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Chapter 2 Solutions of 19

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Problem 2.1

a) See Appendix A and Appendix B for weights of roofing, sheathing, framing, insulation, and gypsum wallboard.

Asphalt shingles = 2.0 psf3/8 in. plywood sheathing (3/8 in.) (3.0 psf/in) = 1.1 psf2x6 @ 16 in. o.c. = 1.4 psfFiberglass loose insulation (5.5 in.) (0.5 psf/in) = 2.75 psfGypsum wallboard (1/2 in.) (5.0 psf/in) = 2.5 psf

Roof Dead Load (D) along roof slope = 9.75 psf

Convert D to load on a horizontal plane: Roof slope = 3:12Hypotenuse = (9 + 144)½ = 12.37Don 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 psf2x4 @ 16 in. o.c. = 0.9 psfGypsum wallboard (½ in.) = 2.5 psf

Wall Dead Load (D) = 13.4 psf

c) Wall Dead Load: wall height = 8 ftD = (13.4 psf) (8 ft) = 107.2 lb/ft

d) R1 = 1.0 since not considering tributary areaR2 = 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 areaR2 = 1.0 for slope of 4:12Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf

______________________________________________________________________

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Problem 2.2

a) See Appendix A and Appendix B for weights of concrete roof tiles, lumber sheathing, and framing.

Concrete tile (950 lb / 100 ft2) = 9.5 psf15/32 in. structural panel (plywood) sheathing (15/32) (3.0 psf/in.) = 1.4 psf2x8 @ 16 in. o.c. = 1.9 psf

Roof Dead Load (D) along roof slope = 12.8 psf

Convert D to load on a horizontal plane: Roof slope = 6:12Hypotenuse = (36 + 144)½ = 13.42Don 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 (10 in.) (0.5 psf/in) = 5.0 psf2x6 @ 16 in. o.c. = 1.4 psfGypsum wallboard (½ in.) (5 psf/in) = 2.5 psf

Ceiling Dead Load (D) = 8.9 psf

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) = 0.9Basic Roof Live Load: Lr = 20 R1 R2 = 18 psf

______________________________________________________________________


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) = 6.5 psf½ in. plywood sheathing (½ in.) (3.0 psf/in) = 1.5 psfRoof trusses @ 24 in. o.c. (9 lb/ft)÷(2 ft) = 4.5 psfSuspended acoustic ceiling: Acoustical fiber tile = 1.0 psfSuspended acoustic ceiling: Channel-suspended system = 1.0 psf

Roof Dead Load (D) = 14.5 psf

b) See Appendix A and Appendix B for weights of framing, sheathing, and suspended ceiling.

Concrete (150 lb/ft3) (0.125 ft) = 18.8 psf2x10 @ 16 in. o.c. = 2.4 psf5/8 in. plywood sheathing (5/8 in.) (3.0 psf/in) = 1.9 psfAir duct = 0.5 psfSuspended acoustic ceiling: Acoustical fiber tile = 1.0 psfSuspended acoustic ceiling: Channel-suspended system = 1.0 psf

2nd Floor Dead Load (D) = 25.6 psf

c) roof slope = 0.25:12R1 = 1.0 since not considering tributary area R2 = 1.0 for roof slope less than 4 in 12Basic Roof Live Load: Lr = 20 R1 R2 = 20 psf

______________________________________________________________________


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. 3density = 62.4 1 + = 33 lb/ft

1001 + G(0.009)(m.c.)[Note that 33 lb/ft3 is a reasonable (and typically conservative) estimate of unit weight for most softwood species of lumber and glulam. A unit weight of 33 lb/ft3 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. (33)(6.75/12)(33/12)/(20) = 2.55 psf4x14 purlin s.w. (33)(3.5/12)(13.25/12)/(8) = 1.33 psfBuilt-up roof (5 ply w/o gravel) = 2.50 psf15/32 in. plywood sheathing (15/32 in.) (3.0 psf/in) = 1.41 psf2x4 @ 24 in. o.c. = 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 4 in 12Basic 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______________________________________________________________________


Problem 2.5

a) Tributary area (on a horizontal plane): AT = (20 ft)(13 ft) = 260 ft2 > 200 ft2

b) R1 = 1.2 – 0.001 AT = 0.94R2 = 1.0 for slope less than 4 in 12Roof Live Load: Lr = 20 R1 R2 = 18.8 psf wLr = (18.8 psf)(13 ft) = 244 lb/ft

______________________________________________________________________

Problem 2.6

a) Tributary area (on a horizontal plane): AT = (22 ft)(13 ft) = 286 ft2 > 200 ft2

b) R1 = 1.2 – 0.001 AT = 0.914R2 = 1.0 for slope less than 4 in 12Roof Live Load: Lr = 20 R1 R2 = 18.3 psfwLr = (18.3 psf)(13 ft) = 238 lb/ft

______________________________________________________________________

Problem 2.7

roof slope = 6/12; θ = arctan (6/12) = 26.57°pg = 70 psf basic ground snow loadI = 1.0 residential occupancyCe = 0.9 Exposure C; fully exposed roofCt = 1.0 heated structureCs = 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°pg = 90 psf basic ground snow loadI = 1.0 residential occupancyCe = 1.2 Exposure B; sheltered roofCt = 1.0 heated structureCs = 0.908 for roof slope of 33.69°

(linear interpolation between Cs = 1 for θ = 30° and Cs = 0 for θ = 70°) Design snow load: S = (0.7 Ce Ct I pg) Cs = 68.6 psf______________________________________________________________________


Problem 2.9

a) Subpurlin: AT = (2 ft)(8 ft) = 16 ft2 < 200 ft2

R1 = 1.0R2 = 1.0 (flat roof)Lr = 20 R1 R2 = 20 psf

Purlin: AT = (8 ft)(20 ft) = 160 ft2 < 200 ft2

R1 = 1.0R2 = 1.0 (flat roof)Lr = 20 R1 R2 = 20 psf

Glulam Beam: AT = (20 ft)(50 ft) = 1000 ft2 > 600 ft2

R1 = 0.6R2 = 1.0 (flat roof)

Lr = 20 R1 R2 = 12 psfb) Subpurlin : wLr = (20 psf)(2 ft) = 40 lb/ft Purlin:

wLr = (20 psf)(8 ft) = 160 lb/ftGlulam 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 Unit Floor Live Load (2nd Floor) ConcentratedLive Load

a) Offices 50 psf 2000 lbb) Light Storage 125 psf --c) Retail Store 75 psf 1000 lbd) Apartments 40 psf (private rooms and corridors serving them) --

(Residential, 100 psf (public rooms and corridors serving them)Multiple-family)

e) Hotel Restrooms 40 psf (private rooms and corridors serving them) --(Residential) 100 psf (public rooms and corridors serving them)

f) School Classrooms 40 psf 1000 lb


Problem 2.12

a) L0 = 50 psf office floor

b) AT = 240 ft2

KLL = 4 interior column KLL A T = (4)(240) = 960 ft2 > 400 ft2

0.25 + 15 = 36.7 psfL = LK A

LL T

c) (35 psf + 36.7 psf)(240 ft2) = 17,200 lb. = 17.2 k______________________________________________________________________


Problem 2.13

D = 20 psf s = 16 ft L = 26 ft

a) L0 = 40 psf classroom occupancy

b) AT = s L = (16 ft) (26 ft) = 416 ft 2 KLL = 2 interior beamKLL AT = (2)(416) = 832 ft2 > 400 ft2

0.25 + 15 = 30.8 psfL = LK 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)Moment: Mmax = wD L2/8 + PL L/4 = (320)(26)2/8 + (1000)(26)/4 = 33,500 lb-ft

(for point load placed at mid-span)Deflection: ∆L = PL L3/48EI = (1000)(26)3/48EI = 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.Moment: Mmax = w(D+L)L2/8 = (813)(26)2/8 = 68,700 lb-ft Deflection: ∆L = 5wL L4/384EI = (5)(493)(26)4/384EI = 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.______________________________________________________________________


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 psfAssembly Areas & Theaters – Lobbies 100 psfAssembly Areas & Theaters – Movable Seats 100 psfAssembly Areas & Theaters – Stages & Platforms 125 psfExterior Balconies (except for one- & two-family residences) 100 psfCorridors 100 psfDance Halls & Ballrooms 100 psfDining Rooms & Restaurants 100 psfFire Escapes (except for single-family residences) 100 psfGarages 40 psfGymnasiums 100 psfLibrary Stack Rooms 150 psfManufacturing Facilities (Light) 125 psfManufacturing Facilities (Heavy) 250 psfOffice Buildings – Lobbies & First Floor Corridors 100 psfPenal Institutions – Corridors 100 psfHotels & Multi-Family Dwellings – Public Rooms & Corridors 100 psfSchools – First Floor Corridors 100 psfSidewalks, Yards & Driveways subject to vehicular traffic 250 psfSkating Rinks 100 psfStadiums & Arenas – Bleachers 100 psfStairs and Exits (except for one- & two-family residences) 100 psfStorage Warehouses (Light) 125 psfStorage Warehouses (Heavy) 250 psfRetail Stores – First Floor 100 psfWholesale Stores 125 psfPedestrian 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.]______________________________________________________________________


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.

______________________________________________________________________


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 2nd floor residence; L = 20 ft.; s = 4 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 ft2

KLL = 2 interior beamKLL AT = (2)(80) = 160 ft2 < 400 ft2 (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 psfw(KD+L) = (56 psf)(4 ft) = 224 lb/ft

d) Girder in 2nd 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 ft2KLL = 2 interior beamKLL AT = (2)(320) = 640 ft2 > 400 ft2 (live load reduction is applicable)L0 = 75 psf (retail; 2nd floor)

+ 15L = L 0.25 = 63.2 psf0 K

LLA

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 psfw(KD+L) = (73.2 psf)(10 ft) = 732 lb/ft

______________________________________________________________________


Problem 2.17

a) ps = λ Kzt I ps30 for main wind-force resisting systemspnet = λKzt I pnet 30 for components and cladding

b) ASCE 7 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 1 (roofs) or Zone 4 (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 2 or 3 (roofs) or Zone 5 (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 do 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.______________________________________________________________________


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 7 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 mph ASCE 7 Figure 6-1B for Tampa, FL

b) I = 1.15 ASCE 7 Table 6-1 and IBC Table 1604.5 for essential facility (Category IV)

c) λ = 1.0ASCE 7 Figure 6-2 for Exposure B and mean roof height of 30 ft.

d) ASCE 7 Figure 6-2 for flat roofZone ps30 (psf) ps = λ Kzt I ps30 (psf)

Wall A 22.8 26.2Wall B - 11.9 - 13.7Wall C 15.1 17.4Wall D - 7.0 - 8.05Roof E - 27.4 - 31.5Roof F - 15.6 - 17.9Roof G - 19.1 - 22.0Roof H - 12.1 - 13.9

Roof Overhang EOH - 38.4 - 44.2Roof Overhang GOH - 30.1 - 34.6

e) ASCE 7 Figure 6-3 for flat roof. Assume 10 ft2 tributary area (effective wind area). Wind pressures would be lower for larger tributary areas.

Zone pnet30 (psf) pnet = λ Kzt I pnet30 (psf)Roof 1 10.5 - 25.9 12.1 - 29.8Roof 2 10.5 - 43.5 12.1 - 50.0Roof 3 10.5 - 65.4 12.1 - 75.2Wall 4 25.9 - 28.1 29.8 - 32.3Wall 5 25.9 - 34.7 29.8 - 39.9

Roof Overhang 2 - 37.3 - 42.9Roof Overhang 3 - 61.5 - 70.7


Problem 2.20 Kzt = 1.0V = 90 mph ASCE 7 Figure 6-1 for Denver, COI = 1.15 ASCE 7 Table 6-1 and IBC Table 1604.5 for essential facility (Category IV)Ridge height = 22 ft + (3/12)(50 ft/2) = 28.25 fthmean = (22 ft + 28.25 ft)/2 = 25.1 ftλ = 1.35 ASCE 7 Figure 6-2 for Exposure C and hmean = 25.1 ft.

0.4hmean = 0.4(25.1 ft) = 10.1 ft0.1b = 0.1(50 ft) = 5 ft

a = lesser of {0.4hmean or 0.1b} = 5 ft 2a = 10 ftRoof angle = arctan (3/12) = 14.0 degrees

a) ASCE 7 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%.]

Zone ps30 (psf) ps = λ Kzt I ps30 (psf)Wall A 16.1 25.0Wall B - 5.4 - 8.4Wall C 10.7 16.6Wall D - 3.0 - 4.7Roof E - 15.4 - 23.9Roof F - 10.1 - 15.7Roof G - 10.7 - 16.6Roof H - 7.7 - 12.0

a) ASCE 7 Figure 6-2 for roof slope of 0° (Wind direction parallel to gable ridge)Zone ps30 (psf) ps = λ Kzt I ps30 (psf)

Wall A 12.8 19.9Wall C 8.5 13.2Roof E - 15.4 - 23.9Roof F - 8.8 - 13.7Roof G - 10.7 - 16.6Roof H - 6.8 - 10.6

b) ASCE 7 Figure 6-3 for roof angle of 14° and 20 ft2 tributary area (effective wind area).Zone pnet30 (psf) pnet = λ Kzt I pnet30 (psf)

Roof 1 7.7* - 13.0 12.0* - 20.2Wall 4 13.9 - 15.1 21.6 - 23.4

c) ASCE 7 Figure 6-3 for roof angle of 14° and 50 ft2 tributary area (effective wind area).Zone pnet30 (psf) pnet = λ Kzt I pnet30 (psf)

Roof 2 6.7* - 18.9 10.4* - 29.3Roof 3 6.7* - 29.1 10.4* - 45.2Wall 5 13.0 - 16.5 20.2 - 25.6

* Per ASCE 7 Section 6.1.4.2, a minimum value of pnet30 = 10 psf will result in pnet = 15.5 psf.


Problem 2.21

ASCE 7-05 seismic force requirements

a. The formulas for base shear. Give section reference.

V = Cs W

Where Cs is taken as:

Cs = (S DS )R I

The following minimum and maximum values apply for Cs:

Cs ≥ 0.01

(ASCE 7 Eq. 12.8-1)

(ASCE 7 Eq. 12.8-2)

(ASCE 7 Eq. 12.8-5)

Cs ≥

0.5S

when S1 ≥ 0.6g (ASCE 7 Eq. 12.8-6)(R I 1)

Cs ≤

SD1

when T ≤ TL (ASCE 7 Eq. 12.8-3)T (R I )

Cs ≤

SD1

TL

when T > TL (ASCE 7 Eq. 12.8-4)T 2 (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 7 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, 124Oregon & Washington 200, 75Montana, Wyoming, Idaho, Utah 125, 60Missouri, Illinois, Kentucky, Tennessee, Mississippi, Arkansas 300, 125South Carolina 258, 73[NOTE: See ASCE 7 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 2 SolutionsPage 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 7 Table 11.4-1, the maximum value of Fa is 2.5 for Site Class E and SS ≤ 0.25.

From ASCE 7 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 7 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 7 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 7 Table 12.2-1.


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 hnx (ASCE 7 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 7 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 7 design response spectrum. Additional damping is also considered in determining R-factors.


Problem 2.23

ASCE 7 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 7 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 7 Equation 12.10-1, and Example 2.14 of this text.

c. Describe forces for out-of-plane design of wall components. Cite ASCE 7 provisions.

The formula for out-of-plane design of structural walls is Fp = 0.4SDS I Wp (see ASCE 7 Sec. 12.11).

For non-structural wall components with discrete attachments to the structure, and for structural parapets, the design formula comes from ASCE 7 Sec. 13.3 (Eqs. 13.3-1 through 13.3-3):

0.3SDS

I pW

p ≤ F

p

0.4a p SDSWp z= 1 + 2 ≤ 1.6SDS I pWpRp h

I p


Problem 2.24 (ASD)

D = 10 psf (assume girder self-weight is included in the 10 psf dead load)Lr = 20 psf

H = 0S = 35 psfR = 30 psf F = 0W = 18 psf (acting downward) L = 0E = 2 psf (acting downward) T = 0

ASCE 7 IBC1 16-8: D + F = 10 psf2 16-9: D + H + F + L + T = 10 psf3 16-10: D + H + F + (Lr or S or R) = 10 + 35 = 45 psf4 16-11: D + H + F + 0.75(L + T) + 0.75(Lr or S or R) = 10 + 0.75(35) = 36.25 psf5 16-12: D + H + F + (W or 0.7E) = 10 + 18 = 28 psf6 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 psf7 & 8 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 7 load combination 6 (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 4 of the textbook.]

Problem 2.25 (LRFD)

Assume girder self-weight is included in the 10 psf dead load.

ASCE 7 IBC 1 16-1: 1.4(D + F) = 1.4(10) = 14 psf2 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 psf3 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 psf4 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 psf5 16-5: 1.2D + E + L + 0.2S = 1.2(10) + 2 + 0.2(35) = 21 psf

6 & 7 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 7 load combination 3 (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 4 of the textbook.]

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