Question 3.3: Use the principle of minimum potential energy to find the di...
Use the principle of minimum potential energy to find the displacement at the free end in Fig. 3.17.
The solid domain in Fig. 3.17a is decomposed into an FEA model in Fig. 3.17b, which consists of 4 elements and 4 nodes. The details of FEA model are summarized in Table 3.6.
| Element | Node i | Node j | ε | A(in^2) | Strain energy (lbf.in) |
| 1 | ➊ | ➋ | \frac{u_{2}-u_{1} }{10} | 10 | \int_{0}^{10}{\frac{EA_{1} }{2}\left(\frac{u_{2}-u_{1} }{10} \right)^2dx } |
| 2 | ➋ | ➌ | \frac{u_{3}-u_{2} }{10} | 2 | \int_{0}^{10}{\frac{EA_{2} }{2}\left(\frac{u_{3}-u_{2} }{10} \right)^2dx } |
| 3 | ➋ | ➌ | \frac{u_{3}-u_{2} }{10} | 2 | \int_{0}^{10}{\frac{EA_{3} }{2}\left(\frac{u_{3}-u_{2} }{10} \right)^2dx } |
| 4 | ➌ | ➍ | \frac{u_{4}-u_{3} }{5} | 10 | \int_{0}^{10}{\frac{EA_{4} }{2}\left(\frac{u_{4}-u_{3} }{5} \right)^2dx } |
Using Eq. (3.14) gets the total potential energy as
\Pi =\Lambda -W=\int_{v}{\frac{1}{2}\sigma (x,y,z) . \varepsilon (x,y,z)dv – \sum\limits_{i=1}^{n}{F_{i}}.u_{i} } (3.14)
\Pi =1.45 \times 10^{7}(u_{2}-u_{1} )^2 +5.8\times 10^{6}(u_{3}-u_{2})^2 +5.8 \times 10^{7}(u_{4}-u_{3})^2-(R)u_{1}-(5000)u_{4} (3.16)
Using Eqs. (3.15) and (3.16) determine the conditions for minimum potential
energy as
\frac{∂\Pi }{∂u_{1}} = 0 \rightarrow -2.9\times 10^{7}(u_{2}-u_{1})=R
\frac{∂\Pi}{∂u_{2}}=0 \rightarrow 2.9\times 10^{7}(u_{2}-u_{1})-11.6\times 10^{6}(u_{3}-u_{2})=0 (3.17)
\frac{∂\Pi}{∂u_{3}}=0 \rightarrow 11.6\times 10^{6}(u_{3}-u_{2})-11.6\times 10^{7}(u_{4}-u_{3})=0
\frac{∂\Pi }{∂u_{4}} = 0 \rightarrow 11.6\times 10^{7}(u_{4}-u_{3})=5000
Equation (3.17) can further be simplified as
2.9\times 10^{6} \begin{bmatrix} 10 & -10 & 0 & 0 \\ -10 & 14 & -4 & 0 \\ 0 & -4 & 44 & -40 \\ 0 & 0 & -40 & 40 \end{bmatrix} \begin{bmatrix} u_{1} \\ u_{2} \\ u_{3} \\ u_{4} \end{bmatrix} = \begin{bmatrix} R \\ 0 \\ 0 \\ 5000 \end{bmatrix} (3.18)
Applying the boundary condition that the left side is fixed, i.e.u_1 = 0, Eq. (3.18) leads to the solution of [u] as
[u]=[0.0 1.72\times 10^{-4} 6.03 \times 10^{-4} 6.47 \times 10^{-4} ]^T
