Reinforced Concrete Manual

INSTRUCTOR’S SOLUTIONS MANUAL

REINFORCEDCONCRETEA FUNDAMENTAL APPROACH

S I X T H E D I T I O N

EDWARD G. NAWY

Pearson Education International

Vice President and Editorial Director, ECS: Marcia J. HortonSenior Editor: Holly StarkAssociate Editor: Dee BernhardEditorial Assistant: Jennifer Lonschein/Alicia LucciDirector of Team-Based Project Management: Vince O’BrienSenior Managing Editor: Scott DisannoArt Director: Kenny BeckCover Designer: Kristine CarneyArt Editor: Greg DullesManufacturing Manager: Alan FischerManufacturing Buyer: Lisa McDowell

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

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 priorto 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 PermissionsDepartment.

Pearson Education Ltd., London Pearson Education Singapore, Pte. Ltd.Pearson Education Canada, Inc.Pearson Education–JapanPearson Education Australia PTY, LimitedPearson Education North Asia, Ltd., Hong KongPearson 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: 978-0-13-136170-6ISBN-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

iiiv

5.1. For the beam cross-section shown in Fig. 5.33 determine whether the failure of the beam will be initiated bycrushing of concrete or yielding of steel. Given:

Also determine whether the section satisfies ACI Code requirements.

fy � 60,000 psi 1414 MPa2

f ¿c � 6000 psi 141.4 MPa2 for case 1b2, As � 6 in.2 f ¿c � 3000 psi 120.7 MPa2 for case 1a2, As � 1 in.2

1

Figure 5.33

Solution:

(a) The following information is given:

b = 8 in. section width

d = 18 in. section depth

dt = 16 in. depth to reinforcement

cf = 3000 psi required compression strength

fy = 60,000 psi steel strength

As = 1 in2 steel area

First, determine the value for 1 using equation 5.9.

1 = 0.85 ( )40002500 < cf

Then calculate the depth of the compression block.

a = bf

fA

c

ys

85.0

= )8)(3000(85.0

)000,60)(1(

= 2.94 in

1

2

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

c = 1

a

= 85.0

94.2

= 3.46 in

Then find the ratio of c and dt.

td

c =

16

46.3

= 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.

= bd

As

= )16)(8(

1

= 0.0078

This value must be greater than the larger of y

c

f

f3 and

yf

200.

000,60

30003 = 0.0027

000,60

200 = 0.0033

Since 0.0078 > 0.0033, the section satisfies the ACI Code.

(b) The following information is given:

b = 8 in. section width

d = 18 in. section depth

dt = 16 in. depth to reinforcement

cf = 6000 psi required compression strength

fy = 60,000 psi steel strength

As = 6 in.2 steel area

2

3

First, determine the value for 1 using equation 5.9.

1 = 1000

400005.085.0 cf

, ( )80004000 < cf

= 1000

4000600005.085.0

= 0.75

Then calculate the depth of the compression block.

a = bf

fA

c

ys

85.0

= )8)(6000(85.0

)000,60)(6(

= 8.82 in


c = 1

a

= 75.0

82.8

= 11.76 in


td

c =

16

76.11

= 0.735

Since this value is greater than 0.6, the flexure is compression controlled and the concrete crushes

before the steel yields.

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 simplysupported over a span of 12 ft. Given:

fy � 60,000 psi 1414 MPa2

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

4

Solution:

Design as a 1 ft wide, singly reinforced section.

The minimum depth for deflection is L

20= 12 12

20= 7.2 in, so try a depth of 8 in. Assume for

flexure an effective depth d = 7 in. Calculate the self weight.

self weight of a 12 in. strip = 150144

128

= 100 lb/ft

Then calculate the factored load.

factored external load wu = )load live(6.1)load dead weightself(2.1 ++

= )100(6.1)50100(2.1 ++

= 340 lb/ft

factored external moment Mu = 8

)( 2Lwu

= 8

)12)(340( 2

= 6120 ft-lb = 73,440 in-lb

The required nominal strength for the slab is Mn = 73,440

0.9= 81,600 in-lb .

Assume a moment arm of in 3.679.090.0 ==d and calculate the area of steel per 12 in. strip.

nM = )armmoment (ys fA

sA = )3.6)(000,60(

600,81

= 0.1889 in2

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

8

5

1 = 0031.0000,60

400033==

y

c

f

f

2 = 0033.0000,60

200200 ==yf

min sA = 2in 28.0)7)(12)(0033.0( =

Determine the value for 1.

1 = 0.85 ( )40002500 < cf

Calculate the depth of the compression block.

a = bf

fA

c

ys

85.0

= )12)(4000(85.0

)000,60)(28.0(

= 0.4118 in


c = 1

a

= 85.0

4118.0

= 0.4844 in


td

c =

7

4844.0

= 0.0692

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

nM = 2

adfA ys

= 2

4118.07)000,60)(28.0(

= 114,141 in-lb

9

6

The strength is sufficient, so accept the design. Now design the shrinkage and temperature

reinforcement.

The minimum required steel fraction is 0.0018.

Area = )8)(12(0018.0

= 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 40)8(5 = , so the maximum spacing is 18 in.

For a slab with a depth of 8 inches and using #4 bars, the maximum spacing is in 57.8)12(28.0

20.0 =

c-c for the main reinforcement and in 89.13)12(1728.0

20.0 = c-c (for the shrinkage and temperature

reinforcement).

10

5.5. Design the simply supported beams shown in Fig. 5.36 as rectangular sections. Given:

fy � 60,000 psi 1414 MPa2


7

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 = LD WW 6.12.1 +

= )1500(6.1)600(2.1 +

= 3120 lb/ft

factored external moment Mu = 8

)( 2Lwu

= 8

)20)(3120( 2

= 156,000 ft-lb = 1,872,000 in-lb

The required nominal strength for the beam is Mn = 1,872,000

0.9= 2,080,000 in-lb .

Figure 5.36

(a)

(b) (c)

1500 21.9

6008.7

15,000(66.7 kN)

7500 750033.4 33.4

11

8

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

1 = 3 f c

fy

= 3 6000

60,000= 0.0039

2 = 0033.0000,60

200200 ==yf

min sA = 2in 488.0)14)(9)(0039.0( =

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 = bf

fA

c

ys

85.0

= )9)(6000(85.0

)000,60)(3(

= 3.92 in


c = 1

a

= 75.0

92.3

= 5.23 in

Then find the ratio of c and d.

d

c =

15

23.5

= 0.348

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

16= 20(12)

16=15 in. Try h = 18 in.,

b = 0.5 h = 9 in., and d = 15 in. Assume c/d = 0.30.

Calculate the required steel area.

c = (0.30)(15) = 4.5 in.

1 = 1000

4000600005.085.0 = 0.75

a = 1 c = (0.75)(4.5) = 3.375 in.

Areq = y

c

f

baf)85.0( =

000,60

)375.3)(9)(6000)(85.0( = 2.58 in2

12

9

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

16= 20(12)

16=15 in. Try h = 16 in.,

b = 0.5 h = 8 in., and d = 14 in. Assume c/d = 0.30.

Calculate the factored external moment. The self weight is 144

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

Dead load Mu = 8

)20)(133)(2.1( 2

= 8000 ft-lb

Live load Mu = 2

)10)(000,15)(6.1( = 120,000 ft-lb

Mu = 128,000 ft-lb = 1,536,000 in-lb

The required nominal strength for the beam is 9.0

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

Calculate the required steel area.

c = (0.30)(14) = 4.2 in

1 = 1000

4000600005.085.0 = 0.75

a = 1 c = (0.75)(4.5) = 3.15 in

Areq = y

c

f

baf)85.0( =

000,60

)15.3)(8)(6000)(85.0( = 2.142 in2

0.348 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least

2,080,000 in-lb)

nM = 2

adfA ys

= 2

92.314)000,60)(3(

= 2,347,058 in-lb

13

10

a = bf

fA

c

ys

85.0 =

)8)(6000(85.0

)000,60)(37.2( = 3.49 in.

c = 1

a =

75.0

49.3 = 4.65 in.

d

c =

14

65.4 = 0.332

0.332 < 0.375, so the section is tension controlled. Verify the strength is sufficient (at least

1,706,667 in-lb)

nM = 2

adfA ys

= 2

49.314)000,60)(37.2(

= 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 144

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

Dead load Mu = 8

)20)(133)(2.1( 2

= 8000 ft-lb

Live load Mu = )5)(7500)(6.1( = 60,000 ft-lb

Mu = 68,000 ft-lb = 816,000 in-lb

The required nominal strength for the beam is 9.0

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

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

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

1 = 0039.0000,60

600033==

y

c

f

f

2 = 0033.0000,60

200200 ==yf

min sA = )14)(8)(0039.0( = 0.434 in2

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.

14

15

16

17

18

5.7. Compute the stresses in the compression steel, , for the cross sections shown in Fig. 5.38. Also compute thenominal moment strength for the section in part (b). Given:

fy � 60,000 psi 1414 MPa2


f ¿s

11

2.050.8

10

18(4

57.2

mm

)9 in.

(228.6 mm)2.0 in.

(50.8 mm)15 in.

(381 mm)

10

30 in

.(7

62 m

m)

Figure 5.38

(a) Calculate the steel areas.

sA = (3)(1.27) = 3.81 in2

sA = (2)(0.31) = 0.62 in2

Calculate 1 = 1000

4000700005.085.0 = 0.7.

Calculate the strain assuming the compression steel has yielded.

a = bf

fAA

c

yss

85.0

)( =

)9)(7000(85.0

)000,60)(62.081.3( = 3.57 in.

c = 1

a =

7.0

57.3 = 5.11 in.

s = c

dc003.0 =

11.5

211.5003.0 = 0.00182 in./in.

Check if the compression steel has actually yielded.

y = s

y

E

f =

61029

000,60 = 0.002 in./in.

19

12

s = c

dc003.0 =

22.5

222.5003.0 = 0.00185 in./in.

Repeat this calculation until the computed values converge. After several trials, sf = 53,605 psi,

a = 3.65 in., c = 5.21 in., and s = 0.00185 in./in.

(b) Calculate the steel areas

sA = (4)(1.27) = 5.08 in2

sA = (2)(0.79) = 1.58 in2

Calculate 1 = 1000

4000700005.085.0 = 0.7.

Calculate the strain assuming the compression steel has yielded.

a = bf

fAA

c

yss

85.0

)( =

)15)(7000(85.0

)000,60)(58.108.5( = 2.35 in.

c = 1

a =

7.0

35.2 = 3.36 in.

s = c

dc003.0 =

36.3

5.236.3003.0 = 0.000769 in./in.

Check if the compression steel has actually yielded.

y = s

y

E

f =

61029

000,60 = 0.002 in./in.

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.

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.

sf = ss E = (0.00182)(29 106) = 52,923 psi

a = bf

fAfA

c

ssys

85.0 =

)9)(7000(85.0

)923,52)(62.0()000,60)(81.3( = 3.66 in.

c = 1

a =

7.0

66.3 = 5.22 in.

20

13

sf = ss E = (0.000769)(29 106) = 22,294 psi

a = bf

fAfA

c

ssys

85.0 =

)15)(7000(85.0

)294,22)(58.1()000,60)(08.5( = 3.02 in.

c = 1

a =

7.0

02.3 = 4.31 in.

s = c

dc003.0 =

31.4

5.231.4003.0 = 0.00126 in./in.

Repeat this calculation until the computed values converge. After several trials, sf = 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.

t = c

cdt003.0 = 04.4

04.45.27003.0 = 0.0174 in./in.

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.

= )5.27)(15(

08.5=t

s

bd

A = 0.0123

1 = 0042.0000,60

700033==

y

c

f

f

2 = 0033.0000,60

200200 ==yf

0.0123 > max(0.0042,0.0033) so the ACI Code is satisfied.

Calculate the nominal strength of the section.

nM = )(2

)( ddfAa

dfAfA ssssys +

= (5.08)(60,000) (1.58)(33,149)[ ] 27.5 2.83

2

+ (1.58)(33,149)(27.5 2.5)

= 7,894,097 in-lb

M = Mn = (0.9)(7,894,097) = 7,104,867 in-lb

21

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 sectionsor as flanged sections. Given:

fy � 60,000 psi 1414 MPa2


14

20 508

2 in.(50.8 mm)

12 in

.(3

04.8

mm

)

8

8 in.(203.2 mm)

20 5084 in.

(101.6 mm)

9

8

30 in

.(7

63 m

m)

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

sA = (3)(0.79) = 2.37 in2

a = bf

fA

c

ys

85.0 =

)20)(4000(85.0

)000,60)(37.2( = 2.09 in.

c = 1

a =

85.0

09.2 = 2.46 in.

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


29

15

1 = 0.85

sA = (5)(0.79) = 3.95 in2

a = bf

fA

c

ys

85.0 =

)30)(4000(85.0

)000,60)(95.3( = 2.32 in.

c = 1

a =

85.0

32.2 = 2.73 in.

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


1 = 0.85

sA = (4)(1) = 4 in2

a = bf

fA

c

ys

85.0 =

)20)(4000(85.0

)000,60)(4( = 3.53 in.

c = 1

a =

85.0

53.3 = 4.15 in.

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

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 deadload WD = 1000 lb per ft (14.6 kN/m) and live load wL = 1500 lb per ft (21.9 kN/m). Determine the maximum fac-tored vertical shear Vu at the critical section. Determine the nominal shear resistance Vc by both the short methodand by the more refined method of taking the contribution of the flexural steel into account. Design the size andspacing of the diagonal tension reinforcement. Given:

Assume that no torsion exists.

fy � 60,000 psi 1413.7 MPa2

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

As � 6.0 in.2 13780 mm22

h � 20 in. 1508 mm2

d � 18 in. 1457.2 mm2

bw � 12 in. 1305 mm2

1

Calculate the factored loads.

self-weight = 144

)20)(12(150 = 250 lb/ft

wu = 1.2(250 + 1000) + 1.6(1500) = 3900 lb/ft

Vu at support = 2

)20)(3900( = 39,000 lb

Vu at d = 39,000 3900 (18 / 12) = 33,150 lb

Mu at d = 2

)12/18)(3900()12/18(000,39

2

= 54,112.5 ft-lb

= 649,350 in.-lb

Use the simplified method to calculate the shear loads.

Vc = bdfc2 = )18)(12(4000)0.1(2 = 27,322 lb

Vn = uV

= 75.0

150,33 = 44,200 lb

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 = s

yV

V

dfA =

878,16

)18)(000,60)(22.0( = 14.08 in.

Since Vs < 4 f c bd = 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

43

44

45

46

47

48

49

2

The minimum thickness required for deflection is 16

)12(18

16=L

= 13.5 in., which is quite small.

Assume beam dimensions of h = 24 in., d = 21 in., and b = 12 in.

Now calculate the factored loads.

self-weight = 144

)24)(12(150 = 300 lb/ft

wu = 1.2(300 + 400) + 1.6(800) = 2120 lb/ft

Pu = 1.2(25,000) + 1.6(30,000) = 78,000 lb

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.

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 ser-vice concentrated dead load PD of 25,000 lb and an external service concentrated live load PL of 30,000 lb (133kN) are applied to one midspan only. Design the diagonal tension reinforcement necessary. Given:

fy � 60,000 psi 1413.7 MPa2


50

3

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 = bf

fA

c

ys

85.0 =

)12)(5000(85.0

)000,60)(08.5( = 5.98 in.

c = 1

a =

8.0

98.5 = 7.47 in.

d

c =

21

47.7 = 0.3557 < 0.375 tension controlled, = 0.90

Check the minimum reinforcement area, >yy

cs

ff

f

bd

A 200,

3max , and the strength,

=2

adfAM ysn .

0202.0)21)(12(

08.5 = > == 0033.0000,60

200,0035.0

000,60

50003max

2

98.521)000,60)(08.5(9.0 = 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 = bf

fA

c

ys

85.0 =

)12)(5000(85.0

)000,60)(81.3( = 4.48 in.

c = 1

a =

8.0

48.4 = 5.60 in.

d

c =

21

60.5 = 0.2668 < 0.375 tension controlled, = 0.90

Check the minimum reinforcement area, >yy

cs

ff

f

bd

A 200,

3max , and the strength,

=2

adfAM ysn .

0151.0)21)(12(

81.3 = > == 0033.0000,60

200,0035.0

000,60

50003max

51

4

2

48.421)000,60)(81.3(9.0 = 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 = bdfc2 = )21)(12(5000)0.1(2 = 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

12

2121205.587,47 = 43,877.5 lbf.

Vn = uV

= 75.0

5.877,43 = 58,503 lb (Vn > Vc / 2, support required)

Vs = Vn Vc = 58,503 35,638 = 22,865 lb

s = s

yV

V

dfA =

865,22

)21)(000,60)(22.0( = 12.12 in.

Since Vs < )21)(12(500044 =bdfc = 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 = uV

= 75.0


Vs = Vn Vc = 68,817 35,638 = 33,178 lb

s = s

yV

V

dfA =

178,33

)21)(000,60)(22.0( = 8.35 in.

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

5

Vn = uV

= 75.0


Vs = Vn Vc = 97,083 35,638 = 61,445 lb

s = s

yV

V

dfA =

445,61

)21)(000,60)(22.0( = 4.51 in.

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

6.6. Design the vertical stirrups for a beam having the shear diagram shown in Figure 6.37 assuming that Vc =2 bwd. Given:

fy � 60,000 psi 1414 MPa2


Vu3 � 30,000 lb 1133 kN2

Vu2 � 50,000 lb 1222 kN2

Vu1 � 80,000 lb 1356 kN2

dw � 20 in. 1508 mm2

bw � 12 in. 1304.8 mm2

2f ¿c

6

Figure 6.37

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

Vc = bdfc2 = )20)(12(5000)0.1(2 = 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.

Vn = uV

= 75.0

000,80 = 106,667 lb (Vn > Vc / 2, support required)

Vs = Vn Vc = 106,667 33,941 = 72,726 lb

s = s

yV

V

dfA =

726,72

)20)(000,60)(4.0( = 6.60 in.

Since Vs > )20)(12(500044 =bdfc = 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.

Vn = uV

= 75.0


Vs = Vn Vc = 66,667 33,941 = 32,726 lb

s = s

yV

V

dfA =

726,32

)20)(000,60)(4.0( = 14.67 in.

54

7

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.

Vn = uV

= 75.0


Vs = Vn Vc = 40,000 33,941 = 6059 lb

s = s

yV

V

dfA =

6059

)20)(000,60)(4.0( = 79.2 in.

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

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 = 6in. (152.4 mm) from the column face; horizontal factored force Nuc = 25,000 lb (111 kN). Given:

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 wasnot roughened before casting the bracket. Detail the reinforcing arrangements for the bracket.

fy � fyt � 60,000 psi 1414 MPa2


b � 24 in. 1609.6 mm2

8

Shear friction approach:

First, choose dimensions for the corbel. Try h = 18 in. and d = 15 in. Then check the vertical

load.

Vn = uV

= 75.0

000,100 = 133,333 lb

V1 = bdfc2.0 = 0.2(5000)(24)(15) = 360,000 lb

V2 = bdfc )08.0480( + = (480 + 0.08 5000)(24)(15) = 316,800 lb

V3 = 1600bd = (1600)(24)(15) = 576,000 lb

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 = y

u

f

V =

)6.0)(000,60(

333,133 = 3.70 in2.

Af = )85.0(

)(

df

dhNaV

y

ucu +

= )15)(85.0)(000,60(75.0

)1518)(000,25()6)(333,133( +

= 1.18 in2.

An = y

uc

f

N =

)000,60)(75.0(

000,25 = 0.56 in2.

The primary tension steel area is the maximum of nvf AA +3

2, nf AA + , and bd

f

f

y

c04.0 .

Asc1 = nvf AA +3

2 = 56.0)70.3(

3

2 + = 3.02 in2.

Asc2 = nf AA + = 1.18 + 0.56 = 1.73 in2.

Asc3 = bdf

f

y

c04.0 = )15)(24(000,60

500004.0 = 1.2 in2.

75

9

Therefore, the required steel areas are Asc = 3.02 in2. and Ah = )(5.0 nsc AA = 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.

The required bearing plate area is )5000)(85.0(7.0

000,100

)85.0(7.0=

c

u

f

V = 33.6 in2. so use a square 6 in.

by 6 in. plate thick enough to be rigid under the load.

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.

BCF = d

LV BC

u = 113,333 lb

ABT = uc

BC

BCBC N

L

xF + = 78,333 lb

ACF = AC

ACAB

x

LT = 166,458 lb

ADT = AC

ACL

dF

CEF = BC

BC

AC

ACL

dF

L

dF + = 246,875 lb

CDT = BC

BCBC

AC

ACAC

L

xF

L

xF = 25,000 lb

The required bearing plate area is )5000)(85.0(7.0

000,100

)85.0(75.0=

c

u

f

V = 31.4 in2. so use a square

plate that is 6 in. 6 in. and thick enough to be rigid under the load.

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 ABT . 76

10

AAB = )000,60(75.0

333,78

75.0=

y

AB

f

T = 1.74 in2.

ACD = )000,60(75.0

000,25

75.0=

y

CD

f

T = 0.56 in2.

A portion of the reinforcement resists the internal friction force.

An = )000,60(75.0

000,25

75.0=

y

uc

f

N = 0.56 in2.

Therefore, the required steel areas are AAB = 1.74 in2., ACD = 0.56 in2., and 2in59.0)(5.0 == nABh AAA . 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.

Strut AC: AC

h

L

d

bs

tieA

)0.3)(24(

)2.0(2sin

/= = 0.002279 < 0.003 so increase the bar size to No. 4 and

use 3 No. 4 closed stirrups spaced 3 in. c-c.

ACL

d

)0.3)(24(

)31.0(2 = 0.003799 > 0.003, OK

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: )3400)(24(

458,166 = 2.04 in.

Strut CE: )3400)(24(

875,246 = 3.02 in.

Strut BC: )3400)(24(

333,113 = 1.39 in.

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

77

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 dis-tance of 3 ft 6 in. (1.07 m) from the wall support. In addition, the beam has to resist an equilibrium fac-tored torsion Tu = 450,000 in.-lb (50.8 kN–m). The beam cross section is 15 in. × 30 in. (381 mm × 762mm) with an effective depth of 27 in. (686 mm). Design the stirrups and the additional longitudinal steelneeded.Given:

As � 4.0 in.2 12580.64 mm22

fy � fyt � 60,000 psi

f ¿c � 4000 psi

1

The length of the beam is not given, so ignore the self weight. Use the following beam

dimensions and loads.

Beam height h = 30 in.

Effective depth d = 27 in.

Beam width b = 15 in.

Force position L = 3.5 ft

Clear cover = 1.5 in.

Factored force uV = LP6.1 = 1.6(30,000) = 48,000 lb

Factored moment nM = LPL6.1 = 1.6(30,000)(3.5 12) = 2,016,000 in-lb

Factored torsion uT = 450,000 in-lb

Determine the minimum steel required for shear support. The load factor for shear is 75.0= ,

and 0.1= for normal weight concrete.

Nominal load nV = uV

= 75.0

000,48 = 64,000 lb

Concrete support cV = bdfc2 = )27)(15(4000)0.1(2 = 51,229 lb

Required support sV = nV cV = 64,000 51,229 = 12,771 lb

Required area s

A =

df

V

yt

s = )27)(000,60(

771,12 = 0.00788 in.2 / in.

Determine the minimum steel required for torsion. The load factor for torsion is also 75.0= .

Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions 1x and 1y .

Nominal load nT = uT

= 75.0

000,450 = 600,000 in.-lb

Outer box 0x = b = 15 in.

0y = h = 30 in.

cpA = 00 yx = (15)(30) = 450 in.2

cpp = )(2 00 yx + = 2(15 + 30) = 90 in.

85

2

Strut angle = 45 degrees

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

No support if uT < cp

cpc

p

Af 2

= 90

)450(4000)0.1(75.0 2

= 106,727 in.-lb (false)

Then verify the section meets the requirement +<+ cc

oh

huu fbd

V

A

pT

bd

V8

7.1

2

2

2

.

2

2

2

7.1+

oh

huu

A

pT

bd

V =

2

2

2

)75.304(7.1

)76)(000,450(

)27)(15(

000,48 + = 246.92 psi

+ cc f

bd

V8 = + 4000)0.1(8

)27)(15(

229,5175.0 = 474.34 psi

246.92 psi < 474.34 psi, so the section is adequate.

Determine the minimum required area of transverse steel.

Torsion s

At = cot2 0 yt

n

fA

T =

)1)(000,60)(04.259(2

000,600 = 0.0193 in.2 / in.

Torsion and shear s

A =

s

A

s

At +2 = 2(0.0193) + 0.00788 = 0.0465 in.2 / in. (controls)

Minimum s

A = b

f

f

y

c75.0 = )15(

000,60

400075.0 = 0.0119 in.

Determine the minimum required longitudinal reinforcement.

Torsion LA = 2)(cot

y

yt

ht

f

fp

s

A = 2)1)(1)(76)(0193.0( = 1.47 in.2

Minimum s

At s

At ytf

b25 =

000,60

1525 = 0.0063 in. (does not control)

Inner box 1x = 0x 2(clear cover) (stirrup d) = 15 2(1.5) 0.5 = 11.5 in.

1y = 0y 2(clear cover) (stirrup d) = 30 2(1.5) – 0.5 = 26.5 in.

ohA = 11 yx = (11.5)(26.5) = 304.75 in.2

hp = )(2 11 yx + = 2(11.5 + 26.5) = 76 in.

Torsion area 0A = 0.85 ohA = 0.85(304.75) = 259.04 in.2

86

3

Minimum LA y

yt

ht

y

cpc

f

fp

s

A

f

Af5

)1)(76)(0193.0(000,60

)450(4000)0.1(5 = 0.9 in.2

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

the smallest of in. 5.98

76

8==hp

, 12 in., and 0465.0

)2.0(2 = 4.3 in. (controls)

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

88

89

90

91

92

93

94

95

96

97

98

99

100

101

4

7.5. Design the rectangular beam shown in Figure 7.34 for bending, shear, and torsion. Assume that thebeam width b = 12 in. (305 mm). Given:

fy � fyt � 60,000 psi 1413.8 MPa2

f ¿c � 4000 psi 127.6 MPa2

10,000 44.5

4 ft (1.22 m)

12 ft (3.66 m)

Figure 7.34

To include the self-weight, a height must be assumed. Try the following dimensions:




Beam length L = 12 ft

“L” length 2L = 2 ft

Clear cover = 1.5 in.

Start by calculating the support force and moment.

Self weights 1w = 150bh = )150(144

)24)(12( = 300 lb / ft

2w = 1502hL = )150(12

)24)(2( = 600 lb / ft

Factored force AF = LPbwbLw 6.12.1)(2.1 21 ++

= 1.2(300)(12 1) + 1.2(600)(1) + 1.6(10,000)

= 20,680 lb

Factored moment AM = ++2

6.12

2.12

)(2.12

2

1 bLP

bLbw

bLwL

= )5.11)(000,10(6.1)5.11)(1)(600(2.12

)11)(300(2.1 2

++

= 214,060 ft-lb = 2,568,720 in.-lb

Factored torsion AT = 222 6.1

2

2.1LP

bLwL+

= )2)(000,10(6.12

)2)(1)(600(2.1 +

= 32,720 ft-lb = 392,640 in.-lb

102

5

Then calculate the design loads.

Factored shear uV = dwFA 12.1 = 20,680 1.2(300)(21/12) = 20,050 lb

Factored moment nM = AM = 2,568,720 in.-lb

Factored torsion uT = AT = 392,640 in.-lb

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

bending.

Nominal load nM = uM

= 9.0

720,568,2 = 2,854,133 in.-lb

Section strength nM 2

adfA ys and a =

bf

fA

c

ys

85.0

Solve for sA sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c

2)21)(12)(4000(85.0

)133,854,2(211

000,60

)21)(12)(4000(85.0

2.48 in.2 (controls)

Minimum area 1 sA bdf

f

y

c3 = )21)(12(

000,60

40003 = 0.8 in.2

Minimum area 2 sA bdf y

200 = )21)(12(

000,60

200 = 0.84 in.2

Determine the minimum steel required for shear support. The load factor for shear is 75.0= ,

and 0.1= for normal weight concrete.

Nominal load nV = uV

= 75.0

050,20 = 26,733 lb

Concrete support cV = bdfc2 = )21)(12(4000)0.1(2 = 31,876 lb

Since cV > nV , the stirrup size will be determined by the torsion design.

Determine the minimum steel required for torsion. The load factor for torsion is also 75.0= .

Assume #4 bars, diameter = 0.5 in., for the stirrups to calculate the dimensions 1x and 1y .

Nominal load nT = uT

= 75.0

640,392 = 523,520 in.-lb

103

6

Torsion area 0A = 0.85 ohA = 0.85(174.25) = 148.11 in.2

Strut angle = 45 degrees

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

No support if uT < cp

cpc

p

Af 2

= 72

)288(4000)0.1(75.0 2

= 54,644 in.-lb (false)

Then verify the section meets the requirement +<+ cc

oh

huu fbd

V

A

pT

bd

V8

7.1

2

2

2

.

2

2

2

7.1+

oh

huu

A

pT

bd

V =

2

2

2

)25.174(7.1

)58)(640,392(

)21)(12(

050,20 + = 448.31 psi

+ cc f

bd

V8 = + 4000)0.1(8

)21)(12(

876,3175.0 = 474.34 psi

448.31 psi < 474.34 psi, so the section is adequate.

Determine the minimum required area of transverse steel.

Torsion s

At = cot2 0 yt

n

fA

T =

)1)(000,60)(11.148(2

520,523 = 0.0295 in.

Torsion and shear s

A =

s

A

s

At +2 = 2(0.0295) + 0 = 0.0589 in. (controls)

Minimum s

A = b

f

f

y

c75.0 = )12(

000,60

400075.0 = 0.0095 in.

Outer box 0x = b = 12 in.

0y = h = 24 in.

cpA = 00 yx = (12)(24) = 288 in.2

cpp = )(2 00 yx + = 2(12 + 24) = 72 in.

Inner box 1x = 0x 2(clear cover) (stirrup d) = 12 2(1.5) 0.5 = 8.5 in.

1y = 0y 2(clear cover) (stirrup d) = 24 2(1.5) – 0.5 = 20.5 in.

ohA = 11 yx = (8.5)(20.5) = 174.25 in.2

hp = )(2 11 yx + = 2(8.5 + 20.5) = 58 in.

104

7

)1)(58)(0295.0(000,60

)288(4000)0.1(5 = 0.19 in.2

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

is the smallest of in. 25.78

58

8==hp

, 12 in., and 0589.0

)31.0(2 = 5.26 in. (controls). Therefore, use #5

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.

Determine the minimum required longitudinal reinforcement.

Torsion LA = 2)(cot

y

yt

ht

f

fp

s

A = 2)1)(1)(58)(0295.0( = 1.71 in.2

Minimum s

At s

At ytf

b25 =

000,60

1225 = 0.005 in. (does not control)

Minimum LA y

yt

ht

y

cpc

f

fp

s

A

f

Af5

105

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 spanningover 15 ft. The service dead and live loads are 60 psf (28.7 kPa) and 150 psf (71.8 kPa), respectively. The rein-forcement consists of No. 5 bars (16-mm diameter) at 6 in. center to center (154 mm center to center). Alsocheck which limitations, if any, need to be placed on its usage. Assume that 50% of the live load is sustainedover a 30-month period. Given:

Es � 29 � 106 psi 1200,000 MPa2

fy � 60,000 psi 1414 MPa2

f ¿c � 5000 psi 134.5 MPa2

1

Design the one-way slab as a rectangular beam with a width of 12 in. Assume the depth of the

steel is 6 in.




Span L = 15 ft

First, estimate the properties of the concrete.

Weight per foot w = 150144

)12)(8( = 100 lb / ft

Modulus cE = cf000,57 = 5000000,57 = 4,030,509 psi

Tensile rupture rf = cf5.7 = 50005.7 = 530 psi

Stiffness ratio n = c

s

E

E =

509,030,4

1029 6

= 7.19

Then calculate the properties of the elastic beam.

Center of gravity y = 2

h =

2

8 = 4 in.

Moment of inertia gI = 3

12

1bh = 3)8)(12(

12

1 = 512 in.4

Strength crM = y

fI rg

r = 4

)530)(512( = 67,882 in-lb

Calculate the properties of the beam during the first phase of cracking.

Center of gravity equation dnAcnAcb

ss+2

2 = 0

Solve for c c = 1.7728 in.

Moment of inertia crI = 23 )(

3

1cdnAbc s+

= 23 )77.16)(31.02)(19.7()77.1)(12(

3

1 +

= 102 in.4

Calculate the load moments. 115

2

Dead load DM = 8

)( 2LwbLD + =

[ ]8

)1215(100)12)(60( 2+ = 54,000 in-lb

Live load LM = 8

2bLLL = 8

)1215()12)(150( 2

= 50,625 in-lb

Now calculate the moments for the deflections.

Dead M = DM = 54,000 in-lb

Dead plus 50% live M = DM + 0.5 LM = 54,000 + 0.5(50,625) = 79,312.5 in-lb

Dead plus live M = DM + LM = 54,000 + 50,625 = 104,625 in-lb

Calculate the effective moment of inertia for each case.

Equation eI = crcrgcr III

M

M+)(

3

if M > crM

Dead eI = gI = 512 in.4 ( M < crM )

Dead plus 50% live eI = 102)102512(5.312,79

882,673

+ = 359 in.4

Dead plus live eI = 102)102512(625,104

882,673

+ = 214 in.4

Calculate the deflections, use t = 1.75 and = 2.

Equation = eEI

ML

48

5 2

(deflection of simply supported beam)

Immediate DL D = )512)(1003.4(48

)1215)(000,54(56

2

= 0.0883 in.

Immediate LL L = )214)(1003.4(48

)1215)(625,104(56

2

D = 0.3211 in.

Immediate 50% LL LS = )359)(1003.4(48

)1215)(5.312,79(56

2

D = 0.0967 in.

Long term deflection LT = LStDL ++

= (0.3211) + (2)(0.0883) + (1.75)(0.0967) = 0.6669 in.

Now check the deflection requirements.

Roof with no supported or attached nonstructural elements

likely to be damaged by large deflections 180

L =

180

)12(15 = 1 in. > L

116

3

Floor with no supported or attached nonstructural elements

likely to be damaged by large deflections 360

L =

360

)12(15 = 0.5 in. > L

Roof or floor with supported or attached nonstructural

elements likely to be damaged by large deflections 480

L =

480

)12(15 = 0.375 in. < LT

Roof or floor with supported or attached nonstructural

elements not likely to be damaged by large deflections 480

L =

240

)12(15 = 0. 75 in. > LT

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

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 aggres-sive chemical environment. Calculate the maximum expected flexural crack width and whether the beam sat-isfies the serviceability criteria for crack control. Given:

minimum clear cover � 112

in. 138.1 mm2

fy � 60,000 psi 1414 MPa2

f ¿c � 4500 psi 131.0 MPa2

4

24 in.

8

b = 10 in.

d =

20

in.

t = 4 in.

Figure 8.21 Beam geometry.




First, estimate the properties of the concrete.

Modulus cE = cf000,57 = 4500000,57 = 3.824 106 psi

Tensile rupture rf = cf5.7 = 45005.7 = 503 psi

Stiffness ratio n = c

s

E

E =

6

6

10824.3

1029 = 7.58

Then calculate the properties of the elastic beam.

Center of gravity y = 2

h =

2

24 = 12 in.


12

1bh = 3)12)(10(

12

1 = 11,520 in.4

Strength crM = y

fI rg

r = 12

)503)(520,11( = 482,991 in-lb

Calculate the center of gravity for the beam during the first phase of cracking.

Center of gravity equation b

2c 2 + nAs + (n 1) A s[ ] c nAsd + (n 1) A s d [ ] = 0

Solve for c c = 7.54 in.

140

5

maxw = 3076.0 Adf cs

= )20)(5.2()36)(32.1(076.0 = 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.

Then calculate the values for the crack equation, 3076.0 Adf cs .

Depth factor = cd

ch =

54.720

54.724 = 1.32

Steel stress sf = 0.6 yf = 0.6(60,000) = 36,000 psi = 36 ksi

Number of bars bc = 4

Tension area A = bc

bdh )(2 =

4

10)2024(2 = 20 in.2

First layer depth cd = (clear cover) + (stirrup d) + 0.5(bar diameter)

= 1.5 + 0.5 + 0.5(1) = 2.5 in.

Finally, calculate the maximum crack size.

141

142

8.9. Find the maximum web of a beam reinforced with bundled bars to satisfy the crack-control criteria for interiorexposure conditions. Given:

No. 4 stirrups used 113-mm diameter2

As � three bundles of three No. 9 bars each 1three bars of 28.6-mm diameter each in a bundle2

fy � 60,000 psi 1414 MPa2

f ¿c � 5000 psi 134.5 MPa2

6

Since the beam dimensions are unknown, assume = 1.2, sf = 0.6 yf , and a clear cover of 1.5

in. The crack equation for a beam with bundles bars is

maxw = 3076.0 Adf cs

= 1.2

sf = 0.6 yf = 0.6(60,000 psi) = 36 ksi

Calculate the center of gravity of the bundles from the bottom of the beam.

cd = (clear cover) + (stirrup d) + (bundle CG)

= )1(6

335.05.1

+++ = 2.890 in.

Write the area of concrete in tension in terms of the web width b.

Number of bars bc = 9

Effective number bc = 0.650 bc = 0.650(9) = 5.85

Effective area A = bc

cbd2 = b

85.5

)890.2(2 = 0.9879b

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

known values into the equation.

Substitute values 016.0 = 1000

)9879.0)(890.2()36)(2.1(076.0 3 b

Solve for b b = )9879.0)(890.2(

1

)36)(2.1)(076.0(

)016.0)(1000(3

= 40.54 in.

Therefore, the maximum beam web width is 40.5 in.

143

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

1

9

8000

8 in.(203.2 mm)

12 in

.(3

04.8

mm

)Figure 9.46 Column sections.

144

2

Beam area gA = bh = (8)(12) = 96 in.2

Steel area stA = 6(1) = 6 in.2

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

Nominal strength nP = (0.80) 0.85 f c (Ag Ast ) + Ast fy[ ]

= )]000,60)(6()696)(8000(85.0)[80.0)(65.0( +

= 505,400 lb

145

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 ifPu = 400,000 lb. Use the trial-and-adjustment method satisfying the compatibility of strains.

3

Beam area gA = bh = (8)(12) = 96 in.2

Top steel area stA = 3(1) = 3 in.2

Bottom steel area stA = 3(1) = 3 in.2

For the column = 0.65 and for cf = 8000 psi, 1 = 0.65. Guess that the neutral axis passes

through the center of the section, c = 6 in.

Compression block a = c1 = (0.65)(6) = 3.9 in.

Concrete force CF = bafc85.0 = (0.85)(8000)(8)(3.9) = 212,160 lb

Top steel strain s = c

dc003.0 =

6

36003.0 = 0.0015 (compression)

Top steel stress sf = ssE = (29 106)(0.0015) = 43,500 psi < 60,000 psi OK

Top steel force SF = ss fA = (3)(43,500) = 130,500 lb

Bottom steel strain s = c

cd003.0 =

6

69003.0 = 0.0015 (tension)

Bottom steel stress sf = ssE = (29 106)(0.0015) = 43,500 psi < 60,000 psi OK

Bottom steel force SF = ss fA = (3)(43,500) = 130,500 lb

Nominal strength nP = CF + SF SF = 212,160 lb

Strength uP = nP = 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 a = c1 = (0.65)(11) = 7.15 in.



dc003.0 =

6

311003.0 = 0.002182 (compression)

Top steel stress sf = ssE = (29 106)(0.002182) = 63,273 psi > 60,000 psi

= 60,000 psi

Top steel force SF = ss fA = (3)(60,000) = 180,000 lb


dc003.0 =

6

911003.0 = 0.000545 (compression)


Bottom steel force SF = ss fA = (3)(15,818) = 47,455 lb


157

4

Strength uP = nP = 0.65(616,415) = 400,669 lb

The strength is sufficient, so calculate the eccentricity.

Nominal moment nM = )()(2

dyFdyFa

yF ssC ++

= (388,960)(6 0.5 7.15) + (180,000)(6 3) + (47,455)(6 9)

= 1,340,864 in.-lb

Eccentricity e = n

n

P

M =

415,616

864,340,1 = 2.18 in.

158

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 factoredultimate axial force Pu = 200,000 lb. The eccentricity e to geometric centroid = 8 in. Given:

fy � 60,000 psi

f ¿c � 4000 psi

5

Use the following design parameters.

Force uP = 200,000 lb

Eccentricity e = 7 in.

Moment uM = ePu = (200,000)(7) = 1,400,000 in.-lb

Top steel depth d = 3 in.

Bottom steel depth d = h 3 = 15 – 3 = 12 in.

Reinforcement ratios = = 0.015

Choose the longitudinal reinforcement (the same for both top and bottom).

Required area stA = bd = (0.015)(12)(12) = 2.16 in.2

Use 3 No. 8 bars stA = (3)(0.79) = 2.37 in.2

Check for compression failure using = 0.65 and for cf = 4000 psi, 1 = 0.85.

Neutral axis depth c = 0.6 d = 0.6(12) = 7.2 in.

Compression block a = c1 = (0.85)(7.2) = 6.12 in.



dc003.0 =

2.7

32.7003.0 = 0.00175 (compression)


Top steel force SF = ss fA = (2.37)(50,750) = 120,278 lb


cd003.0 =

2.7

2.712003.0 = 0.002 (tension)


Bottom steel force SF = ss fA = (2.37)(58,000) = 137,460 lb


Strength uP = nP = 0.65(232,514) = 151,134 lb

Nominal moment nM = )()(

2dyFdyF

ayF ssC ++

= )125.7)(460,137()35.7)(278,120(2

12.615)696,249( +

= 2,268,469 in-lb

Eccentricity e = n

n

P

M =

514,232

469,268,2 = 9.76 in.

169

6

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.





dc003.0 =

08.8

308.8003.0 = 0.00189 (compression)




cd003.0 =

08.8

08.812003.0 = 0.00146 (tension)




Strength uP = nP = 0.65(309,488) = 201,167 lb


2dyFdyF

ayF ssC ++

= )125.7)(184,100()35.7)(596,129(2

87.615)076,280( +

= 2,173,277 in.-lb

Eccentricity e = n

n

P

M =

488,309

277,173,2 = 7.02 in.

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

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 ser-vice dead-load moments M2 = 4,000,000 on top and M1 = 2,000,000 in.-lb at the bottom. The service live-loadmoments are 80% of the dead-load moments. The column carries a service axial dead load PD = 200,000 lb anda service live load PL = 400,000. Design the cross-section size and reinforcement for this column. Given:

d¿ � 3 in.

�A � 1.3, �B � 0.9

fy � 60,000 psi

f ¿c � 8000 psi

7

First, calculate the factored loads on the column.

Force uP = 1.2(200,000) + 1.6(400,000) = 880,000

Moment 1 uM1 = 1.2(3,000,000) + 1.6(0.8)(3,000,000) = 7,440,000 in.-lb

Moment 2 uM1 = 1.2(4,000,000) + 1.6(0.8)(4,000,000) = 9,920,000 in.-lb

Assume the column dimensions.

Section height h = 25 in.

Section width b = 25 in.

Check if slenderness is an issue.

Effective length 1 k )(05.07.0 BA ++ = 0.7 + 0.05(1.3 + 0.9) = 0.81

Effective length 2 k min05.085.0 + = 0.85 + 0.05(0.9) = 0.895

Effective length k = 0.81

Radius of gyration r = h3.0 = 0.3(25) = 7.5 in.

Slenderness ratio r

kL =

5.7

)1230)(81.0( = 38.88

Max slenderness ratio r

kL

2

11234M

M =

000,920,9

000,440,71234 = 25

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 2M = )03.06.0( hPu + = 880,000(0.6 + 0.03 25) = 1,188,000 in.-

2M = 9,920,000 in.-lb

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

Concrete strength cE = cfw 5.133 = 8000)150(33 5.1 = 5.422 106 psi


3bh =

12

)25(25 3

= 32,552 in.4

179

8

EI EI = dns

gc IE

+1

40.0 =

5.01

)552,32)(10422.5(40.0 6

+ = 4.707 1010 in.2-lb

Euler load cP = 2

2

)(kL

EI =

2

102

)123081.0(

)10707.4( = 5,463,464 lb

mC = +2

14.06.0M

M = +

000,920,9

000,440,74.06.0 = 0.9

Magnification factor ns = )75.0/(1 cu

m

PP

C =

)464,463,575.0/(000,8801

mC = 1.146

Design moment uM = 2Mns = 1.146(9,920,000) = 11,369,772 in.-lb

Therefore, design the section as a non-slender column with uP = 880,000 lb, uM = 11,369,772

in.-lb, and e = u

u

P

M = 12.92 in.

Top steel depth d = 3 in.

Bottom steel depth d = h 3 = 25 – 3 = 22 in.

Reinforcement ratios = = 0.015

Choose the longitudinal reinforcement (the same for both top and bottom).

Required area stA = bd = (0.015)(25)(22) = 8.25 in.2

Use 7 No. 10 bars stA = (7)(1.27) = 8.89 in.2

Check for compression failure using = 0.65 and for cf = 8000 psi, 1 = 0.65.



Concrete force CF = bafc85.0 = (0.85)(8000)(25)(8.58) = 1,458,600 lb


dc003.0 =

2.13

32.13003.0 = 0.00232 (compression)


sf = 60,000 psi



cd003.0 =

2.13

2.1312003.0 = 0.002 (tension)



Nominal strength nP = CF + SF SF = 1,476,380 lb

Strength uP = nP = 0.65(1,476,380) = 959,647 lb

180

9


2dyFdyF

ayF ssC ++

= )35.12)(400,533(2

58.825)600,458,1( +

)225.12)(620,515(

= 21,940,796 in.-lb

Eccentricity e = n

n

P

M =

380,476,1

796,940,21 = 14.86 in.

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.



Concrete force CF = bafc85.0 = (0.85)(8000)(25)(9.17) = 1,558,271 lb


dc003.0 =

10.14

310.14003.0 = 0.00236 (compression)


sf = 60,000 psi



cd003.0 =

10.14

10.1422003.0 = 0.00168 (tension)




Strength uP = nP = 0.65(1,658,502) = 1,078,026 lb


2dyFdyF

ayF ssC ++

= )35.12)(400,533(2

17.925)271,558,1( +

)225.12)(169,433(

= 21,519,003 in.-lb

Strength uM = nM = 0.65(21.519,003) = 13,987,353 in.-lb

Eccentricity e = n

n

P

M =

502,658,1

003,519,21 = 12.97 in.

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

10

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

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 fac-tored load Pu = 500,000 lb acting at eccentricities ex = ey = 6 in. Design the column size and reinforcementneeded to resist the applied stresses. Given:

Solve by using all the three methods.

d¿ � 2.5 in.

gross reinforcement percentage �g � 0.03

fy � 60,000 psi

f ¿c � 6000 psi

11

Determine the design loads.

Factored load uP = 500,000 lb

Nominal load nP = uP

= 65.0

000,500 = 769,231 lb

Factored x moment uxM = yueP = (500,000)(4) = 2,000,000 in.-lb

Nominal x moment nxM = uxM

= 65.0

000,000,2 = 3,076,923 in.-lb

Factored y moment uyM = xueP = (500,000)(4) = 2,000,000 in.-lb

Nominal y moment nyM = uyM

= 65.0

000,000,2 = 3,076,923 in.-lb

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 b

h = 1.

Axial column area axialA = )(45.0 gyc

n

ff

P

+ =

)03.0000,606000(45.0

231,769

+ = 219 in.2

Twice the area gA = 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 0.03 22 22 / 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 c = 0.6 d = 11.7 in.

Compression block a = c1 = 8.775 in.

Concrete force CF = bafc85.0 = 984,555 lb


dc003.0 = 0.00236 (compression)

Top steel stress sf = ssE = 68,410 psi > 60,000 psi so use 60,000 psi

Top steel force SF = ss fA = 300,000 lb

195

12

Bottom steel force SF = ss fA = 290,000 lb



2dyFdyF

ayF ssC ++ = 11,525,370 in-lb

Eccentricity e = n

n

P

M = 11.59 in.

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 c = 0.857 d = 16.72 in.

Compression block a = c1 = 12.54 in.

Concrete force CF = bafc85.0 = 280,076 lb


dc003.0 = 0.002551 (compression)

Top steel stress sf = ssE = 73,991 psi > 60,000 psi so use 60,000 psi

Top steel force SF = ss fA = 300,000 lb


cd003.0 = 0.0005 (tension)

Bottom steel stress sf = ssE = 14,467 psi < 60,000 psi OK

Bottom steel force SF = ss fA = 72,335 lb



2dyFdyF

ayF ssC ++ = 9,819,937 in.-lb

Eccentricity e = n

n

P

M = 6.01 in.

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 oxM , where b = h = 22 in., nynx MM = = 4,615,385 in.-lb, and

start with = 0.5.

Required strength, x oxM = + 1

h

bMM nynx = 9,230,769 in.-lb


cd003.0 = 0.002 (tension)

Bottom steel stress sf = ssE 58,000 psi < 60,000 psi

196

13

From the section calculations, oxnM = 9,819,937 in.-lb > 9,230,769 in.-lb, so the strength of the

section is sufficient. Now check nyM . Using oxn

nx

M

M = 0.5 and = 0.5 in the figure,

oyn

ny

M

M = 0.5.

Therefore, nyM = 0.5 oynM , which is 4,909,968 in.-lb. Since nyM = 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, lb 630,634,1== nynx PP . Calculate

the strength of the column for an axial load, where Ast = 16(1.0) = 16 in.2

Axial strength 0nP = yststgc fAAAf +)(85.0 = 3,346,800 lb

Then calculate the nominal load strength.

Axial strength nP

1 =

0

111

nnynx PPP+

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

Modified Load Contour Method

Assume the same design and calculate the value of

5.15.1

0

++nby

ny

nbx

nx

nbn

nbn

M

M

M

M

PP

PP, where

nbP = 994,555 lb, nbxM = nbyM = 11,525,370 in.-lb, nxM = nyM = 4,615,385 in.-lb, and 0nP is

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

5.15.1

0

++nby

ny

nbx

nx

nbn

nbn

M

M

M

M

PP

PP = 0.411, which is significantly less than 1, which indicates

that the section is stronger than it needs to be.

197

198

199

200

201

202

203

204

205

10.1. Calculate the basic development lengths in tension for the following deformed bars embedded in normal-weight concrete.(a) No. 4, No. 10. Given:

(b) No. 14, No. 18. Given:

fy � 80,000 psi

fy � 60,000 psi

f ¿c � 4000 psi

fy � 60,000 psi 1413.7 MPa2

f ¿c � 6000 psi 141.4 MPa2

1

For a No. 4 bar:

Bar diameter bd = 0.5 in.

Bar area bA = 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, 1== et .

For a No. 4 bar, 8.0=s . 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(2 0.5 + 0.5) = 0.75 in. (controls)

Check the values.

cf = 6000 = 77.5 psi < 100 psi OK

b

tr

d

Kc + =

5.0

075.0 + = 1.5 OK

Calculate the development length.

dL = +

b

tr

ste

c

y

b

d

Kcf

fd

40

3 =

)5.1)(5.77(40

)8.0)(0.1)(0.1)(000,60(3)5.0(

= 15.49 in.

For a No. 10 bar:



206

(a)

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, 1== et .

For a No. 10 bar, 0.1=s . 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(3 1.27) = 1.91 in. (controls)

Check the values.

cf = 6000 = 77.5 psi < 100 psi OK

b

tr

d

Kc + =

27.1

091.1 + = 1.5 OK

Calculate the development length.

dL = +

b

tr

ste

c

y

b

d

Kcf

fd

40

3 =

)5.1)(5.77(40

)0.1)(0.1)(0.1)(000,60(3)27.1(

= 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 (452kN-m) and a factored shear Vu = 50,000 lb (222 kN) at the face of the support. Design the top reinforcementand the appropriate embedment of 90° hook into the concrete wall to sustain the external shear and moment.Given:

fy � 60,000 psi

f ¿c � 5000 psi

4

For a cantilever beam, the minimum depth for deflection is L / 8 or (12 15) / 8 = 22.5 in.




Beam length L = 15 ft

Design the beam for flexure.

Nominal moment nM =

Mu = 4,000,000

0.9= 4,444,444 in.-lb


adfA ys and a =

bf

fA

c

ys

85.0


211

85.0

bdf

M

f

bdf

c

n

y

c = 3.86 in.2 (controls)

Minimum area 1 sA bdf

f

y

c3 = 1.11 in.2

Minimum area 2 sA bdf y

200 = 1.05 in.2

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

Steel area sA = (4)(1) = 4 in.2

Compression block a = bf

fA

c

ys

85.0 = 3.76 in.

Nominal strength nM = 2

adfA ys = 4,129,411 in.-lb > 4,000,000 in.-lb.

Design the anchor.


Multiplier e = 1.0 for uncoated bars

Aggregate factor = 1.0 for normal concrete

Basic length hbL = b

c

yed

f

f02.0 = 19.14 in.

215

5

Multiplier d = st

req

A

A = 0.9653

Development length dhL = hbd L = 18.48 in. (> 8 bd and > 6 in. so OK)

Hook length hookL = 12 bd = 13.54 in.

Therefore, use a hook with dhL = 18.5 in and hookL = 14 in.

216

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 psfand an external dead load wD = 20 psf in addition to its self-weight. The center-line dimensions of the panel are20 ft × 30 ft (the dimension of the discontinuous side is 20 ft). Design the panel and the size and spacing of the re-inforcement using the ACI Code direct design method. Given:

= 4000 psi, normal-weight concrete

fy = 60,000 psi

column sizes 24 in. × 24 in.

width of the supporting beam webs = 15 in.

f ¿c

1

Assume the following dimensions.

Slab thickness sh = 8 in.

Beam section height bh = 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.

1. The maximum length ratio is 20

30 = 1.5 which is less than 2.

2. Assume there are more than three panels in each direction.

3. Check the live load to dead load ratio.

dw = 150(8/12) + 20 = 120 lb / ft2

lw = 80 lb/ft2 < 2 dw

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 y = )53)(8()19)(15(

)2/819)(53)(8()2/19)(19)(15(

+++

= 17.57 in. (from bottom)

1bI = 23

23

)57.1723)(53)(8(12

)8)(53()57.175.9)(19)(15(

12

)19)(15( +++

= 41,897 in.4

222

2

L section y = )34)(8()19)(15(

)2/819)(34)(8()2/19)(19)(15(

+++

= 16.09 in.

2bI = 23

23

)09.1623)(34)(8(12

)8)(34()09.165.9)(19)(15(

12

)19)(15( +++

= 35,389 in.4

Then calculate the moments of inertia for the slabs.

Strip Width Height Moment of Inertia

N-S strip on west edge 2

24

2

1230 + = 192 in. 8 in. 12

)8)(192( 3

= 8192 in.4

N-S strip on east edge 1230 = 360 in. 8 in. 12

)8)(360( 3

= 15,360 in.4

E-W strips 1220 = 240 in. 8 in. 12

)8)(240( 3

= 10,240 in.4

Use these values to find the ratios of the beam to slab stiffness along each edge of the panel.

Edge Beam Slab Ratio

North 41,897 in.4 10,240 in.4 = 4.09

South 41,897 in.4 10,240 in.4 = 4.09

East 41,897 in.4 15,360 in.4

= 2.73

West 35,389 in.4 8192 in.4 = 4.32

Average m = 3.81

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

Length ratio = 241220

241230 = 1.56

Minimum h = 936

)000,200/8.0(

++ yfL

= 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 uw = ld ww 6.12.1 + = 272 lb/ft2

Factored shear uV = 2

15.1 minLwu = 2815.2 lb/ft

Nominal shear nV = 75.0

uV = 3753.6 lb/ft

Shear capacity cV = dfc2 = )7)(5000(2 = 11,879 lb/ft

223

3

cV > nV , so the capacity is sufficient with the given thickness. Now calculate and distribute the

moments.

East-West Direction

Column width cb 24 in.

Span 1S 30 ft

Perpendicular span 2S 20 ft

Strip length 1L = cbS1 336 in.

Minimum strip length 1L = 165.0 S 234 in.

Static moment oM = 8

2

12LSwu 6,397,440

in-lb

Exterior negative moment uM = oM16.0 1,023,590 in-lb

Interior negative moment uM = oM7.0 4,478,208 in-lb

Positive moment uM+ = oM57.0 3,646,541 in-lb

Column strip

Exterior negative moment cuM = )(85.0 uM 870,052 in-lb

Interior negative moment cuM = )(85.0 uM 3,806,477 in-lb

Positive moment cuM+ = )(85.0 uM+ 3,099,560 in-lb

Positive design moment cnM+ = cuM+

3,443,955

in-lb

Negative design moment cnM = )max( cuM

4,229,419

in-lb

Strip width cw = 25.0 S 10 ft

Beam width cbw 53 in.

Slab width csw = cbc ww 67 in.

Positive beam moment )(85.0 cnM+ 2,927,362 in-lb

Negative beam moment )(85.0 cnM 3,595,006 in-lb

Positive slab moment / foot cs

cn

w

M )(15.0 + 92,524

in-lb/ft

Negative slab moment / foot cs

cn

w

M )(15.0 113,626

in-lb/ft

Middle strip

Exterior negative moment muM = )(15.0 uM 153,538 in-lb

Interior negative moment muM = )(15.0 uM 671,731 in-lb

Positive moment muM+ = )(15.0 uM+ 546,981 in-lb

Strip width msw = cwS2 10 ft

Positive design moment / foot mnM+ = ms

mu

w

M+ 60,776

in-lb/ft

224

4

Negative design moment / foot mnM = ms

mu

w

M )max( 74,637

in-lb/ft

Calculate the required reinforcement.

Beam: uM+ = 2,927,362 in-lb, uM = 3,595,006 in-lb, d = 24 in., b = 15 in.

Minimum 1 sA y

c

f

bdf3 1.27

in.2

Minimum 2 sA yf

bd200 1.2

in.2

Bottom sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c 2.12

in.2

Top sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c 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: uM+ = 92,524 in-lb, uM = 113,626 in-lb, d = 7 in., b = 12 in., h = 8 in.

Minimum sA bh0018.0 0.1728 in.2/ft

Bottom sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.2476

in.2/ft

Top sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 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: uM+ = 60,776 in-lb, uM = 74,637 in-lb, d = 7 in., b = 12 in., h = 8 in.


Bottom sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.1613

in.2/ft

Top sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.1988

in.2/ft


12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the E-W middle strips.

225

5

North-South Direction

Column width cb 24 in.

Span 1S 20 ft

Perpendicular span 2S 30 ft

Strip length 1L = cbS1 216 in.

Minimum strip length 1L = 165.0 S 156 in.

Static moment oM = 8

2

12LSwu 3,965,760

in-lb

Interior negative moment uM = oM65.0 2,577,744 in-lb

Positive moment uM+ = oM35.0 1,388,016 in-lb

Column strip

Interior negative moment cuM = )(6.0 uM 1,546,646 in-lb

Positive moment cuM+ = )(6.0 uM+ 832,810 in-lb

Positive design moment cnM+ = cuM+

925,344

in-lb

Negative design moment cnM = )max( cuM

1,718,496

in-lb

Strip width cw = 25.0 S 10 ft

Beam width cbw 53 in.

Slab width csw = cbc ww 67 in.

Positive beam moment )(85.0 cnM+ 786,542 in-lb

Negative beam moment )(85.0 cnM 1,460,722 in-lb

Positive slab moment / foot cs

cn

w

M )(15.0 + 24,860

in-lb/ft

Negative slab moment / foot cs

cn

w

M )(15.0 46,169

in-lb/ft

Middle strip

Interior negative moment muM = )(4.0 uM 1,031,098 in-lb

Positive moment muM+ = )(4.0 uM+ 555,206 in-lb

Strip width msw = cwS2 10 ft

Positive design moment / foot mnM+ = ms

mu

w

M+ 30,845

in-lb/ft

Negative design moment / foot mnM = ms

mu

w

M )max( 57,283

in-lb/ft

226

6

Calculate the required reinforcement.

Beam: uM+ =786,542 in-lb, uM = 1,460,722 in-lb, d = 24 in., b = 15 in.

Minimum 1 sA y

c

f

bdf3 1.27

in.2

Minimum 2 sA yf

bd200 1.2

in.2

Bottom sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c 0.55

in.2

Top sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c 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: uM+ = 24,860 in-lb, uM = 46,169 in-lb, d = 7 in., b = 12 in., h = 8 in.


Bottom sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.0595

in.2/ft

Top sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.1110

in.2/ft


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: uM+ = 30,845 in-lb, uM = 57,283 in-lb, d = 7 in., b = 12 in., h = 8 in.


Bottom sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.0739

in.2/ft

Top sA 285.0

211

85.0

df

M

f

bdf

c

n

y

c 0.1380

in.2/ft


12 in. Use No. 5 bars at 12 in. c-c for the top and bottom steel of the N-S middle strips.

227

7

Therefore, the reinforcement design is as shown below.

East-West Reinforcement

North-South Reinforcement

228

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

8

Slab thickness h = 16 in.

Steel depth d = 14 in.

Hinge field radius r = 24 in.

Calculate the load and effective circle.

Dead load dw = 150(16/12) + 20 = 220 lb / ft2

Live load lw = 80 lb/ft2

Factored load uw = 1.2wd +1.6wl = 392 lb/ft 2

Panel area A = (20)(30) = 600 ft2

Panel load P = Awu = 235,200 lb

Radius = A

4 = 27.63 ft

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

Factored moment uM = r

P

3

21

4 = 1,844,675 in-lb / ft

Nominal moment nM = uM

= 2,049,639 in-lb / ft


adfA ys and a =

bf

fA

c

ys

85.0


211

85.0

bdf

M

f

bdf

c

n

y

c = 2.87 in.2

Choose 3 No. 9 bars for 3.0 in.2 Check the strength and failure mode.

Compression block a = bf

fA

c

ys

85.0 = 4.41 in.

Neutral axis depth c = 1

a = 5.19 in.

Ratio d

c = 0.370 < 0.375 Tension controlled, OK

Strength nM = 2

adfA ys = 2,122,941 in-lb > 2,049,639 in-lb, OK

11.7 Use the yield-line theory to evaluate the slab thickness needed in the column zone of the flat plate in Problem11.5 for flexure, assuming that the hinge field would have a radius of 24 in.

256

257

258

259

260

261

11.10 Calculate the maximum crack width in a two-way interior panel of a reinforced concrete floor system. Theslab 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 andspacing of the reinforcement necessary for crack control assuming that (a) the floor is exposed to normal en-vironment; (b) the floor is part of a parking garage. Given: fy = 60.0 ksi (414 MPa).

9

Slab thickness h = 8 in.

Clear cover cc = 0.75 in.

Ratio r = 24

30 = 0.8

For a normal environment, try No. 4 bars ( bd = 0.5 in.)

Max crack width maxw = 0.012 in.

Coefficient K = 2.1 10 5 in.

2 / lb

Standard value = 1.2

Steel depth cd = bdcc 5.0+ = 1 in.

Calculate 1G 1G =

2

max

4.0 yfK

w = 393.6 in.2

Assume sss == 21 s = c

b

d

Gd

8

1 = 8.79 in.

Therefore, use No. 4 bars at 8.5 in. c-c for the crack reinforcement.

For a parking garage, try No. 4 bars ( bd = 0.5 in.)

Max crack width maxw = 0.016 in.

Coefficient K = 2.1 10 5 in.

2 / lb

Standard value = 1.2

Steel depth cd = bdcc 5.0+ = 1 in.

Calculate 1G 1G =

2

max

4.0 yfK

w = 699.9 in.2

Assume sss == 21 s = c

b

d

Gd

8

1 = 11.72 in.

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

262

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 gravelysands, poorly graded. The frost line is assumed to be 3 ft below grade. Given:

surcharge � 90 psf 14.3 kPa2

fy � 60,000 psi 1413.7 MPa2

column f ¿c � 4000 psi 127.6 MPa2

footing f ¿c � 3000 psi 120.7 MPa2

average weight of soil and concrete above the footing, � � 140 pcf 122.0 kN>m32

1

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 h = 3 ft

Clear cover at base cc = 3 in.



Average steel depth d = bdcch = 32.125 in.

For the given soil and concrete types:

Aggregate factor = 1

Soil capacity soilP = 2.5 ton/ft2 = 5000 psf

Calculate the required footing area using the service load and the allowable soil pressure.

Allowable soil load nP = surcharge)frost( +hPsoil = 4070 psf

Service load P = DP + LP = 900,000 lbf

Required area fA = nP

P = 221.13 ft2

Choose size (square) b = 15 ft ( fA = 225 ft2)

Calculate the factored load intensity.

Factored load UP = LD PP 6.12.1 + = 1,240,000 lbf

Load intensity Uq = f

U

A

P = 5511.11 psf

263

2

Consider the shear capacity for beam action.

Factored shear UV = )15(12

125.32

122

22

2

15Uq = 322,917 lbf


UV = 430,556 lbf

Shear capacity cV = bdfc2 = 633,441 lbf > 430,556 lbf, OK

Consider the shear capacity for two-way action.

Factored shear UV = ++

12

125.3226

12

125.3222)15)(15(Uq = 1,119,597 lbf


UV = 1,492,796 lbf

Perimeter 0b = )125.3226125.3222(2 +++ = 224.5 in.

Footing size ratio = 1

Column type factor s = 40 for an interior column

Shear capacity 1 cV = dbfc 0

42 + = 2,370,126 lbf

Shear capacity 2 cV = dbfb

dc

s0

0

2+ = 3,051,075 lbf

Shear capacity 3 cV = dbfc 04 = 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.

Factored moment UM =

2

2

22

2

1215)15(

2

1Uq = 21,496,778 in-lb

Nominal moment nM = 9.0

UM = 23,885,309 in-lb

Minimum 1 sA bd0018.0 = 10.4085 in.2

Minimum 2 sA 285.0

211

85.0

bdf

M

f

bdf

c

n

y

c = 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 15 12/22 = 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 1=== est and 0=trK .

264

3

Effective cover factor b

trb

d

Kc + = 3.428 > 2.5 so use 2.5

Strength factor cf = 54.77 < 100, OK

Development length dl = best

c

yd

f

f

5.240

3 = 28.75 in.

Available space 2

26

2

1215 = 79 in. > 28.75 in. OK

Check the bearing strength for the column and the footing.

Column area 1A = 144

2622 = 3.9722 ft

2

Column bearing UP = 1)85.0(7.0 Afc = 1,361,360 lbf > 1,240,000 lbf, OK

Footing similar area 2A =

2

126

1215A = 190.38 ft

2

Area factor 1

2

A

A = 6.923 > 2.0 so use 2.0

Footing bearing UP = 1)85.0)(7.0(0.2 Afc = 2,042,040 lbf > 1,240,000 lbf, OK

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

Minimum area dA = 1005.0 A = 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 dl yb fd0003.0 = 15.75 in.

Footing dl = c

yb

f

fd02.0 = 19.17 in. (controls)

Column dl = c

yb

f

fd02.0 = 16.60 in. (controls)

The available space in the footing is bdcch 3 = 30.375 in. > 19.17 in. OK

265

4

266

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. 13.5using the portal method of analysis. Use a height between floors of 10′ – 6″ (3.20 m).

1

Given:

Floor height: uL = 10.5 ft

Roof parapet: pL = 2.5 ft

Section width: L = 20 ft

Wind load: uw = 30 psf

Calculate the force at each floor and at the roof.

Roof roofF = upu LwLL

+2

7.1 = 7905 lb

Floor floorF = uu LwL7.1 = 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.

Shear (lb) Moment (ft-lb)

Floor Shear Interior Exterior Interior Exterior

1 3(10,710) + 7905 = 40,035 lb 13,345 6672.5 70,061 35,031

2 2(10,710) + 7905 = 29,325 lb 9775 4887.5 51,319 25,659

3 1(10,710) + 7905 = 18,615 lb 6205 3102.5 32,576 16,288

4 7905 lb 2635 1317.5 13,834 6917

297

2

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) Shear (lb)

Floor Interior Exterior Interior Exterior

2 0.5(70,061 + 51,319) = 60,690 35,031 + 25,659 = 60,690 5057.5 6936

3 41,948 41,948 3495.6 4794

4 23,205 23,205 1933.8 2652

R 6917 6917 576.4 790.5

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

298

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). De-termine the required �� -in.-diameter, stress-relieved, seven-wire strands to resist the applied gravity load andthe 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:

The section properties, given these stresses, are

WL � 4800 plf WD � 568 plf

St � 4531 in.3 Sb � 5900 in.3 cb � 18.24 in.

r 2 �Ic

Ac� 197.5 in.2

Ig � 107,645 in.4 Ac � 545 in.2

fpe � 145,000 psi 11000 MPa2 fpi � 189,000 psi 11303 MPa2 fpu � 270,000 psi 11862 MPa2

ft � 122f ¿c � 930 psi 16.4 MPa2 � 2700 psi 126.7 MPa2

fc � 0.45f ¿c f ¿c � 6000 psi 141.4 MPa2

1

6

4

85

8

8

2 0

Figure 14.18

Given:

Concrete area: cA = 545 in.2

Strands: 1/2 od, 7 wire strand, sA = 0.153 in.2

Final pretension stress: pef = 145,000 psi

Dead load: Dw = 568 lb/ft

Live load: Lw = 4800 lb/ft

Mid span eccentricity: ce = 18.42 in.

Section center of gravity: c = 18.24 in.

Section depth: h = 42 in.

Section radius of gyration: 2r = 197.5 in.

2

Section bottom modulus: bS = 5900 in.3

Section top modulus: tS = 4531 in.3

Maximum tensile stress: tf = 930 psi

Maximum compressive stress: cf = 2700 psi

313

2

Let s be the number of strands needed.

Total steel area psA = sAs = 0.153s in.2

Final pretension eP = peps fA = 21,380.28s lb

Mid span moment TM = 88

22 LwLw LD + = 12,882,500 in-lb

Stress at top tf = t

Ttc

c

e

S

M

r

ce

A

P2

1 = 47.68s 2843 psi

Stress at bottom bf = b

Tbc

c

e

S

M

r

ce

A

P+

21 = 105.98s + 2183.38 psi

Calculate the minimum number of strands.

Strands for top s 68.47

2843+tf = 2.9994

Strands for bottom s 98.105

38.2183cf = 11.8315

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

314

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 overturningmoment at each story level in terms of the weight Ws of each floor. Use the Equivalent Lateral Force Methodin the solution. Given:

Ws per floor � 2,000,000 lb 18896 kN2

R � 5

S1 � 0.34 sec., Ss � 0.90 sec.

1

Building height bh = 12 + (10)(10) = 112 ft

Site coefficient af = 1.0 for SS = 0.90

Site coefficient vf = 1.0 for 1S = 0.34

Importance factor I = 1 for class II

Spectral response

Maximum response MSS = SaSf = 0.9

1MS = 1Sfv = 0.34

Damped response DSS = MSS3

2 = 0.6

1DS = 13

2MS = 0.2267

Calculate the seismic response coefficient.

Nominal value SC = IR

SDS

/ = 0.15

Period coefficient TC = 0.020 for generic building

Approximate period aT = 4/3

bT hC = 0.6886 s

Limit coefficient UC = 1.37 for 1DS = 0.2267

Maximum period aT = aUTC = 0.9456 s

Building period T = aT2.1 = 1.1348 s

Maximum value SC TIR

SD

)/(

1 = 0.049938 (controls)

Minimum value SC DSS044.0 = 0.0264

Calculate the total shear.

Weight at base W = sW11 = 22,000,000 lb

Shear at base V = WCS = 0.549313 sW = 1,098,625 lb

369

2

The coefficient for the force at each floor is vxC =

=

n

i

k

is

k

xs

hW

hW

1

, where ih is the height of floor i

above the ground and k = 1)12(5.05.2

5.0 +T = 1.317376.

The force on each floor is xF = VCvx , the story shear is xV = =

x

i

iFV0

, and the overturning

moment is xM = )( xi

n

xi

i hhF=

. For the top ten stories, = 1.

Denominator for vxC =

n

i

k

ishW1

= 2672.35 sW

Floor xh vxC xF xV xV (lb) xM (ft-lb)

Base 0 0 0 0.549313 sW 1,098,625 89,759,342

1 12 0.009881 0.005428 sW 0.543885 sW 1,087,770 76,575,838

2 22 0.021957 0.012061 sW 0.531824 sW 1,063,647 65,698,137

3 32 0.035971 0.019759 sW 0.512064 sW 1,024,128 55,061,666

4 42 0.051468 0.028272 sW 0.483792 sW 967,584 44,820,384

5 52 0.068191 0.037458 sW 0.446334 sW 892,668 35,144,540

6 62 0.085973 0.047226 sW 0.399108 sW 798,216 26,217,859

7 72 0.104691 0.057508 sW 0.341600 sW 683,200 18,235,695

8 82 0.124256 0.068256 sW 0.273344 sW 546,689 11,403,697

9 92 0.144595 0.079428 sW 0.193917 sW 387,833 5,936,809

10 102 0.165648 0.090993 sW 0.102924 sW 205,848 2,058,477

11 112 0.187368 0.102924 sW 0 0 0

370

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 bear-ing length of 20 in. on the supporting wall to support a roof transmitting 350 plf load to the lintelGiven:

Effective t 7.63 in.

Unit weight 80 psf

Wall height to roof 14 ft.

Height of lintel

Parapet height

Service live load on roof 200 plf

Governing factored loading

163 psi

1500 psi

60,000 psify

fm

fr

1.2D � 1.6L

3¿-6–5¿-3–

1

Given:

Lintel width b = 7.63 in.

Lintel height h = 5 -3 = 63 in.

Lintel span L = Clear span + 1/2 bearing = 20 + 10 = 250 in.

Dead load Dw = bh80350 + = 770 lb/ft

Live load Lw = 200 lb/ft

Masonry strength mf = 1500 psi

Rupture strength rf = 163 psi

Steel strength yf = 60,000 psi

Steel modulus sE = 29 106 psi

Calculate the loads.

Factored load Uw = LD ww 6.12.1 + = 1244 lb/ft

Factored shear UV = 2

LwU = 12,958.33 lb


UV = 16,197.92 lb

Factored moment UM = 8

2LwU = 809,895.83 in.-lb

Nominal moment nM = 90.0

UM = 899,884.26 in.-lb

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

Shear plane area nA = bd = 457.8 in.2

Masonry strength mV = mn fA)75.14( = 39,893.66 lb (controls)

Maximum strength nV = mn fA4 = 70,922.07 lb

396

2

Since 39,893.66 lb > 16,197.92 lb, the shear strength is sufficient.

Determine the required reinforcement steel area.

Required area nM = 2

adfA ys and a =

bf

fA

m

ys

80.0

Solve for sA sA = 280.0

211

80.0

bdf

M

f

bdf

m

n

y

m = 0.2535 in.2

Check the cracking strength.

Critical moment 1.3Mcr = rfbh

63.1

2

= 1,069,511 in.-lb > 899,884.26 in.-lb

Required steel area sA = s

n

cr AM

M3.1 = 0.3013 in.2

Therefore, choose 2 No. 4 bars for an area of 2(0.2) = 0.4 in.2 Check the reinforcement ratio.

Actual ratio = bd

As = 0.000874

Max steel strain y = s

y

E

f = 0.002069

Max masonry strain mu = 0.0025 for concrete

Load factor = 1.5

Maximum ratio + ymu

mu

y

m

f

f64.0 = 0.007138 > 0.000874 OK

397

398

399

400

401

402

403

404

405

406

407

408

409

410

411

412

413

414

415

416

417

418

419

420420

17.8. Design a tension anchor for a grouted masonry wall to withstand a tensile force of magnitudeand a shear force without causing any masonry breakout. Consider the di-

rection of the shear force to be along the wall length.Given

fm � 1500 psi

fye � 27,000 psi

fy � 36,000 psi

Pv � 3000 lbPu � 14,000 lb

3

Given:

Bolt axial strength yf = 36,000 psi

Bolt shear strength yvf = 27,000 psi

Masonry strength mf = 1500 psi

Axial load aP = 14,000 lb

Shear load vP = 3000 lb

Try a design using bN = 2 bolts with a diameter of bd = 0.75 in. and an embedment length of bI

= 6.5 in. Space the bolts 13 in. apart to avoid overlap of the anchorage areas.

Bolt area bA = 4

2

bb dN = 0.8836 in.2

Masonry axial area ptA = 2

bb IN = 265.46 in.2

Masonry shear area pvA = 2

2

bb IN = 132.73 in.2

Masonry axial failure anB = mpt fA4)5.0( = 20,562.8 lb (controls) > 14,000 lb OK

Bolt axial failure anB = yb fA)9.0( = 28,627.8 lb

Masonry shear failure vnB = mpv fA4)5.0( = 10,281.4 lb (controls) > 3000 lb OK

Bolt shear failure vnB = yvb fA)6.0)(9.0( = 12,882.5 lb

Combined load vn

v

an

a

B

P

B

P+ = 0.97263 < 1 OK

Therefore, the design is sufficient.

421