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Problems

vary to produce a moment of 320 ft-k. Click OK, and the value of As will change to 3.55. This means that a steel area of 3.55 in.2 is required to produce a moment capacity Mn of 320 ft-k. The Goal Seek feature can be used in a similar manner for most of the spreadsheets provided in this text.

PROBLEMS Cracking Moments For Problems 2.1 to 2.5, determine the cracking moments for the sections shown if fc = 4000 psi and fr = 7.5 fc .

Problem 2.4 6 in.

Problem 2.1 (Ans. 34.8 ft-k)

20 in. 26 in.

18 in. 21 in.

4 #9

4 #8

3 in.

3 in. 3 in.

18 in.

12 in.

Problem 2.2

Problem 2.5 (Ans. 85.3 ft-k) 18 in.

9 in. 21 in. 27 in.

2 #9

9 in.

3 in. 3 #10 3 in.

14 in.

6 in. 6 in. 6 in.

Problem 2.3 (Ans. 31.6 ft-k)

18 in.

30 in. 4 in.

17 in. 24 in. 1 #11 3 in. 6 in.

9 in.

1

For Problems 2.6 and 2.7, calculate the uniform load (in addi- Problem 2.9 Repeat Problem 2.8 if four #6 bars are used. tion to the beam weight) that will cause the sections to begin (Ans. f = 1356 psi, f = 26,494 psi) c s to crack if they are used for 28-ft simple spans. fc = 4000 psi, fr = 7.5 fc , and reinforced concrete weight = 150 lb/ft3 .

Problem 2.10

Problem 2.6

21 in. 24 in.

21 in.

27 in.

8 #9

4 #7 3 in.

3 in. 3 in.

14 in. 18 in.

Problem 2.7 (Ans. 0.343 k/ft)

M = 120 ft-k n=9

Problem 2.11 (Ans. fc = 1258 psi, fs = 14,037 psi in bottom layer, fs = 12,889 psi at steel centroid) 4 in. 18 in. 6 #9

4 in.

24 in.

22 in. 30 in. 3 in. 2 in. 2 in.

3 #9

M = 110 ft-k n=8

12 in.

Transformed-Area Method For Problems 2.8 to 2.14, assume the sections have cracked and use the transformed-area method to compute their flexural stresses for the loads or moments given. Problem 2.8

17 in. 20 in. 4 #8

14 in.

3 in. 14 in.

3 in. M = 60 ft-k n=8

Problem 2.12

17

1 2 in.

1.5 k/ft (including beam weight)

n = 10 20 in.

4 #8 1

2 2 in.

24 ft 12 in.

Problem 2.13 (Ans. fc = 2369 psi, fs = 32,574 psi at the steel centroid, 36,255 psi in the bottom layer)

30k 2 k/ft (including beam weight)

10 ft

28 in.

n=8 32 in.

6 #9

20 ft 30 ft

4 in.

16 in.

Problem 2.16 Compute the resisting moment of the beam of Problem 2.13 if eight #10 bars are used and n = 10, fs = 20,000 psi, and fc = 1125 psi. Use the transformedarea method.

Problem 2.14 5 in. 5 in. 5 in. 4 in.

23 in.

30 in. M = 70 ft-k n=9

Problem 2.17 Using transformed area, what allowable uniform load can this beam support in addition to its own weight for a 28-ft simple span? Concrete weight = 150 lb/ft3 , fs = 24,000 psi, and fc = 1800 psi. (Ans. 2.757 k/ft) 6 in. 8 in. 6 in. 4 in.

4 #9

8 in. 3 in.

32 in.

15 in. 17 in. 5 #10

Problem 2.15 Using the transformed-area method, compute the resisting moment of the beam of Problem 2.10 if fs = 24,000 psi and fc = 1800 psi. (Ans. 258.8 ft-k)

3 in. 20 in.

n=8

For Problems 2.18 to 2.21, determine the flexural stresses in these members using the transformed-area method. Problem 2.18 48 in. 4 in. 14 in. 21 in.

M = 100 ft-k n = 10

3 #9

3 in.

12 in.

Problem 2.19 (Ans. fc = 1374 psi, fs = 32,611 psi at the steel centroid)

3 in. 1

15 2 in. n=8

2 #8

2 #8

28 in.

5 in.

5 in.

2 in. 1

2 2 in.

Problem 2.20

Problem 2.21 (Ans. fc = 1406 psi, fs = 16,886 psi, fs = 36,217 psi) 15 in.

1

2 2 in. 24 in.

2 #8

3 in.

4 #8

6 in. 10 in. 10 in.

M = 90 ft-k n=9

32 in. 1 26 2 in.

4 #9 3 in. 18 in.

M = 320 ft-k n=9

Problem 2.22 Compute the allowable resisting moment of the section shown using transformed area if allowable stresses are fc = 1800 psi, fs = fs = 24,000 psi, and n = 8.

12 in. 1 #10

2 in. 4 in. 2 in.

4 in.

2 #10

16 in.

2 in. 4 in. 2 in.

10 in.

For Problems 2.23 to 2.25, using the transformed-area method, determine the allowable resisting moments of the sections shown. Problem 2.23 (Ans. 140.18 ft-k)

E = 29 × 106 psi, fallow tension or compression = 30,000 psi

1 in. E = 20 × 106 psi, fallow tension or compression = 20,000 psi 8 in.

1 in. 4 in.

tension or compression = 24,000 psi) a l

wood beams dressed dimensions 1

1 4

1

in. × 9 2 in.

6

(Ew = 1.76 × 10 psi, fallow tension or compression = 1875 psi)

Problem 2.25 (Ans. 124.4 ft-k) 1 in. 1-in. × 5-in. steel plate (Es = 29 × 106 psi, fallow tension or compression = 24,000 psi)

1

4

1 in.

1

four wood planks dressed dimensions 1 4 in. × 11 4 in.

11 1 in.

6

(Ew = 1.76 × 10 psi, fallow tension or compression = 1800 psi)

5 in.

Nominal Strength Analysis For Problems 2.26 to 2.29, determine the nominal or theoretical moment capacity Mn of each beam if fy = 60,000 psi and fc = 4000 psi. Problem 2.26

Problem 2.27 (Ans. 688.2 ft-k)

21 in.

25 in.

24 in .

30 in.

3 #8 6 #9

1

2 2 in. 1

16 in.

2 2 in. 3 in. 16 in.

Problem 2.28

Problem 2.29 (Ans. 845.5 ft-k)

24 in. 30 in.

25 in. 28 in.

6 #10 4 #10

3 in.

16 in.

3 in. 3 in.

18 in.

For Problems 2.30 to 2.34, determine the nominal moment capacity Mn for each of the rectangular beams. Problem No.

b (in.)

d (in.)

Bars

fc (ksi)

f y (ksi)

Ans.

2.30

14

21

3 #9

4.0

60

—

2.31

16

27

8 #9

4.0

60

903.6 ft-k

2.32

14

20.5

4 #10

5.0

60

—

2.33

21

28

4 #10

5.0

75

818.3 ft-k

2.34

22

36

6 #11

3.0

60

—

For Problems 2.35 to 2.39, determine Mn if fy = 60,000 psi and fc = 4000 psi. Problem 2.35 (Ans. 704 ft-k)

Problem 2.36 10 in.

24 in.

14 in.

10 in.

4 in.

4 in. 8 in. 33 in.

4 #8

7 in.

26 in.

5 #9

16 in.

24 in. 3 in.

Problem 2.37 Repeat Problem 2.35 if four #11 bars are used. (Ans. 865 ft-k) Problem 2.38 Compute Mn for the beam of Problem 2.36 if six #8 bars are used.

Problem 2.39 (Ans. 763.3 ft-k) 3 in. 3 in. 6 in. 3 in. 3 in.

3 in. 3 in.

24 in.

33 in.

4 #11 3 in. 18 in.

Problem 2.40 Determine the nominal uniform load wn (including beam weight) that will cause a bending moment equal to Mn . fy = 60,000 psi and fc = 4000 psi.

wn k/ft

23 in. 26 in.

3 #9

18 ft

3 in.

14 in.

Problem 2.41 Determine the nominal uniform load wn (including beam weight) that will cause a bending moment equal to Mn . fc = 3000 psi and fy = 60,000 psi. (Ans. 6.77 k/ft)

wn k/ft 23 in.

27 in.

4 #10 4 in.

16 in.

24 ft

Problems Units

in

SI Problem 2.44

For Problems 2.42 to 2.44, determine the cracking moments for the sections shownc if f = 28 MPa and the modulus of

600 mm

rupture is fr = 0.7 fc with fc in MPa.

100 mm

Problem 2.42

500 mm

320 mm 520 mm 600 mm

2 #19

200 mm

3 #19

For Problems 2.45 to 2.47, compute the flexural stresses in the concrete and steel for the beams shown using the transformed-area method. = 109.31 MPa) Problem 2.45 (Ans. f = 7.785 MPa, f

80 mm 350 mm

c

Problem 2.43 (Ans. 46.30 kN-m)

2 #25

4 #29 80 mm

70 mm

300 mm

350 mm

Problem 2.46

20 kN/m (including beam weight)

420 mm 500 m m 4 #36 80 m m

n=9

s

530 mm 600 mm

420 mm 500 mm

8m

80 mm

300 mm

·

M = 130 kN m n=9

Problem 2.47 (Ans. fc = 10.20 MPa, fs = 103.10 MPa, fs = 188.56 MPa)

2 #25

70 mm

·

M = 275 kN m

560 mm

n=8

700 mm 4 #29 70 mm

400 mm

For Problems 2.48 to 2.55, compute Mn values. Problem No.

b (mm)

d (mm)

Bars

fc (MPa)

f y (MPa)

2.48

300

600

3 #36

35

350

—

2.49

320

600

3 #36

28

350

560.5 kN-m

2.50

350

530

3 #25

24

420

—

2.51

370

530

3 #25

24

420

313 kN-m

Problem 2.52

fy = 420 MPa fc' = 24 MPa

460 mm 600 mm 6 #25

70 mm 70 mm 350 mm

Problem 2.53 Repeat Problem 2.48 if four #36 bars are used. (Ans. 734 kN • m)

Ans.

Problem 2.54 1.200 m 100 mm

330 mm

fc' = 28 MPa fy = 350 MPa

500 mm

2 #36 70 mm 250 mm

Problem 2.55 (Ans. 689.7 kN • m) 800 mm 80 mm

fy = 300 MPa fc' = 28 MPa

300 mm 560 mm

6 #36

100 mm 80 mm 350 mm

Problem 2.56 Repeat Problem 2.27 using Chapter 2 spreadsheets. Problem 2.57 Repeat Problem 2.28 using Chapter 2 spreadsheets. (Ans. 561.9 ft-k) Problem 2.58 Prepare a flowchart for the determination of Mn for a rectangular tensilely reinforced concrete beam.