Plastic analysis
Plastic analysis finds the load at which a ductile structure collapses, rather than the load at which its most stressed fibre first yields. As load rises, the most highly stressed section reaches its plastic moment and forms a plastic hinge. In an indeterminate structure the load can still increase, because moments redistribute to other sections, until enough hinges form to turn the structure into a mechanism.
Two theorems frame the calculation. The upper bound, or kinematic, theorem says that any assumed mechanism gives a collapse load equal to or greater than the true one, so the lowest value over all mechanisms is the answer. The lower bound, or static, theorem says that any moment distribution in equilibrium with the load and nowhere exceeding the plastic moment gives a safe load. When a mechanism's moment diagram also satisfies the yield condition everywhere, both bounds meet and the true collapse load is found.
The virtual work method is the usual hand route. Choose a mechanism, give it a small rotation, and equate the external work of the loads moving through their displacements to the internal work at the hinges, the plastic moment times the hinge rotation. Beams have beam mechanisms; frames add sway and combined mechanisms, and the combined one usually governs when vertical and horizontal loads act together.
Plastic design gives economical, robust steel frames, and the ideas carry over to the yield-line method for concrete slabs. It requires compact sections that can rotate without local buckling, lateral restraint at hinges, and checks that axial force and shear do not reduce the plastic moment. Serviceability, deflection and second-order effects are checked separately.
List every possible mechanism and take the smallest collapse load; then check the moment diagram nowhere exceeds Mp.
Hinge work: internal work = Σ M_p θ_iFixed-ended beam, uniform load: w_c = 16 M_p / L²Simply supported beam, uniform load: w_c = 8 M_p / L²Pinned-base portal, sway mechanism: H_c = 2 M_p / h (hinges at both column tops)
Stopping at the first mechanism tried, which only gives an upper bound.
Problem. A fixed-ended steel beam of span 8 m has M_p = 200 kNm. Find the uniform collapse load and compare it with the load at which the first hinge forms.
- Mechanism: hinges at both supports and at mid-span. Give the ends a rotation θ; mid-span deflects L θ / 2 and its hinge rotates 2θ.
- External work = w L × (average deflection L θ / 4) = w L² θ / 4. Internal work = M_p (θ + 2θ + θ) = 4 M_p θ.
- w_c = 16 M_p / L² = 16 × 200 / 64 = 50 kN/m.
- Elastic support moment w L² / 12 first reaches M_p at w = 12 × 200 / 64 = 37.5 kN/m. Redistribution gives a reserve of 50 / 37.5 = 1.33.
Answer. Collapse at 50 kN/m; first hinge at 37.5 kN/m
Quick quiz
1What does the upper bound theorem guarantee?
Any assumed mechanism gives a collapse load equal to or above the true one.
2What is needed for a plastic hinge to work?
A compact section that can rotate without local or lateral buckling.
3Which mechanism usually governs a portal with vertical and horizontal loads?
The combined mechanism.
- IS 800 (India): General construction in steel: limit state design of steel members and connections.
- EN 1993 (Eurocode 3) (Europe): Design of steel structures.
- AISC 360 (USA): Specification for structural steel buildings.
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- R. C. Hibbeler, Structural Analysis, Pearson, latest edition.
- C. S. Reddy, Basic Structural Analysis, McGraw Hill Education (India), latest edition.
- Devdas Menon, Structural Analysis, Narosa, latest edition.
- A. K. Chopra, Dynamics of Structures, Pearson, latest edition.
Free and official sources
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