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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 $\sigma =E\epsilon$,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.^[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 areThe empty bore contributes no material area. An axial force $F=210\pi \mathrm{N}$ gives strain $\epsilon =0.001$. 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: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 $dF/d\epsilon$ 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.

Axial stiffness measures resistance to extension or shortening along a specified loading axis. For a proportional response, $F=K_{\mathrm{ax}}\Delta L$; for a nonlinear response, specify whether the reported value is a secant stiffness or tangent stiffness. A straight uniform bar in small-strain linear elasticity haswhere $E$ is Young's modulus, $A$ is the load-bearing cross-sectional area and $L$ the deforming length.^[ref] Axial rigidity $EA$ has units of force; axial stiffness $EA/L$ has units of force per length.

An original stepped-bar example shows why a single area is sometimes insufficient. Join two ideal straight segments end to end, with no distributed axial loading. Both carry the same force $F$, so their extensions add:Choose $E_1=E_2=70\mathrm{GPa}$, $L_1=20\mathrm{mm}$, $L_2=30\mathrm{mm}$, $A_1=1\mathrm{m}{\mathrm{m}}^2$ and $A_2=2\mathrm{m}{\mathrm{m}}^2$. The total mechanical compliance is $0.0005\mathrm{mm}{\mathrm{N}}^{-1}$, giving $K_{\mathrm{ax}}=2000\mathrm{N}\mathrm{m}{\mathrm{m}}^{-1}$. A $100\mathrm{N}$ force extends the bar by $0.050\mathrm{mm}$, divided into $0.0286$ and $0.0214\mathrm{mm}$ in the two segments. The thinner segment extends more despite being shorter.

In parallel, components sharing the same extension instead add their axial stiffnesses, as in a straight unbonded material-fibre bundle with every strand engaged. These are different constraints, described by series and parallel spring stiffness.

A physical test should measure extension between defined specimen points. Crosshead displacement can include machine compliance and grip motion, so the slope of the complete apparatus is not automatically the specimen's axial stiffness. Structural geometry and loading constraints must be separated from the material's Young's modulus.