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


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

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.

(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

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)


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


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

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

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)

= 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.

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

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

[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

- 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)

(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

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.


(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

≤ 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