Figure 2.1 Risk Category vs. Ground Motion (Ref Figure C11.5-1) 1-22 Steven T. Hiner, MS, SE

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1 Chater 2 Organizations, Codes & Standards SDR Workbook 2012 IBC Version While PBSD has yet to be fully develoed, it is exected to be used in future building codes to rovide a methodology to more reliably redict seismic risk in all buildings in terms more useful to building owners and building users. Under the ASCE 7-10 seismic design rovisions, there exist imlied erformance levels as demonstrated by Figure 2.1. For examle, Risk Category II (i.e., I e = 1.0) structures designed to ASCE 7-10 rovisions are exected to meet the following erformance levels: Collase revention for MCE R ground motions Life-safety for Design ground motions (i.e., 2/3 MCE R ) Immediate occuancy for Frequent ground motions (i.e., ~ 50% in 50 years / 100 year return interval) Risk Category IV (i.e., I e = 1.5) structures designed to ASCE 7-10 rovisions are exected to meet the following erformance levels: Life-safety for MCE R ground motions Immediate occuancy for Design ground motions (i.e., 2/3 MCE R ) Oerational for Frequent ground motions Similarly, Risk Category III (i.e., I e = 1.25) structures designed to ASCE 7-10 rovisions are exected to meet erformance levels that fall between the Risk Category IV and Risk Category II structures as shown in Figure 2.1 below. Figure 2.1 Risk Category vs. Ground Motion (Ref Figure C11.5-1) 1-22 Steven T. Hiner, MS, SE

2 Chater 3 General Provisions & Seismic Design Criteria Site Class Adjusted MCE R Acceleration Parameters IBC S MS & S M1 reresent the site class adjusted MCE R sectral resonse acceleration arameters at short eriods and at 1-second eriod resectively, and are determined by the following equations: S MS = F a S S IBC (16-37) S M1 = F v S 1 IBC (16-38) where: F a = Site coefficient er IBC Table (1) F v = Site coefficient er IBC Table (2) Site coefficient F a is function of the Site Class and the short eriod maed sectral resonse acceleration arameter (S S ). Site coefficient F v is a function of the Site Class and the 1-second eriod maed sectral resonse acceleration arameter (S 1 ). F a & F v are based on the results of emirical analyses of strong-motion data and analytical studies of site resonse. In general, softer soils exhibit greater amlification of earthquake ground motions than stiffer soils, and that amlification can be even more significant for longer eriod (e.g., taller) structures than for shorter eriod structures (e.g., F v values are greater than F a values for Site Class C, D & E). Design Sectral Resonse Acceleration Parameters IBC S DS & S D1 reresent the 5% damed design sectral resonse acceleration arameters at short eriods and at 1-second eriod resectively and they are determined by the following equations: S DS = 2/3 S MS IBC (16-39) S D1 = 2/3 S M1 IBC (16-40) NOTE: Table 3.2 (. 1-33) and Table 3.3 (. 1-34) are rovided as short cut rocedures in determining S DS & S D1 resectively when S S, S 1 & Site Class are known (Site Class D may be assumed unless given). Figure 3.1 Design Resonse Sectrum (Ref. 9 - ASCE 7 Figure ) where: S D1 TS SDS T0 0. 2T S T L = er ASCE 7 Figure Steven T. Hiner, MS, SE 1-31

3 Chater 3 General Provisions & Seismic Design Criteria SDR Workbook 2012 IBC Version Alternative Seismic Design Category Determination IBC Where S 1 < 0.75, the Seismic Design Category is ermitted to be determined from IBC Table (1) alone (i.e, using S DS only) when all of the following aly: T a < 0.8 T S in each of the two orthogonal directions, and the fundamental eriod of the structure used to calculate the story drift - T < T S in each of the two orthogonal directions, and SDS ASCE 7 (12.8-2) is used to determine the seismic resonse coefficient CS, and ( R Ie) The diahragms are rigid (er ASCE ) or for diahragms that are flexible, the distance between vertical elements of the seismic force-resisting system (SFRS) 40 feet Simlified Design Procedure IBC Where the alternate Simlified Design Procedure of ASCE is used, the Seismic Design Category shall be determined in accordance with ASCE ASCE 7 Seismic Design Criteria ASCE 7 Chater 11 Scoe ASCE Every structure (e.g., buildings and nonbuilding structures), and ortion thereof, including nonstructural comonents, shall be designed and constructed to resist the effects of earthquake motions as rescribed by the seismic requirements of ASCE 7. Alicability ASCE Structures and their nonstructural comonents shall be designed and constructed in accordance with the requirements of the following chaters based on the tye of structure or comonent: Buildings: ASCE 7 Chater 12 Nonbuilding Structures: ASCE 7 Chater 15 Nonstructural Comonents: ASCE 7 Chater 13 Seismically Isolated Structures: ASCE 7 Chater 17 Structures with Daming Systems: ASCE 7 Chater 18 Seismic Imortance Factor, I e ASCE Each structure shall be assigned an imortance factor (I e ) in accordance with ASCE 7 Table based on the Risk Category of the building (or other structure) from IBC Table Risk Category I I = 1.0 Risk Category II I = 1.0 Risk Category III (high occuancy) I = 1.25* Risk Category IV (essential facilities) I = 1.5* The seismic imortance factor (I e ) is used in the Seismic Resonse Coefficient (C S ) equations with the intent to raise the yield level for imortant structures (e.g., hositals, fire stations, emergency oeration centers, hazardous facilities, etc.). Use of an imortance factor greater than one is intended to rovide for a lower inelastic demand on a structure which should result in lower levels of structural and nonstructural damage Steven T. Hiner, MS, SE

4 Chater 4 Seismic Design Requirements for Building Structures SDR Workbook 2012 IBC Version 4.9 Equivalent Lateral Force Procedure ASCE Seismic Base Shear, V Strength Design force leveli ASCE The seismic base shear in a given direction shall be determined in accordance with the following: V C W ASCE 7 (12.8-1) S Figure 4.12 Seismic Base Shear Seismic Resonse Coefficient, C S ASCE The seismic resonse coefficient (C S ) shall be determined as follows: SDS CS ASCE 7 (12.8-2) ( R I e ) NOTE: ASCE 7 (12.8-2) will always governs when T T S which tyically occurs with low rise and/or short eriod structures (i.e., 3 stories). C S need not exceed the following: SD 1 CS for T T L ASCE 7 (12.8-3) T ( R I ) e NOTE: ASCE 7 (12.8-3) tyically governs for longer eriod structures when T S T T L but C S minimum er ASCE 7 (12.8-5) and (12.8-6) need to be considered. SD 1TL CS for T > T 2 L ASCE 7 (12.8-4) T ( R I ) e NOTE: ASCE 7 (12.8-4) can aly for very long eriod (i.e., very tall) structures, when T > T L but C S minimum er ASCE 7 (12.8-5) and (12.8-6) will tyically govern over ASCE 7 (12.8-4) Steven T. Hiner, MS, SE

5 Chater 5 Earthquake Loads and Load Combinations T = self-straining load (e.g., arising from contraction or exansion resulting from temerature change, shrinkage, moisture change, cree in comonent materials, movement due to differential settlement or combinations thereof) W = load due to wind ressure Strength Design or Load & Resistance Factor Design IBC All alicable Strength Design (SD or LRFD) load combinations must be considered since the most critical load effect may occur when one or more of the contributing loads (e.g., D, E, L, L r, S ) are not acting. Basic (SD or LRFD) Load Combinations IBC Where Strength Design (or Load and Resistance Factor Design) is used, structures and ortions thereof shall be designed to resist the most critical effects resulting from the equations of IBC There are a total of seven SD/LRFD basic load combinations equations in the IBC. Only the two load combinations which include the earthquake load (E) are noted below: 1.2(D + F) + 1.0E + f 1 L + 1.6H + f 2 S IBC (16-5) or ( S DS )D + Q E + f 1 L + f 2 S when F = 0 & H = 0 0.9(D + F) + 1.0E + 1.6H IBC (16-7) or ( S DS )D Q E when F = 0 & H = 0 Excetion: Where other factored load combinations are secifically required by the rovisions of the IBC (i.e., Chater 19 Concrete) such combinations shall take recedence. Allowable Stress Design IBC All alicable Allowable Stress Design (ASD) load combinations must be considered since the most critical load effect may occur when one or more of the contributing loads (e.g., D, E, L, L r, S ) are not acting. Basic (ASD) Load Combinations IBC Where Allowable Stress Design (or Working Stress Design) is used, structures and ortions thereof shall be designed to resist the most critical effects resulting from the equations of IBC There are a total of nine Basic (ASD) Load Combination equations in the IBC. Only the three load combinations which include the earthquake load (E) are noted below: D + H + F + (0.6W or 0.7E) IBC (16-12) or ( S DS )D Q E when F = 0 & H = 0 D + H + F (0.7E) L S IBC (16-14) or ( S DS )D (0.7 Q E ) L S 0.6(D + F) + 0.7E + H IBC (16-16) or ( S DS )D 0.7 Q E when F = 0 & H = 0 Excetions: see IBC for excetions to crane hook loads, to flat roof snow loads 30 sf, roof live loads 30 sf, flat roof snow loads > 30 sf, etc. Steven T. Hiner, MS, SE 1-75

6 Chater 5 Earthquake Loads and Load Combinations SDR Workbook 2012 IBC Version 5.3 Equivalent Lateral Force Procedure Overview I, II, III or IV Risk Category I e Seismic Imortance Factor S s & S 1 Maed MCE R sectral resonse acceleration arameters A, B, C, D*, E or F Site Class F a & F v Site Coefficients S MS & S M1 Soil modified MCE R sectral resonse acceleration arameters S DS & S D1 A*, B*, C, D, E or F Design sectral resonse acceleration arameters Seismic Design Category T a Aroximate Fundamental Period C S V = C S W Seismic Resonse Coefficient Seismic Base Shear at Strength Level (i.e. SD/LRFD) F x Vertical Distribution of Seismic Force 1-78 Steven T. Hiner, MS, SE

7 Chater 6 Seismic Design Requirements for Nonstructural Comonents Chater 6 Seismic Design Requirements for Nonstructural Comonents 6.1 ASCE 7 Chater 13 Overview Scoe ASCE ASCE 7 Chater 13 establishes minimum design criteria for nonstructural comonents that are ermanently attached to structures, and for their suorts and attachments. A nonstructural comonent is a art or element of an architectural, electrical or mechanical system. Seismic Design Category ASCE Nonstructural comonents shall be assigned to the same Seismic Design Category (SDC) as the structure that they occuy, or to which they are attached. Comonent Imortance Factor, I ASCE All comonents shall be assigned a comonent imortance factor (I ), which will be equal to 1.5 or 1.0. Use an I = 1.5 if any of the following conditions aly: 1. The comonent is required to function for life-safety uroses after an earthquake, including fire rotection srinkler systems and egress stairways 2. The comonent conveys, suorts, or otherwise contains toxic, highly toxic, or exlosive substances 3. The comonent is in (or attached to) a Risk Category IV structure (i.e., essential facility), and it is needed for continued oeration of the facility or its failure could imair the continued oeration of the facility 4. The comonent conveys, suorts, or otherwise contains hazardous substances All other comonents shall be assigned an I = 1.0 Exemtions ASCE The following nonstructural comonents are exemt from the requirements of ASCE 7 Chater 13: 1. Furniture (excet storage cabinets > 5 feet tall) 2. Temorary or movable equiment 3. Architectural comonents in SDC = B (other than araets suorted by bearing walls or shear walls) rovided I = Mechanical and electrical comonents in SDC = B 5. Mechanical and electrical comonents in SDC = C rovided that I = Mechanical and electrical comonents in SDC = D, E, or F where I = 1.0, the comonent is ositively attached to the structure, flexible connections are rovided between comonent and associated ductwork, iing, and conduit and either: Comonent weighs 400 lbs and has a C.M. located 4 feet above the adjacent floor level; or Steven T. Hiner, MS, SE 1-81

8 Chater 6 Seismic Design Requirements for Nonstructural Comonents Secific Requirements & References Table 6.2 Secific Mechanical & Electrical Comonent References Item Section Reference Mechanical Comonents ASCE Electrical Comonents ASCE Comonent Suorts ASCE Utility and Service Lines ASCE HVAC Ductwork ASCE Piing Systems ASCE Boilers and Pressure Vessels ASCE Elevator & Escalator Design Requirements ASCE Other Mechanical & Electrical Comonents ASCE Structural Walls and their Anchorage ASCE Design for Out-of-Plane Forces ASCE Structural walls and their anchorage shall be designed for a force normal to the surface (e.g., out-of-lane) equal to: F 0. 4S I W DS e W w 0.10 minimum w Figure 6.2 Out-of-Plane Forces on Structural Walls Wall with Cantilever Paraet Wall without Paraet Steven T. Hiner, MS, SE 1-87

9 Chater 6 Seismic Design Requirements for Nonstructural Comonents SDR Workbook 2012 IBC Version Anchorage of Structural Walls and Transfer of Design Forces into Diahragms ASCE Wall Anchorage Forces, F ASCE The anchorage of structural walls to suorting construction (e.g., roof or floor diahragms) shall rovide a direct connection caable of resisting the following: F 0. 4S K I W ASCE 7 ( ) DS a e 0.2KaI ew minimum where: L f Ka maximum ASCE 7 ( ) 100 F = the design force in the individual anchors S DS = short eriod design sectral resonse acceleration arameter I e = seismic imortance factor L f = flexible diahragm san (feet), use 0 for rigid diahragms W = the weight of the masonry or concrete wall tributary to the anchor = W wall (h w / 2 + h ) (anchor sacing) for (one-story) walls with a araet = W wall (h w / 2) (anchor sacing) for (one-story) walls without a araet Where the anchorage is not located at the roof and all diahragms are not flexible, the value from ASCE 7 ( ) is ermitted to be multilied by (1 + 2z/h) / 3, where z is the height of the anchor above the base of the structure and h is the height of the roof above the base (e.g., h = h n ). NOTE: Structural walls shall be designed to resist bending between anchors where anchor sacing exceeds 4 feet (i.e., urlin anchors > 4-0 o.c.). Refer to Figure 6.3 below for examles of the L f and K a values that would be aroriate to determine the wall anchorage force for the structural walls on each wall line deicted for Building A & Building B. Figure 6.3 Wall Anchorage Examles a. Building A b. Building B 1-88 Steven T. Hiner, MS, SE

10 Chater 6 Seismic Design Requirements for Nonstructural Comonents The following values of L f and K a would aly for structural wall anchorage for Building A shown in Figure 6.3a (Building A) - As Flexible Diahragm: Wall anchorage for line A & B: L f = 150 feet K a = 2.0 maximum Wall anchorage for line 1 & 2: L f = 80 feet K a = 1.8 As Rigid Diahragm: Wall anchorage for ALL lines: L f = 0 feet K a = 1.0 The following values of L f and K a would aly for structural wall anchorage for Building B shown in Figure 6.3b (Building B) - As Flexible Diahragm: Wall anchorage for line A & B (line 1 to 2): L f = 100 feet K a = 2.0 Wall anchorage for line A & B (line 2 to 3): L f = 50 feet K a = 1.5 Wall anchorage for line 1, 2 & 3: L f = 80 feet K a = 1.8 As Rigid Diahragm: Wall anchorage for ALL lines: L f = 0 feet K a = 1.0 Additional Requirements for SDC = C, D, E, or F ASCE Diahragms for structures assigned to SDC = C, D, E, or F shall meet the following additional requirements: 1. Transfer of Anchorage Forces - Diahragms shall be rovided with continuous ties (or struts) between diahragm chords to distribute these anchorage forces into the diahragm(s) Diahragm connections shall be ositive, mechanical, or welded Added chords are ermitted to be used to form subdiahragms (see ) to transmit the anchorage forces to the main continuous cross-ties The maximum length-to-width ratio of the structural subdiahragm shall be 2.5 to 1 Connections and anchorages caable of resisting the rescribed forces shall be rovided between the diahragm and the attached comonents Connections shall extend into the diahragm a sufficient distance to develo the force transferred into the diahragm 2. Steel Elements - the strength design forces for steel elements (i.e., urlin anchors), with the excetion of anchor bolts and reinforcing steel, shall be increased by 1.4 times the forces otherwise required by ASCE (i.e., use 1.4 F anchorage force for stra urlin anchors) 3. Wood Diahragms - The continuous ties or struts shall be in addition to the diahragm sheathing Anchorage shall not be accomlished by use of toe nails or nails subject to withdrawal Wood ledgers or framing shall not be used in cross-grain bending or cross-grain tension 4. Metal Deck Diahragms - metal deck shall not be used as the continuous ties (or struts) required by ASCE in the direction erendicular to the deck san. Steven T. Hiner, MS, SE 1-89

11 Chater 7 Seismic Design Requirements for Nonbuilding Structures Determination of Seismic Factors - R, 0 & C d R, 0 & C d are to be determined from ASCE 7 Table for nonbuilding structures NOT similar to buildings. In general, R & 0 values assigned to these nonbuilding structures are less than those assigned to building structures (and nonbuilding structures similar to buildings). This is because buildings tend to have structural redundancy due to multile bays and frame lines and contain nonstructural and non-considered resisting elements that effectively give buildings greater daming, strength and ductility during an earthquake. Seismic Base Shear, V Strength Design force level ASCE When the fundamental eriod is greater than or equal to 0.06 second (i.e., T 0.06 second), the seismic base shear in a given direction shall be determined in accordance with the following: V C W ASCE 7 (12.8-1) S Seismic Resonse Coefficient, C S ASCE The Seismic Resonse Coefficient (C S ) shall be determined as follows: SDS CS ASCE 7 (12.8-2) ( R I e ) NOTE: ASCE 7 (12.8-2) tyically governs for nonbuilding structures, when T < T S C S need not exceed the following: and C S shall not be less than: SD 1 CS for T T L ASCE 7 (12.8-3) T ( R I) SD1TL CS for T > T 2 L ASCE 7 (12.8-4) T ( R I ) e C 0.044S I 0.03 minimum ASCE 7 (15.4-1) S DS e (see ASCE 7-05 Sulement No. 2) In addition, for nonbuilding structures where S C S shall not be less than: 0.8S1 CS * minimum ASCE 7 (15.4-2) ( R I ) e *NOTE: See ASCE , item 2 for an excetion for Tanks and Vessels. Steven T. Hiner, MS, SE 1-101

12 Chater 8 Diahragm Design & Wall Rigidity SDR Workbook 2012 IBC Version 8.2 Diahragm Tyes Diahragm Flexibility ASCE The structural analysis shall consider the relative stiffnesses of diahragms (floor and/or roof), and the vertical elements of the seismic force-resisting system (e.g., shear walls, braced frames, moment frames). Unless a diahragm can be idealized as either flexible or rigid the structural analysis shall exlicitly include consideration of the stiffness of the diahragm (i.e., semi-rigid modeling assumtion). Flexible Diahragm Condition ASCE Diahragms constructed of untoed steel decking (i.e., metal deck without concrete fill) or wood structural anels (WSP) are ermitted to be idealized as flexible if any of the following conditions exist: Structures where the vertical elements are steel braced frames; steel and concrete comosite braced frames; or concrete, masonry, steel, or steel and concrete comosite shear walls One- and two-family dwellings Structures of light-frame construction (i.e., wood studs or metal studs) where all of the following conditions are met: 1. Toing of concrete or similar materials is not laced over wood structural anel (WSP) diahragms excet for nonstructural toings 1½ inches thick 2. Each line of vertical elements of the seismic force-resisting system (SFRS) comlies with the allowable story drift of ASCE 7 Table Rigid Diahragm Condition ASCE Diahragms of concrete slabs (or concrete filled metal deck) with san-to-deth ratios of 3 to 1 or less (i.e., L/d 3) in structures having no horizontal irregularities are ermitted to be idealized as rigid. Calculated Flexible Diahragm Condition ASCE Diahragms not satisfying the conditions to be considered flexible or rigid are ermitted to be idealized as flexible where the comuted maximum in-lane deflection of the diahragm (under lateral load) is more than two times the average story drift of adjoining vertical elements of the seismic-force-resisting system of the associated story under equivalent tributary lateral load (see Figure 8.4). Figure 8.4 Calculated Flexible Diahragm Condition max 2 max min 2 Diahragm Deflection ASCE The deflection of the diahragm shall not exceed the ermissible deflection of the attached elements (e.g., bearing walls) which shall be that deflection that will ermit the attached element to maintain its structural integrity under the individual loading and continue to suort the rescribed loads Steven T. Hiner, MS, SE

13 Chater 8 Diahragm Design & Wall Rigidity Figure 8.8 below demonstrates the comarison that is often made between the analysis of a uniformly loaded flexible diahragm (on the left) and a uniformly loaded simly suorted beam (on the right). For the design of flexible diahragms, the shear diagram can be used to determine the maximum unit shear at the end suorts (e.g., shear walls). The moment diagram can be used to determine the maximum chord force, or the chord force at a secific oint on the chord boundary member (see Figure 8.10, ). Figure 8.8 Flexible Diahragm Loading (Ref. 19) Steven T. Hiner, MS, SE 1-111

14 Chater 8 Diahragm Design & Wall Rigidity The Drag Force should always be equal to zero at each end of the collector. Numerical rounding of the calculated unit diahragm shear and unit wall shear may result in Drag Forces that do not quite zero at the ends of the collector. Tyically, the wall to lates (at the diahragm boundary) are used as the chords and collectors for wood framed stud wall construction. Collector Elements ASCE Collector elements shall be rovided that are caable of transferring the seismic forces originating in other ortions of the structure to the element roviding resistance to those forces (e.g., shear walls, braced frames, moment resisting frames, etc.). Seismic Design Category C, D, E or F ASCE In structures assigned to SDC = C, D, E or F - collector elements and their connections including connections to vertical elements (e.g., shear walls, braced frames, moment resisting frames, etc.) shall be designed to resist the maximum of the forces determined from ASCE items 1, 2 and 3 (see. 1-77), which considers the seismic load effects including overstrength factor ( 0 ) of ASCE ASCE allows two excetions: (drag) forces calculated from F x maximum er ASCE 7 equation ( ) using ASCE load combinations; and for structures braced entirely by light-frame shear walls (i.e., wood studs or cold formed steel studs) with drag forces determined from the governing F x force er ASCE 7 ( ), ( ) and ( ). 8.6 Masonry & Concrete Wall Rigidity Cantilever Shear Wall Deflection Total Flexure Shear Figure 8.12 Cantilever Shear Wall / Pier 3 F H 1.2F H C 3EI A G where: F = force at to of wall H = height of wall to force, F E = modulus of elasticity G = shear modulus A = area = t D I = moment of inertia = t D 3 /12 Wall Rigidity, R Rigidity is roortional to the recirocal of deflection, and is essentially the relative stiffness of the lateral resisting element (i.e., shear wall). Masonry & concrete shear walls resist lateral loads in roortion to their Rigidities (R), therefore only "relative rigidities" are needed. The relative cantilever wall rigidities are determined for each shear wall using their resective H/D ratios and Table D1 - Relative Rigidity of Cantilever Shear Walls / Piers (Aendix D,. 5-20). Steven T. Hiner, MS, SE 1-115

15 Chater 8 Diahragm Design & Wall Rigidity SDR Workbook 2012 IBC Version Figure 8.13 Cantilever Shear Wall Relative Rigidity V R 1 R 1 F R1 R2 2 V 2 F R1 R2 Fixed Shear Wall Deflection Figure 8.14 Fixed Shear Wall / Pier Total Flexure Shear 3 F H 1.2F H F 12EI A G where: F = force at to of wall H = height of wall to force, F E = modulus of elasticity G = shear modulus A = area = t D I = moment of inertia = t D 3 /12 Shear Wall with Oenings This method can be used to aroximate the shear force in a articular ier when the lateral force (F ) to the total wall is known, such as for a flexible diahragm building. The assumtion is that the lighter shaded ortion of wall above and below the oenings will rovide fixity of the darker shaded individual iers. Determine the "Fixed" Rigidity (R F ) of each of the individual iers using their resective H/D ratios and Table D2 - Relative Rigidity of Fixed Shear Walls / Piers (Aendix D,. 5-21). Figure 8.15 Shear Wall with Fixed Piers Force to Pier 1, F 1 F R F1 R F 2 R F1 R F 3 R F Steven T. Hiner, MS, SE

16 Chater 9 IBC Chater 23 Wood where: E E h E v The weight of the wall (W W ) can be considered a dead load effect (D). While the weight of the wall might be considered in the determination of the total seismic force on the shear wall (i.e., V 1 + C S W W ), it will not be used to determine E v because the dead load effect (D) does not contribute to the determination of the calculated unit wall shear. NOTE: The Redundancy factor () shall be considered in the design of the shear walls. Therefore, E = Eh QE V1 So the ASD calculated unit wall shear: (0.7)(total wall shear) (0.7 V1 ) ASD wall (units of lf) shear wall width b NOTE: The equation above is used to determine the drag force, and may be used for shear wall design when the wall weight (W w ) is not significant, not given in a roblem statement, or when the diahragm design force (e.g., w s = f x = F x / L) includes all erimeter walls of the building essentially when the base shear (V ) is used to design the diahragm. Otherwise, the weight of the shear wall (W w ) can be included in the ASD calculated unit wall shear: ( 0.7)( V1 CSWW ) ASD calculated wall (units of lf) b Using the ASD calculated wall and SDPWS Table 4.3A (see Table 9.5), choose the aroriate: WSP anel grade WSP nominal anel thickness Fastener (nail) size Fastener (nail) sacing Such that the ASD allowable wall ASD calculated wall Unit Wall Shear Reduction SDPWS Table 4.3.4, footnote 1 The nominal unit shear caacities of SDPWS Table 4.3A will require a reduction for walls resisting seismic loads for any individual shear wall asect ratio that exceeds 2:1 (but is less than 3.5:1) h/b 2:1 use nominal unit shear caacities from SDPWS Table 4.3A 3 with no reduction 2:1 < h/b 3.5:1 use nominal unit shear caacities from SDPWS Table 4.3A 3 multilied by 2b/h NOTE: For h/b = 3.5:1 this will result in 2b/h = 0.57, or a 43% reduction in the nominal unit shear caacities noted in SDPWS Table 4.3A when resisting seismic loads (not wind loads). For 2:1 < h/b 3½:1 the reduction factor 2b/h will be somewhere between 1.00 and 0.57, resulting in a 0% to 43% reduction in the nominal unit wall shear values noted in SDPWS Table 4.3A when resisting seismic loads (not wind loads). Steven T. Hiner, MS, SE 1-139

17 Chater 10 Other Material Chaters NOTE: Following the January 17, 1994 Northridge Earthquake, over 100 steel buildings with welded moment-resisting frames were found to have beam-to-column connection fractures. Usually the fractures initiated at the comlete joint enetration weld between the beam bottom flange and the column flange. Investigators have suggested that the fractures may be due to a number of factors such as notch effects created by a left in lace weld backing bar; sub-standard welding (excessive orosity, slag inclusions, incomlete fusion); and/or re-earthquake fractures due to initial shrinkage of the weld during cool-down. In Setember 1994, the International Conference of Building Officials (ICBO) adoted an emergency code change to the 1994 Uniform Building Code (UBC). This code change omitted the re-qualified connection (see Figure 10.7) and required that connections be designed to sustain inelastic rotation and develo the strength criteria as demonstrated by cyclic testing or calculation. In November 2000, the SAC Joint Venture finalized the welded steel moment frame issues which were ublished by the Federal Emergency Management Agency (FEMA) in the following documents: FEMA 350 Recommended Seismic Design Criteria for New Steel Moment-Frame Buildings FEMA 351 Recommended Seismic Evaluation and Ugrade Criteria for Existing Welded Steel Moment-Frame Buildings FEMA 352 Recommended Post-Earthquake Evaluation and Reair Criteria for Welded Steel Moment-Frame Buildings FEMA 353 Recommended Secifications and Quality Assurance Guidelines for Steel Moment- Frame Construction for Seismic Alications Doubler Plates Doubler lates are sometimes added to one (or both sides) of the column web and are intended to increase the columns web shear strength and web criling caacity within the anel zone. Continuity Plates Continuity lates are sometimes added to rovide continuity of the intersecting beam (or girder) flanges across the column web. The requirement for continuity lates is a function of the column yield strength, flange width, and flange thickness, and the intersecting beam yield strength, flange width, and flange thickness. Intermediate Moment Frames (IMF) AISC E.2 R = 4½ for steel IMF* IMF s are exected to withstand limited inelastic deformations (in their members and connections) when subjected to the Design Basis Earthquake ground motions. Limited secial detailing of beam-column joint, therefore limited ductility (i.e., lower R). Ordinary Moment Frames (OMF) AISC E.1 R = 3½ for steel OMF* OMF s are exected to withstand minimal inelastic deformations (in their members and connections) when subjected to the Design Basis Earthquake ground motions. Very limited secial detailing of beam-column joint, therefore very limited ductility (i.e., very low R). *NOTE: Tyically steel IMF s and OMF s will not be ermitted in structures assigned to Seismic Design Category D, E or F with the excetion of steel OMF s meeting the requirements of ASCE for SDC = D or E or for SDC = F, and steel IMF s meeting requirements of ASCE for SDC = D or for SDC = E or for SDC = F. Steven T. Hiner, MS, SE 1-165

18 Part 2 Examle Problems S M1 = F v S 1 IBC (16-38) = 1.3 (0.52) = 0.68 S D1 = 2/3 S M1 IBC (16-40) = 2/3 (0.68) = 0.45i NOTE: Alternatively, S DS & S D1 can be quickly determined using Tables 3.2 & 3.3 ( & 34): Table 3.2 (. 1-33) S S = 1.25 Site Class C S DS = 0.83 Table 3.3 (. 1-34) S 1 = 0.52 Site Class C S D1 = 0.45 B.) Seismic Design Category, SDC S 1 = 0.52 < 0.75 therefore, use IBC Table (1) & Table (2) to determine SDC S DS = 0.83 & RC = II 2012 IBC Table (1) SDC = D S D1 = 0.45 & RC = II 2012 IBC Table (2) SDC = D isdc = Dii C.) Aroximate Fundamental Period, T a T C h ASCE 7 (12.8-7) a t x n Steel SCBF ASCE 7 Table : all other structural systems C t = 0.02 & x = 0.75 T a = 0.02 (h n ) 0.75 = 0.02 (36 feet) 0.75 = 0.29 secondi Or using Table C1 (Aendix C,. 5-18) CBF & h n = 36 feet T a = 0.29 second NOTE: T = 0.29 second < T S S S D 1 DS = 0.54 second ASCE (12.8-2) will govern C s D.) Seismic Resonse Coefficient, C S SDS CS = ( R I e ) (0.83) (6 1.0) C S need not exceed the following, SD1 CS = T ( R I ) e (0.45) 0.29(6 1.0) = governs ASCE 7 (12.8-2) = ASCE 7 (12.8-3) NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not alicable since T < T S (and T << T L ) C S = 0.138i E.) Seismic Base Shear, V V = C S W ASCE 7 (12.8-1) = (280 kis) = 38.6 kisi Steven T. Hiner, MS, SE 2-7

19 Part 2 Examle Problems S D1 = 2/3 S M1 IBC (16-40) = 2/3 (0.28) = 0.19i Table 3.2 (. 1-33) S S = 0.73 Site Class B S DS = 0.49 by interolation Table 3.3 (. 1-34) S 1 = 0.28 Site Class B S D1 = 0.19 B.) Seismic Design Category, SDC S 1 = 0.28 < 0.75 therefore, use IBC Table (1) & Table (2) to determine SDC S DS = 0.49 & RC = IV 2012 IBC Table (1) SDC = D S D1 = 0.19 & RC = IV 2012 IBC Table (2) SDC = D isdc = Dii C.) Aroximate Fundamental Period, T a T C h ASCE 7 (12.8-7) a t x n Concrete SMF ASCE 7 Table : Concrete moment-resisting frames C t = & x = 0.9 T a = (h n ) 0.9 = (67 feet) 0.9 = 0.70 secondi Using Table C1 (Aendix C,. 5-18) Concrete MRF & h n = 67 feet T a = 0.70 second NOTE: T = 0.70 second > T S S S D 1 DS = 0.39 second ASCE 7 (12.8-2) will not govern C s D.) Seismic Resonse Coefficient, C S SDS CS = ( R I e ) e (0.49) (8 1.5) = ASCE 7 (12.8-2) C S need not exceed the following, SD1 (0.19) CS = = T ( R I ) 0.70(8 1.5) governs ASCE 7 (12.8-3) C S shall not be less than the following, C S I = 0.044(0.49)(1.5) = ASCE 7 (12.8-5) S DS e 0.01 minimum NOTE: ASCE 7 (12.8-4) is not alicable since T = 0.70 second << T L = 12 seconds & ASCE 7 (12.8-6) is not alicable since S 1 = 0.28 < 0.6 C S = 0.051i E.) Seismic Base Shear, V V = C s W ASCE 7 (12.8-1) = (5,000 kis) = 255 kisi Steven T. Hiner, MS, SE 2-9

20 Part 2 Examle Problems S D1 = 2/3 S M1 IBC (16-40) = 2/3 (1.35) = 0.90i Table 3.2 (. 1-33) S S = 1.75 Site Class D (assumed) S DS = 1.17 Table 3.3 (. 1-34) S 1 = 0.90 Site Class D (assumed) S D1 = 0.90 B.) Seismic Design Category, SDC S 1 = 0.90 > 0.75 & RC = II 2012 IBC SDC = E isdc = Ei C.) Aroximate Fundamental Period, T a T C h ASCE 7 (12.8-7) a t x n WSP shear wall ASCE 7 Table : all other structural systems C t = 0.02 & x = 0.75 T a = 0.02 (h n ) 0.75 = 0.02 (25 feet) 0.75 = 0.22 secondi Using Table C1 (Aendix C,. 5-18) Shear Walls & h n = 25 feet T a = 0.22 second NOTE: T = 0.22 second < S D T S = 0.77 second ASCE 7 (12.8-2) will govern C s. S 1.17 DS D.) Seismic Resonse Coefficient, C S SDS CS = ( R I e ) (1.17) ( ) C S need not exceed the following, SD1 CS = T ( R I ) e (0.90) 0.22( ) = governs ASCE 7 (12.8-2) = ASCE 7 (12.8-3) NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not alicable since T < T S (and T << T L ) C S = 0.180i E.) Seismic Base Shear, V V = C S W ASCE 7 (12.8-1) = (135 kis) = 24.3 kisi NOTE: For low-rise buildings (i.e., 3 stories), ASCE 7 (12.8-2) will tyically govern since these structures have relatively short eriods such that T < T S therefore, it is tyically not necessary to calculate the structures eriod from equation ASCE 7 (12.8-7), or seismic resonse coefficient from ASCE 7 (12.8-3), (12.8-4), (12.8-5) & (12.8-6) for low-rise buildings. Steven T. Hiner, MS, SE 2-11

21 Part 2 Examle Problems Table 3.2 (. 1-33) S S = 1.52 Site Class E S DS = 0.91 by interolation Table 3.3 (. 1-34) S 1 = 0.76 Site Class E S D1 = 1.22 by interolation B.) Seismic Design Category, SDC S 1 = 0.76 > 0.75 & RC = IV 2012 IBC SDC = F isdc = Fi C.) Aroximate Fundamental Period, T a T C h ASCE 7 (12.8-7) a t x n Dual System D.1 - steel EBF & steel SMF ASCE 7 Table : C t = 0.03 & x = 0.75 T a = 0.03 (h n ) 0.75 = 0.03 (108 feet) 0.75 = 1.01 secondi Using Table C1 (Aendix C,. 5-18) Dual System D.1 & h n = 108 feet T a = 1.01 second NOTE: T = 1.00 second < T S S S D = 1.34 second ASCE 7 (12.8-2) will govern C s 0.91 DS D.) Seismic Resonse Coefficient, C S SDS CS = ( R I e ) (0.91) (8 1.5) C S need not exceed the following, SD1 CS = T ( R I ) e (1.22) 1.00(8 1.5) = governs ASCE 7 (12.8-2) = ASCE 7 (12.8-3) NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not alicable since T < T S (and T << T L ) C S = 0.170i E.) Seismic Base Shear, V V = C S W ASCE 7 (12.8-1) = (24,000 kis) = 4,080 kisi NOTE: This scenario is somewhat unusual in that ASCE 7 (12.8-2) ended u as the governing equation for the seismic resonse coefficient (C S ) even though the building was an 8-story structure simly because T = T a < T S. The reason that ASCE 7 (12.8-3) did not end u governing is because the structure is suorted by soft soil (Site Class E). The resence of soft soil will ush T S toward the longer eriod range of the design resonse sectrum to where taller/longer eriod structures will still be governed by ASCE 7 (12.8-2). Steven T. Hiner, MS, SE 2-13

22 Part 2 Examle Problems B.) N-S DIRECTION: Diahragm san = L 1 = L 2 = L / 2 = 62.5, d = Maximum Unit (Roof) Diahragm Shear, r The addition of the interior shear wall on line 2 will create two diahragms that san equal amounts (e.g., L 1 = L 2 = L / 2). One diahragm sans from line 1 to line 2, and the other diahragm sans from line 2 to line 3. w s = V / L = (35,000 lbs) / 125 = 280 lf V 1 = V 3 = w s (L 1 / 2) = w s (L 2 / 2) = (280 lf )(62.5 ) / 2 = 8,750 lbs roof 1 = 3 = V 1 / d = (8,750 lbs) / 50 = 175 lfi (SD force level) From Part A.1 - roof 1 = 3 = 350 lf 50% reduction in max. unit (roof) diahragm sheari 2. Maximum Unit Wall Shear, w By insection, the maximum unit wall shear will occur on line 2 & 3 - Wall Line 1: V 1 = V / 4 & total shear wall length, Σb 1 = 30 wall 1 = V 1 / Σw = (8,750 lbs) / 30 = 291 lfi Wall Line 2: V 2 = V / 2 & total shear wall length, Σb 2 = 50 wall 1 = V 1 / Σw = (17,500 lbs) / 50 = 350 lfi governs (SD force level) (SD force level) Wall Line 3: V 3 = V / 4 & total shear wall length, Σb 3 = 25 wall 3 = V 3 / Σw = (8,750 lbs) / 25 = 350 lfi governs From Part A.2 - wall 3 = 700 lf 50% reduction in max. unit wall sheari (SD force level) 3. Maximum Chord Force on lines A & B, CF max. CF = w s (L 1 ) 2 / 8d = (280 lf )(62.5 ) 2 / 8(50 ) = 2,730 lbsi (SD force level) From Part A.3 - max. chord force CF = 10,940 lbs 4. Maximum Drag Force, F d 75% reduction in maximum chord forcei Drag Force - Line 1 Drag Force - Line 2 Drag Force - Line 3 Steven T. Hiner, MS, SE 2-33

23 Part 2 Examle Problems 4. Drag Force Diagram on lines A & B, F d roof A = B = 494 lf Wall Line A: F d = 0 lbs Wall Line B: F d = (494 lf)(20 ) = 9,880 lbs (SD/LRFD force level) Drag Force Line A Drag Force Line B B.) E-W DIRECTION: L = 40, d = Design Seismic Force to Diahragm, w s = f 1 = F 1 /L roof DL + 20% snow exterior walls w s = f 1 = [(16 sf + 20% 100 sf)(70 ) + (85 sf)(14 /2 + 2 )(2 walls)] = 770 lfi 2. Unit Roof Shear on lines 1 & 2, r V 1 = V 2 = w s L / 2 = (770 lf)(40 /2) = 15,400 lbs Roof 1 = 2 = V 1 / d = (15,400 lbs) / 70 = 220 lfi 3. Maximum Chord Force on lines A & B, CF max. M = w s L 2 / 8 = (770 lf)(40 ) 2 / 8 = 154,000 lb-ft max. CF = (154,000 lb-ft) / 70 = 2,200 lbsi (SD/LRFD force level) (SD/LRFD force level) 4. Shear Force to walls 1A & 1B Relative Rigidities: assume cantilever walls, Table D1 - Relative Rigidity of Cantilever Shear Walls / Piers (Aendix D,. 5-20) Wall 1A: H/D = 14 /11 = 1.27 Table D1 (. 5-20) R 1A = Wall 1B: H/D = 14 /22 = 0.64 Table D1 (. 5-20) R 1B = R = R 1A + R 1B = = Steven T. Hiner, MS, SE 2-37

24 Part 2 Examle Problems SDR Workbook 2012 IBC Version B.) Center of Rigidity, CR Shear Wall Rigidities: (assume cantilever walls, Table D1 - Relative Rigidity of Cantilever Shear Walls / Piers, Aendix D,. 5-20) Wall A : H/D = 15 /30 = 0.50 Table D1 (. 5-20) R A = 5.0 Wall B : H/D = 15 /20 = 0.75 Table D1 (. 5-20) R B = 2.54 RY R A RB = = 7.54 Wall C : H/D = 15 /40 = Table D1 (. 5-20) R C = 7.49 Wall D : R D = R B = 2.54 Wall E : R E = R D = 2.54 R X RC RD RE = = RY x X CR = [5.0 (0 ) (80 )] / (7.54) = 26.9 feeti R Y RX y Y CR = [7.49 (40 ) (10 ) (0 )] / (12.57) = 25.9 feeti R X 2-54 Steven T. Hiner, MS, SE

25 Part 3 Multile Choice Problems 4.19 What is the aroximate fundamental eriod (T a ) of a 110 foot tall steel eccentrically braced frame (steel EBF) building? a second b second c second d second 4.20 What is the aroximate fundamental eriod (T a ) of a 5 story building using entirely intermediate steel moment frames (steel IMF), with all story heights greater than 10 feet? a second b second c second d second 4.21 What is the aroximate fundamental eriod (T a ) of a 195 foot tall secial steel moment frame (steel SMF) building? a second b second c second d second 4.22 What is the aroximate fundamental eriod (T a ) of a 35 foot tall Dual System building (w/ steel SMF s & reinforced concrete shear walls)? a second b second c second d second 4.23 Given T S = 0.6 second & Seismic Design Category D (SDC = D), which of the following structures would not be ermitted the use of the Equivalent Lateral Force (ELF) rocedure? a. A 200 foot tall regular structure with T = 2.4 seconds b. A 3-story light-frame irregular structure of Risk Category II c. A 150 feet tall structure with T = 2.0 seconds and reentrant corner irregularity d. All of the above e. Both a & c An owner rooses to construct an office building of Seismic Design Category D using steel secial concentrically braced frames (SCBF) as the vertical seismic force-resisting elements (in both rincial directions). Answer questions 4.24 to What resonse modification coefficient (R) should be used for seismic design? a. 8 b. 7 c. 6 d. 3¼ Steven T. Hiner, MS, SE 3-19

26 Part 3 Multile Choice Problems SDR Workbook 2012 IBC Version 5.3 A lateral analysis of a 2-story office building determines that a steel braced frame column has the following axial loads: D = 20 kis, L = 15 kis, L r = 0 kis and E = 25 kis. Assume = 1.0 to determine the maximum axial comression force in this column using the Strength Design (SD or LRFD) load combinations of IBC a kis b kis c kis d kis 5.4 In ASCE , the symbol Q E reresents which of the following? a. The effects of the horizontal seismic forces from the seismic base shear (V ) b. The effects of the horizontal seismic forces from the nonstructural comonent seismic design force (F ) c. The seismic design base shear (V ) d. Both a & b 5.5 Which of the following conditions would require the use of the load combinations with overstrength factor (Ω 0 ) of ASCE ? a. I b. I & III c. II & III d. I, II & III I. Vertical structural irregularity tye 4 (ASCE 7 Table ) II. Vertical structural irregularity tye 5a (ASCE 7 Table ) III. Horizontal structural irregularity tye 4 (ASCE 7 Table ) 5.6 Use of a redundancy factor ( ρ) greater than 1.0 is intended to: a. reduce the inelastic resonse and ductility demand of a structure b. increase the seismic base shear (V) on a structure c. decrease the calculated story drift within a structure d. Both a & b A column of a steel secial concentrically braced frame (SCBF), of a single-story Medical Office building (SDC = D), is determined to suort the following axial load effects: dead load - D = 35 kis, floor live load - L = 0 kis, roof live load - L r = 15 kis, and horizontal seismic load effect - Q E = 15 kis. Given S DS = 1.25, overstrength factor Ω 0 = 2, and redundancy factor = 1.3, answer questions 5.7 through What is the vertical seismic load effect (E v ) axial force in this column? a. 0 kis b. ± 7.0 kis c. ± 8.8 kis d. ± 11.7 kis 5.8 What is the horizontal seismic load effect (E h ) axial force in this column? a. ± 15.0 kis b. ± 19.5 kis c. ± 24.4 kis d. ± 11.7 kis 3-34 Steven T. Hiner, MS, SE

27 Part 3 Multile Choice Problems 5.9 What is the maximum axial comression force in this column using the Strength Design (SD or LRFD) load combinations of IBC ? a kis b kis c kis d kis 5.10 What is the minimum axial comression force in this column using the Strength Design (SD or LRFD) load combinations of IBC ? a. 3.2 kis b. 7.3 kis c. 9.1 kis d kis 5.11 What is the horizontal seismic load effect with overstrength factor (E mh ) axial force in this column? a. ± 15.0 kis b. ± 20.0 kis c. ± 30.0 kis d. ± 37.5 kis 5.12 What is the maximum axial comression force in this column using the strength design load combinations with overstrength factor of ASCE ? a kis b kis c kis d kis 5.13 What is the minimum axial comression force in this column using the strength design load combinations with overstrength factor of ASCE ? a kis b. 5.0 kis c. 1.5 kis d. 7.2 kis 5.14 What is the maximum redundancy factor ( ρ) that needs to be considered for a structure assigned to Seismic Design Category E (SDC = E)? a. 1.0 b c. 1.3 d. 1.5 Steven T. Hiner, MS, SE 3-35

28 Part 3 Multile Choice Problems SDR Workbook 2012 IBC Version 6.3 A 2,000 ound sign is to be attached to a three-story medical office building. The sign is to be attached to the exterior 8 thick concrete wall, 5 feet above the to of the 3 rd floor (i.e., Level 2) as shown in the figure below. Given an S DS = 0.93, what would be the horizontal seismic force for anchorage? a. 500 lbs b. 1,000 lbs c. 1,700 lbs d. 2,400 lbs Section 6.4 Using the figure in Problem 6.3 (above), what would be the horizontal seismic force for anchorage if the sign was to be anchored at the to of the araet (i.e., 4 feet above the roof level)? a. 500 lbs b. 1,900 lbs c. 2,100 lbs d. 2,400 lbs 6.5 What comonent resonse modification factor (R ) should be used for the seismic bracing of a susended ceiling system in a building? a. 1 b. 1½ c. 2½ d A heat exchanger attached to a building structure is considered a/an: a. Element of a structure b. Architectural comonent c. Mechanical comonent d. Electrical comonent 6.7 A chimney of a single-family wood framed house is considered a/an: a. Element of a structure b. Architectural comonent c. Mechanical comonent d. Electrical comonent 3-38 Steven T. Hiner, MS, SE

29 Part 3 Multile Choice Problems 6.8 Given a 2000 lbs storage tank mounted on the roof of a Risk Category II building where S DS = 0.73, a = 1 & R = 2½ what is the seismic design force, F? a. 700 lbs b lbs c lbs d lbs 6.9 Below is a cross section through a large ie suorted from the concrete roof slab of a Hosital where S DS = 0.95, I = 1.5, a = 2½ & R = 9. The ie and its contents weigh 100 lf and have lateral bracing at a sacing of 8 feet (i.e., 8'-0" o.c.). Calculate the axial force in the brace due to the seismic design force, F? a. 50 lbs b. 80 lbs c. 380 lbs d. 540 lbs Section 6.10 What comonent amlification factor (a ) should be used to design the required steel reinforcement size and sacing in a 4 foot tall concrete araet, braced 3 feet above the roof level? a. 1 b. 1¼ c. 1½ d. 2½ 6.11 Given a mechanical comonent with an oerating weight (W ) of 1500 lbs and comonent stiffness (K ) of 3000 lbs/inch, what is the comonent eriod (T )? a second b second c second d second 6.12 For masonry or concrete wall buildings, anchor stras (i.e., urlin anchors) at the boundaries of floor and roof diahragms are utilized to: a. transfer gravity loads to the wall b. transfer diahragm shear to the shear walls c. reduce shrinkage of wood joists d. secure the walls to the diahragm Steven T. Hiner, MS, SE 3-39

30 Part 4 Multile Choice Solutions SDR Workbook 2012 IBC Version Problem Answer Reference / Solution 0.9( D F) 1.0E 1. 6H IBC (16-7) minimum axial, P = 0.9(35 kis + 0) + 1.0( 28.3 kis) + 1.6(0) = 3.2 kis 5.11 c. 1-74, Horizontal Seismic Load Effect w/ Overstrength Factor & ASCE , E mh = ± Ω 0 Q E ASCE 7 (12.4-7) E mh = ± 2 (15 kis) = ± 30.0 kis 5.12 d. 1-76, Basic (SD or LRFD) Load Combinations w/ Overstrength Factor & ASCE , S DS = 0.2(1.25) = 0.25 Ω 0 Q E = kis use + to determine maximum comression. The following strength design load combination including overstrength factor (Ω 0 ) will govern for maximum axial comression 5. ( SDS ) D 0QE L 0. 2S ( )(35 kis) kis (0) = 80.8 kis 5.13 d. 1-76, Basic (SD or LRFD) Load Combinations w/ Overstrength Factor & ASCE , S DS = 0.2(1.25) = 0.25 Ω 0 Q E = 30.0 kis use to determine minimum comression. The following strength design load combination including overstrength factor (Ω 0 ) will govern for mimimum axial comression 7. ( SDS ) D 0QE ( )(35 kis) + ( 30.0 kis) = 7.2 kis 5.14 c to 73, Redundancy Factor & ASCE , For structures assigned to SDC = D E or F = 1.3 shall be used unless one of the following two conditions is met, where = 1.0 is ermitted maximum ρ = d. 1-59, Overturning Moment OTM base = 6 kis (12) + 12 kis (24) = 360 ki-ft T A = C B = (360 ki-ft) / (16 ) = 22.5 kis T A = 22 kis 5.16 b By observation, the lateral forces shown in the given direction will result in the frame deflecting to the right which will stretch the braces X1 & X2, and comress the braces Y1 & Y2. Therefore braces X1 & X2 will be in tension and braces Y1 & Y2 will be in comression. Because the rod braces cannot resist comression, only braces X1 & X2 will be effective to resist the forces in the given direction. X1 & X Steven T. Hiner, MS, SE

31 Part 4 Multile Choice Solutions Problem Answer Reference / Solution 5.17 c X2 is the brace is the 2 nd story and it will carry the entire 12 ki force as the horizontal comonent in that brace. The resultant axial (tension) force can be determined from Trigonometry. With a 12 story height and 16 bay sacing, the brace length will be 20 (i.e., triangle or ). T X2 = (12 kis)(20) / (16) = 15 kis 5.18 c. 1-72, Horizontal Seismic Load Effect & ASCE , Q E = effects of horizontal seismic forces from V or F The axial force in brace X2 is due to F 2 (but not equal to F 2 ), and the F x forces are due to the base shear V, therefore the axial force in brace X2 is from V (but certainly not equal to V ). Q E 5.19 d X1 is the brace is the 1 st story and it will carry the entire (12 ki + 6 ki) force as the horizontal comonent in that brace. The resultant axial (tension) force can be determined from Trigonometry. With a 12 story height and 16 bay sacing, the brace length will be 20 (i.e., triangle or ). T X1 = (18 kis)(20) / (16) = 22.5 kis 22 kis 5.20 c Per 5.19, brace X1 will carry the entire 18 kis of horizontal seismic force in the given direction, and that entire force will be transferred to suort A. Shear at A, V A = 18 kis 5.21 a Per 5.19, brace Y1will carry none of the horizontal seismic force in the given direction because brace Y1 cannot resist comression forces. Shear at B, V B = 0 kis 5.22 a. 1-72, Vertical Seismic Load Effect & ASCE , E v = ± 0.2S DS D ASCE 7 (12.4-4) E v = ± 0.2(0.72)(110 kis) = ± 15.8 kis ± 16 kis 5.23 d. 1-72, Horizontal Seismic Load Effect & ASCE , Per 5.18 and 5.19, the axial force in brace X1 (T X1 ) is a Q E force and is equal to 22.5 kis. Q E = T X1 = 22.5 kis ρ = 1.3 (given) E h = ρ Q E ASCE 7 (12.4-3) E h = 1.3(22.5 kis) = 29.2 kis 29 kis 6.1 d & ASCE , Table Architectural Comonent Cantilever Elements (Unbraced ) - araets a = 2½ 6.2 b ASCE & 93, Tables and Footnote c overtsrength as required for anchorage to concrete 6.3 c. 1-82, Seismic Design Force & ASCE & 89, (continued) Steven T. Hiner, MS, SE 4-29