Sample Reinforced concrete a fundamental approach Edward Nawy 6th edition solutions manual pdf PDF

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Authors: Edward Nawy
Published: Pearson 2008
Edition: 6th
Pages: 427
Type: pdf
Size: 11.5MB
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FOLFNKHUHWRGRZQORDG

IN ST R U C T O R ’S SO LU T IO N S MAN U AL

REINFORCED CONCRETE A FUNDAMENTAL APPROACH SIXTH EDITION

EDWARD G. NAWY

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FOLFNKHUHWRGRZQORDG 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

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CONTENTS FOLFNKHUHWRGRZQORDG

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

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https://gioumeh.com/product/reinforced-concrete-a-fundamental-approach-solutions/ 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 orFOLFNKHUHWRGRZQORDG yielding of steel. Given: fc¿ ⫽ 3000 psi 1 20.7 MPa 2 for case 1a2 , A s ⫽ 1 in.2 fc¿ ⫽ 6000 psi 1 41.4 MPa 2 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 = 8 in. d = 18 in. dt = 16 in. = 3000 psi c fy = 60,000 psi As = 1 in2

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

(

fc

Then calculate the depth of the compression block. a =

As f y

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

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FOLFNKHUHWRGRZQORDG a

c =

1

2.94 0.85 = 3.46 in

=

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

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 = 8 in. d = 18 in. dt = 16 in. = 6000 psi c fy = 60,000 psi As = 6 in.2

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

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FOLFNKHUHWRGRZQORDG 0.85 0.05

0.85 0.05

=

fc

4000

,

(

fc

)

6000 4000

= 0.75 Then calculate the depth of the compression block. As f y

a =

fc b (6)(60,000) = 0 .85(6000)(8) = 8.82 in

a

c =

1

8.82 = 0.75 = 11.76 in

Then find the ratio of c and dt.

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.

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https://gioumeh.com/product/reinforced-concrete-a-fundamental-approach-solutions/ 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 ofFOLFNKHUHWRGRZQORDG 12 ft. Given: f c¿ ⫽ 4000 psi 1 27.6 MPa 2 , normal-weight concrete fy ⫽ 60,000 psi 1414 MPa2

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

L

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

The minimum depth for deflection is

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 Mn =

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

Assume a moment arm of

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 in 2

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

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3 f c 3 4000 FOLFNKHUHWRGRZQORDG = = 0 .0031 = 1 f

2

As min

=

60,000 200 200 = = 0.0033 60,000 fy

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

(

)

fc

Calculate the depth of the compression block. As f y

a =

fc b (0 .28)(60,000) = 0 .85(4000)(12) = 0.4118 in

c =

a 1

0.4118 = 0.85 = 0.4844 in

Then find the ratio of c and dt.

c dt

0.4844 7 = 0.0692

=

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

=

As f y d

a

= (0 .28)(60,000) 7 = 114,141 in-lb

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0.4118...