Solution Manual For
Structural Analysis
By: Russell C. Hibbeler
11th Edition
1
,© 2012 Pearson Education, Inc., Upper Saddle River, Nj. All Rights Reserved. This Material Is Protected Under All Copyright Laws As They Currently Exist.
No Portion Of This Material May Be Reproduced, In Any Form Or By Any Means, Without Permission In Writing From The Publisher.
1–1. The Floor Of A Heavy Storage Warehouse Building Is
Made Of 6-In.-Thick Stone Concrete. If The Floor Is A Slab
Having A Length Of 15 Ft And Width Of 10 Ft, Determine The
Resultant Force Caused By The Dead Load And The Live
Load.
From Table 1–3
Dl = [12 Lb/Ft2 . In.(6 In.)] (15 Ft)(10 Ft) = 10,800 Lb
From Table 1–4
Ll = (250 Lb/Ft2)(15 Ft)(10 Ft) = 37,500 Lb
Total Load
F = 48,300 Lb = 48.3 K Ans.
1–2. The Floor Of The Office Building Is Made Of 4-In.-
Thick Lightweight Concrete. If The Office Floor Is A Slab
Having A Length Of 20 Ft And Width Of 15 Ft, Determine
The Resultant Force Caused By The Dead Load And The
Live Load.
From Table 1–3
Dl = [8 Lb/Ft2 . In. (4 In.)] (20 Ft)(15 Ft) = 9600 Lb
From Table 1–4
Ll = (50 Lb/Ft2)(20 Ft)(15 Ft) = 15,000 Lb
Total Load
Ans.
F = 24,600 Lb = 24.6 K
1–3. The T-Beam Is Made From Concrete Having A 40 In.
Specific Weight Of 150 Lb/Ft3. Determine The Dead Load
Per Foot Length Of Beam. Neglect The Weight Of The
Steel Reinforcement. 8 In.
26 In.
W = (150 Lb/Ft ) [(40 In.)(8 In.) + (18 In.) (10 In.)] A B
144 In2
Ans.
W = 521 Lb/Ft
10 In.
2
,*1–4. The “New Jersey” Barrier Is Commonly Used
During Highway Construction. Determine Its Weight Per
Foot Of Length If It Is Made From Plain Stone Concrete. 4 In.
1 1
Cross-Sectional Area = 6(24) + A B(24 + 7.1950)(12) + A B(4 + 7.1950)(5.9620) 75°
2 2
12 In. 55°
= 364.54 In2
6 In.
Use Table 1–2.
3 2 1 Ft2 24 In.
W = 144 Lb/Ft (364.54 In ) A 2B = 365 Lb/Ft Ans.
144 In
1–5. The Floor Of A Light Storage Warehouse Is Made Of
150-Mm-Thick Lightweight Plain Concrete. If The Floor Is A
Slab Having A Length Of 7 M And Width Of 3 M, Determine The
Resultant Force Caused By The Dead Load And The Live Load.
From Table 1–3
Dl = [0.015 Kn/M2 . Mm (150 Mm)] (7 M) (3 M) = 47.25 Kn
From Table 1–4
Ll = (6.00 Kn/M2) (7 M) (3 M) = 126 Kn
Total Load
F = 126 Kn + 47.25 Kn = 173 Kn Ans.
3
, © 2012 Pearson Education, Inc., Upper Saddle River, Nj. All Rights Reserved. This Material Is Protected Under All Copyright Laws As They Currently Exist.
No Portion Of This Material May Be Reproduced, In Any Form Or By Any Means, Without Permission In Writing From The Publisher.
1–6. The Prestressed Concrete Girder Is Made From Plain
Stone Concrete And Four 3-In. Cold Form Steel
Reinforcing 4 8 In.
Rods. Determine The Dead Weight Of The Girder Per Foot Of Its
Length. 6 In.
1 32 20 In.
Area Of Concrete = 48(6) + 4 C (14 + 8)(4)D - 4(Π) A B = 462.23 In2
2 8
3 2 2
6 In.
Area Of Steel = 4(Π) A B = 1.767 In
8 8 In.
From Table 1–2,
3 2 1 Ft2 3 2 1 Ft2 4 In. 6 In. 4 In.
W = (144 Lb/Ft )(462.23 In ) A B + 492 Lb/Ft (1.767 In ) A
B 144 In2 144 In2 Ans.
= 468 Lb/Ft
1–7. The Wall Is 2.5 M High And Consists Of 51 Mm × 102 Mm
Studs Plastered On One Side. On The Other Side Is 13 Mm
Fiberboard, And 102 Mm Clay Brick. Determine The Average
Load In Kn/M Of Length Of Wall That The Wall Exerts On The Floor.
2.5 M
Use Table 1–3.
For Studs
Weight = 0.57 Kn/M2 (2.5 M) = 1.425 Kn/M
For Fiberboard
Weight = 0.04 Kn/M2 (2.5 M) = 0.1 Kn/M
For Clay Brick
Weight = 1.87 Kn/M2 (2.5 M) = 4.675 Kn/M
Total Weight = 6.20 Kn/M Ans.
4
Structural Analysis
By: Russell C. Hibbeler
11th Edition
1
,© 2012 Pearson Education, Inc., Upper Saddle River, Nj. All Rights Reserved. This Material Is Protected Under All Copyright Laws As They Currently Exist.
No Portion Of This Material May Be Reproduced, In Any Form Or By Any Means, Without Permission In Writing From The Publisher.
1–1. The Floor Of A Heavy Storage Warehouse Building Is
Made Of 6-In.-Thick Stone Concrete. If The Floor Is A Slab
Having A Length Of 15 Ft And Width Of 10 Ft, Determine The
Resultant Force Caused By The Dead Load And The Live
Load.
From Table 1–3
Dl = [12 Lb/Ft2 . In.(6 In.)] (15 Ft)(10 Ft) = 10,800 Lb
From Table 1–4
Ll = (250 Lb/Ft2)(15 Ft)(10 Ft) = 37,500 Lb
Total Load
F = 48,300 Lb = 48.3 K Ans.
1–2. The Floor Of The Office Building Is Made Of 4-In.-
Thick Lightweight Concrete. If The Office Floor Is A Slab
Having A Length Of 20 Ft And Width Of 15 Ft, Determine
The Resultant Force Caused By The Dead Load And The
Live Load.
From Table 1–3
Dl = [8 Lb/Ft2 . In. (4 In.)] (20 Ft)(15 Ft) = 9600 Lb
From Table 1–4
Ll = (50 Lb/Ft2)(20 Ft)(15 Ft) = 15,000 Lb
Total Load
Ans.
F = 24,600 Lb = 24.6 K
1–3. The T-Beam Is Made From Concrete Having A 40 In.
Specific Weight Of 150 Lb/Ft3. Determine The Dead Load
Per Foot Length Of Beam. Neglect The Weight Of The
Steel Reinforcement. 8 In.
26 In.
W = (150 Lb/Ft ) [(40 In.)(8 In.) + (18 In.) (10 In.)] A B
144 In2
Ans.
W = 521 Lb/Ft
10 In.
2
,*1–4. The “New Jersey” Barrier Is Commonly Used
During Highway Construction. Determine Its Weight Per
Foot Of Length If It Is Made From Plain Stone Concrete. 4 In.
1 1
Cross-Sectional Area = 6(24) + A B(24 + 7.1950)(12) + A B(4 + 7.1950)(5.9620) 75°
2 2
12 In. 55°
= 364.54 In2
6 In.
Use Table 1–2.
3 2 1 Ft2 24 In.
W = 144 Lb/Ft (364.54 In ) A 2B = 365 Lb/Ft Ans.
144 In
1–5. The Floor Of A Light Storage Warehouse Is Made Of
150-Mm-Thick Lightweight Plain Concrete. If The Floor Is A
Slab Having A Length Of 7 M And Width Of 3 M, Determine The
Resultant Force Caused By The Dead Load And The Live Load.
From Table 1–3
Dl = [0.015 Kn/M2 . Mm (150 Mm)] (7 M) (3 M) = 47.25 Kn
From Table 1–4
Ll = (6.00 Kn/M2) (7 M) (3 M) = 126 Kn
Total Load
F = 126 Kn + 47.25 Kn = 173 Kn Ans.
3
, © 2012 Pearson Education, Inc., Upper Saddle River, Nj. All Rights Reserved. This Material Is Protected Under All Copyright Laws As They Currently Exist.
No Portion Of This Material May Be Reproduced, In Any Form Or By Any Means, Without Permission In Writing From The Publisher.
1–6. The Prestressed Concrete Girder Is Made From Plain
Stone Concrete And Four 3-In. Cold Form Steel
Reinforcing 4 8 In.
Rods. Determine The Dead Weight Of The Girder Per Foot Of Its
Length. 6 In.
1 32 20 In.
Area Of Concrete = 48(6) + 4 C (14 + 8)(4)D - 4(Π) A B = 462.23 In2
2 8
3 2 2
6 In.
Area Of Steel = 4(Π) A B = 1.767 In
8 8 In.
From Table 1–2,
3 2 1 Ft2 3 2 1 Ft2 4 In. 6 In. 4 In.
W = (144 Lb/Ft )(462.23 In ) A B + 492 Lb/Ft (1.767 In ) A
B 144 In2 144 In2 Ans.
= 468 Lb/Ft
1–7. The Wall Is 2.5 M High And Consists Of 51 Mm × 102 Mm
Studs Plastered On One Side. On The Other Side Is 13 Mm
Fiberboard, And 102 Mm Clay Brick. Determine The Average
Load In Kn/M Of Length Of Wall That The Wall Exerts On The Floor.
2.5 M
Use Table 1–3.
For Studs
Weight = 0.57 Kn/M2 (2.5 M) = 1.425 Kn/M
For Fiberboard
Weight = 0.04 Kn/M2 (2.5 M) = 0.1 Kn/M
For Clay Brick
Weight = 1.87 Kn/M2 (2.5 M) = 4.675 Kn/M
Total Weight = 6.20 Kn/M Ans.
4
