Springs in series: the bearing sets the period too

Picture a swing. With a short rope it goes back and forth quickly. Lengthen the rope and it swings slower. A bridge in an earthquake behaves in a similar way: the deck mass is the seat, and whatever holds it up decides how fast it oscillates. That “how fast” is the period, and it controls how much force and displacement the earthquake asks for.

When we compute a bridge’s period, we tend to look only at the column. But the deck is not glued to the column: elastomeric bearings sit between them, and under the column there is a foundation that also moves. In this note we will:

1. see why those three pieces work “in series”;
2. run an example with round numbers;
3. look at what changes in design when you stop ignoring them.

Why flexibilities add up

Think of three springs hanging one below the other. Pull the bottom end and the force goes through all three; each one stretches a little. The total stretch is the sum. So in series, what adds up is not stiffness but its inverse, flexibility:

$$\frac{1}{K_{\mathrm{total}}} = \frac{1}{K_{\mathrm{col}}} + \frac{1}{K_{\mathrm{brg}}} + \frac{1}{K_{\mathrm{fdn}}}$$

A practical consequence: the softest spring rules. You can make the column very stiff, but if the bearing is soft, the whole system stays soft.

Each spring on its own

The column acts as a cantilever, fixed at the bottom and free on top:

$$K_{\mathrm{col}} = \frac{3 E_c I_{\mathrm{eff}}}{H^3}$$

Notice the $H^3$: a column twice as tall is eight times more flexible.

Bearings work in shear. With $n$ bearings, each of area $A$ and total rubber thickness $h_{\mathrm{rt}}$:

$$K_{\mathrm{brg}} = n \, \frac{G A}{h_{\mathrm{rt}}}$$

The foundation (piles or footing) has its own lateral stiffness. Here I take it as given; another day we will see where it comes from.

An example with round numbers

Input	Value
Column	circular, $D = 1.20$ m, $H = 8$ m
Concrete	$f'_c = 28$ MPa → $E_c \approx 24{,}900$ MPa
Cracked inertia	$I_{\mathrm{eff}} \approx 0.5\,I_g \approx 0.051$ m⁴
Bearings	4 × 400 × 400 mm, $G = 0.9$ MPa, $h_{\mathrm{rt}} = 80$ mm
Foundation	$K_{\mathrm{fdn}} \approx 30{,}000$ kN/m (assumed)
Vibrating weight	$W = 3{,}000$ kN → $m \approx 306$ t

Each spring:

$$K_{\mathrm{col}} \approx \frac{3 \cdot 1.27\times10^{6}}{8^3} \approx 7{,}400 \text{ kN/m} \qquad K_{\mathrm{brg}} = 4 \cdot \frac{900 \cdot 0.16}{0.08} = 7{,}200 \text{ kN/m}$$

Together:

$$K_{\mathrm{total}} = \left(\frac{1}{7{,}400} + \frac{1}{7{,}200} + \frac{1}{30{,}000}\right)^{-1} \approx 3{,}300 \text{ kN/m}$$

With $T = 2\pi\sqrt{m/K}$:

What we model	$K$ (kN/m)	$T$ (s)
Fixed-base column only	≈ 7,400	≈ 1.3
Column and bearings	≈ 3,650	≈ 1.8
Column, bearings and foundation	≈ 3,300	≈ 1.9

What stands out: the four bearings are almost as stiff as the column. Adding them raises the period by about 40%, and with the foundation it reaches almost 50%.

What changes in design

Force goes down. In most spectra, past a certain period the acceleration drops. Longer period, less force. So if you ignore the bearings, your force is conservative and you feel safe.

Displacement goes up. That is the trap. Displacement grows with period, and so do the demands on the joint, the seat width and the bearing itself. Those checks do land on the unsafe side with rigid supports.

The split between piers changes. With piers of different heights, the short stiff one takes almost all the force. Bearings, whose stiffness does not depend on pier height, even that out.

Before you argue about column diameter, compute the three springs and find the softest one. That is where the bridge’s response lives.

YARM

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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