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.
Chapter No. 01: Types of Structures and Loads
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.
SOLUTION
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 wall is 15 ft high and consists of 2 × 4 in. studs, plastered
on one side. On the other side there is 4-in. clay brick.
Determine the average load in lb ft of length of wall that
the wall exerts on the floor.
SOLUTION
Using the data tabulated in Table 1.3,
4@in. clay brick: (39 lb>ft2)(15 ft) = 585 lb>ft
2 * 4 in. studs plastered
on one side: (12 lb>ft2) (15 ft) = 180 lb>ft
wD = 765 lb>ft Ans.
1–3.
The second floor of a light manufacturing building is
constructed from a 4-in.-thick stone concrete slab with an
added 3-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.
SOLUTION
Ans.
From Table 1.3, F = 48.3 k
4 in. concrete slab = 4 ( 12 ) = 48 lb ft2
3 in. cinder fill = 3( 9 ) = 27 lb ft2
Ceiling = 10 lb ft2
Total DL = 85 lb ft2 Ans.
, *1–4. The “New Jersey” barrier is commonly used during
highway
*1–4. construction. Determine its weight per foot of length
if it is made from plain stone concrete. 4 in.
The
*1–4.“New TheJersey”
“New barrier
Jersey”is barrier
commonly used duringused
is commonly highway
during
construction. Determine
highway construction. its weightits
Determine per foot of
weight perlength if itlength
foot of is
made
if it is from
madeplain
fromstone
plain concrete.
stone concrete. 4 in.
758
12 in. 558
6 in. 758
12 in. 558
24 in.
6 in.
24 in.
SOLUTION
1 1
Cross@sectional area = 6(24) + a b(24 + 7.1950)(12) + a b(4 + 7.1950)(5.9620)
2 2
SOLUTION
= 364.54 in2
1 1
Use Table 1.2. area = 6(24) + a b(24 + 7.1950)(12) + a b(4 + 7.1950)(5.9620)
Cross@sectional
2 2
l ft2
w = 144 lb>ft3(364.54=in364.54
2
) a in2 2 b = 365 lb>ft Ans.
144 in
Use Table 1.2.
l ft2
w = 144 lb>ft3(364.54 in2) a b = 365 lb>ft Ans.
144 in2
1–5.
The precast floor beam is made from concrete having a
specific weight of 23.6 kN m3 . If it is to be used for a floor 1.5 m
of an office building, calculate its dead and live loadings per
foot length of beam.
0.15 m
0.3 m
0.15 m
SOLUTION
The dead load is caused by the self-weight of the beam.
wD = [(1.5 m)(0.15 m) + (0.15 m)(0.3 m)](23.6 kN>m3)
= 6.372 kN>m = 6.37 kN>m Ans.
.
. 2.39 kN>m2. Thus,
wL = (2.39 kN>m2)(1.5 m) = 3.58 kN>m Ans.
Ans.
w = 365 lb>ft
Ans.
w = 365 lb>ft
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.
Chapter No. 01: Types of Structures and Loads
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.
SOLUTION
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 wall is 15 ft high and consists of 2 × 4 in. studs, plastered
on one side. On the other side there is 4-in. clay brick.
Determine the average load in lb ft of length of wall that
the wall exerts on the floor.
SOLUTION
Using the data tabulated in Table 1.3,
4@in. clay brick: (39 lb>ft2)(15 ft) = 585 lb>ft
2 * 4 in. studs plastered
on one side: (12 lb>ft2) (15 ft) = 180 lb>ft
wD = 765 lb>ft Ans.
1–3.
The second floor of a light manufacturing building is
constructed from a 4-in.-thick stone concrete slab with an
added 3-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.
SOLUTION
Ans.
From Table 1.3, F = 48.3 k
4 in. concrete slab = 4 ( 12 ) = 48 lb ft2
3 in. cinder fill = 3( 9 ) = 27 lb ft2
Ceiling = 10 lb ft2
Total DL = 85 lb ft2 Ans.
, *1–4. The “New Jersey” barrier is commonly used during
highway
*1–4. construction. Determine its weight per foot of length
if it is made from plain stone concrete. 4 in.
The
*1–4.“New TheJersey”
“New barrier
Jersey”is barrier
commonly used duringused
is commonly highway
during
construction. Determine
highway construction. its weightits
Determine per foot of
weight perlength if itlength
foot of is
made
if it is from
madeplain
fromstone
plain concrete.
stone concrete. 4 in.
758
12 in. 558
6 in. 758
12 in. 558
24 in.
6 in.
24 in.
SOLUTION
1 1
Cross@sectional area = 6(24) + a b(24 + 7.1950)(12) + a b(4 + 7.1950)(5.9620)
2 2
SOLUTION
= 364.54 in2
1 1
Use Table 1.2. area = 6(24) + a b(24 + 7.1950)(12) + a b(4 + 7.1950)(5.9620)
Cross@sectional
2 2
l ft2
w = 144 lb>ft3(364.54=in364.54
2
) a in2 2 b = 365 lb>ft Ans.
144 in
Use Table 1.2.
l ft2
w = 144 lb>ft3(364.54 in2) a b = 365 lb>ft Ans.
144 in2
1–5.
The precast floor beam is made from concrete having a
specific weight of 23.6 kN m3 . If it is to be used for a floor 1.5 m
of an office building, calculate its dead and live loadings per
foot length of beam.
0.15 m
0.3 m
0.15 m
SOLUTION
The dead load is caused by the self-weight of the beam.
wD = [(1.5 m)(0.15 m) + (0.15 m)(0.3 m)](23.6 kN>m3)
= 6.372 kN>m = 6.37 kN>m Ans.
.
. 2.39 kN>m2. Thus,
wL = (2.39 kN>m2)(1.5 m) = 3.58 kN>m Ans.
Ans.
w = 365 lb>ft
Ans.
w = 365 lb>ft
