STUDY NOTES & REFERENCE GUIDE
All 13 Chapters Covered
SOLUTIONS
@LECTJULIESOLUTIONSSTUVIA
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Contents
Chapter 1 ........................................................................................................................ 1
Chapter 2 ........................................................................................................................ 6
Chapter 3 ........................................................................................................................ 23
Chapter 4 ........................................................................................................................ 41
Chapter 5 ........................................................................................................................ 55
Chapter 6 ........................................................................................................................ 62
Chapter 7 ........................................................................................................................ 74
Chapter 8 ........................................................................................................................ 84
Chapter 9 ........................................................................................................................ 92
Chapter 10 ...................................................................................................................... 109
Chapter 11 ...................................................................................................................... 119
Chapter 12 ...................................................................................................................... 137
Chapter 13 ...................................................................................................................... 154
Appendix B....................................................................................................................... 159
@LECTJULIESOLUTIONSSTUVIA
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Problem 1-4
The size and cross-sectional areas are obtained from Part 1 of the AISCM as follows:
Size Self-weight (lb/ft.) Cross-sectional area (in2)
W14x22 22 6.49
W21x44 44 13.0
HSS 6x6x½ 35.11 9.74
L6x4x½ 16.2 4.75
C12x30 30 8.81
WT18x128 128 37.7
Problem 1-5
a)
Element A y Ay I d = y- y I + Ad2
top flange 21 26.25 551.25 3.94 -12.75 3418
web 21 13.5 283.5 1008 0 1008
bot flange 21 0.75 15.75 3.94 12.75 3418
= 63 in.2 850.5 I = 7844 in.4
Ay 850.5
y = 13.5 in.
A 63
Self weight = (63/144)(490 lb/ft3) = 214 lb/ft.
b)
Element A y Ay I d = y- y I + Ad2
top plate 2.63 18.26 47.93 0.03 -9.04 214.3
beam 10.3 9.23 95.02 510 0 510
bot plate 2.63 0.188 0.49 0.03 9.04 214.3
= 15.55 in.2 143.4 I = 939 in.4
Ay 143.4
y = 9.23 in.
A 15.55
Self weight = (15.55/144)(490 lb/ft3) = 52.9 lb/ft.
c) From AISCM Table 1-20, Ix = 314 in.4
Area = 13.8 in2
Self weight = 47.1 lb/ft.
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Problem 1-7
Plot the idealized stress-strain diagram for a 6-in. wide by ½-in. thick plate and a 6-in. wide by 1-in.
thick plate of ASTM A36 steel. Assume that the original length between two points on the specimen
over which the elongation will be measured (i.e. the gage length) is 2-in.
Solution:
Gage length, Lo = 2 in.
For 6 x ½-in. plate, Area = (6 in.)(½ in.) = 3 in2
E = 29,000 ksi
P Stress = P/A Strain, = P/EA Elongation, Lo = Strain x gage length = Lo
(kips) (ksi) (in.)
0 0 0 0
20 6.67 0.00023 0.00046
40 13.33 0.00046 0.00092
60 20.0 0.00069 0.00138
80 26.67 0.00092 0.00184
100 33.33 0.00111 0.00222
108 36.0 0.00124 0.00248
For 6 x 1-in. plate, Area = (6 in.)(1 in.) = 6 in2
P Stress = P/A Strain, = P/EA Elongation, Lo = Strain x gage length = Lo
(kips) (ksi) (in.)
0 0 0 0
40 6.67 0.00023 0.00046
80 13.33 0.00046 0.00092
120 20.0 0.00069 0.00138
160 26.67 0.00092 0.00184
200 33.33 0.00111 0.00222
216 36.0 0.00124 0.00248
Problem 1-8
Determine the most economical layout of the roof framing (joists and girders) and the
gage (thickness) of the roof deck for a building with a 25 ft x 35 ft typical bay size. The
total roof dead load is 25 psf and the snow load is 35 psf. Assume deep
galvanized wide rib deck and an estimated weight of roof framing of 6 psf.
*Assume beams (or joists) span the 35’ direction
* Assume 3-span condition
*Total roof load = (25psf + 35psf) – 6psf = 54psf
nolanjack - Stuvia | Page 4 of 216
All 13 Chapters Covered
SOLUTIONS
@LECTJULIESOLUTIONSSTUVIA
nolanjack - Stuvia | Page 1 of 216
, STUDY NOTES & REFERENCE GUIDE
Contents
Chapter 1 ........................................................................................................................ 1
Chapter 2 ........................................................................................................................ 6
Chapter 3 ........................................................................................................................ 23
Chapter 4 ........................................................................................................................ 41
Chapter 5 ........................................................................................................................ 55
Chapter 6 ........................................................................................................................ 62
Chapter 7 ........................................................................................................................ 74
Chapter 8 ........................................................................................................................ 84
Chapter 9 ........................................................................................................................ 92
Chapter 10 ...................................................................................................................... 109
Chapter 11 ...................................................................................................................... 119
Chapter 12 ...................................................................................................................... 137
Chapter 13 ...................................................................................................................... 154
Appendix B....................................................................................................................... 159
@LECTJULIESOLUTIONSSTUVIA
nolanjack - Stuvia | Page 2 of 216
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Problem 1-4
The size and cross-sectional areas are obtained from Part 1 of the AISCM as follows:
Size Self-weight (lb/ft.) Cross-sectional area (in2)
W14x22 22 6.49
W21x44 44 13.0
HSS 6x6x½ 35.11 9.74
L6x4x½ 16.2 4.75
C12x30 30 8.81
WT18x128 128 37.7
Problem 1-5
a)
Element A y Ay I d = y- y I + Ad2
top flange 21 26.25 551.25 3.94 -12.75 3418
web 21 13.5 283.5 1008 0 1008
bot flange 21 0.75 15.75 3.94 12.75 3418
= 63 in.2 850.5 I = 7844 in.4
Ay 850.5
y = 13.5 in.
A 63
Self weight = (63/144)(490 lb/ft3) = 214 lb/ft.
b)
Element A y Ay I d = y- y I + Ad2
top plate 2.63 18.26 47.93 0.03 -9.04 214.3
beam 10.3 9.23 95.02 510 0 510
bot plate 2.63 0.188 0.49 0.03 9.04 214.3
= 15.55 in.2 143.4 I = 939 in.4
Ay 143.4
y = 9.23 in.
A 15.55
Self weight = (15.55/144)(490 lb/ft3) = 52.9 lb/ft.
c) From AISCM Table 1-20, Ix = 314 in.4
Area = 13.8 in2
Self weight = 47.1 lb/ft.
nolanjack - Stuvia | Page 3 of 216
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Problem 1-7
Plot the idealized stress-strain diagram for a 6-in. wide by ½-in. thick plate and a 6-in. wide by 1-in.
thick plate of ASTM A36 steel. Assume that the original length between two points on the specimen
over which the elongation will be measured (i.e. the gage length) is 2-in.
Solution:
Gage length, Lo = 2 in.
For 6 x ½-in. plate, Area = (6 in.)(½ in.) = 3 in2
E = 29,000 ksi
P Stress = P/A Strain, = P/EA Elongation, Lo = Strain x gage length = Lo
(kips) (ksi) (in.)
0 0 0 0
20 6.67 0.00023 0.00046
40 13.33 0.00046 0.00092
60 20.0 0.00069 0.00138
80 26.67 0.00092 0.00184
100 33.33 0.00111 0.00222
108 36.0 0.00124 0.00248
For 6 x 1-in. plate, Area = (6 in.)(1 in.) = 6 in2
P Stress = P/A Strain, = P/EA Elongation, Lo = Strain x gage length = Lo
(kips) (ksi) (in.)
0 0 0 0
40 6.67 0.00023 0.00046
80 13.33 0.00046 0.00092
120 20.0 0.00069 0.00138
160 26.67 0.00092 0.00184
200 33.33 0.00111 0.00222
216 36.0 0.00124 0.00248
Problem 1-8
Determine the most economical layout of the roof framing (joists and girders) and the
gage (thickness) of the roof deck for a building with a 25 ft x 35 ft typical bay size. The
total roof dead load is 25 psf and the snow load is 35 psf. Assume deep
galvanized wide rib deck and an estimated weight of roof framing of 6 psf.
*Assume beams (or joists) span the 35’ direction
* Assume 3-span condition
*Total roof load = (25psf + 35psf) – 6psf = 54psf
nolanjack - Stuvia | Page 4 of 216
