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Fundamentals of Structural Analysis (6th Edition) by William M. C. Leet, C. Hendry, and Kenneth P. Salmon – Complete Solutions Manual for Structural Analysis Problems

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This solutions manual provides detailed, step-by-step solutions to the exercises in Fundamentals of Structural Analysis, 6th Edition by William M. C. Leet, C. Hendry, and Kenneth P. Salmon. It covers fundamental topics such as statically determinate and indeterminate structures, shear and bending moment diagrams, deflection of beams, analysis of trusses and frames, influence lines, and approximate methods of analysis. The guide helps civil and structural engineering students understand problem-solving techniques, mathematical derivations, and practical applications in structural analysis and design. It is designed to support homework practice, exam preparation, and deeper mastery of structural analysis principles.

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All 16 Chapters Covered




SOLUTION MANUAL

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

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

Figure P2.1. Beam is constructed with 48ʺ 8ʺ 24ʺ
3
lightweight concrete which weighs 120 lbs/ft .
12ʺ

18ʺ
Section

P2.1




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




Compute cross sectional area:
æ 1 ö
Area = (0.5¢´6¢)+ 2 ç ´0.5¢´2.67¢÷ +(0.67¢´2.5¢)+(1.5¢´1¢)
ç2 ÷
è ø
= 7.5 ft2
Weight of member per foot length:

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




2-2
Copyright © 2018 McGraw-Hill Education. All rights reserved.
No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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, P2.2. Determine the deadweight of a 1-ft-long three ply felt
2ʺ insulation 3/4ʺ plywood
tar and gravel
segment of a typical 20-in-wide unit of a roof
supported on a nominal 2 × 16 in. southern pine
beam (the actual dimensions are 12 in. smaller).
2
The 3 -in. plywood weighs 3 lb/ft . 1 1/2ʺ 15 1/2ʺ
4




20ʺ 20ʺ
Section

P2.2




See Table 2.1 for weights

wt /20¢¢ unit
20¢¢
Plywood: 3 psf´ ´1¢ = 5 lb
12
20¢¢
Insulation: 3 psf ´ ´1¢ = 5 lb
12
20¢¢ 9.17 lb
Roof’g Tar & G: 5.5 psf ´ ´1¢ =
12 19.17 lb
lb (1.5¢¢´15.5)¢¢ ´1¢ = 5.97 lb
Wood Joist = 37 3
ft 14.4 in2/ ft3
Total wt of 20¢¢ unit = 19.17 + 5.97
= 25.14 lb. Ans.




2-3
Copyright © 2018 McGraw-Hill Education. All rights reserved.
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
weight of 90 psf. The composite concrete floor slab mechanical
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:
Wall Load:
9.5¢(0.09 ksf) = 0.855 klf
Floor Slab:
10¢(0.05 ksf) = 0.50 klf
Steel Frmg, Fireproof’g, Arch’l Features, Floor Finishes, & Ceiling:
10¢(0.024 ksf) = 0.24 klf
Mech’l, Piping & Electrical Systems:
10¢(0.006 ksf) = 0.06 klf
Total WDL = 1.66 klf




2-4
Copyright © 2018 McGraw-Hill Education. All rights reserved.
No reproduction or distribution without the prior written consent of McGraw-Hill Education.
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Connected book
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Kenneth M. Leet, Emeritus, Chia-Ming Uang Loose Leaf for Fundamentals of Structural Analysis
Publisher: 2020 ISBN: 9781260588668 Edition: Unknown

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