Ultimate Load of Beams

Animation 12.3.1: a prismatic simply supported rectangular-section beam under a midspan point load $F_{\mathrm P}$ growing from zero. The method that uses equilibrium at the ultimate state is the static method; alternatively, apply the virtual-work principle to the collapse mechanism — the kinematic method.

Elastic limit → plastic hinge → ultimate load

With increasing load:

• Outermost fibers at the midspan section yield first — the elastic-limit load is $F_{\mathrm{Ps}} = 4 M_{\mathrm s}/l$;
• A plastic zone forms at midspan (see Animation 12.3.1) and grows with load;
• Finally the midspan moment reaches $M_{\mathrm u}$ — a plastic hinge forms (see Animation 12.3.1) and the beam becomes a mechanism.

Animation 12.3.2: a fixed-pinned prismatic beam under uniform load $q$. Key property of single-span indeterminate beams: enough plastic hinges must form before the beam becomes a mechanism (collapses).

① Elastic stage · $M$ diagram (see Animation 12.3.2)

Elastic analysis gives $M_{A} = -q l^{2}/8$ at the fixed end, positive max near midspan $= q l^{2}/14.22$, and $M_{B} = 0$ at the pinned support. The fixed-end moment has the largest magnitude — it reaches $M_{\mathrm u}$ first.

② First plastic hinge at A

At $q \to q_{\mathrm s}^{A}$, the fixed end reaches $M_{A} = M_{\mathrm u}$ and the first plastic hinge forms. The beam now acts as a simply supported beam with moment $M_{\mathrm u}$ at $A$ — still can carry more load.

③ Second plastic hinge · mechanism forms

Further loading: some midspan section $C$ also reaches $M_{\mathrm u}$ — second plastic hinge; the system becomes a mechanism. From the ultimate-state $M$ diagram, $A$ carries $-M_{\mathrm u}$ and $C$ carries $+M_{\mathrm u}$.

Static-method solution

Kinematic-method solution

Using the collapse mechanism of the figure:

$W_{\mathrm{ext}} = W_{\mathrm{int}}$ gives $q_{\mathrm u} = 16 M_{\mathrm u}/l^{2}$ — identical result.

At first glance, ultimate analysis of an indeterminate beam seems to require tracking "how the plastic hinges develop". In fact, we can jump directly to the ultimate state — thanks to two key properties:

Find the ultimate load $q_{\mathrm u}$ of the fixed-pinned prismatic beam of under uniform load. Section ultimate moment $M_{\mathrm u}$.

① Mechanism · expression for $M_{C}$

From the elastic $M$ diagram, the fixed end $A$ forms a plastic hinge first. Assume the second hinge lies at section $C$ a distance $x$ from the pinned support $B$. From the equilibrium we get the reaction at $B$:

So the moment at $C$ (distance $x$ from $B$):

② Maximum positive moment (stationary condition)

$C$ is the maximum positive-moment location — $\mathrm d M_{C}/\mathrm d x = 0$ gives

Rearranging: $q_{\mathrm u} = 2 M_{\mathrm u}\,/\,[\, l(l - 2x)\,]$.

③ Exact hinge location

Substituting back into the $M_{C}$ expression and simplifying: $x^{2} + 2lx - l^{2} = 0$, giving

Therefore

Find the ultimate load $F_{\mathrm{Pu}}$ of the stepped beam of Animation 12.3.5. Segment $AB$ has ultimate moment $M_{\mathrm{u1}}$, segment $BC$ has $M_{\mathrm{u2}}$ with $M_{\mathrm{u1}} > M_{\mathrm{u2}}$.

① Possible plastic-hinge locations

For a stepped beam, since $AB$ and $BC$ have different ultimate moments, plastic hinges may form not only at the maximum-moment sections $A$ and $D$, but also at the step $B$ — because the $BC$ side has smaller $M_{\mathrm{u2}}$, so $B$'s $BC$ side is more easily capped.

② Mechanism with hinges at $B$ and $D$

If hinges form at $B, D$, both reach $M_{\mathrm{u2}}$ in the figure. The moment at $A$ is then

For this mechanism to be valid, $A$ must be able to carry this moment: $M_{\mathrm{u1}} \ge M_{A}$. Hence

③ Kinematic method

For the mechanism of the figure, the virtual-work equation is

The ultimate load:

A multi-span continuous beam may collapse in two mechanism types: single-span mechanism (one span collapses alone) or combined mechanism (two or more spans collapse together).

Unequal positive & negative ultimate moments

In practice (e.g. reinforced-concrete beams) the positive / negative ultimate moments may differ — $M_{\mathrm u}^{+}$ vs $M_{\mathrm u}^{-}$. Each plastic-hinge moment must be substituted individually in the mechanism equation:

Find the ultimate load of the equi-section continuous beam (spans $AB$ and $BC$, both carrying uniform load $q$). Section ultimate moment $M_{\mathrm u}$.

① AB span mechanism alone

By §12.3.6's rule (equi-section + same-direction loads), only the single-span mechanism is possible. $AB$-span mechanism: two plastic hinges — negative at $A$ ($M_{A} = -M_{\mathrm u}$) and positive at some interior point in $AB$.

By kinematic method with hinges at $A$, midspan, and $B$, let midspan deflection be $\delta_{1}$ and rotation $\theta_{1} = 2\delta_{1}/l_{1}$:

Giving $q_{\mathrm u}^{AB} \cdot l_{1}^{2}/4 = 3 M_{\mathrm u}\cdot (2\delta_{1}/l_{1})$, i.e.

② BC span mechanism alone

Similarly:

③ Take the smaller as the true ultimate load

Take the smaller of the two — i.e. the longer span fails first:

Once that span collapses, the whole continuous beam loses capacity — so the real ultimate load is the smaller of the two — a direct application of the minimum theorem (rigorously proved in §12-4).