Exam (elaborations)
Solutions Manual for Structural Analysis 6th Edition By Aslam Kassimali
- Course
- Nursing
- Institution
- Allegany College Of Maryland
Solutions Manual for Structural Analysis 6th Edition By Aslam Kassimali
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SOLUTIONS MANUAL FORSTRUCTURAL ANALYSIS 6TH
EDITION BY ASLAM KASSIMALI
, P2.1. Determine the deadweight of a
1-ft-long segment of the prestressed,
reinforced concrete tee-beam whose
cross section is shown in
Figure P2.1. Beam is constructed with
lightweight concrete which weighs 120
lb/ft3.
P2.1
Compute the weight/ft. of cross section @ 120 lb/ft 3.
Compute cross sectional area:
1
Area (0.5)(6) 2 (0.5)(2.67) (0.67)(2.5) 15(1)
2
7.5ft2
Weight of member per foot length:
wt/ft 7.5 120 900 lb/ft
2-2
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P2.2. A wide flange steel beam shown
in
Figure P2.2 supports a permanent
concrete masonry wall, floor slab,
architectural finishes, mechanical and
electrical systems. Determine the
uniform dead load in kips per linear
foot acting on the beam.
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 weight of 90 psf. The
composite concrete floor slab
construction spans over simply
P2.2
supported steel beams, with a
tributary width of 10 ft, and weighs 50
psf.
The estimated uniform dead load
Uniform Dead Load WDL Acting on the Wide Flange Beam:
Wall Load:
9.5(0.09) 0.855 klf
Floor Slab:
10(0.05) 0.50 klf
Steel Framing, Fireproo昀椀ng, Architectur al Features, Floor Finish, and Ceiling Tiles:
10(0.024) 0.24 klf
Mechanical Ducting, Piping, and Electrical Systems:
10(0.006) 0.06 klf
Total WDL 1.66 klf
2-3
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P2.3. Consider the floor plan shown in
Figure P2.3. Compute the tributary
areas for (a) floor beam B1, (b) floor
beam B2, (c) girder G1,
(d) girder G2, (e) corner column C1,
and
(f ) interior column C2.
P2.3
8 8
(a) Method 1: AT 40 AT 320 ft2
2 2
1
Method 2: AT 320 4 4(4) AT 288 ft2
2
6.67
(b) Method 1: AT 20 AT 66.7 ft
2
2
1
Method 2: AT 66.7 2 3.33(3.33) AT 55.6 ft2
2
6.67
(c) Method 1: AT 20 10(10)
2
AT 166.7 ft2
1 1
Method 2: AT 166.7 2 3.33(3.33) 2 5(5)
2 2
AT 180.6 ft2
40 20
(d) Method 1: AT 36
2 2
AT 1080 ft2
1
Method 2: AT 1080 2 4(4)
2
AT 1096 ft2
40 20
(e) AT ; AT 200 ft
2
2 2
40 20 40 20
(f ) AT ; AT 900 ft2
2 2 2 2
2-4
emmy (clarekemmy@gmail.com)
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