Structural Steel Design 3rd Edition (Aghayere, 2020)
| All 13 Chapters | Complete Solutions Manual with
Step-by-Step Answers
, Contents
Chapter 1 ........................................................................................................................................... 1
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Chapter 2 ........................................................................................................................................... 6
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Chapter 3 ........................................................................................................................................... 23
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Chapter 4 ........................................................................................................................................... 41
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Chapter 5 ........................................................................................................................................... 55
x
Chapter 6 ........................................................................................................................................... 62
x
Chapter 7 ........................................................................................................................................... 74
x
Chapter 8 ........................................................................................................................................... 84
x
Chapter 9 ........................................................................................................................................... 92
x
Chapter 10 ..........................................................................................................................................109
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Chapter 11 ..........................................................................................................................................119
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Chapter 12 ..........................................................................................................................................137
x
Chapter 13 ..........................................................................................................................................154
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Appendix B ........................................................................................................................................ 159
x
,Problem 1-4 x
The size and cross-sectional areas are obtained from Part 1 of the AISCM as follows:
x x x x x x x x x x x x x x
Size Self-weight (lb/ft.) x Cross-sectional area (in2) x x
W14x22 22 6.49
W21x44 44 13.0
HSS 6x6x½ x 35.11 9.74
L6x4x½ 16.2 4.75
C12x30 30 8.81
WT18x128 128 37.7
Problem 1-5 x
a)
Element A y Ay I d = y- y
x x x I + Ad2
x x
top flange x 21 26.25 551.25 3.94 -12.75 3418
web 21 13.5 283.5 1008 0 1008
bot flange x 21 0.75 15.75 3.94 12.75 3418
= x 63 in.2 x 850.5 I = 7844 in.4
x x x
Ay 850.5
y= =
x x
x x = 13.5 in. x x x
A 63
Self weight = (63/144)(490 lb/ft3) = 214 lb/ft.
x x x x x x x
b)
Element A y Ay I d = y- y
x x x I + Ad2
x x
top plate x 2.63 18.26 47.93 0.03 -9.04 214.3
beam 10.3 9.23 95.02 510 0 510
bot plate x 2.63 0.188 0.49 0.03 9.04 214.3
= x 15.55 in.2 x 143.4 I = 939 in.4
x x x
Ay 143.4
y= =
x x
x x = 9.23 in. x
x
x x
A 15.55
Self weight = (15.55/144)(490 lb/ft3) = 52.9 lb/ft.
x x x x x x x
c) From AISCM Table 1-20, Ix = 314 in.4
x x x x x x x x
Area = 13.8 in2 x x x
Self weight = 47.1 lb/ft. x x x x
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, Problem 1-7 x
Plot the idealized stress-strain diagram for a 6-in. wide by ½-in. thick plate and a 6-in. wide by 1-in.
x x x x x x x x x x x x x x x x x x
thick plate of ASTM A36 steel. Assume that the original length between two points on the specimen
x x x x x x x x x x x x x x x x x
over which the elongation will be measured (i.e. the gage length) is 2-in.
x x x x x x x x x x x x x
Solution:
Gage length, Lo = 2 in.
x x x x x
For 6 x ½-in. plate, Area = (6 in.)(½ in.) = 3 in2
x x x x x x x x x x x x
E = 29,000 ksi
x x x x
P Stress = P/A x x Strain, = P/EA x x x Elongation, Lo = Strain x gage length = Lo x x x x x x x x
(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
x x x x x x x x x x x x
P Stress = P/A x x Strain, = P/EA x x x Elongation, Lo = Strain x gage length = Lo x x x x x x x x
(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 x
Determine the most economical layout of the roof framing (joists and girders) and the gage
x x x x x x x x x x x x x x
(thickness) of the roof deck for a building with a 25 ft x 35 ft typical bay size. The total roof dead
x x x x x x x x x x x x x x x x x x x x x x
load is 25 psf and the snow load is 35 psf. Assume a 1½” deep galvanized wide rib deck and an
x x x x x x x x x x x x x x x x x x x x x
estimated weight of roof framing of 6 psf.
x x x x x x x x
*Assume beams (or joists) span the 35’ direction
x x x x x x x
* Assume 3-span condition x x
*Total roof load = (25psf + 35psf) – 6psf = 54psf
x x x x x x x x x x
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| All 13 Chapters | Complete Solutions Manual with
Step-by-Step Answers
, Contents
Chapter 1 ........................................................................................................................................... 1
x
Chapter 2 ........................................................................................................................................... 6
x
Chapter 3 ........................................................................................................................................... 23
x
Chapter 4 ........................................................................................................................................... 41
x
Chapter 5 ........................................................................................................................................... 55
x
Chapter 6 ........................................................................................................................................... 62
x
Chapter 7 ........................................................................................................................................... 74
x
Chapter 8 ........................................................................................................................................... 84
x
Chapter 9 ........................................................................................................................................... 92
x
Chapter 10 ..........................................................................................................................................109
x
Chapter 11 ..........................................................................................................................................119
x
Chapter 12 ..........................................................................................................................................137
x
Chapter 13 ..........................................................................................................................................154
x
Appendix B ........................................................................................................................................ 159
x
,Problem 1-4 x
The size and cross-sectional areas are obtained from Part 1 of the AISCM as follows:
x x x x x x x x x x x x x x
Size Self-weight (lb/ft.) x Cross-sectional area (in2) x x
W14x22 22 6.49
W21x44 44 13.0
HSS 6x6x½ x 35.11 9.74
L6x4x½ 16.2 4.75
C12x30 30 8.81
WT18x128 128 37.7
Problem 1-5 x
a)
Element A y Ay I d = y- y
x x x I + Ad2
x x
top flange x 21 26.25 551.25 3.94 -12.75 3418
web 21 13.5 283.5 1008 0 1008
bot flange x 21 0.75 15.75 3.94 12.75 3418
= x 63 in.2 x 850.5 I = 7844 in.4
x x x
Ay 850.5
y= =
x x
x x = 13.5 in. x x x
A 63
Self weight = (63/144)(490 lb/ft3) = 214 lb/ft.
x x x x x x x
b)
Element A y Ay I d = y- y
x x x I + Ad2
x x
top plate x 2.63 18.26 47.93 0.03 -9.04 214.3
beam 10.3 9.23 95.02 510 0 510
bot plate x 2.63 0.188 0.49 0.03 9.04 214.3
= x 15.55 in.2 x 143.4 I = 939 in.4
x x x
Ay 143.4
y= =
x x
x x = 9.23 in. x
x
x x
A 15.55
Self weight = (15.55/144)(490 lb/ft3) = 52.9 lb/ft.
x x x x x x x
c) From AISCM Table 1-20, Ix = 314 in.4
x x x x x x x x
Area = 13.8 in2 x x x
Self weight = 47.1 lb/ft. x x x x
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, Problem 1-7 x
Plot the idealized stress-strain diagram for a 6-in. wide by ½-in. thick plate and a 6-in. wide by 1-in.
x x x x x x x x x x x x x x x x x x
thick plate of ASTM A36 steel. Assume that the original length between two points on the specimen
x x x x x x x x x x x x x x x x x
over which the elongation will be measured (i.e. the gage length) is 2-in.
x x x x x x x x x x x x x
Solution:
Gage length, Lo = 2 in.
x x x x x
For 6 x ½-in. plate, Area = (6 in.)(½ in.) = 3 in2
x x x x x x x x x x x x
E = 29,000 ksi
x x x x
P Stress = P/A x x Strain, = P/EA x x x Elongation, Lo = Strain x gage length = Lo x x x x x x x x
(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
x x x x x x x x x x x x
P Stress = P/A x x Strain, = P/EA x x x Elongation, Lo = Strain x gage length = Lo x x x x x x x x
(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 x
Determine the most economical layout of the roof framing (joists and girders) and the gage
x x x x x x x x x x x x x x
(thickness) of the roof deck for a building with a 25 ft x 35 ft typical bay size. The total roof dead
x x x x x x x x x x x x x x x x x x x x x x
load is 25 psf and the snow load is 35 psf. Assume a 1½” deep galvanized wide rib deck and an
x x x x x x x x x x x x x x x x x x x x x
estimated weight of roof framing of 6 psf.
x x x x x x x x
*Assume beams (or joists) span the 35’ direction
x x x x x x x
* Assume 3-span condition x x
*Total roof load = (25psf + 35psf) – 6psf = 54psf
x x x x x x x x x x
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