All 13 Chapters Covered
SOLUTIONS
, 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
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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, ΔLo = 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, ΔLo = 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 a 1½” 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
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, # of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 12.5 none - -
3 8.33 16 10’-3” 85psf
4 6.25 22 6’-11” 76psf select
5 5 24 5’-10” 130psf
*Vulcraft deck assumed
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,1-11 Determine the most economical layout of the floor framing (beams and girders), the total
depth of the floor slab, and the gage (thickness) of the floor deck for a building with a 30 ft x
47 ft typical bay size. The total floor dead load is 110 psf and the floor live load is 250 psf.
Assume normal weight concrete, a 3” deep galvanized composite wide rib.
*Assume beams span the 47’ direction
* Assume 3-span condition
* Assume weight of the framing = 10psf
*Total floor load = (110psf +250psf) – 10psf = 350psf
t =2.5” (superimposed load = 350psf – 50psf – 2psf) = 298psf)
# of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 15 16 15’-5” none N.G.
3 10 16 15’-5” 218psf N.G.
4 7.5 18 13’-11” 298psf select
t =3” (superimposed load = 350psf – 57psf – 2psf) = 291psf)
# of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 15 none - -
3 10 16 14’-11” 245psf N.G.
4 7.5 18 13’-4” 334psf select
*Vulcraft deck assumed
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,From Equation 1-1, the carbon content is
CE = 0.16 + (0.20 + 0.25)/15 + (0.10 + 0.15 + 0.06)/5 + (0.80 + 0.20)/6 = 0.419 < 0.5
Therefore, the steel member is weldable.
Problem 1-13
Anticipated expansion or contraction = (6.5 x 10-6 in./in.)(300 ft.)(12 in./ft.)(70 oF) = 1.64 in.
Expansion joint width = (2)(1.64 in.) = 3.28 in.
Therefore, use a 3¼ in. wide expansion joint.
The width of the required expansion joint appears large, and one way to reduce this width is to
reduce the length between expansion joints from 300 ft to say 200 ft. That will bring the required
expansion joint width down to (200/300)(3.28 in.) = 2.2 in. (i.e. 2¼ in. expansion joint)
(a) Determine the factored axial load or the required axial strength, Pu of a column in an
office building with a regular roof configuration. The service axial loads on the column are as
follows
PD = 200 kips (dead load)
PL = 300 kips (floor live load)
PS = 150 kips (snow load)
PW = ±60 kips (wind load)
PE = ±40 kips (seismic load)
(b) Calculate the required nominal axial compression strength, Pn of the column.
1: Pu = 1.4 PD = 1.4 (200k) = 280 kips
2: Pu = 1.2 PD + 1.6 PL + 0.5 PS
= 1.2 (200) + 1.6 (300) + 0.5 (150) = 795 kips (governs)
3 (a): Pu = 1.2 PD + 1.6 PS + 0.5PL
= 1.2 (200) + 1.6 (150) + 0.5(300) = 630 kips
3 (b): Pu = 1.2 PD + 1.6 PS + 0.8 PW
= 1.2 (200) + 1.6 (150) + 0.8 (60) = 528 kips
4: Pu = 1.2 PD + 1.6 PW + 0.5 PL + 0.5 PS
= 1.2 (200) + 1.6 (60) + 0.5(300) + 0.5 (150) = 561 kips
5: Pu = 1.2 PD + 1.0 PE + 0.5 PL + 0.2 PS
= 1.2 (200) + 1.0 (40) + 0.5 (300) + 0.2 (150) = 460 kips
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,Note that PD must always oppose PW and PE in load combination 6
6: Pu = 0.9 PD + 1.6 PW
= 0.9 (200) +1.6 (-60) = 84 kips (no net uplift)
7: Pu = 0.9 PD + 1.0 PE
= 0.9 (200) + 1.0 (-40) = 140 kips (no net uplift)
Pn > Pu
c = 0.9
(0.9)(Pn) = (795 kips)
Pn = 884 kips
Problem 2-4
(a) Determine the ultimate or factored load for a roof beam subjected to the following
service loads:
Dead Load = 29 psf (dead load)
Snow Load = 35 psf (snow load)
Roof live load = 20 psf
Wind Load = 25 psf upwards / 15 psf downwards
(b) Assuming the roof beam span is 30 ft and tributary width of 6 ft, determine the
factored moment and shear.
Since, S = 35psf > Lr = 20psf, use S in equations and ignore Lr.
1: pu = 1.4D = 1.4 (29) = 40.6 psf
2: = 1.2 D + 1.6 L + 0.5 S
p = 1.2 (29) + 1.6 (0) + 0.5 (35) = 52.3 psf
u
3 (a): = 1.2D + 1.6S + 0.8W
p = 1.2 (29) + 1.6 (35) + 0.8 (15) = 102.8 psf (governs)
u
3 (b): = 1.2D + 1.6S + 0.5L
p = 1.2 (29) + 1.6 (35) + (0) = 90.8 psf
u
4: = 1.2 D + 1.6 W + L + 0.5S
p = 1.2 (29) + 1.6 (15) + (0) + 0.5 (35) = 76.3 psf
u
5:
p
u
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,6: = 1.2 D + 1.0 E + 0.5L + 0.2S
p = 1.2 (29) + 1.0 (0) + 0.5(0) + 0.2 (35) = 41.8 psf
u
= 0.9D + 1.6W (D must always oppose W in load combinations 6 and 7)
= 0.9 (29) + 1.6(-25) (upward wind load is taken as negative)
= -13.9 psf (net uplift)
7: = 0.9D + 1.0E (D must always oppose E in load combinations 6 and 7)
p = 0.9 (29) + 1.6(0) (upward wind load is taken as negative)
u = 26.1 psf (no net uplift)`
wu = (102.8psf)(6ft) = 616.8 plf (downward)
wu = (-13.9psf)(6ft) = -83.4 plf (upward)
downward uplift
w u L (616.8)(30) w u L ( 83.4)(30)
Vu = 9252 lb. Vu = 1251 lb.
2 2 2 2 2
w u L2 (616.8)(30)2 wuL ( 83.4)(30)2
Mu = 69.4 ft-kips Mu = 9.4 ft-kips
8 8 8 8
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, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
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SOLUTIONS
, 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
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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, ΔLo = 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, ΔLo = 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 a 1½” 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
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, # of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 12.5 none - -
3 8.33 16 10’-3” 85psf
4 6.25 22 6’-11” 76psf select
5 5 24 5’-10” 130psf
*Vulcraft deck assumed
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,1-11 Determine the most economical layout of the floor framing (beams and girders), the total
depth of the floor slab, and the gage (thickness) of the floor deck for a building with a 30 ft x
47 ft typical bay size. The total floor dead load is 110 psf and the floor live load is 250 psf.
Assume normal weight concrete, a 3” deep galvanized composite wide rib.
*Assume beams span the 47’ direction
* Assume 3-span condition
* Assume weight of the framing = 10psf
*Total floor load = (110psf +250psf) – 10psf = 350psf
t =2.5” (superimposed load = 350psf – 50psf – 2psf) = 298psf)
# of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 15 16 15’-5” none N.G.
3 10 16 15’-5” 218psf N.G.
4 7.5 18 13’-11” 298psf select
t =3” (superimposed load = 350psf – 57psf – 2psf) = 291psf)
# of beam beam spacing Selected deck max. constr. Deck
spaces (ft.) gage span Load
capacity
*
2 15 none - -
3 10 16 14’-11” 245psf N.G.
4 7.5 18 13’-4” 334psf select
*Vulcraft deck assumed
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,From Equation 1-1, the carbon content is
CE = 0.16 + (0.20 + 0.25)/15 + (0.10 + 0.15 + 0.06)/5 + (0.80 + 0.20)/6 = 0.419 < 0.5
Therefore, the steel member is weldable.
Problem 1-13
Anticipated expansion or contraction = (6.5 x 10-6 in./in.)(300 ft.)(12 in./ft.)(70 oF) = 1.64 in.
Expansion joint width = (2)(1.64 in.) = 3.28 in.
Therefore, use a 3¼ in. wide expansion joint.
The width of the required expansion joint appears large, and one way to reduce this width is to
reduce the length between expansion joints from 300 ft to say 200 ft. That will bring the required
expansion joint width down to (200/300)(3.28 in.) = 2.2 in. (i.e. 2¼ in. expansion joint)
(a) Determine the factored axial load or the required axial strength, Pu of a column in an
office building with a regular roof configuration. The service axial loads on the column are as
follows
PD = 200 kips (dead load)
PL = 300 kips (floor live load)
PS = 150 kips (snow load)
PW = ±60 kips (wind load)
PE = ±40 kips (seismic load)
(b) Calculate the required nominal axial compression strength, Pn of the column.
1: Pu = 1.4 PD = 1.4 (200k) = 280 kips
2: Pu = 1.2 PD + 1.6 PL + 0.5 PS
= 1.2 (200) + 1.6 (300) + 0.5 (150) = 795 kips (governs)
3 (a): Pu = 1.2 PD + 1.6 PS + 0.5PL
= 1.2 (200) + 1.6 (150) + 0.5(300) = 630 kips
3 (b): Pu = 1.2 PD + 1.6 PS + 0.8 PW
= 1.2 (200) + 1.6 (150) + 0.8 (60) = 528 kips
4: Pu = 1.2 PD + 1.6 PW + 0.5 PL + 0.5 PS
= 1.2 (200) + 1.6 (60) + 0.5(300) + 0.5 (150) = 561 kips
5: Pu = 1.2 PD + 1.0 PE + 0.5 PL + 0.2 PS
= 1.2 (200) + 1.0 (40) + 0.5 (300) + 0.2 (150) = 460 kips
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,Note that PD must always oppose PW and PE in load combination 6
6: Pu = 0.9 PD + 1.6 PW
= 0.9 (200) +1.6 (-60) = 84 kips (no net uplift)
7: Pu = 0.9 PD + 1.0 PE
= 0.9 (200) + 1.0 (-40) = 140 kips (no net uplift)
Pn > Pu
c = 0.9
(0.9)(Pn) = (795 kips)
Pn = 884 kips
Problem 2-4
(a) Determine the ultimate or factored load for a roof beam subjected to the following
service loads:
Dead Load = 29 psf (dead load)
Snow Load = 35 psf (snow load)
Roof live load = 20 psf
Wind Load = 25 psf upwards / 15 psf downwards
(b) Assuming the roof beam span is 30 ft and tributary width of 6 ft, determine the
factored moment and shear.
Since, S = 35psf > Lr = 20psf, use S in equations and ignore Lr.
1: pu = 1.4D = 1.4 (29) = 40.6 psf
2: = 1.2 D + 1.6 L + 0.5 S
p = 1.2 (29) + 1.6 (0) + 0.5 (35) = 52.3 psf
u
3 (a): = 1.2D + 1.6S + 0.8W
p = 1.2 (29) + 1.6 (35) + 0.8 (15) = 102.8 psf (governs)
u
3 (b): = 1.2D + 1.6S + 0.5L
p = 1.2 (29) + 1.6 (35) + (0) = 90.8 psf
u
4: = 1.2 D + 1.6 W + L + 0.5S
p = 1.2 (29) + 1.6 (15) + (0) + 0.5 (35) = 76.3 psf
u
5:
p
u
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,6: = 1.2 D + 1.0 E + 0.5L + 0.2S
p = 1.2 (29) + 1.0 (0) + 0.5(0) + 0.2 (35) = 41.8 psf
u
= 0.9D + 1.6W (D must always oppose W in load combinations 6 and 7)
= 0.9 (29) + 1.6(-25) (upward wind load is taken as negative)
= -13.9 psf (net uplift)
7: = 0.9D + 1.0E (D must always oppose E in load combinations 6 and 7)
p = 0.9 (29) + 1.6(0) (upward wind load is taken as negative)
u = 26.1 psf (no net uplift)`
wu = (102.8psf)(6ft) = 616.8 plf (downward)
wu = (-13.9psf)(6ft) = -83.4 plf (upward)
downward uplift
w u L (616.8)(30) w u L ( 83.4)(30)
Vu = 9252 lb. Vu = 1251 lb.
2 2 2 2 2
w u L2 (616.8)(30)2 wuL ( 83.4)(30)2
Mu = 69.4 ft-kips Mu = 9.4 ft-kips
8 8 8 8
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, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
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