Seismic Pushover Analysis Using AASHTO Guide Specifications for LRFD Seismic Bridge Design


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1 Seismic Pushover Analysis Using AASHTO Guide Specifications for LRFD Seismic Bridge Design Elmer E. Marx, Alaska Department of Transportation and Public Facilities Michael Keever, California Department of Transportation 2015 TRB Pushover Webinar 1
2 Learning Objective Given the plans for a conventional bridge pier, calculate the lateral forcedisplacement (pushover) response of the system necessary for the seismic design of bridges in accordance with the AASHTO Guide Specifications for LRFD Seismic Bridge Design (SGS). Focus is pushover analysis not modeling, seismic displacement demand or capacity design 2
3 Overview Flexural mechanics refresher Generation of MomentCurvature ( M  φ ) Material models and failure strains Hand checking M  φ Analytical plastic hinge length, L p ForceDisplacement pushover example Hand checking forcedisplacement Design Example Two circular column bent 3
4 Why Pushover? SGS is primarily a displacementbased approach More rational approach than forcebased method of AASHTO LRFD Seismic Design Category D (SDC D) requires pushover analysis 4
5 Why Pushover? Need to verify that the seismic displacement demand, L D, exceed the displacement capacity, L C Results can be used to help create the seismic model and subsequent displacement demands Provides for a better understanding of bridge response and behavior 5
6 Flexural Mechanics Relationship between force, shear, moment, curvature, slope and deflection Integration Moment area Energy methods Stiffness methods 6
7 Force Displacement Relationship P = load V = P dx M = V dx 7
8 Force Displacement Relationship M = V dx ϕ = M EI 8
9 Force Displacement Relationship ϕ = M E I θ = ϕ dx = θ dx 9
10 Example  Cantilever P V = P θ L o = ϕ 2 L L = ϕo 3 2 L L M o = P L ϕ o = M o E I = P E L I 10
11 Compatibility BernoulliEuler assumption sections that are plane (linear) before bending remain plane after bending Perfect bond between the steel reinforcing bars and surrounding concrete Material (constitutive σε) models are representative of actual material response (confinement, strain hardening, buckling, spalling, cracking, creep, strainrate, shear, etc.) 11
12 Moment Curvature ( M  φ ) 12
13 Moment Curvature ( M  φ ) For small angles, curvature (φ ) may be calculated as: φ = ε c / c = ε t / (D c) φ = strain / distance from neutral axis 13
14 Moment Curvature ( M  φ ) From equilibrium: P = F SC + F CC + F CU F ST Summing moments about the neutral axis M = CG SC *F SC +CG CC *F CC +CG CU *F CU +CG ST *F ST +P*(D/2 c) 14
15 Generation of MomentCurvature Relationship 15
16 Moment Curvature ( M  φ ) Predicted Response Idealized Bilinear Response used in SGS SGS
17 Idealized M φ Response Used for design purposes First yield Moment and Curvature, M y and φ y Effective stiffness, E ce * I eff = M y / φ y Idealized yield Moment and Curvature, M p and φ yi Expected nominal moment, M ne at ε c = Balance the area above and below the curve 17
18 SGS Material Models Concrete: Mander et al model for unconfined and confined condition Reinforcing steel: elastic  perfectly plastic with strain hardening Other: prestressing steel (see SGS) Any function / model that accurately captures the material σε relationship is acceptable 18
19 SGS Material Models  Concrete Stressstrain relationship for concrete where: f c = r ' fccxr 1+ x r f ' cc r = E ε c x = ε ' ' ' 7.94 fl 2 fl = f ce ' ' fce fce ce cc E ce E sec Mander, Priestley, Park
20 SGS Material Models  Concrete (continue) ε cc ' f cc = ' fce E = ce ' 1900 fce E sec = ' f cc ε cc Mander, Priestley, Park
21 SGS Material Models  Concrete Confinement of circular columns ρ s = 4 A sp ' D s f = K f ' l e l f l f l = 2A sp ' f D s yh = ρ f s 2 yh Mander, Priestley, Park
22 SGS Material Models  Concrete Confinement of rectangular columns f ' l K e ( + ρ ) ρ x y f yh 0.8( ρ x + ρ y ) 2 2 f yh ρ x = A sh sx c ρ y = A sy sb c Mander, Priestley, Park
23 SGS Material Models  Concrete ε c = strain in concrete (IN/IN) f c = stress in concrete corresponding to strain ε c (KSI) f ce = expected nominal compressive strength (KSI) f yh = nominal yield stress of transverse reinforcing (KSI) ε R su = reduced ultimate tensile strain of transverse bars (IN/IN A sx = total area of transverse bars in the x axis (IN 2 ) A sy = total area of transverse bars in the y axis (IN 2 ) A sp = area of hoop /spiral bar for circular column (IN 2 ) H c = confined core dimension in the y axis (IN) B c = confined core dimension in the x axis (IN) D = diameter of spiral or hoop for circular column (IN) s = pitch of spiral or spacing of hoops or ties (IN) Mander, Priestley, Park
24 SGS Material Models  Concrete Chen
25 Effective confinement factor for circular columns Effective confinement factor for rectangular columns SGS Material Models  Concrete ~ ' 1 2 ' 1 6 ) ( ' cc c c N i c d i e d s b s d b w K ρ = = ~ ' 2 ' 1 cc n e D s K ρ = Chen
26 SGS Material Models  Concrete s = clear distance between transverse bars = s d bh (IN) d bh = diameter of transverse reinforcing bar (IN) D = centerline diameter of hoop or spiral (IN) ρ cc = A st / A cc A st = total area of longitudinal reinforcement (IN 2 ) A cc = area of confined concrete core (IN 2 ) n = 1 for continuous spiral n = 2 for individual hoops w i = clear distance between adjacent tied longitudinal bars(in) b d = confined core dimension in the longer direction(in) d c = confined core dimension in the shorter direction (IN) N = number of spaces between longitudinal bars Chen
27 SGS Material Models  Concrete Stressstrain curves for concrete Spalling strain, ε sp, not to exceed Confined concrete crushing strain, ε cu, critical component ε cu Recommend ε cu < 0.02 for design purposes SGS
28 SGS Failure Strain  Concrete Confined concrete crushing strain limit, ε cu important ε cu = 1.4ρs f yhε su < ' f cc where: ρ s = transverse reinforcement ratio (ρ x +ρ y for rect.) f yh = nominal yield stress of transverse steel ε su = rupture strain of transverse steel ~ 0.09 to 0.12 f cc = confined concrete compressive stress Mander, Priestley, Park
29 SGS Material Models  Steel Kowalsky 2013 and SGS 29
30 SGS Material Models  Steel Stressstrain relationship for reinforcing steel Elastic ε s ε ye f = s E s ε s Plastic ε ye ε s ε sh f s = f ye Strain hardening ε sh ε s ε R su f s = f ue 1 ( f ue f ye ε ) ε su su ε s ε sh 2 Priestley, Calvi, Kowalsky
31 SGS Failure Strains  Steel Reinforcing steel failure strain, ε R su Property Notation Bar Size ASTM A706 ASTM A615 Grade 60 Specified minimum yield stress (ksi) f y #3  # Expected yield stress (ksi) f ye #3  # Expected tensile strength (ksi) f ue #3  # Expected yield strain ε ye #3  # Onset of strain hardening #3  # # ε sh #10  # # # Reduced ultimate tensile strain #4  # R ε su #11  # Ultimate tensile strain ε su #4  # #11  # SGS
32 SGS Material Models  Steel ASTM A 706 Grade 60 v. ASTM A 615 Grade 60 ASTM 2011,
33 Quasistatic v. Cyclic Discussion Quasistatic constitutive models for cyclic response Kowalsky et al
34 Moment Curvature Example Find the Mφ for the column shown below 940 KIP 2 IN clr. #5 4IN 20 #11 L =H o = 20 FT D = 48 IN ASTM A 706 Grade 60 f c = 4 KSI 34
35 Moment Curvature Example Diameter, D = 48 IN Gross Area, A g = π*d 2 /4 = 1810 IN^2 20 #11 A st = 20*1.56 = 31.2 IN^2 ρ l = A st / A g = = 1.724% #5 hoop (d sp = IN and A sp = 0.31 IN^2) Hoop spacing = 4 IN pitch (s = 4 IN) Clear cover, cov = 2 IN over transverse bars Core diameter, D = D 2*cov d sp = IN ρ s = 4*A sp / (D * s) = = 0.715% ASTM A 706 Grade 60 so f ye = 68 KSI, f ue = 95 KSI f c = 4 KSI so f ce = 5.2 KSI P = 940 K Axial Load Ratio, ALR = P / (f ce *A g ) =
36 Moment Curvature Example Confined concrete crushing strain limit, ε cu ε cu 1.4ρs f yhε su = = ' f cc where: ρ s = 4*A sp /(D *s) = f l = K e *ρ s *f yh /2 ~ 0.95* *60/2 = KSI f yh = 60 KSI (use nominal for horizontal steel) ε su = ε R su = 0.09 for #5 hoops f ' cc = f ' ce ' fl 2 f ' f ' ce f 6.49 Spreadsheet demo then check with commercial ce ' l = 36
37 M φ Spreadsheet Example 37
38 Moment Curvature Example Spreadsheet summary results for ALR =
39 Idealized Moment Curvature Elasticperfectly plastic relationship is defined as: M i = idealized moment values = E ce I eff *φ < M p M p = idealized plastic moment (solving for this) φ = actual calculated curvature from Mφ results E ce I eff = effective stiffness at first yield = M y / φ y (Mφ) = actual  idealized = ( M)*( φ) Σ (Mφ) = sum of difference = 0 by changing M p Spreadsheet demo then check with commercial 39
40 Idealized Moment Curvature 40
41 Moment Curvature Example Idealized M  φ relationship using more data points 41
42 Moment Curvature Example 42
43 Moment Curvature Example P = 940 K φ yi = /IN 2% φ u = /IN 0.2% M p = KIN 2% E ce I eff = E6 KIN 2 0.3% Spreadsheet Comparison φ yi = /IN φ u = /IN M p = KIN E ce I eff = 456.3E6 KIN 2 43
44 Approximate Methods the use of curvature ( φ ) in design is uncommon don t have a good feel for curvature values simple method to check the computer results can help with iterative processes use these approximations with due skepticism 44
45 Approximate Methods  φ yi For a rough check of conventional circular reinforced concrete column sections: where: φ yi ~ 2.25*ε ye /12B o ~ 1/2300B o φ yi ~ 2.25*ε ye /D ~ 1/190D φ yi = idealized yield curvature (1/IN) B o = column diameter (FT) D = column diameter (IN) ε ye = expected yield strain ~ (IN/IN) Priestley, Calvi, Kowalsky
46 Approximate Methods  φ yi For a rough check of conventional rectangular reinforced concrete column sections: φ yi ~ 2.1*ε ye /12B o ~ 1/2500B o where: φ yi = idealized yield curvature (1/IN) B o = column width in loaded direction (FT) ε ye = expected yield strain ~ (IN/IN) Priestley, Calvi, Kowalsky
47 Approximate Methods  φ u And for a very rough check of conventional circular or rectangular reinforced concrete column sections with good confinement: where: φ u = min (ε cu /c c, ε sur /dc) ~ ε R su /12B o φ u = ultimate curvature (1/IN) ε cu = ultimate confined concrete strain (1/IN) ε R su = reduced ultimate tensile strain (IN/IN) c c = neutral axis to edge of confined core (IN) dc = neutral axis to extreme tension bar (IN) B o = column diameter / width (FT) 47
48 Effective Stiffness  E ce *I eff E ce *I eff = M y / φ y where: E ce = Expected modulus of elasticity for concrete M y = moment at first yield (expected materials at first yield) φ y = curvature at first yield (expected materials at first yield) So, And typically, I eff = M y /( φ y *E ce ) 0.7 < I eff / I g <
49 Approximate Methods I eff / I g where: I eff /I g ~ *ρ l + 0.5*(P/f ce *A g ) ρ l = longitudinal reinforcement ratio in % = A st / A g * 100 < 3% practical limit P = axial load keep below 0.2*A g *f ce SGS
50 Approximate Methods  M p where: M p ~ D 3 * [ *ρ l + P/(f ce *A g )] M p = idealized plastic moment for circular section (KIN) D = column diameter (IN) > 30 IN ρ l = longitudinal reinforcement ratio in % = A st / A g * 100 < 4% but 3% is practical limit P = axial load on column (K) < 0.2 * f ce * A g f ce = expected concrete strength (KSI) A g = gross area of column (IN^2) Assumes ASTM A 706 Grade 60 reinforcing steel and f ce =
51 Approximate Methods  M p where: M p ~ b*d 2 * [ *ρ l + 1.5*P/(f ce *A g )] M p = idealized plastic moment for rectangular section (KIN) d b = column depth in direction of loading (IN) > 30 IN = column width (IN) > 30 IN ρ l = A st / A g * 100 = longitudinal reinforcement ratio in % P = axial load on column (K) < 0.2 * f ce * A g f ce = expected concrete strength (KSI) A g = b * d = gross area of column (IN^2) Assumes ASTM A 706 Grade 60 reinforcing steel and f ce =
52 Moment Curvature Check Diameter, D = 48 IN I g = D 4 π/64 = IN 4 ρ l = A st / A g = = 1.724% Axial Load Ratio, ALR = P / (f ce *A g ) = 0.1 [P ~ 940 K] ε ye = f ye /E s = 68/29000 = ε R su = 0.06 (#11) φ yi ~2.25* /48 = /IN v /IN φ u ~ 0.06/48 = /IN v /IN I eff ~ (0.2+.1* *0.1)* = 0.42* I eff ~ IN 4 v IN 4 4% M p ~ 48 3 ( * ) M p ~ KIN v KIN 6% 52
53 Moment Curvature Check Approximate Method Predicted Response Idealized Response XTRACT Response 53
54 Force  Displacement Calculate deflections using Mφ result Elastic deformation component, yi Plastic deformation component, p Foundation deformation component important but not specifically addressed in this workshop Conservative to neglect shear deformations 54
55 Force  Displacement Caltrans SDC Version
56 Analytical Plastic Hinge Length  L p Approximation used to simplify analysis Converts (integrates) curvature to rotation Includes a moment gradient part (integration), tension shift (diagonal cracking) and a strain penetration part (yielding into cap, footing or shaft) Calibrated to the failure condition only and modification may be needed for full straindisplacement response 56
57 Analytical Plastic Hinge Length  L p L p = k * L + L sp > 2 * L sp where: L p = 0.08 * L * f ye * d bl > 0.3 * f ye * d bl k = 0.2*(f ue /f ye 1) 0.08 L = length of column from point of maximum moment to the point of moment contraflexure (IN) L sp = strain penetration component = 0.15*f ye *d bl (IN) f ye = expected yield stress of longitudinal bars (KSI) f ue = expected tensile strength (KSI) d bl = diameter of longitudinal column bars (IN) SGS
58 Predicted Force Displacement y ~ 1/3*φ y *(L+L sp ) 2 (M, φ) ~ y *(M/M y )+ (φ  φ y )*L p *(LL p /2) where: φ y L L p L sp φ M y M F = curvature at first yield = column height = analytical plastic hinge length = strain penetration = curvature at point of interest = moment at first yield = moment associated with φ = force associated with = M / L Priestley, Calvi, Kowalsky
59 Idealized Force Displacement yi ~ 1/3*φ yi *(L+L sp ) 2 L C = yi + p ~ yi + (φ u  φ yi )*L p *(LL p /2) where: φ yi yi p L L p L sp φ u L C F p = idealized yield curvature = idealized yield displacement = plastic displacement capacity = column height = analytical plastic hinge length = strain penetration = ultimate curvature = ultimate displacement = plastic force = M p / L Priestley, Calvi, Kowalsky
60 Idealized Force  Displacement So the deformation values for the example problem are: yi = 1/3* *( ) 2 = 2.44 IN L C = ( )*33.58*( /2) = 9.62 IN where: φ yi L L sp L p φ u M p F p = /IN = 20 FT = 240 IN = 0.15*68*1.41 = IN = 0.08* = IN > 28.8 IN = /IN = M yi = M u = KIN = M p / L = / 240 = 215 KIP µ D = L C / yi = 9.62 / 2.44 =
61 Force  Displacement The idealized response for the single column example Approximate M p, φ yi and φ u Idealized Spreadsheet M p, φ yi and φ u Predicted Response from Spreadsheet calculated Mφ 61
62 Approximate Methods As with Mφ, we prefer a simplified method to check the computer results simple method to check the computer results could use the closedform AASHTO equations to start but they are developed for specific target ductility / strain limits use these approximations with due skepticism 62
63 Approximate Methods  yi For a rough check of conventional circular reinforced concrete column sections: yi ~ 1/3*φ yi *(12*L+0.15*f ye *d b ) 2 ~ L 2 / 42B o where: yi = idealized yield displacement (IN) L = contraflexure to plastic hinge distance (FT) d b = diameter of longitudinal column bar (IN) φ yi = idealized yield curvature ~ 2.25*ε ye /B o (1/IN) B o = column diameter (FT) f ye = expected yield stress (KSI) 63
64 Approximate Methods  yi For a rough check of conventional rectangular reinforced concrete column sections: yi ~ 1/3*φ yi *(12*L+0.15*f ye *d b ) 2 ~ L 2 / 45B o where: yi = idealized yield displacement (IN) L = contraflexure to plastic hinge distance (FT) d b = diameter of longitudinal column bar (IN) φ yi = idealized yield curvature ~ 2.1*ε ye /12B o (1/IN) B o = column width in direction of loading (FT) f ye = expected yield stress (KSI) 64
65 Approximate Methods  c L And for a very rough check of conventional reinforced concrete columns with L/B o > 4 : L C ~ yi + (φ u  φ yi )*L p *(12*LL p /2) ~ L 2 / 10B o where: L C = local displacement capacity (IN) L = contraflexure to plastic hinge distance (FT) L p = analytical plastic hinge length (IN) φ u = ultimate curvature ~ ε R su /12B o (1/IN) φ yi = idealized yield curvature (1/IN) B o = column diameter / width (FT) 65
66 ForceDisplacement Check Diameter, D = 48 IN => B o = 4 FT Height, L = H o = 20 FT M p ~ 54713KIN/12 = 4560 KFT (see previous check) yi ~ L 2 / 42B o = 20 2 /(42*4) = 2.4 IN v. 2.4 IN L C ~ L 2 / 10B o = 20 2 /(10*4) = 10 IN v. 9.6 IN F p = M p / L = 4560/20 = 228 KIP v. 215 KIP 66
67 Force  Displacement Idealized response for the single column example Approximate M p, yi and L C Spreadsheet calculated M p, φ yi and φ u 67
68 What about Double Curvature? Effective column height, L, is taken from the maximum moment location to the contraflexure point Then add the displacement results for each part L 1 L 2 SGS
69 What about P Effects? From equilibrium and summing moments about the base of the column: M p = F p * H o + P * Caltrans SDC Version
70 What about P Effects? The moment left to resist lateral forces becomes, M p = M p P * F P = M P / H o But when using an idealized elastic perfectlyplastic forcedisplacement relationship check (SGS method) P dl * r 0.25 * M p 70
71 What about P Effects? Adjusted response for the single column example Unadjusted for P Adjusted for P 71
72 What about P Effects? Rearranging the P limit the maximum permissible deflection for the single column example r 0.25 * M p / P dl r 0.25 * KIN / 940 K = 13.7 IN In this case, the P limit is greater than the calculated L C value With more confinement, the P limit may govern 72
73 Local Ductility v. Global Ductility Use local member displacements D L < C L µ D = L D / yi where: L D = Local member deformation demand L C = Local member deformation capacity µ D = local member displacement ductility demand yi = idealized yield deformation 73
74 Local Ductility v. Global Ductility Must remove the deformation components associated with noncolumn deformation demands Caltrans SDC Version
75 Design Example 2 IN clr. #5 4IN d = 4 FT 940 KIP 940 KIP H o = 24 FT Rigid cap and footings 20 #11 D = 48 IN Z = 32 FT 75
76 Design Example Assume that the point of contraflexure is at column midheight L 1 =L = H o / 2 L 2 = L = H o / 2 Caltrans SDC Version
77 Design Example Predict Response Use approximate methods to predict expected response Diameter, D = 48 IN => B o = 4 FT Height, H o = 24 FT so for the transverse direction, L = 12 FT P = P DL ± P EQ P EQ = (M pl + M pr ) * (1 + d / H o ) / Z why? Use P = P DL = 940 K for first iteration M p ~ 54720KIN/12 = 4560 KFT (see previous calculations) yi ~ 2 * L 2 /42B o = 2 * 12 2 /(42*4) = 1.71 IN L C ~ 2 * L 2 /10B o = 2 * 12 2 /(10*4) = 7.2 IN F p = (2*M pl + 2*M pr ) / H o ~ 4 * M p /24 = 760 KIP why? 77
78 Design Example  FBD F p M pl *(1 + d/h o ) M pr *(1 + d/h o ) d /2 L = H o / 2 M pl M pr Moment contraflexure M pl M pr Moment Diagram slope of line is the shear shear in cap beam is axial force in column 78
79 Design Example Predict Response Perform a second iteration to verify initial assumptions P EQ = (M pl +M pr )*(1+d/H o )/ Z ~ 2*54720*1.167 /12/32 ~ 332 K P = P DL ± P EQ = 940 /+ 332 = 608K and 1272K [compression] M pl ~ ( * ) * 48 3 = KIN M pr ~ ( * ) * 48 3 = KIN P EQ = (M pl + M pr )*(1 + d / H o ) / Z P EQ = ( )(1.167)/32/12 = 332 K F p = (2 * M pl + 2 * M pr ) / H o = 760 KIP 79
80 Design Example  Process Now use the refined analysis to determine the forcedisplacement response of the pier First calculation idealized Mφ Then calculate L sp and L p Then calculate the forcedisplacement for each column Then add the results of each column to find the total pier response 80
81 Design Example Refined Analysis 81
82 Design Example Refined Analysis Verify that the axial forces are reasonably close to the initial estimate P EQ = (M pl + M pr ) * (1 + d / H o ) / Z P EQ = ( )*1.167 /(12*32) ~ 314 K v. 332k P = P DL  P EQ = = 626K v. 608K close 3% P = P DL + P EQ = = 1254K v. 1272K close 2% Use the initial values since they are close 82
83 Design Example Hinge Formation Order of hinge formation and hinge failure 83
84 Design Example Displacement For the left / trailing side column: yil = 1/3* *( ) 2 = 0.92 IN L CL = ( )*28.8*( /2) = 4.92 IN where: φ yi L L sp L p φ u M p F pl = /IN (from Mφ analysis) = H o / 2 = 24 FT / 2 * 12 = 144 IN = 0.15*68*1.41 = IN = 0.08* = 25.9 IN < 2*L sp = 28.8 IN = /IN (from Mφ analysis) = KIN (from Mφ analysis) = M p / L = / 144 = 335 KIP 84
85 Design Example Displacement For the right / right side column: yir = 1/3* *( ) 2 = 0.94 IN L CR = ( )*28.8*( /2) = 4.21 IN where: φ yi L L sp L p φ u M p F pr = /IN (from Mφ analysis) = H o / 2 = 24 FT / 2 * 12 = 144 IN = 0.15*68*1.41 = IN = 0.08* = 25.9 IN < 2*L sp = 28.8 IN = /IN (from Mφ analysis) = KIN (from Mφ analysis) = M p / L = / 144 = 383 KIP 85
86 Design Example Left Column Right Column Axial compression, P Effective stiffness, (EI) eff 4.36E9 4.91E9 Plastic moment, M p Idealized yield curvature, φ yi Ultimate curvature, φ u Plastic curvature, φ p Analytical Plastic Hinge Length, L p yi (per half of column height) p (per half of column height) u (per half of column height) Plastic shear, F p
87 Design Example Combined Response Total effective lateral force F p = F pl + F pr = 335K + 383K = 718 K Idealized yield displacement yi ~ (2* yil + 2* yir ) / 2 = = 1.86 IN Ultimate displacement (first hinge failure on right) L c = 2 * L cr = 2 * 4.21 IN = 8.42 IN 87
88 Design Example Idealized Response Combined Right / Leading Column Left / Trailing Column 88
89 Design Example Refined Response Idealized force displacement for the two column pier Approximate M p, yi and L C Spreadsheet Idealized M p, φ yi and φ u 89
90 Design Example Verification The maximum deflection based on P limits r 0.25 * KIN / 940 K = 12.8 IN (per half) And the displacement ductility capacity µ D < L c / yi = 8.42 / 1.86 =
91 Capacity Design Designer dictates the bridge response (e.g., column hinge response mechanism Type I) Preclude all failure modes that would prevent the formation of the predicted forcedeflection response including: Shear failure inside and outside the plastic hinge region (not the same as analytical plastic hinge length, L p ) Columncap and columnfooting joint failure Cap beam, footing or shaft moment, shear, and axial overload / hinging Footing overturning, sliding, uplift and rocking failure Any other failure that occurs prior to reaching L C 91
92 Thank you  Questions 92
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