Plastic Analysis of Beams: 4 Collapse Mechanisms and Solved Examples (2026)

1. Introduction to Plastic Analysis of Beams

In structural design, elastic beam analysis calculates the bending moment field up to the instant the most heavily stressed fiber reaches yield stress. However, statically indeterminate ductile steel and reinforced concrete beams possess substantial additional load-carrying capacity beyond this initial elastic boundary.

Through the plastic analysis of beams, structural engineers evaluate the true ultimate collapse load $P_c$ or $w_c$. As high-moment regions yield, cross-sections transform into localized ductile plastic hinges. These hinges rotate under a constant plastic moment $M_p$, shedding excess load to less-stressed regions of the span until a kinematic failure mechanism forms.

2.3 Complete, Partial, and Over-Complete Collapse Mechanisms

• Complete Collapse: The entire beam participates in the kinematic mechanism, and the full system becomes statically determinate at incipient collapse ($N = r + 1$).

• Partial Collapse: A local sub-span forms a mechanism with fewer than $r + 1$ hinges ($N < r + 1$), while adjacent spans remain rigid and indeterminate.

• Over-Complete Collapse: Two or more independent collapse mechanisms develop simultaneously under identical collapse loads.

3. Virtual Work Method (Kinematic Approach) for Beam Collapse

3.1 Principle of Virtual Displacements and Virtual Velocities

The kinematic method evaluates beam collapse by postulating an admissible failure mechanism and applying the Principle of Virtual Work. For an assumed kinematically compatible displacement field characterized by small virtual hinge rotations $\theta_j$ and transverse displacements $\Delta_i$:

$$W_E = W_I$$

Where:

• $W_E$ = External work done by applied design loads.

• $W_I$ = Internal plastic energy dissipated by rotating plastic hinges.

3.2 Internal Work Dissipation vs. External Work Done

Internal energy dissipation $W_I$ is the sum of products of the fully plastic moment $M_p$ and absolute plastic rotation angle $\theta_j$ at each plastic hinge:

$$W_I = \sum_{j=1}^{N} M_{p,j} \cdot |\theta_j|$$

Because plastic work is strictly dissipative, internal work is always positive ($W_I > 0$).

External work done $W_E$ by concentrated loads $P_i$ and distributed loads $w(x)$ equals:

$$W_E = \sum_{i} P_i \cdot \Delta_i + \int_{0}^{L} w(x) \cdot \delta(x) \, dx$$

Equating $W_E = W_I$ gives the kinematic collapse load $\lambda_k$. By the Upper Bound Theorem, this value is greater than or equal to the exact collapse load $\lambda_c$:

$$\lambda_k \ge \lambda_c$$

3.3 Locating Critical Intermediate Plastic Hinges under Distributed Loads

Under concentrated point loads, plastic hinges form directly beneath load points. Under uniformly distributed loads (UDL), the internal bending moment field varies parabolically. The intermediate plastic hinge does not form at midspan; it forms at the section of maximum bending moment where shear force equals zero ($V = \frac{dM}{dx} = 0$).

Engineers find the critical hinge position $x_{cr}$ by expressing the collapse load parameter $w(x)$ as a function of hinge position $x$ and minimizing $w(x)$:

$$\frac{dw}{dx} = 0$$

4. Moment Redistribution and Reserve Capacity Beyond Elastic Limits

In elastic analysis, indeterminate beams reach peak moments at stiff interior supports first. In a fixed-ended beam under UDL:

• Elastic Support Moment: $M_{supp} = \frac{w L^2}{12}$

• Elastic Midspan Moment: $M_{span} = \frac{w L^2}{24}$

When $w$ increases, the fixed supports reach $M_p$ first at load:

$$w_y = \frac{12 M_p}{L^2}$$

In elastic design, loading stops here. In plastic analysis of beams, support hinges continue to rotate, holding constant moment $M_p$. Further load increments transfer entirely to the midspan region until midspan also reaches $M_p$:

$$w_c = \frac{16 M_p}{L^2}$$

Plastic analysis unlocks $33.3\%$ additional load capacity by accounting for full moment redistribution.

5. Step-by-Step Solved Engineering Examples

5.1 Example 1: Propped Cantilever Beam Under Uniformly Distributed Load (UDL)

 Fixed End A                                              Roller B
 |============================================================^
 |<------------------------- Length L ----------------------->|
             Uniformly Distributed Load w (kN/m)

Problem Statement:

A propped cantilever beam of span $L = 6.0 \text{ m}$ carries a uniformly distributed load $w$ ($\text{kN/m}$). The beam has constant fully plastic moment capacity $M_p = 120 \text{ kN}\cdot\text{ m}$.
1. Derive the exact location of the span plastic hinge $x$.
2. Calculate the exact ultimate plastic collapse load $w_c$.
3. Calculate the elastic first-yield load $w_y$ and determine the percentage reserve capacity.

 Assumed Collapse Mechanism:
 Hinge 1 (Fixed End A)                 Hinge 2 (Distance x from A)
    (θ)                                   (θ + β)
 |---|---------------------------------------o------------------------^
     \                                      /
      \                                    /
       \                                  /
        \________________________________/ Δ = x*θ = (L - x)*β

Step 1: Kinematic Kinematics Formulation

Let the intermediate plastic hinge form at distance $x$ from the fixed support $A$.
Under a virtual rotation $\theta$ at fixed support $A$:

• Transverse deflection at intermediate hinge: $\Delta = x \cdot \theta$

• Rotation angle at simple support $B$: $\beta = \frac{\Delta}{L – x} = \frac{x \theta}{L – x}$

• Total rotation at intermediate hinge 2: $\theta_2 = \theta + \beta = \theta \left( 1 + \frac{x}{L – x} \right) = \theta \left( \frac{L}{L – x} \right)$

Step 2: Internal Work Dissipation ($W_I$)

Plastic hinges form at fixed support $A$ (rotation $\theta$) and span location $x$ (rotation $\theta + \beta$):

$$W_I = M_p \cdot \theta + M_p \cdot (\theta + \beta) = M_p \cdot \theta \left[ 1 + \frac{L}{L – x} \right] = M_p \cdot \theta \left[ \frac{2L – x}{L – x} \right]$$

Step 3: External Work Done ($W_E$)

The total area under the triangular virtual deflection diagram:

$$\text{ Area} = \frac{1}{2} \cdot L \cdot \Delta = \frac{1}{2} L x \theta$$

$$W_E = \int_0^L w \cdot \delta(x) \, dx = w \cdot \text{ Area} = \frac{1}{2} w L x \theta$$

Step 4: Equate External and Internal Work

$$\frac{1}{2} w L x \theta = M_p \theta \left( \frac{2L – x}{L – x} \right)$$

Canceling $\theta$ and solving for $w$:

$$w = \frac{2 M_p (2L – x)}{L x (L – x)} = \frac{2 M_p}{L} \left[ \frac{2L – x}{L x – x^2} \right]$$

Step 5: Minimize $w$ to Find Critical Hinge Location ($x_{cr}$)

Setting $\frac{dw}{dx} = 0$:

$$\frac{d}{dx} \left[ \frac{2L – x}{Lx – x^2} \right] = \frac{(-1)(Lx – x^2) – (2L – x)(L – 2x)}{(Lx – x^2)^2} = 0$$

$$-(Lx – x^2) – (2L^2 – 4Lx – Lx + 2x^2) = 0$$

$$-Lx + x^2 – 2L^2 + 5Lx – 2x^2 = 0 \implies -x^2 + 4Lx – 2L^2 = 0$$

$$x^2 – 4Lx + 2L^2 = 0$$

Applying the quadratic formula:

$$x = \frac{4L \pm \sqrt{(4L)^2 – 4(1)(2L^2)}}{2} = \frac{4L \pm \sqrt{16L^2 – 8L^2}}{2} = \frac{4L \pm 2\sqrt{2}L}{2} = (2 – \sqrt{2})L$$

$$x_{cr} = (2 – 1.414214)L = 0.585786 L \approx 0.414214 L \text{ from roller support B}$$

Measured from fixed support $A$:
$$x_{cr} = 0.585786 \times 6.0 \text{ m} = 3.5147 \text{ m}$$

Step 6: Compute Exact Ultimate Collapse Load ($w_c$)

Substituting $x_{cr} = (2 – \sqrt{2})L$ into the collapse equation:

$$w_c = (6 + 4\sqrt{2}) \frac{M_p}{L^2} \approx 11.65685 \frac{M_p}{L^2}$$

Substituting numerical parameters ($M_p = 120 \text{ kN}\cdot\text{ m}$, $L = 6.0 \text{ m}$):

$$w_c = 11.65685 \times \frac{120}{(6.0)^2} = 11.65685 \times 3.3333 = 38.856 \text{ kN/m}$$

Step 7: Compare Against Elastic Yield Load ($w_y$)

In elastic theory, peak support moment is $M_A = \frac{w L^2}{8}$:
$$w_y = \frac{8 M_p}{L^2} = \frac{8 \times 120}{36} = 26.667 \text{ kN/m}$$

$$\text{Plastic Reserve Capacity} = \frac{w_c – w_y}{w_y} \times 100 = \frac{38.856 – 26.667}{26.667} \times 100 = +45.71\%$$

5.2 Example 2: Two-Span Continuous Steel Girder Under Concentrated Loads

      P1 = 1.5 P                           P2 = P
          |                                  |
          v                                  v
 A +--------------+--------------+ B +-------+-------+ C
 ^   L1/2 = 3.0 m   L1/2 = 3.0 m   ^   L2/2=4m L2/2=4m ^
 |<--------- Span 1 = 6.0 m ------>|<-- Span 2 = 8.0 m ->|
 Plastic Moment Capacity = Mp = 180 kN·m throughout

Problem Statement:

A continuous two-span beam $ABC$ rests on simple supports at $A$, $B$, and $C$.

• Span 1: $L_1 = 6.0 \text{ m}$ carrying midspan point load $P_1 = 1.5 P$

• Span 2: $L_2 = 8.0 \text{ m}$ carrying midspan point load $P_2 = 1.0 P$

• Constant plastic moment capacity $M_p = 180 \text{ kN}\cdot\text{ m}$.

• Determine the governing collapse mechanism, collapse load $P_c$, and support reactions.

 Mechanism 1: Collapse of Span 1 (Span 2 remains rigid)
    A (0)          Midspan 1 (2θ)         B (θ)               C (0)
    +---------------------o-----------------+-----------------+
    \                    /
     \                  /
      \________________/ Δ1 = 3θ

 Mechanism 2: Collapse of Span 2 (Span 1 remains rigid)
    A (0)               B (θ)          Midspan 2 (2θ)         C (0)
    +-------------------+---------------------o---------------+
                        \                    /
                         \                  /
                          \________________/ Δ2 = 4θ

Step 1: Evaluate Mechanism 1 (Span 1 Collapse)

Plastic hinges form at midspan 1 and intermediate support $B$.

• Virtual displacement at load $P_1$: $\Delta_1 = 3.0 \cdot \theta$

• Hinge rotation at support $B$: $\theta_B = \theta$

• Hinge rotation at midspan 1: $\theta_1 = \theta + \theta = 2\theta$

• Internal work: $W_I = M_p \cdot (2\theta) + M_p \cdot (\theta) = 3 M_p \theta$

• External work: $W_E = P_1 \cdot \Delta_1 = (1.5 P) \cdot (3.0 \theta) = 4.5 P \theta$

Equating $W_E = W_I$:
$$4.5 P \theta = 3 M_p \theta \implies P_1 = \frac{3 M_p}{4.5} = \frac{2}{3} M_p$$

$$P^{(1)} = \frac{2}{3} \times \frac{180 \text{ kN}\cdot\text{ m}}{1.0} = 120.0 \text{ kN}$$

Step 2: Evaluate Mechanism 2 (Span 2 Collapse)

Plastic hinges form at intermediate support $B$ and midspan 2.

• Virtual displacement at load $P_2$: $\Delta_2 = 4.0 \cdot \theta$

• Hinge rotation at support $B$: $\theta_B = \theta$

• Hinge rotation at midspan 2: $\theta_2 = \theta + \theta = 2\theta$

• Internal work: $W_I = M_p \cdot (\theta) + M_p \cdot (2\theta) = 3 M_p \theta$

• External work: $W_E = P_2 \cdot \Delta_2 = (1.0 P) \cdot (4.0 \theta) = 4.0 P \theta$

Equating $W_E = W_I$:
$$4.0 P \theta = 3 M_p \theta \implies P^{(2)} = \frac{3 M_p}{4.0} = 0.75 M_p = 0.75 \times 180 = 135.0 \text{ kN}$$

Step 3: Identify Governing Collapse Load ($P_c$)

By the Kinematic Theorem, the true collapse load is the minimum of all kinematically admissible mechanisms:

$$P_c = \min(P^{(1)}, P^{(2)}) = \min(120.0 \text{ kN}, 135.0 \text{ kN}) = 120.0 \text{ kN}$$

Collapse occurs strictly in Span 1 at $P_c = 120.0 \text{ kN}$ ($P_1 = 180.0 \text{ kN}$), while Span 2 remains elastic and uncollapsed.

6. Practical Structural Code Applications (AISC 360-22 and Eurocode 3)

Design specifications regulate the use of plastic analysis of beams:

• AISC 360-22 Appendix 1 (Design by Inelastic Analysis): Permits plastic analysis provided cross-sections satisfy Class 1 (compact) web and flange slenderness requirements and unbraced lengths satisfy:
  $$L_b \le L_{pd} = \left[ 0.12 + 0.076 \left(\frac{M_1}{M_2}\right) \right] \left(\frac{E}{F_y}\right) r_y$$

• Eurocode 3 (EN 1993-1-1 Section 5.4.3): Mandates Class 1 sections for plastic global analysis and requires adequate rotation capacity at plastic hinges ($\theta_{Cd} \ge \theta_{Ed}$).

• Moment Redistribution Limits (ACI 318-19): For reinforced concrete continuous beams, ACI 318 allows up to $20\%$ redistribution of negative moments at supports, provided net tensile strain $\var\epsilon_t \ge 0.0075$.

7. Core Engineering Conclusions on Beam Plastic Analysis

The plastic analysis of beams provides a direct approach to determining the true collapse capacity of indeterminate flexural members. By idealizing material response into rigid-plastic hinges, engineers use the Principle of Virtual Work to bracket collapse loads.

Evaluating competing mechanisms confirms whether full or partial span collapse governs. Adhering to AISC 360-22 and Eurocode 3 lateral bracing and compact section limits ensures steel beams achieve complete plastic redistribution without premature instability.

References & Standards Cited

1. American Institute of Steel Construction (AISC). (2022). Specification for Structural Steel Buildings (ANSI/AISC 360-22). Chicago, IL: AISC.
2. American Concrete Institute (ACI). (2019). Building Code Requirements for Structural Concrete (ACI 318-19). Farmington Hills, MI: ACI.
3. Baker, J. F., Horne, M. R., & Heyman, J. (1956). The Steel Skeleton (Vol. 2: Plastic Behaviour and Design). Cambridge: Cambridge University Press.
4. British Standards Institution / European Committee for Standardization. (2005). Eurocode 3: Design of Steel Structures – Part 1-1 (EN 1993-1-1). Brussels: CEN.
5. Horne, M. R. (1979). Plastic Theory of Structures (2nd Edition). Oxford: Pergamon Press.
6. Neal, B. G. (1977). The Plastic Methods of Structural Analysis (3rd Edition). London: Chapman and Hall.

Frequently Asked Questions (FAQ)

What is plastic analysis of beams?

Plastic analysis of beams determines the ultimate collapse load of flexural members by modeling ductile plastic hinges that sustain constant plastic moment capacity ($M_p$) while redistributing moments to form a kinematic collapse mechanism.

How many plastic hinges are needed for a beam to collapse?

For a beam with degree of statical indeterminacy $r$, complete collapse requires $N = r + 1$ plastic hinges. Partial collapse can occur in single spans of continuous beams with fewer total hinges.

Why do intermediate hinges not form at midspan under uniformly distributed loads?

Under distributed loading, the moment diagram is parabolic and asymmetric in propped or fixed spans. The maximum moment and zero shear point occur offset from midspan (e.g., at $0.414L$ from the simple support in a propped cantilever).

What is the difference between elastic first yield and plastic collapse load?

Elastic first yield occurs when the single extreme fiber reaches yield stress ($sigma_y$). Plastic collapse occurs only after every fiber across multiple critical cross-sections yields completely, unlocking $20% - 50%$ additional reserve capacity.

What cross-section requirements are necessary to perform plastic beam analysis?

Beams must satisfy Class 1 (compact) slenderness criteria under AISC 360 or Eurocode 3 and maintain adequate lateral-torsional bracing to prevent premature local or global buckling prior to full mechanism formation.