You do not design the cap beam with the reduced force: capacity protection

Your home’s electrical panel has fuses or breakers. They are built to fail first: if something overheats, the fuse blows and the wires stay safe. Nobody would design a fuse stronger than the wire it protects.

Seismic bridge design follows the same logic. We do not aim for the bridge to come out of a strong earthquake untouched, because that would be very expensive. We aim for damage to happen where we choose: in the column, in a zone we can confine, inspect and repair. The column is the fuse. The cap beam, the column joint and the foundation are the wires: they must stay intact.

In this note we will see why the force from the seismic analysis is the wrong one for designing those “wires”, what the right force is, and an example with round numbers.

The problem with the reduced force

In the usual seismic analysis, the elastic force is divided by a factor $R$ . The idea is that the column will yield and dissipate energy, so it does not need to be designed for the full force. That reduced force sizes the column.

The catch is that the real column almost always ends up stronger than the analysis asked for:

often its steel is set by minimum reinforcement or confinement, not by demand;
real materials are stronger than nominal ones;
steel keeps gaining some strength after it yields.

When the earthquake comes, the column does not deliver your analysis force: it delivers what it can actually resist. If the cap beam was designed for the reduced force, the fuse is stronger than the wire, and damage shows up where you did not want it.

The right force: column overstrength

The AASHTO Guide Specifications for LRFD Seismic Bridge Design (§4.11.2) ask for the column’s maximum probable moment:

M_{po} = \lambda_{mo}\, M_p

$M_p$ is the column strength with the steel it actually has and expected material properties. $\lambda_{mo}$ covers strain hardening: 1.2 for A706 steel and 1.4 for A615. Caltrans uses the same idea with $M_o = 1.2\,M_p$ .

That moment also gives the largest shear the column can transmit. In double curvature:

V_o = \frac{2\, M_{po}}{H}

An example with round numbers

A two-column bent, 8 m columns, fixed at the footing and at the cap beam.

Input	Value
Analysis moment (seismic force divided by $R$ )	≈ 2,600 kN·m at each end
Column strength with its real steel, $M_p$	≈ 4,000 kN·m
Steel	A706 → $\lambda_{mo} = 1.2$

M_{po} = 1.2 \cdot 4{,}000 \approx 4{,}800 \text{ kN·m} \qquad V_o = \frac{2 \cdot 4{,}800}{8} \approx 1{,}200 \text{ kN per column}

At the exterior joint, the whole column moment goes into the cap beam. So the cap beam must resist about 4,800 kN·m of seismic moment at the joint face, plus dead load, not the 2,600 from the analysis. Almost double.

What else gets designed with this value

The column’s own shear. Use $V_o$ , not the analysis shear. Shear failure is brittle and cannot come before the hinge.
The cap beam-column joint. Large forces cross there in a small space. Caltrans checks its stresses and asks for extra steel when principal tension exceeds $0.29\sqrt{f'_c}$ (MPa).
The foundation. It receives whatever the column delivers. Designed with the seismic force divided by $R$ , it falls short by a factor close to $R$ .

Decide first where you want the damage. Then design everything else for what that damage can transmit.

About the author

Yordan Rocio Maldonado

Structural engineer, seismic and bridge specialist. CIP 215845

Educational, reference-only content. Opinions are my own and do not represent any employer. On a real project, the engineer of record and the governing code decide.

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