written 6.8 years ago by | • modified 2.8 years ago |
Subject : Structural Analysis 1
Topic : General Principles
Difficulty : Low
written 6.8 years ago by | • modified 2.8 years ago |
Subject : Structural Analysis 1
Topic : General Principles
Difficulty : Low
written 6.8 years ago by | • modified 6.6 years ago |
Types of Energy Theorems:
Castigliano's Second Theorem states that:
The displacement at the point in a body is equal to the first partial derivative of the strain energy in the structure with respect to a force acting at the point, in the direction of the displacement.
$\triangle point i = \frac{\partial U}{\partial P_{i}}$ of the system
$\theta point i = \frac{\partial U}{\partial M_{i}}$ of the system
For example, we can find the deflexion at the end of the axially loaded bar using Castiliano's theorem as follows.
$\nu =\frac{1}{2}\frac{P^{2}L}{AE}$
$\triangle_{i}=\frac{\partial v}{\partial p}=\frac{\partial }{\partial b}\left[\frac{1}{2}\frac{P^{2}L}{AE}\right]=\frac{PL}{AE}$
$\vee=v\left(P_{1},P_{2}.....P_{n},M_{1},M_{2}....M_{n}\right)$
$\therefore \partial v=\frac{\partial u}{\partial P_{1}}dP_{1}+\frac{\partial u}{\partial P_{2}}dP_{2}+...+\frac{\partial u}{\partial P_{k}}dP_{k}+..+\frac{\partial u}{\partial m}dm$
$\therefore \partial v=\frac{\partial u}{\partial P_{k}}dP_{k}$
Work of the external force:
$\therefore W_{e}+dW_{e}$ $\therefore W_{e}+(dP_{k})\triangle_{k}$ (Defination of complementary work)
$\therefore W_{e}+(dP_{k})\triangle_{k}=\nu+\frac{\partial u}{\partial P_{k}}dP_{k}$
$\therefore (dP_{k})\triangle_{k}=\frac{\partial u}{\partial P_{k}}dP_{k}$
$\therefore \triangle_{k}=\frac{\partial u}{\partial P_{k}}$
Similarlly, $ \triangle_{k}=\frac{\partial u}{\partial M_{k}}$