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Bridges & Structures

LRFD Steel Girder SuperStructure Design Example

Bearing Design Example Design Step 6

Table of Contents

Design Step 6.1 - Obtain Design Criteria
Design Step 6.2 - Select Optimum Bearing Typ
Design Step 6.3 - Select Preliminary Bearing Properties
Design Step 6.4 - Select Design Method (A or B)
Design Step 6.5 - Compute Shape Factor
Design Step 6.6 - Check Compressive Stress
Design Step 6.7 - Check Compressive Deflection
Design Step 6.8 - Check Shear Deformation
Design Step 6.9 - Check Rotation or Combined Compression and Rotation
Design Step 6.10 - Check Stability
Design Step 6.11 - Check Reinforcement
Design Step 6.12 - Design for Anchorage
Design Step 6.13 - Design Anchorage for Fixed Bearing
Design Step 6.14 - Draw Schematic of Final Bearing Design

Design Step 6.1 - Obtain Design Criteria

For this bearing design example, an abutment bearing was chosen. It was decided that the abutment would have expansion bearings. Therefore, the bearing design will be for an expansion bearing.

Refer to Design Step 1 for introductory information about this design example. Additional information is presented about the design assumptions, methodology, and criteria for the entire bridge, including the bearing design.

The following units are defined for use in this design example:

Formula: K = 1000lb Formula: ksi = Kips per square inch

For bearing design, the required design criteria includes:

  1. Longitudinal and transverse movement
  2. Longitudinal, transverse, and vertical rotation
  3. Longitudinal, transverse, and vertical loads

Most of the above information is typically obtained from the superstructure design software output, which is the case for this bearing design (first trial of girder design):

Formula: DL subscript serv = 78 point 4K Service I limit state dead load

Formula: LL subscript serv = 110 point 4K Service I limit state live load (including dynamic load allowance)

Formula: theta subscript sx = 0 point 0121rad Service I limit state total rotation about the transverse axis (see Figure 6-1)

Formula: P subscript sd = 67 point 8K Strength limit state minimum vertical force due to permanent loads (used in Design Step 6.12)

Design Step 6.2 - Select Optimum Bearing Type

Selecting the optimum bearing type depends on the load, movement capabilities, and economics. Refer to STable 14.6.2-1 and SFigure 14.6.2-1 for guidance on selecting the most practical bearing type. For the abutment bearing design, a steel-reinforced elastomeric bearing was selected. If the loads were considerably larger, pot bearings, which are more expensive than elasomeric bearings, would be an option.

S14.6.2

Design Step 6.3 - Select Preliminary Bearing Properties

Once the most practical bearing type has been selected, the preliminary bearing properties must be defined. The bearing properties are obtained from the Specifications, as well as from past experience. The following preliminary bearing properties were selected:

Bearing Pad Configuration

Pad length (bridge longitudinal direction): Formula: L subscript pad = 14 inches

Pad width (bridge transverse direction): Formula: W subscript pad = 15 inches

Elastomer cover thickness: Formula: h subscript rcover = 0 point 25 inches

Elastomer internal layer thickness: Formula: h subscript rinternal = 0 point 375 inches

Number of steel reinforcement layers: Formula: N subscript stlayers = 9

Steel reinforcement thickness: Formula: h subscript reinf = 0 point 1196 inches

Material Properties

S14.7.6.2 & S14.7.5.2

Elastomer hardness: Formula: H subscript shoreA = 50

Elastomer shear modulus: Formula: G = 0 point 095ksi

STable 14.7.5.2-1

Elastomer creep deflection at 25 years divided by the instantaneous deflection: Formula: C subscript d = 0 point 25

STable 14.7.5.2-1

Steel reinforcement yield strength: Formula: F subscript y = 50ksi

Design Step 6.4 - Select Design Method (A or B)

For this design example, Method A will be used. Method A usually results in a bearing with a lower capacity than a bearing designed with Method B. However, Method B requires additional testing and quality control. Method A is described in S14.7.6, while Method B is described in S14.7.5.

C14.7.5.1

Design Step 6.5 - Compute Shape Factor

The shape factor for individual elastomer layers is the plan area divided by the area of perimeter free to bulge.

S14.7.6.1 & S14.7.5.1

For steel-reinforced elastomeric bearings, the following requirements must be met prior to calculating the shape factor:

  1. All internal layers of elastomer must be the same thickness.
  2. The thickness of the cover layers cannot exceed 70 percent of the thickness of the internal layers.

S14.7.6.1 & S14.7.5.1

From Design Step 6.3, all internal elastomer layers are the same thickness, which satisfies Requirement 1. The following calculation verifies that Requirement 2 is satisfied:

Formula: 0 point 70 times h subscript rinternal = 0 point 26 inches

Formula: h subscript rcover = 0 point 25 inches OK

For rectangular bearings without holes, the shape factor for the ith layer is:

S14.7.5.1

Formula: S subscript i = numerator (L times W) divided by denominator (2 times h subscript ri times ( L + W)) . Equation not used

The shape factor for the cover layers is then:

Formula: S subscript cov = numerator (L subscript pad times W subscript pad) divided by denominator (2 times h subscript rcover times ( L subscript pad + W subscript pad ))

Formula: S subscript cov = 14 point 48

The shape factor for the internal layers is then:

Formula: S subscript int = numerator (L subscript pad times W subscript pad) divided by denominator (2 times h subscript rinternal times ( L subscript pad + W subscript pad ))

Formula: S subscript int = 9 point 66

Design Step 6.6 - Check Compressive Stress

The compressive stress check limits the compressive stress in the elastomer at the service limit state as follows:

S14.7.6.3.2

Formula: sigma subscript s less than or equal to 1 point 0ksi and Formula: sigma subscript s less than or equal to 1 point 0 times G times S

The compressive stress is taken as the total reaction at one of the abutment bearings for the service limit state divided by the elastomeric pad plan area. The service limit state dead and live load reactions are obtained from the Opis superstructure output. The shape factor used in the above equation should be for the thickest elastomer layer.

Service I limit state dead load: Formula: DL subscript serv = 78 point 4 K

Service I limit state live load (including dynamic load allowance):

Formula: LL subscript serv = 110 point 4 K

Formula: sigma subscript s = numerator (DL subscript serv + LL subscript serv) divided by denominator (( L subscript pad times W subscript pad ))

Formula: sigma subscript s = 0 point 899 ksi

Formula: 1 point 0 times G times S subscript int = 0 point 917 ksi OK

The service average compressive stress due to live load only will also be computed at this time. It will be needed in Design Step 6.11. Again, the service limit state live load value was obtained from Opis superstructure output.

Formula: sigma subscript L = numerator (LL subscript serv) divided by denominator (( L subscript pad times W subscript pad ))

Formula: sigma subscript L = 0 point 526 ksi

Design Step 6.7 - Check Compressive Deflection

The compressive deflection due to the total load at the service limit state is obtained from the following equation:

S14.7.5.3.3

Formula: delta = Sigma epsilon subscript i times h subscript ri. Equation not used

For this design example, the instantaneous compressive strain was approximated from CTable 14.7.5.3.3-1 for 50 durometer reinforced bearings using a compressive stress of 0.899 ksi and a shape factor of 9.66.

Formula: epsilon subscript int = 0 point 04

CTable 14.7.5.3.3-1

The instantaneous deflection is then:

Formula: delta subscript inst = 2 times epsilon subscript int times h subscript rcover + 8 times epsilon subscript int times h subscript rinternal

S14.7.5.3.3

Formula: delta subscript inst = 0 point 140 inches

The effects of creep should also be considered. For this design example, material-specific data is not available. Therefore, calculate the creep deflection value as follows:

STable 14.7.5.2-1

Formula: delta subscript creep = C subscript d times delta subscript inst

Formula: delta subscript creep = 0 point 035 inches

The total deflection is then:

Formula: delta subscript total = delta subscript inst + delta subscript creep

Formula: delta subscript total = 0 point 175 inches

The initial compressive deflection in any layer of a steel-reinforced elastomeric bearing at the service limit state without dynamic load allowance shall not exceed 0.07hri.

S14.7.6.3.3

In order to reduce design steps, the above requirement will be checked using the deflection calculated for the service limit state including dynamic load allowance. If the compressive deflection is greater than 0.07hri, then the deflection without dynamic load allowance would need to be calculated.

Formula: delta subscript int1layer = epsilon subscript int times h subscript rinternal

Formula: delta subscript int1layer = 0 point 015 inches

Formula: 0 point 07h subscript rinternal = 0 point 026 inches OK

Design Step 6.8 - Check Shear Deformation

The shear deformation is checked to ensure that the bearing is capable of allowing the anticipated horizontal bridge movement. Also, the shear deformation is limited in order to avoid rollover at the edges and delamination due to fatigue caused by cyclic expansion and contraction deformations. The horizontal movement for this bridge design example is based on thermal effects only. The thermal movement is taken from Design Step 7.6 for the controlling movement, which is contraction. Other criteria that could add to the shear deformation include construction tolerances, braking force, and longitudinal wind if applicable. One factor that can reduce the amount of shear deformation is the substructure deflection. Since the abutment height is relatively short and the shear deformation is relatively small, the abutment deflection will not be taken into account.

S14.7.6.3.4

C14.7.5.3.4

The bearing must satisfy:

Formula: h subscript rt greater than or equal to 2 times Delta subscript s

Formula: h subscript rt = 2 times h subscript rcover + 8 times h subscript rinternal

Formula: h subscript rt = 3 point 50 inches

Formula: Delta subscript contr = 0 point 636 inches from Design Step 7.6 for thermal contraction

Formula: gamma subscript TU = 1 point 20 for the service limit state

STable 3.4.1-1 & S3.4.1

Formula: Delta subscript s = gamma subscript TU times Delta subscript contr

Formula: Delta subscript s = 0 point 76 inches

Formula: 2 times Delta subscript s = 1 point 53 inches

Formula: 3 point 50 inches greater than or equal to 1 point 53 inches OK

Design Step 6.9 - Check Rotation or Combined Compression and Rotation

Since Design Method A was chosen, combined compression and rotation does not need to be checked. The rotation check ensures that no point in the bearing undergoes net uplift and is as follows:

S14.7.6.3.5

Formula: sigma subscript s greater than or equal to 0 point 5G times S times ( numerator (L) divided by denominator (h subscript ri) ) squared times numerator ( theta subscript sx) divided by denominator (n) . Equation not used (associated with rotation about transverse axis)

S14.7.6.3.5d

and

Formula: sigma subscript s greater than or equal to 0 point 5G times S times ( numerator (W) divided by denominator (h subscript ri) ) squared times numerator ( theta subscript sz) divided by denominator (n) . Equation not used (associated with rotation about longitudinal axis)

Formula: sigma subscript s = 0 point 899 ksi

The service rotation due to the total load about the transverse axis was taken from Opis:

Formula: theta subscript sx = 0 point 0121 Formula: rad

S14.7.6.3.5d

Formula: MathCad tip logo

Construction Tolerance

For spans over approximately 100 feet, it is good engineering practice to include an additional 0.005 radians of rotation about both pad axes to account for construction tolerances.

The number of interior layers is:

Formula: n = 8 + 0 point 5 + 0 point 5

Formula: 0 point 5 times G times S subscript int times ( numerator (L subscript pad) divided by denominator (h subscript rinternal) ) squared times numerator ( theta subscript sx) divided by denominator (( 8 + 1)) = 0 point 859 ksi OK

The service rotation due to the total load about the longitudinal axis is negligible compared to the service rotation about the transverse axis. Therefore, the check about the longitudinal axis will be assumed to be negligible and is not computed in this bearing design example.

Design Step 6.10 - Check Stability

The total thickness of the pad shall not exceed the least of L/3 or W/3.

S14.7.6.3.6

Formula: numerator (L subscript pad) divided by denominator (3) = 4 point 67 inches Formula: numerator (W subscript pad) divided by denominator (3) = 5 point 00 inches

The total thickness of the pad based on the preliminary

dimensions is:

Formula: h subscript total = 2 times h subscript rcover + 8 times h subscript rinternal + N subscript stlayers times h subscript reinf

Formula: h subscript total = 4 point 5764 inches OK

Design Step 6.11 - Check Reinforcement

The thickness of the steel reinforcement must be able to sustain the tensile stresses induced by compression in the bearing. The reinforcement thickness must also satisfy the requirements of the AASHTO LRFD Bridge Construction Specifications.

S14.7.6.3.7

S14.7.5.3.7

For the service limit state:

Formula: h subscript s greater than or equal to numerator (3h subscript max times sigma subscript s) divided by denominator (F subscript y)

Formula: h subscript max = h subscript rinternal Formula: h subscript max = 0 point 375 inches

Formula: sigma subscript s = 0 point 899 ksi Formula: F subscript y = 50 ksi

Formula: numerator (3 times h subscript max times sigma subscript s) divided by denominator (F subscript y) = 0 point 0202 inches

Formula: h subscript reinf = 0 point 1196 inches OK

For the fatigue limit state:

Formula: h subscript s greater than or equal to numerator (2h subscript max times sigma subscript L) divided by denominator ( Delta F subscript TH)

From Design Step 6.6, the service average compressive stress due to live load only is:

Formula: sigma subscript L = 0 point 526 ksi

Formula: Delta F subscript TH = 24 point 0ksi

STable 6.6.1.2.5-3

Formula: numerator (2 times h subscript max times sigma subscript L) divided by denominator ( Delta F subscript TH) = 0 point 0164 inches

Formula: h subscript reinf = 0 point 1196 inches OK

Design Step 6.12 - Design for Anchorage

The bearing pad must be secured against transverse horizontal movement if the factored shear force sustained by the deformed pad at the strength limit state exceeds one-fifth of the minimum vertical force due to permanent loads, Psd.

S14.7.6.4

Formula: P subscript sd = 67 point 8K taken from Opis output

The maximum factored shear force sustained by the deformed pad at the strength limit state is obtained from Design Step 7.6, adding wind on superstructure and wind on live load. The maximum shear force will occur when wind is taken at 0 degrees.

The shear force due to wind on superstructure is taken from Table 7-1:

Formula: WS = 30 point 69K

The shear force due to wind on live load is taken from Table 7-2:

Formula: WL = 6 point 00K

The controlling shear force is either from Strength III or Strength V:

Factored shear force per bearing for Strength III:

Formula: gamma subscript WS = 1 point 40

STable 3.4.1-1

Formula: gamma subscript WL = 0 point 00

STable 3.4.1-1

Formula: V subscript windstrIII = numerator (( gamma subscript WS times WS + gamma subscript WL times WL )) divided by denominator (5)

Formula: V subscript windstrIII = 8 point 59 K

Factored shear force per bearing for Strength V:

Formula: gamma subscript WS = 0 point 40

STable 3.4.1-1

Formula: gamma subscript WL = 1 point 00

STable 3.4.1-1

Formula: V subscript windstrV = numerator (( gamma subscript WS times WS + gamma subscript WL times WL )) divided by denominator (5)

Formula: V subscript windstrV = 3 point 66 K

Use: Formula: V subscript max = max ( V subscript windstrIII , V subscript windstrV )

Formula: V subscript max = 8 point 59 K

Formula: numerator (1) divided by denominator (5) times P subscript sd = 13 point 56 K

Since the maximum shear force at the strength limit state does not exceed one-fifth of the minimum vertical force due to permanent dead loads, the pad does not need to be secured against horizontal movement.

Design Step 6.13 - Design Anchorage for Fixed Bearings

The abutment bearings are expansion in the longitudinal direction but fixed in the transverse direction. Therefore, the bearings must be restrained in the transverse direction. Based on Design Step 6.12, the expansion bearing pad does not need to be secured against horizontal movement. However, based on S3.10.9.2, the horizontal connection force in the restrained direction cannot be less than 0.1 times the vertical reaction due to the tributary permanent load and the tributary live loads assumed to exist during an earthquake. In addition, since all abutment bearings are restrained in the transverse direction, the tributary permanent load can be taken as the reaction at the bearing. Also, γEQ is assumed to be zero. Therefore, no tributary live loads will be considered. This transverse load will be used to design the bearing anchor bolts for this design example.

S14.8.3.1

S3.10.9.2

C3.4.1

For the controlling girder (interior):

Formula: DL subscript serv = 78 point 4 K

The maximum transverse horizontal earthquake load per bearing is then:

Formula: H subscript EQ = 0 point 1 times DL subscript serV

Formula: H subscript EQ = 7 point 84 K

The factored shear resistance of the anchor bolts per bearing is then:

S14.8.3.1 & S6.13.2.7

Assume two 5/8" diameter A 307 bolts with a minimum tensile strength of 60 ksi:

S6.4.3

Formula: R subscript n = 0 point 48 times A subscript b times F subscript ub times N subscript s. Equation not used for threads excluded from shear plane

S6.13.2.7

Formula: phi subscript s = 0 point 65 resistance factor for A 307 bolts in shear

S6.5.4.2

Formula: A subscript b = numerator (pi times ( 0 point 625 inches ) squared ) divided by denominator (4)

Formula: A subscript b = 0 point 31 inches squared

Formula: F subscript ub = 60ksi

Formula: N subscript s = 2 (number of bolts)

Formula: R subscript n = 0 point 48 times A subscript b times F subscript ub times N subscript s Formula: R subscript n = 17 point 67 K

Formula: R subscript r = phi subscript s times R subscript n Formula: R subscript r = 11 point 49 K

Formula: R subscript r greater than or equal to H subscript EQ OK

Once the anchor bolt quantity and size are determined, the anchor bolt length must be computed. As an approximation, the bearing stress may be assumed to vary linearly from zero at the end of the embedded length to its maximum value at the top surface of the concrete. The bearing resistance of the concrete is based on S5.7.5.

S14.8.3.1 & C14.8.3.1

Formula: phi subscript b times P subscript n = phi subscript b times 0 point 85 times f subscript c times A subscript 1 times m. Equation not used

S5.7.5

Formula: Stress subscript brg = numerator ( phi subscript b times P subscript n) divided by denominator (A subscript 1) . Equation not used

Formula: Stress subscript brg = phi subscript b times 0 point 85 times f subscript c times m. Equation not used

Assume: Formula: m = 0 point 75 (conservative assumption)

Formula: phi subscript b = 0 point 70 for bearing on concrete

S5.5.4.2.1

Formula: Stress subscript brg = phi subscript b times 0 point 85 times ( 4ksi) times m

Formula: Stress subscript brg = 1 point 78 ksi

The total transverse horizontal load is:

Formula: H subscript EQ = 7 point 84 K

The transverse load per anchor bolt is then:

Formula: P subscript 1bolt = numerator (H subscript EQ) divided by denominator (2)

Formula: P subscript 1bolt = 3 point 92 K

Using the bearing stress approximation from above, the required anchor bolt area resisting the transverse horizontal load can be calculated.

Formula: A subscript 1 = numerator (P subscript 1bolt) divided by denominator (( numerator (Stress subscript brg + 0) divided by denominator (2) ))

Formula: A subscript 1 = 4 point 39 inches squared

A1 is the product of the anchor bolt diameter and the length the anchor bolt is embedded into the concrete pedestal/beam seat. Since we know the anchor bolt diameter, we can now solve for the required embedment length.

Formula: L subscript embed = numerator (A subscript 1) divided by denominator (0 point 625 inches )

Formula: L subscript embed = 7 point 03 inches

Individual states and agencies have their own minimum anchor bolt embedment lengths. For this design example, a minimum of 12 inches will be used.

Use: Formula: L subscript embed = 12 point 0 inches

Design Step 6.14 - Draw Schematic of Final Bearing Design

The bearing pad plan view shows the rectangular pad with the centerline of girder in the vertical direction, or longitudinal axis, and the centerline of bearing in the transverse direction in the horizontal direction. The axes are 90 degrees to each other. The pad length is 14 point 00 inches and runs in the direction of the longitudinal axis or in the vertical direction. The pad width is given at 15 point 00 inches along the transverse axis in the horizontal direction.

Figure 6-1 Bearing Pad Plan View

The bearing pad elevation shows a laminated pad with a 15 point 00 inch width and an overall height of 4 point 5764 inches. The total height consists of elastomeric material and steel reinforcement layers. There are 8 interior elastomeric layers of zero point 375 inches each, 2 exterior layers of elastomeric zero point 25 inches thickness and 9 steel reinforcement layers at zero point 1196 inches thick each.

Figure 6-2 Bearing Pad Elevation View

The section is through the girder showing the bottom portion of the web and bottom flange. Also shown is the bearing pad, sole plate, anchor bolts and beam seat area. The bearing pad height of 4 point 5764 inches. The two 5 eights inch diameter A307 anchor bolts project through the sole plate with a washer and two hex nuts above the sole plate. An embedment length of 12 point 00 inches typical for the anchor bolts is called out.

Figure 6-3 Anchor Bolt Embedment

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Updated: 07/18/2013
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