FUNDAMENTALS OF
STRUCTURAL ANALYSIS
5th Edition
Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, and Anne M. Gilbert
SOLUTIONS MANUAL
CHAPTER 2: DESIGN LOADS AND
STRUCTURAL FRAMING
2-1
Copyright © 2018 McGraw-Hill Education. All rights reserved.
No reproduction or distribution without the prior written consent of McGraw-Hill Education.
, 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
Compute the weight/ft. of cross section @ 120 lb/ft3.
Compute cross sectional area:
æ1 ö
Area = (0.5¢ ´6 ¢) + 2 çç ´ 0.5¢ ´ 2.67 ¢÷÷ + (0.67¢ ´ 2.5¢) + (1.5¢ ´1¢)
çè 2 ÷ø
= 7.5 ft 2
Weight of member per foot length:
wt/ft = 7.5 ft 2 ´120 lb/ft 3 = 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.
, P2.2. Determine the deadweight of a 1-ft-long three ply felt
2ʺ insulation tar and gravel 3/4ʺ plywood
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 43 -in. plywood weighs 3 lb/ft . 1 1/2ʺ 15 1/2ʺ
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 in 2 / ft 3
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.
, 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.
STRUCTURAL ANALYSIS
5th Edition
Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, and Anne M. Gilbert
SOLUTIONS MANUAL
CHAPTER 2: DESIGN LOADS AND
STRUCTURAL FRAMING
2-1
Copyright © 2018 McGraw-Hill Education. All rights reserved.
No reproduction or distribution without the prior written consent of McGraw-Hill Education.
, 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
Compute the weight/ft. of cross section @ 120 lb/ft3.
Compute cross sectional area:
æ1 ö
Area = (0.5¢ ´6 ¢) + 2 çç ´ 0.5¢ ´ 2.67 ¢÷÷ + (0.67¢ ´ 2.5¢) + (1.5¢ ´1¢)
çè 2 ÷ø
= 7.5 ft 2
Weight of member per foot length:
wt/ft = 7.5 ft 2 ´120 lb/ft 3 = 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.
, P2.2. Determine the deadweight of a 1-ft-long three ply felt
2ʺ insulation tar and gravel 3/4ʺ plywood
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 43 -in. plywood weighs 3 lb/ft . 1 1/2ʺ 15 1/2ʺ
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 in 2 / ft 3
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.
, 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.
