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 Figure 6ʺ
P2.1. Beam is constructed with lightweight concrete 8ʺ
3 48ʺ 24ʺ
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:
æ1 ö
Area = (0.5 ´6 ) + 2 ´0.5 ´2.67 +(0.67 ´2.5 )+(1.5 ´1 )
¢ ¢ ¢ ¢÷ ¢ ¢ ¢ ¢
ç ÷ø
è2
= 7.5 ft 2
Weight of member per foot length:
wt/ft = 7.5 ft 2 ´120 lb/ft3 =
900 lb/ft.
2-2
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, three ply felt
P2.2. Determine the deadweight of a 1-ft-long 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 1 in. smaller).
2
The 34-in. plywood weighs 3 lb/ft . 2 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 ft3 14.4 in2 / ft3
Total wt of 20¢¢ unit =19.17 + 5.97
=
25.14 lb. Ans.
2-3
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, P2.3. A wide flange steel beam shown in Figure P2.3
supports a permanent concrete masonry wall, floor slab,
architectural finishes, mechanical and electrical 8ʺ concrete masonry
partition
systems. Determine the uniform dead load in kips per 9.5ʹ
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 weight of 90 psf. The piping
mechanical
composite concrete floor slab construction spans over duct
simply supported steel beams, with a tributary width of 10 wide flange steel
beam with fireproofing
ft, and weighs 50 psf.
ceiling tile and suspension hangers
The estimated uniform dead load for structural steel
Section
framing, fireproofing, architectural features, floor finish,
and ceiling tiles equals 24 psf, and for mechanical P2.3
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
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, P2.4. Consider the floor plan shown in Figure P2.4. 1 2 3
6 @ 6.67ʹ = 40ʹ
Compute the tributary areas for (a) floor beam B1, A
C4
(b) floor beam B2, (c) girder G1, B2 G1
B3
(d) girder G2, (e) corner column C1, and (f ) B4 2 @ 10ʹ = 20ʹ
C2
interior column C
B
G4
G3 G2
B1 5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
4 ft 36 ft 4 ft
8 8
40
2
(a) Method 1: A AT 320 ft
T
2 2
1 B1 B1
320 4 4(4)
2
Method 2: A AT 288 ft
T
6.67 ft 6.66 ft 6.67 ft
2
6.67
(b) Method 1: A
T
20 AT 66.7 ft
2
2 1
10 ft 10 ft
AT 55.6 ft
2
T
5 ft 5 ft
Method 2: A 66.7 2 3.33(3.33) Right
Side
2
Left
6.67
(c) Method 1: A
T
20 10(10)
2 Side 6.67 ft 6.66 ft 6.67 ft
AT 166.7 ft G1 G1
2
1
1 5(5)
Method 2: A 166.7 2 3.33(3.33) 2 36 ft 36 ft
T
4 ft 4 ft
2 2
A 180.6 ft
2
T
40 20 G2 G2
(d) Method 1: A 36
T
2 2
2
AT 1080 ft
1
Method 2: A 1080 2 4(4)
T
2 B4
A 1096 ft
2
T
AT,C2
40 20
(e) A ; AT 200 ft
2
T
2 2 AT,C1
40 20 40 20
(f) A ; AT 900 ft
2
T
2 2 2 2
2-5
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, P2.5. Refer to Figure P2.4 for the floor plan. 1 2 3
6 @ 6.67ʹ = 40ʹ
Calculate the tributary areas for (a) floor beam B3, C4
A
(b) floor beam B4, (c) girder G3, (d) girder
G4, (e) edge column C3, and (f ) corner column C4. B2 G1 2 @ 10ʹ = 20ʹ
B4 B3
C2
B G4
G3 G2
B1 5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
5 ft 10 ft 5 ft
10 20
(a) Method 1: A
T
2
AT 200 ft B3 B3
1 2
Method 2: A 200 4 5 6.67 ft 6.66 ft 6.67 ft
T
2
2
AT 150 ft B4 B4
36 ft 36 ft
2
(b) Method 1: A 6.67 20 AT 133.4 ft
4 ft 4 ft
T
1
Method 2: A
2
133.4 4 3.33
T
2
AT 111.2 ft
2
G3 G3
33.33 ft 33.33 ft
36 20 AT 720 ft Right 3.33 ft 3.33 ft
(c) Method 1: A
T 1 2 2
Side
720 2 4 AT 736 ft 2
Method 2: A
T
Left
2 Side
4 ft 36 ft 4 ft
4 40 33.33(10) G4 G4
(d) Method 1: A
T
AT 493.4 ft
2
Method 2: A 493.4 2 14 2 1 3.33 AT,C4
T
2 2
B4
A 488.5 ft
2
T
30 20 ;
(e) A A 600 ft 2
T T
10 10 ; AT,C3
(f) A A 100 ft 2
T T
2-6
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, P2.6. The uniformly distributed live load on the floor 1
6 @ 6.67ʹ = 40ʹ
2 3
2
plan in Figure P2.4 is 60 lb/ft . Establish C4
A
the loading for members (a) floor beam B1,
(b) floor beam B2, (c) girder G1, and (d) girder G2. B2 G1
B3 2 @ 10ʹ = 20ʹ
B4
Consider the live load reduction if permitted by the C2
ASCE standard. B
G4
G3 G2
B1
5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
(a) A = 8(40) = 320 ft , K
2
= 2, A K = 640 > 400 w
ö÷
T LL T LL
æ 15 60
L = 60ç0.25 + = 50.6 psf > , ok
èç 640 ø
÷ 2 B1 and B2
2
(b) A 6.67 (20) = 66.7 ft , K = 2, A K = 133.4 < 400, No Reduction
= w = 8(50.6) = 404.8 lb/ft = 0.40 kips/ft
T LL T LL
2
6.67
w= (60) = 200.1 lb/ft = 0.20 kips/ft
2
6.67
(c) A =
2
(20) +10(10) = 166.7 ft , K = 2, A K = 333.4 < 400, No Reduction
T LL T LL
2
6.67 P
w= (60) = 200.1 lb/ft = 0.20 kips/ft
2 w
q(W )(L ) 60(10)(20)
trib beam
P= = = 6000 lbs = 6 kips
2 2
G1
æ40 20 ö÷
(d) A = ç
2
+
çè 2 ÷36 = 1080 ft , K = 2, A K = 2160 > 400
ø ö÷
T LL T LL
2 60 P P P P
æ 15 = 34.4 > , ok
L = 60ç0.25 +
2160 ø÷
5 spaces @ 8’ each
çè 2
G2
L = 34.4 psf
æ 40 20 ö÷
P = 8(34.4) + = 8256 lbs = 8.26 kips
ççè 2
2
ø÷
2-7
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, P2.7. The uniformly distributed live load on the floor 1
6 @ 6.67ʹ = 40ʹ
2 3
2
plan in Figure P2.4 is 60 lb/ft . Establish C4
A
the loading for members (a) floor beam B3,
(b) floor beam B4, (c) girder G3, and girder G4. B2 G1
B3 2 @ 10ʹ = 20ʹ
B4
Consider the live load reduction if permitted by the C2
ASCE standard. B
G4
G3 G2
B1
5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
2
(a) A = 10(20) = 200 ft , K = 2, A K = 400 > 400
w
ö÷
T LL T LL
æ 15
L = 60ç0.25 + = 60 psf
èç 400 ø÷
w = 10(60) = 600 lb/ft = 0.60 kips/ft B3 and B4
2
(b) A = 6.67(20) = 133.4 ft , K = 2, A K = 266.8 < 400, No Reduction
T LL T LL
w = 6.67(60) = 400.2 lb/ft = 0.40 kips/ft
(c) A = 36(20) = 720 ft 2 , K = 2, A K =1440 > 400 P P P P
T LL T LL
æ 15 ÷ö = 38.7 psf > 60
L = 60ç0.25 + , ok
÷
èç 1440 ø 2
5 spaces @ 8’ each
q(Wtrib )(L beam ) 38.7(8)(40)
P= = = 6192 lbs = 6.19 kips G3
2 2
(d) AT = ç
æ 8 ö÷
÷ 40 + 33.33(10) = 493.3 ft , K = 2, A K 2
= 986.6 > 400
ç
è2 ø ö
LL T LL
15 ÷
æ = 43.7 > 60
L = 60ç0.25 + , ok P P P P P
èç 986.6ø÷ 2 w
w = 43.7(4) = 174.8 lb/ft = 0.17 kips/ft
43.7(6.67(20) 6 spaces @ 6.67’ each
P= =2914.8 lbs =2.91 kips
2 G4
2-8
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, P2.8. The building section associated with the floor
plan in Figure P2.4 is shown in Figure P2.8. Assume
3 @ 12ʹ = 36ʹ
2
a live load of 60 lb/ft on all three floors. Calculate the
axial forces produced by the live load in column C1
in the third and first stories. Consider any live load
reduction if permitted by the ASCE standard.
C3 C1
40ʹ 20ʹ
Building Section
P2.8
æ40 20 ÷öæ 40 20 ö
ç + ÷÷ = 900 ft , K
2
(a) A = ç + = 4, A K = 3600 > 400
T çè 2 ÷ ç
2 øè 2 2ø
LL T LL
æ 15 ö÷ 60
L = 60ç0.25 + = 30 psf = , ok (minimum permitted)
èç 3600 ø÷ 2
P3rd = 900(30) = 27000 lbs = 27 kips
P1st = (3)900(30) = 27000 lbs = 81 kips
B4
AT,C2
AT,C1
PLAN
P3rd
P1st
C2
ELEVATION
2-9
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2-10
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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 Figure 6ʺ
P2.1. Beam is constructed with lightweight concrete 8ʺ
3 48ʺ 24ʺ
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:
æ1 ö
Area = (0.5 ´6 ) + 2 ´0.5 ´2.67 +(0.67 ´2.5 )+(1.5 ´1 )
¢ ¢ ¢ ¢÷ ¢ ¢ ¢ ¢
ç ÷ø
è2
= 7.5 ft 2
Weight of member per foot length:
wt/ft = 7.5 ft 2 ´120 lb/ft3 =
900 lb/ft.
2-2
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, three ply felt
P2.2. Determine the deadweight of a 1-ft-long 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 1 in. smaller).
2
The 34-in. plywood weighs 3 lb/ft . 2 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 ft3 14.4 in2 / ft3
Total wt of 20¢¢ unit =19.17 + 5.97
=
25.14 lb. Ans.
2-3
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, P2.3. A wide flange steel beam shown in Figure P2.3
supports a permanent concrete masonry wall, floor slab,
architectural finishes, mechanical and electrical 8ʺ concrete masonry
partition
systems. Determine the uniform dead load in kips per 9.5ʹ
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 weight of 90 psf. The piping
mechanical
composite concrete floor slab construction spans over duct
simply supported steel beams, with a tributary width of 10 wide flange steel
beam with fireproofing
ft, and weighs 50 psf.
ceiling tile and suspension hangers
The estimated uniform dead load for structural steel
Section
framing, fireproofing, architectural features, floor finish,
and ceiling tiles equals 24 psf, and for mechanical P2.3
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
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, P2.4. Consider the floor plan shown in Figure P2.4. 1 2 3
6 @ 6.67ʹ = 40ʹ
Compute the tributary areas for (a) floor beam B1, A
C4
(b) floor beam B2, (c) girder G1, B2 G1
B3
(d) girder G2, (e) corner column C1, and (f ) B4 2 @ 10ʹ = 20ʹ
C2
interior column C
B
G4
G3 G2
B1 5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
4 ft 36 ft 4 ft
8 8
40
2
(a) Method 1: A AT 320 ft
T
2 2
1 B1 B1
320 4 4(4)
2
Method 2: A AT 288 ft
T
6.67 ft 6.66 ft 6.67 ft
2
6.67
(b) Method 1: A
T
20 AT 66.7 ft
2
2 1
10 ft 10 ft
AT 55.6 ft
2
T
5 ft 5 ft
Method 2: A 66.7 2 3.33(3.33) Right
Side
2
Left
6.67
(c) Method 1: A
T
20 10(10)
2 Side 6.67 ft 6.66 ft 6.67 ft
AT 166.7 ft G1 G1
2
1
1 5(5)
Method 2: A 166.7 2 3.33(3.33) 2 36 ft 36 ft
T
4 ft 4 ft
2 2
A 180.6 ft
2
T
40 20 G2 G2
(d) Method 1: A 36
T
2 2
2
AT 1080 ft
1
Method 2: A 1080 2 4(4)
T
2 B4
A 1096 ft
2
T
AT,C2
40 20
(e) A ; AT 200 ft
2
T
2 2 AT,C1
40 20 40 20
(f) A ; AT 900 ft
2
T
2 2 2 2
2-5
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, P2.5. Refer to Figure P2.4 for the floor plan. 1 2 3
6 @ 6.67ʹ = 40ʹ
Calculate the tributary areas for (a) floor beam B3, C4
A
(b) floor beam B4, (c) girder G3, (d) girder
G4, (e) edge column C3, and (f ) corner column C4. B2 G1 2 @ 10ʹ = 20ʹ
B4 B3
C2
B G4
G3 G2
B1 5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
5 ft 10 ft 5 ft
10 20
(a) Method 1: A
T
2
AT 200 ft B3 B3
1 2
Method 2: A 200 4 5 6.67 ft 6.66 ft 6.67 ft
T
2
2
AT 150 ft B4 B4
36 ft 36 ft
2
(b) Method 1: A 6.67 20 AT 133.4 ft
4 ft 4 ft
T
1
Method 2: A
2
133.4 4 3.33
T
2
AT 111.2 ft
2
G3 G3
33.33 ft 33.33 ft
36 20 AT 720 ft Right 3.33 ft 3.33 ft
(c) Method 1: A
T 1 2 2
Side
720 2 4 AT 736 ft 2
Method 2: A
T
Left
2 Side
4 ft 36 ft 4 ft
4 40 33.33(10) G4 G4
(d) Method 1: A
T
AT 493.4 ft
2
Method 2: A 493.4 2 14 2 1 3.33 AT,C4
T
2 2
B4
A 488.5 ft
2
T
30 20 ;
(e) A A 600 ft 2
T T
10 10 ; AT,C3
(f) A A 100 ft 2
T T
2-6
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, P2.6. The uniformly distributed live load on the floor 1
6 @ 6.67ʹ = 40ʹ
2 3
2
plan in Figure P2.4 is 60 lb/ft . Establish C4
A
the loading for members (a) floor beam B1,
(b) floor beam B2, (c) girder G1, and (d) girder G2. B2 G1
B3 2 @ 10ʹ = 20ʹ
B4
Consider the live load reduction if permitted by the C2
ASCE standard. B
G4
G3 G2
B1
5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
(a) A = 8(40) = 320 ft , K
2
= 2, A K = 640 > 400 w
ö÷
T LL T LL
æ 15 60
L = 60ç0.25 + = 50.6 psf > , ok
èç 640 ø
÷ 2 B1 and B2
2
(b) A 6.67 (20) = 66.7 ft , K = 2, A K = 133.4 < 400, No Reduction
= w = 8(50.6) = 404.8 lb/ft = 0.40 kips/ft
T LL T LL
2
6.67
w= (60) = 200.1 lb/ft = 0.20 kips/ft
2
6.67
(c) A =
2
(20) +10(10) = 166.7 ft , K = 2, A K = 333.4 < 400, No Reduction
T LL T LL
2
6.67 P
w= (60) = 200.1 lb/ft = 0.20 kips/ft
2 w
q(W )(L ) 60(10)(20)
trib beam
P= = = 6000 lbs = 6 kips
2 2
G1
æ40 20 ö÷
(d) A = ç
2
+
çè 2 ÷36 = 1080 ft , K = 2, A K = 2160 > 400
ø ö÷
T LL T LL
2 60 P P P P
æ 15 = 34.4 > , ok
L = 60ç0.25 +
2160 ø÷
5 spaces @ 8’ each
çè 2
G2
L = 34.4 psf
æ 40 20 ö÷
P = 8(34.4) + = 8256 lbs = 8.26 kips
ççè 2
2
ø÷
2-7
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, P2.7. The uniformly distributed live load on the floor 1
6 @ 6.67ʹ = 40ʹ
2 3
2
plan in Figure P2.4 is 60 lb/ft . Establish C4
A
the loading for members (a) floor beam B3,
(b) floor beam B4, (c) girder G3, and girder G4. B2 G1
B3 2 @ 10ʹ = 20ʹ
B4
Consider the live load reduction if permitted by the C2
ASCE standard. B
G4
G3 G2
B1
5 @ 8ʹ = 40ʹ
C
C3 C1
40ʹ 20ʹ
P2.4
2
(a) A = 10(20) = 200 ft , K = 2, A K = 400 > 400
w
ö÷
T LL T LL
æ 15
L = 60ç0.25 + = 60 psf
èç 400 ø÷
w = 10(60) = 600 lb/ft = 0.60 kips/ft B3 and B4
2
(b) A = 6.67(20) = 133.4 ft , K = 2, A K = 266.8 < 400, No Reduction
T LL T LL
w = 6.67(60) = 400.2 lb/ft = 0.40 kips/ft
(c) A = 36(20) = 720 ft 2 , K = 2, A K =1440 > 400 P P P P
T LL T LL
æ 15 ÷ö = 38.7 psf > 60
L = 60ç0.25 + , ok
÷
èç 1440 ø 2
5 spaces @ 8’ each
q(Wtrib )(L beam ) 38.7(8)(40)
P= = = 6192 lbs = 6.19 kips G3
2 2
(d) AT = ç
æ 8 ö÷
÷ 40 + 33.33(10) = 493.3 ft , K = 2, A K 2
= 986.6 > 400
ç
è2 ø ö
LL T LL
15 ÷
æ = 43.7 > 60
L = 60ç0.25 + , ok P P P P P
èç 986.6ø÷ 2 w
w = 43.7(4) = 174.8 lb/ft = 0.17 kips/ft
43.7(6.67(20) 6 spaces @ 6.67’ each
P= =2914.8 lbs =2.91 kips
2 G4
2-8
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, P2.8. The building section associated with the floor
plan in Figure P2.4 is shown in Figure P2.8. Assume
3 @ 12ʹ = 36ʹ
2
a live load of 60 lb/ft on all three floors. Calculate the
axial forces produced by the live load in column C1
in the third and first stories. Consider any live load
reduction if permitted by the ASCE standard.
C3 C1
40ʹ 20ʹ
Building Section
P2.8
æ40 20 ÷öæ 40 20 ö
ç + ÷÷ = 900 ft , K
2
(a) A = ç + = 4, A K = 3600 > 400
T çè 2 ÷ ç
2 øè 2 2ø
LL T LL
æ 15 ö÷ 60
L = 60ç0.25 + = 30 psf = , ok (minimum permitted)
èç 3600 ø÷ 2
P3rd = 900(30) = 27000 lbs = 27 kips
P1st = (3)900(30) = 27000 lbs = 81 kips
B4
AT,C2
AT,C1
PLAN
P3rd
P1st
C2
ELEVATION
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