Earthquake Resstat Desg Accordg to UBC 1997 Major Chages from UBC 1994 (1) Sol Profle Types: The four ste coeffcets S 1 to S 4 of the UBC 1994, whch are depedet of the level of groud shakg, were epaded to s sol profle types, whch are depedet o the sesmc zoe factors, the 1997 UBC (S A to S F ) based o prevous earthquake records. The ew sol profle types are based o sol characterstcs for the top 30 m of the sol. The shear wave velocty, stadard peetrato test ad udraed shear stregth are used to detfy the sol profle types. (2) Structural Framg Systems: I addto to the four basc framg systems (bearg wall, buldg frame, mometresstg frame, ad dual), two ew structural system classfcatos were troduced: catlevered colum systems ad shear wall-frame teracto systems. (3) Load Combatos: The 1997 UBC sesmc desg provsos are based o stregth-level desg rather tha servce-level desg. (4) Earthquake Loads: I the 1997 UBC, the earthquake load (E) s a fucto of both the horzotal ad vertcal compoets of the groud moto. (5) Desg Base Shear: The desg base shear the 1997 UBC vares verse proporto to the perod T, rather tha T 2/3 prescrbed prevously. Also, the mmum desg base shear lmtato for Sesmc Zoe 4 was troduced as a result of the groud moto that was observed at stes ear the fault rupture 1994 Northrdge earthquake (USA). (6) Smplfed Desg Base Shear: I the 1997 UBC, a smplfed method for determg the desg base shear (V) was troduced for buldgs ot more tha three stores heght (ecludg basemets). (7) Dsplacemet ad Drft: I the 1997 UBC, dsplacemets are determed for the stregth-level earthquake forces. (8) Lateral Forces o Elemets of Structures: New equatos for determg the sesmc forces (F p ) for elemets of structures, ostructural compoets ad equpmet are gve. 71
Applcablty: The Statc Lateral Force Procedure The statc lateral force procedure may be used for the followg structures: 1- All structures, regular or rregular (Table A-1), Sesmc Zoe o. 1 (Table A- 2) ad Occupacy Categores 4 ad 5 (Table A-3) Sesmc Zoe 2. 2- Regular structures uder 73 m heght wth lateral force resstace provded by systems gve Table (A-4) ecept for structures located o sol profle type S F, that have a perod greater tha 0.70 sec. (see Table A-5 for sol profles). 3- Irregular structures ot more tha fve stores or 20 m heght. 4- Structures havg a fleble upper porto supported o a rgd lower porto where both portos of the structure cosdered separately ca be classfed as beg regular, the average story stffess of the lower porto s at least te tmes the average stffess of the upper porto ad the perod of the etre structure s ot greater tha 1.10 tmes the perod of the upper porto cosdered as a separate structure fed at the base. Regular Structures: Regular structures are structures havg o sgfcat physcal dscotutes pla or vertcal cofgurato or ther lateral force resstg systems. Irregular Structures: Irregular structures are structures havg sgfcat physcal dscotutes cofgurato or ther lateral force resstg systems (See Table A-1.a, Table A-1.b, Fgures A-2 ad A-3 for detaled descrpto of such structures). Desg Base Shear: The total desg base shear a gve drecto s to be determed from the followg formula. Cv I W V= (1) RT The total desg base shear eed ot eceed the followg: 2.5 Ca I W V = (2) R 72
The total desg base shear shall ot be less tha the followg: V = 0.11 C I W (3) a I addto, for Sesmc Zoe 4, the total base shear shall ot be less tha the followg: 0.8 Z Nv I W V = (4) R The mmum desg base shear lmtato for Sesmc Zoe 4 was troduced as a result of the groud moto effects observed at stes ear fault rupture 1994 Northrdge earthquake. Where V = total desg lateral force or shear at the base. W = total sesmc dead load - I storage ad warehouse occupaces, a mmum of 25 % of floor lve load s to be cosdered. - Total weght of permaet equpmet s to be cluded. - Where a partto load s used floor desg, a load of ot less tha 50 kg/m 2 s to be cluded. I Z R C a = Buldg mportace factor gve Table (A-3). = Sesmc Zoe factor, show Table (A-2) ad Fgure (A.4). = respose modfcato factor for lateral force resstg system, show Table (A-4). = accelerato-depedet sesmc coeffcet, show Table (A-6). C v = velocty-depedet sesmc coeffcet, show Table (A-7). N a = ear source factor used determato of C a Sesmc Zoe 4, show Table (A-8). v N = ear source factor used determato of v C Sesmc Zoe 4, show Table (A-9). 73
T = elastc fudametal perod of vbrato, secods, of the structure the drecto uder cosderato evaluated from the followg equatos: For reforced cocrete momet-resstg frames, T = 0.073( h ) 3 / 4 (5) For other buldgs, ( h ) 3 / 4 T = 0.0488 Alteratvely, for shear walls, Where T = 0.0743 ( h ) 3 / 4 h = total heght of buldg meters A c (6) (7) A c = combed effectve area, m 2, of the shear walls the frst story of the structure, gve by 2 D e A c = A 0.2 + D e / h 0. 9 (8) h Where D e s the legth, meters, of each shear wall the frst story the drecto parallel to the appled forces. A = cross-sectoal area of dvdual shear walls the drecto of loads m 2 Earthquake Loads: Based o UBC 1630.1.1, horzotal earthquake loads to be used the above-stated load combatos are determed as follows: E = ρ E h + E v (9) E = Ω (10) m o E h Where: E = earthquake load resultg from the combato of the horzotal compoet E h, ad the vertcal compoet, E v E h = the earthquake load due to the base shear, V E m = the estmated mamum earthquake force that ca be developed the structure 74
E v = the load effects resultg from the vertcal compoet of the earthquake groud moto ad s equal to the addto of 0.50 C a I D to the dead load effects D. Ω o = sesmc force amplfcato factor as gve Table (A-4), ad accouts for structural over-stregth ρ = relablty/redudacy factor, to crease the effects of earthquake loads o structures wth few lateral force resstg elemets, gve by 6. 10 ρ = 2 (11) r ma A B AB = groud floor area of structure m 2 to clude area covered by all overhags ad projectos. rma = the mamum elemet-story shear rato For a gve drecto of loadg, the elemet story shear rato s the rato of desg story shear the most heavly loaded sgle elemet dvded by the total desg story shear. rma s defed as the largest of the elemet story shear rato, r, whch occurs ay of the story levels at or below two-thrds heght level of the buldg. For momet-resstg frames, r s take as the mamum of the sum of the shears ay two adjacet colums a momet-resstg frame bay dvded by the story shear For shear walls, r s take as the mamum of the product of the wall shear multpled by 3.05/ lw ad dvded by the total story shear, where l w s the legth of the wall meters. For dual systems ρ 80 % of the values calculated above. Whe calculatg drft, or whe the structure s located Sesmc Zoes 0, 1, or 2, ρ shall be take as 1.0. ρ ca't be smaller tha 1.0 ad ca't be grater tha 1.5. Vertcal Dstrbuto of Force: The base shear evaluated from Eq. (12) s dstrbuted over the heght of the buldg accordg to the followg Eq. F ( V F ) t = = 1 w w h h (12) 75
Fg. (A-1) Vertcal Dstrbuto of Force Where F =0 for T 0. 7 sec. t F t = 0.07TV 0. 25V for T > 0. 7 sec. The shear force at each story s gve by Eq. (13) Where V = F + F (13) t = = umber of stores above the base of the buldg F t = the porto of the base shear, cocetrated at the top of the structure to accout for hgher mode effects F, F, F = lateral forces appled at levels,, or, respectvely h, h, h = heght above the base to levels,, or, respectvely V = desg shear story 76
Horzotal Dstrbuto of Force: The desg story shear ay drecto V, s dstrbuted to the varous elemets of the lateral force-resstg system proporto to ther rgdtes, cosderg the rgdty of the daphragm. Horzotal Torsoal Momet: To accout for the ucertates locatos of loads, the mass at each level s assumed to be dsplaced from the calculated ceter of mass each drecto a dstace equal to 5 % of the buldg dmeso at that level perpedcular to the drecto of the force uder cosderato. The torsoal desg momet at a gve story s gve by momet resultg from eccetrctes betwee appled desg lateral forces appled through each story s ceter of mass at levels above the story ad the ceter of stffess of the vertcal elemets of the story, addto to the accdetal torso. Overturg Momets: Buldgs must be desged to resst the overturg effects caused by the earthquake forces. The overturg momet M = F t ( h h ) + F ( h h ) = + 1 M at level s gve by Eq. (14). Overturg momets are dstrbuted to the varous elemets of the vertcal lateral forceresstg system proporto to ther rgdtes. Dsplacemet ad Drft: The calculated story drfts are computed usg the mamum elastc respose dsplacemet drft ( m ), whch s a estmate of the dsplacemet that occurs whe the structure s subjected to the desg bass groud moto. Accordg to UBC 1630.9.2, m = 0. 7 R s (15) Where: s = desg level respose dsplacemet, whch s the total drft or total story drft that occurs whe the structure s subjected to the desg sesmc forces. Calculated story drft m shall ot eceed 0.025 tmes the story heght for structures havg a fudametal perod of less tha 0.70 secods. (14) 77
Calculated story drft m shall ot eceed 0.020 tmes the story heght for structures havg a fudametal perod equal to or greater tha 0.70 secods. P Effects: P effects are eglected whe the rato gve by Eq. (16) s 0.1. Msec odary P = (16) M V h prmary P = total ufactored gravty load at ad above level = sesmc story drft by desg sesmc forces ( ) V = sesmc shear betwee levels ad 1 h = story heght below level s s I sesmc zoes o. 3 ad 4, P eed ot be cosdered whe the story drft ) ( 0.02 / R) tmes the story heght. ( s h s s 78
Smplfed Desg Base Shear Applcablty: Buldgs of ay occupacy ot more tha three stores heght, ecludg basemets. Other buldgs ot more tha two stores heght, ecludg basemets. Base Shear: The total desg base shear a gve drecto s determed from the followg formula: 3.0 Ca W V = (17) R Whe the sol propertes are ot kow suffcet detal to determe the sol profle type, type S D s used Sesmc Zoes 3 ad 4. Whe the sol propertes are ot kow suffcet detal to determe the sol profle type, type S E s used Sesmc Zoes 1, 2A ad 2B. Vertcal Dstrbuto of Force: The forces at each level are calculated from the followg formula: 3.0 Ca w F = (18) R 79
Table (A-1.a) Vertcal Structural Irregulartes Irregularty Type ad Defto 1- Stffess Irregularty-Soft Story A soft story s oe whch the lateral stffess s less tha 70 percet of that the story above or less tha 80 percet of the average stffess of the three stores above. 2- Weght (mass) Irregularty Mass rregularty s cosdered to est where the effectve mass of ay story s more tha 150 percet of the effectve mass of a adjacet story. A roof that s lghter tha the floor below eed ot be cosdered. 3- Vertcal Geometrc Irregularty Vertcal geometrc rregularty shall be cosdered to est where the horzotal dmeso of the lateral force-resstg system ay story s more tha 130 percet of that a adjacet story. Oe-story pethouses eed ot be cosdered. 4- I-Plae Dscotuty Vertcal Lateral Force-resstg Elemet A -plae offset of the lateral loadresstg elemets greater tha the legth of those elemets. 5- Dscotuty Capacty-Weak Story A weak story s oe whch the story stregth s less tha 80 percet of that the story above. The story stregth s the total stregth of all sesmc-resstg elemets sharg the story shear for the drecto uder cosderato. Requremet Use dyamc aalyss to determe lateralforce dstrbuto. Use dyamc aalyss to determe lateralforce dstrbuto. Use dyamc aalyss to determe lateralforce dstrbuto. Use specal sesmc load combatos for members below dscotuty. Icrease sesmc loads for members below dscotuty by a factor = Ω. o 80
Fgure (A.2): Vertcal Irregulartes 81
Table (A-1.b) Pla Structural Irregulartes Irregularty Type ad Defto 1- Torsoal Irregularty Torsoal rregularty s to be cosdered to est whe the mamum story drft, computed cludg accdetal torso, at oe ed of the structure trasverse to a as s more tha 1.2 tmes the average of the story drfts of the two eds of the structure. 2- Re-etrat Corers Pla cofguratos of a structure ad ts lateral force-resstg system cota reetrat corers, where both projectos of the structure beyod a re-etrat corer are greater tha 15 % of the pla dmeso of the structure the gve drecto. 3- Daphragm Dscotuty Daphragms wth abrupt dscotutes or varatos stffess, cludg those havg cutout or ope areas greater tha 50 % of the gross eclosed area of the daphragm, or chages effectve daphragm stffess of more tha 50 % from oe story to the et. 4- Out-of-plae Offsets Dscotutes a lateral force path, such as out-of-plae offsets of the vertcal elemets. 5- Noparallel Systems The vertcal lateral load-resstg elemets are ot parallel to or symmetrc about the major orthogoal aes of the lateral forceresstg system. Requremet Icrease torsoal forces by a amplfcato factor A. Provde structural elemets daphragms to resst flappg actos. Provde structural elemets to trasfer forces to the daphragm ad structural system. Reforce boudares at opegs. Use specal sesmc load combatos. Oethrd crease stress s ot permtted. The requremet that orthogoal effects be cosdered may be satsfed by desgg such elemets for 100 % of the prescrbed sesmc forces oe drecto plus 30 % of the prescrbed forces the perpedcular drecto. Alterately, the effects of the two orthogoal drectos may be combed o a square root of the sum of the squares bass. 82
Fgure (A.3): Pla Irregulartes 83
Table (A-2) Sesmc Zoe Factor Z Zoe 1 2A 2B 3 4 Z 0.075 0.15 0.20 0.30 0.40 Note: The zoe shall be determed from the sesmc zoe map. Fgure (A.4): Sesmc map of Paleste 84
Table (A-3) Occupacy Importace Factors (cocse) Occupacy Category Sesmc Importace Factor, I 1-Essetal facltes 1.25 2-Hazardous facltes 1.25 3-Specal occupacy structures 1.00 4-Stadard occupacy structures 1.00 5-Mscellaeous structures 1.00 85
Table (A-4) Structural Systems Basc Structural System Lateral- force resstg system descrpto R Ω o Heght lmt Zoes 3 &4. (meters) Bearg Wall Cocrete shear walls 4.5 2.8 48 Buldg Frame Cocrete shear walls 5.5 2.8 73 Momet- Resstg Frame SMRF IMRF 8.5 5.5 2.8 2.8 N.L ---- Dual OMRF Shear wall + SMRF Shear wall + IMRF 3.5 8.5 6.5 2.8 2.8 2.8 ---- N.L 48 Catlevered Colum Buldg Shear-wall Frame Iteracto Catlevered colum elemets 2.2 2.0 10 5.5 2.8 48 86
Table (A-5) Spol Profle Types Sol Profle Type Sol Profle Name/Geerc Descrpto Average Sol Propertes For Top 30 m Of Sol Profle Shear Wave Stadard Udraed Velocty, v Peetrato Shear s m/s Test, N Stregth, S u (blows/foot) kpa S A Hard Rock > 1,500 --- --- S B Rock 760 to 1,500 S C Very Dese Sol ad 360 to 760 > 50 > 100 Soft Rock S Stff Sol Profle 180 to 360 15 to 50 50 to 100 D S E Soft Sol Profle < 180 < 15 < 50 S F Sol Requrg Ste-specfc Evaluato Table (A-6) Sesmc Coeffcet Sol Type C a Profle Sesmc Zoe Factor, Z Z =0.075 Z = 0.15 Z = 0.2 Z = 0.3 Z = 0.4 S A 0.06 0.12 0.16 0.24 0.32 N a S B 0.08 0.15 0.20 0.30 0.40 N a S C 0.09 0.18 0.24 0.33 0.40 N a S D 0.12 0.22 0.28 0.36 0.44 N a S E 0.19 0.30 0.34 0.36 0.36 N a S F See Footote Footote: Ste-specfc geotechcal vestgato ad dyamc respose aalyss shall be performed to determe sesmc coeffcets for sol Profle Type S F. 87
C Table (A-7) Sesmc Coeffcet v Sol Profle Sesmc Zoe Factor, Z Type Z =0.075 Z = 0.15 Z = 0.2 Z = 0.3 Z = 0.4 S A 0.06 0.12 0.16 0.24 0.32 N v S B 0.08 0.15 0.20 0.30 0.40 N v S C 0.13 0.25 0.33 0.45 0.56 N v S D 0.18 0.32 0.40 0.54 0.64 Nv S E 0.26 0.50 0.64 0.84 0.96 N v S F See Footote Footote: Ste-specfc geotechcal vestgato ad dyamc respose aalyss shall be performed to determe sesmc coeffcets for sol Profle Type S F. Table (A-8) Near-Source Factor Sesmc Type N a Source Closest Dstace to Kow Sesmc Source 2 km 5 km 10 km A 1.5 1.2 1.0 B 1.3 1.0 1.0 C 1.0 1.0 1.0 Table (A-9) Near-Source Factor Sesmc Type N v Source Closest Dstace to Kow Sesmc Source 2 km 5 km 10 km 15 km A 2.0 1.6 1.2 1.0 B 1.6 1.2 1.0 1.0 C 1.0 1.0 1.0 1.0 88
Eample (1): Usg UBC 97, evaluate the sesmc base shear actg o a regular twelve-story buldg frame system wth reforced cocrete shear walls the prcpal drectos, as the ma lateral force-resstg system. The buldg whch s located Gaza Cty s 31.2 m by 19 m pla ad 32.8 m heght (Stadard Occupacy). It s costructed o a sady sol profle wth SPT values ragg from 20 to 50 blows/foot. Soluto: From Table A-2 ad for Zoe 1, Z = 0.075 From Table A-3 ad for Stadard Occupacy, I = 1.0 From Table A-5, Sol Profle Type s From Table A-4, R = 5.5 From Table A-6, C a = 0.12 From Table A-7, C v = 0.18 From Eq. (6), T = 0.0488 = 0.75 sec. ( 38.28) 3/ 4 From Eq. (A-1), the total base shear s Cv I W 0.18 W V= = = 0. 0436W RT 5.5 ( 0.75) S D From Eq. (A-2), the total base s ot to eceed 2.5 C I W 2.5( 0.12) W = = = 0. W O.K a V 0545 R 5.5 From Eq. (A-3), the total desg base s ot to be less tha V= 0.11 Ca I W = 0.11(0.12) W = 0. 0132W O.K Thus, V= 0. 0436 W 89
Eample (2): For the 8-storey buldg frame system wth shear walls, show Fgure (A.5), (1) Evaluate the base shear V usg UBC-97 provsos (both drectos). (2) Dstrbute the forces the vertcal drecto (both drectos). Provded Data: - The buldg s used for resdetal purposes, ad located Gaza Cty. - Story heght s 3.0 m. - Sol profle s classfed as S D. 2 2 - Use f ' c = 350 Kg / cm ad f y = 4200 Kg / cm. - Floor sustaed dead load = 1200 kg/m 2. - Floor lve load = 200 kg/m 2. - Colums are 40 cm 40 cm cross secto. Soluto: From Table A-2 ad for Zoe 1, Z = 0.075 Pla From Table A-3 ad for Stadard Occupacy, I = 1.0 Sol Profle Type s gve as S D 90
From Table A-4, R = 5.5 From Table A-6, C a = 0.12 From Table A-7, C v = 0.18 (1) Drecto of shear walls A, B ad C: 0.0743 T = A ( h ) c 3 4 2 D e A c = A 0.2 + D e / h 0. 9 h 2 3 2 3 A c = ()( 3 0.20)( 3) 0.2 + = 0.3881m, = 0.125 < 0. 9 O.K 24 24 T = 0.0743 4 ( h ) 3/ ( 24) A c = 0.0743 3/ 4 = 1.293sec 0.3881 From Eq. (A-1), the total base shear s Cv I W 0.18W V= = = 0. 0253W RT 5.5 ( 1.293) From Eq. (A-2), the total base s ot to eceed ( 0.12) 2.5 Ca I W 2.5 W V= = = 0. 0545W O.K R 5.5 From Eq. (A-3), the total desg base s ot to be less tha V= 0.11 Ca I W = 0.11(0.12) W = 0. 0132W O.K Thus, V= 0. 0253W= 0.0253*(1200/1000)*15*15*8= 54.65 tos Vertcal Dstrbuto of Force: Sce T > 0.7 secod, Ft = 0.07 T V = 0.07 ( 1.293)( 54.65) = 4. 95 tos F ( V F ) w h ( 54.65 4.95) t = 8 = 1 w h = 29160 w h < 0.25 (54.65) tos 91
(2) Drecto of shear walls D ad E: 3 0.0743( h ) 4 T = A c 2 D e A c = A 0.2 + D e / h 0. 9 h 2 2 4 3 A c = = 24 24 2 ( 4)( 0.20) 0.2 + + ()( 3 0.20) 0.2 + 0.3116 m T = 0.0743 4 ( h ) 3 / ( 24) A c = 0.0743 3 / 4 = 1.443sec 0.3116 From Eq. (A-1), the total base shear s Cv I W 0.18W V= = = 0. 0227W RT 5.5 ( 1.443) From Eq. (A-2), the total base s ot to eceed ( 0.12) 2.5 Ca I W 2.5 W V= = = 0. 0545W O.K R 5.5 92
From Eq. (A-3), the total desg base s ot to be less tha V= 0.11 Ca I W = 0.11(0.12) W = 0. 0132W O.K Thus, V= 0. 0227W = 0.0227*(1200/1000)*15*15*8= 49.03 tos Vertcal Dstrbuto of Force: Sce T > 0.7 secod, Ft = 0.07 T V = 0.07 ( 1.443) ( 49.03) = 4. 95 tos F ( V F ) w h ( 49.03 4.95) t = 8 = 1 w h = 29160 w h < 0.25 (49.03) tos 93