Collapse Load Analysis of Propped Cantilever and Fixed Beams (2026 Guide)

1. Introduction to Collapse Load Analysis in Structural Beams

Determining the true ultimate capacity of statically indeterminate beams is a foundational requirement in structural engineering. A classical elastic evaluation restricts the allowable capacity to the moment causing initial yield at the most critical fiber. However, ductile steel and reinforced concrete members possess substantial reserve strength. Performing a rigorous collapse load analysis enables structural engineers to calculate the exact load factor causing catastrophic structural failure through plastic mechanism formation.

Applying collapse load analysis to standard indeterminate elements—such as propped cantilevers and fixed-ended beams—demonstrates how internal bending moments redistribute from heavily stressed support regions to underutilized span regions. By evaluating the equilibrium of internal plastic moments and kinematics of plastic hinges, engineers unlock the true ultimate load-carrying capacity of the structure.

Whether evaluating industrial crane runways, continuous bridge decks, or offshore platform girders, conducting a thorough collapse load analysis ensures that structural designs are both economically optimized and robustly protected against catastrophic overload.

2. Governing Principles of Beam Plastic Collapse

The structural mechanics governing beam collapse depend on material ductility, cross-sectional compactness, and the degree of static indeterminacy.

2.1 Static Indeterminacy and Required Plastic Hinges ($N_h = R + 1$)

For any statically indeterminate beam with degree of redundancy $R$, the formation of a single plastic hinge reduces the redundancy by 1. To transform the structure into a single-degree-of-freedom plastic collapse mechanism, the required number of active plastic hinges $N_h$ is:

$$N_h = R + 1$$

• Simply Supported Beam ($R = 0$): Requires $N_h = 0 + 1 = 1$ plastic hinge (at mid-span or load point) to collapse.

• Propped Cantilever ($R = 1$): Requires $N_h = 1 + 1 = 2$ plastic hinges (one at fixed support, one within the span).

• Fixed-Ended Beam ($R = 2$ for symmetrical loads): Requires $N_h = 2 + 1 = 3$ plastic hinges (two at end supports, one within the span).

2.2 Elastic First Yield vs Ultimate Plastic Collapse

In elastic design, the first-yield load $W_y$ is governed by the elastic section modulus $S$ and yield strength $\sigma_y$: $M_y = S \sigma_y$. In plastic limit design, the ultimate capacity $W_c$ is governed by the plastic section modulus $Z_p$ ($M_p = Z_p \sigma_y$) and structural moment redistribution.

The overall load factor margin is expressed as:

$$\text{Total Reserve Ratio} = \frac{W_c}{W_y} = \left( \frac{M_p}{M_y} \right) \times \left( \frac{\text{Plastic Redistribution Factor}}{\text{Elastic Factor}} \right) = S_f \times K_{\text{ redist}}$$

where $S_f$ is the section shape factor.

3. Propped Cantilever Beams: Rigorous Collapse Load Formulations

A propped cantilever of span $L$ is fixed at support $A$ and roller-supported at $B$. The degree of indeterminacy is $R = 1$. Consequently, an ultimate collapse load is reached when exactly $N_h = 2$ plastic hinges develop.

    A                                     D                         B
    |=====================================*=========================O
    ///                                 P \/                      /\
    <----------------- a -----------------><---------- b ----------->
    <------------------------------ L ------------------------------>

3.1 Case 1: Point Load at Arbitrary Location ($a, b$)

Let a concentrated load $P$ be placed at distance $a$ from fixed end $A$ and distance $b$ from propped end $B$ ($a + b = L$).

Kinematic Mechanism Formulation:

Plastic hinges form at fixed support $A$ and directly beneath load point $D$. Roller $B$ undergoes free rotation without resisting moment ($M_B = 0$).

1. Let the downward virtual displacement under load $P$ be $\delta$.
2. The virtual rotation of segment $AD$ is $\theta_1 = \delta / a$.
3. The virtual rotation of segment $DB$ is $\theta_2 = \delta / b$.
4. Hinge rotation at support $A$: $\theta_A = \theta_1 = \delta / a$.
5. Relative hinge rotation at load point $D$: $\theta_D = \theta_1 + \theta_2 = \delta \left( \frac{1}{a} + \frac{1}{b} \right) = \delta \left( \frac{a+b}{ab} \right) = \frac{L \delta}{ab}$.

Equating External and Internal Virtual Work:

$$W_{ext} = P \cdot \delta$$

$$W_{int} = M_p \theta_A + M_p \theta_D = M_p \left( \frac{\delta}{a} \right) + M_p \left( \frac{L \delta}{ab} \right) = M_p \delta \left( \frac{b + L}{ab} \right)$$

Setting $W_{ext} = W_{int}$:

$$P \cdot \delta = M_p \delta \left( \frac{L + b}{ab} \right) \implies P_c = \frac{M_p (L + b)}{ab}$$

For a symmetrical mid-span load ($a = b = L/2$):

$$P_c = \frac{M_p (L + L/2)}{(L/2)(L/2)} = \frac{M_p (1.5 L)}{0.25 L^2} = \frac{6 M_p}{L}$$

In a collapse load analysis, the propped cantilever supports $P_c = 6.00 M_p / L$, compared to the elastic first yield load of $P_y = \frac{32 M_y}{11 L} \approx 2.91 M_y / L$.

3.2 Case 2: Uniformly Distributed Load (UDL) and Exact Hinge Location

Under a uniformly distributed load $w$ (force per unit length), the internal plastic hinge does not form at mid-span because shear force distribution shifts the maximum moment toward the propped end.

             w (kN/m)
    ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
    A                                 D                        B
    |=================================*========================O
    ///                             (x_0)                     /\
    <-------------------------- L ----------------------------->

Let the positive plastic hinge form at distance $x_0$ from propped end $B$ ($0 < x_0 < L$).

1. Hinge rotation at $A$: $\theta_A = \theta \cdot \left(\frac{x_0}{L – x_0}\right)$
2. Hinge rotation at $D$: $\theta_D = \theta + \theta_A = \theta \left( 1 + \frac{x_0}{L – x_0} \right) = \theta \left( \frac{L}{L – x_0} \right)$
3. Virtual displacement at $D$: $\delta = x_0 \theta$.
4. External virtual work done by UDL $w$:
   $$W_{ext} = w \times \text{Area of displacement triangle} = w \cdot \left( \frac{1}{2} L \delta \right) = \frac{1}{2} w L x_0 \theta$$
5. Internal plastic work:
   $$W_{int} = M_p \theta_A + M_p \theta_D = M_p \theta \left( \frac{x_0 + L}{L – x_0} \right)$$

Equating $W_{ext} = W_{int}$:

$$\frac{1}{2} w L x_0 \theta = M_p \theta \left( \frac{L + x_0}{L – x_0} \right) \implies w = \frac{2 M_p (L + x_0)}{L x_0 (L – x_0)}$$

By the upper bound theorem, the true collapse state minimizes $w$ with respect to $x_0$:

$$\frac{dw}{dx_0} = 0 \implies \frac{d}{dx_0} \left[ \frac{L + x_0}{L x_0 – x_0^2} \right] = 0$$

Executing the derivative:

$$(1)(L x_0 – x_0^2) – (L + x_0)(L – 2 x_0) = 0$$

$$L x_0 – x_0^2 – (L^2 – 2 L x_0 + L x_0 – 2 x_0^2) = 0 \implies x_0^2 + 2 L x_0 – L^2 = 0$$

Solving this quadratic equation for $x_0 > 0$:

$$x_0 = \frac{-2L \pm \sqrt{4L^2 – 4(1)(-L^2)}}{2} = \frac{-2L + \sqrt{8} L}{2} = (\sqrt{2} – 1) L \approx 0.4142 L$$

Substituting $x_0 = 0.4142 L$ into the collapse load expression:

$$w_c = \frac{2 M_p (L + 0.4142 L)}{L (0.4142 L)(0.5858 L)} = \frac{2.8284 M_p}{0.2426 L^2} = (6 + 4\sqrt{2}) \frac{M_p}{L^2} \approx 11.6568 \frac{M_p}{L^2}$$

$$w_c = 11.657 \frac{M_p}{L^2}$$

Read the AISC Specification for Structural Steel Buildings (AISC 360-22)

4. Fixed-Ended (Encastre) Beams: Comprehensive Analysis

A beam fixed at both ends $A$ and $B$ possesses a static indeterminacy of $R = 2$ for transverse flexure. Collapse occurs when $N_h = R + 1 = 3$ fixed beam plastic hinges form.

    A                                 C                                 B
    |=================================*=================================|
    ///                               P                               ///
    <------------------------------ L ---------------------------------->

4.1 Case 3: Fixed Beam Under Central Point Load

For a fixed beam of span $L$ and constant capacity $M_p$ subjected to a mid-span point load $P$:

1. Hinges form at fixed support $A$, fixed support $B$, and mid-span $C$.
2. Kinematics: $\theta_A = \theta$, $\theta_B = \theta$, $\theta_C = 2\theta$, $\delta = (L/2)\theta$.
3. External work: $W_{ext} = P \delta = P (L/2)\theta$.
4. Internal work: $W_{int} = M_p(\theta + 2\theta + \theta) = 4 M_p \theta$.

$$P \left(\frac{L}{2}\right) \theta = 4 M_p \theta \implies P_c = \frac{8 M_p}{L}$$

Elastic first yield occurs at $P_y = \frac{8 M_y}{L}$. The collapse load ratio is $P_c / P_y = M_p / M_y = S_f$.

4.2 Case 4: Fixed Beam Under Off-Center Point Load

Let concentrated load $P$ be applied at distances $a$ and $b$ from supports $A$ and $B$ ($a + b = L$):

Hinge rotations:

- Support A: \theta_A = \delta / a

- Support B: \theta_B = \delta / b

- Load Point: \theta_C = \delta(1/a + 1/b) = L\delta / (ab)

$$W_{ext} = P \cdot \delta$$

$$W_{int} = M_p \left( \frac{\delta}{a} \right) + M_p \left( \frac{\delta}{b} \right) + M_p \left( \frac{L \delta}{ab} \right) = M_p \delta \left( \frac{b + a + L}{ab} \right) = \frac{2 L M_p \delta}{ab}$$

$$P \cdot \delta = \frac{2 L M_p \delta}{ab} \implies P_c = \frac{2 L M_p}{a b}$$

When $a = b = L/2$, $P_c = \frac{2 L M_p}{(L/2)^2} = \frac{8 M_p}{L}$, matching the symmetric case.

4.3 Case 5: Fixed Beam Under Uniformly Distributed Load (UDL)

Under a uniform load $w$ per unit length over span $L$:

1. By symmetry, plastic hinges form at support $A$, support $B$, and span centerline $x = L/2$.
2. Virtual displacement profile: Symmetric triangle with peak deflection $\delta = (L/2)\theta$.
3. External work:
   $$W_{ext} = w \cdot \left( \frac{1}{2} L \delta \right) = \frac{w L^2 \theta}{4}$$
4. Internal work:
   $$W_{int} = M_p(\theta + 2\theta + \theta) = 4 M_p \theta$$

$$\frac{w L^2 \theta}{4} = 4 M_p \theta \implies w_c = \frac{16 M_p}{L^2}$$

Total collapse load on the span is $W_c = w_c L = \frac{16 M_p}{L}$.

5. Master Design Comparison: Elastic vs Plastic Limits

The following synthesis table highlights the comparative capacity increases obtained across indeterminate beam configurations through collapse load analysis.

6. Step-by-Step Practical Calculation Example: Non-Symmetric Fixed Girder

Consider an industrial crane bridge girder modeled as a fixed-ended beam with span $L = 12.0\text{ m}$. Due to haunched support details, the plastic moment capacities are non-uniform:

• Left Support $A$: $M_{pA} = 450\text{ kN}\cdot\text{ m}$

• Right Support $B$: $M_{pB} = 350\text{ kN}\cdot\text{ m}$

• Span Section: $M_{pS} = 300\text{ kN}\cdot\text{ m}$

The girder carries an asymmetric wheel load $P$ located at $a = 4.0\text{ m}$ from left support $A$ ($b = 8.0\text{ m}$).

    A (MpA = 450)                 D (Load P)                   B (MpB = 350)
    |=============================*=============================|
    ///                         P \/                          ///
    <----------- 4.0 m ----------><------------- 8.0 m --------->

Step 1: Establish Kinematic Mechanism

Let the downward displacement under load $P$ at $D$ be $\delta$.

• Rotation of segment $AD$: $\theta_1 = \delta / 4.0 = 0.250 \delta$

• Rotation of segment $DB$: $\theta_2 = \delta / 8.0 = 0.125 \delta$

• Support $A$ hinge rotation: $\theta_A = \theta_1 = 0.250 \delta$

• Support $B$ hinge rotation: $\theta_B = \theta_2 = 0.125 \delta$

• Load point $D$ hinge rotation: $\theta_D = \theta_1 + \theta_2 = (0.250 + 0.125) \delta = 0.375 \delta$

Step 2: Formulate Virtual Work Equation

1. External Virtual Work:
   $$W_{ext} = P \cdot \delta$$

2. Internal Plastic Work:
   $$W_{int} = M_{pA} |\theta_A| + M_{pS} |\theta_D| + M_{pB} |\theta_B|$$
   $$W_{int} = (450)(0.250 \delta) + (300)(0.375 \delta) + (350)(0.125 \delta)$$
   $$W_{int} = (112.5 + 112.5 + 43.75) \delta = 268.75 \delta\text{ kN}\cdot\text{ m}$$

Step 3: Compute Ultimate Collapse Load

$$W_{ext} = W_{int} \implies P_c \cdot \delta = 268.75 \delta \implies P_c = 268.75\text{ kN}$$

The exact ultimate collapse load for the non-uniform girder is $P_c = 268.8\text{ kN}$.

7. Structural Code Compliance and Deflection Considerations

Modern engineering design codes permit collapse load analysis subject to serviceability and stability verifications.

• AISC 360-22 (Appendix 1): Inelastic analysis is permitted when lateral bracing satisfies stability requirements and cross-sections meet compactness criteria for high-ductility members.

• Eurocode 3 (EN 1993-1-1 Section 5.4.3): Limits plastic global analysis to Class 1 sections with verified rotational ductility.

• Serviceability Limit State (SLS): Although plastic analysis predicts ultimate collapse, deflections under service loads must still be verified using standard elastic deflection formulas to prevent excessive sagging or architectural cracking.

Consult Eurocode 3 Steel Design Guides at European Standards

8. Synthesis and Engineering Wrap-Up

Executing a rigorous collapse load analysis reveals the true load-carrying potential of indeterminate beams. By accounting for the progressive formation of plastic hinges and internal moment redistribution, structural engineers transition beyond conservative elastic bounds to achieve optimized, high-performance structural systems. Understanding both static equilibrium and kinematic mechanisms ensures that the collapse limit state is accurately predicted and safely controlled in engineering practice.

References & Standards Cited

1. AISC (2022). Specification for Structural Steel Buildings (ANSI/AISC 360-22), American Institute of Steel Construction, Chicago, IL.
2. CEN (2005). Eurocode 3: Design of steel structures — Part 1-1: General rules and rules for buildings (EN 1993-1-1), European Committee for Standardization, Brussels.
3. Horne, M. R. (1979). Plastic Theory of Structures, 2nd Edition, Pergamon Press, Oxford.
4. Baker, J. F., & Heyman, J. (1969). Plastic Design of Frames: Volume 1, Fundamentals, Cambridge University Press.
5. Neal, B. G. (1977). The Plastic Methods of Structural Analysis, 3rd Edition, Chapman and Hall, London.
6. ASCE (2022). Minimum Design Loads and Associated Criteria for Buildings and Other Structures (ASCE/SEI 7-22), American Society of Civil Engineers, Reston, VA.

Frequently Asked Questions (FAQ)

Why does a propped cantilever under UDL form its span plastic hinge at $0.414L$ instead of mid-span?

Under a uniformly distributed load, the bending moment distribution in a propped cantilever is asymmetric due to the rotational restraint at the fixed support. The point of zero shear force (which corresponds mathematically to the point of maximum span bending moment) shifts toward the simple support, occurring precisely at $(sqrt{2}-1)L approx 0.4142L$ from the propped end.

Can plastic collapse occur if only support hinges form without a span hinge?

No. Forming hinges solely at the supports reduces the degree of indeterminacy to zero (transforming a fixed beam into a simply supported beam). However, a simply supported beam remains a stable, statically determinate structure. An additional plastic hinge within the span is required to transform the beam into a kinematic mechanism with one degree of freedom.

How does shear force affect the ultimate plastic collapse load?

High coincident shear forces reduce the plastic moment capacity of a cross-section due to plastic yield surface interaction (Von Mises yield criterion). Under Eurocode 3 and AISC 360, when the applied shear force $V_{Ed}$ exceeds $50%$ of the plastic shear resistance $V_{pl,Rd}$, the effective plastic moment capacity $M_{y,V,Rd}$ is reduced by a shear-moment interaction factor.

What is the difference between complete and partial collapse in continuous beams?

In a multi-span continuous beam, **partial collapse** occurs when sufficient plastic hinges form within a single span (or sub-assemblage) to cause localized collapse while adjacent spans remain stable. **Complete collapse** occurs when hinges develop across all spans simultaneously under proportional loading.

Why is plastic analysis more advantageous for fixed beams under UDL than under point loads?

For a fixed beam under UDL, the elastic distribution gives $M_{support} = w L^2 / 12$ and $M_{span} = w L^2 / 24$ (a $2:1$ ratio). Plastic redistribution equalizes these moments to $M_p = w L^2 / 16$, yielding a substantial $33%$ increase in load capacity. For a central point load, elastic moments are already equal ($P L / 8$), meaning the plastic capacity increase comes purely from the section shape factor without additional redistribution benefit.