Chapter 9 Seismic Isolation Systems. Chapter 10 9 Seismic Isolation Systems

Transcription

1 Chapter 9 Seismic Isolation Systems 1

2 CONTENT 1. Introduction 2. Laminated Rubber Bearings 3. Lead-rubber Bearings 4. Friction Pendulum System 5. Other Seismic Isolation Systems 6. Example of adequacy assessment of a lead-rubber bearing under Maximum Credible Earthquake (MCE) 2

3 Major References Chapter 10 Sections 10.1 to D-Based-Analysis-and-Design-Procedures-for-Bridge- Bearings-and-Seismic-Isolators-MCEER html 3

4 1. Introduction Overview of various isolator components developed, successfully tested and implemented. Emphasizes two main types of systems: Laminated rubber bearing systems Friction Pendulum System Overview of other systems and recent developments in isolation hardware also presented. 4

5 2. Laminated Rubber Bearings Laminated rubber bearings (elastomeric bearings) used extensively for bridge superstructures to accommodate temperature-induced movements deformations. In last 20 years, use extended to seismic isolation of buildings and other structures. Lead-rubber (lead-plug) bearing: Elastomeric bearing with central lead plug designed to yield under lateral deformation and to dissipate supplemental energy. Discussed in next section. 5

6 2. Laminated Rubber Bearings Photo: Courtesy of M. Constantinou 6

7 2. Laminated Rubber Bearings Image T. Saito 7

8 2. Laminated Rubber Bearings Elastomeric Bearings for Sakhalin I Orlan Platform. Tested at University at Buffalo. Photos: Courtesy of M. Constantinou 8

9 2. Laminated Rubber Bearings Force-Displacement relationship for various types of elastomeric bearings Shear strain defined as lateral displacement/total height of rubber High Damping Rubber Low Damping Scragging Lead Rubber (From Thompson et al. 2000) 9

10 2. Laminated Rubber Bearings Full-Scale Isolated Bridge Testing Video 10

11 2. Laminated Rubber Bearings Full-Scale Isolated Bridge Testing 4 3 Force-Displacement Hysteresis - Side A LC6 vs D5 4 3 Force-Displacement Hysteresis-Side B LC2 vs D F (kip) 0-1 F (kip) Uy (in) Uy (in) 11

12 2. Laminated Rubber Bearings Disadvantage of Laminated Rubber Bearings: Relatively low damping provided by the rubber. High damping rubbers: Developed for laminated rubber bearings. Used mainly in Japan (Pan et al. 2004). Significant more energy dissipation than low damping rubbers. 20% damping at shear strains of 300%. More susceptible to heat related property changes during cyclic loading and to aging effects. Increases complexity to predict short and long term properties for bounding analysis. Isolator damping external components: Lead plug inserted in center of the bearing (lead-rubber bearings). External supplemental damping by hysteretic or viscous dampers. 12

13 2. Laminated Rubber Bearings Key Parameters in Design of Laminated Rubber Bearings: Gravity load carrying capacity. Bearings must not be overloaded under gravity loads and vertical loads induced by lateral response. Rotational effects between the top and bottom of bearings. Maximum achievable relative displacement between top and base of bearing. Limited by allowable rubber strain or bearing stability Minimum thickness of steel shims. 13

14 2. Laminated Rubber Bearings Gravity Load Carrying Capacity of Laminated-Rubber Bearings. 14

15 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Overlapping Area Circular Bearings: 2 D Ar = ( δ sin δ) 4 δ = D 1 2cos ( ) A r 15 15

16 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Overlapping Area Rectangular bearings: A r Ar = BL ( ) B L 16 16

17 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Overlapping Area Cylindrical hollow bearings: A r A r A δ = ( δ sin δ) π D 1 2cos ( ) o 17 17

18 2. Laminated Rubber Bearings Gravity Load Carrying Capacity For a given compression load, P, the maximum shear strain, γ c, in the rubber is given by: γ = c P AGS f 1 A is the rubber area bonded to the shim plates (the area must be reduced to A r to take into account the lateral displacement). G is the shear modulus of the rubber (0.5 to 1.0 MPa). S is the shape factor of each rubber layer. f 1 is a numerical factor that depends on the shape of the bearing, the compressibility of the rubber and the locations of the maximum shear strain (1.0 f 1 3.4)

19 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Shape factor, S: circular γ = c P AGS f 1 Note: the shape factor, S, must be calculated for each rubber layer of thickness t and not for the total rubber thickness

20 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Shape factor, S: γ = c P AGS f 1 The shape factor for a circular hollow bearing of outside diameter D o and inside diameter D i and made of rubber layers of thickness t is given by: Note: the shape factor, S, must be calculated for each rubber layer of thickness t and not for the total rubber thickness

21 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for circular bearings: γ = c P AGS f 1 f 1 S K/G K is the bulk modulus of rubber( 2000 MPa). Location of maximum shear strain caused by a compression load. 21

22 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for rectangular bearings: γ = c P AGS f 1 K/G = 2000 L/B S Location of maximum shear strain caused by a compression load. 22 K is the bulk modulus of rubber ( 2000 MPa). 22

23 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for rectangular bearings: γ = c P AGS f 1 K/G = 4000 L/B S Location of maximum shear strain caused by a compression load. 23 K is the bulk modulus of rubber( 2000 MPa). 23

24 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for rectangular bearings: γ = c P AGS f 1 K/G = 6000 L/B S Location of maximum shear strain caused by a compression load. 24 K is the bulk modulus of rubber ( 2000 MPa). 24

25 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for rectangular bearings: γ = c P AGS f 1 K/G = L/B S K is the bulk modulus of rubber( 2000 MPa). Location of maximum shear strain caused by a compression load. 25

26 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Coefficient f 1 for cylindrical hollow bearings: γ = c P AGS f 1 D o /D i = 10 D o /D i = 5 S K/G K/G K is the bulk modulus of rubber( 2000 MPa). Location of maximum shear strain caused by a compression load. 26

27 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Critical buckling load in compression P cr = π λgsar T r A is the rubber area attached to the shim plates (the area must be reduced to A r to take into account the lateral displacement). G is the shear modulus of the rubber (0.5 à 1.0 MPa). S is the shape factor of each rubber layer. r is the radius of gyration of the bonded rubber (r 2 = I / A, where I is the moment of inertia around the weak axis of the bearing). T r is the total rubber thickness. λ is a numerical factor that depends on the rotational stiffness of the rubber (λ=2.25 for circular or rectangular bearings). Valid only for bolted bearings

28 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Critical buckling load in compression Circular bearings: P cr GD = tt r 4 28 Square bearings: Cylindrical hollow bearings: GL = tt t is the thickness of each rubber layer. Also valid for lead-rubber bearings since lead does not contribute to the stability of the rubber. P P cr cr r 4 Di ( D ) B = L D GD o o = ttr Di 1 + D 2 o 2 i D 2 o 28

29 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Critical displacement for simply supported bearings Instability by overturning occurs when the overturning moment is larger than the stabilizing moment

30 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Critical displacement for simply supported bearings Critical displacement, D cr, causing overturning: If D cr D 1 : If D cr > D 1 : 30 30

31 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Critical displacement for simply supported bearings Critical displacement, D cr, causing overturning : If the behavior is represented by the effective stiffness, K eff : 31 31

32 2. Laminated Rubber Bearings Gravity Load Carrying Capacity Tension force causing cavitation for bolted bearings. Delamination of the rubber and steel shims. Cavitation force for each bolt: P CAV = 3G eff A r 32 P CAV 32

33 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings For a rotation θ relative to the bottom part, the maximum shear strain, γ r. Is given by: γ = r 2 L θ tt r f 2 L is the dimension perpendicular to the rotational plan (L for rectangular or square bearings, D for circular bearings and D o for cylindrical hollow bearings). t is the thickness of each rubber layer. T r is the total rubber thickness. f 2 is a numerical factor that depends on the shape of the bearing, the compressibility of the rubber and the location of the maximum shear strain (1.0 f 2 4.0) 33 33

34 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 for circular bearings: S K/G Location of maximum shear strain caused by a rotation between the top and bottom of bearings. K is the bulk modulus of rubber ( 2000 MPa)

35 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 rectangular bearings: K/G = 2000 L/B S Location of maximum shear strain caused by a rotation between the top and bottom of bearings. K is the bulk modulus of rubber ( 2000 MPa)

36 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 for rectangular bearings: K/G = 4000 L/B S Location of maximum shear strain caused by a rotation between the top and bottom of bearings.. K is the bulk modulus of rubber ( 2000 MPa)

37 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 for rectangular bearings: K/G = 6000 L/B S Location of maximum shear strain caused by a rotation between the top and bottom of bearings. K is the bulk modulus of rubber ( 2000 MPa)

38 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 for rectangular bearings: K/G = L/B S Location of maximum shear strain caused by a rotation between the top and bottom of bearings.. K is the bulk modulus of rubber ( 2000 MPa)

39 2. Laminated Rubber Bearings Rotations between the top and bottom of bearings Coefficient f 2 for cylindrical hollow bearings: Exterior Surface D o /D i = 10 D o /D i = 5 S K/G K/G Interior Surface D o /D i = 10 D o /D i = 5 S K/G K/G K is the bulk modulus of rubber( 2000 MPa). Location of maximum shear strain caused by a rotation between the top and bottom of bearings 39 39

40 2. Laminated Rubber Bearings Maximum allowable lateral relative displacement For a lateral displacement, Δ, of the top part of the bearing relative to the bottom part of the bearing, the maximum shear strain, γ s,is given by: γ s = T r T r is the total rubber thickness

41 2. Laminated Rubber Bearings Minimum thickness of steel shims Stress state in steel shims of circular bearings: Radial and circumferential (hoop stress) tension caused by the shear stresses at the steel-rubber interface; Compression caused by the vertical pressure

42 2. Laminated Rubber Bearings Minimum thickness of steel shims Solution for elastic stress distribution developed by Roeder et al. (1987). Axial pressure is maximum at the center of the shim: σ = 2 s z P A t P 3 + ν t P σr = σθ = = 1.65 t A 2 t A ν is Poisson s ratio of steel (0.3). Minus sign indicates compression. s 42 42

43 Minimum thickness of steel shims For design, Tresca s yield criterion is used to limit the maximum shear stress: 2. Laminated Rubber Bearings σr σz P t τ max = = A ts The maximum shear stress, τ max, caused by the factored load is limited to: τmax = φ(0.6 Fy) = 0.54F Therefore, the thickness of the steel shims is selected such that: t s 1.65t A 1.08Fy 2 P u P u is the factored compression load. The 1.65 factor applies for shims without holes. When holes are present, this factor should be increased to 3.0. y 43 43

44 2. Laminated Rubber Bearings Lateral Stiffness of Laminated Rubber Bearings 44

45 2. Laminated Rubber Bearings Natural Period of Vibration of Laminated- Rubber Bearings Supporting a Rigid Structure 45

46 2. Laminated Rubber Bearings Damping Provided by Laminated Rubber Bearings Experiments showed energy dissipation through shear deformations in rubber layers of laminatedrubber bearings proportional to velocity. Damping modeled by equivalent viscous damping. Natural rubber bearings: 5% to 10% damping. High damping rubber bearing: up to 25% damping. 46

47 2. Laminated Rubber Bearings Vertical Stiffness of Laminated Rubber Bearings Vertical stiffness much larger than lateral stiffness. Often assumed rigid in vertical direction. In some applications, vertical deflection of laminated rubber bearings may be important and vertical stiffness must be known. 47

48 2. Laminated Rubber Bearings Vertical Stiffness of Laminated Rubber Bearings 48

49 2. Laminated Rubber Bearings Adequacy assessment of bearings Analysis-and-Design-Procedures-for-Bridge-Bearings-and-Seismic- Isolators-MCEER html 49 49

50 2. Laminated Rubber Bearings Adequacy assessment of bearings Three evaluation criteria: 1. Verification for service loads; 2. Verification for the Design Earthquake (DE); and 3. Verification for Maximum Considered Earthquake (MCE) Generally: MCE = 1.5 x DE The analyses are conducted for the upper and lower bounds of the mechanical properties of the isolation system (see Chapter 11)

51 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for service loads Shear strain caused by compression load: P u u γ Cs = ArGS f 1 P u = Factored axial load from applicable code with the cyclic component of the live load multiplied by P = γ P + γ P γ P u D D L Lst L Lcy P D is the dead load; P Lst is the static component of the live load and P lcy is the cyclic component of the live load. γ D et γ L are load factors for the dead and live loads. A r = Overlapped area for a lateral displacement,, equal to: 51 = + Sst Non-seismic lateral displacement : Sst (static), Scy (cyclic). Scy 51

52 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for service loads Shear strain due to lateral displacement: Shear strain due to rotation: γ u S S Sst = T r Scy γ u r s 2 L ( θsst θscy ) = tt r f 2 Non-seismic rotation : θ Sst (static), θ Scy (cyclic) 52 52

53 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for service loads Buckling load at the lateral displacement caused by the service loads: = Sst + Scy A ' r Pcr = P s cr A 53 53

54 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for service loads A bearing design is considered adequate if: γ P + γ P A GS D D L Lst r γ + γ + γ u u u C S r s s s f Compression capacity Maximum shear strain P ' cr s γ P + γ ( P + P ) D D L Lst Lcy 2.0 Stability of bearings The amplification factor of 1.75 on the cyclic component of the live loads applies only for the calculation of the rubber deformation but does not apply for the evaluation of the compression capacity and in the verification of the stability

55 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for service loads Furthermore, the thickness of the steel shims, t s, must be at least equal to: t s αt 1.9 mm (0.075inch) Ar 1.08Fy 2 γ DPD + γ L( PLst + PLcy ) The amplification factor of 1.75 on the cyclic component of the live loads applies only for the calculation of the rubber deformation but does not apply for the evaluation of the steel shims. α = 1.65 for shims without holes otherwise a value of 3.0 must be used. The minimum thickness required for the steel shims correspond to a gage 14 sheet

56 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Design Earthquake (DE) Shear strain due to compression: P u u γ C = DE ArGS P u = Factored axial load; This load is determined according to the extreme load Combination I of the AASHTO LRFD Bridge Design Code (AASHTO, 2007, 2010). P = γ P + P + P u D D SL E DE f 1 DE P D is the dead load; P SLDE is the part of the live load, P L, assumed present during the design earthquake (recommended to use PSL = 0.5P DE L ) and P EDE is the axial load caused by the design earthquake. γ D is the dead load factor applicable for a seismic load combination. A r = Overlapping area at a displacement,, equal to: 56 ( ΔS + Δ ) st S Δ cy EDE Δ = γ + Non-seismic lateral displacement: Sst (static), Scy (cyclic) Displacement caused by the design earthquake: EDE γ =

57 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Design Earthquake (DE) Shear strain due to lateral displacement: γ u S DE γ S + = T r E DE with: γ = γ( + ) S Sst Scy 57 57

58 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Design Earthquake (DE) A bearing design is considered adequate if: u u u γ + γ + 0.5γ rs 7.0 CDE SDE Total shear strain u γ r s is the shear strain due to the rotation caused by the service loads as calculated earlier. No stability verification is required for the design earthquake. A stability verification will be considered for the MCE

59 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Design Earthquake (DE) Furthermore, the thickness of the steel shims, t s, must be at least equal to: t s 1.65t Ar 1.08Fy 2 P u 1.9 mm (0.075inch) The minimum thickness required for the steel shims correspond to a gage 14 sheet

60 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Maximum Considered Earthquake (MCE) Shear strain due to compression: u u γ C = MCE ArGS P u Factored axial load; This load is determined according to the extreme load Combination I of the AASHTO LRFD Bridge Design Code (AASHTO, 2007, 2010). P = γ P + P + P P D is the dead load; P SLMCE is the part of the live load, P L, assumed to be PSL 0.5P P MCE SLDE E = MCE present during the MCE (recommended to use ) et γ D is the dead load factor applicable to the seismic load combination. A r = Overlapped area for a displacement,, equal to: MCE P u D D SL E ( ΔS + Δ ) st S Δ cy EMCE Δ = γ + MCE f 1 = 1.5P E DE 60 Non-seismic displacement: Sst (static), Ssy (cyclic) Displacement caused by the MCE: EMCE γ =

61 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Maximum Considered Earthquake (MCE) Shear strain due to lateral displacement: γ u S MCE = 0.5γ + S T r E MCE with: 0.5γ = 0.5 γ( + ) = 0.25( + ) S Sst Scy Sst Scy 61 61

62 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Maximum Considered Earthquake (MCE) Critical buckling load at the MCE lateral displacement: Δ = 0. 5γΔ s + Δ EMCE P = P A 0.15P MCE A ' r cr cr cr 62 62

63 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Maximum Considered Earthquake (MCE) A bearing design is considered adequate if: u γ r u u u γ + γ γ rs 9.0 CMCE SMCE Total shear strain is the shear strain caused by the service loads rotation as calculated earlier. s P ' cr MCE P u 1.1 Bearing stability The amplification factor of 1.75 on the cyclic component of the live loads applies only for the calculation of the rubber deformation but does not apply for the evaluation of the compression capacity and in the verification of the stability

64 2. Laminated Rubber Bearings Adequacy assessment of bearings Verification for the Maximum Considered Earthquake (MCE) Furthermore, the thickness of the steel shims, t s, must be at least equal to: t s 1.65t Ar 1.08Fye 2 P u 1.9 mm (0.075inch) F ye is the probable yield strength of the steel shims = R y F y with R y = 1.3 for ASTM A36 steel and 1.1 for ASTM A573 Grade 50 steel. The minimum thickness required for the steel shims correspond to a gage 14 sheet

65 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Very difficult for conventional steel storage rack to meet seismic performance objectives of FEMA-460. Evaluation of a novel base isolation system for storage racks. Patented by Ridg-U-Rak Inc., Erie, PA. Uni-axial and tri-axial shake table tests performed on directly bolted and base isolated storage racks loaded with simulated and real merchandise. 65

66 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Lateral load-resisting systems of steel storage racks Moment-resisting frames in the down-aisle (longitudinal) direction. Braced frames in cross-aisle (transverse) direction. 66

67 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Requirements of Base Isolation System for Storage Racks Provide base isolation in the cross-aisle direction only. Reduce horizontal accelerations in cross-aisle direction to reduce content spillage and structural damage. Range of down-aisle natural periods of typical rack structures already similar to typical base isolated structures ( 1.5 sec). Horizontal accelerations in down-aisle direction do not contribute substantially to content spillage. No interference with normal material handling operations. 67

68 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Upright Down-Aisle Rubber Mount Welded Stud Cross-Aisle Box Horizontal Support Low Friction Bearing Material Courtesy of Ridg-U-Rak Inc All Dimensions in mm 68

69 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Rubber Durometer Horizontal Stiffness (kn/m) Equivalent Viscous Damping Ratio Rubber Layers Steel Plate All Dimensions in mm 69

70 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings 70 Courtesy of Ridg-U-Rak Inc

71 2. Laminated Rubber Bearings Seismic Isolation of Pallet-Type Steel Storage Racks with Laminated Rubber Bearings Isolated Rack Video Conventional Rack Courtesy of Ridg-U-Rak Inc 71

72 3. Lead-rubber Bearings Lead-rubber bearing composed of a laminated-rubber bearing with a cylindrical lead plug inserted in it center. Lead plug introduced to increase damping by hysteretic shear deformations of the lead. 72 Photo: Courtesy of M. Constantinou

73 3. Lead-rubber Bearings Reasons to use lead for central plug: At room temperature, lead behaves as elastic-plastic solid. Yields in shear at low stress of about 10 MPa. Lead is hot-worked at room temperature. Properties continuously restored when cycled in inelastic range. Very good fatigue resistance properties. Lead commonly available since used in batteries at purity level of more than 99.9% 73

74 3. Lead-rubber Bearings Properties of Lead-Rubber Bearings 74

75 3. Lead-rubber Bearings SRMD Testing Machine, UC-San Diego DIS LR Bearing, Vertical Load = 558 kips, Displacement = 22 in, velocity = 60 in/s 75

76 3. Lead-rubber Bearings SRMD Testing Machine, UC-San Diego DIS LR Bearing, 400% strain 76

77 3. Lead-rubber Bearings Failure Test, NIED, Tsukuba, Japan 77

78 3. Lead-rubber Bearings Modeling of Lead-Rubber Bearings F y k 2 k 1 78

79 3. Lead-rubber Bearings Modeling of Lead-Rubber Bearings F y k 2 k 1 79

80 3. Lead-rubber Bearings Modeling of Lead-Rubber Bearings F y k 2 k 1 80

81 3. Lead-rubber Bearings Modeling of Lead-Rubber Bearings 81

82 4. Friction Pendulum System General Description FPS manufactured by Earthquake Protection Systems (EPS), Richmond, California. Friction-type sliding bearing using gravity as restoring force. 82

83 4. Friction Pendulum System General Description Articulated friction slider traveling on spherical concave lining surface. 83

84 4. Friction Pendulum System General Description Photo: Courtesy of M. Constantinou 84

85 4. Friction Pendulum System General Description Sliders on smooth flat surface that dissipates energy by friction with parallel linear springs to provide re-centering capabilities Typically PTFE on polished stainless steel surface Slider on concave surface to provide re-centering capabilities through gravity Friction Pendulum bearing shown below 85

86 4. Friction Pendulum System Salkhalin II offshore gas platform bearings. Largest seismic isolators. 700mm displacement. 87,400kN vertical load. Full-scale testing Reduced scale dynamic testing (load of up to 13,000kN, velocity of 1m/sec). Photo: Courtesy of M. Constantinou 86

87 FPS Isolator, Vertical Load = 3490 kips, Displacement = 29 in, velocity = 53 in/s 87

88 4. Friction Pendulum System New International Terminal San Francisco International Airport 88

89 4. Friction Pendulum System HAYWARD CITY HALL, CALIFORNIA NEXT TO HAYWARD FAULT 53 FP BEARINGS AND 15 NONLINEAR VISCOUS DAMPING DEVICES 600 mm DISPLACEMENT CAPACITY Courtesy of M. Constantinou 89

90 4. Friction Pendulum System KODIAK, ALASKA COLD TEMPERATURE APPLICATION -40 DEG TEMPERATURE, STRONG WIND Courtesy of M. Constantinou 90

91 Properties of Frictionless Pendulum System 91

92 Properties of Frictionless Pendulum System 92

93 93 4. Friction Pendulum System Partial list of standard dimensions for concave surfaces of friction pendulums Radius of Curvature, mm (inch) Diameter of concave surface, mm (inch) 356 (14) 457 (18) 1555 (61) 559 (22) 787 (31) 914 (36) 686 (27) 787 (31) 914 (36) 991 (39) 2235 (88) 1041 (41) 1118 (44) 1168 (46) 1295 (51) 1422 (56) 686 (27) 3048 (120) 1422 (56) 1600 (63) 1778 (70) 3962 (156) 2692 (106) 3150 (124) 1981 (78) 2388 (94) 6045 (238) 2692 (106) 3327 (131) 3632 (143) 93

94 4. Friction Pendulum System Properties of Pendulum System including Friction 94

95 4. Friction Pendulum System SAKHALIN II PLATFORMS PROTOTYPE BEARING PR1, LOAD=6925kN, DISPLACEMENT=240mm, VELOCITY=0.9 m/sec EPS BEARING TESTING MACHINE, OCTOBER 2005 Courtesy of M. Constantinou 95

96 4. Friction Pendulum System Electrical 3-phase disconnect switch isolated by friction pendulum system UNIV. AT BUFFALO,

97 Double curvature friction pendulum system Combination of two friction pendulum systems 97

98 4. Friction Pendulum System Double curvature friction pendulum system TWO CONCAVE PLATES, EACH WITH EQUAL RADII OF CURVATURE AND EQUAL COEFFICIENTS OF FRICTION BEHAVIOR NEARLY IDENTICAL TO SINGLE CONCAVE FP BEARING- RIGID-LINEAR HYSTERETIC BUT OFFERS ADVANTAGE OF LARGE DISPLACEMENT CAPACITY Courtesy of M. Constantinou 98

99 4. Friction Pendulum System Double curvature friction pendulum system TWO CONCAVE PLATES, EACH WITH EQUAL RADII OF CURVATURE AND UNEQUAL COEFFICIENTS OF FRICTION RIGID-BILINEAR HYSTERETIC BEHAVIOR OFFERS ADVANTAGE OF REDUCTION OF SECONDARY SYSTEM RESPONSE Courtesy of M. Constantinou 99

100 4. Friction Pendulum System Double curvature friction pendulum system UNIV. AT BUFFALO, 2004 Courtesy of M. Constantinou 100

101 4. Friction Pendulum System Triple curvature friction pendulum system Courtesy of M. Constantinou 101

102 4. Friction Pendulum System Triple curvature friction pendulum system

103 Triple curvature friction pendulum system Friction Pendulum System Assumptions: (1) Reff 1 = Reff 4 Reff 2 = Reff 3, (2) * * (4) > ( µ µ ), (5) > ( µ µ ) d R eff d R eff Regime Description Force-Displacement Relationship I II III Sliding on surfaces 2 and 3 only Motion stops on surface 2; Sliding on surfaces 1 and 3 Motion is stopped on surfaces 2 and 3; Sliding on surfaces 1 and 4 W F R + F R F = u+ R + R R + R * µ =µ <µ <µ, (3) ( ) f 2 eff 2 f 3 eff 3 eff 2 eff 3 eff 2 eff 3 Valid F F 1 until: f =, = = ( µ µ ) + ( µ µ ) u u R R ( ) 1 2 eff eff 3 W F R R + F R + F R F = u+ R + R R + R f 1 eff 1 eff 2 f 2 eff 2 f 3 eff 3 eff 1 eff 3 eff 1 eff 3 Valid F F 4 until: f W F = u+ R + R eff 1 eff 4 =, u u = = u + ( µ 4 µ 1)( Reff 1+ Reff 3) ( ) ( ) F R R F R F R F R R f 1 eff 1 eff 2 f 2 eff 2 f 3 eff 3 f 4 eff 4 eff 3 Valid until: R + R eff 1 eff 4 W F = F = d + F, * dr1 1 f 1 Reff 1 d1 > µ 4 µ 1 R eff 1, u = u = u + d R 1+ µ µ R + R ( )( ) * eff 4 dr eff 1 eff 4 R eff 1 103

104 4. Friction Pendulum System Triple curvature friction pendulum system Assumptions: (1) Reff 1 = Reff 4 Reff 2 = Reff 3, (2) * * (4) > ( µ µ ), (5) > ( µ µ ) d R eff d R eff * µ =µ <µ <µ, (3) ( ) d1 > µ 4 µ 1 R eff 1, W W * F = ( u udr ) + d + F R + R R Regime Description Force-Displacement ff Relationship IV V Slider contacts restrainer on surface 1; Motion remains stopped on surface 3; Sliding on surfaces 2 and 4 Slider bears on restrainer of surface 1 and 4; Sliding on surfaces 2 and 3 Valid until: 1 1 f 1 eff 2 eff 4 eff 1 W F = F = d + F, * dr4 4 f 4 Reff 4 d d u = u = u + +µ +µ R + R * * 4 1 dr4 dr1 4 1 eff 2 eff 4 R eff 4 R eff 1 W W * F = ( u u ) + d + F R R R dr4 4 f 4 eff 2 + eff 3 eff 4 ( )

105 105

106 4. Friction Pendulum System Shake table tests of a full-scale 5-story steel moment frame building (PI: K. Ryan, Reno; S. Mahin, Berkeley; G. Mosqueda, San Diego) triple friction pendulum isolators lead rubber bearing/cross linear slider Fixed base o Simulations designed to impose large displacement demands in isolation systems o Simulations both with and without vertical component of ground motion o 4 th and 5 th floor included nonstructural systems

107 4. Friction Pendulum System

108 4. Friction Pendulum System

109 4. Friction Pendulum System E-Defense Experiments: 5 story steel moment frame Level Peak Acceleration Profile Measured response X-direction Fixed Base TPB Isolated LRB Isolated Base Shear Coefficient Peak Acc. (g) Y-direction Time (sec) 109

110 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Uniform Moment Bending-Beam Damper 110

111 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Tapered-Cantilever Bending-Beam Damper 111

112 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Tapered-Cantilever Bending-Beam Damper 112

113 Seismic Isolation Systems Incorporating Metallic Dampers The Tapered-Cantilever Bending-Beam Damper 113

114 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Tapered-Cantilever Bending-Beam Damper 114

115 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Torsional-Beam Damper 115

116 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers The Torsional-Beam Damper South Rangitikei River Railroad Bridge, New Zealand, built in

117 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers Lead-Extrusion Bearings 117

118 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers Sliding Bearing with C-Shaped Yielding Steel Devices - ALGA C-element 118

119 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers Sliding Bearing with C-Shaped Yielding Steel Devices - ALGA 119

120 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers Sliding Bearing with C-Shaped Yielding Steel Devices - ALGA Courtesy of M. Constantinou 120

121 Seismic Isolation Systems Incorporating Metallic Dampers Sliding Bearing with C-Shaped Yielding Steel Devices - ALGA BOLU VIADUCT, TURKEY 2.3 km LONG DAMAGED IN DUCZE EARTHQUAKE OF NOV CROSSED BY ANATOLIAN FAULT BEARING DISPL. CAPACITY 210 mm REQUIRED CAPACITY PER AASHTO OVER 1000 mm LIKELY DEMAND IN EARTHQUAKE 1400mm Courtesy of M. Constantinou 121

122 5. Other Seismic Isolation Systems Seismic Isolation Systems Incorporating Metallic Dampers Rubber Bearings and U-Shaped Yielding Steel Devices 122

123 6. Example of adequacy assessment of a lead-rubber bearing under Maximum Credible Earthquake (MCE) 123

124 Bearing Description and Loading Conditions Circular bolted bearing. Factored axial load: P = 6000 kn u = γ DPD + PSL + P MCE EMCE Lateral displacement under MCE: Δ γ ( Δ + Δ ) + Δ = 0.5γΔ + Δ = 555mm = Sst Scy EMCE S E MCE Lateral displacement caused by rotation is negligible. Effective bounded diameter: D = 813 mm 29 rubber layers, each 7 mm thick Rubber moduli: G = 0.5 MPa K = 2000 MPa Steel shims: Probable yield strength: F ye =380 MPa Thickness: 3.04 mm

125 Design Requirements Assess the adequacy of the bearing for the Maximum Considered Earthquake (MCE). See Section

126 Adequacy Assessment for MCE Δ Calculation of reduced overlapping area, A r γ A r = Overlapping area for a displacement,, equal to: ( Δ + Δ ) + Δ = 0.5γΔ + Δ = 555mm = Sst Scy EMCE S E MCE δ = 2cos A r = D D = 2cos mm 813mm = ( δ sinδ ) = ( sin1.6388) = mm A r

127 Adequacy Assessment for MCE Shear strains calculations Shear strain due to lateral deformation: γ u S T γ MCE r u S = MCE = 29 = 0.5γ + S T r ( 7 mm) = 555mm 203mm E MCE 203mm = 2.73 A r

128 Adequacy Assessment for MCE Buckling load calculation 128 Buckling load at MCE lateral displacement: P = P A 0.15P MCE A P P ' r cr cr cr Ar A cr λ = 2.25 S = r = cr mm mm = = 2 π mm π = D 4t I A π = ( 813mm) λgsar T r 813mm = 4 = ( 7 mm) 2.25 = 29 ( 813mm) π mm = mm 2 2 ( 0.5 N / mm )( 29)( mm )( mm) 203mm = 0.20 ( 35515kN) = 7103kN Pcr ' Pcr = MCE = 35515kN A r 128

129 Adequacy Assessment for MCE Shear strains calculations Shear strain due to compression: f 1 S K/G f P u u γ C = MCE ArGS 1 = 1.35 u γ = ( )( )( ) ( 1.35) = C MCE 2 2 f N mm 0.5 N/mm 1 29 A r

130 Adequacy Assessment for MCE Verification of total shear strain A bearing design is considered adequate if: u u u γ + γ γ rs 9.0 CMCE SMCE = A r

131 Adequacy Assessment for MCE Stability verification A bearing design is considered adequate if: P ' cr MCE P u 1.1 A r 7103kN 6000 kn =

132 Adequacy Assessment for MCE Minimum steel shims thickness. The thickness of the steel shims, t s, must be at least: A r ( 7 mm) 1.65 t s mm 1.08( 380 N/mm ) N 3.04 mm 2.2 mm 2 2 = 2.2 mm 1.9 mm

133 Questions/Discussions 133