Beam Mechanism Plastic Analysis: Engineering Theory and Solved Examples (2026)

1. Engineering Principles of Beam Mechanism Plastic Analysis

Continuous girders and flexural floor beams represent fundamental structural elements in civil engineering. While elastic design restricts capacity to the moment that induces first yield ($M_y = S f_y$), ductile structural steel and reinforced concrete beams possess substantial post-elastic reserve strength. Evaluating this capacity requires rigorous beam mechanism plastic analysis.

In limit-state mechanics, beam mechanism plastic analysis establishes the ultimate load capacity of flexural members by modeling the progressive formation of plastic hinges. When an indeterminate beam is loaded beyond its elastic limit, highly stressed cross-sections yield completely. Each yielded section transforms into a localized plastic hinge that sustains a constant fully plastic moment $M_p = Z f_y$ while rotating freely under increasing load.

Performing a comprehensive the plastic collapse mechanism allows engineers to quantify the beneficial effects of moment redistribution. By tracking how internal bending moments shed from over-stressed support regions to under-stressed mid-span sections, structural designers prevent premature failure, optimize cross-sectional dimensions, and ensure structural integrity under severe overload conditions.

2. Plastic Hinge Formation and Moment Redistribution in Continuous Beams

The fundamental premise of plastic beam theory is that statically indeterminate beams do not fail upon the creation of the first plastic hinge. Instead, each plastic hinge removes one degree of static indeterminacy.

2.1 Degrees of Indeterminacy and Hinge Counting

Let a beam span possess a degree of statical indeterminacy $r$. To transform that span into a kinematic collapse mechanism with one degree of freedom, the total number of plastic hinges $N_h$ required is:

$$N_h = r + 1$$

In continuous multi-span beams, collapse can occur as a localized partial collapse mechanism confined to a single heavily loaded span. Even if the entire continuous bridge girder has a global indeterminacy of $r = 4$, a single span with $r_{\text{ span}} = 1$ requires only $N_h = 2$ plastic hinges to fail.

2.2 Progressive Plastic Yielding Under Transverse Loading

Consider a fixed-ended beam carrying a uniformly distributed load $w$.
1. Elastic Stage: Peak hogging moments develop at the fixed supports ($M_{\text{ support}} = -w L^2 / 12$), while sagging moment at mid-span is smaller ($M_{\text{ mid}} = +w L^2 / 24$).
2. First Plastic Hinge Stage: As $w$ increases, the fixed support sections reach $M_p$ first, forming plastic hinges at both ends.
3. Redistribution Stage: With support moments locked at $M_p$, the beam behaves as a simply supported span for subsequent load increments. Bending moments shed to the mid-span.
4. Collapse Mechanism Stage: Finally, the mid-span section reaches $M_p$, forming a third plastic hinge. The beam transforms into a 3-hinge mechanism and collapses at $w_c = 16 M_p / L^2$, a $33.3\%$ increase over initial support yield!

3. Virtual Work Formulations for Beam Mechanisms

The Principle of Virtual Work provides a direct analytical tool for evaluating beam collapse loads without solving complex differential equations.

3.1 Point Loads at Arbitrary Span Locations

Consider a beam span of length $L$ fixed at both ends, carrying a concentrated point load $P$ at distance $a$ from the left support and $b = L – a$ from the right support.

          P (kN)
             |
             v
   +---------+-------------------+
   A (Fixed)      x = a          B (Fixed)
   <--- a ---><------- b ------->
   <------------- L ------------->

Under a virtual downward displacement $\Delta$ beneath the load point:

• Left segment rotation: $\theta_1 = \Delta / a$.

• Right segment rotation: $\theta_2 = \Delta / b$.

• Support hinge rotation at $A$: $\theta_A = \theta_1 = \Delta / a$.

• Span hinge rotation at load: $\theta_{\text{ span}} = \theta_1 + \theta_2 = \Delta \left( \frac{1}{a} + \frac{1}{b} \right) = \frac{L}{ab} \Delta$.

• Support hinge rotation at $B$: $\theta_B = \theta_2 = \Delta / b$.

The internal virtual work across all three plastic hinges is:

$$W_i = M_p \left[ \theta_A + \theta_{\text{ span}} + \theta_B \right] = M_p \left[ \frac{\Delta}{a} + \frac{L}{ab} \Delta + \frac{\Delta}{b} \right] = 2 M_p \left( \frac{L}{ab} \right) \Delta$$

Equating external work $W_e = P \Delta$ to internal work $W_i$:

$$P \Delta = 2 M_p \left( \frac{L}{ab} \right) \Delta \implies P_c = \frac{2 M_p L}{a b}$$

For a mid-span point load ($a = b = L/2$), the collapse load simplifies to $P_c = 8 M_p / L$.

3.2 Uniform and Non-Uniform Distributed Loading

For distributed transverse loads $w(x)$, the external virtual work integrates the product of load intensity and displacement function:

$$W_e = \int_{0}^{L} w(x) \delta(x) \, dx$$

Equating $W_e = W_i$ establishes the collapse load multiplier $\lambda$ for arbitrary load distributions.

4. Step-by-Step Solved Problem 1: Continuous Two-Span Steel Bridge Girder

Let us solve a practical problem for a continuous two-span highway bridge girder with non-uniform cross-sections.

            V = 200 kN              w = 40 kN/m
                |             vvvvvvvvvvvvvvvvvvvvvvv
   +------------+------------+-----------------------+
   A (Pinned)   E (midspan)  B (Interior)            C (Roller)
   <--- 6 m ---><--- 6 m ---><--------- 14 m -------->
   Span 1: L1 = 12 m         Span 2: L2 = 14 m
   Capacity: Mp1 = 360 kNm   Capacity: Mp2 = 480 kNm

4.1 Girder Geometry and Section Capacities

Consider the continuous girder $ABC$ supported on three piers:

• Span 1 ($AB$): Length $L_1 = 12.0\text{ m}$, carrying a concentrated load $V = 200\text{ kN}$ at midspan $E$. Section plastic capacity: $M_{p1} = 360\text{ kNm}$.

• Span 2 ($BC$): Length $L_2 = 14.0\text{ m}$, carrying a uniformly distributed load $w = 40\text{ kN/m}$. Section plastic capacity: $M_{p2} = 480\text{ kNm}$.

• Pier $B$: Continuous interior support. The smaller adjacent section capacity governs negative hinge resistance ($M_{p,B} = \min(M_{p1}, M_{p2}) = 360\text{ kNm}$).

We evaluate each span independently to find which span mechanism controls global collapse.

4.2 Collapse Mechanism in Span 1

Span 1 is propped continuous (pinned at $A$, continuous over $B$). A beam mechanism requires $N_h = 2$ plastic hinges (one at midspan $E$, one over pier $B$).

• Downward displacement at midspan $E$: $\Delta_1 = (L_1 / 2) \theta = 6.0 \theta$.

• Hinge rotation at $E$: $\theta_E = \theta + \theta = 2\theta$.

• Hinge rotation over pier $B$: $\theta_B = \theta$.

Virtual work balance:

$$W_e = \lambda V \Delta_1 = \lambda (200\text{ kN})(6.0\theta) = 1200 \lambda \theta$$

$$W_i = M_{p1} |\theta_E| + M_{p,B} |\theta_B| = (360)(2\theta) + (360)(\theta) = 720\theta + 360\theta = 1080\theta$$

Equating $W_e = W_i$:

$$1200 \lambda \theta = 1080 \theta \implies \lambda_1 = \frac{1080}{1200} = 0.900$$

4.3 Collapse Mechanism in Span 2

Span 2 carries distributed load $w = 40\text{ kN/m}$ over length $L_2 = 14.0\text{ m}$.
A propped continuous beam under UDL develops its span hinge at distance $x = 0.4142 L_2$ from the continuous support $B$:

• Span hinge position: $x = 0.4142(14.0) = 5.80\text{ m}$ from $B$.

• The collapse load for a propped cantilever under UDL is:
  $$w_c = 11.6568 \frac{M_{p2}}{L_2^2} = 11.6568 \frac{480\text{ kNm}}{(14.0\text{ m})^2} = 11.6568 \frac{480}{196} = 28.547\text{ kN/m}$$

The load multiplier for Span 2 is:

$$\lambda_2 = \frac{w_c}{w} = \frac{28.547\text{ kN/m}}{40.0\text{ kN/m}} = 0.7137 \approx 0.714$$

4.4 Controlling Global Collapse Load Factor

Comparing the collapse multipliers for both spans:

• Span 1 (Point Load Mechanism): $\lambda_1 = 0.900$.

• Span 2 (Distributed Load Mechanism): $\lambda_2 = 0.714$.

Span 2 controls global failure. The continuous bridge girder undergoes partial collapse in Span 2 at load factor $\lambda_c = 0.714$. Performing this the plastic collapse mechanism demonstrates that the distributed load in Span 2 precipitates structural failure well before Span 1 reaches its capacity.

+-------------------------------------------------------------------------------+
|                    TWO-SPAN GIRDER PLASTIC ANALYSIS SUMMARY                   |
+----------------------+-------------------+--------------------+---------------+
| Span Assessment      | Applied Loading   | Section Capacity   | Load Factor λ |
+----------------------+-------------------+--------------------+---------------+
| Span 1 ($AB$) Mode   | Point $V = 200\text{ kN}$| $M_{p1} = 360\text{ kNm}$| $\lambda = 0.900$     |
| Span 2 ($BC$) Mode * | UDL $w = 40\text{ kN/m}$| $M_{p2} = 480\text{ kNm}$| $\lambda = 0.714$ *   |
+----------------------+-------------------+--------------------+---------------+

* Controlling partial collapse mechanism for the continuous girder.

5. Step-by-Step Solved Problem 2: Fixed-Ended Beam Under Triangular Loading

Consider a fixed-ended beam carrying a linearly varying hydrostatic load increasing from zero at $A$ to $w_0$ at $B$.

• Span length $L$, plastic capacity $M_p$.

• Load function: $w(x) = w_0 (x / L)$.

• Critical sections: Support $A$, span location $x$, and support $B$.

Formulating the virtual work balance and minimizing with respect to hinge position $x$ reveals that the maximum span moment shifts toward the heavier load end ($x \approx 0.55 L$). The resulting collapse load factor is $w_{0,c} = 23.48 M_p / L^2$, demonstrating how load asymmetry shifts plastic hinge formation.

6. High Shear Force and Moment Interaction ($M-V$) at Support Hinges

In continuous beams carrying heavy point loads near supports, large shear forces $V_{Ed}$ coincide with plastic hogging moments $M_p$. Per Eurocode 3 (EN 1993-1-1 Section 6.2.8) and AISC 360-22 Chapter G:

• When design shear $V_{Ed} \le 0.5 V_{pl,Rd}$, shear has no detrimental effect on plastic moment capacity.

• When $V_{Ed} > 0.5 V_{pl,Rd}$, yielding of the web reduces the effective plastic section modulus:
  $$M_{V,Rd} = M_{pl,Rd} – \frac{\rho A_w^2}{4 t_w} f_y \le M_{pl,Rd}, \quad \text{where } \rho = \left( \frac{2 V_{Ed}}{V_{pl,Rd}} – 1 \right)^2$$

Engineers must verify shear-moment interaction at continuous support hinges to prevent premature web shear yielding.

AISC 360-22 Specification for Structural Steel Buildings
Eurocode 3 Design of Steel Structures EN 1993-1-1

7. Modern Design Code Provisions and Compactness Criteria

Design codes mandate strict rules when applying plastic beam analysis:

• AISC 360-22 (Appendix 1): Members developing plastic hinges must be compact Class 1 sections, satisfying flange slenderness $b / t \le 0.38 \sqrt{E / f_y}$ and web slenderness $h / t_w \le 3.76 \sqrt{E / f_y}$.

• Eurocode 3 (EN 1993-1-1): Enforces lateral torsional bracing spacing $L_m \le L_{cr}$ adjacent to plastic hinges to prevent out-of-plane lateral buckling before full mechanism mobilization.

• ACI 318-19 Building Code: Regulates moment redistribution in reinforced concrete continuous beams, permitting up to $20\%$ redistribution provided net tensile steel strain $\var\epsilon_t \ge 0.0075$.

8. Synthesis and Engineering Wrap-Up

Mastering the plastic collapse mechanism bridges the gap between isolated section strength and true system-level continuous beam resilience. By capturing plastic hinge rotation and post-yield moment redistribution, engineers optimize structural material efficiency while ensuring robust safety margins.

In continuous flexural frameworks, structural capacity is a collaborative mechanism. Carrying out a complete the plastic collapse mechanism ensures that every span is safely proportioned to resist extreme overload demands.

References & Standards Cited

1. AISC 360-22: Specification for Structural Steel Buildings, American Institute of Steel Construction, Chicago, IL, 2022.
2. EN 1993-1-1: Eurocode 3: Design of Steel Structures – Part 1-1: General Rules and Rules for Buildings, CEN, Brussels, 2020.
3. ACI 318-19: Building Code Requirements for Structural Concrete, American Concrete Institute, Farmington Hills, MI, 2019.
4. Horne, M. R.: Plastic Theory of Structures, 2nd Edition, Pergamon Press, Oxford, UK.
5. ASCE 7-22: Minimum Design Loads and Associated Criteria for Buildings and Other Structures, ASCE, Reston, VA, 2022.

Frequently Asked Questions (FAQ)

How many plastic hinges are required to collapse a single span of a continuous beam?

For a single span with statical indeterminacy $r_{text{span}}$, a complete beam mechanism requires $N_h = r_{text{span}} + 1$ plastic hinges. A propped continuous span requires 2 hinges (one over the interior support and one within the span), while a fixed-fixed span requires 3 hinges.

What is the difference between complete collapse and partial collapse?

A complete collapse mechanism involves the entire continuous structural framework failing simultaneously. A partial collapse mechanism occurs when plastic hinges form exclusively within a single span, transforming that individual span into an unstable mechanism while adjacent spans remain stable.

How does high shear force affect plastic hinge formation?

When concurrent shear force exceeds $50%$ of the plastic shear capacity ($V_{Ed} > 0.5 V_{pl,Rd}$), shear yielding in the web reduces the available flexural yield capacity, lowering the effective plastic moment capacity $M_p$.

Can plastic beam analysis be applied to non-uniform cross-sections?

Yes. When beams feature stepped flanges or cover plates, the virtual work equation assigns each specific plastic moment capacity $M_{p,j}$ to its corresponding hinge rotation $theta_j$.

Why do design codes limit moment redistribution in concrete beams?

Reinforced concrete possesses limited rotational ductility compared to structural steel. If steel reinforcement strains are low or concrete crushes prematurely in compression, hinges cannot rotate sufficiently to achieve full theoretical moment redistribution.