Ultimate Moment and Plastic Hinge of a Pure-Bending Beam
Using a rectangular-section pure-bending beam as the canonical example, we trace the evolution of cross-sectional stress through elastic → elasto-plastic → plastic stages and obtain the key quantities: yield moment $M_{\mathrm s}$, ultimate moment $M_{\mathrm u}$, shape factor $\alpha$ — the foundation for the rest of the chapter.
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Three stages of stress distribution in a pure-bending beam
Three Stagesthe figure: a pure-bending beam made of ideal elasto-plastic material with a rectangular section of width $b$ and height $h$. As the moment $M$ grows, the beam passes through elastic, elasto-plastic, and plastic stages in turn. Throughout, the plane-sections assumption holds — strain $\varepsilon$ is linear in the fiber distance from the neutral axis, and relates to curvature $\kappa$ by
Cross-sectional equilibrium $\sum F_{x} = 0$ and $\sum M = 0$:
$M \le M_{\mathrm s}$
Stress over the section is linear, $\sigma = E\kappa y$ (see Animation 12.2.1); the peak has not yet reached $\sigma_{\mathrm s}$.
$M_{\mathrm s} < M < M_{\mathrm u}$
The edges $|y| \ge y_{0}$ have yielded ($\sigma = \pm\sigma_{\mathrm s}$); the elastic core inside remains linear.
$M = M_{\mathrm u}$
The elastic core $y_{0} \to 0$. The whole section carries two rectangles of $\pm \sigma_{\mathrm s}$ — the plastic hinge forms.
① Elastic stage — linear distribution
$\sigma = E\varepsilon = E\kappa y$ (Eqs. 12-4, 12-5), substituted in Eq. (12-3) gives
At the end of the elastic stage, the outermost fiber stress reaches $\sigma_{\mathrm s}$ with strain $\varepsilon_{\mathrm s}$. The corresponding moment $M_{\mathrm s}$ and curvature $\kappa_{\mathrm s}$ are
Ultimate moment and shape factor
Plastic Moment② Elasto-plastic stage
For $M > M_{\mathrm s}$, the edges form a plastic zone ($\sigma = \pm\sigma_{\mathrm s}$); the inner core $|y| \le y_{0}$ remains elastic. Within the elastic core, $\sigma = \sigma_{\mathrm s}\cdot y/y_{0}$ stays linear. Integrating $\int \sigma y\, \mathrm d A = M$ piecewise:
At the core edge $y = y_{0}$, $\sigma = \sigma_{\mathrm s}$; hence the curvature
Eliminating $y_{0}$ gives the nonlinear $M$-$\kappa$ relation:
③ Plastic stage · Ultimate moment $M_{\mathrm u}$
As $M$ grows further, the elastic core $y_{0}$ shrinks to the limit $y_{0} \to 0$. The stress distribution becomes two rectangles (see Animation 12.2.2.1); the corresponding ultimate moment is
Shape factors for common sections
| Cross-section | $\alpha = M_{\mathrm u}/M_{\mathrm s}$ |
|---|---|
| Rectangle | 1.5 |
| Solid circle | 1.70 |
| Thin-walled tube | 1.27 |
| I-section (strong axis) | 1.10 ~ 1.17 |
$\alpha$ depends only on cross-section shape — not on material, size, or loading. A rectangle's $\alpha = 1.5$ means a 50% capacity gain over elastic design; an I-section with flanges already concentrated far from the neutral axis uses material efficiently in elastic design, so the plastic margin is smaller.
The true $M$-$\kappa$ curve for an ideal elasto-plastic material is a line + a curve + a plateau (segments $OA$ + $ACB$ + $BG$ in the figure). For simplicity we often use a bilinear $OFG$: linear $OA$ for $M < M_{\mathrm s}$, instantaneous jump to $M_{\mathrm u} = 1.5 M_{\mathrm s}$ at yield, then plastic flow at constant $M = M_{\mathrm u}$. The subsequent analyses in this chapter use this bilinear model.
Residual stress on unloading · Ultimate moment of general sections
Unloading · General Section④ Unloading · residual stress
If the beam reaches the elasto-plastic stage (point $C$ on the figure) and is then unloaded, the unloading stress/strain increments are linear, so when $M$ returns to zero the beam retains a residual curvature. Also, because loading and unloading follow different laws, the section carries a residual stress distribution.
Animation 12.2.3 shows the unloading process: (a) stress distribution at $C$ before unloading, (b) linear stress increments during unloading, (c) residual stress $\sigma_{\mathrm r} = \Delta\sigma - \sigma_{\mathrm s}$. After unloading, internal forces on the section vanish — the residual stress is a self-equilibrating self-stress state.
Post-unloading residual stresses matter in engineering — welded steel beams, cold-bent shapes, etc. retain residual stresses from the manufacturing plastic deformation, affecting the onset of yielding and the fatigue life under subsequent loads.
⑤ Ultimate moment of a general section
For non-symmetric sections, the neutral axis passes through the section's centroid in the elastic stage — but in the plastic stage, with stress $\pm\sigma_{\mathrm s}$ split into two rectangles, where does the dividing line lie?
Answer: the equal-area axis — here $A_{1} = A_{2} = A/2$. This follows immediately from $\int\sigma\,\mathrm d A = 0$: for non-symmetric sections, the centroidal axis and the equal-area axis do not coincide.
For a section with one axis of symmetry:
- Elastic stage: linear distribution $\sigma = M y/I$ (12-13), neutral axis through the centroid;
- At the ultimate state: stress as two rectangles of $\pm \sigma_{\mathrm s}$ — the neutral axis lies at the equal-area line:
Ultimate moment = force couple of the two resultants about the neutral axis:
Two essential differences from a regular hinge
Regular hinge Regular Hinge
- Cannot carry moment ($M = 0$) on either side;
- Bidirectional — free to rotate in both directions.
Plastic hinge Plastic Hinge
- Each side carries a couple of magnitude $M_{\mathrm u}$;
- Unidirectional — free only in the direction of $M_{\mathrm u}$; the reverse restores rigid continuity.
- Pure-bending stress passes through elastic → elasto-plastic → plastic — linear → elastic core + plastic edges → two-rectangle distribution;
- Yield moment $M_{\mathrm s} = b h^{2} \sigma_{\mathrm s}/6$; ultimate moment $M_{\mathrm u} = b h^{2} \sigma_{\mathrm s}/4$;
- Shape factor $\alpha = M_{\mathrm u}/M_{\mathrm s}$ depends only on section shape (rectangle 1.5, circle 1.7, I-section ≈1.1);
- Residual stress after unloading is self-equilibrating; the ultimate moment of a general section uses the equal-area axis;
- Plastic hinge vs regular hinge: the former carries $M_{\mathrm u}$ and is unidirectional;
- Next, §12-3 solves the ultimate load of statically determinate and indeterminate beams.