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Chapter 2 Solutions

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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 psf

3/8 in. plywood sheathing (3/8 in.) (3.0 psf/in) = 1.1 psf

2x6 @ 16 in. o.c. = 1.4 psf

Fiberglass loose insulation (5.5 in.) (0.5 psf/in) = 2.75 psf

Gypsum 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 = 3:12

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 = 8 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

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Chapter 2 Solutions

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


______________________________________________________________________

of 19

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 psf

15/32 in. structural panel (plywood) sheathing (15/32) (3.0 psf/in.) = 1.4 psf

2x8 @ 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 = 6:12

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 (10 in.) (0.5 psf/in) = 5.0 psf

2x6 @ 16 in. o.c. = 1.4 psf

Gypsum 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

Chapter 2 Solutions of 19

R2 = 1.2 – 0.05 F = 1.2 – (0.05)(6) = 0.9


______________________________________________________________________

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 psf

½ in. plywood sheathing (½ in.) (3.0 psf/in) = 1.5 psf

Roof trusses @ 24 in. o.c. (9 lb/ft)÷(2 ft) = 4.5 psf

Suspended acoustic ceiling: Acoustical fiber tile = 1.0 psf

Suspended 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 psf

2x10 @ 16 in. o.c. = 2.4 psf


Air duct = 0.5 psf

Suspended acoustic ceiling: Acoustical fiber tile = 1.0 psf

Suspended acoustic ceiling: Channel-suspended system = 1.0 psf

2nd

Floor Dead Load (D) = 25.6 psf

c) roof slope = 0.25:12

R1 = 1.0 since not considering tributary area

R2 = 1.0 for roof slope less than 4 in 12


______________________________________________________________________

Page 4 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:

Chapter 2 Solutions

G m.c. 3

density = 62.4

1 + = 33 lb/ft

1 + G(0.009)(m.c.) 100

[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 psf 4x14 purlin s.w.

(33)(3.5/12)(13.25/12)/(8) = 1.33 psf Built-up roof (5 ply w/o gravel)

= 2.50 psf


2x4 @ 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 12


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 ft

2


b) R1 = 1.2 – 0.001 AT = 0.94


Roof 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 ft

2

b) R1 = 1.2 – 0.001 AT = 0.914


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° pg

= 70 psf basic ground snow load

I = 1.0 residential occupancy

Ce = 0.9 Exposure C; fully exposed roof

Ct = 1.0 heated structure

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°

pg = 90 psf basic ground snow load I =

1.0 residential occupancy

Ce = 1.2 Exposure B; sheltered roof

Ct = 1.0 heated structure

Chapter 2 Solutions

Cs = 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 ft

2

R1 = 1.0

R2 = 1.0 (flat roof)

Lr = 20 R1 R2 = 20 psf

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

2

R1 = 1.0


Lr = 20 R1 R2 = 20 psf

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

> 600 ft2

R1 = 0.6


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.)


nd

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 ft

2

L = L 0.25 + 15

= 36.7 psf

LL T

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

______________________________________________________________________

Occupancy/Use

Unit Floor Live Load (2 Floor) Concentrated

Live Load

a) Offices 50 psf 2000 lb

b) Light Storage 125 psf --

c) Retail Store 75 psf 1000 lb

d) Apartments 40 psf (private rooms and corridors serving them) --

(Residential,

Multiple-family)

100 psf (public rooms and corridors serving them)

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

K A


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 beam

KLL AT = (2)(416) = 832 ft2 > 400 ft

2

L = L 0.25 + 15

= 30.8 psf

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:

K A


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.]

______________________________________________________________________

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


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


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 beam

KLL AT = (2)(80) = 160 ft2 < 400 ft

2 (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

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 ft2

KLL = 2 interior beam

KLL AT = (2)(320) = 640 ft2 > 400 ft

2 (live load reduction is applicable)

L0 = 75 psf (retail; 2nd

floor)

L = L 0.25 + 15 = 63.2 psf

0 K A

LL 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

______________________________________________________________________ Problem

2.17

a) ps = λ Kzt I ps30 for main wind-force resisting systems pnet = 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 roof

Zone ps30 (psf) ps = λ Kzt I ps30 (psf)

Wall A 22.8 26.2

Wall B - 11.9 - 13.7

Wall C 15.1 17.4

Wall D - 7.0 - 8.05

Roof E - 27.4 - 31.5

Roof F - 15.6 - 17.9

Roof G - 19.1 - 22.0

Roof H - 12.1 - 13.9

Roof Overhang EOH - 38.4 - 44.2

Roof 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 - 29.8

Roof 2 10.5 - 43.5 12.1 - 50.0


Roof 3 10.5 - 65.4 12.1 - 75.2

Wall 4 25.9 - 28.1 29.8 - 32.3

Wall 5 25.9 - 34.7 29.8 - 39.9

Roof Overhang 2 - 37.3 - 42.9

Roof Overhang 3 - 61.5 - 70.7

Problem 2.20 Kzt = 1.0

V = 90 mph ASCE 7 Figure 6-1 for Denver, CO

I = 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 ft hmean = (22 ft + 28.25 ft)/2

= 25.1 ft λ = 1.35 ASCE 7 Figure 6-2 for Exposure C and hmean = 25.1 ft.

0.4hmean = 0.4(25.1 ft) = 10.1 ft

0.1b = 0.1(50 ft) = 5 ft a =

lesser of {0.4hmean or 0.1b} = 5

ft 2a = 10 ft

Roof 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%.]


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 7 Figure 6-2 for roof slope of 0° (Wind direction parallel to gable ridge)


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 7 Figure 6-3 for roof angle of 14° and 20 ft2 tributary area (effective wind area).


Roof 1 7.7* - 13.0 12.0* - 20.2

Wall 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).


Roof 2 6.7* - 18.9 10.4* - 29.3

Roof 3 6.7* - 29.1 10.4* - 45.2

Wall 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 (ASCE 7 Eq. 12.8-1)

Where C s is taken as:

Cs = (S DS ) (ASCE 7 Eq. 12.8-2)

R I

The following minimum and maximum values apply for Cs:

Cs ≥ 0.01 (ASCE 7 Eq. 12.8-5)

0.5S

Cs ≥ (R I 1) when S1 ≥ 0.6g (ASCE 7 Eq. 12.8-6) S D1

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


S T D1 L

Cs ≤ T (R I ) when T > TL (ASCE 7 Eq.

12.8-4)

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, 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 7 maps for peak California values, since the 2006 IBC maps include a

typographical error for peak California values.]

2

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 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 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.

Problem 2.22

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.

Chapter 2 Solutions Page 17 of 19

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 2.24

Chapter 2 Solutions

Page 18 of 19

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):

Problem 2.25

0.3S I W ≤ 0.4a p SDSWp z SDS I pWp

Rp h DS p p

F

p I

p

Chapter 2 Solutions

Page 19 of 19

(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 = 2 psf (acting downward) T = 0

ASCE 7 IBC

1 16-8: D + F = 10 psf

2 16-9: D + H + F + L + T = 10 psf

3 16-10: D + H + F + (Lr or S or R) = 10 + 35 = 45 psf

4 16-11: D + H + F + 0.75(L + T) + 0.75(Lr or S or R) = 10 + 0.75(35) = 36.25 psf

5 16-12: D + H + F + (W or 0.7E) = 10 + 18 = 28 psf

6 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

= 1 + 2

≤ 1.6

Problem 2.26

7 & 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 psf

2 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

3 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

4 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

5 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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