ISEGORIA / MATH ENCYCLOPEDIA
Continuum mechanics: how solids stretch, bend and buckle
Continuum mechanics treats a solid as a smooth body whose every small piece can stretch, shear and turn. The deformation gradient says exactly how each piece moves, and separating its stretch from its rotation is the key to measuring strain correctly. Applied to a slender beam, the same ideas give the Euler–Bernoulli equation for bending; applied to a slender column under compression, they predict a load at which straightness itself becomes unstable.
Before you begin: Linear algebra, calculus and Newton’s laws
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Deformation gradient, stretch and rotation
Follow a small square of material as the body deforms. If the deformation is smooth, close enough to any point it is linear: a vector of material \(dX\) becomes \(dx = F\,dX\), where F is the deformation gradient. Its columns are where the two edges of the square go. Every such F splits uniquely into a pure stretch U along two perpendicular directions followed by a rotation R, and only the stretch should count as strain. Drag the edges, then add a rigid rotation and see which strain measure notices.
Worked example. Simple shear by \(\gamma=0.6\), \(F=\begin{pmatrix}1&0.6\\0&1\end{pmatrix}\), keeps the area (\(J=1\)) but stretches by \(\lambda_1=\gamma/2+\sqrt{1+\gamma^2/4}=1.344\) and shortens by \(\lambda_2=1/\lambda_1=0.744\), along axes that are then turned by \(-16.7^\circ\). The Green–Lagrange principal strains are 0.403 and \(-0.223\); the small-strain estimate gives \(\pm0.3\). A rigid turn by 30° has \(E=0\), yet its small strain is \(\cos30^\circ-1=-0.134\) in every direction.
Watch out. The deformation here is homogeneous: the same F everywhere, so straight lines stay straight. In a real body F varies from point to point and this picture applies to each infinitesimal neighbourhood. The Mohr diagram plots, for each plane through the point, its normal strain against half its shear strain, as in the strain tensor’s components.
Why is the small strain ε wrong under a rotation, when it is used everywhere in engineering?
Because it is the first-order part of E. Writing \(F=I+H\), \(E=\varepsilon+\tfrac12H^{\mathsf T}H\), and for a rotation by θ the quadratic term is of size \(\theta^2/2\), which exactly cancels the spurious \(\cos\theta-1\). When strains and rotations are both small, say under 1%, the dropped term is under \(10^{-4}\) and ε is excellent; for rubber, soft tissue or a slender rod that rotates a lot, it is not.
2. Bending of beams
A beam carries load by bending: one side stretches, the other compresses, and the stress grows linearly with distance from the neutral axis. Bernoulli’s assumption that cross-sections stay plane and perpendicular to the axis turns that into one equation: the curvature equals the bending moment divided by the bending stiffness EI, which with equilibrium gives \(EI\,w^{\prime\prime\prime\prime}=-q\). Choose how the bar is held, place the load, and read the deflection, the shear and moment diagrams and the stress.
Worked example. For the 40 × 60 mm steel bar, \(I=bh^3/12=7.2\times10^{-7}\) m⁴ and \(EI=144\) kN·m². A 3 kN load at the tip of a 1 m cantilever deflects it by \(PL^3/3EI=6.94\) mm and makes a root moment of 3 kN·m, so the outer fibres carry \(\sigma=Mc/I=3000\times0.03/7.2\times10^{-7}=125\) MPa, half the yield stress. The same load at the middle of the bar on simple supports gives only 0.43 mm, and clamped at both ends 0.11 mm.
Watch out. Euler–Bernoulli theory ignores shear deformation and assumes small slopes; it is accurate when the beam is more than about ten times longer than it is deep, as here. The deflections are exact solutions of the equation for these loads and supports, written in closed form; the drawing exaggerates them and the depth so that both can be seen.
Why does doubling a beam’s depth make it eight times stiffer, but doubling its width only twice as stiff?
Material far from the neutral axis does more work: it is strained in proportion to its distance y, carries stress in proportion to y, and that stress has a lever arm y, so its contribution to stiffness goes as \(y^2\). Integrating \(y^2\) over a depth h gives \(h^3\), while width just adds more of the same strips. This is why I-beams put almost all their material in the flanges.
3. Buckling and Euler’s elastica
Push on the ends of a ruler and it stays straight, until suddenly it bows sideways. Euler found the load at which this happens in 1744: below it the straight shape is stable, above it the straight shape still balances but the slightest disturbance makes the rod bend. The critical load depends only on the bending stiffness and the effective length KL, set by how the ends are held. Beyond it, the exact shapes are the elastica curves Euler also described. Choose the end conditions and the length, add a small crookedness, and load the column.
Worked example. A 20 mm steel rod has \(I=\pi d^4/64=7.85\times10^{-9}\) m⁴ and radius of gyration \(r=d/4=5\) mm. Pinned at both ends and 1.2 m long, its slenderness is 240 and it buckles at \(P_{\rm cr}=\pi^2EI/L^2=10.8\) kN, a stress of only 34 MPa, a seventh of yield. Past that load the elastica rises slowly: 1% more load bows the rod by 8.9% of its length with end slopes of 16°, and 10% more load by 25% with end slopes of 50°.
Watch out. The elastica is exact for a perfectly straight, inextensible, elastic rod; real steel yields long before such large deflections. For a crooked rod the formula \(\delta_0/(1-P/P_{\rm cr})\) holds only for small deflections, so the drawing stops at 0.985 \(P_{\rm cr}\); above \(P_{\rm cr}\) it shows the perfect rod. The Johnson parabola is an empirical design curve, not a law of mechanics.
Why does freeing one end make the column four times weaker, and clamping both ends four times stronger?
Because the buckled shapes are pieces of the same curve. A fixed–free column of length L bends like one half of a pinned–pinned column of length 2L, so its effective length is 2L; a fixed–fixed column has inflection points a quarter of the way from each end, so its middle half bends like a pinned–pinned column of length L/2. Since \(P_{\rm cr}\propto1/(KL)^2\), doubling or halving the effective length divides or multiplies the load by four.