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


Chapter 2 Solutions
Page 2 of 19

Problem 2.2

a) See Appendix A and Appendix B for weights of concrete roof tiles, lumber sheathing,
and framing.
2
(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 2 Solutions
Page 3 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.

3
(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
2 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 4 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 2 Solutions
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:
G
m.c.
3
density = 62.4

1

+

= 33 lb/ft

1 + G(0.009)(m.c.) 100
3
[Note that 33 lb/ft is a reasonable (and typically conservative) estimate of unit weight for
3
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 4 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 2 Solutions
Page 5 of 19

Problem 2.5

2
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 4 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
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 4 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 = 1 for θ
70°) Design snow load: S = (0.7 Ce Ct I pg) Cs = 68.6 psf

= 30° and Cs = 0 for θ =


Chapter 2 Solutions
Page 6 of 19

Problem 2.9

2
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
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
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 2 Solutions
Page 7 of 19

Problem 2.12
a) L0 = 50 psf

office floor

2
b) AT = 240 ft
KLL = 4 interior column KLL A T =
2
2
(4)(240) = 960 ft > 400 ft
= 36.7 psf
L = L 0.25 + 15
K

A

LL T

2
c) (35 psf + 36.7 psf)(240 ft ) = 17,200 lb. = 17.2 k


Chapter 2 Solutions
Page 8 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
2
KLL = 2 interior beam
2
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
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
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
2
Moment: Mmax = w(D+L)L /8 = (813)(26) /8 = 68,700 lb-ft Deflection:
4
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 2 Solutions
Page 9 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 2 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 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.
Recommended Total Load Deflection: L/180 = (40 ft)(12 in/ft)/180 = 2.67 in.
c) Floor joist in 2

nd

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.

2
AT = (20 ft)(4 ft) = 80 ft
KLL = 2
interior beam
2
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 2 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.
2
AT = (32 ft)(10 ft) = 320 ft
KLL = 2
interior beam
2
2
KLL AT = (2)(320) = 640 ft > 400 ft (live load reduction is applicable)

L0 = 75 psf (retail; 2
L = L 0.25

0

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 2 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 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.


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
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

2
e) ASCE 7 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
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


Chapter 2 Solutions
Page 14 of 19

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
ASCE 7 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) = 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%.]
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 7 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

2
b) ASCE 7 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
1
7.7*
- 13.0
12.0*
- 20.2
Wall
4
13.9
- 15.1
21.6
- 23.4
2
c) ASCE 7 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
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.


Chapter 2 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 7 Eq. 12.8-1)

Where Cs is taken as:

S DS

Cs = (

)

(ASCE 7 Eq. 12.8-2)

RI
The following minimum and maximum values apply for Cs:

Cs ≥ 0.01

(ASCE 7 Eq. 12.8-5)

0.5S

(

Cs ≥ R I

) when S1 ≥ 0.6g

(ASCE 7 Eq. 12.8-6)

when T ≤ TL

(ASCE 7 Eq. 12.8-3)

when T > TL

(ASCE 7 Eq. 12.8-4)

1

S
D1

Cs ≤ T (R I )

ST
D1 L

Cs ≤ 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, 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.]


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.

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

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 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.


Chapter 2 Solutions
Page 18 of 19

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):

0.3 S I W ≤
DS

= 0.4 a p SDSWp
Rp

pp

1+2

F
p

z

I
p

h

≤ 1.6SDS I pWp


Chapter 2 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 = 2 psf (acting downward)
T=0
ASCE 7
1
2
3
4
5
6

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
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.]