© 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兾ft2 # in.(6 in.)] (15 ft)(10 ft) = 10,800 lb
From Table 1–4
LL = (250 lb兾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兾ft2 # in. (4 in.)] (20 ft)(15 ft) = 9600 lb
From Table 1–4
LL = (50 lb兾ft2)(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 40 in.
weight of 150 lb兾ft3. Determine the dead load per foot length
of beam. Neglect the weight of the steel reinforcement.
8 in.
1 ft2
w = (150 lb兾ft3) [(40 in.)(8 in.) + (18 in.) (10 in.)] a b 26 in.
144 in2
w = 521 lb兾ft Ans.
10 in.
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.
*1–4. 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.
Cross-sectional area = 6(24) + a b (24 + 7.1950)(12) + a b (4 + 7.1950)(5.9620)
1 1
2 2 75°
12 in. 55°
= 364.54 in2
6 in.
Use Table 1–2.
1 ft2
w = 144 lb兾ft3 (364.54 in2) a b = 365 lb兾ft
24 in.
Ans.
144 in2
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兾m2 # mm (150 mm)] (7 m) (3 m) = 47.25 kN
From Table 1–4
LL = (6.00 kN兾m2) (7 m) (3 m) = 126 kN
Total Load
F = 126 kN + 47.25 kN = 173 kN Ans.
2
,© 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 34 -in. cold form steel reinforcing
rods. Determine the dead weight of the girder per foot of its 8 in.
length.
6 in.
3 2
Area of concrete = 48(6) + 4 c (14 + 8)(4) d - 4() a b = 462.23 in2
1
20 in.
2 8
3 2
Area of steel = 4() a b = 1.767 in2 6 in.
8
8 in.
From Table 1–2,
1 ft2 1 ft2
w = (144 lb兾ft3)(462.23 in2) a 2
b + 492 lb兾ft3(1.767 in2) a b 4 in. 6 in.
144 in 144 in2 4 in.
= 468 lb兾ft Ans.
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兾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兾m2 (2.5 m) = 1.425 kN兾m
For fiberboard
Weight = 0.04 kN兾m2 (2.5 m) = 0.1 kN兾m
For clay brick
Weight = 1.87 kN兾m2 (2.5 m) = 4.675 kN兾m
Total weight = 6.20 kN兾m 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.
*1–8. 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 kN兾m that it exerts on the
floor.
For stud wall with brick veneer.
w = (2.30 kN兾m2)(4 m) = 9.20 kN兾m
For Fiber board
w = (0.04 kN兾m2)(4 m) = 0.16 kN兾m
Total weight = 9.2 + 0.16 = 9.36 kN兾m 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 lb兾ft of length of wall that it exerts on
the floor.
From Table 1–3
w = (20 lb兾ft2)(12 ft) = 240 lb兾ft Ans.
1–10. The second floor of a light manufacturing building is 4 in. cinder fill
constructed from a 5-in.-thick stone concrete slab with an 5 in. concrete slab
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.
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 lb兾ft2 Ans.
4
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兾ft2 # in.(6 in.)] (15 ft)(10 ft) = 10,800 lb
From Table 1–4
LL = (250 lb兾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兾ft2 # in. (4 in.)] (20 ft)(15 ft) = 9600 lb
From Table 1–4
LL = (50 lb兾ft2)(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 40 in.
weight of 150 lb兾ft3. Determine the dead load per foot length
of beam. Neglect the weight of the steel reinforcement.
8 in.
1 ft2
w = (150 lb兾ft3) [(40 in.)(8 in.) + (18 in.) (10 in.)] a b 26 in.
144 in2
w = 521 lb兾ft Ans.
10 in.
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.
*1–4. 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.
Cross-sectional area = 6(24) + a b (24 + 7.1950)(12) + a b (4 + 7.1950)(5.9620)
1 1
2 2 75°
12 in. 55°
= 364.54 in2
6 in.
Use Table 1–2.
1 ft2
w = 144 lb兾ft3 (364.54 in2) a b = 365 lb兾ft
24 in.
Ans.
144 in2
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兾m2 # mm (150 mm)] (7 m) (3 m) = 47.25 kN
From Table 1–4
LL = (6.00 kN兾m2) (7 m) (3 m) = 126 kN
Total Load
F = 126 kN + 47.25 kN = 173 kN Ans.
2
,© 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 34 -in. cold form steel reinforcing
rods. Determine the dead weight of the girder per foot of its 8 in.
length.
6 in.
3 2
Area of concrete = 48(6) + 4 c (14 + 8)(4) d - 4() a b = 462.23 in2
1
20 in.
2 8
3 2
Area of steel = 4() a b = 1.767 in2 6 in.
8
8 in.
From Table 1–2,
1 ft2 1 ft2
w = (144 lb兾ft3)(462.23 in2) a 2
b + 492 lb兾ft3(1.767 in2) a b 4 in. 6 in.
144 in 144 in2 4 in.
= 468 lb兾ft Ans.
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兾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兾m2 (2.5 m) = 1.425 kN兾m
For fiberboard
Weight = 0.04 kN兾m2 (2.5 m) = 0.1 kN兾m
For clay brick
Weight = 1.87 kN兾m2 (2.5 m) = 4.675 kN兾m
Total weight = 6.20 kN兾m 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.
*1–8. 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 kN兾m that it exerts on the
floor.
For stud wall with brick veneer.
w = (2.30 kN兾m2)(4 m) = 9.20 kN兾m
For Fiber board
w = (0.04 kN兾m2)(4 m) = 0.16 kN兾m
Total weight = 9.2 + 0.16 = 9.36 kN兾m 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 lb兾ft of length of wall that it exerts on
the floor.
From Table 1–3
w = (20 lb兾ft2)(12 ft) = 240 lb兾ft Ans.
1–10. The second floor of a light manufacturing building is 4 in. cinder fill
constructed from a 5-in.-thick stone concrete slab with an 5 in. concrete slab
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.
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 lb兾ft2 Ans.
4
