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

2: Design Loads Chapter 2 Solutions Page 1 of 19 Problem 2.1 a See Appendix A and Appendix B for weights of roofing, sheathing, framing, insulation, and gypsum wallboard.. Problem 2.2

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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 2 Solutions

Page 1 of 19

Problem 2.1

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

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

[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

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

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Problem 2.2

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

) = 9.5 psf 15/32 in structural panel (plywood) sheathing (15/32) (3.0 psf/in.) = 1.4 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

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

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) = 0.9 Basic

Roof Live Load: Lr = 20 R1 R2 = 18 psf

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Page 3 of 19

Problem 2.3

a) See Appendix A and Appendix B for weights of roofing, sheathing, and suspended ceiling

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 = 1.0psf

Roof Dead Load (D) = 14.5 psf b) See Appendix A and Appendix B for weights of framing, sheathing, and suspended ceiling

) (0.125 ft) = 18.8 psf

Suspended acoustic ceiling: Acoustical fiber tile = 1.0 psf

Suspended acoustic ceiling: Channel-suspended system = 1.0psf

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 Basic

Roof Live Load: Lr = 20 R1 R2 = 20 psf

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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:

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

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

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Page 5 of 19

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 = 0.94

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 ft2

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 p g ) C s = 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

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 p g ) C s = 68.6 psf

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Problem 2.9

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

< 200 ft2

R1 =

1.0

R2 = 1.0 (flat roof)

Lr = 20 R1 R2 = 20 psf

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

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.)

Occupancy/Use Unit Floor Live Load (2nd

Live Load

d) 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)

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Page 7 of 19

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

L=L 0.25 + 15 = 36.7 psf

LL T

c) (35 psf + 36.7 psf)(240 ft2

) = 17,200 lb = 17.2 k

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Problem 2.13

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 ft2

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

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See IBC Table 1607.1 and Sections 1607.9.1.1 through 1607.9.1.3:

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

Fire Escapes (except for single-family residences) 100 psf

Office Buildings – Lobbies & First Floor Corridors 100 psf

Hotels & Multi-Family Dwellings – Public Rooms & Corridors 100 psf

Sidewalks, Yards & Driveways subject to vehicular traffic 250 psf

Stairs and Exits (except for one- & two-family residences) 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.]

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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

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Chapter 2 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

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 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

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 AT = (2)(320) = 640 ft2 > 400 ft2 (live load reduction is applicable)

L0 = 75 psf (retail; 2nd floor)

0

K A

LL T

wL = (63.2 psf)(10 ft) = 632 lb/ft

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

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Problem 2.17

a) p s = λ K zt I p s30for main wind-force resisting systems

p net = λK zt I p net30for 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 p s 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 p net along with tabulated values of p net30 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 p net along with tabulated values of p net30 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

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Chapter 2 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 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 mphASCE 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 K zt I ps30 (psf)

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 K zt I pnet30 (psf)

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Problem 2.20 K zt = 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

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%.]

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

b) ASCE 7 Figure 6-3 for roof angle of 14° and 20 ft2

c) ASCE 7 Figure 6-3 for roof angle of 14° and 50 ft2

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

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

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