1
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 lbft 2 # in.(6 in.)] (15 ft)(10 ft) = 10,800 lb
From Table 1–4
LL = (250 lbft2)(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 lbft 2 # in. (4 in.)] (20 ft)(15 ft) = 9600 lb
From Table 1–4
LL = (50 lbft2)(20 ft)(15 ft) = 15,000 lb
Total Load
F = 24,600 lb = 24.6 k Ans.
1–3. The T-beam is made from concrete having a specific
weight of 150 lbft3. Determine the dead load per foot length
of beam. Neglect the weight of the steel reinforcement.
w = (150 lbft3) [(40 in.)(8 in.) + (18 in.) (10 in.)]
w = 521 lbft Ans.
a 1 ft2
144 in 2 b 26 in.
40 in.
8 in.
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.
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 kNm2 # mm (150 mm)] (7 m) (3 m) = 47.25 kN
From Table 1–4
LL = (6.00 kNm2) (7 m) (3 m) = 126 kN
Total Load
F = 126 kN + 47.25 kN = 173 kN Ans.
12 in.
4 in.
24 in.
6 in.
55°
75°Cross-sectional area = 6(24) + (24 + 7.1950)(12) + (4 + 7.1950)(5.9620)
= 364.54 in2
Use Table 1–2.
w = 144 lbft3 (364.54 in2) = 365 lbft Ans.a 1 ft2
144 in 2 b
a 1
2 ba 1
2 b

3
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 kNm of length of wall that the wall exerts on the floor.
8 in.
8 in.
4 in. 4 in.6 in.
6 in.
6 in.
20 in.
Use Table 1–3.
For studs
Weight = 0.57 kNm2 (2.5 m) = 1.425 kNm
For fiberboard
Weight = 0.04 kNm2 (2.5 m) = 0.1 kNm
For clay brick
Weight = 1.87 kNm2 (2.5 m) = 4.675 kNm
Total weight = 6.20 kNm Ans.
1–6. The prestressed concrete girder is made from plain
stone concrete and four -in. cold form steel reinforcing
rods. Determine the dead weight of the girder per foot of its
length.
3
4
Area of concrete = 48(6) + 4 (14 + 8)(4) - 4() = 462.23 in2
Area of steel = 4() = 1.767 in2
From Table 1–2,
w = (144 lbft3)(462.23 in2) + 492 lbft3(1.767 in2)
= 468 lbft
a 1 ft2
144 in2 ba 1 ft2
144 in2 b
a 3
8 b
2
a 3
8 b
2
d
1
2
c
2.5 m
Ans.

4
1–10. The second floor of a light manufacturing building is
constructed from a 5-in.-thick stone concrete slab with an
added 4-in. cinder concrete fill as shown. If the suspended
ceiling of the first floor consists of metal lath and gypsum
plaster, determine the dead load for design in pounds per
square foot of floor area.
4 in. cinder fill
5 in. concrete slab
ceiling
From Table 1–3,
5-in. concrete slab = (12)(5) = 60.0
4-in. cinder fill = (9)(4) = 36.0
metal lath & plaster = 10.0
Total dead load = 106.0 lbft2 Ans.
1–9. The interior wall of a building is made from 2 4
wood studs, plastered on two sides. If the wall is 12 ft high,
determine the load in lbft of length of wall that it exerts on
the floor.
From Table 1–3
w = (20 lbft2)(12 ft) = 240 lbft Ans.
*1–8. A building wall consists of exterior stud walls with
brick veneer and 13 mm fiberboard on one side. If the wall
is 4 m high, determine the load in kNm that it exerts on the
floor.
For stud wall with brick veneer.
w = (2.30 kNm2)(4 m) = 9.20 kNm
For Fiber board
w = (0.04 kNm2)(4 m) = 0.16 kNm
Total weight = 9.2 + 0.16 = 9.36 kNm Ans.

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Solution Manual For Structural Analysis, 8th Edition

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