Young's modulus

Remarks

The ratio does not depend on $s$, because the law is linear. Example: inverting $\sigma = \lambda (\tr \varepsilon) I + 2\mu \varepsilon$ for $\sigma = s \, e_1 \otimes e_1$ gives $\tr \varepsilon = s/(3\lambda + 2\mu)$ and $\varepsilon_{11} = \frac{s(\lambda + \mu)}{\mu(3\lambda + 2\mu)}$, which is the formula above. With the steel values $\lambda \approx 120$ GPa and $\mu \approx 80$ GPa it gives $E \approx 80 \cdot 520 / 200 = 208$ GPa.

Non-example: in uniaxial strain the sides are held fixed, so that only $\varepsilon_{11} \ne 0$. Then $\sigma_{11} / \varepsilon_{11} = \lambda + 2\mu$, the P-wave modulus, which is larger than $E$ when $\lambda > 0$. The definition requires the sides to be free of stress.

For a bar of length $L$ and cross-section $A$ pulled by a force $F$, the elongation is $\Delta L = F L / (E A)$. For anisotropic materials Young's modulus depends on the direction of the pull.