This post will expand the partial derivative method to consider deflections due to axial strains.
Deriving Partial Derivative Method Due to Axial Strains
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From our previous post, recall that we need an expression for the strain energy \(U\) to fit our general equation. We will focus on strains due to axial action.
This energy we need is familiar because we already have derived it in another post. In our discussion with real work strain energy due to axial strains, we recall that strain energy is:
\(U=\frac{1}{2}\int\frac{S^2}{AE}dx\)
We highly recommend reading the said post if we need a refresher.
Key Idea: Partial Derivative Method Due to Axial Strains
With axial strain energy derived, we can apply the general equation. As a result, we will obtain the following equations:
Castigliano’s Theorem - Axial Strains - Translation: \(\Delta=\frac{\partial}{\partial{F}}\left[ \frac{1}{2}\int\frac{S^2}{AE}dx \right]\)
Castigliano’s Theorem - Axial Strains - Rotation: \(\theta=\frac{\partial}{\partial{M_P}}\left[ \frac{1}{2}\int\frac{S^2}{AE}dx \right]\)
Here is the meaning of these variables:
- \(F\) or \(M_P\) refers to the placeholder loads.
- \(\Delta\) is to translation of joint (vertical or horizontal).
- \(\theta\) refers to the rotation of a member.
- \(S\) refers to the real internal axial force made by the imposed loading and the placeholder loads.
- \(L\) is the length of the member.
- \(AE\) is axial rigidity.
These equations are what we use to solve for the translation and rotation of a structure.
Simplified Approach
Like our discussion on flexural strains, you can approach the two equations above in a simplified approach.
\(\Delta=\int\frac{S}{AE}\times\frac{\partial{S}}{\partial{F}}dx\)
\(\theta=\int\frac{S}{AE}\times\frac{\partial{S}}{\partial{M_P}}dx\)
These equations are simpler to evaluate. Later, we shall use an example to demonstrate how to use these equations.
Summary
Let's summarize:
The expression for the strain energy due to axial strains is \(U=\frac{1}{2}\int\frac{S^2}{AE}dx\).
The equations we will use to solve for the deflections are the following: (1) \(\Delta=\int\frac{S}{AE}\times\frac{\partial{S}}{\partial{F}}dx\), and (2) \(\theta=\int\frac{S}{AE}\times\frac{\partial{S}}{\partial{M_P}}dx\).
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