8.bearing Design

CHAPTER 8
Bearing Design
1/20/2017 Chalachew A. 1
8. Bearing
8.1 General
 Bearings are structural devices positioned between the bridge superstructure and
the substructure to transmit loads from the superstructure to substructure, and
accommodate relative movements between the superstructure and the
substructure.
 The forces applied to a bridge bearing mainly include superstructure self-weight,
traffic loads, wind loads, and earthquake loads.
• Movements in bearings include translations and rotations.
• Creep, shrinkage, and temperature effects are the most common causes of the
translational movements.
• Traffic loading, construction  Expansion bearings allow
tolerances, and uneven settlement both rotational and
of the foundation are the common translational movements.
causes of the rotations.
8.2 Types of Bearings
 Bearings may be classified as

 fixed bearing
 Fixed bearings allow rotations but Fixed Bearing
restrict translational movements. Expansion
Bearing
 Expansion bearing
• There are numerous types of bearings available. The following are the
principal types of bearings currently in use.
• Rocker Bearing
• Pin Bearing
• Roller Bearing
• Slider Bearing
• Elastomeric Bearing
• Curved Bearing
• Pot Bearing
• Disk Bearing
1. Pin Bearing
• A pin bearing is a type of fixed bearings that accommodates rotations
through the use of a steel
• Translational movements are not allowed.
• The pin at the top is composed of upper and lower semi circularly
recessed surfaces with a solid circular pin placed between.
• Usually, there are caps at both ends of the pin to keep the pin from sliding
off the seats and to resist uplift loads if required.
• The upper plate is connected to the sole plate by either bolting or
welding. The lower curved plate sits on the masonry plate.
.
2. Roller Type of Bearing
2. Roller Type of Bearing
3. Rocker Type of Bearing
4. Sliding Bearings a widely used brand of PTFE
•A sliding bearing utilizes one
plane metal plate sliding against
another to accommodate
translations.
•The sliding bearing surface
produces a frictional force that is
applied to the superstructure,
substructure, and the bearing
itself.
•To reduce this friction force, PTFE
(polytetrafluoroethylene) is often
used as a sliding lubricating
material. PTFE is sometimes
referred to as Teflon, named after
5.1 Plain Elastomer Bearings 5.2 Laminated Elastomeric Bearing
.
FIXED ELASTOMER BEARING AND COVER PLATE
.
EXPANSION BEARINGS AND COVER PLATE
.
.
.
.
8.3 Design of Steel-Reinforced Elastomeric Bearings
 Method A Design Methods
 Method B Design Methods
 The Method A procedure in LFRD Article 14.7.6 shall be used for steel-
reinforced elastomeric bearings.
 The Method B procedure in LRFD Article 14.7.5 may be used for high-capacity
bearings, but only with the approval of the Chief Structures Engineer.
 High-capacity elastomeric bearings should be used only where very tight

geometric constraints, extremely high loads, or special conditions or
circumstances require the use of higher grade material.
 The Method B design procedure allows significantly higher average
compressive stresses.
 These higher allowable stress levels are justified by an additional

acceptance test, specifically a long-duration compression test.
 Designers must prepare a unique Special Provision for inclusion in

the contract documents if a high-capacity elastomeric bearing is used.
Consider the following factor when selecting a bearing to use:
Vertical and Horizontal Loads
Translational and Rotational Movements
Available Clearance
Environmental (Corrosion/temperature)
Initial Cost
Maintenance Cost
Availability
Owner’s Preference
Bearing Capacity of Common Bearings
Example :- Steel-Elastomer Bearing Design
I. Bearing Pad Configuration:
Pad Length (in the Bridge Longitudinal Direction) = Lpad 450mm
Pad Width (in the Bridge Transverse Direction) = Wpad 380mm

Elastomer Cover Thickness h_e(cover)= 4mm
Elastomer Internal Layer Thickness h_i(internal)= 10mm
Number of Steel Reinforcement Layers = N_steel layers 6
Number of Internal Elastomer Layers = N_steel layers-1 =5-1= 4
Steel reinforcement Thickness = h_s (Steel)= 4mm
Total Thickness of the Bearing Pad (ho) =2*4+6*4+5*10= 82mm
.
.
II. Material Properties
 Elastomer Hardness Durometer=60
 Elastomer Shear Modulus=G=1.07Mpa
𝐶𝑟𝑒𝑒𝑝 𝑑𝑒𝑓𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 25 𝑦𝑒𝑟𝑎𝑠

 For Durometer 60 elastomer 𝑐𝑑 = = 0.35
𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑑𝑒𝑓𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛
 Steel Reinforcement yield strength=fy=300Mpa
III. Design Method

 Method A bearing Design will be used.
IV. Compute Shape Factor
 For steel reinforced Elastomer bearing,
1. All interior layers of Elastomer need to have the same thickness
2. The thickness of the cover layers shall not exceed 70% of
thickness of internal layers.
0.7*hi= 0.7*10=7mm
he=4mm
7mm > Cover thickness (4mm)………………………….OK!!!
 For rectangular bearing without holes, Shape Factor is:
 Shape factor for internal layer=
450 ∗ 380
S 4mm = = 25.8
2 ∗ 4 ∗ 450 + 380
450 ∗ 380
S(10mm) = = 10.3
2 ∗ 10 ∗ (450 + 380)
V. Check Compressive Stress
𝜎𝑠 ≤ 7.00 𝑀𝑝𝑎 𝑎𝑛𝑑 𝜎𝑠 ≤ 1 ∗ 𝐺 ∗ 𝑆 −−−− −14.7.6.3.2 − 4

 Where the value of S(shape factor) used shall be that for the thickest layer
of bearing.
 These stress limits may be increased by 10 percent where shear

deformation is prevented
 Service Limit State Dead Load =165.95KN+18.88KN=184.83KN
 Service Limit State Live Load with Impact = LL_Serv =325.45KN
 Total Service Limit State Load per bearing =Total Serv. Load= 510.28KN
𝑇𝑜𝑡𝑎𝑙 𝑆𝑒𝑟𝑣𝑖𝑐𝑒 𝐿𝑜𝑎𝑑
𝜎𝑠 =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝑝𝑎𝑑
510.28∗1000
 𝜎𝑠 = = 2.98𝑀𝑝𝑎 = 𝟑𝑴𝒑𝒂
380∗450
 1*G*S=1*1.07*10.3=11.021Mpa
 𝜎𝑠 ≤ 7𝑀𝑝𝑎 𝑎𝑛𝑑 𝜎𝑠 ≤ 1 ∗ 𝐺 ∗ 𝑆
 3𝑀𝑝𝑎 < 7𝑀𝑝𝑎 𝑎𝑛𝑑 3𝑀𝑝𝑎 < 11.021𝑀𝑝𝑎

OK!!!
VI. Check Shear Deformation
 The horizontal movement for this bridge design is based on thermal effects
only.
o Temperature Range (Between Max and Min. Temperatures) = ∆T= 40℃
o Coefficient of Thermal Expansion =ε = 11.7 × 10− 6 (mm/mm/oC) =0.000117
o Length of Superstructure under Thermal Stress = Total Length of Girder = 16200mm
o Width of Superstructure under Thermal Stress = Superstructure Width =8900mm
o Shear deformation of the elastomer at service limit state in the longitudinal direction of the
bridge
o Δsz = ε ⋅ΔT⋅(Lspan)=40*0.0000117*16200=7.58mm
o Shear deformation of the elastomer at service limit state in the transverse direction of
the bridge
o Δsx = ε ⋅ΔT⋅(Wbridge)=40*0.0000117*8900=4.17mm
o max. total shear deformation of the elastomer at service limit state =
o ∆𝑠 = 7.582 + 4.172 =8.65mm
o The pad elastomer material (steel plate thickness not included) total thickness must be
twice the expected thermal movement at the bearing.
hiMin = Minimum Allowable Total Elastomer Height >2* ∆𝑠 =17.3mm
o Thickness of the Elastomer excluding the steel laminates =82-6*4=58>17.3 OK!!

VII Bearing Pad Slip Check
Δs Max = Maximum Allowable total shear deformation of the elastomer at service

limit state Where Gr = Roadway Gradient on the Bridge
Gr =0 [Bridge is with zero Gradient.]
Go=1.07Mpa
0.2−0 ∗184.83∗1000∗58
ΔsMax = = 11.72𝑚𝑚
1.07∗450∗380
11.72𝑚𝑚 > ∆𝑠 = 8.65𝑚𝑚 … … … … … . . 𝑂𝐾‼!
VIII. Check Compressive Deflection
 Instantaneous Compressive Strain for 60 Durometer reinforced bearings using a

compressive stress shown here below;
 𝜎𝑠 = 3𝑀𝑝𝑎 𝑎𝑛𝑑 𝑆ℎ𝑎𝑝𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑒 𝑆𝑖𝑛𝑡 = 10.3 = 10
 𝑓𝑟𝑜𝑚 𝑔𝑟𝑎𝑝ℎ 𝜀 𝑖𝑛𝑡 = 0.02
 Thickness of the Elastomer excluding the steel laminates hint =58mm
 Instantaneous deflection = 𝛿𝐢𝐧𝐬𝐭 = 𝜀 ∗ 𝐡𝐢𝐧𝐭 = 𝟓𝟖 ∗ 𝟎. 𝟎𝟐 = 𝟏. 𝟏𝟔𝐦𝐦
 𝛿creep=Cd*𝛿ints=0.35*1.16=0.41mm
 Total Deflection =𝛿creep+ 𝛿𝐢𝐧𝐬𝐭=1.57mm
.
 𝛿 int_1_layer= 𝜀 int*h_internal≤0.07h_internal ------------Art. 14.7.6.3.3
 𝛿 int_1_layer=0.02*10=0.2mm
 0.07h_internal=0.07*10=0.7mm
 0.2 mm< 0.7mm ------------------------------------------------OK!!!
IX. Check Stability
 Total Thickness of Pad shall not exceed minimum of (Lpad/3 or Wpad/3)
 Lpad/3 =150mm
 Wpad/3 =126.67mm
 Minimum of the two =126.67mm>82mm----------------------------------OK!!!

X. Check Reinforcement
• The thickness of the steel reinforcement, hs, shall satisfy the following:
• At the service limit state:
• hmax = Max (h_internal, h_rcover) =10mm
𝒉𝒔 = 𝟒𝒎𝒎
𝟑 ∗ 𝟏𝟎 ∗ 𝟑
𝒉𝒔(𝒂𝒍𝒍𝒐𝒘) = = 𝟎. 𝟑𝒎𝒎
𝟑𝟎𝟎
4mm>0.3mm------------------------------------------------------------OK!!!
• The thickness of the steel reinforcement, hs, shall satisfy the following:
• At the fatigue limit state:
325.45 ∗ 1000
𝜎𝐿𝐿 = = 1.9𝑀𝑝𝑎
450 ∗ 380
hmax = Max (h_int, h_rcover) =10mm
𝒉𝒔 = 𝟒𝒎𝒎
𝟐 ∗ 𝟏𝟎 ∗ 𝟏. 𝟗
𝒉𝒔(𝒂𝒍𝒍𝒐𝒘) = = 𝟎. 𝟐𝟑𝒎𝒎
𝟏𝟔𝟓
4mm>0.23mm--------------------------OK!!!

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8.bearing Design

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CHAPTER 8

 Bearings may be classified as

FIXED ELASTOMER BEARING AND COVER PLATE

 Method A Design Methods

 Method B Design Methods

 High-capacity elastomeric bearings should be used only where very tight

 These higher allowable stress levels are justified by an additional

 Designers must prepare a unique Special Provision for inclusion in

Pad Width (in the Bridge Transverse Direction) = Wpad 380mm

𝐶𝑟𝑒𝑒𝑝 𝑑𝑒𝑓𝑒𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑡 25 𝑦𝑒𝑟𝑎𝑠

 Steel Reinforcement yield strength=fy=300Mpa

III. Design Method

1. All interior layers of Elastomer need to have the same thickness

2. The thickness of the cover layers shall not exceed 70% of

thickness of internal layers.

7mm > Cover thickness (4mm)………………………….OK!!!

𝜎𝑠 ≤ 7.00 𝑀𝑝𝑎 𝑎𝑛𝑑 𝜎𝑠 ≤ 1 ∗ 𝐺 ∗ 𝑆 −−−− −14.7.6.3.2 − 4

 These stress limits may be increased by 10 percent where shear

 Service Limit State Dead Load =165.95KN+18.88KN=184.83KN

 Service Limit State Live Load with Impact = LL_Serv =325.45KN

 3𝑀𝑝𝑎 < 7𝑀𝑝𝑎 𝑎𝑛𝑑 3𝑀𝑝𝑎 < 11.021𝑀𝑝𝑎

o Coefficient of Thermal Expansion =ε = 11.7 × 10− 6 (mm/mm/oC) =0.000117

o Length of Superstructure under Thermal Stress = Total Length of Girder = 16200mm

o Width of Superstructure under Thermal Stress = Superstructure Width =8900mm

o max. total shear deformation of the elastomer at service limit state =

o ∆𝑠 = 7.582 + 4.172 =8.65mm

o Thickness of the Elastomer excluding the steel laminates =82-6*4=58>17.3 OK!!

Δs Max = Maximum Allowable total shear deformation of the elastomer at service

Gr =0 [Bridge is with zero Gradient.]

 Instantaneous Compressive Strain for 60 Durometer reinforced bearings using a

 𝜎𝑠 = 3𝑀𝑝𝑎 𝑎𝑛𝑑 𝑆ℎ𝑎𝑝𝑒 𝑓𝑎𝑐𝑡𝑜𝑟𝑒 𝑆𝑖𝑛𝑡 = 10.3 = 10

 𝑓𝑟𝑜𝑚 𝑔𝑟𝑎𝑝ℎ 𝜀 𝑖𝑛𝑡 = 0.02

 Thickness of the Elastomer excluding the steel laminates hint =58mm

 Instantaneous deflection = 𝛿𝐢𝐧𝐬𝐭 = 𝜀 ∗ 𝐡𝐢𝐧𝐭 = 𝟓𝟖 ∗ 𝟎. 𝟎𝟐 = 𝟏. 𝟏𝟔𝐦𝐦

 Total Deflection =𝛿creep+ 𝛿𝐢𝐧𝐬𝐭=1.57mm

 0.2 mm< 0.7mm ------------------------------------------------OK!!!

IX. Check Stability

 Total Thickness of Pad shall not exceed minimum of (Lpad/3 or Wpad/3)

 Minimum of the two =126.67mm>82mm----------------------------------OK!!!

• hmax = Max (h_internal, h_rcover) =10mm

hmax = Max (h_int, h_rcover) =10mm

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