Elastomeric bearings per AASHTO §14.7.5.3.3: shear strain, step by step

If you have ever seen a bridge bearing off site, it looks like an unremarkable block of rubber. Inside it is something else: a sandwich of thin rubber layers separated by steel plates. The plates stop the rubber from bulging sideways, which lets the bearing carry a lot of weight while staying flexible horizontally.

What ends up damaging a bearing is shear strain in those rubber layers, and three things cause it at the same time:

• vertical load, which squeezes each layer and pushes rubber out at the edges;
• girder rotation, which squeezes one side more than the other;
• horizontal movement of the deck from temperature, shrinkage or braking.

AASHTO LRFD §14.7.5.3.3 puts all three into one check. In this note we will see what the equation says, where each term comes from, and work it out for a real bearing with round numbers.

The idea behind the equation

AASHTO adds the three strains but separates static (loads that are always there: self-weight, temperature, shrinkage) from cyclic (traffic, repeated millions of times). Cyclic loading fatigues rubber more, so it gets a 1.75 multiplier:

$$(\gamma_{a,st} + \gamma_{r,st} + \gamma_{s,st}) + 1.75\,(\gamma_{a,cy} + \gamma_{r,cy} + \gamma_{s,cy}) \le 5.0$$

Plus a separate limit, only for static compression:

$$\gamma_{a,st} \le 3.0$$

A side note: the report behind this clause (NCHRP 596, Stanton et al., 2008) proposed 2.0 instead of 1.75. AASHTO calibrated and kept 1.75.

Earthquake stays out of this sum. It is an extreme event checked separately.

Where each term comes from

First, a number that shows up everywhere: the shape factor $S$. It tells how “flat” a rubber layer is, the ratio between the loaded area and the edge where rubber can escape:

$$S = \frac{L W}{2 h_{\mathrm{ri}}(L + W)}$$

A thin, wide layer has a high $S$: it takes a lot of compression with little strain.

Compression. Grows with stress, drops with shape factor:

$$\gamma_a = D_a \frac{\sigma}{G S}$$

Rotation. Grows with the square of the bearing length, because a long bearing that rotates squeezes its edge much harder:

$$\gamma_r = D_r \left(\frac{L}{h_{\mathrm{ri}}}\right)^2 \frac{\theta}{n}$$

Displacement. The most intuitive one: how far the deck moves divided by total rubber thickness:

$$\gamma_s = \frac{\Delta}{h_{\mathrm{rt}}}$$

$L$ is the bearing dimension along the bridge, $h_{\mathrm{ri}}$ one layer thickness, $n$ the number of layers and $h_{\mathrm{rt}}$ total rubber thickness. For rectangular bearings $D_a$ depends on proportions; for preliminary design I use 1.4. $D_r = 0.5$.

An example

Input	Value
Plan	$L = 350$ mm, $W = 450$ mm
Layers	5 × 12 mm + 2 covers × 6 mm → $h_{\mathrm{rt}} = 72$ mm
Rubber	$G = 0.9$ MPa
Dead load	600 kN → $\sigma \approx 3.8$ MPa
Live load	250 kN → $\sigma \approx 1.6$ MPa
Rotation	static 0.005 rad (incl. erection tolerance), cyclic 0.002 rad
Displacement	static 20 mm (temperature, shrinkage), cyclic 3 mm (braking)

The shape factor comes out at $S \approx 8.2$. Then:

Term	Static	Cyclic
Compression $\gamma_a$	0.72	0.30
Rotation $\gamma_r$	0.43	0.17
Displacement $\gamma_s$	0.28	0.04
Sum	≈ 1.4	≈ 0.5

$$1.43 + 1.75 \cdot 0.51 \approx 2.3 \le 5.0 \qquad \gamma_{a,st} \approx 0.7 \le 3.0$$

It passes with margin. Compression too: $\sigma \approx 5.4$ MPa against a $1.66\,G S \approx 12$ MPa limit (§14.7.5.3.2).

Look at the table again: in this bearing, compression is half the total. If the check failed, the first move would be raising the shape factor with thinner layers.

Three mistakes I keep seeing

1. Earthquake in the cyclic part. Seismic displacement has its own check. Adding it here with the 1.75 oversizes the bearing for nothing.
2. Forgetting the 3.0 compression limit. With heavy dead load and thick layers, this limit governs before the sum does.
3. Using total height in the rotation term. Rotation spreads over the layers: use one layer thickness and $\theta/n$. With total height, in this example $\gamma_r$ comes out 7 to 36 times smaller and the check passes falsely.

Before you enlarge the bearing, see which term dominates. If it is rotation, a longer $L$ makes it worse because $\gamma_r$ grows with $L^2$. Add layers or widen $W$ instead.

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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