= Axial rigidity
{title2=$EA$}
For a uniform bar in uniaxial <linear elasticity>, <axial rigidity> is the product $EA$ of <Young's modulus> and load-bearing <cross-sectional area>. Since $\sigma=F/A$ and $\sigma=E\varepsilon$,
$$
F=EA\varepsilon.
$$
Thus $EA$ has units of <force>. It relates force to strain; <axial stiffness> $EA/L$ relates force to extension over a specified length $L$. The uniform-bar force-extension relation makes this distinction explicit.<Structural Stiffness Concept by TU Delft OpenCourseWare>[Axial member relation delta=PL/(EA), structural stiffness and geometry]{ref}
An original hollow-tube example keeps the area convention visible. Let the outer radius be $2\,\mathrm{mm}$, the inner radius $1\,\mathrm{mm}$ and the stipulated modulus $70\,\mathrm{GPa}$. The material area and rigidity are
$$
A=\pi(2^2-1^2)\,\mathrm{mm^2}=3\pi\,\mathrm{mm^2},\qquad
EA=210000\pi\,\mathrm N\approx6.60\times10^5\,\mathrm N.
$$
The empty bore contributes no material area. An axial force $F=210\pi\,\mathrm N$ gives strain $\varepsilon=0.001$. Tubes of lengths $50$ and $100\,\mathrm{mm}$ have this same rigidity, but extensions $0.050$ and $0.100\,\mathrm{mm}$ respectively. Their axial stiffnesses differ by a factor of two.
For a slowly varying area or modulus and constant axial force, extension instead accumulates locally:
$$
\Delta L=\int_0^L\frac{F}{E(x)A(x)}\,dx.
$$
A single arithmetic average of $EA$ generally cannot replace that compliance integral. The stepped-bar calculation in <axial stiffness> is its piecewise-constant counterpart.
A useful experiment measures axial force and strain between defined specimen points, over the proportional regime. The slope $dF/d\varepsilon$ estimates rigidity under that loading model; converting a <crosshead displacement> directly into strain can mix in <machine compliance> or grip motion. The model also assumes that lateral constraints and end effects do not replace the intended uniaxial material response.
Back to article page
