Reinforced Concrete by Edward G. Nawy - Solutions Manual

INSTRUCTOR’S SOLUTIONS MANUAL

REINFORCED CONCRETE A FUNDAMENTAL APPROACH SIXTH EDITION

EDWARD G. NAWY

Pearson Education International

Vice President and Editorial Director, ECS: Marcia J. Horton Senior Editor: Holly Stark Associate Editor: Dee Bernhard Editorial Assistant: Jennifer Lonschein/Alicia Lucci Director of Team-Based Project Management: Vince O’Brien Senior Managing Editor: Scott Disanno Art Director: Kenny Beck Cover Designer: Kristine Carney Art Editor: Greg Dulles Manufacturing Manager: Alan Fischer Manufacturing Buyer: Lisa McDowell

This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials.

Copyright © 2009 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. All rights reserved. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Education Ltd., London Pearson Education Singapore, Pte. Ltd. Pearson Education Canada, Inc. Pearson Education–Japan Pearson Education Australia PTY, Limited Pearson Education North Asia, Ltd., Hong Kong Pearson Educación de Mexico, S.A. de C.V. Pearson Education Malaysia, Pte. Ltd. Pearson Education, Upper Saddle River, New Jersey

10 9 8 7 6 5 4 3 2 1 ISBN-13: 978-0-13-136170-6 ISBN-10: 0-13-136170-8

CONTENTS

Please note that there are no solutions for Chapters 1 through 4. Solutions begin with Chapter 5.

Chapter 5

Flexure in Beams, 1–41

Chapter 6

Shear and Diagonal Tension in Beams, 42–82

Chapter 7

Torsion, 83–111

Chapter 8

Serviceability of Beams and One-Way Slabs, 112–143

Chapter 9

Combined Compression and Bending: Columns, 144–205

Chapter 10

Bond Development of Reinforcing Bars, 206–221

Chapter 11

Design of Two-Way Slabs and Plates, 222–262

Chapter 12

Footings, 263–281

Chapter 13

Continuous Reinforced Concrete Structures, 282–312

Chapter 14

Introduction to Prestressed Concrete, 313–329

Chapter 15

LRFD AASHTO Design of Concrete Bridge Structures, 330–368

Chapter 16

Seismic Design of Concrete Structures, 369–395

Chapter 17

Strength Design of Masonry Structures, 396–421

v iii

5.1. For the beam cross-section shown in Fig. 5.33 determine whether the failure of the beam will be initiated by crushing of concrete or yielding of steel. Given: f c¿ 3000 psi 120.7 MPa2 for case 1a2, A s 1 in.2

f c¿ 6000 psi 141.4 MPa2 for case 1b2, A s 6 in.2 fy 60,000 psi 1414 MPa2

Also determine whether the section satisfies ACI Code requirements.

Figure 5.33

Solution: (a) The following information is given: b d dt f c fy As

= = = = = =

8 in. 18 in. 16 in. 3000 psi 60,000 psi 1 in2

section width section depth depth to reinforcement required compression strength steel strength steel area

First, determine the value for 1 using equation 5.9. 1 = 0.85 (2500 < f c 4000 ) Then calculate the depth of the compression block. a =

As f y

0.85 f c b (1)(60,000) = 0.85(3000)(8) = 2.94 in

1 1

Calculate the depth to the neutral axis using 1 and a.

a 1 2.94 = 0.85 = 3.46 in

c =

Then find the ratio of c and dt.

c dt

3.46 16 = 0.216

=

Since this value is less than 0.375, the flexure is tension controlled and the steel yields before the concrete crushes. To determine if the section meets ACI Code requirements, calculate the reinforcement ratio. =

As bd

1 (8)(16) = 0.0078

=

This value must be greater than the larger of

3 3000 60,000 200 60,000

3 f c fy

and

200 . fy

= 0.0027 = 0.0033

Since 0.0078 > 0.0033, the section satisfies the ACI Code. (b) The following information is given: b d dt f c fy As

= = = = = =

8 in. 18 in. 16 in. 6000 psi 60,000 psi 6 in.2

section width section depth depth to reinforcement required compression strength steel strength steel area 2 2

First, determine the value for 1 using equation 5.9. 1 =

f 4000 0.85 0.05 c , (4000 < f c 8000) 1000

6000 4000 0.85 0.05 1000 = 0.75

=

Then calculate the depth of the compression block. a =

As f y

0.85 f c b (6)(60,000) = 0.85(6000)(8) = 8.82 in


a 1 8.82 = 0.75 = 11.76 in

c =


c dt

11.76 16 = 0.735

=

Since this value is greater than 0.6, the flexure is compression controlled and the concrete crushes before the steel yields.

3 3

4

5

6

7

5.4. Design a one-way slab to carry a live load of 100 psf and an external dead load of 50 psf. The slab is simply supported over a span of 12 ft. Given: f c¿ 4000 psi 127.6 MPa2, normal-weight concrete fy 60,000 psi 1414 MPa2

Solution: Design as a 1 ft wide, singly reinforced section.

L 12 12 = = 7.2 in, so try a depth of 8 in. Assume for 20 20 flexure an effective depth d = 7 in. Calculate the self weight.

The minimum depth for deflection is

8 12 150 144 = 100 lb/ft

self weight of a 12 in. strip =

Then calculate the factored load. factored external load wu = 1.2(self weight + dead load) + 1.6(live load) = 1.2(100 + 50) + 1.6(100) = 340 lb/ft wu ( L) 2 factored external moment Mu = 8 (340)(12) 2 = 8 = 6120 ft-lb = 73,440 in-lb The required nominal strength for the slab is M n =

73,440 = 81,600 in-lb . 0.9

Assume a moment arm of 0.90d = 0.9 7 = 6.3 in and calculate the area of steel per 12 in. strip. Mn

=

As

=

As f y (moment arm)

81,600 (60,000)(6.3) = 0.1889 in2

Make sure the area is large enough to meet the minimum reinforcement ratio.

8 4

3 f c

=

3 4000 = 0.0031 60,000

1

=

2

=

200 200 = = 0.0033 60,000 fy

As min

=

(0.0033)(12)(7) = 0.28 in 2

fy

Determine the value for 1. 1 = 0.85 (2500 < f c 4000 ) Calculate the depth of the compression block. As f y

a =

0.85 f c b (0.28)(60,000) = 0.85(4000)(12) = 0.4118 in


a 1 0.4118 = 0.85 = 0.4844 in

c =


c dt

0.4844 7 = 0.0692

=

Verify the strength is sufficient (at least 73,440 in-lb) Mn

=

a As f y d 2

=

0.4118 (0.28)(60,000) 7 2

= 114,141 in-lb

9 5

The strength is sufficient, so accept the design. Now design the shrinkage and temperature reinforcement. The minimum required steel fraction is 0.0018. Area = 0.0018(12)(8) = 0.1728 in2 The maximum spacing is the smaller of 5 times the depth and 18 inches. In this case, 5 times the depth is 5(8) = 40 , so the maximum spacing is 18 in. For a slab with a depth of 8 inches and using #4 bars, the maximum spacing is c-c for the main reinforcement and

0.20 (12) = 8.57 in 0.28

0.20 (12) = 13.89 in c-c (for the shrinkage and temperature 0.1728

reinforcement).

10 6

5.5. Design the simply supported beams shown in Fig. 5.36 as rectangular sections. Given: f c¿ 6000 psi 141.4 MPa2, normal-weight concrete fy 60,000 psi 1414 MPa2

1500

21.9

600 8.7

(a)

7500 33.4

7500 33.4

15,000 (66.7 kN)

(b)

(c)

Figure 5.36

span = 20 ft (a) Distributed dead load (including self weight) = 600 lb/ft. Distributed live load of 1500 lb/ft. (b) Point load at mid span of 15,000 lb (c) Point loads at 5 and 15 ft of 7500 lb Solution: (a) Calculate the factored load. factored external load wu = 1.2WD + 1.6WL = 1.2(600) + 1.6(1500) = 3120 lb/ft wu ( L) 2 factored external moment Mu = 8 (3120)(20) 2 = 8 = 156,000 ft-lb = 1,872,000 in-lb The required nominal strength for the beam is M n = 11 7

1,872,000 = 2,080,000 in-lb . 0.9

Determine the minimum depth from the ACI Code: min h =

L 20(12) = = 15 in. Try h = 18 in., 16 16

b = 0.5h = 9 in., and d = 15 in. Assume c/d = 0.30. Calculate the required steel area. c = (0.30)(15) = 4.5 in. 6000 4000 = 0.75 1 = 0.85 0.05 1000 a = 1c = (0.75)(4.5) = 3.375 in. (0.85) f cba (0.85)(6000)(9)(3.375) = = 2.58 in2 Areq = 60,000 fy Make sure the steel area meets the required minimum reinforcement ratio.

1

=

2

=

As min

=

3 f c 3 6000 = = 0.0039 fy 60,000 200 200 = = 0.0033 60,000 fy (0.0039)(9)(14) = 0.488 in 2

The code is satisfied because 2.58 in2 > 0.488 in2. The area can be provided by 3 #9 bars (the area is 3(1.0) = 3.0 in2). Now check the design using the actual steel area. a =

As f y

0.85 f c b (3)(60,000) = 0.85(6000)(9) = 3.92 in


a 1 3.92 = 0.75 = 5.23 in

c =

Then find the ratio of c and d. c d

5.23 15 = 0.348

=

12 8

0.348 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least 2,080,000 in-lb) Mn

a As f y d 2 3.92 = (3)(60,000)14 2 = 2,347,058 in-lb =

The strength is sufficient, so accept the design. (b) Assume some dimensions in order to calculate the self weight. Determine the minimum depth from the ACI Code: min h =

L 20(12) = = 15 in. Try h = 16 in., 16 16

b = 0.5h = 8 in., and d = 14 in. Assume c/d = 0.30. Calculate the factored external moment. The self weight is

(150)(8)(16) = 133 lb/ft. 144

(1.2)(133)(20) 2 = 8000 ft-lb 8 (1.6)(15,000)(10) = 120,000 ft-lb = 2 = 128,000 ft-lb = 1,536,000 in-lb

Dead load Mu = Live load Mu Mu

The required nominal strength for the beam is M n =

1,536,000 = 1,706,667 in-lb. 0.9

Calculate the required steel area. c = (0.30)(14) = 4.2 in 6000 4000 = 0.75 1 = 0.85 0.05 1000 a = 1c = (0.75)(4.5) = 3.15 in (0.85) f cba (0.85)(6000)(8)(3.15) = = 2.142 in2 Areq = 60 , 000 fy

13 9

Make sure the steel area meets the required minimum reinforcement ratio.

3 f c

=

3 6000 = 0.0039 60,000

1

=

2

=

200 200 = = 0.0033 60,000 fy

As min

=

(0.0039)(8)(14) = 0.434 in2

fy

The code is satisfied because 2.142 in2 > 0.434 in2. The area can be provided by 3 #8 bars (area is 3(0.79) = 2.37 in2). Now check the design using the actual steel area. a = c = c d

=

As f y

=

(2.37)(60,000) = 3.49 in. 0.85(6000)(8)

0.85 f c b 3.49 a = = 4.65 in. 0.75 1 4.65 = 0.332 14

0.332 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least 1,706,667 in-lb) Mn

=

a As f y d 2

3.49 (2.37)(60,000)14 2 = 1,742,995 in-lb =

The strength is sufficient, so accept the design. (c) Assume the same dimensions as in part (b). Calculate the factored external moment. The self weight is

(150)(8)(16) = 133 lb/ft. 144

(1.2)(133)(20) 2 = 8000 ft-lb 8 = (1.6)(7500)(5) = 60,000 ft-lb = 68,000 ft-lb = 816,000 in-lb

Dead load Mu = Live load Mu Mu

The required nominal strength for the beam is M n =

1,536,000 = 906,667 in-lb. 0.9

From part (b) the strength of the beam is 1,742,995 in-lb, so the design is sufficient. 14 10

15

16

17

18

5.7. Compute the stresses in the compression steel, f s¿, for the cross sections shown in Fig. 5.38. Also compute the nominal moment strength for the section in part (b). Given: f c¿ 7000 psi 148.3 MPa2, normal-weight concrete fy 60,000 psi 1414 MPa2

18 (457.2 mm)

10

9 in. (228.6 mm)

30 in. (762 mm)

2.0 50.8

10

2.0 in. (50.8 mm)

15 in. (381 mm)

Figure 5.38

(a) Calculate the steel areas. As As

= (3)(1.27) = 3.81 in2 = (2)(0.31) = 0.62 in2

7000 4000 Calculate 1 = 0.85 0.05 = 0.7. 1000 Calculate the strain assuming the compression steel has yielded. a = c =

s

=

( As As ) f y

=

(3.81 0.62)(60,000) = 3.57 in. 0.85(7000)(9)

0.85 f c b 3.57 a = = 5.11 in. 0.7 1 c d 5.11 2 0.003 = 0.003 = 0.00182 in./in. c 5.11

Check if the compression steel has actually yielded.

y

=

fy Es

=

19 11

60,000 = 0.002 in./in. 29 10 6

0.002 in./in. > 0.00182 in./in. so the compression steel has not yielded. Calculate the strain again using the actual compression steel stress found from the first calculation. f s

=

a = c =

s

=

s E s = (0.00182)(29106) = 52,923 psi As f y As f s (3.81)(60,000) (0.62)(52,923) = = 3.66 in. 0.85(7000)(9) 0.85 f c b 3.66 a = = 5.22 in. 0.7 1 c d 5.22 2 0.003 = 0.003 = 0.00185 in./in. c 5.22

Repeat this calculation until the computed values converge. After several trials, f s = 53,605 psi, a = 3.65 in., c = 5.21 in., and s = 0.00185 in./in. (b) Calculate the steel areas As As

= (4)(1.27) = 5.08 in2 = (2)(0.79) = 1.58 in2

7000 4000 Calculate 1 = 0.85 0.05 = 0.7. 1000 Calculate the strain assuming the compression steel has yielded. a = c =

s

=

( As As ) f y

=

(5.08 1.58)(60,000) = 2.35 in. 0.85(7000)(15)

0.85 f c b 2.35 a = = 3.36 in. 0.7 1 c d 3.36 2.5 0.003 = 0.003 = 0.000769 in./in. c 3.36

Check if the compression steel has actually yielded.

y

=

fy Es

=

60,000 = 0.002 in./in. 29 10 6

0.002 in./in. > 0.000769 in./in. so the compression steel has not yielded. Calculate the strain again using the actual compression steel stress found from the first calculation.

20 12

f s

=

a = c =

s

=

s E s = (0.000769)(29106) = 22,294 psi As f y As f s (5.08)(60,000) (1.58)(22,294) = = 3.02 in. 0.85(7000)(15) 0.85 f c b 3.02 a = = 4.31 in. 0.7 1 c d 4.31 2.5 0.003 = 0.003 = 0.00126 in./in. c 4.31

Repeat this calculation until the computed values converge. After several trials, f s = 33,149 psi, a = 2.83 in., c = 4.04 in., and s = 0.00114 in./in. To compute the nominal strength, first verify that the section is tension controlled. d c 27.5 4.04 t = 0.003 t = 0.003 = 0.0174 in./in. 4.04 c

The strain 0.0174 in./in. > 0.005 in./in. so the section is tension controlled and = 0.9. Now verify that the section meets code by checking the minimum reinforcement ratio.

=

1

=

2

=

As 5.08 = 0.0123 = bd t (15)(27.5)

3 f c fy

=

3 7000 = 0.0042 60,000

200 200 = = 0.0033 60,000 fy

0.0123 > max(0.0042,0.0033) so the ACI Code is satisfied. Calculate the nominal strength of the section.

a ( As f y As f s) d + As f s(d d ) 2 2.83 + (1.58)(33,149)(27.5 2.5) = [(5.08)(60,000) (1.58)(33,149)] 27.5 2 = 7,894,097 in-lb M = Mn = (0.9)(7,894,097) = 7,104,867 in-lb

Mn

=

21 13

22

23

24

25

26

27

28

5.10. At failure, determine whether the precast sections shown in Fig. 5.39 will act similarly to rectangular sections or as flanged sections. Given: f c¿ 4000 psi 127.6 MPa2, normal-weight concrete fy 60,000 psi 1414 MPa2 2 in. (50.8 mm)

8

8

30 in. (763 mm)

508

12 in. (304.8 mm)

20

8 in. (203.2 mm)

20

4 in. (101.6 mm)

508

9

Figure 5.39

(a) Calculate the compression block dimensions assuming the sections behaves as a rectangular section with a width equal to the flange width. 1 = 0.85 As = (3)(0.79) = 2.37 in2 As f y (2.37)(60,000) = = 2.09 in. a = 0.85(4000)(20) 0.85 f c b 2.09 a = = 2.46 in. c = 0.85 1 Since both a and c are greater than the flange thickness, the section must be treated as a T-section. (b) Calculate the compression block dimensions assuming the sections behaves as a rectangular section with a width equal to the flange width.

29 14

1 = 0.85 As = (5)(0.79) = 3.95 in2 As f y (3.95)(60,000) = = 2.32 in. a = 0.85(4000)(30) 0.85 f c b 2.32 a = = 2.73 in. c = 0.85 1 Since both a and c are less than the flange thickness, the section can be treated as a rectangular section. (c) Calculate the compression block dimensions assuming the sections behaves as a rectangular section with a width equal to the flange width. 1 = 0.85 As = (4)(1) = 4 in2 As f y (4)(60,000) = = 3.53 in. a = 0.85(4000)(20) 0.85 f c b 3.53 a = = 4.15 in. c = 0.85 1 Since a is less than the flange thickness and c is greater than the flange thickness, the section can be treated as either a rectangular section or an L-section.

30 15

31

32

33

34

35

36

37

38

39

40

41

6.1. A simply supported beam has a clear span ln = 20 ft (6.10 m) and is subjected to an external uniform service dead load WD = 1000 lb per ft (14.6 kN/m) and live load wL = 1500 lb per ft (21.9 kN/m). Determine the maximum factored vertical shear Vu at the critical section. Determine the nominal shear resistance Vc by both the short method and by the more refined method of taking the contribution of the flexural steel into account. Design the size and spacing of the diagonal tension reinforcement. Given: bw 12 in. 1305 mm2

d 18 in. 1457.2 mm2 h 20 in. 1508 mm2

A s 6.0 in.2 13780 mm2 2

f¿c 4000 psi 127.6 MPa2, normal-weight concrete

fy 60,000 psi 1413.7 MPa2 Assume that no torsion exists.

Calculate the factored loads. self-weight = wu = Vu at support = Vu at d = Mu at d = =

150(12)(20) = 250 lb/ft 144 1.2(250 + 1000) + 1.6(1500) = 3900 lb/ft (3900)(20) = 39,000 lb 2 39,000 3900 (18 / 12) = 33,150 lb (3900)(18 / 12) 2 39,000(18 / 12) = 54,112.5 ft-lb 2 649,350 in.-lb

Use the simplified method to calculate the shear loads. Vc =

2 f cbd = 2(1.0) 4000 (12)(18) = 27,322 lb

Vn =

Vu 33,150 = = 44,200 lb 0.75

Since Vn > Vc / 2, shear reinforcement is necessary. The required shear support is Vs = Vn Vc = 44,200 27,322 = 16,878 lb Choose a size for the stirrup steel. Try No. 3 bars, Av = 2(0.11) = 0.22 in2. Calculate the minimum spacing based on the steel strength. s =

AV f y d Vs

=

(0.22)(60,000)(18) = 14.08 in. 16,878

Since Vs < 4 f cbd = 4 4000(12)(18) = 54,644, use d/2. The minimum required spacing is the minimum of d/2 = 9 in. and 14.08 in. Therefore, use No. 3 stirrups at 9 in. c-c spacing. 42 1

43

44

45

46

47

48

49

6.5. A continuous beam has two equal spans ln = 20 ft (6.10 m) and is subjected to an external service dead load wD of 400 lb per ft. (5.8 kN/m) and a service live load wL of 800 lb per ft. (11.7 kN/m). In addition, an external service concentrated dead load PD of 25,000 lb and an external service concentrated live load PL of 30,000 lb (133 kN) are applied to one midspan only. Design the diagonal tension reinforcement necessary. Given: f¿c 5000 psi 134.5 MPa2, normal-weight concrete

fy 60,000 psi 1413.7 MPa2

L 18(12) = = 13.5 in., which is quite small. 16 16 Assume beam dimensions of h = 24 in., d = 21 in., and b = 12 in.

The minimum thickness required for deflection is

Now calculate the factored loads. 150(12)(24) = 300 lb/ft 144 = 1.2(300 + 400) + 1.6(800) = 2120 lb/ft = 1.2(25,000) + 1.6(30,000) = 78,000 lb

self-weight = wu Pu

Then analyze the beam to find the shear and moment distribution. Split the loading into two separate problems. wu = 2120 lb/ft

Combine the two beams.

50 2

Design the longitudinal support for the largest positive moment, M = 4,438,500 in.-lb. Try using 4 #10 bars at the bottom, As = (4)(1.27) = 5.08 in2. a = c = c d

=

As f y

(5.08)(60,000) = 5.98 in. 0.85(5000)(12)

=

0.85 f c b 5.98 a = = 7.47 in. 0.8 1 7.47 = 0.3557 < 0.375 tension controlled, = 0.90 21

Check the minimum reinforcement area,

3 f c 200 As , and the strength, , > max fy bd f y

a M n = As f y d . 2

5.08 = 0.0202 (12)(21) 5.98 0.9(5.08)(60,000) 21 2

>

3 5000 200 max = 0.0035, = 0.0033 60,000 60,000

= 4,940,987 in.-lb > 4,438,500 in.-lb

Design the longitudinal support for the largest negative moment, M = 3,027,000 in.-lb. Try using 3 #10 bars at the top, As = (3)(1.27) = 3.81 in2. a = c = c d

=

As f y

(3.81)(60,000) = 4.48 in. 0.85(5000)(12)

=

0.85 f c b 4.48 a = = 5.60 in. 0.8 1 5.60 = 0.2668 < 0.375 tension controlled, = 0.90 21

Check the minimum reinforcement area,

3 f c 200 As , and the strength, , > max fy bd f y

a M n = As f y d . 2

3.81 = 0.0151 (12)(21)

>

51 3

3 5000 200 max = 0.0035, = 0.0033 60,000 60,000

4.48 0.9(3.81)(60,000) 21 2

= 3,859,440 in.-lb > 3,027,000 in.-lb

So use 4 #10 bars at the bottom and 3 #10 bars at the top of the section. Now design the shear or diagonal reinforcement. Check to see if stirrups are necessary. Use the simplified method to determine the shear capacity of the beam. Vc =

2 f cbd = 2(1.0) 5000 (12)(21) = 35,638 lb

Determine the minimum stirrup spacing at three points: the left end, at the point load, and at the center support. Choose a size for the stirrup steel and then determine the required spacing. Try No. 3 bars, Av = 2(0.11) = 0.22 in2. At the left end, choose a section a distance d from the end. The shear loading at that point is 21 47,587.5 2120 = 43,877.5 lbf. 12

Vu 43,877.5 = = 58,503 lb (Vn > Vc / 2, support required) 0.75 Vs = Vn Vc = 58,503 35,638 = 22,865 lb AV f y d (0.22)(60,000)(21) = = 12.12 in. s = 22,865 Vs

Vn =

Since Vs < 4 f cbd = 4 5000 (12)(21) = 71,276 lb, use d/2. The minimum required spacing is the minimum of d/2 = 10.5 in. and 12.12 in. Therefore, use No. 3 stirrups at 10.5 in. c-c spacing. Then determine the minimum spacing under the point load. The shear load is 51,612.5 lb.


Vn =

The minimum required spacing is the minimum of d/2 = 10.5 in. and 8.35 in. Therefore, use No. 3 stirrups at 8.35 in. c-c spacing. Finally, determine the minimum spacing at the center support. The shear load is 72,812,5 lb.

52 4


Vn =

The minimum required spacing is the minimum of d/2 = 10.5 in. and 4.51 in. Therefore, use No. 3 stirrups at 4.51 in. c-c spacing.

53 5

6.6. Design the vertical stirrups for a beam having the shear diagram shown in Figure 6.37 assuming that Vc = 2 2f c¿ bwd. Given: bw 12 in. 1304.8 mm2 dw 20 in. 1508 mm2

Vu1 80,000 lb 1356 kN2 Vu2 50,000 lb 1222 kN2 Vu3 30,000 lb 1133 kN2


fy 60,000 psi 1414 MPa2

Figure 6.37

Use the simplified method to determine the shear capacity of the beam. Vc =

2 f cbd = 2(1.0) 5000 (12)(20) = 33,941 lb

Try No. 4 bars for the stirrups, Av = 2(0.2) = 0.4 in2. Determine the spacing for Vu1. The shear loading at that point is 80,000 lbf.

Vu 80,000 = = 106,667 lb (Vn > Vc / 2, support required) 0.75 Vs = Vn Vc = 106,667 33,941 = 72,726 lb AV f y d (0.4)(60,000)(20) = = 6.60 in. s = 72,726 Vs

Vn =

Since Vs > 4 f cbd = 4 5000 (12)(20) = 67,882 lb, use d/4. The minimum required spacing is the minimum of d/4 = 5 in and 6.60 in. Therefore, use No. 4 stirrups at 5 in. c-c spacing. Then determine the spacing for Vu2. The shear load is 50,000 lb.

Vu 50,000 = = 66,667 lb (Vn > Vc / 2, support required) 0.75 Vs = Vn Vc = 66,667 33,941 = 32,726 lb AV f y d (0.4)(60,000)(20) = = 14.67 in. s = 32,726 Vs

Vn =

54 6

Since Vs < 67,882 lb, use d/2. The minimum required spacing is the minimum of d/2 = 10 in. and 14.67 in. Therefore, use No. 4 stirrups at 10 in. c-c spacing. Finally, determine the spacing for Vu3. The shear load is 30,000 lb.

Vu 30,000 = = 40,000 lb (Vn > Vc / 2, support required) 0.75 Vs = Vn Vc = 40,000 33,941 = 6059 lb AV f y d (0.4)(60,000)(20) = = 79.2 in. s = 6059 Vs

Vn =

Since Vs < 67,882 lb, use d/2. The minimum required spacing is the minimum of d/2 = 10 in. and 79.2 in. Therefore, use No. 4 stirrups at 10 in. c-c spacing.

55 7

56

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

72

73

74

6.10. Design a bracket to support a concentrated factored load Vu = 100,000 lb (444.8 kN) acting at a lever arm a = 6 in. (152.4 mm) from the column face; horizontal factored force Nuc = 25,000 lb (111 kN). Given: b 24 in. 1609.6 mm2


fy fyt 60,000 psi 1414 MPa2

Column size = 12 24 in. (305 609.6 mm); Corbel width = 24 in. Use both the shear-friction approach and the strut-and-tie method in your solution. Assume that the bracket was cast after the supporting column cured and that the column surface at the bracket location was not roughened before casting the bracket. Detail the reinforcing arrangements for the bracket.

Shear friction approach: First, choose dimensions for the corbel. Try h = 18 in. and d = 15 in. Then check the vertical load.

Vu 100,000 = = 133,333 lb 0.75 = 0.2 f c bd = 0.2(5000)(24)(15) = 360,000 lb = (480 + 0.08 f c) bd = (480 + 0.085000)(24)(15) = 316,800 lb = 1600bd = (1600)(24)(15) = 576,000 lb

Vn = V1 V2 V3

Vn is less than all three values, so the design is okay so far. For a corbel cast on an unroughened hardened column, the value for - is 0.6 = 0.6. Calculate the required areas for the steel reinforcement. Avf = Af =

Vu 133,333 = = 3.70 in2. (60,000)(0.6) fy Vu a + N uc (h d ) f y (0.85d )

(133,333)(6) + (25,000)(18 15) 0.75(60,000)(0.85)(15) = 1.18 in2. N uc 25,000 = = 0.56 in2. = (0.75)(60,000) fy =

An

The primary tension steel area is the maximum of

f 2 Avf + An , A f + An , and 0.04 c bd . 3 fy

Asc2 =

2 2 Avf + An = (3.70) + 0.56 = 3.02 in2. 3 3 A f + An = 1.18 + 0.56 = 1.73 in2.

Asc3 =

0.04

Asc1 =

f c 5000 (24)(15) = 1.2 in2. bd = 0.04 60,000 fy 75 8

Therefore, the required steel areas are Asc = 3.02 in2. and Ah = 0.5( Asc An ) = 1.23 in2. Select the bar sizes. For the primary steel, use 4 No. 8 bars. For the shear reinforcement, use 4 No. 5 bars spaced 2.5 in. c-c. Also use 4 No. 5 framing bars and 1 No. 5 anchor bar to complete the cage.

Vu 100,000 = 33.6 in2. so use a square 6 in. = 0.7(0.85) f c 0.7(0.85)(5000) by 6 in. plate thick enough to be rigid under the load. The required bearing plate area is

Solution using the strut and tie approach:

Assume the following dimensions: h = 18 in. d = 15 in. Points C and D inset 2 in. from the surface Use statics to calculate the forces. LBC = 113,333 lb d x BC + N uc = 78,333 lb = FBC LBC L AC = 166,458 lb = T AB x AC d = FAC L AC

FBC = Vu

T AB FAC

T AD

FCE

=

FAC

TCD

=

FAC

d d = 246,875 lb + FBC L AC LBC x AC x FBC BC = 25,000 lb L AC LBC

Vu 100,000 = 31.4 in2. so use a square = 0.75(0.85) f c 0.7(0.85)(5000) plate that is 6 in.6 in. and thick enough to be rigid under the load. The required bearing plate area is

Use the tie loads with a load factor of = 0.75 to design the steel reinforcement cage. The primary tension steel must support the tension T AB . 76

9

AAB =

T AB 78,333 = 1.74 in2. = 0.75 f y 0.75(60,000)

ACD =

TCD 25,000 = 0.56 in2. = 0.75 f y 0.75(60,000)

A portion of the reinforcement resists the internal friction force. An =

N uc 25,000 = 0.56 in2. = 0.75 f y 0.75(60,000)

Therefore, the required steel areas are AAB = 1.74 in2., ACD = 0.56 in2., and Ah = 0.5( AAB An ) = 0.59 in 2 . Select the bar sizes. Use 3 No. 7 bars for the top steel, Area = 3(0.6) = 1.8 in2. Use 3 No. 4 bars for the bottom steel, Area = 3(0.2) = 0.6 in2. Use 3 No. 4 closed stirrups for the reinforcement, Area = 3(0.2) = 0.6 in2. The bars should be spaced to fill (2/3)(15) = 10 in., so space them 3 in. c-c. Check the shear reinforcement for struts AB and AC.

Ah / tie 2(0.2) d = 0.002279 < 0.003 so increase the bar size to No. 4 and sin = bs (24)(3.0) L AC use 3 No. 4 closed stirrups spaced 3 in. c-c.

Strut AC:

2(0.31) d = 0.003799 > 0.003, OK (24)(3.0) L AC In addition, use 3 No. 4 framing bars and 1 No. 4 anchor bar to complete the cage. Finally, the strut portions need to be checked for adequate strength. The allowable concrete strength in the nodal zones is fce = 0.85(0.8)(5000) = 3400 psi. So the required widths for the struts are Strut AC:

166,458 = 2.04 in. (24)(3400)

246,875 = 3.02 in. (24)(3400) 113,333 Strut BC: = 1.39 in. (24)(3400)

Strut CE:

They are all within the available space in the corbel, so accept the design. 77 10

78

79

80

81

82

83

84

7.2. A cantilever beam is subjected to a concentrated service live load of 30,000 lb (133.5 kN) acting at a distance of 3 ft 6 in. (1.07 m) from the wall support. In addition, the beam has to resist an equilibrium factored torsion Tu = 450,000 in.-lb (50.8 kN–m). The beam cross section is 15 in. × 30 in. (381 mm × 762 mm) with an effective depth of 27 in. (686 mm). Design the stirrups and the additional longitudinal steel needed. Given: f c¿ 4000 psi fy fyt 60,000 psi

A s 4.0 in.2 12580.64 mm2 2

The length of the beam is not given, so ignore the self weight. Use the following beam dimensions and loads. Beam height Effective depth Beam width Force position Clear cover Factored force Factored moment Factored torsion

h d b L

= = = = = Vu = Mn = Tu =

30 in. 27 in. 15 in. 3.5 ft 1.5 in. 1.6 PL = 1.6(30,000) = 48,000 lb 1.6 PL L = 1.6(30,000)(3.512) = 2,016,000 in-lb 450,000 in-lb

Determine the minimum steel required for shear support. The load factor for shear is = 0.75 , and = 1.0 for normal weight concrete. Nominal load Concrete support Required support Required area

Vu 48,000 = = 64,000 lb 0.75 Vc = 2 f cbd = 2(1.0) 4000 (15)(27) = 51,229 lb

Vn =

Vs = Vn Vc = 64,000 51,229 = 12,771 lb Vs 12,771 A = = 0.00788 in.2 / in. = (60,000)(27) f yt d s

Determine the minimum steel required for torsion. The load factor for torsion is also = 0.75 . Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions x1 and y1 . Nominal load Outer box

Tu 450,000 = = 600,000 in.-lb 0.75 x0 = b = 15 in. y 0 = h = 30 in. Acp = x0 y 0 = (15)(30) = 450 in.2 p cp = 2( x0 + y 0 ) = 2(15 + 30) = 90 in. Tn =

85 1

x1 y1 Aoh ph A0

Inner box

Torsion area

= = = = =

x0 2(clear cover) (stirrup d) = 15 2(1.5) 0.5 = 11.5 in. y 0 2(clear cover) (stirrup d) = 30 2(1.5) – 0.5 = 26.5 in.

x1 y1 = (11.5)(26.5) = 304.75 in.2 2( x1 + y1 ) = 2(11.5 + 26.5) = 76 in. 0.85 Aoh = 0.85(304.75) = 259.04 in.2

= 45 degrees

Strut angle

Check that torsion support is required for Tu = 450,000 in.-lb.

Tu <

No support if

f c Acp2 p cp

=

0.75(1.0) 4000 (450) 2 = 106,727 in.-lb (false) 90 2

Vu Tu p h + 2 bd 1.7 Aoh

Then verify the section meets the requirement

2

2

2

2

V < c + 8 f c . bd 2

48,000 (450,000)(76) Vu Tu p h = = 246.92 psi + + 2 2 ( 15 )( 27 ) 1 . 7 ( 304 . 75 ) bd 1.7 Aoh 51,229 V + 8(1.0) 4000 = 474.34 psi c + 8 f c = 0.75 bd (15)(27)

246.92 psi < 474.34 psi, so the section is adequate. Determine the minimum required area of transverse steel. Torsion Torsion and shear Minimum

Tn 600,000 At = = 0.0193 in.2 / in. = 2(259.04)(60,000)(1) 2 A0 f yt cot s A A A = 2 t + = 2(0.0193) + 0.00788 = 0.0465 in.2 / in. (controls) s s s 0.75 f c 0.75 4000 A (15) = 0.0119 in. b = = 60,000 s fy

Determine the minimum required longitudinal reinforcement.

Minimum

f yt At (cot ) 2 = (0.0193)(76)(1)(1) 2 = 1.47 in.2 ph s fy 15 b = 25 = 0.0063 in. (does not control) 25 60,000 f yt

AL =

Torsion At s

At s

86 2

Minimum

AL

5 f c Acp fy

f yt At ph fy s

5(1.0) 4000 (450) (0.0193)(76)(1) = 0.9 in.2 60,000

Therefore, the required transverse steel area is 0.0465 in.2 /in., the specified flexural steel is 6.0 in.2, and the required longitudinal reinforcement is 1.47 in.2. Assume that one quarter of the reinforcement is added to the bottom steel, one quarter goes in the top corners of the stirrups, and the remaining steel is distributed along the vertical sides of the stirrups. Choose the bar sizes and spacings. For the transverse steel, try #4 closed stirrups. The bar area is 0.2 in.2, so the required spacing is p 2(0.2) 76 the smallest of h = = 4.3 in. (controls) = 9.5 in. , 12 in., and 0.0465 8 8 Therefore, use #4 closed stirrups spaced 4 in. c-c. For the bottom steel, the required area is 4 + 0.25(1.47) = 4.37 in.2 Use 5 #9 bars (1 in.2 per bar) for a total area of 4(1) = 5 in.2 For the top steel, the required area is 0.25(1.47) = 0.37 in.2, so use 2 #4 bars in the corners for an area of 2(0.2) = 0.4 in.2 For the sides, each side requires 0.25(1.47) = 0.37 in.2 of steel distributed in 25 in. Use 2 #4 bars spaced equally for an area of 2(0.2) = 0.4 in.2 per side. The final design is shown below.

87 3

88

89

90

91

92

93

94

95

96

97

98

99

100

101

7.5. Design the rectangular beam shown in Figure 7.34 for bending, shear, and torsion. Assume that the beam width b = 12 in. (305 mm). Given: f c¿ 4000 psi 127.6 MPa2

fy fyt 60,000 psi 1413.8 MPa2

10,000

12

ft (

44.5

3.6

6m

)

4 ft (1.22 m)

Figure 7.34

To include the self-weight, a height must be assumed. Try the following dimensions: Beam height Effective depth Beam width Beam length “L” length Clear cover

h d b L L2

= = = = = =

24 in. 21 in. 12 in. 12 ft 2 ft 1.5 in.

Start by calculating the support force and moment. Self weights

w2 = Factored force

Factored moment

FA = = = MA = = =

Factored torsion

(12)(24) (150) = 300 lb / ft 144 (2)(24) (150) = 600 lb / ft L2 h 150 = 12 1.2 w1 ( L b) + 1.2 w2 b + 1.6 PL 1.2(300)(12 1) + 1.2(600)(1) + 1.6(10,000) 20,680 lb 1.2 w1 ( L b) 2 b b + 1.2 w2 b L + 1.6 PL L 2 2 2 1.2(300)(11) 2 + 1.2(600)(1)(11.5) + 1.6(10,000)(11.5) 2 214,060 ft-lb = 2,568,720 in.-lb 1.2 w2 bL2 + 1.6 PL L2 2 1.2(600)(1)(2) + 1.6(10,000)(2) 2 32,720 ft-lb = 392,640 in.-lb

w1 = bh 150 =

TA = = =

102 4

Then calculate the design loads. Factored shear Factored moment Factored torsion

Vu = FA 1.2 w1 d = 20,680 1.2(300)(21/12) = 20,050 lb M n = M A = 2,568,720 in.-lb Tu = T A = 392,640 in.-lb

Now design the beam for bending. Determine the minimum required area of steel, = 0.9 for bending.

Mu 2,568,720 = = 2,854,133 in.-lb 0.9 As f y a As f y d and a = 2 0.85 f cb

Nominal load

Mn =

Section strength

Mn

Solve for As

As

Minimum area 1 Minimum area 2

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2

0.85(4000)(12)(21) 2(2,854,133) 1 1 60,000 0.85(4000)(12)(21) 2

2.48 in.2 (controls) 3 f c 3 4000 As (12)(21) = 0.8 in.2 bd = 60,000 fy As

200 200 (12)(21) = 0.84 in.2 bd = 60,000 fy

Determine the minimum steel required for shear support. The load factor for shear is = 0.75 , and = 1.0 for normal weight concrete. Nominal load Concrete support

Vu 20,050 = = 26,733 lb 0.75 Vc = 2 f cbd = 2(1.0) 4000 (12)(21) = 31,876 lb

Vn =

Since Vc > Vn , the stirrup size will be determined by the torsion design. Determine the minimum steel required for torsion. The load factor for torsion is also = 0.75 . Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions x1 and y1 . Nominal load

Tn =

Tu 392,640 = = 523,520 in.-lb 0.75

103 5

x0 = b = 12 in. y 0 = h = 24 in. Acp = x0 y 0 = (12)(24) = 288 in.2

Outer box

p cp = 2( x0 + y 0 ) = 2(12 + 24) = 72 in. x1 = x0 2(clear cover) (stirrup d) = 12 2(1.5) 0.5 = 8.5 in. y1 = y 0 2(clear cover) (stirrup d) = 24 2(1.5) – 0.5 = 20.5 in.

Inner box

Aoh = x1 y1 = (8.5)(20.5) = 174.25 in.2 p h = 2( x1 + y1 ) = 2(8.5 + 20.5) = 58 in. A0 = 0.85 Aoh = 0.85(174.25) = 148.11 in.2 = 45 degrees

Torsion area Strut angle

Check that torsion support is required for Tu = 392,640 in.-lb.

Tu <

No support if

f c Acp2 p cp

=

0.75(1.0) 4000 (288) 2 = 54,644 in.-lb (false) 72 2

Vu Tu p h + 2 bd 1.7 Aoh

Then verify the section meets the requirement

2

2

2

2

V < c + 8 f c . bd 2

20,050 (392,640)(58) Vu Tu p h = = 448.31 psi + + 2 2 ( 12 )( 21 ) 1 . 7 ( 174 . 25 ) bd 1.7 Aoh 31,876 V + 8(1.0) 4000 = 474.34 psi c + 8 f c = 0.75 bd (12)(21)

448.31 psi < 474.34 psi, so the section is adequate. Determine the minimum required area of transverse steel. Torsion Torsion and shear Minimum

Tn 523,520 At = = 0.0295 in. = 2(148.11)(60,000)(1) 2 A0 f yt cot s A A A = 2 t + = 2(0.0295) + 0 = 0.0589 in. (controls) s s s 0.75 f c 0.75 4000 A (12) = 0.0095 in. b = = 60,000 s fy

104 6

Determine the minimum required longitudinal reinforcement.

Minimum Minimum

f yt At (cot ) 2 = (0.0295)(58)(1)(1) 2 = 1.71 in.2 ph s fy 12 b = 25 = 0.005 in. (does not control) 25 60,000 f yt

AL =

Torsion At s

At s

AL

5 f c Acp fy

f yt At ph s fy

5(1.0) 4000 (288) (0.0295)(58)(1) = 0.19 in.2 60,000

Therefore, the required transverse steel area is 0.0589 in.2 / in., the required flexural steel is 2.48 in.2, and the required longitudinal reinforcement is 1.71 in.2. Assume that one quarter of the reinforcement is added to the bottom steel, one quarter goes in the top corners of the stirrups, and the remaining steel is distributed along the vertical sides of the stirrups. Choose the bar sizes and spacings. For the transverse steel, try #5 closed stirrups. The bar area is 0.31 in.2, so the required spacing p 2(0.31) 58 is the smallest of h = = 5.26 in. (controls). Therefore, use #5 = 7.25 in. , 12 in., and 0.0589 8 8 closed stirrups spaced 5 in. c-c. For the bottom steel, the required area is 2.48 + 0.25(1.71) = 2.91 in.2. Use 4 #8 bars (0.79 in.2 per bar) for a total area of 4(0.79) = 3.16 in.2. For the top steel, the required area is 0.25(1.71) = 0.43 in.2, so use 2 #5 bars in the corners for an area of 2(0.31) = 0.62 in.2. For the sides, each side requires 0.25(1.71) = 0.43 in.2 of steel distributed in 18.88 in. Use 2 #5 bars spaced equally for an area of 2(0.31) = 0.62 in.2 per side. The final design is shown below.

105 7

106

107

108

109

110

111

112

113

114

8.2. Calculate the maximum immediate and long-term deflection for a 8-in.-thick slab on simple supports spanning over 15 ft. The service dead and live loads are 60 psf (28.7 kPa) and 150 psf (71.8 kPa), respectively. The reinforcement consists of No. 5 bars (16-mm diameter) at 6 in. center to center (154 mm center to center). Also check which limitations, if any, need to be placed on its usage. Assume that 50% of the live load is sustained over a 30-month period. Given: f c¿ 5000 psi 134.5 MPa2

fy 60,000 psi 1414 MPa2

Es 29 106 psi 1200,000 MPa2

Design the one-way slab as a rectangular beam with a width of 12 in. Assume the depth of the steel is 6 in. h d b L

Beam height Effective depth Beam width Span

= = = =

8 in. 6 in. 12 in. 15 ft

First, estimate the properties of the concrete.

Modulus

(8)(12) 150 = 100 lb / ft 144 E c = 57,000 f c = 57,000 5000 = 4,030,509 psi

Tensile rupture

f r = 7.5 f c = 7.5 5000 = 530 psi

Stiffness ratio

Es 29 10 6 = = 7.19 n = 4,030,509 Ec

w =

Weight per foot

Then calculate the properties of the elastic beam.

Ig

Moment of inertia Strength

h 8 = = 4 in. 2 2 1 1 3 bh = (12)(8) 3 = 512 in.4 = 12 12 I g fr (512)(530) = 67,882 in-lb = r = y 4

y =

Center of gravity

M cr

Calculate the properties of the beam during the first phase of cracking. Center of gravity equation Solve for c Moment of inertia

c = I cr =

= = Calculate the load moments.

b 2 c + nAs c nAs d = 0 2 1.7728 in. 1 3 bc + nAs (d c) 2 3 1 (12)(1.77) 3 + (7.19)(2 0.31)(6 1.77) 2 3 102 in.4 115 1

( LD b + w) L2 [ (60)(12) + 100](15 12) 2 = = 54,000 in-lb 8 8 L bL2 (150)(12) (15 12) 2 = L = = 50,625 in-lb 8 8

Dead load

MD =

Live load

ML

Now calculate the moments for the deflections. Dead Dead plus 50% live Dead plus live

M = M D = 54,000 in-lb M = M D + 0.5 M L = 54,000 + 0.5(50,625) = 79,312.5 in-lb M = M D + M L = 54,000 + 50,625 = 104,625 in-lb

Calculate the effective moment of inertia for each case. 3

Equation

Ie

Dead

Ie

M = cr ( I g I cr ) + I cr if M > M cr M 4 = I g = 512 in. ( M < M cr ) 3

67,882 = (512 102) + 102 = 359 in.4 79 , 312 . 5

Dead plus 50% live

Ie

Dead plus live

67,882 Ie = (512 102) + 102 = 214 in.4 104,625

3

Calculate the deflections, use t = 1.75 and = 2. Equation Immediate DL Immediate LL

5ML2 (deflection of simply supported beam) = 48 EI e

D

5(54,000)(15 12) 2 = 0.0883 in. = 48(4.03 10 6 )(512)

L

5(104,625)(15 12) 2 D = 0.3211 in. = 48(4.03 10 6 )(214)

Immediate 50% LL

LS

Long term deflection

LT

5(79,312.5)(15 12) 2 D = 0.0967 in. = 48(4.03 10 6 )(359) = L + D + t LS = (0.3211) + (2)(0.0883) + (1.75)(0.0967) = 0.6669 in.

Now check the deflection requirements. 15(12) Roof with no supported or attached nonstructural elements L = 1 in. > L = likely to be damaged by large deflections 180 180 116 2

Floor with no supported or attached nonstructural elements likely to be damaged by large deflections Roof or floor with supported or attached nonstructural elements likely to be damaged by large deflections Roof or floor with supported or attached nonstructural elements not likely to be damaged by large deflections

L 15(12) = 0.5 in. > L = 360 360 L 15(12) = 0.375 in. < LT = 480 480 L 15(12) = 0. 75 in. > LT = 480 240

Therefore, the slab should be restricted to roofs or floors without attached nonstructural elements, or with attached nonstructural elements not likely to be damaged by large deflections.

117 3

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

8.7. A rectangular beam under simple bending has the dimensions shown in Fig. 8.21. It is subjected to an aggressive chemical environment. Calculate the maximum expected flexural crack width and whether the beam satisfies the serviceability criteria for crack control. Given: f c¿ 4500 psi 131.0 MPa2

fy 60,000 psi 1414 MPa2

24 in. 8

d = 20 in.

1 minimum clear cover 1 in. 138.1 mm2 2

t = 4 in. b = 10 in.

Figure 8.21

Beam geometry.

h = 24 in. d = 20 in. b = 10 in.

Beam height Effective depth Beam width

First, estimate the properties of the concrete. Modulus

E c = 57,000 f c = 57,000 4500 = 3.824106 psi

Tensile rupture

f r = 7.5 f c = 7.5 4500 = 503 psi

Stiffness ratio

n =

Es 29 10 6 = = 7.58 Ec 3.824 10 6

Then calculate the properties of the elastic beam.

Ig

Moment of inertia Strength

h 24 = = 12 in. 2 2 1 1 3 bh = (10)(12) 3 = 11,520 in.4 = 12 12 I g fr (11,520)(503) = 482,991 in-lb = r = y 12

y =

Center of gravity

M cr

Calculate the center of gravity for the beam during the first phase of cracking. Center of gravity equation Solve for c

b 2 c + [ nAs + (n 1) As] c [ nAsd + (n 1) Asd ] = 0 2 c = 7.54 in. 140 4

Then calculate the values for the crack equation, 0.076 f s 3 d c A . hc 24 7.54 = = 1.32 d c 20 7.54 = 0.6 f y = 0.6(60,000) = 36,000 psi = 36 ksi

Depth factor

=

Steel stress

fs

Number of bars Tension area First layer depth

bc = 4 2(24 20)10 2(h d )b = = 20 in.2 A = 4 bc d c = (clear cover) + (stirrup d) + 0.5(bar diameter) = 1.5 + 0.5 + 0.5(1) = 2.5 in.

Finally, calculate the maximum crack size. wmax = 0.076 f s 3 d c A

= 0.076(1.32)(36) (2.5)(20) = 13.3 mils = 0.0133 in. The maximum tolerable crack for an aggressive chemical environment is 0.007 in., so the beam does not satisfy the ACI code for crack control tolerance.

141 5

142

8.9. Find the maximum web of a beam reinforced with bundled bars to satisfy the crack-control criteria for interior exposure conditions. Given: f c¿ 5000 psi 134.5 MPa2

fy 60,000 psi 1414 MPa2

A s three bundles of three No. 9 bars each 1three bars of 28.6-mm diameter each in a bundle2 No. 4 stirrups used 113-mm diameter2

Since the beam dimensions are unknown, assume = 1.2, f s = 0.6 f y , and a clear cover of 1.5 in. The crack equation for a beam with bundles bars is wmax = 0.076 f s 3 d c A = 1.2 f s = 0.6 f y = 0.6(60,000 psi) = 36 ksi

Calculate the center of gravity of the bundles from the bottom of the beam. d c = (clear cover) + (stirrup d) + (bundle CG)

= 1.5 + 0.5 +

3+ 3 (1) = 2.890 in. 6

Write the area of concrete in tension in terms of the web width b. Number of bars Effective number Effective area

bc = 9 bc = 0.650 bc = 0.650(9) = 5.85 2d c b 2(2.890) b = 0.9879b = A = 5.85 bc

For a beam in interior exposure, the maximum tolerable crack width is 0.016 in. Substitute the known values into the equation. Substitute values

0.016 =

0.076(1.2)(36)3 (2.890)(0.9879b) 1000 3

Solve for b

(1000)(0.016) 1 = 40.54 in. b = (0.076)(1.2)(36) (2.890)(0.9879)

Therefore, the maximum beam web width is 40.5 in.

143 6

12 in. (304.8 mm)

9.1. Calculate the axial load strength Pn for columns having the cross-sections shown in Figure 9.46. Assume zero eccentricity for all cases. Cases (a), (b), (c), and (d) are tied columns; case (e) is spirally reinforced.

8 in. (203.2 mm) 8000 9

Figure 9.46

Column sections.

144 1

Beam area Steel area

Ag = bh = (8)(12) = 96 in.2 Ast = 6(1) = 6 in.2

Since the column is compression controlled (axial load), = 0.65. For a tied column: Nominal strength

Pn = (0.80)[0.85 f c(Ag Ast ) + Ast f y ] = (0.65)(0.80)[0.85(8000)(96 6) + (6)(60,000)] = 505,400 lb

145 2

146

147

148

149

150

151

152

153

154

155

156

9.4. For the cross section shown in Figure 9.46c of Problem 9.1, determine the safe eccentricity e if Pu = 400,000 lb. Use the trial-and-adjustment method satisfying the compatibility of strains.

Beam area Top steel area Bottom steel area

Ag = bh = (8)(12) = 96 in.2 Ast = 3(1) = 3 in.2 Ast = 3(1) = 3 in.2

For the column = 0.65 and for f c = 8000 psi, 1 = 0.65. Guess that the neutral axis passes through the center of the section, c = 6 in. Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Strength

a = 1c = (0.65)(6) = 3.9 in. FC = 0.85 f cba = (0.85)(8000)(8)(3.9) = 212,160 lb c d 6 3 s = 0.003 = 0.003 = 0.0015 (compression) c 6 f s = E s s = (29106)(0.0015) = 43,500 psi < 60,000 psi OK FS = As f s = (3)(43,500) = 130,500 lb

d c 96 s = 0.003 = 0.003 = 0.0015 (tension) c 6 f s = E s s = (29106)(0.0015) = 43,500 psi < 60,000 psi OK FS = As f s = (3)(43,500) = 130,500 lb Pn = FC + FS FS = 212,160 lb Pu = Pn = 0.65(212,160) = 137,904 lb

Since the strength is less than the specified load of 400,000 lb, the size of the compression block must be increased. This can be achieved by increasing c and iterating until the calculated strength is equal to the load. After several iterations, c = 11.0 in. Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength

a = 1c = (0.65)(11) = 7.15 in. FC = 0.85 f cba = (0.85)(8000)(8)(7.15) = 388,960 lb c d 11 3 s = 0.003 = 0.003 = 0.002182 (compression) c 6 f s = E s s = (29106)(0.002182) = 63,273 psi > 60,000 psi = 60,000 psi FS = As f s = (3)(60,000) = 180,000 lb

cd 11 9 s = 0.003 = 0.003 = 0.000545 (compression) c 6 f s = E s s = (29106)(0.000545) = 15,818 psi < 60,000 psi OK FS = As f s = (3)(15,818) = 47,455 lb Pn = FC + FS FS = 616,415 lb

157 3

Strength

Pu = Pn = 0.65(616,415) = 400,669 lb

The strength is sufficient, so calculate the eccentricity. Nominal moment

Eccentricity

a M n = FC y + Fs( y d ) + Fs ( y d ) 2 = (388,960)(6 0.57.15) + (180,000)(6 3) + (47,455)(6 9) = 1,340,864 in.-lb Mn 1,340,864 = = 2.18 in. e = 616,415 Pn

158 4

159

160

161

162

163

164

165

166

167

168

9.8. Design the reinforcement for a nonslender 12 in. × 15 in. column to carry the following loading. The factored ultimate axial force Pu = 200,000 lb. The eccentricity e to geometric centroid = 8 in. Given: f c¿ 4000 psi fy 60,000 psi

Use the following design parameters. Force Eccentricity Moment Top steel depth Bottom steel depth Reinforcement ratios

Pu e Mu d d

= = = = = =

200,000 lb 7 in. Pu e = (200,000)(7) = 1,400,000 in.-lb 3 in. h 3 = 15 – 3 = 12 in. = 0.015

Choose the longitudinal reinforcement (the same for both top and bottom). Required area Use 3 No. 8 bars

Ast = bd = (0.015)(12)(12) = 2.16 in.2 Ast = (3)(0.79) = 2.37 in.2

Check for compression failure using = 0.65 and for f c = 4000 psi, 1 = 0.85. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Strength Nominal moment

c = 0.6 d = 0.6(12) = 7.2 in. a = 1c = (0.85)(7.2) = 6.12 in. FC = 0.85 f cba = (0.85)(4000)(12)(6.12) = 249,696 lb c d 7.2 3 s = 0.003 = 0.003 = 0.00175 (compression) c 7.2 f s = E s s = (29106)(0.00175) = 50,750 psi < 60,000 psi OK FS = As f s = (2.37)(50,750) = 120,278 lb

d c 12 7.2 s = 0.003 = 0.003 = 0.002 (tension) c 7.2 f s = E s s = (29106)(0.002) = 58,000 psi < 60,000 psi OK FS = As f s = (2.37)(58,000) = 137,460 lb Pn = FC + FS FS = 232,514 lb Pu = Pn = 0.65(232,514) = 151,134 lb a M n = FC y + Fs( y d ) + Fs ( y d ) 2 15 6.12 + (120,278)(7.5 3) (137,460)(7.5 12) = (249,696) 2

Eccentricity

= 2,268,469 in-lb Mn 2,268,469 = = 9.76 in. e = 232,514 Pn 169 5

Since 9.76 in. > 7 in., the column is compression controlled. Choose a larger ratio and iterate until the calculated eccentricity is 7 in. After a few iterations, the ratio is about 0.673. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Strength Nominal moment

Eccentricity

c = 0.64 d = 0.673(12) = 8.08 in. a = 1c = (0.85)(8.08) = 6.87 in. FC = 0.85 f cba = (0.85)(4000)(12)(6.87) = 280,076 lb c d 8.08 3 s = 0.003 = 0.003 = 0.00189 (compression) c 8.08 f s = E s s = (29106)(0.00189) = 54,682 psi < 60,000 psi OK FS = As f s = (2.37)(54,682) = 129,596 lb

d c 12 8.08 s = 0.003 = 0.003 = 0.00146 (tension) c 8.08 f s = E s s = (29106)(0.00146) = 42,272 psi < 60,000 psi OK FS = As f s = (2.37)(42,272) = 100,184 lb Pn = FC + FS FS = 309,488 lb Pu = Pn = 0.65(309,488) = 201,167 lb a M n = FC y + Fs( y d ) + Fs ( y d ) 2 15 6.87 + (129,596)(7.5 3) (100,184)(7.5 12) = (280,076) 2 = 2,173,277 in.-lb Mn 2,173,277 = = 7.02 in. e = 309,488 Pn

The strength of the column is sufficient (201,167 lb > 200,000 lb and 2,173,277 in.-lb > 1,400,000 in.-lb). Now design the ties. Since the longitudinal bars are No. 8 bars, use No. 3 ties. The spacing is the smallest of 1. 48 times No. 3 bar diameter = 48(0.375) = 18 in. 2. 16 times No. 8 bar diameter = 16(1) = 16 in. 3. Smaller column dimension = 12 in. Therefore, use No. 3 ties spaced 12 in. c-c.

170 6

171

172

173

174

175

176

177

178

9.12. A rectangular braced column of a multistory frame building has a floor height lu = 30 ft. It is subjected to service dead-load moments M2 = 4,000,000 on top and M1 = 2,000,000 in.-lb at the bottom. The service live-load moments are 80% of the dead-load moments. The column carries a service axial dead load PD = 200,000 lb and a service live load PL = 400,000. Design the cross-section size and reinforcement for this column. Given: f c¿ 8000 psi fy 60,000 psi A 1.3,

B 0.9

d¿ 3 in.

First, calculate the factored loads on the column. Force Moment 1 Moment 2

Pu = 1.2(200,000) + 1.6(400,000) = 880,000 M 1u = 1.2(3,000,000) + 1.6(0.8)(3,000,000) = 7,440,000 in.-lb M 1u = 1.2(4,000,000) + 1.6(0.8)(4,000,000) = 9,920,000 in.-lb

Assume the column dimensions. Section height Section width

h = 25 in. b = 25 in.

Check if slenderness is an issue. Effective length 1 Effective length 2 Effective length Radius of gyration Slenderness ratio Max slenderness ratio

k k k r kL r kL r

0.7 + 0.05( A + B ) = 0.7 + 0.05(1.3 + 0.9) = 0.81 0.85 + 0.05 min = 0.85 + 0.05(0.9) = 0.895 0.81 0.3h = 0.3(25) = 7.5 in. (0.81)(30 12) = 38.88 = 7.5 M 7,440,000 = 25 34 12 1 = 34 12 9,920,000 M2 = =

Since 38.88 > 25, slenderness needs to be considered. The frame is braced, so there is no side sway. Check that the moment is greater than the minimum required. Minimum moment

M 2 = Pu (0.6 + 0.03h) = 880,000(0.6 + 0.0325) = 1,188,000 in.M 2 = 9,920,000 in.-lb

Now calculate the magnified moment for a non-sway frame.

f c = 33(150)1.5 8000 = 5.422106 psi

Concrete strength

E c = 33w1.5

Moment of inertia

3 3 I g = bh = 25(25) = 32,552 in.4 12 12

179 7

EI =

EI

Pc =

Euler load

Magnification factor Design moment

0.40 E c I g 1 + dns

=

0.40(5.422 10 6 )(32,552) = 4.7071010 in.2-lb 1 + 0.5

2 EI 2 (4.707 1010 ) = = 5,463,464 lb (kL) 2 (0.81 30 12) 2

M 7,440,000 C m = 0.6 + 0.4 1 = 0.6 + 0.4 = 0.9 9,920,000 M2 Cm Cm = = 1.146 ns = 1 Pu /(0.75 Pc ) 1 880,000 /(0.75 5,463,464) M u = ns M 2 = 1.146(9,920,000) = 11,369,772 in.-lb

Therefore, design the section as a non-slender column with Pu = 880,000 lb, M u = 11,369,772 in.-lb, and e =

Mu = 12.92 in. Pu

Top steel depth Bottom steel depth Reinforcement ratios

d = 3 in. d = h 3 = 25 – 3 = 22 in. = = 0.015

Choose the longitudinal reinforcement (the same for both top and bottom). Required area Use 7 No. 10 bars

Ast = bd = (0.015)(25)(22) = 8.25 in.2 Ast = (7)(1.27) = 8.89 in.2

Check for compression failure using = 0.65 and for f c = 8000 psi, 1 = 0.65. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Strength

c = 0.6 d = 0.6(22) = 13.2 in. a = 1c = (0.65)(13.2) = 8.58 in. FC = 0.85 f cba = (0.85)(8000)(25)(8.58) = 1,458,600 lb c d 13.2 3 s = 0.003 = 0.003 = 0.00232 (compression) c 13.2 f s = E s s = (29106)(0.00232) = 67,227 psi > 60,000 psi f s = 60,000 psi FS = As f s = (8.89)(60,000) = 533,400 lb d c 12 13.2 s = 0.003 = 0.003 = 0.002 (tension) c 13.2 f s = E s s = (29106)(0.002) = 58,000 psi < 60,000 psi OK FS = As f s = (8.89)(58,000) = 515,620 lb Pn = FC + FS FS = 1,476,380 lb Pu = Pn = 0.65(1,476,380) = 959,647 lb 180 8

Nominal moment

Eccentricity

a M n = FC y + Fs( y d ) + Fs ( y d ) 2 25 8.58 + (533,400)(12.5 3) = (1,458,600) 2 (515,620)(12.5 22) = 21,940,796 in.-lb Mn 21,940,796 = = 14.86 in. e = 1,476,380 Pn

Since 14.86 in. > 12.92 in., the column is compression controlled. Choose a larger ratio and iterate until the calculated eccentricity is 12.92 in. After a few iterations, the ratio is about 0.641. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Strength Nominal moment

Strength Eccentricity

c = 0.641 d = 0.641(22) = 14.10 in. a = 1c = (0.65)(14.10) = 9.17 in. FC = 0.85 f cba = (0.85)(8000)(25)(9.17) = 1,558,271 lb c d 14.10 3 s = 0.003 = 0.003 = 0.00236 (compression) c 14.10 f s = E s s = (29106)(0.00236) = 68,492 psi > 60,000 psi f s = 60,000 psi FS = As f s = (8.89)(60,000) = 533,400 lb d c 22 14.10 s = 0.003 = 0.003 = 0.00168 (tension) c 14.10 f s = E s s = (29106)(0.00168) = 48,725 psi < 60,000 psi OK FS = As f s = (8.89)(48,725) = 433,169 lb Pn = FC + FS FS = 1,658,502 lb Pu = Pn = 0.65(1,658,502) = 1,078,026 lb

a M n = FC y + Fs( y d ) + Fs ( y d ) 2 25 9.17 + (533,400)(12.5 3) = (1,558,271) 2 (433,169)(12.5 22) = 21,519,003 in.-lb M u = M n = 0.65(21.519,003) = 13,987,353 in.-lb Mn 21,519,003 = = 12.97 in. e = 1,658,502 Pn

The strength of the column is sufficient (1,078,026 lb > 880,000 lb and 13,987,353 in-lb > 11,369,772 in-lb). Now design the ties. Since the longitudinal bars are No. 10 bars, use No. 4 ties. The spacing is the smallest of 1. 48 times No. 4 bar diameter = 48(0.5) = 24 in. 181 9

2. 16 times No. 8 bar diameter = 16(1.25) = 20.32 in. 3. Smaller column dimension = 25 in. Therefore, use No. 4 ties spaced 20 in. c-c.

182 10

183

184

185

186

187

188

189

190

191

192

193

194

9.16. A nonslender square corner column is subjected to biaxial bending about its x and y axes. It supports a factored load Pu = 500,000 lb acting at eccentricities ex = ey = 6 in. Design the column size and reinforcement needed to resist the applied stresses. Given: f c¿ 6000 psi fy 60,000 psi gross reinforcement percentage g 0.03 d¿ 2.5 in. Solve by using all the three methods.

Determine the design loads.

Factored x moment

Pu = 500,000 lb P 500,000 Pn = u = = 769,231 lb 0.65 M ux = Pu e y = (500,000)(4) = 2,000,000 in.-lb

Nominal x moment

M nx =

Factored y moment

M uy

Nominal y moment

M ny

Factored load Nominal load

M ux 2,000,000 = = 3,076,923 in.-lb 0.65 = Pu e x = (500,000)(4) = 2,000,000 in.-lb M uy 2,000,000 = = 3,076,923 in.-lb = 0.65

Then make a reasonable guess about the required dimensions of the column. Choose a total reinforcement ratio of g = 0.03 and try 4 bars per side. Since the moments are equal, use

Axial column area Twice the area

Aaxial =

h = 1. b

Pn 769,231 = = 219 in.2 0.45(6000 + 60,000 0.03) 0.45( f c + f y g )

Ag = 2(219) = 438 in.2

Choose a section with h = b = 22 in. Try using 5 bars per side for a total of 16 bars. For the chosen reinforcement ratio, the steel area required per bar is 0.032222 / 16 = 0.90 in.2, so try 5 No. 9 bars per side. Each side has 5(1.0) = 5.0 in.2 of reinforcement steel. Each method requires the nominal load strength and bending moment of the section. Calculate the load and moment strength for bending along the x-axis. First check for compression failure. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force

c = 0.6 d = 11.7 in. a = 1c = 8.775 in. FC = 0.85 f cba = 984,555 lb c d s = 0.003 = 0.00236 (compression) c f s = E s s = 68,410 psi > 60,000 psi so use 60,000 psi FS = As f s = 300,000 lb 195 11

Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Nominal moment Eccentricity

d c s = 0.003 = 0.002 (tension) c f s = E s s 58,000 psi < 60,000 psi FS = As f s = 290,000 lb Pn = FC + FS FS = 994,555 lb

a M n = FC y + Fs( y d ) + Fs ( y d ) = 11,525,370 in-lb 2 Mn = 11.59 in. e = Pn

Since 11.59 in. > 6 in., the column is compression controlled for bending along the x-axis. Choose a larger ratio and iterate until the calculated eccentricity is 6 in. After a few iterations, the ratio is about 0.857. Neutral axis depth Compression block Concrete force Top steel strain Top steel stress Top steel force Bottom steel strain Bottom steel stress Bottom steel force Nominal strength Nominal moment Eccentricity

c = 0.857 d = 16.72 in. a = 1c = 12.54 in. FC = 0.85 f cba = 280,076 lb c d s = 0.003 = 0.002551 (compression) c f s = E s s = 73,991 psi > 60,000 psi so use 60,000 psi FS = As f s = 300,000 lb

d c s = 0.003 = 0.0005 (tension) c f s = E s s = 14,467 psi < 60,000 psi OK FS = As f s = 72,335 lb Pn = FC + FS FS = 1,634,630 lb a M n = FC y + Fs( y d ) + Fs ( y d ) = 9,819,937 in.-lb 2 Mn = 6.01 in. e = Pn

In this case the column is square, so the same calculations apply to bending about the y-axis. Now check the design. Load Contour Method Since the loads are equal, use M ox , where b = h = 22 in., M nx = M ny = 4,615,385 in.-lb, and start with = 0.5. Required strength, x

M ox = M nx + M ny

b 1 = 9,230,769 in.-lb h

19612

From the section calculations, M oxn = 9,819,937 in.-lb > 9,230,769 in.-lb, so the strength of the M ny M nx section is sufficient. Now check M ny . Using = 0.5 and = 0.5 in the figure, = 0.5. M oyn M oxn Therefore, M ny = 0.5 M oyn , which is 4,909,968 in.-lb. Since M ny = 3,076,923 in.-lb, the strength is sufficient about this axis as well. Reciprocal Load Method Assume the same design as the previous method. Therefore, Pnx = Pny = 1,634,630 lb . Calculate the strength of the column for an axial load, where Ast = 16(1.0) = 16 in.2 Pn 0 = 0.85 f c( Ag Ast ) + Ast f y = 3,346,800 lb

Axial strength

Then calculate the nominal load strength.

1 1 1 1 + = Pnx Pny Pn 0 Pn

Axial strength

Pn = 1,081,401 lb > 769,230 lb

Modified Load Contour Method 1.5

1.5

M P Pnb M nx + ny , where + Assume the same design and calculate the value of n M Pn 0 Pnb M nbx nby Pnb = 994,555 lb, M nbx = M nby = 11,525,370 in.-lb, M nx = M ny = 4,615,385 in.-lb, and Pn 0 is

the same as the value found for the Reciprocal Load Method. 1.5

1.5

M Pn Pnb M nx + ny = 0.411, which is significantly less than 1, which indicates + M Pn 0 Pnb M nbx nby that the section is stronger than it needs to be.

197 13

198

199

200

201

202

203

204

205

10.1. Calculate the basic development lengths in tension for the following deformed bars embedded in normalweight concrete. (a) No. 4, No. 10. Given: f c¿ 6000 psi 141.4 MPa2

fy 60,000 psi 1413.7 MPa2

(b) No. 14, No. 18. Given: f c¿ 4000 psi fy 60,000 psi fy 80,000 psi

(a) For a No. 4 bar: Bar diameter Bar area

d b = 0.5 in. Ab = 0.2 in.2

Assume the clear cover is 1.5 in. and the clear spacing between the bars is 2db. Assume the bars are not being used for top reinforcement, and that they are not coated. Therefore, t = e = 1 . For a No. 4 bar, s = 0.8 . Use Ktr = 0. Determine the value for c.

c = 0.5db + (clear cover) = 0.5(0.5) + 1.5 = 1.75 in. c = 0.5(clear spacing + db) = 0.5(20.5 + 0.5) = 0.75 in. (controls) Check the values.

f c =

6000 = 77.5 psi < 100 psi

c + K tr 0.75 + 0 = 1.5 = db 0.5

OK

OK

Calculate the development length.

Ld

3 f y e t s = (0.5) 3(60,000)(1.0)(1.0)(0.8) = db 40(77.5)(1.5) 40 f c c + K tr d b = 15.49 in.

For a No. 10 bar: Bar diameter Bar area

d b = 1.27 in. Ab = 1.27 in.2

206 1

Assume the clear cover is 1.5 in. and the clear spacing between the bars is 2db. Assume the bars are not being used for top reinforcement, and that they are not coated. Therefore, t = e = 1 . For a No. 10 bar, s = 1.0 . Use Ktr = 0. Determine the value for c.

c = 0.5db + (clear cover) = 0.5(1.27) + 1.5 = 2.14 in. c = 0.5(clear spacing + db) = 0.5(31.27) = 1.91 in. (controls) Check the values.

f c =

6000 = 77.5 psi < 100 psi

c + K tr 1.91 + 0 = 1.5 = db 1.27

OK

OK

Calculate the development length.

Ld

3 f y e t s = (1.27) 3(60,000)(1.0)(1.0)(1.0) = db 40(77.5)(1.5) 40 f c c + K tr d b = 49.19 in.

207

208

209

210

211

212

213

214

10.5. An 15-ft (4.57-m) normal-weight concrete cantilever beam is subjected to a factored Mu = 4,000,000 in-lb (452 kN-m) and a factored shear Vu = 50,000 lb (222 kN) at the face of the support. Design the top reinforcement and the appropriate embedment of 90° hook into the concrete wall to sustain the external shear and moment. Given: f c¿ 5000 psi fy 60,000 psi

For a cantilever beam, the minimum depth for deflection is L / 8 or (1215) / 8 = 22.5 in. Beam height Effective depth Beam width Beam length

h d b L

= = = =

24 in. 21 in. 15 in. 15 ft

Design the beam for flexure. M u 4,000,000 = = 4,444,444 in.-lb 0.9 As f y a As f y d and a = 2 0.85 f cb

Nominal moment

Mn =

Section strength

Mn

Solve for As

As

Minimum area 1

As

Minimum area 2

As

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2

3 f c fy

= 3.86 in.2 (controls)

bd = 1.11 in.2

200 bd = 1.05 in.2 fy

The required area is 3.86 in.2 Use 4 #9 bars. Verify that the strength is sufficient. Steel area Compression block Nominal strength

As = (4)(1) = 4 in.2 As f y = 3.76 in. a = 0.85 f cb

M n = As f y d

a = 4,129,411 in.-lb > 4,000,000 in.-lb. 2

Design the anchor. Bar diameter Multiplier Aggregate factor Basic length

d b = 1.128 in. e = 1.0 for uncoated bars = 1.0 for normal concrete 0.02 e f y d b = 19.14 in. Lhb = f c 215 4

Multiplier Development length Hook length

d =

Areq Ast

= 0.9653

Ldh = d Lhb = 18.48 in. (> 8 d b and > 6 in. so OK) Lhook = 12 d b = 13.54 in.

Therefore, use a hook with Ldh = 18.5 in and Lhook = 14 in.

216 5

217

218

219

220

221

11.1 An end panel of a floor system supported by beams on all sides carries a uniform service live load wL = 80 psf and an external dead load wD = 20 psf in addition to its self-weight. The center-line dimensions of the panel are 20 ft × 30 ft (the dimension of the discontinuous side is 20 ft). Design the panel and the size and spacing of the reinforcement using the ACI Code direct design method. Given: f c¿ = 4000 psi, normal-weight concrete fy = 60,000 psi column sizes 24 in. × 24 in. width of the supporting beam webs = 15 in.

Assume the following dimensions. Slab thickness Beam section height

hs = 8 in. hb = 27 in.

Use a depth of 24 in. for the beam steel, a depth of 7 in. for the East-West slab steel, and a depth of 6.375 in. for the North-South slab steel. Check the conditions for using the Direct Design Method. 30 = 1.5 which is less than 2. 20 2. Assume there are more than three panels in each direction. 3. Check the live load to dead load ratio.

1. The maximum length ratio is

wd = 150(8/12) + 20 = 120 lb / ft2 wl = 80 lb/ft2 < 2 wd

Now calculate the stiffness properties. The beams on the north, south, and east sides are all the same, which the beam on the west side is L shaped. because it is the edge of the panel.

Calculate the moments of inertia for each beam. T section

(15)(19)(19 / 2) + (8)(53)(19 + 8 / 2) = 17.57 in. (from bottom) (15)(19) + (8)(53) (15)(19) 3 (53)(8) 3 = + (15)(19)(9.5 17.57) 2 + + (8)(53)(23 17.57) 2 12 12 = 41,897 in.4

y = I b1

222 1

y =

L section

(15)(19)(19 / 2) + (8)(34)(19 + 8 / 2) = 16.09 in. (15)(19) + (8)(34)

3 3 I b 2 = (15)(19) + (15)(19)(9.5 16.09) 2 + (34)(8) + (8)(34)(23 16.09) 2 12 12 4 = 35,389 in.

Then calculate the moments of inertia for the slabs. Strip

Width 30 12 24 + = 192 in. 2 2

Height

N-S strip on east edge

30 12 = 360 in.

8 in.

E-W strips

20 12 = 240 in.

8 in.

N-S strip on west edge

8 in.

Moment of Inertia (192)(8) 3 = 8192 in.4 12 (360)(8) 3 = 15,360 in.4 12 (240)(8) 3 = 10,240 in.4 12

Use these values to find the ratios of the beam to slab stiffness along each edge of the panel. Edge North South East West

Beam 41,897 in.4 41,897 in.4 41,897 in.4 35,389 in.4

Slab 10,240 in.4 10,240 in.4 15,360 in.4 8192 in.4 Average

m

= = = = =

Ratio 4.09 4.09 2.73 4.32 3.81

Now check the minimum slab thickness, using the appropriate equation for m > 2.5. Length ratio

=

Minimum

h =

30 12 24 = 1.56 20 12 24 L(0.8 + f y / 200,000)

36 + 9

= 7.39 in.

The slab thickness 8 in. > 7.39 in. Check the shear capacity. Use the depth for the steel in the N-S direction since the shortest span is in the N-S direction. Factored load Factored shear Nominal shear Shear capacity

wu = 1.2 wd + 1.6 wl = 272 lb/ft2 1.15wu Lmin Vu = = 2815.2 lb/ft 2 Vu Vn = = 3753.6 lb/ft 0.75 Vc = 2 f cd = 2(5000)(7) = 11,879 lb/ft 223 2

Vc > Vn , so the capacity is sufficient with the given thickness. Now calculate and distribute the

moments.

Column width Span Perpendicular span Strip length Minimum strip length Static moment Exterior negative moment Interior negative moment Positive moment Column strip Exterior negative moment Interior negative moment Positive moment Positive design moment Negative design moment Strip width Beam width Slab width Positive beam moment Negative beam moment

East-West Direction bc S1 S2 L1 = S1 bc

L1 = 0.65S1 2 M o = wu S 2 L1 8 M u = 0.16 M o M u = 0.7 M o + M u = 0.57 M o M cu = 0.85( M u ) M cu = 0.85( M u ) + M cu = 0.85(+ M u ) + M cu + M cn = max( M cu ) M cn = wc = 0.5S 2 wcb wcs = wc wcb 0.85(+ M cn ) 0.85( M cn )

0.15(+ M cn ) wcs 0.15( M cn ) wcs

Positive slab moment / foot Negative slab moment / foot Middle strip Exterior negative moment Interior negative moment Positive moment Strip width Positive design moment / foot

M mu = 0.15( M u ) M mu = 0.15( M u ) + M mu = 0.15(+ M u ) wms = S 2 wc + M mu + M mn = wms 224 3

24 30 20 336 234

in. ft ft in. in.

6,397,440 in-lb 1,023,590 in-lb 4,478,208 in-lb 3,646,541 in-lb 870,052 in-lb 3,806,477 in-lb 3,099,560 in-lb 3,443,955 in-lb 4,229,419 in-lb 10 ft 53 67 2,927,362 3,595,006

in. in. in-lb in-lb

92,524 in-lb/ft 113,626 in-lb/ft

153,538 671,731 546,981 10

in-lb in-lb in-lb ft

60,776 in-lb/ft

Negative design moment / foot

max( M mu ) wms

M mn =

74,637 in-lb/ft

Calculate the required reinforcement. Beam: + M u = 2,927,362 in-lb, M u = 3,595,006 in-lb, d = 24 in., b = 15 in.

Minimum 1

As

Minimum 2

As

Bottom

As

Top

3 f cbd

1.27 in.2

fy 200bd fy

1.2 in.2

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2 0.85 f cbd 2M n 1 1 As fy 0.85 f cbd 2

2.12 in.2 2.63 in.2

Use 3 No. 9 bars (3.0 in.2) for both the top and bottom steel of the E-W beams. Column strip: + M u = 92,524 in-lb, M u = 113,626 in-lb, d = 7 in., b = 12 in., h = 8 in. Minimum

As 0.0018bh

Bottom

As

Top

0.1728 in.2/ft

0.85 f cbd 2M n 1 1 fy 0.85 f cd 2 0.85 f cbd 2M n 1 1 As fy 0.85 f cd 2

0.2476 in.2/ft 0.3057 in.2/ft

Using No. 5 bars, the required spacings are 15.03 in., and 12.17 in., but the maximum spacing is 12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the E-W column strips. Middle strip: + M u = 60,776 in-lb, M u = 74,637 in-lb, d = 7 in., b = 12 in., h = 8 in. Minimum

As 0.0018bh

Bottom

As

Top

0.1728 in.2/ft

0.85 f cbd 2M n 1 1 fy 0.85 f cd 2

0.85 f cbd 2M n 1 1 As 2 fy 0 . 85 f d c

0.1613 in.2/ft 0.1988 in.2/ft

Using No. 5 bars, the required spacings are 21.53 in., and 18.71 in., but the maximum spacing is 12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the E-W middle strips. 225 4

Column width Span Perpendicular span Strip length Minimum strip length Static moment Interior negative moment Positive moment Column strip Interior negative moment Positive moment Positive design moment Negative design moment Strip width Beam width

North-South Direction bc

S1 S2 L1 = S1 bc L1 = 0.65S1 2 M o = wu S 2 L1 8 M u = 0.65M o + M u = 0.35M o M cu = 0.6( M u ) + M cu = 0.6(+ M u ) + M cu + M cn = max( M cu ) M cn = wc = 0.5S 2 wcb wcs = wc wcb 0.85(+ M cn )

Slab width Positive beam moment Negative beam moment

0.85( M cn )

0.15(+ M cn ) wcs 0.15( M cn ) wcs

Positive slab moment / foot Negative slab moment / foot Middle strip Interior negative moment Positive moment Strip width Positive design moment / foot Negative design moment / foot

M mu = 0.4( M u ) + M mu = 0.4(+ M u ) wms = S 2 wc + M mu + M mn = wms max( M mu ) M mn = wms

226 5

24 20 30 216 156

in. ft ft in. in.

3,965,760 in-lb 2,577,744 in-lb 1,388,016 in-lb 1,546,646 in-lb 832,810 in-lb 925,344 in-lb 1,718,496 in-lb 10 ft 53 67 786,542 1,460,722

in. in. in-lb in-lb

24,860 in-lb/ft 46,169 in-lb/ft

1,031,098 in-lb 555,206 in-lb 10 ft 30,845 in-lb/ft 57,283 in-lb/ft

Calculate the required reinforcement. Beam: + M u =786,542 in-lb, M u = 1,460,722 in-lb, d = 24 in., b = 15 in.

Minimum 1

As

Minimum 2

As

Bottom

As

Top

3 f cbd

1.27 in.2

fy 200bd fy

1.2 in.2

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2

0.85 f cbd 2M n 1 1 As 2 fy 0 . 85 f bd c

0.55 in.2 1.03 in.2

Use 2 No. 9 bars (2.0 in.2) for both the top and bottom steel of the N-S beams. Column strip: + M u = 24,860 in-lb, M u = 46,169 in-lb, d = 7 in., b = 12 in., h = 8 in. Minimum

As 0.0018bh

Bottom

As

Top

0.1728 in.2/ft

0.85 f cbd 2M n 1 1 fy 0.85 f cd 2

0.85 f cbd 2M n 1 1 As 2 fy 0 . 85 f d c

0.0595 in.2/ft 0.1110 in.2/ft

Using No. 5 bars, the required spacings are 21.53 in., and 21.53 in., but the maximum spacing is 12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the N-S column strips. Middle strip: + M u = 30,845 in-lb, M u = 57,283 in-lb, d = 7 in., b = 12 in., h = 8 in. Minimum

As 0.0018bh

Bottom

As

Top

0.1728 in.2/ft

0.85 f cbd 2M n 1 1 fy 0.85 f cd 2

0.85 f cbd 2M n 1 1 As 2 fy 0 . 85 f d c

0.0739 in.2/ft 0.1380 in.2/ft

Using No. 5 bars, the required spacings are 21.53 in., and 21.53 in., but the maximum spacing is 12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the N-S middle strips.

227 6

Therefore, the reinforcement design is as shown below. East-West Reinforcement

North-South Reinforcement

228 7

229

230

231

232

233

234

235

236

237

238

239

240

241

242

243

244

245

246

247

248

249

250

251

252

253

254

255

11.7 Use the yield-line theory to evaluate the slab thickness needed in the column zone of the flat plate in Problem 11.5 for flexure, assuming that the hinge field would have a radius of 24 in.

Slab thickness Steel depth Hinge field radius

h = 16 in. d = 14 in. r = 24 in.

Calculate the load and effective circle. Dead load Live load Factored load Panel area Panel load Radius

wd = 150(16/12) + 20 = 220 lb / ft2 wl = 80 lb/ft2 wu = 1.2w d + 1.6w l = 392 lb/ft 2 2 A = (20)(30) = 600 ft P = wu A = 235,200 lb

4 = 27.63 ft A

=

Assume M = M and calculate the required steel for a 12 in. strip (b = 12 in.)

P 2 1 = 1,844,675 in-lb / ft 4 3r Mu = 2,049,639 in-lb / ft = As f y a As f y d and a = 2 0.85 f cb

Factored moment

Mu =

Nominal moment

Mn

Section strength

Mn

Solve for As

As

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2

= 2.87 in.2

Choose 3 No. 9 bars for 3.0 in.2 Check the strength and failure mode. Compression block Neutral axis depth Ratio Strength

As f y

= 4.41 in. 0.85 f cb a = 5.19 in. c = 1 c = 0.370 < 0.375 Tension controlled, OK d a M n = As f y d = 2,122,941 in-lb > 2,049,639 in-lb, OK 2

a =

256 8

257

258

259

260

261

11.10 Calculate the maximum crack width in a two-way interior panel of a reinforced concrete floor system. The slab thickness is 7 in. (177.8 mm) and the panel size is 24 ft × 30 ft (7.32 m × 9.14 m). Also design the size and spacing of the reinforcement necessary for crack control assuming that (a) the floor is exposed to normal environment; (b) the floor is part of a parking garage. Given: fy = 60.0 ksi (414 MPa).

Slab thickness Clear cover Ratio

h = 8 in. cc = 0.75 in. 30 = 0.8 r = 24

For a normal environment, try No. 4 bars ( d b = 0.5 in.) Max crack width Coefficient Standard value Steel depth

wmax = 0.012 in. 5 2 K = 2.110 in. / lb = 1.2 d c = cc + 0.5d b = 1 in. 2

Calculate G1 Assume s1 = s 2 = s

wmax = 393.6 in.2 G1 = 0.4 Kf y d b G1 = 8.79 in. s = 8d c

Therefore, use No. 4 bars at 8.5 in. c-c for the crack reinforcement. For a parking garage, try No. 4 bars ( d b = 0.5 in.) Max crack width Coefficient Standard value Steel depth

wmax = K = = dc =

0.016 in. 2.110 5 in.2 / lb 1.2 cc + 0.5d b = 1 in. 2

Calculate G1 Assume s1 = s 2 = s

wmax = 699.9 in.2 G1 = 0.4 Kf y d b G1 = 11.72 in. s = 8d c

Therefore, use No. 4 bars at 11.5 in. c-c for the crack reinforcement.

262 9

12.1. Design a reinforced concrete, square, isolated footing to support an axial column service live load PL = 400,000 lb (1779 kN) and service dead load PD = 500,000 lb (2224 kN). The size of the column is 26 in. × 22 in. (0.66 m × 0.56 m). The soil test borings indicate that it is composed of medium compacted sands and gravely sands, poorly graded. The frost line is assumed to be 3 ft below grade. Given: average weight of soil and concrete above the footing, 140 pcf 122.0 kN>m3 2 footing f c¿ 3000 psi 120.7 MPa2

column f c¿ 4000 psi 127.6 MPa2

fy 60,000 psi 1413.7 MPa2

surcharge 90 psf 14.3 kPa2

Given:

Column dimensions: 26 in. 22 in. Combined soil and concrete weight: = 140 pcf Surcharge: 90 psf Frost line depth: 3 ft

Assume the following dimensions, and try using No. 7 bars throughout. Slab thickness Clear cover at base Bar diameter Bar area Average steel depth

h = 3 ft cc = 3 in. d b = 0.875 in. Ab = 0.6 in.2 d = h cc d b = 32.125 in.

For the given soil and concrete types: Aggregate factor Soil capacity

= 1 Psoil = 2.5 ton/ft2 = 5000 psf

Calculate the required footing area using the service load and the allowable soil pressure. Allowable soil load Service load Required area Choose size (square)

Pn = Psoil (h + frost ) surcharge = 4070 psf P = PD + PL = 900,000 lbf P = 221.13 ft2 Af = Pn 2 b = 15 ft ( A f = 225 ft )

Calculate the factored load intensity. Factored load Load intensity

PU = 1.2 PD + 1.6 PL = 1,240,000 lbf P qU = U = 5511.11 psf Af

263 1

Consider the shear capacity for beam action. Factored shear Nominal shear Shear capacity

22 32.125 15 VU = qU (15) = 322,917 lbf 12 2 2 12 VU Vn = = 430,556 lbf 0.75 Vc = 2 f cbd = 633,441 lbf > 430,556 lbf, OK

Consider the shear capacity for two-way action. Factored shear Nominal shear Perimeter Footing size ratio Column type factor Shear capacity 1 Shear capacity 2 Shear capacity 3

22 + 32.125 26 + 32.125 VU = qU (15)(15)

= 1,119,597 lbf 12 12 VU Vn = = 1,492,796 lbf 0.75 b0 = 2(22 + 32.125 + 26 + 32.125) = 224.5 in. = 1 s = 40 for an interior column 4 Vc = 2 + f cb0 d = 2,370,126 lbf d Vc = s + 2 f cb0 d = 3,051,075 lbf b0 Vc = 4 f cb0 d = 1,580,083 lbf (controls) > 1,492,796 lbf, OK

Design the reinforcement for the moment at the critical section. The critical section is plane parallel to the wider column face. 2

Factored moment Nominal moment Minimum 1 Minimum 2

M U = 1 qU (15) 15 12 22 = 21,496,778 in-lb 2 2 2 MU Mn = = 23,885,309 in-lb 0.9 As 0.0018bd = 10.4085 in.2 As

0.85 f cbd 2M n 1 1 fy 0.85 f cbd 2

= 12.7211 in.2

That requires 22 No. 7 bars for an area of 13.2 in.2 For a footing width of 15 ft, the required spacing is 1512/22 = 8.18 in. so use 22 No. 7 bars spaced 8 in. c-c. for both directions. Check the development length for the reinforcement bars. Use t = s = e = 1 and K tr = 0 .

264 2

cb + K tr = 3.428 > 2.5 so use 2.5 db

Effective cover factor

f c = 54.77 < 100, OK

Strength factor Development length

ld =

3 f y t s e d b = 28.75 in. 40 f c 2.5

15 12 26 = 79 in. > 28.75 in. OK 2 2

Available space

Check the bearing strength for the column and the footing. 22 26 = 3.9722 ft2 144 0 . = 7(0.85) f cA1 = 1,361,360 lbf > 1,240,000 lbf, OK

Column area

A1 =

Column bearing

PU

2

Footing similar area Area factor Footing bearing

15 12 2 A2 = A1 = 190.38 ft 26 A2 = 6.923 > 2.0 so use 2.0 A1 PU = 2.0(0.7)(0.85) f cA1 = 2,042,040 lbf > 1,240,000 lbf, OK

Since the bearing strength is adequate, only the minimum dowel area is required. Minimum area

Ad = 0.005 A1 = 2.86 in.2

Using No. 7 bars, 5 are required. Use 6 No. 7 dowels for symmetry. Space 3 dowels 8.5 in. c-c along each 26 in. face of the column. Check the development length in the column and the footing Minimum Footing Column

l d 0.0003d b f y = 15.75 in. db f y = 19.17 in. (controls) l d = 0.02 f c l d = 0.02

db f y f c

= 16.60 in. (controls)

The available space in the footing is h cc 3d b = 30.375 in. > 19.17 in. OK

265 3

266 4

267

268

269

270

271

272

273

274

275

276

277

278

279

280

281

282

283

284

285

286

287

288

289

290

291

292

293

294

295

296

13.3. Find the moments and shears caused by a wind intensity of 30 psf acting on the structural system in Ex. 13.5 using the portal method of analysis. Use a height between floors of 10′ – 6″ (3.20 m).

Given: Floor height: Lu = 10.5 ft Roof parapet: L p = 2.5 ft Section width: L = 20 ft Wind load: wu = 30 psf Calculate the force at each floor and at the roof. Roof Floor

L Froof = 1.7 u + L p Lwu = 7905 lb 2 F floor = 1.7 Lu Lwu = 10,710 lb

Therefore the wind load on the building is as shown.

The shear in the each floor supports the wind forces for all the floors above, and the interior columns carry twice the shear as the exterior columns. The moment is the shear multiplied by half the floor height.

Floor 1

Shear 3(10,710) + 7905 = 40,035 lb

2 3 4

2(10,710) + 7905 = 29,325 lb 1(10,710) + 7905 = 18,615 lb 7905 lb

Shear (lb) Interior Exterior 13,345 6672.5 9775 6205 2635

297 1

4887.5 3102.5 1317.5

Moment (ft-lb) Interior Exterior 70,061 35,031 51,319 32,576 13,834

25,659 16,288 6917

For the beams, the moment at each beam/column joint must sum to zero, and the unbalanced moment from the columns is assumed to be split evenly between the beams for the interior joints. The vertical shear in the beams is the moment divided by half the span between the beams. Moment (ft-lb) Floor Interior 2 0.5(70,061 + 51,319) = 60,690 3 41,948 4 23,205 R 6917

Exterior 35,031 + 25,659 = 60,690 41,948 23,205 6917

The shear forces in the columns and beams are shown in the figure below.

298 2

Shear (lb) Interior Exterior 5057.5 6936 3495.6 4794 1933.8 2652 576.4 790.5

299

300

301

302

303

304

305

306

307

308

309

310

311

312

14.1. An AASHTO prestressed simply supported I beam has a span of 40 ft (12.9 m) and is 42 in. (106.7 cm) deep. Its cross section is shown in Figure 14.18. It is subjected to a live-load intensity WL = 4800 plf (58.4 kN/m). Determine the required -in.-diameter, stress-relieved, seven-wire strands to resist the applied gravity load and the self-weight of the beam, assuming that the tendon eccentricity at midspan is ec = 18.42 in. (467.9 mm). Maximum permissible stresses are as follows: f c¿ 6000 psi 141.4 MPa2 fc 0.45f c¿ 2700 psi 126.7 MPa2 ft 122f c¿ 930 psi 16.4 MPa2 fpu 270,000 psi 11862 MPa2 fpi 189,000 psi 11303 MPa2 fpe 145,000 psi 11000 MPa2 The section properties, given these stresses, are A c 545 in.2 I g 107,645 in.4 Ic 197.5 in.2 r2 Ac cb 18.24 in. Sb 5900 in.3 S t 4531 in.3 WD 568 plf WL 4800 plf 4 8

6

5

8

8 2 0

Figure 14.18

Given: Concrete area: Ac = 545 in.2 Strands: 1/2 od, 7 wire strand, As = 0.153 in.2 Final pretension stress: f pe = 145,000 psi Dead load: wD = 568 lb/ft Live load: wL = 4800 lb/ft Mid span eccentricity: ec = 18.42 in. Section center of gravity: c = 18.24 in. Section depth: h = 42 in. Section radius of gyration: r 2 = 197.5 in.2 Section bottom modulus: S b = 5900 in.3 Section top modulus: S t = 4531 in.3 Maximum tensile stress: f t = 930 psi Maximum compressive stress: f c = 2700 psi 313 1

Let s be the number of strands needed.

Final pretension

A ps = As s = 0.153s in.2 Pe = A ps f pe = 21,380.28s lb

Mid span moment

MT =

Total steel area

wD L2 wL L2 + = 12,882,500 in-lb 8 8 Pe ec ct M T 1 2 t = 47.68s 2843 psi = Ac r S

Stress at top

ft

Stress at bottom

fb =

Pe ec cb M T = 105.98s + 2183.38 psi 1 2 + Ac r Sb

Calculate the minimum number of strands. Strands for top Strands for bottom

f t + 2843 = 2.9994 47.68 f 2183.38 = 11.8315 s c 105.98

s

Therefore, use 12 1/2” O.D. 7 wire strands for pretensioning.

314 2

315

316

317

318

319

320

321

322

323

324

325

326

327

328

329

330

331

332

333

334

335

336

337

338

339

340

341

342

343

344

345

346

347

348

349

350

351

352

353

354

355

356

357

358

359

360

361

362

363

364

365

366

367

368

16.1. A 3 × 18 panel, ductile, moment-resistant category II, site class B frame building has a ground story 12 ft high (3.66 m) and 10 upper stories of equal height of 10 ft (3.05 m). Calculate the base shear V and the overturning moment at each story level in terms of the weight Ws of each floor. Use the Equivalent Lateral Force Method in the solution. Given: S1 0.34 sec., Ss 0.90 sec. R5

Ws per floor 2,000,000 lb 18896 kN2

Building height Site coefficient Site coefficient Importance factor

hb = 12 + (10)(10) = 112 ft f a = 1.0 for S S = 0.90 f v = 1.0 for S1 = 0.34 I = 1 for class II

Spectral response Maximum response

Damped response

S MS = f a S S = 0.9 S M 1 = f v S1 = 0.34 2 S DS = S MS = 0.6 3 2 S D1 = S M 1 = 0.2267 3

Calculate the seismic response coefficient. Nominal value Period coefficient Approximate period Limit coefficient

S DS = 0.15 R/I CT = 0.020 for generic building Ta = CT hb 3 / 4 = 0.6886 s CU = 1.37 for S D1 = 0.2267

CS =

Maximum period Building period

Ta = CU Ta = 0.9456 s T = 1.2Ta = 1.1348 s

Maximum value

CS

Minimum value

CS

S D1 = 0.049938 (controls) ( R / I )T 0.044 S DS = 0.0264

Calculate the total shear. Weight at base Shear at base

W = 11Ws = 22,000,000 lb V = C S W = 0.549313 Ws = 1,098,625 lb

369 1

The coefficient for the force at each floor is C vx =

Ws hxk n

, where hi is the height of floor i

W h i =1

k s i

T 0.5 (2 1) + 1 = 1.317376. above the ground and k = 2.5 0.5 x

The force on each floor is Fx = C vxV , the story shear is V x = V Fi , and the overturning i =0

n

moment is M x = Fi (hi hx ) . For the top ten stories, = 1. i= x

n

W h

Denominator for C vx

k s i

i =1

= 2672.35 Ws

Floor Base 1 2 3

hx

C vx

Fx

Vx

V x (lb)

M x (ft-lb)

0 12 22 32

0 0.009881 0.021957 0.035971

0 0.005428 Ws 0.012061 Ws 0.019759 Ws

0.549313 Ws 0.543885 Ws 0.531824 Ws 0.512064 Ws

1,098,625 1,087,770 1,063,647 1,024,128

89,759,342 76,575,838 65,698,137 55,061,666

4

42

0.051468

0.028272 Ws

0.483792 Ws

967,584

44,820,384

5

52

0.068191

0.037458 Ws

0.446334 Ws

892,668

35,144,540

6

62

0.085973

0.047226 Ws

0.399108 Ws

798,216

26,217,859

7

72

0.104691

0.057508 Ws

0.341600 Ws

683,200

18,235,695

8

82

0.124256

0.068256 Ws

0.273344 Ws

546,689

11,403,697

9

92

0.144595

0.079428 Ws

0.193917 Ws

387,833

5,936,809

10

102

0.165648

0.090993 Ws

0.102924 Ws

205,848

2,058,477

11

112

0.187368

0.102924 Ws

0

0

0

370 2

371

372

373

374

375

376

377

378

379

380

381

382

383

384

385

386

387

388

389

390

391

392

393

394

395

17.1. Design for flexure and shear an 8 in. thick a grouted CMU masonry lintel having a 20 ft span and a bearing length of 20 in. on the supporting wall to support a roof transmitting 350 plf load to the lintel Given: Effective t

7.63 in.

Unit weight

80 psf

Wall height to roof

14 ft.

Height of lintel

5¿-3–

Parapet height

3¿-6–

Service live load on roof

200 plf

Governing factored loading

1.2D 1.6L

fr

163 psi

fm

1500 psi

fy

60,000 psi

Given: Lintel width Lintel height Lintel span Dead load Live load Masonry strength Rupture strength Steel strength Steel modulus

b h L wD wL fm

= = = = = = =

7.63 in. 5-3 = 63 in. Clear span + 1/2 bearing = 20 + 10 = 250 in. 350 + 80bh = 770 lb/ft 200 lb/ft 1500 psi 163 psi

fr f y = 60,000 psi E s = 29106 psi

Calculate the loads. Factored load Factored shear Nominal shear Factored moment Nominal moment

wU = 1.2 wD + 1.6 wL = 1244 lb/ft w L VU = U = 12,958.33 lb 2 VU Vn = = 16,197.92 lb 0.80 w L2 MU = U = 809,895.83 in.-lb 8 MU Mn = = 899,884.26 in.-lb 0.90

Check the design for shear. Assume 3” cover for the steel reinforcement so d = h 3 = 60 in. Shear plane area

An = bd = 457.8 in.2

Masonry strength

Vm = (4 1.75) An

Maximum strength

Vn = 4 An

f m = 39,893.66 lb (controls)

f m = 70,922.07 lb

396 1

Since 39,893.66 lb > 16,197.92 lb, the shear strength is sufficient. Determine the required reinforcement steel area. Required area Solve for As

As f y a M n = As f y d and a = 2 0.80 f m b As =

0.80 f m bd 2M n 1 1 fy 0.80 f m bd 2

= 0.2535 in.2

Check the cracking strength. Critical moment Required steel area

bh 2 f r = 1,069,511 in.-lb > 899,884.26 in.-lb 6 1.3M cr As = 0.3013 in.2 As = Mn

1.3M cr = 1.3

Therefore, choose 2 No. 4 bars for an area of 2(0.2) = 0.4 in.2 Check the reinforcement ratio. Actual ratio Max steel strain Max masonry strain Load factor Maximum ratio

= As = 0.000874 bd f y = y = 0.002069 Es mu = 0.0025 for concrete = 1.5 0.64 f m mu = 0.007138 > 0.000874 OK f y mu + y

397 2

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420

17.8. Design a tension anchor for a grouted masonry wall to withstand a tensile force of magnitude Pu 14,000 lb and a shear force Pv 3000 lb without causing any masonry breakout. Consider the direction of the shear force to be along the wall length. Given fy 36,000 psi fye 27,000 psi fm 1500 psi

Given: Bolt axial strength Bolt shear strength Masonry strength Axial load Shear load

f y = 36,000 psi f yv = 27,000 psi f m = 1500 psi

Pa = 14,000 lb Pv = 3000 lb

Try a design using N b = 2 bolts with a diameter of d b = 0.75 in. and an embedment length of I b = 6.5 in. Space the bolts 13 in. apart to avoid overlap of the anchorage areas. Bolt area Masonry axial area

N b d b2 = 0.8836 in.2 4 2 = N b I b = 265.46 in.2

Ab = A pt

Masonry axial failure

2 A pv = N b I b = 132.73 in.2 2 Ban = (0.5)4 A pt f m = 20,562.8 lb (controls) > 14,000 lb OK

Bolt axial failure

Ban = (0.9) Ab f y = 28,627.8 lb

Masonry shear failure

Bvn = (0.5)4 A pv f m = 10,281.4 lb (controls) > 3000 lb OK Bvn = (0.9)(0.6) Ab f yv = 12,882.5 lb

Masonry shear area

Bolt shear failure Combined load

Pa P + v = 0.97263 < 1 OK Ban Bvn

Therefore, the design is sufficient.

421 3

Reinforced Concrete by Edward G. Nawy - Solutions Manual

Recommend Documents