Ultimate Load of Plane Frames

Animation 12.5.1: a rectangular-section bar carries both axial force $F_{\mathrm N}$ and bending moment $M$. With increasing load, the section passes through elastic (see Animation 12.5.1), elasto-plastic (see Animation 12.5.1), and plastic stages (see Animation 12.5.1).

Ultimate-state stress distribution (see Animation 12.5.1)

In the plastic stage the section splits into a $+\sigma_{\mathrm s}$ block and a $-\sigma_{\mathrm s}$ block of unequal area — the dividing distance $y_{1}$ depends on the load balance. From Eqs. (12-1) and (12-2):

Eliminate $y_{1}$ — yield locus

Eliminating $y_{1}$ from the two equations:

Define two pure-load limits:

• Pure axial ultimate: $F_{\mathrm{Nu}} = \sigma_{\mathrm s}\, b\, h$
• Pure bending ultimate: $M_{\mathrm u} = \sigma_{\mathrm s}\, b\, h^{2}/4$

Using the two pure-load limits, rewrite Eq. (a) in non-dimensional form:

Eq. (12-16) expresses the $M$-$F_{\mathrm N}$ relation at section yield. If the coordinate $(M, F_{\mathrm N})$ is below the locus, the section has not reached its ultimate state; on the locus, plastic flow can occur. For ideal elasto-plastic material, the coordinate cannot lie above the locus.

Quantifying the axial-force effect on $M_{\mathrm u}$

So when the axial force is small ($F_{\mathrm N}/F_{\mathrm{Nu}} \le 0.3$) its effect on $M_{\mathrm u}$ is minor; in frame ultimate-load calculations we usually ignore the axial-force effect.

Ignoring the axial-force effect, a frame's ultimate-load analysis is similar to continuous beams but with more and more complex possible mechanisms. A mechanism formed by a single member or joint collapsing is a basic mechanism:

Find the ultimate load of the frame of Animation 12.5.4. Columns and beam have ultimate moments $M_{\mathrm u}$ and $2 M_{\mathrm u}$, respectively. Horizontal force $F_{\mathrm P}$ and midspan concentrated force $2 F_{\mathrm P}$; column height $1.5\, l$, beam span $2\, l$.

① Sway mechanism (see Animation 12.5.4)

Both columns rotate by $\theta$ in the same direction; plastic hinges at $B$-column top, $A$-column base, $D$-column top, $E$-column base. Virtual work:

② Beam mechanism

The beam sags at $C$; beam-end rotations $\theta$, midspan hinge rotates $2\theta$. Virtual work:

③ Combined mechanism

Superposing the two: columns sway + beam sags simultaneously. Plastic hinges at column bases $A$, $E$, and midspan $C$; $B$ joint no longer forms a hinge (the rotations cancel):

④ Compare and confirm the ultimate load

The combined mechanism has the smallest $F_{\mathrm P}^{+}$. By the upper-bound theorem:

A combined mechanism superposes two or more basic mechanisms. Superposition: displacements add; rotations at certain positions cancel — so some plastic hinges "vanish" (where the two mechanisms' rotations are opposite), reducing the total plastic-deformation work while external work remains the sum of the two — hence the combined mechanism's $F_{\mathrm P}^{+}$ is typically less than any basic mechanism's.

Construction procedure

1. List all independent basic mechanisms (one beam mechanism per span + one sway mechanism per story);
2. Stepwise combine two, three, … basic mechanisms, each time checking for plastic-hinge cancellation;
3. For each possible combination, compute $F_{\mathrm P}^{+}$ via the virtual-work method;
4. Take the smallest $F_{\mathrm P}^{+\min}$ — by the upper-bound theorem $F_{\mathrm{Pu}} \le F_{\mathrm P}^{+\min}$;
5. Verify the corresponding ultimate-moment diagram satisfies $|M| \le M_{\mathrm u}$ — if so, by the uniqueness theorem $F_{\mathrm{Pu}} = F_{\mathrm P}^{+\min}$.

A joint mechanism is one in which a rigid joint rotates independently — all member ends meeting at the joint reach their ultimate moments simultaneously, forming plastic hinges. It can occur only when an external moment acts at the joint (or it contributes indirectly in combined mechanisms).

Role in Example 12-7

For multi-bay frames (Ex. 12-7, the figure), in addition to beam and sway mechanisms, joint mechanisms must be counted as independent basic mechanisms — otherwise some real combined mechanisms cannot be expressed. Systematic enumeration requires completeness.

Find the ultimate load of the two-bay frame of Animation 12.5.7.1. Section ultimate moments labeled in the figure ($M_{\mathrm u}$, $2 M_{\mathrm u}$, $3 M_{\mathrm u}$). Horizontal force $F_{\mathrm P}$; midspan concentrated forces $F_{\mathrm P}$ and $2 F_{\mathrm P}$.

Four independent basic mechanisms

Combined mechanism (see Animation 12.5.7.4)

Combining (c), (d), (e) gives the combined mechanism of Animation 12.5.7.4. Virtual-work equation:

Simplifying: $3 F_{\mathrm P}\, l\, \theta = 14\, M_{\mathrm u}\, \theta$, so

Determining the ultimate load

Comparing all mechanisms:

The combined mechanism (f) has the smallest $F_{\mathrm P}^{+}$. The corresponding moment diagram satisfies the yield condition. By the uniqueness theorem:

For complex frames, enumerating mechanisms is tedious — the number of possibilities grows rapidly. The incremental-stiffness method splits the nonlinear problem into a series of piecewise-linear sub-problems, easy to program, and also reveals the order in which plastic hinges form.

Basic assumptions

1. Ideal elasto-plastic material with small deformations before the ultimate state;
2. Before a plastic hinge forms, the section is elastic; once it forms, the plastic zone degenerates to a hinge (maintaining $M = M_{\mathrm u}$) and the rest of the member remains elastic;
3. Proportional loading + joint loads only — plastic hinges occur only at joints;
4. Ignore shear and axial-force effects on $M_{\mathrm u}$.

Algorithm

1. Perform an elastic analysis under $F_{\mathrm P} = 1$ to get joint displacements and member-end moments;
2. Find the location of first reach to $M_{\mathrm u}$ (maximum $|M|/M_{\mathrm u}$); compute the load increment $\Delta F_{\mathrm P}$ and accumulated $F_{\mathrm P}^{(k)}$; update displacements and internal forces;
3. Modify the element stiffness matrix at the new plastic hinge (set the end-moment increment to 0); reassemble the structural stiffness $\boldsymbol K$;
4. Return to step ① — "elastic" analysis continues until the next plastic hinge forms;
5. Repeat until a collapse mechanism forms — signaled by $|\boldsymbol K|$ becoming singular or the appearance of zero diagonal entries.

Modifying the element stiffness

From the general frame-element stiffness equation (Ch. 8 matrix displacement method):

If one or both ends of a member develop a plastic hinge, modify that end's row (setting $\bar M_{i} = 0$), eliminating the rotation DOF — yielding a softer element stiffness matrix. The three end-hinge cases:

• Hinge at $i$ end — set $\bar M_{i} = 0$ to eliminate $\bar\theta_{i}$;
• Hinge at $j$ end — set $\bar M_{j} = 0$ to eliminate $\bar\theta_{j}$;
• Hinges at both ends — the member degenerates to a truss member (axial only).