Chapter 4 Seismic Design Requirements for Building Structures

Chapter 4 Seismic Design Requirements for Building Structures where: F a = 1.0 for rock sites which may be assumed if there is 10 feet of soil between the rock surface and the bottom of spread footings or mat foundation = 1.4 for soil sites S S = the mapped MCE R short-period spectral response acceleration per ASCE 7 11.4.1 but S S need not exceed 1.5 Seismic Base Shear, V Strength Design force leveli ASCE 7 12.14.8.1 The seismic base shear in a given direction shall be determined in accordance with the following: FSDS V W ASCE 7 (12.14-12) R where: F = 1.0 for one-story buildings (above grade plane) = 1.1 for two-story buildings (above grade plane) = 1.2 for three-story buildings (above grade plane) R = Response modification factor per ASCE 7 Table 12.14-1 (not Table 12.2-1) W = effective seismic weight per ASCE 7 12.14.8.1 items 1 to 5 One-Story Building - 1.0S V R DS W Two-Story Building - 1.1S V R DS W Three-Story Building - 1.2S V R DS W NOTE: When using the Simplified Design Procedure of ASCE 7 12.14, it is not necessary to determine the Redundancy Factor ( ) when calculating the earthquake forces on the elements of the structure (e.g., shear walls, braces, columns, etc.) since is not used to determine E h : i.e., E h = Q E ASCE 7 (12.14-5) Vertical Distribution, F x ASCE 7 12.14.8.2 The force at each level shall be calculated using the following equation: wx Fx V ASCE 7 (12.14-13) W or FSDS Fx wx R Steven T. Hiner, MS, SE 1-67

Chapter 9 IBC Chapter 23 Wood Table 9.2 Blocked WSP Diaphragms (Ref. 13 Table 4.2A) courtesy, American Wood Council, Leesburg, VA Steven T. Hiner, MS, SE 1-131

Part 2 Example Problems Solution: A.) N-S DIRECTION: L = 125, d = 50 1. Maximum Unit (Roof) Diaphragm Shear, r w s = V / L = (35,000 lbs) / 125 = 280 plf V 1 = V 3 = w s L / 2 = (280 plf )(125 ) / 2 = 17,500 lbs roof 1 = 3 = V 1 / d = (17,500 lbs) / 50 = 350 plfi (SD force level) 2. Maximum Unit Wall Shear, w By inspection, the maximum unit wall shear will occur on line 3 (i.e., V 1 = V 3, Σb 3 = 25 < Σb 1 = 30 ) Wall Line 1: V 1 = V / 2 & total shear wall width (i.e., length) - Σb 1 = 30 wall 1 = V 1 / Σb 1 = (17,500 lbs) / 30 = 583 plfi Wall Line 3: V 3 = V / 2 & total shear wall width (i.e., length) - Σb 3 = 25 wall 3 = V 3 / Σb 3 = (17,500 lbs) / 25 = 700 plfi governs (SD force level) (SD force level) 3. Maximum Chord Force on lines A & B, CF max. CF = w s L 2 / 8d = (280 plf )(125 ) 2 / 8(50 ) = 10,940 lbsi (SD force level) 4. Maximum Drag Force, F d Drag Force - Line 1 Drag Force - Line 3 By inspection, the maximum drag force will occur on line 3 - Wall Line 1: roof 1 = 350 plf F d = 1 (10 ) = (350 plf )(10 ) = 3,500 lbs (SD force level) Wall Line 3: roof 3 = 350 plf F d = 3 (25 ) = (350 plf )(25 ) = 8,750 lbs governs (SD force level) 2-32 Steven T. Hiner, MS, SE

Part 2 Example Problems B.) N-S DIRECTION: Diaphragm span = L 1 = L 2 = L / 2 = 62.5, d = 50 1. Maximum Unit (Roof) Diaphragm Shear, r The addition of the interior shear wall on line 2 will create two diaphragms that span equal amounts (e.g., L 1 = L 2 = L / 2). One diaphragm spans from line 1 to line 2, and the other diaphragm spans from line 2 to line 3. w s = V / L = (35,000 lbs) / 125 = 280 plf V 1 = V 3 = w s (L 1 / 2) = w s (L 2 / 2) = (280 plf )(62.5 ) / 2 = 8,750 lbs roof 1 = 3 = V 1 / d = (8,750 lbs) / 50 = 175 plfi (SD force level) From Part A.1 - roof 1 = 3 = 350 plf 50% reduction in max. unit (roof) diaphragm sheari 2. Maximum Unit Wall Shear, w By inspection, 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 / Σb 1 = (8,750 lbs) / 30 = 291 plfi Wall Line 2: V 2 = V / 2 & total shear wall length, Σb 2 = 50 wall 2 = V 2 / Σb 2 = (17,500 lbs) / 50 = 350 plfi 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 / Σb 3 = (8,750 lbs) / 25 = 350 plfi governs From Part A.2 - wall 3 = 700 plf 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 plf )(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

Part 2 Example Problems Problem #16 Given: Single-story Office building S DS = 0.95 Seismic Design Category D Building Frame System o Special reinforced masonry shear walls o Flexible (metal deck) roof diaphragm Roof dead load, D = 16 psf Roof snow load, S = 100 psf 8 masonry wall dead load, W w = 85 psf Plan Wall Section Find: A.) N-S DIRECTION: 1. Design seismic force to diaphragm, w s 2. Unit roof shear on lines A & B, ν r 3. Maximum chord force on lines 1 & 2, CF 4. Drag force diagram on lines A & B, F d B.) E-W DIRECTION: 1. Design seismic force to diaphragm, w s 2. Unit roof shear on lines 1 & 2, ν r 3. Maximum chord force on lines A & B, CF 4. Shear Force to walls 1A & 1B 5. Drag force diagram on lines 1 & 2, F d NOTE: Neglect the wall openings (i.e., dashed lines) shown in the Plan when determining the weight of the masonry walls tributary to the roof diaphragm. Therefore, assume the entire perimeter is surrounded by solid walls, as shown on the Wall Section, with no openings. Steven T. Hiner, MS, SE 2-35

Part 2 Example Problems Solution: Office building = Risk Category II 2012 IBC Table 1604.5 S DS = 0.95 given I e = 1.0 R = 5½ ASCE 7 Table 11.5-1: Risk Category II ASCE 7 Table 12.2-1, item B.16: Building Frame System special reinforced masonry shear walls NOTE: by observation, for a single story building T < T S ASCE 7 (12.8-2) will govern C S Seismic Response Coefficient, C S SDS C S = = ( R Ie) (0.95) (5.5 1.0) Seismic Base Shear, V = 0.173 ASCE 7 (12.8-2) V = C S W ASCE 7 (12.8-1) = 0.173 W use for shear wall design Diaphragm Design Force at Roof, F px F px n Fi i= x = w n px ASCE 7 (12.10-1) w i= x i For a single-story building (i.e., F 1 = V): CSW F p1 = wp1= C S w p1 = 0.173 w p1 W F p1 0.4 S DS I e w p1 = 0.4 (0.95)(1.0) w p1 = 0.380 w p1 maximum F p1 0.2 S DS I e w p1 = 0.2 (0.95)(1.0) w p1 = 0.190 w p1 minimum use for roof diaphragm design A.) N-S DIRECTION: L = 70, d = 40 1. Design Seismic Force to Diaphragm, w s = f p1 = F p1 /L roof DL + 20% snow exterior walls w s = f p1 = 0.190 [(16 psf + 20% 100 psf)(40 ) + (85 psf)(14 /2 + 2 )(2 walls)] = 564 plfi 2. Unit Roof Shear on lines A & B, ν r V A = V B = w s L / 2 = (564 plf)(70 /2) = 19,740 lbs Unit roof shear, ν A = ν B = V A / d = (19,740 lbs) / 40 = 494 plfi 3. Maximum Chord Force on lines 1 & 2, CF max. M = w s L 2 / 8 = (564 plf)(70 ) 2 / 8 = 345,450 lb-ft max. CF = (345,450 lb-ft) / 40 = 8,636 lbsi (SD/LRFD force level) (SD/LRFD force level) 2-36 Steven T. Hiner, MS, SE

Part 2 Example Problems 4. Drag Force Diagram on lines A & B, F d roof ν A = ν B = 494 plf Wall Line A: F d = 0 lbs Wall Line B: F d = (494 plf)(20 ) = 9,880 lbs (SD/LRFD force level) Drag Force Line A Drag Force Line B NOTE: The actual design of the collector elements and their connections would need to consider the overstrength factor as required by ASCE 7-12.10.2.1 for this structure assigned to SDC = D (i.e., design for Ω 0 Q E where Q E is the drag force determined above see p. 1-115). B.) E-W DIRECTION: L = 40, d = 70 1. Design Seismic Force to Diaphragm, w s = f p1 = F p1 /L roof DL + 20% snow exterior walls w s = f p1 = 0.190 [(16 psf + 20% 100 psf)(70 ) + (85 psf)(14 /2 + 2 )(2 walls)] = 770 plfi 2. Unit Roof Shear on lines 1 & 2, ν r V 1 = V 2 = w s L / 2 = (770 plf)(40 /2) = 15,400 lbs Roof ν 1 = ν 2 = V 1 / d = (15,400 lbs) / 70 = 220 plfi 3. Maximum Chord Force on lines A & B, CF max. M = w s L 2 / 8 = (770 plf)(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 (Appendix D, p. 5-20) Wall 1A: H/D = 14 /11 = 1.27 Table D1 (p. 5-20) R 1A = 0.833 Wall 1B: H/D = 14 /22 = 0.64 Table D1 (p. 5-20) R 1B = 3.369 ΣR = R 1A + R 1B = 0.833 + 3.369 = 4.202 Steven T. Hiner, MS, SE 2-37

Part 2 Example Problems NOTE: For the purpose of determining the Drag Force Diagrams in part 5, the force level based on a C S = 0.190 will continue to be used. BUT, for shear wall design use of C S = 0.173 would be appropriate, which is equal to (0.173 / 0.190) times the force level being used below. Wall 1A: V 1A = V 1 (R 1A / ΣR) = (15,400 lbs)(0.833 / 4.202) = 3,050 lbsi for drag force (SD/LRFD force level) [ = (3,050 lbs) (0.173 / 0.190) = 2,780 lbs for shear wall design ] Wall 1B: V 1B = V 1 (R 1B / ΣR) = (15,400 lbs)(3.369 / 4.202) = 12,350 lbsi for drag force (SD/LRFD force level) [ = (12,350 lbs) (0.173 / 0.190) = 11,240 lbs for shear wall design ] 5. Drag Force Diagram on lines 1 & 2, F d roof ν 1 = ν 2 = 220 plf Wall Line 1: F d = (220 plf)(11 ) 3,050 lbs = 630 lbs = (220 plf)(31 ) 3,050 lbs = 3,770 lbs max = (220 plf)(53 ) (3,050 lbs + 12,350 lbs) = 3,740 lbs = (220 plf)(70 ) (3,050 lbs + 12,350 lbs) = 0 lbs OK Drag Force Line 1 Wall Line 2: F d = (220 plf)(11 ) = 2,420 lbs = (220 plf)(53 ) 15,400 lbs = 3,740 lbs max = (220 plf)(70 ) 15,400 lbs = 0 lbs OK Drag Force Line 2 See NOTE on p. 2-37 regarding use of Ω 0 Q E for design of collector elements and their connections 2-38 Steven T. Hiner, MS, SE

Part 2 Example Problems Problem #21 Given: Reinforced concrete cantilever retaining wall Soil weight, γ = 110 pcf Allowable soil bearing pressure, q s : o 3,000 psf dead + live load o 4,000 psf total load (including seismic or wind) Sliding friction factor, µ = 0.40 Resisting passive pressure: o K P = 3.18 o Resultant height, y P = D 3 Assume that concrete stem wall is included in static plus seismic pressure (P AE ) Static active pressure coefficient: o K A = 0.318 o Resultant height, y A = H 3 Static plus seismic active pressure coefficient: o K AE = 0.538 o Resultant height, y AE = 0. 45H Assume backfill is fully drained (i.e., no hydrostatic pressure). Neglect vertical component of seismic force. Assume static plus seismic active pressure is given at an allowable stress level (ASD) Find: A.) STATIC CONDITION, K A : 1. Sliding factor of safety 2. Overturning factor of safety 3. Maximum soil bearing pressure (at toe) Cantilever Retaining Wall Section B.) STATIC PLUS SEISMIC CONDITION, K AE : 1. Sliding factor of safety 2. Overturning factor of safety 3. Maximum soil bearing pressure (at toe) 2-58 Steven T. Hiner, MS, SE

Part 2 Example Problems B.) STATIC PLUS SEISMIC CONDITION, K AE : Static plus Seismic Active Soil Pressure Total static plus seismic active force, P AE 2 P AE = ½ K AE γ H = ½ (0.538)(110 pcf)(11.25 ) 2 = 3,745 lbs/ft resultant height, y AE = 0.45 H = 0.45 (11.25 ) = 5.06 Total static passive (resisting) force, P P (from part A) P P = 885 lbs/ft resultant height, y P = D 3 = 0.75 Weights (from part A) Resultant weight, R = ΣW = 7,284 lbs/ft 1. Sliding Factor of Safety Sliding force, F S = P AE = 3,745 lbs/ft Resisting force, F R = passive force + friction force = P P + µ ΣW = 885 lbs/ft + 0.4 (7,284 lbs/ft) = 3,799 lbs/ft Sliding factor of safety, FS = F F R S 3,799 = 3,745 = 1.01 < 1.1 NG! NOTE: 2012 IBC 1806.1, and most Geotechnical reports, permit a one-third increase in the allowable bearing pressure where the alternative basic load combinations of IBC 1605.3.2 include wind loads (W ) or earthquake loads (E ). Furthermore, IBC 1807.2.3 exception allows a minimum factor of safety of 1.1 (i.e., FS = 1.5 / 1.33 = 1.1) for retaining wall sliding and overturning where earthquake loads are included. Steven T. Hiner, MS, SE 2-61

Part 2 Example Problems 2. Overturning Factor of Safety Overturning moment, OTM = P y AE AE = (3,745lbs/ft)(0.45)(11.25 ) = 18,960 lb-ft/ft Resisting moment, (from part A.2) RM = 27,550 lb-ft/ft Overturning factor of safety, FS = RM OTM 27,550 = 18,960 = 1.45 > 1.1 OK 3. Maximum Soil Bearing Pressure ( 27,550 18,960) RM OTM x = = = 1.18 R 7,284 eccentricity from centerline of footing, e = L 2 x = (6.5 )/2 1.18 = 2.07 Soil Bearing Pressure e = 2.07 > L/6 = 6.5 /6 = 1.08 soil pressure distribution is triangular for a triangular soil pressure distribution, 2R maximum soil pressure, q s = 3x minimum soil pressure, q s = 0 therefore, max. q s = 2 (7,284) = 4,120 psf/ft > 4,000 psf allowable (D + L + seismic) NG! 3 (1.18' ) 2-62 Steven T. Hiner, MS, SE

Part 3 Multiple Choice Problems 4.19 What is the approximate fundamental period (T a ) of a 110 foot tall steel eccentrically braced frame (steel EBF) building? a. 0.68 second b. 1.02 second c. 1.20 second d. 1.54 second 4.20 What is the approximate fundamental period (T a ) of a 5 story building using entirely intermediate steel moment frames (steel IMF), with all story heights greater than 10 feet? a. 0.50 second b. 0.43 second c. 0.64 second d. 0.89 second 4.21 What is the approximate fundamental period (T a ) of a 195 foot tall special steel moment frame (steel SMF) building? a. 1.04 second b. 1.56 second c. 1.90 second d. 2.12 second 4.22 What is the approximate fundamental period (T a ) of a 35 foot tall Dual System building (w/ steel SMF s & reinforced concrete shear walls)? a. 0.48 second b. 0.39 second c. 0.35 second d. 0.29 second 4.23 Given T S = 0.6 second & Seismic Design Category D (SDC = D), which of the following structures would not be permitted the use of the Equivalent Lateral Force (ELF) procedure? a. 200 foot tall regular structure with T = 2.4 seconds b. 3-story light-frame irregular structure of Risk Category II c. 150 foot tall structure with T = 2.0 seconds and reentrant corner irregularity d. All of the above e. Both a & c An owner proposes to construct an office building of Seismic Design Category D using steel special concentrically braced frames (SCBF) as the vertical seismic force-resisting elements (in both principal directions). Answer questions 4.24 to 4.27. 4.24 What response modification coefficient (R) should be used for seismic design? a. 8 b. 7 c. 6 d. 3¼ Steven T. Hiner, MS, SE 3-19

Part 3 Multiple Choice Problems 6.13 What component amplification factor (a p ) & component response modification factor (R p ) would be appropriate for a boiler attached to a building structure? a. a p = 1 & R p = 2½ b. a p = 2½ & R p = 2½ c. a p = 1 & R p = 6 d. a p = 2½ & R p = 6 6.14 When anchoring structural walls to a roof or floor diaphragm assigned to Seismic Design Category D, and where the purlin anchors are provided at a spacing of 4 feet (i.e., 4'-0" o.c.), the design anchorage force (F p ) shall not be less than: a. 280 lbs per anchor b. 1120 lbs per anchor c. 0.4 S DS K a I e W p d. 0.2 K a I e W p 6.15 Structural walls (e.g., concrete or masonry walls) are required to be designed to resist bending between purlin anchors where the anchor spacing exceeds: a. 4 times the thickness of the wall b. 4 feet c. 6 feet d. None of the above 6.16 Given one-story Police Station with 8 thick concrete exterior walls with weight (W Wall ) of 100 psf & S DS = 0.62. These structural walls shall be designed to span vertically from foundation to roof level to resist a force normal to the surface of the wall (F p ) equal to: a. 10 psf b. 25 psf c. 37 psf d. 62 psf 6.17 Given one-story County Jail with 8 thick masonry exterior walls with weight (W Wall ) of 85 psf, wall height of 14 feet with 3 foot parapet, a concrete (rigid) roof diaphragm and S DS = 1.13. Determine the governing wall anchorage force (F p ) at the rigid roof diaphragm that spans 100 feet between shear walls. a. 210 plf b. 380 plf c. 480 plf d. 580 plf 6.18 Given a three-story Office building assigned to Seismic Design Category D with 8 thick masonry exterior walls with weight (W Wall ) of 85 psf, 12 foot story heights, flexible roof and floor diaphragms that span 80 feet between shear walls, and S DS = 1.37. Determine the governing wall anchorage force (F p ) at the Level 1 (and Level 2) flexible floor diaphragms. a. 210 plf b. 370 plf c. 560 plf d. 1010 plf 3-40 Steven T. Hiner, MS, SE

Part 3 Multiple Choice Problems 9.27 What is the maximum length-width ratio for the blocked wood structural panel (WSP) horizontal diaphragms (second floor and roof)? a. 2:1 b. 2½:1 c. 3:1 d. 4:1 9.28 Given a structure assigned to Seismic Design Category C with a typical subdiaphragm span of 20 feet, what would be the minimum required depth of each structural subdiaphragm? a. 10-0 b. 8-0 c. 6-8 d. 5-0 9.29 What is the minimum seismic design force for structural subdiaphragms that are part of a flexible diaphragm in SDC = C, D, E or F? a. F px L plf b. F x L plf c. 0.2KaIeW p d. 280 plf Given a single-story wood frame Police Station assigned to Seismic Design Category F with wood structural panels used for the flexible roof diaphragm and for the shear walls. The roof diaphragm is to use 19/32 rated sheathing with 10d common nails (3 x 0.148 ) fastened to 2x nominal framing members, with blocking omitted at intermediate joints. The shear walls are to use 15/32 Structural I sheathing with 10d common nails (3 x 0.148 ) fastened to 2x nominal framing members. Answer questions 9.30 through 9.33. 9.30 What is the allowable unit shear for the roof diaphragm with seismic loads perpendicular to the continuous panel joints? a. 285 plf b. 255 plf c. 215 plf d. 190 plf 9.31 What is the allowable unit shear for a shear wall with 4 o.c. edge nailing, a height (h) of 12-0 and a width (b) of 8-6 resisting seismic loads? a. 310 plf b. 380 plf c. 460 plf d. 510 plf Steven T. Hiner, MS, SE 3-63

Part 4 Multiple Choice Solutions Problem Answer Reference / Solution 1.11 c p. 1-5, Earthquake intensity The intensity of an earthquake is based on the damage to structures, the ground surface, and observed effects on people and other features. I & III 1.12 d p. 1-7, Attenuation Attenuation is the decrease in seismic energy received at a site with increasing distance from the earthquake epicenter. Distance to the epicenter 1.13 c p. 1-8, Table 1.3 MMI VI 1.14 d p. 1-4, Earthquake Strength n = M5.0 M3.0 = 2 10 n = (10) 2 = 100 1.15 b p. 1-5, Energy Release n = M6.8 M5.8 = 1 32 n = (32) 1 = 32 1.16 c p. 1-7, Earthquake intensity Large magnitude and long duration is likely to cause the most damage M7.0 / 30 seconds 1.17 f p. 1-10, Natural Period T 2 W / K g function of weight (W) & stiffness (K) both b & d 1.18 c p. 1-12, Damping ratio = B / B critical actual damping coefficient to critical damping coefficient 1.19 d p. 1-12, T d is slightly greater than T & p. 1-14, Figure 1.11 - spectral acceleration (S a ) and therefore base shear (V) decreases with increasing damping ratio ( ). both a & c 1.20 a p. 1-12, Damping ratio typical ratios of = 2% (flexible steel frame) to 15% (light wood-frame) Plywood shear wall (i.e., light wood-frame) 1.21 c p. 1-10, Table 1.4 K 1 = 12EI/h 3 = 12(29,000 ksi)(200 in 4 ) / (14 ft) 3 (12 in/ft) 3 = 14.68 kips/in (continued) 4-2 Steven T. Hiner, MS, SE

Part 4 Multiple Choice Solutions Problem Answer Reference / Solution 4.18 a p. 1-55 & ASCE 7-10 p. 56, 11.4.5 SD 1 0.65 T S 0.58 second S 1.12 DS 4.19 b p. 1-56 & ASCE 7-10 p. 72, 12.8.2.1 x Ta Cthn ASCE 7 (12.8-7) Steel EBF ASCE 7-10 Table 12.8-2 C t = 0.03 & x = 0.75 T a = 0.03 (110 ) 0.75 = 1.02 sec Or using Table C1 (p. 5-16) Steel EBF & h n = 110 T a = 1.02 sec T a = 1.02 second 4.20 a p. 1-56 & ASCE 7-10 p. 72, 12.8.2.1 T a = 0.1 N ASCE 7 (12.8-8) Applicable for moment frames (concrete or steel) 12 stories maximum, and with average story height 10 feet DO NOT use Table C1 (p. 5-16) it only applies to ASCE 7 (12.8-7) T a = 0.1 (5 stories) = 0.50 second 4.21 c p. 1-56 & ASCE 7-10 p. 72, 12.8.2.1 x Ta Cthn ASCE 7 (12.8-7) Steel SMF ASCE 7-10 Table 12.8-2 C t = 0.028 & x = 0.8 T a = 0.028 (195 ) 0.8 = 1.90 sec Or using Table C1 (p. 5-16) Steel MRF & h n = 195 T a = 1.90 sec DO NOT use T a = 0.1 N since the number of stories (or levels) is not given, and it is not known if this steel MRF buidling is 12 stories and whether the average story height 10 feet T a = 1.90 second 4.22 d p. 1-56 & ASCE 7-10 p. 72, 12.8.2.1 x Ta Cthn ASCE 7 (12.8-7) Dual systems (except D.1 - SMF combined w/ steel EBF) ASCE 7-10 Table 12.8-2 C t = 0.02 & x = 0.75 T a = 0.02 (35 ) 0.75 = 0.29 sec Or using Table C1 (p. 5-16) Dual Systems & h n = 35 T a = 0.29 sec T a = 0.29 second 4.23 a p. 1-50 & ASCE 7-10 p. 71, Table 12.6-1 3.5 T S = 3.5 (0.6) = 2.10 seconds SDC = D & T = 2.4 seconds > 3.5 T S Table 12.6-1 ELF procedure is noted as NP (not permitted). All others allow use of the ELF procedure. 200 foot tall regular structure with T = 2.4 seconds 4.24 c ASCE 7-10 p. 60, Table 12.2-1, item B.2 - Steel SCBF R = 6 4-14 Steven T. Hiner, MS, SE

Part 4 Multiple Choice Solutions Problem Answer Reference / Solution Structure B By observation, ASCE 7 (12.8-3) will have to govern for Structure B for V A = V B since T B will have to be much larger than T A to counteract the increase in effective seismic weight between structure A to structure B (i.e., W B = 3 W A ). Therefore, assume ASCE 7 (12.8-3) governs for C S C S-B = S D1 / T B (R / I e ) ASCE 7 (12.8-3) V B = C S-B W B ASCE 7 (12.8-1) Set the base shears equal V A = V B C S-A W A = C S-B W B S DS W A / (R / I e ) = S D1 3W A / T B (R / I e ) W A / (R / I e ) cancel from both sides of the equation S DS = 3 S D1 / T B Solve for T B T B = 3 S D1 / S DS T B = 3 (0.30 ) / 0.73 = 1.23 seconds 1.25 seconds 4.85 b p. 1-63, Structural Separation & ASCE 7-10 p. 77, 12.12.3 Cd max M ASCE 7 (12.12-1) Ie Structure 1: C d = 5½ ASCE 7-10 p. 61, Table 12.2.1- item C.1 for steel SMF At Level 4 - δ M1 = C d δ max / I e = (5½)(2.1 ) / (1.00) = 11.55 At Level 7 (roof) - δ M1 = C d δ max / I e = (5½)(3.5 ) / (1.00) = 19.25 Structure 2: C d = 4 ASCE 7-10 p. 60, Table 12.2.1- item B.16 for special masonry SW At Level 4 (roof) - δ M2 = C d δ max / I e = (4)(1.4 ) / (1.00) = 5.60 NOTE: If Structure 1 and Structure 2 are to collide, that collision will occur at Level 4 for both structures, so that will be the level used to determine the required separation. Adjacent buildings on the same property, structural separation - MT 2 2 ASCE 7 (12.12-2) M 1 M 2 2 2 ( 11.55") (5.60") = 12.80 13 inches 4.86 d p. 1-63 to 64, Structural Separation & ASCE 7-10 p. 77, 12.12.3 From Problem 4.85 - Structure 1: At Level 7 (roof) - δ M1 = C d δ max / I e = (5½)(3.5 ) / (1.00) = 19.25 Where a structure adjoins a property line (not common to a public way), the structure shall be set back from the property line by at least the displacement M of that structure. (continued) 4-24 Steven T. Hiner, MS, SE

Part 4 Multiple Choice Solutions Problem Answer Reference / Solution 9.15 a p. 1-113, Chord Force Maximum chord force (CF) occurs at the point of largest moment and least diaphragm depth. This maximum moment occurs on lines A & B (East-West loading) at the center of the diaphragm span (i.e., maximum CF = w s L 2 /8d where L is maximum & d is minimum). a 9.16 c p. 1-130, Wood Structural Panel Diaphragms V 1 = V 5 = w s L / 2 = (350 plf)(120 /2) = 21,000 lbs For ASD, roof 1 = 5 = (0.7 V 1 ) / d = 0.7 (21,000 lbs) / 40 = 370 plf 9.17 c p. 1-131, Table 9.2 & SDPWS p. 18, Table 4.2A (blocked diaphragms) Load perpendicular to continuous panel joints = CASE 1 (strong direction) 15/32 Structural I w/ 10d @ 6 o.c. blocked Table 9.2 (4.2A) s = 640 plf ASD = s / 2.0 = (640 plf) / 2.0 = 320 plf < 400 plf NG! 15/32 Structural I w/ 10d @ 4 o.c. blocked Table 9.2 (4.2A) s = 850 plf ASD = s / 2.0 = (850 plf) / 2.0 = 425 plf > 400 plf OK STOP no need to check 10d @ 2½ o.c. or 10d @ 2 o.c. Diaphragm boundary nailing = 10d @ 4 o.c. 9.18 c p. 1-114, Drag Force Maximum drag force on line 5 occurs at the left end of the 20 shear wall (i.e., at y ) F 5 = (400 plf)(20 ) = 8,000 lbs = 8 kips 9.19 d p. 1-139, Wood Structural Panel Shear Walls V 5 = w s L / 2 = (350 plf)(120 /2) = 21,000 lbs ( 0.7 Vmax ) 1.0(0.7)(21,000) For ASD, wall 5 b 20' = 740 plf 9.20 a p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A 15/32 Structural I w/ 10d common @ 6 o.c. s = 680 plf ASD = s / 2.0 = (680 plf) / 2.0 = 340 plf << 850 plf NG! 15/32 Structural I w/ 10d common @ 4 o.c. s = 1020 plf ASD = s / 2.0 = (1020 plf) / 2.0 = 510 plf << 850 plf NG! 15/32 Structural I w/ 10d common @ 3 o.c. s = 1330 plf ASD = s / 2.0 = (1330 plf) / 2.0 = 665 plf << 850 plf NG! 15/32 Structural I w/ 10d common @ 2 o.c. s = 1740 plf ASD = s / 2.0 = (1740 plf) / 2.0 = 870 plf > 850 plf OK shear wall edge nailing = 10d common @ 2 o.c. 4-48 Steven T. Hiner, MS, SE