11.7 Minimum potential energy

This is related to the principle of virtual work.

\[\begin{equation*} \delta V + \delta U = 0 = \delta (V + U) \end{equation*}\]

Total potential energy \(= U + V = \Pi\) \[\begin{align} \delta \Pi =&\; 0 \\ \Pi =&\; \Pi (q_1, q_2, ..., q_i, ..., q_n ) \\ \end{align}\] where \(q_i\) are independent DOFs
\[\begin{align} \delta \Pi =& {\frac{\partial \Pi}{\partial q_1}} \delta q_1 + {\frac{\partial \Pi}{\partial q_2}} \delta q_2 + ... \\ \delta \Pi =& \displaystyle\sum_{i=1}^n {\frac{\partial \Pi}{\partial q_i}} \delta q_i = 0 \\ \end{align}\]

Since \(\delta q_i\) are virtual displacements, or arbitrary kinematically admissible displacements: \[\begin{equation*} {\frac{\partial \Pi}{\partial q_i}} \delta q_i = 0 \end{equation*}\]
Of all the kinematically admissible displacements, those corresponding to a stable equilibrium position make \(\Pi\) a minimum. (Each and every \(q_i\) must minimize the energy.)
Therefore: \[\begin{equation*} {\frac{\partial \Pi}{\partial q_i}} \delta q_i = 0 \end{equation*}\]

Deformable body virtual work example

\[\begin{equation*} 4A_1 = A_2 = 4 A \end{equation*}\]

The DOFs are the displacements \(u_D\) , \(u_C\) and \(u_B (=0)\)
Write strains first in terms of DOFs: \[\begin{equation*} {\varepsilon}_1 = \frac{(u_C - u_B)}{2l} = \frac{u_C}{2l} \end{equation*}\] \[\begin{equation*} {\varepsilon}_2 = \frac{(u_D - u_C)}{l} \end{equation*}\]
Then strain energy: \[\begin{equation*} U_i = {\frac{1}{2}}\sigma_i \cdot {\varepsilon}_i \cdot A_i L_i = {\frac{1}{2}}(E_i \cdot {\varepsilon}_i) \cdot {\varepsilon}_i \cdot A_i L_i = {\frac{1}{2}}({\varepsilon}_i^2 \cdot E_i \cdot A_i L_i) \end{equation*}\] \[\begin{equation*} U = U_1 + U_2 = {\frac{1}{2}}({\varepsilon}_1^2 \cdot E_1 \cdot A_1 L_1) + {\frac{1}{2}}({\varepsilon}_2^2 \cdot E_2 \cdot A_2 L_2) \end{equation*}\] \[\begin{equation*} = {\frac{1}{2}}\left(\frac{u_C}{2l}\right)^2 EA \cdot 2l + {\frac{1}{2}}\left(\frac{u_D - u_C}{l} \right)^2 4 EA \cdot l \end{equation*}\]

Therefore: \[\begin{equation*} U = (u_C)^2 \cdot \frac{EA}{4l} + (u_D - u_C)^2 \cdot \frac{2EA}{l} \end{equation*}\]
Also, the external work: \[\begin{equation*} V = - P u_D \end{equation*}\]
The energy: \[\begin{equation*} \Pi=U+V \end{equation*}\]
Thus: \[\begin{equation*} {\frac{\partial \Pi}{\partial u_C}} = 0 \end{equation*}\] \[\begin{equation*} 2 u_C \cdot \frac{EA}{4l} + 2 (u_D - u_C) \cdot (-1) \cdot \frac{2 EA}{l} = 0 \end{equation*}\] \[\begin{equation*} {\frac{\partial \Pi}{\partial u_D}} = 0 \end{equation*}\] \[\begin{equation*} 2 (u_D - u_C) \cdot \frac{2 EA}{l} - P = 0 \end{equation*}\]

By adding both equations: \[\begin{equation*} 2 u_C \cdot \frac{EA}{4 l} - P = 0 \end{equation*}\] \[\begin{equation*} u_C = P \cdot \frac{2l}{EA} = P \cdot \frac{2L}{3 EA} \end{equation*}\] \[\begin{equation*} 2 (u_D - u_C) \cdot \frac{2EA}{l} = P \end{equation*}\]
Therefore: \[\begin{equation*} u_D = u_C + P \frac{l}{4 EA} \end{equation*}\] \[\begin{equation*} u_D = \left(P \frac{l}{EA}\right) \cdot (2 + \frac{1}{4}) = 9P \frac{l}{4 EA} = 3P \frac{L}{4 EA} = u_D \end{equation*}\]
Therefore: \[\begin{equation*} u_D= 3P \frac{L}{4 EA} \end{equation*}\]

Due to overlapping material and to provide continuity, the lectures slides for Lecture 1 are included in the materials for Lecture 0