Minimum Weight Plastic Design: Foulkes Optimization Guide (2026)

1. Introduction to Structural Optimization and Plastic Frame Economy

In structural steel design, minimizing total material weight while satisfying ultimate load-carrying capacity is a central engineering objective. Traditional elastic design relies on iterative sizing to ensure that extreme fiber stresses do not exceed allowable limits. However, minimum weight plastic design optimizes member cross-sections by directly leveraging plastic moment redistribution at the collapse limit state.

                  THE OPTIMIZATION DESIGN CONVERGENCE
       [ External Design Actions (Loads P, H) ]
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                          v
       [ Postulate Kinematic Mechanisms (Upper Bound) ]
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                          v
       [ Formulate Linear Constraints (Mp_beam, Mp_col) ]
                          |
                          v
       [ Apply Foulkes Theorem: Minimize Weight Function W ]
                          |
                          v
       [ Optimal Section Allocation (Zero Excess Steel) ]

Applying minimum weight plastic design allows engineers to determine the exact distribution of plastic moment capacities $M_{p,i}$ among beams, columns, and rafters that minimizes the structural weight objective function. Through optimal structural plastic design, structural engineers systematically eliminate excess material while guaranteeing stability under ultimate factored design actions.

By incorporating foulkes theorem, designers achieve an analytical bridge between virtual work kinematics and mathematical linear programming.

2. Mathematical Formulation of the Minimum Weight Objective Function

2.1 Linear Weight-Capacity Relationship

To formulate weight minimization frames mathematically, structural mechanics establishes a relationship between the weight per unit length $w_i$ of a structural steel member and its plastic moment capacity $M_{p,i}$.

Empirical and analytical data for standard hot-rolled structural steel sections (such as AISC W-shapes and European IPE/HEA sections) demonstrate that within practical design ranges, the cross-sectional area $A$ (and therefore weight per unit length $w = ho A$) is approximately linearly proportional to the plastic section modulus $Z$ (and plastic moment capacity $M_p = Z f_y$):

$$w_i \approx k_0 + k_1 M_{p,i}$$

For a given steel grade with constant yield strength $f_y$, the variable portion of the total structural weight $\mathcal{W}$ is directly proportional to the product of member length $L_i$ and plastic moment capacity $M_{p,i}$.

2.2 Objective Function Formulation for Framed Structures

For a structural frame consisting of $n$ member groups, where each member group $i$ has total length $L_i$ and is assigned a uniform plastic moment capacity $M_{p,i}$, the linearized structural weight objective function $\mathcal{W}$ is expressed as:

$$\mathcal{W} = \sum_{i=1}^{n} L_i M_{p,i} = L_1 M_{p1} + L_2 M_{p2} + \dots + L_n M_{pn}$$

The optimization problem requires minimizing $\mathcal{W}$ subject to the kinematic collapse constraints established by the plastic collapse theorems:

$$\text{Minimize } \mathcal{W} = \mathbf{L}^T \mathbf{M}_p$$

$$\text{Subject to: } \mathbf{C} \mathbf{M}_p \ge \mathbf{P}_{ext}, \quad \mathbf{M}_p \ge \mathbf{0}$$

where $\mathbf{C}$ is the kinematic constraint matrix derived from all possible failure mechanisms, and $\mathbf{P}_{ext}$ is the vector of external virtual work contributions.

3. Foulkes Theory: Optimality Criteria for Plastic Frameworks

3.1 Foulkes Theorem and Mechanism Compatibility

In 1954, J. D. Foulkes established the foundational optimality theorem governing the minimum weight plastic design of ductile steel frameworks. Foulkes theorem establishes that an assigned set of plastic moment capacities $M_{p,i}$ corresponds to an absolute minimum weight design if and only if:

1. The frame is on the verge of collapse under at least one kinematically admissible mechanism, or a combination of concurrent mechanisms $k = 1, \dots, m$.
2. Positive weighting multipliers $x_k \ge 0$ exist for each active mechanism such that for every member group $i$:

$$\sum_{k=1}^{m} x_k |\theta_{i,k}| = L_i$$

where $|\theta_{i,k}|$ represents the total absolute plastic hinge rotation occurring within member group $i$ under mechanism $k$, and $L_i$ is the total length of member group $i$.

       FOULKES MECHANISM COMBINATION CRITERIA

       Member Group 1 (Columns, L1):  x1 |θ_col,1| + x2 |θ_col,2| = L1
       Member Group 2 (Beam, L2):     x1 |θ_beam,1| + x2 |θ_beam,2| = L2

3.2 Geometric Interpretation in Plastic Moment Space

In two-variable design space ($M_{p1}$ versus $M_{p2}$), the kinematic collapse constraints form a convex feasible region bounded by linear constraint lines. The objective function $\mathcal{W} = L_1 M_{p1} + L_2 M_{p2}$ represents a family of parallel lines with normal gradient vector:

$$abla \mathcal{W} = \begin{bmatrix} L_1 \ L_2 \end{bmatrix}$$

The optimal minimum weight solution occurs at a convex vertex where the weight gradient vector $abla \mathcal{W}$ lies within the cone formed by the inward normal vectors of the intersecting mechanism constraint lines.

4. Linear Programming Formulation of Plastic Limit Design

The mathematical framework of optimal structural plastic design maps directly into the Primal-Dual formulation of Linear Programming (LP):

Modern computational limit analysis codes use Simplex or Interior-Point algorithms to solve large-scale frame optimization models in milliseconds.

5. Step-by-Step Worked Calculation: Optimization of a Pitched Portal Frame

To demonstrate the application of minimum weight plastic design and foulkes theorem, consider a rectangular portal frame of span $L = 10\text{ m}$ and column height $h = 5\text{ m}$.

                 P = 120 kN (Midspan Vertical Load)
                 v
       B +-------o-------+ C
         |               |
   H = 60 kN             |   Columns: Group 1 (Mp1), Total Length L1 = 2 x 5 = 10 m
   --->  |               |   Beam:    Group 2 (Mp2), Total Length L2 = 10 m
         |               |
       A ///           D ///

• Columns $AB$ and $CD$: Assigned capacity $M_{p1}$, total length $L_1 = 5 + 5 = 10\text{ m}$.

• Beam $BC$: Assigned capacity $M_{p2}$, total length $L_2 = 10\text{ m}$.

• Fixed bases at $A$ and $D$.

• Applied Loads: Lateral wind load $H = 60\text{ kN}$ at $B$; Vertical gravity load $P = 120\text{ kN}$ at beam midspan.

5.1 Frame Geometry and Applied Design Actions

Weight Objective Function to minimize:

$$\mathcal{W} = L_1 M_{p1} + L_2 M_{p2} = 10 M_{p1} + 10 M_{p2} \implies \frac{\mathcal{W}}{10} = M_{p1} + M_{p2}$$

The weight gradient vector is:

$$abla \mathcal{W} = \begin{bmatrix} 10 \ 10 \end{bmatrix} \implies \text{Slope of weight contour lines: } \frac{d M_{p2}}{d M_{p1}} = -\frac{L_1}{L_2} = -\frac{10}{10} = -1.0$$

5.2 Identification of Independent and Combined Mechanisms

We evaluate the three candidate collapse mechanisms using the Principle of Virtual Work:

    (1) BEAM MECHANISM                (2) SWAY MECHANISM               (3) COMBINED MECHANISM
        B    o    C                       B---------C                      B    o    C
        |   / \   |                       |         |                      |   / \   |
        |  /   \  |                      /         /                       |  /   \  |
        | /  P  \ |                     /    H    /                       /  /  P  \ |
        |/   v   \|                    /     v   /                       /  /   v   \|
        A         D                   A         D                   A  /         D

Mechanism 1: Pure Beam Mechanism

• Hinges at $B$ (in beam), midspan $E$, and $C$ (in beam).

• Rotation angles: $\theta_B = \theta$, $\theta_E = 2\theta$, $\theta_C = \theta$.

• External Work: $W_{ext} = P \cdot (5\theta) = 120 \times 5\theta = 600\theta$.

• Internal Dissipation: $D_{int} = M_{p2} (\theta + 2\theta + \theta) = 4 M_{p2} \theta$.

• Constraint 1:
  $$4 M_{p2} \ge 600 \implies M_{p2} \ge 150.0\text{ kN}\cdot\text{ m}$$

Mechanism 2: Pure Sway Mechanism

• Lateral sway displacement: $\Delta = 5\theta$.

• Hinges at base $A$ (column), $B$ (column), $C$ (column), base $D$ (column).

• Rotation angles: $\theta_A = \theta_B = \theta_C = \theta_D = \theta$.

• External Work: $W_{ext} = H \cdot (5\theta) = 60 \times 5\theta = 300\theta$.

• Internal Dissipation: $D_{int} = M_{p1} (\theta + \theta + \theta + \theta) = 4 M_{p1} \theta$.

• Constraint 2:
  $$4 M_{p1} \ge 300 \implies M_{p1} \ge 75.0\text{ kN}\cdot\text{ m}$$

Mechanism 3: Combined Mechanism

• Hinges at base $A$ (column), midspan $E$ (beam), joint $C$ (weaker of column/beam), base $D$ (column).

• Assuming joint $C$ hinge forms in the column ($M_{p1} < M_{p2}$) or beam:

• External Work: $W_{ext} = H(5\theta) + P(5\theta) = 300\theta + 600\theta = 900\theta$.

• Internal Dissipation:
  $$D_{int} = M_{p1}(\theta_A) + M_{p2}(\theta_E) + M_{p,min}(\theta_C) + M_{p1}(\theta_D) = M_{p1}(\theta) + M_{p2}(2\theta) + M_{p1}(\theta) + M_{p1}(\theta) = 3 M_{p1} \theta + 2 M_{p2} \theta$$

• Constraint 3:
  $$3 M_{p1} + 2 M_{p2} \ge 900$$

5.3 Graphical Optimization and Minimum Weight Solution

We determine the intersection points of the constraint boundaries:

• Intersection Point 1 (Beam Mechanism & Combined Mechanism):
  Set $M_{p2} = 150.0\text{ kN}\cdot\text{ m}$.
  $$3 M_{p1} + 2(150.0) = 900 \implies 3 M_{p1} + 300 = 900 \implies 3 M_{p1} = 600 \implies M_{p1} = 200.0\text{ kN}\cdot\text{ m}$$
  Weight value at Point 1:
  $$\mathcal{W}_1 = 10(200.0) + 10(150.0) = 2000 + 1500 = 3500\text{ kN}\cdot\text{ m}^2$$

• Intersection Point 2 (Sway Mechanism & Combined Mechanism):
  Set $M_{p1} = 75.0\text{ kN}\cdot\text{ m}$.
  $$3(75.0) + 2 M_{p2} = 900 \implies 225.0 + 2 M_{p2} = 900 \implies 2 M_{p2} = 675.0 \implies M_{p2} = 337.5\text{ kN}\cdot\text{ m}$$
  Weight value at Point 2:
  $$\mathcal{W}_2 = 10(75.0) + 10(337.5) = 750 + 3375 = 4125\text{ kN}\cdot\text{ m}^2$$

Comparing total weights: $\mathcal{W}_1 = 3500 < \mathcal{W}_2 = 4125$.

The minimum weight optimal design is achieved at Point 1:
$$M_{p1} = 200.0\text{ kN}\cdot\text{m (Columns)}, \quad M_{p2} = 150.0\text{ kN}\cdot\text{m (Beam)}$$

5.4 Foulkes Optimality Verification

To formally verify the solution via foulkes theorem, we check if positive multipliers $x_1$ (Beam Mechanism) and $x_2$ (Combined Mechanism) satisfy the member length conditions:

1. Column Group ($L_1 = 10\text{ m}$):
   * Beam Mechanism column rotations: $|\theta_{col,1}| = 0$.
   * Combined Mechanism column rotations: $|\theta_{col,2}| = \theta_A + \theta_C + \theta_D = 1 + 1 + 1 = 3$.
   $$x_1 (0) + x_2 (3) = 10 \implies 3 x_2 = 10 \implies x_2 = \frac{10}{3} \approx 3.333 > 0$$

2. Beam Group ($L_2 = 10\text{ m}$):
   * Beam Mechanism beam rotations: $|\theta_{beam,1}| = \theta_B + \theta_E + \theta_C = 1 + 2 + 1 = 4$.
   * Combined Mechanism beam rotations: $|\theta_{beam,2}| = \theta_E = 2$.
   $$x_1 (4) + x_2 (2) = 10 \implies 4 x_1 + 2 \left(\frac{10}{3} ight) = 10 \implies 4 x_1 + \frac{20}{3} = 10 \implies 4 x_1 = \frac{10}{3} \implies x_1 = \frac{10}{12} = \frac{5}{6} > 0$$

Because both Foulkes multipliers are strictly positive ($x_1 = 0.833 > 0$ and $x_2 = 3.333 > 0$), foulkes theorem confirms that this configuration is the mathematically unique, global minimum weight design.

6. Practical Structural Engineering Applications and Code Constraints

While theoretical limit optimization provides the lower bound of material usage, practical implementation must respect code-mandated boundaries:

• Discrete Commercial Sections: Optimization yields continuous mathematical capacities $M_p$. Practicing engineers must select the nearest standard commercial rolling (e.g., AISC W-shapes or Eurocode HE profiles) with $M_{p,actual} \ge M_{p,opt}$.

• Serviceability Deflections: Plastic optimization does not restrict service load elastic deflections. Post-optimization checks must verify that $\delta \le L/360$ under un-factored live loads.

• Strong-Column Weak-Beam (SCWB) Seismic Provisions: In seismic design (AISC 341), columns must be stronger than beams ($\sum M_{pc}^* / \sum M_{pb}^* > 1.0$) to avoid soft-story collapse mechanisms.

7. Synthesis and Optimization Wrap-Up

Mastering minimum weight plastic design provides structural engineers with an analytical method to eliminate material waste while preserving full limit-state capacity. By combining kinematic virtual work with foulkes theorem, designers identify the exact point where structural equilibrium meets material economy.

Applying these limit optimization principles produces steel frameworks that are safe, ductile, and cost-effective.

References & Standards Cited

1. American Institute of Steel Construction (AISC). (2022). Specification for Structural Steel Buildings (AISC 360-22). Chicago, IL: AISC.
2. Foulkes, J. D. (1954). “The Minimum Weight Design of Structural Frames.” Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences, 223(1155), 482–494.
3. Heyman, J. (1971). Plastic Design of Frames: Volume 2, Applications. Cambridge: Cambridge University Press.
4. Neal, B. G. (1977). The Plastic Methods of Structural Analysis. 3rd Edition, London: Chapman and Hall.
5. Vanderplaats, G. N. (2007). Multidiscipline Design Optimization. Colorado Springs, CO: VR&D.

Frequently Asked Questions (FAQ)

How does Foulkes theorem determine the absolute minimum weight design?

**Foulkes theorem** establishes that when the vector of structural member lengths $mathbf{L}$ can be expressed as a positive linear combination of the plastic hinge rotation vectors from active collapse mechanisms, the design represents the global minimum of the weight objective function.

Can minimum weight plastic design be applied to pitched roof portal frames?

Yes. Pitched roof portal frames can be optimized by defining column ($M_{p1}$) and rafter ($M_{p2}$) capacity groups and evaluating roof gable mechanisms alongside sway and frame mechanisms.

What happens if serviceability deflection limits govern the member sizes?

If serviceability deflection limits ($delta_{max} le L/360$) govern, the strictly plastic optimal sections must be upsized to increase their elastic moment of inertia $I_x$. In such cases, the design becomes stiffness-governed rather than strength-governed.

Why is the relationship between steel section weight and plastic moment linear?

For standardized structural steel wide-flange shapes of similar depth-to-width geometry, cross-sectional area $A$ correlates linearly with plastic section modulus $Z$ ($Z approx c cdot A$). Therefore, weight per unit length ($w =
ho A$) varies linearly with $M_p$.

How do discrete rolling constraints affect theoretical plastic optimization?

Because structural steel is manufactured in discrete standard sizes, the continuous mathematical optimum serves as an ideal baseline. Designers choose the nearest available standard sections that encompass the theoretical capacities while minimizing surplus capacity.