Sample Reliability of Structures 2nd Edition Andrzej Nowak Solutions Manual pdf PDF

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Authors: Andrzej S. Nowak , Kevin Collins
Published: CRC 2012
Edition: 2nd
Pages: 114
Type: pdf
Size: 3MB
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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR Reliability of Structures, Second Edition

by Andrzej Nowak and Kevin Collins

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FOLFNKHUHWRGRZQORDG

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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR Reliability of Structures, Second Edition

by Andrzej Nowak and Kevin Collins

Boca Raton London New

York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

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FOLFNKHUHWRGRZQORDG

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2013 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20130416 International Standard Book Number-13: 978-1-4665-8014-5 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at ht tp://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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FOLFNKHUHWRGRZQORDG TABLE OF CONTENTS Solutions to Problems in Chapter 2

pp. 1-18

Solutions to Problems in Chapter 3

pp. 19-28

Solutions to Problems in Chapter 4

pp. 29-48

Solutions to Problems in Chapter 5

pp. 49-75

Solutions to Problems in Chapter 6

pp. 76-78

Solutions to Problems in Chapter 8

pp. 79-99

Solutions to Problems in Chapter 9

pp. 100-107

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FOLFNKHUHWRGRZQORDG SOLUTIONS TO PROBLEMS IN CHAPTER 2 Problem 2.1. The results of tests to determine the modulus of rupture (MOR) for a set of timber beams are shown in Table P2.1. A. Plot the relative frequency and cumulative frequency histograms. B. Calculate the sample mean, standard deviation, and coefficient of variation. C. Plot the data on normal probability paper. Solution: A. For the histogram plots, the interval size is chosen to be 250. There are 45 data points. Interval 3250-3500 3500-3750 3750-4000 4000-4250 4250-4500 4500-4750 4750-5000 5000-5250 5250-5500 5500-5750 5750-6000 6000-6250 6250-6500 6500-6750 6750-7000 7000-7250 7250-7500

Relative Cumulative Frequency Frequency 0 0 0.06667 0.066667 0.11111 0.177778 0.02222 0.200000 0.06667 0.266667 0.11111 0.377778 0.08889 0.466667 0.15556 0.622222 0.11111 0.733333 0.04444 0.777778 0.11111 0.888889 0.02222 0.911111 0.04444 0.955556 0 0.955556 0 0.955556 0.04444 1 0 1

Relative Frequency

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7250-7500

6750-7000

6250-6500

5750-6000

5250-5500

4750-5000

4250-4500

3750-4000

3250-3500

0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0

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FOLFNKHUHWRGRZQORDG Cumulative Frequency 1.2 1 0.8 0.6 0.4 0.2 7250-7500

6750-7000

6250-6500

5750-6000

5250-5500

4750-5000

4250-4500

3750-4000

3250-3500

0

B. Using Eqns. 2.25 and 2.26, sample mean = x = 5031 and sample standard deviation = sX = 880.4. The coefficient of variation based on sample parameters is s X / x = 0.175. C. The step-by-step procedure described in Section 2.5 is followed to construct the plot on normal probability paper.

Standard Normal Variate

MOR data on normal probability paper 3 2 1 0 -1 -2 -3 0

2000

4000

6000

8000

MOR ------------------------------------------Problem 2.2. A set of test data for the load-carrying capacity of a member is shown in Table P2.2. A. Plot the test data on normal probability paper. B. Plot a normal distribution on the same probability paper. Use the sample mean and standard deviation as estimates of the true mean and standard deviation. C. Plot a lognormal distribution on the same normal probability paper. Use the sample mean and standard deviation as estimates of the true mean and standard deviation.

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D. Plot the relative frequency and cumulative frequency histograms. FOLFNKHUHWRGRZQORDG Solution: A. Using Eqns. 2.25 and 2.26, the sample mean, sample standard deviation, and sample coefficient of variation are x = 4.127

s X = 0.1770

CoV = s X / x = 0.04289

To plot on normal probability paper, we follow the step-by-step procedure outlined in Section 2.5. Raw Data 3.95 4.07 4.14 3.99 4.21 4.39 4.21 3.90 3.74 4.28 4.15 4.04 4.26 4.41 4.22 4.07 3.97 4.05 4.36

Sorted 3.74 3.90 3.95 3.97 3.99 4.04 4.05 4.07 4.07 4.14 4.15 4.21 4.21 4.22 4.26 4.28 4.36 4.39 4.41

i i/(N+1)=p i 1 0.05 2 0.10 3 0.15 4 0.20 5 0.25 6 0.30 7 0.35 8 0.40 9 0.45 10 0.50 11 0.55 12 0.60 13 0.65 14 0.70 15 0.75 16 0.80 17 0.85 18 0.90 19 0.95

Φ-1(pI) -1.64485 -1.28155 -1.03643 -0.84162 -0.67449 -0.5244 -0.38532 -0.25335 -0.12566 0 0.125661 0.253347 0.385321 0.524401 0.67449 0.841621 1.036433 1.281551 1.644853

Normal Probability Paper

Standard Normal Variate

3 2 1 0 -1 3.5

3.7

3.9

4.1

4.3

4.5

-2 -3 Capacity Raw data

Normal Distribution

Lognormal distribution

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B. See part A. To plot the normal distribution on normal probability paper: (1) Calculate FOLFNKHUHWRGRZQORDG values of the standard normal variable z for some arbitrary values of x (in ascending order) in the range of interest. The formula for z is z=

x − µ X x − 4.127 = 0.1770 σX

(2) Plot z versus x on standard linear graph paper. The plot is shown in the graph in Part A. Note that the relationship between z and x is linear, so only two points are needed to plot the graph. C. To plot a lognormal distribution, we need the lognormal distribution parameters. We will assume that µX= x and σX=sX. 2 2 2 σ ln X = ln(1 + VX ) = ln(1 + (0.04289) ) ⇒ σ ln X = 0.04287

µ ln X = ln(µ X ) − 0.5σ ln X = ln(4.127) − 0.5(0.04287) = 1.417 2

2

To plot the lognormal distribution on normal probability paper: (1) Calculate FX(x) for some arbitrary values of x (in ascending order) in the range of interest. FX(x) is the lognormal distribution, and it can be calculated as shown in Section 2.4.3. (2) Use the values of FX(xi) as the values of pi for plotting points on normal probability paper. (3) Calculate zi = Φ-1(pi). (4) Plot zi versus xi. The plot is shown in the graph in Part A. D. There are 19 data points. The interval size was chosen to be 0.05. Interval 3.70-3.75 3.75-3.8 3.8-3.85 3.85-3.9 3.9-3.95 3.95-4.0 4.0-4.05 4.05-4.1 4.1-4.15 4.15-4.2 4.2-4.25 4.25-4.3 4.3-4.35 4.35-4.4 4.4-4.45

Number 1 0 0 1 1 2 2 2 2 0 3 2 0 2 1

Cumulative % 5.26% 5.26% 5.26% 10.53% 15.79% 26.32% 36.84% 47.37% 57.89% 57.89% 73.68% 84.21% 84.21% 94.74% 100.00%

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FOLFNKHUHWRGRZQORDG

4.4-4.45

4.3-4.35

4.2-4.25

4.1-4.15

4.0-4.05

3.8-3.85

3.9-3.95

0.18 0.16 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 3.70-3.75

Relative frequency

Relative Frequency Plot

Intervals

Cumulative Frequency Plot 100.00% 80.00% 60.00% 40.00%

4.4-4.45

4.3-4.35

4.2-4.25

4.1-4.15

4.0-4.05

3.9-3.95

3.70-3.75

.00%

3.8-3.85

20.00%

Intervals

-------------------------------------Problem 2.3. For the data in Table 2.5, calculate the statistical estimate of the correlation coefficient using Equation (2.99). Solution: The formula is n

ρˆ XY =

1 n −1

∑ (xi − x )(yi − y) i =1

sX sY

 n   ∑ x iy i  − nxy 1  i =1  = n −1 s Xs Y

Note that is doesn’t matter which variable is x and which is y. There are 100 data points. After manipulating the data, you should find: viii @solutionmanual1...