Solution Manual for Fundamentals of Structural Analysis, 6th Edition – (Leet, 2020) | All 16 Chapters Covered

All 16 Chapters Covered
B1 B1 B1




SOLUTION MANUAL
B1

, P2.1. Determine the deadweight of a 1-ft-
B1 B1 B1 B1 B1 B1
72ʺ

long segment of the prestressed, reinforced
B1 B1 B1 B1 B1 6ʺ
concrete tee-beam whose cross section is
B1 B1 B1 B1 B1 B1
6ʺ

shown in Figure P2.1. Beam is
B1 B1 B1 B1 B1 B1
48ʺ 8ʺ 24ʺ
constructed with lightweight concrete
B1 B1 B1 B1

3
which weighs 120 lbs/ft .
B1 B1 B1 B1
12ʺ

18ʺ
B 1 Secti
on

P2.1



3
Compute the weight/ft. of cross section @ 120 lb/ft .
B1 B1 B1 B1 B1 B1 B1 B1




Compute cross sectional area:
B1 B1 B1


æ 1 ö
Area = (0.5¢´6¢)+ 2 ç ´0.5¢´2.67¢÷ +(0.67¢´2.5¢)+(1.5¢´1¢)
B1 B1



ç2 ÷
B1 B1 B1
B1


è ø
= 7.5 ft2
B1 B1




Weight of member per foot length:
B1 B1 B1 B1 B1




wt/ft = 7.5 ft2 ´120 lb/ft3 =900 lb/ft.
B1 B1 B1 B1 B1 B1




2-2
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, P2.2. Determine the deadweight of a 1-ft-long three B1ply
2ʺ insulation 3/4ʺ B1plywood
B1felt B 1 tar
segment of a typical 20-in-wide unit of a roof B1and B1gravel


supported on a nominal 2 × 16 in. southern pine
beam (the actual dimensions are 12 in. smaller).
2
The 34-in. plywood weighs 3 lb/ft .
1 B11/2ʺ 15 B11/2ʺ



20ʺ 20ʺ
Section

P2.2




See Table 2.1 for
B1 B1 B1




B1 weights wt /20¢¢ unit
B1 B1 B1




B 1


20¢¢
´1¢ = 5 lb
B1

Plywood: 3 psf´ B1 B1 B1 B1 B1


12
20¢¢
´1¢ = 5 lb
Insulation: 3 psf ´
B1

B1 B1 B1 B1 B1 B1

12
20¢¢ 9.17 lb
Roof’g Tar & G: 5.5 psf ´ ´1¢ =
B1 B1 B1 B 1
B1 B1 B1 B1 B1 B1 B1 B1


12 19.17 lb B1


lb (1.5¢¢´15.5)¢¢ ´1¢ = 5.97 lb B1
B1
B1 B1 B1



Wood Joist = B1 B1

ft3 14.4 in2/ ft3 B1 B1

37 B1



Total wt of 20¢¢ unit =19.17 + 5.97
B1 B1 B1 B1 B1 B1 B1 B1




= 25.14 lb. Ans.




2-3
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No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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, P2.3. A wide flange steel beam shown in Figure
P2.3 supports a permanent concrete masonry wall,
8ʺ concrete masonry
floor slab, architectural finishes, mechanical and partition
electrical systems. Determine the uniform dead
9.5ʹ
load in kips per linear foot acting on the beam. concrete floor slab
The wall is 9.5-ft high, non-load bearing and
laterally braced at the top to upper floor framing
(not shown). The wall consists of 8-in. lightweight
reinforced concrete masonry units with an average piping
mechanical
weight of 90 psf. The composite concrete floor slab duct
construction spans over simply supported steel wide flange steel
beams, with a tributary width of 10 ft, and weighs beam with fireproofing
50 psf. ceiling tile and suspension hangers

The estimated uniform dead load for structural Section

steel framing, fireproofing, architectural features, P2.3
floor finish, and ceiling tiles equals 24 psf, and for
mechanical ducting, piping, and electrical systems
equals 6 psf.



Uniform Dead Load WDL Acting on the Wide Flange Beam:
B1 B1 B1 B1 B1 B1 B1 B1 B1




Wall Load:B1



9.5¢(0.09 ksf) = 0.855 klf B1 B1 B1 B1



Floor Slab:
B1 B1



10¢(0.05 ksf) = 0.50 klf B1 B1 B1 B1



Steel Frmg, Fireproof’g, Arch’l Features, Floor Finishes, &
B1 B1 B1 B1 B1 B1 B1



Ceiling: 10¢(0.024 ksf) = 0.24 klf
B1 B1 B1 B1 B1 B1



Mech’l, Piping & Electrical Systems:
B1 B1 B1 B1



10¢(0.006 ksf) = 0.06 klf
B1 B 1 B1 B1 B 1



Total WDL =1.66 klf
B1
B1
B
1 B1




2-4
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No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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