Solutions manual for design of wood structures ASD LRFD 6th edition by breyer fridley cobeen pollock

Solutions Manual for Design of Wood Structures-ASD/LRFD 6th edition by Donald E Breyer, Kenneth J Fridley, Kelly E Cobeen, David G Pollock, Jr Ch 2: Design Loads Chapter Solutions Page of 19 Problem 2.1 a) See Appendix A and Appendix B for weights of roofing, sheathing, framing, insulation, and gypsum wallboard Asphalt shingles 3/8 in plywood sheathing (3/8 in.) (3.0 psf/in) 2x6 @ 16 in o.c Fiberglass loose insulation (5.5 in.) (0.5 psf/in) Gypsum wallboard (1/2 in.) (5.0 psf/in) Roof Dead Load (D) along roof slope Convert D to load on a horizontal plane: = 2.0 psf = 1.1 psf = 1.4 psf = 2.75 psf = 2.5 psf = 9.75 psf Roof slope = 3:12 Hypotenuse = (9 + 144)½ = 12.37 Don horizontal plane = (9.75 psf) (12.37/12) = 10.1 psf [NOTE that this does not include an allowance for weight of re-roofing over existing roof For each additional layer of shingles add 2.0 psf along roof slope, or (2.0 psf)(12.37/12)=2.1 psf on horizontal plane.] b) Wall Dead Load: Stucco (7/8 in.) = 10.0 psf 2x4 @ 16 in o.c = 0.9 psf Gypsum wallboard (½ in.) = 2.5 psf Wall Dead Load (D) = 13.4 psf c) Wall Dead Load: wall height = ft D = (13.4 psf) (8 ft) = 107.2 lb/ft d) R1 = 1.0 since not considering tributary area R2 = 1.0 for slope less than 4:12 Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf e) R1 = 1.0 since not considering tributary area R2 = 1.0 for slope of 4:12 Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf Chapter Solutions Page of 19 Problem 2.2 a) See Appendix A and Appendix B for weights of concrete roof tiles, lumber sheathing, and framing (950 lb / 100 ft ) = 9.5 psf Concrete tile 15/32 in structural panel (plywood) sheathing (15/32) (3.0 psf/in.) = 1.4 psf 2x8 @ 16 in o.c = 1.9psf Roof Dead Load (D) along roof slope = 12.8 psf Convert D to load on a horizontal plane: Roof slope = 6:12 Hypotenuse = (36 + 144)½ = 13.42 Don horizontal plane = (12.8 psf) (13.42/12) = 14.3 psf b) See Appendix A and Appendix B for weights of framing, insulation, gypsum lath and plaster Fiberglass loose insulation 2x6 @ 16 in o.c Gypsum wallboard (10 in.) (0.5 psf/in) (½ in.) (5 psf/in) Ceiling Dead Load (D) c) roof slope = 6:12 (F = 6) R1 = 1.0 since not considering tributary area R2 = 1.2 – 0.05 F = 1.2 – (0.05)(6) = 0.9 Basic Roof Live Load: Lr = 20 R1 R2 = 18 psf = 5.0 psf = 1.4 psf = 2.5psf = 8.9 psf Chapter Solutions Page of 19 Problem 2.3 a) See Appendix A and Appendix B for weights of roofing, sheathing, and suspended ceiling Built-up roof (5 ply w/ gravel) ½ in plywood sheathing (½ in.) (3.0 psf/in) Roof trusses @ 24 in o.c (9 lb/ft)÷(2 ft) Suspended acoustic ceiling: Acoustical fiber tile Suspended acoustic ceiling: Channel-suspended system Roof Dead Load (D) = 6.5 psf = 1.5 psf = 4.5 psf = 1.0 psf = 1.0psf = 14.5 psf b) See Appendix A and Appendix B for weights of framing, sheathing, and suspended ceiling (150 lb/ft ) (0.125 ft) Concrete 2x10 @ 16 in o.c 5/8 in plywood sheathing (5/8 in.) (3.0 psf/in) Air duct Suspended acoustic ceiling: Acoustical fiber tile Suspended acoustic ceiling: Channel-suspended system nd Floor Dead Load (D) c) roof slope = 0.25:12 R1 = 1.0 since not considering tributary area R2 = 1.0 for roof slope less than in 12 Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf = 18.8 psf = 2.4 psf = 1.9 psf = 0.5 psf = 1.0 psf = 1.0psf = 25.6 psf Chapter Solutions Page of 19 Problem 2.4 a) See Appendix A and Appendix B for weights of roofing, sheathing, and subpurlins Assume Douglas-Fir/Larch (G = 0.5) at 12% m.c for 4x14 purlin and 6.75x33 glulam girder Using density formula from NDS Supplement: G m.c density = 62.4 + = 33 lb/ft + G(0.009)(m.c.) 100 [Note that 33 lb/ft is a reasonable (and typically conservative) estimate of unit weight for most softwood species of lumber and glulam A unit weight of 33 lb/ft was used to develop the “Equivalent Uniform Weights of Wood Framing” in Appendix A of the textbook.] To determine self- weight (s.w.) of purlin or girder, converted to distributed load in units of psf: distributed s.w = (density)(cross-sectional area)/(width of tributary area) Glulam girder s.w 4x14 purlin s.w Built-up roof (5 ply w/o gravel) 15/32 in plywood sheathing 2x4 @ 24 in o.c (33)(6.75/12)(33/12)/(20) = 2.55 psf (33)(3.5/12)(13.25/12)/(8) = 1.33 psf = 2.50 psf (15/32 in.) (3.0 psf/in) = 1.41 psf = 0.6 psf Average Dead Load of entire Roof= 8.4 psf (NOTE that this does not include an allowance for weight of re-roofing over existing roof.) b) Subpurlin Dead Load: (2.50 + 1.41 + 0.6 psf) (2 ft) = 9.0 lb/ft c) Purlin Dead Load: (2.50 + 1.41 + 0.6 + 1.33 psf) (8 ft) = 46.7 lb/ft d) Girder Dead Load: (2.50 + 1.41 + 0.6 + 1.33 + 2.56 psf) (20 ft) = 168 lb/ft e) Column Dead Load: (2.50 + 1.41 + 0.6 +1.33 + 2.56 psf) (20 ft) (50 ft) = 8400 lb f) R1 = 1.0 since not considering tributary area R2 = 1.0 for slope less than in 12 Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf g) R = 5.2(ds + dh) = 5.2 (5 in + 0.5 in.) = 28.6 psf Chapter Solutions Page of 19 Problem 2.5 2 a) Tributary area (on a horizontal plane): AT = (20 ft)(13 ft) = 260 ft > 200 ft b) R1 = 1.2 – 0.001 AT = 0.94 R2 = 1.0 for slope less than in 12 Roof Live Load: Lr = 20 R1 R2 = 18.8 psf wLr = (18.8 psf)(13 ft) = 244 lb/ft Problem 2.6 2 a) Tributary area (on a horizontal plane): AT = (22 ft)(13 ft) = 286 ft > 200 ft b) R1 = 1.2 – 0.001 AT = 0.914 R2 = 1.0 for slope less than in 12 Roof Live Load: Lr = 20 R1 R2 = 18.3 psf wLr = (18.3 psf)(13 ft) = 238 lb/ft Problem 2.7 roof slope = 6/12; θ = arctan (6/12) = 26.57° basic ground snow load pg = 70 psf I = 1.0 residential occupancy Ce = 0.9 Exposure C; fully exposed roof heated structure Ct = 1.0 Cs = 1.0 for roof slope < 30° Design snow load: S = (0.7 Ce Ct I pg) Cs = 44.1 psf Problem 2.8 roof slope = 8/12; θ = arctan (8/12) = 33.69° basic ground snow load pg = 90 psf I = 1.0 residential occupancy Exposure B; sheltered roof Ce = 1.2 heated structure Ct = 1.0 Cs = 0.908 for roof slope of 33.69° (linear interpolation between Cs = for θ 70°) Design snow load: S = (0.7 Ce Ct I pg) Cs = 68.6 psf = 30° and Cs = for θ = Chapter Solutions Page of 19 Problem 2.9 2 Subpurlin: AT = (2 ft)(8 ft) = 16 ft < 200 ft R1 = 1.0 R2 = 1.0 (flat roof) Lr = 20 R1 R2 = 20 psf 2 Purlin: AT = (8 ft)(20 ft) = 160 ft < 200 ft R1 = 1.0 R2 = 1.0 (flat roof) Lr = 20 R1 R2 = 20 psf 2 Glulam Beam: AT = (20 ft)(50 ft) = 1000 ft > 600 ft R1 = 0.6 R2 = 1.0 (flat roof) Lr = 20 R1 R2 = 12 psf b) Subpurlin: wLr = (20 psf)(2 ft) = 40 lb/ft Purlin: wLr = (20 psf)(8 ft) = 160 lb/ft Glulam Beam:wLr = (12 psf)(20 ft) = 240 lb/ft Problem 2.10 a) Subpurlin: wS = (25 psf)(2 ft) = 50 lb/ft Purlin: wS = (25 psf)(8 ft) = 200 lb/ft Glulam Beam: wS = (25 psf)(20 ft) = 500 lb/ft b) PS = (25 psf)(20 ft)(50 ft) = 25,000 lb = 25 k Problem 2.11 - See IBC Table 1607.1 (Basic values are noted; additional values may be applicable for specific locations/uses.) Occupancy/Use a) b) c) d) e) f) Offices Light Storage Retail Store Apartments (Residential, Multiple-family) Hotel Restrooms (Residential) School Classrooms Unit Floor Live Load (2 nd Floor) 50 psf 125 psf 75 psf 40 psf (private rooms and corridors serving them) 100 psf (public rooms and corridors serving them) 40 psf (private rooms and corridors serving them) 100 psf (public rooms and corridors serving them) 40 psf Concentrated Live Load 2000 lb -1000 lb -1000 lb Chapter Solutions Page of 19 Problem 2.12 a) L0 = 50 psf office floor b) AT = 240 ft KLL = interior column KLL A T = 2 (4)(240) = 960 ft > 400 ft = 36.7 psf L = L 0.25 + 15 K A LL T c) (35 psf + 36.7 psf)(240 ft ) = 17,200 lb = 17.2 k Chapter Solutions Page of 19 Problem 2.13 D = 20 psf s = 16 ft a) L0 = 40 psf L = 26 ft classroom occupancy b) AT = s L = (16 ft) (26 ft) = 416 ft KLL = interior beam 2 KLL AT = (2)(416) = 832 ft > 400 ft = 30.8 psf L = L 0.25 + 15 K A LL T c) w(D+L) = (20 + 30.8 psf) (16 ft) = 813 lb/ft d) IBC Table 1607.1 concentrated load: PL = 1000 lb wD = (20 psf) (16 ft) = 320 lb/ft Point Load + Distributed Dead Load (PL plus wD): Shear: Vmax = wD L/2 + PL = (320)(26)/2 + 1000 = 5160 lb (for point load placed adjacent to support) 2 Moment: Mmax = wD L /8 + PL L/4 = (320)(26) /8 + (1000)(26)/4 = 33,500 lbft (for point load placed at mid-span) 3 Deflection: ∆L = PL L /48EI = (1000)(26) /48EI = 366,000/EI (for point load placed at mid-span) Distributed Dead Load + Distributed Live Load (w(D+L)): Shear: Vmax = w(D+L)L/2 = (813)(26)/2 = 10,600 lb 2 Moment: Mmax = w(D+L)L /8 = (813)(26) /8 = 68,700 lb-ft Deflection: 4 ∆L = 5wL L /384EI = (5)(493)(26) /384EI = 2,932,000/EI where wL = (30.8 psf)(16 ft) = 493 lb/ft ∴ Uniformly distributed total load (w(D+L)) is critical for shear, moment, and deflection Chapter Solutions Page of 19 Problem 2.14 See IBC Table 1607.1 and Sections 1607.9.1.1 through 1607.9.1.3: Occupancy Unit Floor Live Load (psf) Access floor systems – Computer use 100 psf Assembly Areas & Theaters – Lobbies 100 psf Assembly Areas & Theaters – Movable Seats 100 psf Assembly Areas & Theaters – Stages & Platforms 125 psf Exterior Balconies (except for one- & two-family residences) 100 psf Corridors 100 psf Dance Halls & Ballrooms 100 psf Dining Rooms & Restaurants 100 psf Fire Escapes (except for single-family residences) 100 psf Garages 40 psf Gymnasiums 100 psf Library Stack Rooms 150 psf Manufacturing Facilities (Light) 125 psf Manufacturing Facilities (Heavy) 250 psf Office Buildings – Lobbies & First Floor Corridors 100 psf Penal Institutions – Corridors 100 psf Hotels & Multi-Family Dwellings – Public Rooms & Corridors 100 psf Schools – First Floor Corridors 100 psf Sidewalks, Yards & Driveways subject to vehicular traffic 250 psf Skating Rinks 100 psf Stadiums & Arenas – Bleachers 100 psf Stairs and Exits (except for one- & two-family residences) 100 psf Storage Warehouses (Light) 125 psf Storage Warehouses (Heavy) 250 psf Retail Stores – First Floor 100 psf Wholesale Stores 125 psf Pedestrian Yards & Terraces 100 psf [NOTE: Members supporting live loads for two or more floors may be reduced by up to 20% for some of the occupancy categories listed above See IBC Sections 1607.1.1 through 1607.1.3.] Chapter Solutions Page 10 of 19 Problem 2.15 a) Floor beam; L = 22 ft Allowable Live Load Deflection: L/360 = (22 ft)(12 in/ft)/360 = 0.73 in Allowable Total Load Deflection: L/240 = (22 ft)(12 in/ft)/240 = 1.10 in b) Roof rafter supporting plaster ceiling; L = 12 ft Allowable Live Load Deflection: L/360 = (12 ft)(12 in/ft)/360 = 0.40 in Allowable Total Load Deflection: L/240 = (12 ft)(12 in/ft)/240 = 0.60 in Chapter Solutions Page 11 of 19 Problem 2.16 a) Roof rafter supporting a gypsum board ceiling; L = 16 ft Recommended Live Load Deflection: L/240 = (16 ft)(12 in/ft)/240 = 0.80 in Recommended Total Load Deflection: L/180 = (16 ft)(12 in/ft)/180 = 1.07 in b) Roof girder supporting acoustic suspended ceiling; L = 40 ft Recommended Live Load Deflection: L/240 = (40 ft)(12 in/ft)/240 = 2.00 in Recommended Total Load Deflection: L/180 = (40 ft)(12 in/ft)/180 = 2.67 in c) Floor joist in nd floor residence; L = 20 ft.; s = ft.; D = 16 psf Recommended Live Load Deflection: L/360 = (20 ft)(12 in/ft)/360 = 0.67 in AT = (20 ft)(4 ft) = 80 ft KLL = interior beam 2 KLL AT = (2)(80) = 160 ft < 400 ft (live load reduction is not applicable) L = 40 psf (residential) wL = (40 psf)(4 ft) = 160 lb/ft Recommended Total Load Deflection: L/240 = (20 ft)(12 in/ft)/240 = 1.00 in Assume floor joists spanning 20 ft are seasoned sawn lumber with m.c.< 19%: K = 1.0 KD + L = (1.0)(16 psf) + 40 psf = 56 psf w(KD+L) = (56 psf)(4 ft) = 224 lb/ft nd d) Girder in floor retail store; increased stiffness desired; L = 32 ft; s = 10 ft; D = 20 psf Recommended Live Load Deflection: L/420 = (32 ft)(12 in/ft)/420 = 0.91 in AT = (32 ft)(10 ft) = 320 ft KLL = interior beam 2 KLL AT = (2)(320) = 640 ft > 400 ft (live load reduction is applicable) L0 = 75 psf (retail; L = L 0.25 nd floor) + 15 K A LL = 63.2 psf T wL = (63.2 psf)(10 ft) = 632 lb/ft Recommended Total Load Deflection: L/300 = (32 ft)(12 in/ft)/300 = 1.28 in Assume floor girder spanning 32 ft is seasoned glulam with m.c.< 16%: K = 0.5 KD + L = (0.5)(20 psf) + 63.2 psf = 73.2 psf w(KD+L) = (73.2 psf)(10 ft) = 732 lb/ft Chapter Solutions Page 12 of 19 Problem 2.17 a) ps = λ Kzt I ps30for main wind-force resisting systems pnet = λKzt I pnet 30for components and cladding b) ASCE Section 6.4 defines wind load terms and provisions for the Simplified Procedure (Method 1) c) (1) Use the formula for ps to determine loads for main wind-force resisting systems Main wind-force resisting systems are primary structural systems such as diaphragms and shearwalls Wind force areas are the projected vertical or horizontal surface areas of the overall structure that are tributary to the specified structural system (2) Use the formula for pnet along with tabulated values of pnet 30 for Zone (roofs) or Zone (walls) to determine loads for components and cladding away from discontinuities Components and cladding are individual structural components such as rafters, studs, structural panel sheathing, and nails Wind force areas are the surface areas that are tributary to the specified structural component (3) Use the formula for pnet along with tabulated values of pnet 30 for Zones or (roofs) or Zone (walls) to determine loads for components and cladding near discontinuities Discontinuities include corners of walls, roof ridges, roof eaves, gable ends, and roof overhangs Components and cladding are individual structural components such as rafters, studs, sheathing panels, and nails Wind force areas are the surface areas that are tributary to the specified structural component d) Exposure B includes terrain with buildings, wooded areas, or other obstructions approximately the height of a single-family dwelling (Surface Roughness B) extending at least 2600 ft or 20 times the building height (whichever is greater) from the site Exposure B applies to most urban and suburban areas Exposure B is the least severe wind exposure Exposure C applies where Exposures B and D not apply Exposure D applies to unobstructed flat terrain (including mud flats, salt flats and unbroken ice) (Surface Roughness D) extending a distance of 5000 ft or 20 times the building height (whichever is greater) from the site Exposure D also applies to building sites adjacent to large water surfaces outside hurricane prone regions Exposure D is the most severe wind exposure Chapter Solutions Page 13 of 19 Problem 2.18 a) A mean recurrence interval of 50 years (annual probability of exceedence of 0.02) generally applies to basic wind speeds in ASCE Fig 6-1 b) A mean recurrence interval of 100 years applies to wind pressures for essential and hazardous facilities (I = 1.15) c) The mean roof height above ground (hmean) is used to determine the height and exposure factor (λ) Problem 2.19 Kzt = 1.0 a) V = 120 mphASCE Figure 6-1B for Tampa, FL b) I = 1.15 ASCE Table 6-1 and IBC Table 1604.5 for essential facility (Category IV) c) λ = 1.0ASCE Figure 6-2 for Exposure B and mean roof height of 30 ft d) ASCE Figure 6-2 for flat roof Zone Wall A Wall B Wall C Wall D Roof E Roof F Roof G Roof H Roof Overhang EOH Roof Overhang GOH ps30 (psf) 22.8 - 11.9 15.1 - 7.0 - 27.4 - 15.6 - 19.1 - 12.1 - 38.4 - 30.1 ps = λ Kzt I ps30 (psf) 26.2 - 13.7 17.4 - 8.05 - 31.5 - 17.9 - 22.0 - 13.9 - 44.2 - 34.6 e) ASCE Figure 6-3 for flat roof Assume 10 ft tributary area (effective wind area) Wind pressures would be lower for larger tributary areas pnet30 (psf) pnet = λ Kzt I pnet30 (psf) Zone Roof 10.5 - 25.9 12.1 - 29.8 Roof 10.5 - 43.5 12.1 - 50.0 Roof 10.5 - 65.4 12.1 - 75.2 Wall 25.9 - 28.1 29.8 - 32.3 Wall 25.9 - 34.7 29.8 - 39.9 Roof Overhang - 37.3 - 42.9 Roof Overhang - 61.5 - 70.7 Chapter Solutions Page 14 of 19 Problem 2.20 Kzt = 1.0 V = 90 mph ASCE Figure 6-1 for Denver, CO I = 1.15 ASCE Table 6-1 and IBC Table 1604.5 for essential facility (Category IV) Ridge height = 22 ft + (3/12)(50 ft/2) = 28.25 ft hmean = (22 ft + 28.25 ft)/2 = 25.1 ft ASCE Figure 6-2 for Exposure C and hmean = 25.1 ft λ = 1.35 0.4hmean = 0.4(25.1 ft) = 10.1 ft 0.1b = 0.1(50 ft) = ft a = lesser of {0.4hmean or 0.1b} = ft 2a = 10 ft Roof angle = arctan (3/12) = 14.0 degrees a) ASCE Figure 6-2 for roof angle of 15º (Wind direction perpendicular to gable ridge) [Note that this solution is for a tabulated roof angle of 15º Interpolation of tabulated values for a 14º roof slope would provide results within 2%.] ps30 (psf) ps = λ Kzt I ps30 (psf) Zone Wall A 16.1 25.0 Wall B - 5.4 - 8.4 Wall C 10.7 16.6 Wall D - 3.0 - 4.7 Roof E - 15.4 - 23.9 Roof F - 10.1 - 15.7 Roof G - 10.7 - 16.6 Roof H - 7.7 - 12.0 a) ASCE Figure 6-2 for roof slope of 0° (Wind direction parallel to gable ridge) ps30 (psf) ps = λ Kzt I ps30 (psf) Zone Wall A 12.8 19.9 Wall C 8.5 13.2 Roof E - 15.4 - 23.9 Roof F - 8.8 - 13.7 Roof G - 10.7 - 16.6 Roof H - 6.8 - 10.6 b) ASCE Figure 6-3 for roof angle of 14° and 20 ft tributary area (effective wind area) pnet = λ Kzt I pnet30 (psf) Zone pnet30 (psf) Roof 7.7* - 13.0 12.0* - 20.2 Wall 13.9 - 15.1 21.6 - 23.4 c) ASCE Figure 6-3 for roof angle of 14° and 50 ft tributary area (effective wind area) Zone pnet30 (psf) pnet = λ Kzt I pnet30 (psf) Roof 6.7* - 18.9 10.4* - 29.3 Roof 6.7* - 29.1 10.4* - 45.2 Wall 13.0 - 16.5 20.2 - 25.6 * Per ASCE Section 6.1.4.2, a minimum value of pnet30 = 10 psf will result in pnet = 15.5 psf Chapter Solutions Page 15 of 19 Problem 2.21 ASCE 7-05 seismic force requirements a The formulas for base shear Give section reference V = Cs W (ASCE Eq 12.8-1) Where Cs is taken as: S DS Cs = ( ) (ASCE Eq 12.8-2) RI The following minimum and maximum values apply for Cs: Cs ≥ 0.01 (ASCE Eq 12.8-5) 0.5S ( Cs ≥ R I ) when S1 ≥ 0.6g (ASCE Eq 12.8-6) when T ≤ TL (ASCE Eq 12.8-3) when T > TL (ASCE Eq 12.8-4) S D1 Cs ≤ T (R I ) ST D1 L Cs ≤ T (R I ) b The highest mapped spectral response accelerations SS and S1 from the seismic hazard maps of the conterminous U.S (Appendix C) Several regions of the United States have high mapped spectral response accelerations given in ASCE Figures 22-1 through 22-9, or IBC Figures 1613.5(1) through 1613.5(9) Some of the highest values that can be read from these maps include (SS, S1, in %g): California 275, 124 Oregon & Washington 200, 75 Montana, Wyoming, Idaho, Utah 125, 60 Missouri, Illinois, Kentucky, Tennessee, Mississippi, Arkansas 300, 125 South Carolina 258, 73 [NOTE: See ASCE maps for peak California values, since the 2006 IBC maps include a typographical error for peak California values.] The significance of SS and S1 is that they describe the anticipated seismic ground shaking hazard that can be expected, based on available ground motion data, for structures with short and long Chapter Solutions Page 16 of 19 periods (0.2 and 1.0 seconds), respectively, based on Site Class B These parameters serve as the basis for the design response spectrum, from which seismic design forces are determined c The maximum tabulated Site Coefficients Fa and Fv From ASCE Table 11.4-1, the maximum value of Fa is 2.5 for Site Class E and SS ≤ 0.25 From ASCE Table 11.4-2, the maximum value of Fv is 3.5, for Site Class E and S1 ≤ 0.1 The site coefficients modify the mapped spectral response accelerations for the soil profile at a particular building location d The maximum values of SMS, SM1, SDS and SD1 based on previous values Using the maximum mapped value of SS=300% or 3.0g, and multiplying by the highest Fa value of 1.0 for SS>1.25g, the maximum value for SMS is 3.0g Multiplying by 2/3, SDS is 2.0g The other possible combinations of SS and Fa from ASCE Table 11.4-1 can be quickly checked and found to be smaller: 0.25g(2.5)=0.625g, 0.5g(1.7)=0.85g, 0.75g(1.2)=0.90g, 1.00g(0.9)=0.90g Using the maximum mapped value of S1=125% or 1.25g, and multiplying by the highest Fv value of 2.4 for S1>0.5g, the maximum value for SM1 is 3.0g Multiplying by 2/3 SD1 is 2.0g The other possible combinations of S1 and Fv from ASCE Table 11.4-2 can be quickly checked and found to be smaller: 0.1g(3.5)=0.35g, 0.2g(3.2)=0.64g, 0.3g(2.8)=0.84g, 0.4g(2.4)=0.99g e Briefly discuss the purpose of the R-factor What value of R is used for a building with wood-frame bearing walls that are sheathed with wood structural panel sheathing? The R- factor is used to reduce seismic forces from the design response spectrum to design level forces The reduction is based on expected over strength (both in the system and individual elements) and the expectation that elements can perform beyond the elastic stress range A building with light -frame bearing walls and sheathed with wood structural panel sheathing is assigned an R-factor of 6.5 The R-factors are given in ASCE Table 12.2-1 Chapter Solutions Page 17 of 19 Problem 2.22 ASCE 7-05 seismic force requirements a The definition of period of vibration and the methods for estimating the fundamental period The period of vibration is the length of time that it takes for a structure to complete one cycle of free vibration, and is a characteristic of the structure mass and stiffness While other methods involving building modeling may be used, the primary method is an approximate formula: Ta = Ct hn x (ASCE Eq 12.8-7) b How does the period of vibration affect seismic forces? Based on structural dynamics principles, buildings with the same fundamental period and same damping have essentially the same response to an earthquake ground motion record In general, the anticipated seismic forces decrease for longer period buildings For design purposes, wood buildings generally have periods too short to suggest any decrease in force c Describe the effects of the interaction of the soil and structure on seismic forces Local soil conditions, and particularly soft soils can significantly amplify earthquake ground motions d What is damping and how does it affect seismic forces? Do the ASCE criteria take damping into account? Damping is resistance to motion provided by the building materials through mechanisms such as friction, metal yielding and wood crushing A low level of damping is assumed in the ASCE design response spectrum Additional damping is also considered in determining R-factors Chapter Solutions Page 18 of 19 Problem 2.23 ASCE seismic force requirements a Briefly describe the general distribution of seismic forces over the height of a multi-story building For all seismic design categories, for buildings with periods of 0.5 seconds or less, the Fx story forces to the vertical resisting elements (such as shearwalls) will have a roughly triangular force distribution, as per ASCE Equations 12.8-11 and 12.8-12, and Example 2.14 of this text Fpx forces are described in Item b, below b Describe differences in vertical distribution for vertical element and diaphragm forces between Seismic Design Categories B and D Fpx story forces for design of diaphragms exhibit a vertical distribution similar to the distribution of Fx forces for design of shearwalls in all Seismic Design Categories The formula for Fpx is found in ASCE Equation 12.10-1, and Example 2.14 of this text c Describe forces for out-of-plane design of wall components Cite ASCE provisions The formula for out-of-plane design of structural walls is Fp = 0.4SDS I Wp (see ASCE Sec 12.11) For non-structural wall components with discrete attachments to the structure, and for structural parapets, the design formula comes from ASCE Sec 13.3 (Eqs 13.3-1 through 13.3-3): 0.3 S I W ≤ DS = 0.4 a p SDSWp Rp pp 1+2 F p z I p h ≤ 1.6SDS I pWp Chapter Solutions Page 19 of 19 Problem 2.24 (ASD) D = 10 psf (assume girder self-weight is included in the 10 psf dead load) Lr = 20 psf S = 35 psf H=0 R = 30 psf F=0 W = 18 psf (acting downward) L=0 E = psf (acting downward) T=0 ASCE IBC 16-8: D + F = 10 psf 16-9: D + H + F + L + T = 10 psf 16-10: D + H + F + (Lr or S or R) = 10 + 35 = 45 psf 16-11: D + H + F + 0.75(L + T) + 0.75(Lr or S or R) = 10 + 0.75(35) = 36.25 psf 16-12: D + H + F + (W or 0.7E) = 10 + 18 = 28 psf 16-13: D + H + F + 0.75(W or 0.7E) + 0.75L + 0.75(Lr or S or R) = 10 + 0.75(18 + 35) = 49.75 psf & 16-14 & 16-15: not applicable since W and E act in same direction as D Maximum service load (for ASD) on the glulam girder is 49.75 psf based on ASCE load combination (IBC load combination 16-13) [Note that one of the other service load combinations may be critical for design, depending on the applicable load duration factor, CD See discussion of CD in Chapter of the textbook.] Problem 2.25 (LRFD) Assume girder self-weight is included in the 10 psf dead load ASCE IBC 16-1: 1.4(D + F) = 1.4(10) = 14 psf 16-2: 1.2(D + F + T) + 1.6(L + H) + 0.5(Lr or S or R) = 1.2(10) + 0.5(35) = 29.5 psf 16-3: 1.2D + 1.6(Lr or S or R) + (L or 0.8W) = 1.2(10) + 1.6(35) + 0.8(18) = 82.4 psf 16-4: 1.2D + 1.6W + L + 0.5(Lr or S or R) = 1.2(10) + 1.6(18) + 0.5(35) = 58.3 psf 16-5: 1.2D + E + L + 0.2S = 1.2(10) + + 0.2(35) = 21 psf & 16-6 & 16-7: not applicable since W and E act in same direction as D Maximum factored load (for LRFD) on the glulam girder is 82.4 psf based on ASCE load combination (IBC load combination 16-3) [Note that one of the other factored load combinations may be critical for design, depending on the applicable time effect factor, λ See discussion of CD and λ in Chapter of the textbook.]