end{equation} \begin{equation} S = \frac{\textnormal{Shear stress}}{ \textnormal{Shear strain}} = \frac{F_{\parallel} / A}{x/h} = \frac{F_{\parallel}}{A} \frac{h}{x} \qquad (\textnormal{Shear modulus}) \tag{12.7} \end{equation}
Hooke's law is only valid to a certain point. It is valid to the proportional limit. The proportional limit is the maximum stress for which stress and strain will follow the elastic modulus.
The elastic limit is the point beyond which irreversible deformation occurs.
The stress limit is the point at which the material breaks.
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