Introduction to plastic analysis

Shape factor is the ratio of the plastic moment capacity $M_p$ of a cross-section to its yield moment $M_y$ (the moment at which the extreme fibre first reaches the yield stress):

$$S = \frac{M_p}{M_y} = \frac{\sigma_y Z_p}{\sigma_y Z_e} = \frac{Z_p}{Z_e}$$

where $Z_p$ is the plastic section modulus (the first moment of the area above and below the equal-area axis) and $Z_e$ is the elastic section modulus. It depends only on the shape of the section and shows the reserve of strength after first yield.

Typical values: rectangle 1.5, solid circle 1.70, I-section about 1.10 to 1.15, thin-walled circular tube about 1.27, triangle 2.34, T-section about 1.7 to 2.0.

Assumptions

• Portal frame: A (0,0) fixed, B (0,3), C (3,3) at mid-span, D (6,3), E hinged at the base of the right column. The figure gives "3 m + 2 m": the left column AB is taken as 3 m high and the right column DE as 2 m high (so E is 1 m above the level of A). Beam BD is 6 m ($1.5M_p$), AB is $2M_p$ and DE is $M_p$.
• Loads: 1.5W horizontal (to the right) at B and W vertical at C. $W_c$ is found in terms of $M_p$ (kN·m, lengths in m).
• Method: virtual work, $\sum M_p\theta = \sum P\delta$. Reactions: 3 at A and 2 at E, degree of indeterminacy $= 5 - 3 = 2$; critical sections A, B, C, D give 4 - 2 = 2 independent mechanisms (beam and sway), plus the combination.

Mechanisms

Mechanism 1: beam mechanism

No sway. Hinges at B (beam end, $1.5M_p$ is smaller than the column $2M_p$), under W (mid-span) and at D (in the weaker column DE, $M_p$). The rotations are $\theta/2$ at B and D and $\theta$ at C.

Internal work $= \frac{11}{4}\,M_p\,\theta$.

Equation: $W_c\times \frac{3}{2}\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{6}\,M_p = 1.8333\,M_p$.

Mechanism 2: sway mechanism

Left column AB (3 m) rotates $\theta$ about A; the beam moves $3\theta$ to the right; the right column DE (2 m high) rotates $3\theta/2$ about E. Hinges at A, B (beam end) and D (in DE).

Internal work $= 5\,M_p\,\theta$.

Equation: $W_c\times \frac{9}{2}\,\theta = 5\,M_p\theta$, so $W_c = \frac{10}{9}\,M_p = 1.1111\,M_p$.

Mechanism 3: combined mechanism

Mechanisms 1 and 2 are combined so that the hinge at B cancels. Hinges at A, C and D.

Internal work $= \frac{15}{4}\,M_p\,\theta$.

Equation: $W_c\times \frac{15}{4}\,\theta = \frac{15}{4}\,M_p\theta$, so $W_c = 1\,M_p$.

Collapse load

The combined mechanism is the lowest (the sway mechanism gives $1.111M_p$ and the beam mechanism $1.833M_p$), and a search over all hinge positions confirmed it:

Answer: $W_c = M_p$ (hinges at A, C and D; the loads at collapse are 1.5$M_p$ horizontal and $M_p$ vertical, with $M_p$ in kN·m and lengths in m).

Assumptions

• Portal frame: A (0,0) hinged, B (0,L), C (L/2, L), D (L, L), E fixed at the base of the right column, which is $2L/3$ high (so E is at height $L/3$ above A). Column AB $M_p$, beam BD $3M_p$ (span L), column DE $M_p$.
• Loads: $W/2$ horizontal (to the right) at B and $W$ vertical at mid-span C. $W_c$ in terms of $M_p$ and $L$.
• Degree of indeterminacy $= 5 - 3 = 2$; critical sections B, C, D, E: $4 - 2 = 2$ independent mechanisms (beam and sway) and their combination.

Mechanisms (tables with $L = 1$)

Mechanism 1: beam mechanism

Lengths in units of $L$ (take $L = 1$ in the tables). No sway; hinges at B (in the weaker column AB), at C and at D (in the column DE, $M_p$). Rotations $\theta/2$ at B and D, $\theta$ at C.

Internal work $= 4\,M_p\,\theta$.

Equation: $W_c\times \frac{1}{4}\,\theta = 4\,M_p\theta$, so $W_c = 16\,M_p$ per $L$.

Mechanism 2: sway mechanism

The right column DE (height $2L/3$) rotates $\theta$ about the fixed base E, so the beam moves $2L\theta/3$; the left column AB (height $L$, hinged base) rotates $2\theta/3$. Hinges at B (in AB), D (in DE) and E.

Internal work $= \frac{8}{3}\,M_p\,\theta$.

Equation: $W_c\times \frac{1}{3}\,\theta = \frac{8}{3}\,M_p\theta$, so $W_c = 8\,M_p$ per $L$.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels. Hinges at C, D and E.

Internal work $= 5\,M_p\,\theta$.

Equation: $W_c\times \frac{1}{2}\,\theta = 5\,M_p\theta$, so $W_c = 10\,M_p$ per $L$.

Collapse load

The sway mechanism governs (beam 16 and combined 10 are higher):

Answer: $W_c = 8M_p/L$ (hinges at B in column AB, D in column DE and E).

Assumptions

• Portal frame: span $2L$ ($L + L$), height $2L$. A hinged, E fixed (right base). Columns $M_p$, beam $2M_p$.
• Loads: $3W$ horizontal (to the right) at the top-left joint B and $2W$ vertical at mid-span C. Take $L = 1$ in the tables.
• Reactions $2 + 3 = 5$, degree of indeterminacy $= 2$; critical sections B, C, D, E: two independent mechanisms (beam, sway) plus the combined mechanism.

Mechanisms

Mechanism 1: beam mechanism

Taking $L = 1$. No sway; hinges at B (in column AB, $M_p$), at mid-span C ($2M_p$) and at D (in column DE, $M_p$). Rotations $\theta/2$ at B and D, $\theta$ at C.

Internal work $= 3\,M_p\,\theta$.

Equation: $W_c\times 1\,\theta = 3\,M_p\theta$, so $W_c = 3\,M_p$ per $L$.

Mechanism 2: sway mechanism

Both columns are $2L$ high. They rotate $\theta$ (A is hinged), the beam moves $2L\theta$ to the right. Hinges at B (in AB), D (in DE) and E.

Internal work $= 3\,M_p\,\theta$.

Equation: $W_c\times 6\,\theta = 3\,M_p\theta$, so $W_c = \frac{1}{2}\,M_p = 0.5000\,M_p$ per $L$.

Mechanism 3: combined mechanism

Combine 1 and 2 so that the hinge at B cancels. Hinges at C, D and E.

Internal work $= \frac{7}{2}\,M_p\,\theta$.

Equation: $W_c\times 4\,\theta = \frac{7}{2}\,M_p\theta$, so $W_c = \frac{7}{8}\,M_p = 0.8750\,M_p$ per $L$.

Collapse load

The sway mechanism governs:

Answer: $W_c = \dfrac{M_p}{2L} = 0.5\,\dfrac{M_p}{L}$.

Load factor $\lambda$ is the ratio of the collapse load (the load that produces a plastic collapse mechanism) to the working (service) load:

$$\lambda = \frac{W_c}{W_w}$$

In plastic design the working loads are multiplied by this factor (specified by the code, which gives different values for dead, live and wind combinations) and the structure is designed to just collapse under the factored loads. For a section it is related to the shape factor: $\lambda = S\times F$, where $F$ is the factor of safety against first yield (when moments are proportional to the loads).

A plastic hinge is a section of a flexural member where the bending moment has reached the plastic moment $M_p$, so the whole depth has yielded and the section can rotate under a constant moment $M_p$, like a hinge that carries a moment. The yielding actually spreads over a finite length of the member, called the plastic hinge length.

Plastic hinges form at points of maximum moment: fixed supports, joints, and under concentrated loads or at the section of zero shear under a UDL. When enough hinges form to turn the structure into a mechanism, it collapses.

Limit analysis finds the collapse load of a structure made of ductile material with a plastic moment $M_p$. Three conditions govern a collapse state:

1. Equilibrium condition: the bending moment diagram is in equilibrium with the loads.
2. Mechanism condition: enough plastic hinges have formed to make a mechanism.
3. Yield (plastic moment) condition: the moment at no section exceeds $M_p$.

1. Static theorem (lower-bound theorem)

Statement: the load factor $\lambda_s$ obtained from any bending moment distribution that satisfies equilibrium and does not violate the yield condition ($|M|\le M_p$ everywhere, a statically admissible state) is not greater than the true collapse load factor: $\lambda_s \le \lambda_c$.

Explanation: the structure can certainly carry a load for which a safe moment field exists, so the true collapse load is at least this value. The mechanism condition is not required.

Uses: gives a safe (conservative) estimate; used to check the design and to confirm a result found by the kinematic method (if no section exceeds $M_p$ the load is correct).

2. Kinematic theorem (upper-bound theorem)

Statement: the load factor $\lambda_k$ obtained from any assumed collapse mechanism, by equating the work of the external loads to the internal work in the plastic hinges, $\sum M_p\theta = \lambda \sum P\delta$ (it satisfies equilibrium and mechanism conditions), is not less than the true collapse load factor: $\lambda_k \ge \lambda_c$.

Explanation: an assumed mechanism may not be the real one, and the real one needs less load. The correct mechanism gives the lowest value.

Uses: all possible mechanisms are tried (beam, sway, joint and combined) and the smallest load factor is taken. It is simple to apply by hand.

3. Uniqueness theorem

If one load factor satisfies all three conditions (equilibrium, mechanism, yield), it is the true collapse load, $\lambda_s = \lambda_k = \lambda_c$.

Example

Propped cantilever (A fixed, C roller), span $L$, central load $P$.

• Static: choose $M_A = -0.5M_p$, then $M_{mid} = PL/4 - 0.25M_p \le M_p$, so $P = 5M_p/L$ (lower bound).
• Kinematic: hinges at A and at a section $0.4L$ from A: $P = 6.67M_p/L$ (upper bound).
• Correct mechanism (hinges at A and mid-span): $P_c = 6M_p/L$, which lies between the two.

Assumptions

• Portal frame: A (0,0) hinged, B (0,5), C (6,5), D (6,0) fixed. Column AB has plastic moment $M_p$, beam BC and column CD have $2M_p$.
• The loads (two 20 kN point loads at 2 m and 4 m from B, and 10 kN/m on column CD) are applied proportionally with a load factor $\lambda$; the 10 kN/m acts horizontally towards the left on CD. The collapse load factor $\lambda_c$ is found in terms of $M_p$ (in kN·m).
• Method: kinematic (upper-bound) theorem with virtual work, $\sum M_p\theta = \lambda \sum P\delta$.

Possible hinge positions

B (in the weaker member AB), under each 20 kN load, C, D. Degree of indeterminacy = 5 - 3 = 2 (A hinged gives 2 reactions, D fixed gives 3), so with 5 critical sections the number of independent mechanisms is 5 - 2 = 3: the beam mechanism (hinge under either load; both give the same value here) and the sway mechanism. Their combination gives the third.

Mechanism 1: beam mechanism

Joints B and C do not rotate. The beam BC forms hinges at B, P2 (4 m from B) and C. With rotation $\theta$ at P2, the left portion (4 m) rotates $\theta/3$ and the right portion (2 m) rotates $2\theta/3$ (rigid portions); the deflection under the second load is $4\times\theta/3$.

Internal work $= 4\,M_p\,\theta$.

Equation: $\lambda\times40\,\theta = 4\,M_p\theta$, so $\lambda = \frac{1}{10}M_p = 0.1000\,M_p$.

Mechanism 2: sway mechanism

Columns rotate through $\theta$ about A and D, the beam translates horizontally by $5\theta$. Hinges form at B (the weaker section is AB, $M_p$), C and D. The vertical loads do no work. (The column CD is moved to the left by the UDL.)

Internal work $= 5\,M_p\,\theta$.

Equation: $\lambda\times125\,\theta = 5\,M_p\theta$, so $\lambda = \frac{1}{25}M_p = 0.0400\,M_p$.

Mechanism 3: combined mechanism

Combine mechanisms 1 and 2 so that the hinge at C cancels. Both columns rotate $2\theta/3$ (sway to the left), the beam portion from B to the hinge rotates $\theta/3$ in the opposite sense (hinge at B $= 2\theta/3 + \theta/3 = \theta$), and the portion from the hinge to C rotates $2\theta/3$ with the column (hinge at the load $= \theta$, no hinge at C).

Internal work $= \frac{13}{3}\,M_p\,\theta$.

Equation: $\lambda\times\frac{370}{3}\,\theta = \frac{13}{3}\,M_p\theta$, so $\lambda = \frac{13}{370}M_p = 0.03514\,M_p$.

Collapse load

The smallest value governs (combined mechanism, lower than both the beam value 0.1 and the sway value 0.04):

$$\lambda_c = \frac{13}{370}M_p = 0.0351\,M_p$$

So the collapse loads are: point loads $20\lambda_c = 0.703\,M_p$ kN each and UDL $10\lambda_c = 0.351\,M_p$ kN/m (with $M_p$ in kN·m). A check with a linear-programming search over all possible hinge sets gave the same value $0.03514\,M_p$.

Answer: collapse load factor $\lambda_c = 13M_p/370 \approx 0.0351\,M_p$ (combined mechanism with hinges at B, under the 20 kN load at 4 m, and at D).

Assumptions

• Frame ABCDE: A (0,0) hinged, B (0,6), C (4,6), D (8,6), E (8,3) fixed. Column AB $2M_p$ (6 m high), beam BCD $1.5M_p$ (BC = CD = 4 m), column DE $M_p$ (3 m high; the right base is 3 m above A's level, so the columns have different heights).
• Loads: W at C, and W/2 per metre on column AB acting horizontally to the right. $W_c$ is expressed in terms of $M_p$ (in kN·m, lengths in m).
• Method: virtual work, $\sum M_p\theta = \sum P\delta$.

Mechanisms

Mechanism 1: beam mechanism

Hinges at B, C and D (the weaker section at D is the column DE, $M_p$). Beam BC rotates $\theta/2$ at B, the two beam halves rotate $\theta/2$ each in opposite senses, so C deflects $4\times\theta/2 = 2\theta$.

Internal work $= \frac{11}{4}\,M_p\,\theta$.

Equation: $W_c\times 2\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{8}\,M_p = 1.3750\,M_p$.

Mechanism 2: sway mechanism

The beam BD translates horizontally by $3\theta$ (column DE, 3 m high, rotates $\theta$ about E); column AB rotates $\theta/2$ about A. Hinges at B (beam side, $1.5M_p$), D (in DE) and E. W does no work.

Internal work $= \frac{11}{4}\,M_p\,\theta$.

Equation: $W_c\times \frac{9}{2}\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{18}\,M_p = 0.6111\,M_p$.

Mechanism 3: combined mechanism

Combination of 1 and 2 removing the hinge at B: hinges at C, D and E.

Internal work $= 4\,M_p\,\theta$.

Equation: $W_c\times \frac{13}{2}\,\theta = 4\,M_p\theta$, so $W_c = \frac{8}{13}\,M_p = 0.6154\,M_p$.

Refinement: sway with the hinge inside the column AB. For a UDL the plastic hinge in the loaded column does not form at B but at the section of maximum moment, at a height $y$ above A. Let the beam translate by $\Delta = 3\phi$ ($\phi$ = rotation of DE). The part AK rotates $\theta$ about A, so $y\theta = \Delta$, and the part KB translates with the beam. Hinges: K (rotation $\theta$, $2M_p$), D ($\phi$, $M_p$), E ($\phi$, $M_p$), where $\phi = y\theta/3$.

$$\begin{aligned} \text{Internal} &= M_p\theta\,(2 + \tfrac{2y}{3}) \\ \text{External} &= \tfrac W2\left[\tfrac{y^2\theta}{2} + (6-y)\,y\theta\right] = \tfrac W2\,\theta\,y\left(6 - \tfrac y2\right) \end{aligned}$$ $$W = \frac{M_p\,(4 + \tfrac{4y}{3})}{y\,(6 - \tfrac y2)}$$

Setting $dW/dy = 0$ gives $y^2 + 6y - 36 = 0$, so $y = 3(\sqrt5 - 1) = 3.708$ m, and

$$W_c = 0.5818\,M_p$$

(At $y = 6$, a hinge at the top of the column, this expression gives 0.667; mechanism 2, with the hinge in the beam, gives 0.611.)

Collapse load

The smallest value is the sway mechanism with the hinge in the column AB, and it is lower than the beam (1.375) and combined (0.615) mechanisms:

Answer: $W_c \approx 0.582\,M_p$ (in kN when $M_p$ is in kN·m and lengths are in m).

Assumptions

• Portal frame: A (0,0) fixed, B (0,4), C (8,4), D (8,0) hinged. Column AB $2M_p$, beam BC $3M_p$ (span 8 m), column CD $M_p$; both columns are 4 m high.
• Loads: W horizontal (to the right) at B, and two vertical loads W at 2 m and 6 m from B. Find $W_c$ in terms of $M_p$ (kN·m, lengths in m).
• Hinges may form at A, B, under the loads and at C (the weaker of the members meeting at a joint). The fixed base gives 3 and the hinge 2 reactions, so the degree of indeterminacy is 5 - 3 = 2. There are 5 critical sections (A, B, two load points, C), hence 5 - 2 = 3 independent mechanisms (two beam mechanisms, one for each load, and one sway); their combinations are also examined.

Mechanisms

Mechanism 1: beam mechanism

Joints do not rotate or sway. Hinges at B (in the weaker column AB), under the load 2 m from C and at C (in the weaker column CD). The rotations of the beam parts are $\theta/4$ (B to hinge, 6 m) and $3\theta/4$ (hinge to C, 2 m): the hinge under the load has $\theta$.

Internal work $= \frac{17}{4}\,M_p\,\theta$.

Equation: $W_c\times 2\,\theta = \frac{17}{4}\,M_p\theta$, so $W_c = \frac{17}{8}\,M_p = 2.1250\,M_p$.

Mechanism 2: sway mechanism

The columns rotate $\theta$ about the bases and the beam moves horizontally by $4\theta$. Hinges at A, B (in AB) and C (in CD).

Internal work $= 5\,M_p\,\theta$.

Equation: $W_c\times 4\,\theta = 5\,M_p\theta$, so $W_c = \frac{5}{4}\,M_p = 1.2500\,M_p$.

Mechanism 3: combined mechanism

Combining a beam mechanism (hinge under the load 2 m from B) with the sway mechanism removes the hinge at B. Hinges at A, under the first W, and at C (in CD).

Internal work $= \frac{11}{2}\,M_p\,\theta$.

Equation: $W_c\times 5\,\theta = \frac{11}{2}\,M_p\theta$, so $W_c = \frac{11}{10}\,M_p = 1.1000\,M_p$.

A search of all other hinge combinations (including the hinge under the load at 6 m with the sway mechanism, 1.5 $M_p$, and sway with hinge in the beam, 1.5 $M_p$) gave higher values.

Collapse load

The lowest value is from the combined mechanism:

Answer: $W_c = 1.1\,M_p$ (hinges at A, under the first W and at C; $M_p$ in kN·m, $W_c$ in kN with lengths in m).

Assumptions

• Portal frame: A (0,0) and D (6,0) hinged, B (0,4), C (6,4); both columns are 4 m high, the beam is 6 m with the 80 kN at 2 m from B.
• The plastic moments are in proportion to $I$: $M_{AB} = 3M_p$, $M_{BC} = 1.5M_p$, $M_{CD} = 2M_p$, where $M_p$ is the plastic moment capacity to be found. The loads are taken as ultimate (collapse) loads: 20 kN/m on AB and 100 kN on CD (2 m above D), both horizontal towards the right, and 80 kN vertical on the beam.
• Method: virtual work for each mechanism, $M_p\sum(\text{factor}\times\theta) = \sum P\delta$; the required $M_p$ is the largest value obtained from all mechanisms.
• Two hinged bases give 4 reactions, so the degree of indeterminacy is 1 and a mechanism needs 2 hinges (sway) or 3 hinges (beam).

Mechanisms

Mechanism 1: beam mechanism

No sway. Hinges at B, under the 80 kN load (2 m from B) and at C, all in the beam ($1.5M_p$ is smaller than the column moments $3M_p$ and $2M_p$). The beam parts rotate $2\theta/3$ (hinge at B) and $\theta/3$ (hinge at C), with $\theta$ under the load.

Internal work $= 3\,M_p\,\theta$.

Equation: $M_p\times 3\,\theta = \frac{320}{3}\,\theta$ (kN·m), so $M_p = \frac{320}{9} = 35.56$ kN·m.

Mechanism 2: sway mechanism

Both columns rotate $\theta$ about their hinged bases, the beam translates by $4\theta$ to the right. Hinges form at B and C in the beam, each rotating $\theta$.

Internal work $= 3\,M_p\,\theta$.

Equation: $M_p\times 3\,\theta = 360\,\theta$ (kN·m), so $M_p = 120$ kN·m.

Mechanism 3: combined mechanism

Beam and sway mechanisms combined so that the hinge at B cancels; hinges under the 80 kN load and at C.

Internal work $= 3\,M_p\,\theta$.

Equation: $M_p\times 3\,\theta = \frac{1040}{3}\,\theta$ (kN·m), so $M_p = \frac{1040}{9} = 115.56$ kN·m.

Required plastic moment

The largest required $M_p$ comes from the sway mechanism (the others give 35.56 and 115.56 kN·m); a frame with $M_p$ smaller than this would collapse in sway, while a larger value is safe against all mechanisms.

Answer: $M_p = 120$ kN·m (so $M_{AB} = 360$, $M_{BC} = 180$, $M_{CD} = 240$ kN·m).

Assumptions

• Portal frame A-B-D-E with width $l$ and height $l$: A hinged, E fixed. Columns AB and DE have $M_p$, beam BD has $2M_p$; C is the mid-point of BD.
• Loads: UDL $q$ per unit length on AB acting horizontally (towards the right), and $P = 2ql$ vertical at C. Both increase together; find the collapse value of $q$ (and $P$) in terms of $M_p$ and $l$.
• Take $l = 1$ for the working (all lengths in units of $l$); restore $l$ at the end. Reactions: A hinged (2), E fixed (3), so degree of indeterminacy = 2; critical sections B, C, D, E (and the hinge A) give 4 - 2 = 2 independent mechanisms (beam, sway) and their combination.

Mechanisms

Mechanism 1: beam mechanism

No sway. Hinges at B (in the column, the weaker member), under P (at mid-span C) and at D (in the column DE). Joint rotations: $\theta/2$ at B and D, $\theta$ at C.

Internal work $= 3\,M_p\,\theta$.

Equation: $q\times \frac{1}{2}\,\theta = 3\,M_p\theta$, so $q = 6\,M_p$ per $l^2$.

Mechanism 2: sway mechanism

The beam translates by $l\theta$; both columns rotate $\theta$ (A is hinged). Hinges at B, D (in the columns) and E.

Internal work $= 3\,M_p\,\theta$.

Equation: $q\times \frac{1}{2}\,\theta = 3\,M_p\theta$, so $q = 6\,M_p$ per $l^2$.

Mechanism 3: combined mechanism

The beam and sway mechanisms are combined so that the hinge at B cancels. Hinges at C, D and E.

Internal work $= \frac{7}{2}\,M_p\,\theta$.

Equation: $q\times \frac{3}{4}\,\theta = \frac{7}{2}\,M_p\theta$, so $q = \frac{14}{3}\,M_p = 4.6667\,M_p$ per $l^2$.

A mechanism with the sway hinge inside the loaded column (at 0.618 of the height, the section of maximum moment) gives 5.236, which is higher.

Collapse load

The lowest value governs (combined mechanism):

$$q_c = \frac{14}{3}\frac{M_p}{l^2} = 4.667\,\frac{M_p}{l^2},\qquad P_c = 2q_cl = \frac{28}{3}\frac{M_p}{l} = 9.333\,\frac{M_p}{l}$$

Answer: $q_c = 14M_p/(3l^2)$ and $P_c = 28M_p/(3l)$.