Bearings, Expansion Joints, and Restraint Systems

Engineering story

Between the deck and the pier: the smallest, most abused piece of the bridge

A bridge only appears rigid. Every day the deck grows and shrinks by inches with the temperature, the girders rotate under passing trucks, and the substructure settles a hair at a time. Something between the deck and the pier has to absorb all of that motion without letting the superstructure crack or the substructure pry itself apart. That something is a bearing, and the running seam that opens between adjacent deck panels is an expansion joint. Together they are the bridge's release valves. Get them wrong and every other calculation in this course loses its assumptions.

AASHTO §14 devotes an entire chapter to bearings and joints because a neglected 4-inch-tall rubber pad or a leaking strip-seal has toppled more bridges through corrosion of the pier cap and abutment seat than any collision or scour event. Modern practice tries to eliminate joints entirely by making abutments integral; where a joint is unavoidable, the goal is to keep it watertight for the full 75-year service life.

Chapter objectives

What you will be able to do

Learning objectives

By the end of this chapter you will be able to:

1Select an appropriate bearing family (elastomeric, pot, disc, spherical) based on load, rotation, and translation demand.
2Design a steel-reinforced elastomeric bearing pad per AASHTO §14.7.5 (Method A) — compressive stress, shape factor, shear translation, and rotation check.
3Compute the required movement rating of an expansion joint from thermal, shrinkage, creep, and elastic post-tensioning effects.
4Select between finger, strip-seal, and modular joint systems based on total movement.
5Detail anchor bolts and sole plates for a pinned or expansion bearing per §14.8.
6Understand integral and semi-integral abutment concepts and when they eliminate joints entirely.
7Estimate seismic isolation demand and identify when a lead-rubber or friction-pendulum isolator is warranted.

10.1 — Bearing families

Four bearing types cover 95 % of bridges

AASHTO LRFD §14.6

Every bridge bearing must accomplish three things: transfer the vertical reaction, permit the required horizontal translation, and permit the required rotation about at least one horizontal axis. Different products distribute those duties differently, and cost rises rapidly with load and rotation capacity.

Photograph of steel-reinforced elastomeric bridge bearing pads staged on a construction site. — Fig. 10.1Figure 10.1. Steel-reinforced elastomeric bearing pads staged before installation. Method-A designs of this pad type (alongside pot, disc, and spherical bearings) cover roughly 95 % of highway bridges. Photo: Georgia DOT / Wikimedia Commons, CC-BY 2.0.

Type	Vertical load	Translation	Rotation	Cost index
Elastomeric pad	up to 1,000 kip	up to ~4 in.	~0.02 rad	1.0
Pot bearing	up to 5,000 kip	with PTFE slider, unlimited	0.02 rad	3–4
Disc bearing	up to 4,000 kip	with PTFE slider, unlimited	0.02 rad	2–3
Spherical	up to 10,000 kip	with slider, unlimited	≥ 0.05 rad	4–6

10.2 — Load path at the bearing

Reaction, shear, and eccentricity

AASHTO LRFD §14.5, §14.6

Each bearing sees a factored vertical reaction $R_u$ , a factored horizontal shear $H_u$ from braking, wind, and thermal restraint, and a service-load rotation $\theta_s$ from live-load deflection of the girder.

R_u \;=\; \eta \, \bigl[\, 1.25\, R_{DC} + 1.50\, R_{DW} + 1.75\, R_{LL+IM}\,\bigr]

(10.1)

$R_u$: factored bearing reaction, Strength I [kip]
$\eta$: load modifier (typ. 1.0)
$R_{DC}, R_{DW}, R_{LL+IM}$: unfactored reactions from DC, DW, and live+impact

H_u \;=\; H_{BR} + H_{TU} + H_{WS} + H_{EQ}

(10.2)

$H_{BR}$: braking force per bearing (§3.6.4)
$H_{TU}$: thermal restraint force = k_h · Δ_T
$H_{WS}$: wind on superstructure share
$H_{EQ}$: seismic if applicable

Fig. 10.2Figure 10.2. Load path through a single bearing. Vertical reaction

R_u

pushes down through the pad into the pier cap; horizontal shear

H_u

is developed by the pad's shear stiffness or transferred through an anchor / restrainer.

10.3 — Steel-reinforced elastomeric bearings

Method A: rubber sandwiched with steel shims

AASHTO LRFD §14.7.5, §14.7.6

A steel-reinforced elastomeric pad is a stack of thin rubber layers vulcanized to internal steel shims. The rubber's shear flexibility lets the deck translate and rotate; the steel shims prevent the rubber from bulging out sideways under vertical load, dramatically increasing compressive stiffness. AASHTO's Method A is the standard design procedure for pads with a shape factor $S \ge 6$ under 800 psi.

Fig. 10.3Figure 10.3. Steel-reinforced elastomeric pad section. Cover rubber protects the shims from corrosion; internal layer thickness

h_{ri}

is the key design variable — thinner layers ⇒ higher shape factor ⇒ higher compressive capacity.

The shape factor is loaded area over the perimeter free-to-bulge area of one internal layer:

S \;=\; \dfrac{L \, W}{2\,h_{ri}\,(L + W)}

(10.3)

$L, W$: plan dimensions of the pad [in]
$h_{ri}$: thickness of one internal elastomer layer [in]

Compressive stress limit (Method A):

\sigma_s \;=\; \dfrac{P_s}{L\,W} \;\le\; \min\!\bigl(0.80\;\text{ksi},\; 1.00\, G\, S\bigr)

(10.4)

$P_s$: unfactored service reaction (Service I) [kip]
$G$: shear modulus of elastomer, 0.080–0.175 ksi

Translation (shear) check — pad must be thick enough that the shear angle stays modest:

h_{rt} \;\ge\; 2\,\Delta_s

(10.5)

$h_{rt}$: sum of all internal elastomer-layer thicknesses [in]
$\Delta_s$: maximum service horizontal displacement (thermal + shrinkage) [in]

Rotation check — one edge must not lift off. For a rectangular pad rotating about the transverse axis:

\sigma_s \;\ge\; 0.5\, G\, S \left(\dfrac{L}{h_{ri}}\right)^{\!2} \dfrac{\theta_s}{n}

(10.6)

$\theta_s$: service rotation about transverse axis [rad]
$n$: number of internal elastomer layers

Fig. 10.4Figure 10.4. Shear deformation. Total elastomer thickness

h_{rt}

must be at least twice the design translation

\Delta_s

so shear strain stays below 50 %.

10.4 — Pot, disc, and spherical bearings

When rubber alone is not enough

AASHTO LRFD §14.7.4, §14.7.8

When the reaction exceeds ~1,000 kip or the rotation exceeds 0.02 rad, the elastomer must be confined to prevent bulging failure. A pot bearing traps a circular disc of elastomer inside a shallow steel pot. A steel piston sits on top of the elastomer, which behaves like a hydraulic fluid — carrying huge compressive stress while still permitting rotation. A PTFE sliding sheet on top of the piston (against a stainless-steel mating plate) provides the translation freedom.

Photograph of a pot bearing installed under a bridge girder, showing the steel pot, upper piston, and PTFE slider. — Fig. 10.5Figure 10.5. Pot bearing installed under a highway bridge girder (Wiesenbrücke B317, Lörrach). The confined elastomer inside the steel pot behaves nearly as an incompressible fluid, transmitting the vertical reaction while allowing rotation; the PTFE / stainless slider on top provides translation. Photo: Wikimedia Commons, CC-BY-SA.

10.5 — Expansion joint systems

Watertight seams sized to a temperature range

AASHTO LRFD §14.5, §3.12

A joint is characterized by its movement rating — the total opening/closing range it can accommodate. The joint gap must be at least the design movement plus a construction tolerance and must never close to zero at the hottest design temperature.

Photograph of a modular expansion joint installed in a US highway bridge deck. — Fig. 10.6Figure 10.6. A modular expansion joint installed in a US highway bridge deck. Strip-seal, modular, and finger-plate systems are chosen from total-movement thresholds (~4 in., 4–24 in., and 4–24 in. with drainage trough respectively). Photo: Wikimedia Commons, public domain.

Total design movement combines temperature, shrinkage, creep, and elastic-shortening effects:

\Delta_{total} \;=\; \Delta_T \;+\; \Delta_{SH} \;+\; \Delta_{CR} \;+\; \Delta_{PS}

(10.7)

\Delta_T \;=\; \alpha \, L_{tr} \, (T_{max} - T_{min})

(10.8)

$\alpha$: coefficient of thermal expansion (6.0 × 10⁻⁶ /°F concrete, 6.5 × 10⁻⁶ /°F steel)
$L_{tr}$: tributary length of deck contributing to this joint [in]
$T_{max}-T_{min}$: design temperature range per §3.12.2 [°F]

Fig. 10.7Figure 10.7. The joint gap opens in cold weather and closes in hot weather. Design must satisfy both a maximum opening (winter, seal must not tear) and a minimum opening (summer, plates must not bind).

10.6 — Worked example 1

Steel-reinforced elastomeric bearing — 4-girder composite bridge

AASHTO LRFD §14.7.5, §14.7.6 (Method A)

Problem statement

A simply-supported, 4-girder composite steel plate-girder bridge spans 130 ft over a rural highway. Two bearings at each end of each girder rest on the abutment seats. Design the interior-girder bearing pad.

Given

Girder spacing8 ft; 4 girders total; span 130 ft
Service reaction $P_s = 320\;\text{kip}$ (Service I, interior girder)
Strength reaction $R_u = 470\;\text{kip}$
Girder bottom flange18 in. wide × 1.5 in. thick
Thermal rangeΔT = 120 °F (Cold climate, §3.12.2.2)
ElastomerG = 0.130 ksi (60 durometer), Grade 3 neoprene
RotationLive-load rotation $\theta_s = 0.008\;\text{rad}$

Required

Size the pad plan, layer thickness, and shim count; check compression, translation, and rotation per Method A.

Step 1 — Thermal translation. Half the span translates toward each abutment (bearing fixed at one end, expansion at the other; assume fixed at abutment 1, expansion at abutment 2 — full span moves at abutment 2):

Formula

\Delta_T \;=\; \alpha \, L \, \Delta T

Substitute

\Delta_T \;=\; 6.5 \times 10^{-6}\,(130 \times 12)(120)

Result

\Delta_T \;=\; 1.22\;\text{in}

Include 0.20 in. for shrinkage / installation ⇒ $\Delta_s = 1.42\;\text{in}$ .

Step 2 — Trial pad plan. Try $L = 9\;\text{in}$ (parallel to bridge, movement direction), $W = 20\;\text{in}$ (transverse). Check compressive stress limit at 0.80 ksi:

Formula

\sigma_s \;=\; \dfrac{P_s}{L\,W}

Substitute

\sigma_s \;=\; \dfrac{320}{9 \times 20}

Result

\sigma_s \;=\; 1.78\;\text{ksi} \;>\; 0.80\;\text{ksi} \;\times

Undersized. Increase plan to $L = 14\;\text{in}$ , $W = 24\;\text{in}$ :

Formula

\sigma_s \;=\; \dfrac{320}{14 \times 24}

Substitute

\sigma_s \;=\; 0.95\;\text{ksi}

Result

\text{still } > 0.80 \; \times

Increase to $L = 14\;\text{in}$ , $W = 30\;\text{in}$ :

Formula

\sigma_s \;=\; \dfrac{320}{14 \times 30}

Substitute

\sigma_s \;=\; 0.76\;\text{ksi}

Result

0.76 \;<\; 0.80\;\text{ksi} \quad \checkmark

Step 3 — Layer thickness and shape factor. Try 6 internal layers at $h_{ri} = 0.50\;\text{in}$ :

Formula

S \;=\; \dfrac{L\,W}{2\,h_{ri}\,(L+W)}

Substitute

S \;=\; \dfrac{14 \times 30}{2\,(0.50)(14+30)}

Result

S \;=\; 9.5 \;\ge\; 6 \quad \checkmark

Second stress cap:

Formula

1.00\, G\, S

Substitute

(1.00)(0.130)(9.5)

Result

1.24\;\text{ksi} \;>\; 0.76 \;\text{ksi} \quad \checkmark

Step 4 — Translation check.

Formula

h_{rt} \;=\; n \cdot h_{ri}

Substitute

h_{rt} \;=\; 6\,(0.50)

Result

h_{rt} \;=\; 3.0\;\text{in}

Formula

h_{rt} \;\ge\; 2\,\Delta_s

Substitute

3.0 \;\ge\; 2\,(1.42) = 2.84

Result

\checkmark

Step 5 — Rotation check (about transverse axis, L direction).

Formula

0.5\,G\,S \left(\dfrac{L}{h_{ri}}\right)^{\!2}\dfrac{\theta_s}{n}

Substitute

0.5\,(0.130)(9.5)\left(\dfrac{14}{0.50}\right)^{\!2}\dfrac{0.008}{6}

Result

= 0.65\;\text{ksi} \;<\; \sigma_s = 0.76\;\text{ksi} \quad \checkmark

Step 6 — Assemble pad. Add 1/8 in. cover rubber top and bottom, 7 steel shims (14-gage, 0.075 in. each) between the 6 internal layers.

Formula

h_{total} \;=\; 2\,(0.125) + 6\,(0.50) + 7\,(0.075)

Substitute

h_{total} \;=\; 0.25 + 3.0 + 0.525

Result

h_{total} \;=\; 3.78\;\text{in} \;\;\Rightarrow\; \text{specify 3.75 in.}

Fig. 10.8Figure 10.8. Bearing plan for the worked example — 8 bearings total, two per girder line at each abutment.

Fig. 10.9Figure 10.9. Final bearing section — 14 in. × 30 in. plan, 6 internal 1/2 in. rubber layers, 7 internal 14-gage steel shims, 1/8 in. cover top and bottom, top masonry plate welded to girder bottom flange, bottom sole plate anchored to abutment.

Final section detailing (from computed A_s)

Interior-girder elastomeric bearing — Simply-supported 130 ft composite steel bridge

Location	A_s required	Bars provided	Spacing / detail
Plan dimensions	σ_s ≤ 0.80 ksi	14 in. (L) × 30 in. (W)	Long dimension aligned with movement direction
Elastomer layers	S ≥ 6	6 internal layers × 1/2 in. + 1/8 in. cover top & bottom	Grade 3 neoprene, G = 0.130 ksi
Steel shims	per §14.7.5.3.5	7 shims @ 14-gage (0.075 in.)	Fully bonded/vulcanized to rubber
Sole and masonry plates	per §14.8.2	1 in. thick, welded to girder flange; 1 1/4 in. bottom plate with 4 – 1 in. Ø anchor rods 6 in. embedment	Anchor rods engage in 2 in. deep grouted pockets in abutment seat
Movement provided	Δ<sub>s</sub> = 1.42 in.	h<sub>rt</sub> = 3.0 in. (permits Δ = 1.5 in. at 50 % shear strain)	Include ±1 in. installation tolerance in seat design

The pad meets AASHTO Method A on compression, shape factor, translation, and rotation with margin. In hot climates recheck with reduced G at 120 °F. Provide 1 in. minimum clearance between pad edge and any seat feature that could restrain shear translation.

10.7 — Worked example 2

Expansion joint sizing — 3-span continuous PC girder bridge

AASHTO LRFD §14.5, §3.12

Problem statement

A three-span continuous prestressed-concrete girder bridge (spans 120 + 150 + 120 = 390 ft) has integral piers and joints only at the two abutments. Each abutment sees roughly half the movement. Select an appropriate joint system for the north abutment.

Given

Bridge length390 ft (deck total)
Tributary length to N. abutment $L_{tr} = 195\;\text{ft}$
Temperature rangeΔT = 100 °F (moderate climate, concrete)
Shrinkage strainε_SH = 0.0002 (post-age-28 additional)
Creep strain (long-term)ε_CR = 0.0003
Elastic PS shortening (already occurred pre-erection)included in girder length; no additional joint movement
Installation temperature60 °F

Required

Compute total design movement $\Delta_{total}$ , then select a joint from the AASHTO family and size the initial gap width for the 60 °F installation temperature.

Step 1 — Thermal movement.

Formula

\Delta_T \;=\; \alpha \, L_{tr} \, \Delta T

Substitute

\Delta_T \;=\; 6.0\times10^{-6}\,(195 \times 12)(100)

Result

\Delta_T \;=\; 1.40\;\text{in}

Step 2 — Shrinkage and creep.

Formula

\Delta_{SH+CR} \;=\; (\epsilon_{SH} + \epsilon_{CR})\, L_{tr}

Substitute

\Delta_{SH+CR} \;=\; (0.0002 + 0.0003)(195 \times 12)

Result

\Delta_{SH+CR} \;=\; 1.17\;\text{in}

Shrinkage and creep move only in the shortening direction (opening the joint).

Step 3 — Total design movement range.

Formula

\Delta_{open,max} \;=\; \Delta_{T,cold} + \Delta_{SH+CR}

Substitute

\Delta_{open,max} \;=\; \alpha L (T_{inst}-T_{min}) + 1.17 = 6\times10^{-6}(2340)(60-0) + 1.17

Result

= 0.84 + 1.17 = 2.01\;\text{in} \;\;\text{(joint opens further from install)}

Formula

\Delta_{close,max} \;=\; \alpha L (T_{max}-T_{inst})

Substitute

\Delta_{close,max} \;=\; 6\times10^{-6}(2340)(100-60)

Result

= 0.56\;\text{in}

Formula

\text{Total range} \;=\; \Delta_{open,max} + \Delta_{close,max}

Substitute

= 2.01 + 0.56

Result

= 2.57\;\text{in}

Step 4 — Joint selection. Total range 2.57 in. lies well within the 4-in. capacity of a single-cell strip-seal joint. Use a strip-seal system with a 4 in. movement rating.

Step 5 — Installation gap at 60 °F. Choose the gap so the joint fully closes at the hottest design temperature but never opens beyond the seal's tear rating:

Formula

G_{60°F} \;=\; G_{min} + \Delta_{close,max}

Substitute

G_{60°F} \;=\; 0.50 + 0.56

Result

G_{60°F} \;\approx\; 1.06 \;\Rightarrow\; \text{set gap to 1 1/8 in. at 60 °F}

At the coldest design temperature the gap will be $1.125 + 2.01 = 3.14\;\text{in}$ , comfortably below the 4-in. seal capacity.

Final section detailing (from computed A_s)

North abutment expansion joint — 3-span continuous PC bridge, 195 ft tributary

Location	A_s required	Bars provided	Spacing / detail
Joint system	Total range 2.57 in.	Strip-seal joint, 4 in. movement rating	Watertight EPDM V-gland between extruded steel edge rails
Installation gap (at 60 °F)	closes to 0.5 in. at 100 °F	1 1/8 in. gap opening at 60 °F	Adjust ±1/16 in. per 10 °F if installed at other temperature
Edge rail anchorage	shear & pullout, §14.5.3.2	3/4 in. Ø headed studs @ 12 in. o.c., alternating top and side	Embedded in header concrete cast against a formed blockout
Header concrete	high-strength, low-shrinkage	5 ksi silica-fume concrete, epoxy-coated #5 rebar mat	Header edge 1 in. below deck wearing surface
Drainage	self-draining	Rails sloped 1 % to low side; deck drain 12 in. before joint	Prevents debris ponding on gland

10.8 — Guided practice

Size a bearing pad for a lightly loaded pedestrian span

A 60-ft simply-supported pedestrian steel truss delivers a service reaction of $P_s = 45\;\text{kip}$ and requires $\Delta_s = 0.6\;\text{in}$ of thermal movement. Using G = 0.100 ksi neoprene, propose $L$ , $W$ , $h_{ri}$ , and the number of internal layers such that all Method-A limits are met.

Expected result

Try

L = 6\;\text{in},\; W = 10\;\text{in},\; h_{ri} = 3/8\;\text{in},\; n = 3

. Then

\sigma_s = 0.75\;\text{ksi} < 0.80

S = 5.0

(use two thinner layers if S must be ≥ 6),

h_{rt} = 1.125 \ge 2\Delta_s = 1.2

(barely — bump n to 4). Iterate.

10.9 — Mini design challenge

Bearings and joints for a curved 3-span PC bridge

Fig. 10.10Figure 10.10. Design-challenge geometry — curved-alignment three-span bridge, joints at each abutment.

Deliver a complete bearing-and-joint package for the curved bridge shown:

Service and factored reactions for the interior and exterior bearings.
Elastomeric pad size for each bearing type (verify Method A).
Total design movement at each abutment (including curvature-induced radial component).
Joint selection and installation gap for a 55 °F install temperature.
Sole-plate, masonry-plate, and anchor-bolt detail for a fixed and an expansion bearing.
A one-page design memo and a marked bearing/joint plan.

Submit your design challenge

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10.10 — Chapter summary

What you leave with

The three bearing duties — vertical, translation, rotation — and how each of the four bearing families addresses them.
Method A design of an elastomeric pad: shape factor, compressive limit, translation, and rotation.
The four sources of joint movement (T, SH, CR, PS) and how to combine them into a design range.
Strip-seal, modular, and finger-plate selection windows.
Rules for anchor bolts, sole plates, and the value of eliminating joints via integral abutments.

Section 2

Fully Worked Examples

Complete AASHTO LRFD solutions with knowns, assumptions, step calculations, verification, and design commentary. Difficulty rises from basic to consulting-grade.

Worked Example 1

Elastomeric bearing area from compressive stress limit

Basic

Problem

Size the plan dimensions of a steel-reinforced rectangular bearing (aspect ratio ≈ 1.4).

Step-by-Step

P = 120 + 90

Result

P = 210 kip

A \ge P/\sigma _{s} = 210/1.60

Result

A_{min} = 131 in^{2}

Design Verification

Aspect L/W = 1.4 keeps rotation demand roughly equal on both axes and simplifies steel-plate detailing. ✓

Discussion

Method A (§14.7.6) uses lower σ_s = 1.25 ksi but skips shear/rotation checks. Choose Method B when demand governs or when Method A limits force area too large.

Worked Example 2

Shape factor and instantaneous compressive deflection

Intermediate

Problem

Compute the shape factor S and instantaneous compressive deflection Δ_c.

Step-by-Step

S = W\cdot L/(2\cdot h_{ri}\cdot (W+L)) = 10\cdot 14/(2\cdot 0.5\cdot (10+14)) = 140/24

Result

S = 5.83

From §14.7.5.3.3 chart at S = 5.83, \sigma _{s} = 1.5, G = 0.15 \rightarrow \varepsilon _{c} \approx 0.055 per layer

Design Verification

Δ_c ≈ 0.15–0.2 in is normal for typical girder pads. If Δ_c > 0.5·h_rt, revise (either lower stress or thicker plates).

Discussion

S ≥ 4 is required for steel-reinforced pads to be modeled as constrained. Cotton-duck pads use S ≥ 100 conceptually — do not confuse the two.

Worked Example 3

Thermal shear displacement and stability check

Advanced

Problem

Compute thermal displacement Δ_s and verify shear-strain limit γ_s ≤ 0.5.

Step-by-Step

\Delta _{s} = \alpha \cdot L_{th}\cdot \Delta T\cdot \gamma _{TU} = 6.5e-6\cdot 2640\cdot 100\cdot 1.2 (\gamma _{TU} = 1.2 for Service I)

Result

\Delta _{s} = 2.06 in

\gamma _{s} = \Delta _{s}/h_{rt} = 2.06/3.0

Result

\gamma _{s} = 0.687 > 0.5 \rightarrow \ \text{fails}

Design Verification

Total elastomer thickness must scale with expansion length. Never carry a 200-ft segment on a 3-in pad without checking γ_s.

Discussion

If γ_s fails, options are: thicker elastomer, sliding surface (PTFE), or split expansion at intermediate joint. Do not exceed γ_s = 0.5 under service.

Worked Example 4

PTFE sliding-surface area for a pot bearing

Intermediate

Problem

Size the PTFE disc diameter D_p.

Step-by-Step

A_{req} = \dfrac{P}{\sigma_{allow}} = \dfrac{620}{3.5}

Result

A_{req} = 177.1\ \text{in}^{2}

D_p = \sqrt{\dfrac{4\,A_{req}}{\pi}} = \sqrt{\dfrac{4(177.1)}{\pi}} = 15.02\ \text{in}

Result

D_p = 16\ \text{in (standard size, } A = 201\ \text{in}^{2})

Design Verification

3.08/3.5 = 0.88 utilization — leaves headroom for LL under-estimation. Rounding up to a standard diameter is normal shop practice.

Discussion

PTFE creep at sustained stresses above 3.5 ksi accelerates surface wear and can seize the bearing. When live-load fraction is very high, consider dropping σ to 2.5 ksi to extend service life.

Worked Example 5

Rotation check on a steel-reinforced elastomeric pad

Advanced

Problem

Verify the Method-B combined compression + rotation requirement per §14.7.5.3.3.

Step-by-Step

\theta_s/n = 0.010/6 = 1.667\times 10^{-3}\ \text{rad}

\sigma_{s,\text{req}} = 0.5\,G\,S\,\left(\dfrac{L}{h_{ri}}\right)^{2}\left(\dfrac{\theta_s}{n}\right)

Design Verification

The pad has ~2.6× reserve against uplift at the toe edge. If θ_s doubled (e.g. staged construction rotation added), σ_s would need to double to keep the same margin — often prompting a shift to a pot or tall taper-reinforced pad.

Discussion

Rotation drives elastomeric-bearing failure more often than pure compression. Always check §14.7.5.3.3 with the largest realistic rotation, including construction sequence and any thermal gradient bowing.

Section 3

Guided Practice

Complete the missing steps. Use Hints for AASHTO article pointers and setup logic before revealing the full step. Submit at the end to send your work to your instructor.

Guided Problem 1

Elastomeric bearing — Method A shape factor

Steel-laminated elastomeric bearing pad, plan 14×22 in, three interior 0.5-in-thick elastomer layers. Method A design (AASHTO §14.7.6).

Step 1

Loaded plan area $A$ (in²).

Answer

Step 2

Perimeter of loaded plan (in).

Answer

Step 3

Shape factor $S = A/(P \cdot h_{ri})$ for $h_{ri} = 0.5\ \text{in}$ .

Answer

Step 4

Method A compressive stress limit under total load: $\sigma_{s} = \min(1.25 G S,\ 1.25\ \text{ksi})$ . $G = 0.130\ \text{ksi}$ .

Answer

Guided Problem 2

Thermal movement at expansion joint

Steel I-girder bridge with total joint-controlled length $L = 320\ \text{ft}$ . Design temperature range in AASHTO Procedure A: 120°F to 10°F.

Step 1

Temperature range $\Delta T$ (°F).

Answer

Step 2

Steel coefficient $\alpha$ (per °F). Use $6.5\times10^{-6}$ .

Answer

Step 3

Movement $\Delta = \alpha L \Delta T$ (in). $L$ in inches.

Answer

Step 4

Add creep/shrinkage safety factor ( $\times 1.2$ ) to size joint (in).

Answer

Guided Problem 3

Bearing selection by movement

Two-span PS girder bridge, $L = 180\ \text{ft}$ continuous, no intermediate joint. Girder rotation $\approx 0.010\ \text{rad}$ at end supports, longitudinal movement $\pm 1.6\ \text{in}$ .

Step 1

Approx elastomer thickness for 1.6-in shear: $h_{rt} \ge 2\Delta_{\max}$ (in).

Answer

Step 2

Number of interior layers needed for 0.5-in interior layer plan.

Answer

Step 3

Total bearing height incl. two 0.25-in exterior layers and six 0.075-in shims (in).

Answer

Step 4

For rotation $0.010\ \text{rad}$ and $W = 22\ \text{in}$ , differential compression $\Delta h = \theta W$ (in).

Answer

Guided Problem 4

Modular joint sizing

Steel-composite girder, $L = 780\ \text{ft}$ joint-to-joint, $\alpha = 6.5\times10^{-6}$ , $\Delta T = 120\ ^{\circ}\text{F}$ , plus 0.4-in shrinkage. Use 3-in per module gap capacity.

Step 1

Thermal movement (in).

Answer

Step 2

Total movement (thermal + shrinkage) (in).

Answer

Step 3

Number of modules at 3-in each (round up).

Answer

Step 4

Total joint width in fully open position with 2 in per module minimum (in).

Answer

Section 4

Independent Practice

Every problem randomizes its inputs. Work each step, submit for immediate feedback, request new values to practice again.

Practice 1

Shape factor S

Length

L = 19 in

Width

W = 10 in

Interior layer

hri = 0.25 in

Step 1A = L·W.

Answer (in²)

Step 2P = 2(L+W).

Answer (in)

Step 3S = A/(P·h_ri).

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 2

Thermal movement of a joint

Span length

L = 1180 ft

ΔT

dT = 115 °F

α (×10⁻⁶/°F)

alpha = 6.5 -

Step 1α·L·ΔT (in). Enter α as-provided (multiply ×10⁻⁶ internally).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 3

Compressive stress under total load

Load

P = 480 kip

L = 26 in

W = 27 in

Step 1σ = P/(L·W) (ksi).

Answer (ksi)

Randomized inputs, symbolic grading (±2%).

Practice 4

Elastomer shear strain from movement

Δ (in)

d = 2.0500000000000003 in

h_rt

hrt = 3.25 in

Step 1γ_s = Δ/h_rt.

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 5

Number of modules for modular joint

Total movement

d = 18 in

Per-module cap

cap = 3.25 in

Step 1Ceil(Δ/cap).

Answer (-)

Randomized inputs, symbolic grading (±2%).

Practice 6

Method A stress limit

G = 0.08 ksi

Shape factor

S = 5.5 -

Step 1σ = min(1.25·G·S, 1.25).

Answer (ksi)

Randomized inputs, symbolic grading (±2%).

Practice 7

Slab expansion Δ (concrete)

L = 370 ft

ΔT

dT = 105 °F

Step 1α = 6.0×10⁻⁶ · L · ΔT (in).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 8

Sliding pot bearing friction load

Vertical R

R = 400 kip

mu = 0.07 -

Step 1H = μ·R (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 9

Bearing rotation-induced compression

theta = 0.009000000000000001 rad

Bearing L

L = 9 in

Step 1Δh = θ·L (in).

Answer (in)

Randomized inputs, symbolic grading (±2%).

Practice 10

Girder longitudinal wind on bearing

V (mph)

V = 95 mph

Exposed A (ft²)

A = 1350 ft²

Step 1F = 0.00256·V²·A/1000 (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 11

Anchor bolt shear per bolt

Bearing H

H = 135 kip

# bolts

n = 5 -

Step 1v = H/n (kip).

Answer (kip)

Randomized inputs, symbolic grading (±2%).

Practice 12

Finger joint tooth spacing

Total movement

d = 14 in

Step 1Ceil(d/1.5) teeth per 12 in (approx).

Answer (-)

Randomized inputs, symbolic grading (±2%).