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An Instructor’s Solutions Manual to Accompany PRINCIPLES OF GEOTECHNICAL ENGINEERING, 8TH EDITION BRAJA M DAS & KHALED SOBHAN ISBN-13: 978-1-133-11089-7 ISBN-10: 1-133-11089-4 © 2014, 2010 Cengage Learning ALL RIGHTS RESERVED No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher except as may be permitted by the license terms below For product information and technology assistance, contact us at Cengage Learning Academic Resource Center, 1-800-423-0563 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.com Cengage Learning 200 First Stamford Place, Suite 400 Stamford, CT 06902 USA Cengage Learning is a leading provider of customized learning solutions with office locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan Locate your local office at: international.cengage.com/region Cengage Learning products are represented in Canada by Nelson Education, Ltd For your course and learning solutions, visit www.cengage.com/engineering Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN READ IMPORTANT LICENSE INFORMATION Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below Cengage Learning has established these use limitations in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party Test banks and other testing materials may be made available in the classroom and collected at the end of each class session, or Printed in the United States of America 16 15 14 13 12 posted electronically as described herein Any material posted electronically must be through a password-protected site, with all copy and download functionality disabled, and accessible solely by your students who have purchased the associated textbook for the Course You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you not accept these conditions, you must return the Supplement unused within 30 days of receipt All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement We trust you find the Supplement a useful teaching tool INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY PRINCIPLES OF GEOTECHNICAL ENGINEERING Eighth Edition, SI BRAJA M DAS KHALED SOBHAN Contents Chapter Page 11 19 25 31 41 51 57 10 69 11 83 12 99 13 109 14 121 15 127 16 141 17 153 Chapter 2.1 Cu = 0.212 D302 D60 0.42 = = 0.656 ≈ 0.66 = = 2.625 ≈ 2.63 ; Cc = ( D60 )( D10 ) (0.42)(0.16) D10 0.16 2.2 Cu = 0.412 D302 D60 0.81 = = 0.768 ≈ 0.77 = = 3.0 ; Cc = ( D60 )( D10 ) (0.81)(0.27) D10 0.27 2.3 a Sieve no 10 20 40 60 100 200 Pan Mass of soil retained on each sieve (g) 28 42 48 128 221 86 40 24 ∑ 617 g Percent retained on each sieve 4.54 6.81 7.78 20.75 35.82 13.94 6.48 3.89 Percent finer 95.46 88.65 80.88 60.13 24.31 10.37 3.89 0.00 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part b D10 = 0.16 mm; D30 = 0.29 mm; D60 = 0.45 mm 2.4 c Cu = D60 0.45 = = 2.812 ≈ 2.81 D10 0.16 d Cc = D302 0.292 = = 1.168 ≈ 1.17 ( D60 )( D10 ) (0.45)(0.16) a Sieve no 10 20 40 60 100 200 Pan Mass of soil retained on each sieve (g) 30 48.7 127.3 96.8 76.6 55.2 43.4 22 ∑ 500 g Percent retained on each sieve 0.0 6.0 9.74 25.46 19.36 15.32 11.04 8.68 4.40 Percent Finer 100.00 94.0 84.26 58.80 39.44 24.12 13.08 4.40 0.00 b D10 = 0.13 mm; D30 = 0.3 mm; D60 = 0.9 mm © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part c Cu = D60 = = 6.923 ≈ 6.92 D10 0.13 0.32 D302 d Cc = = = 0.769 ≈ 0.77 ( D60 )( D10 ) (0.9)(0.13) 2.5 a Sieve no 10 20 40 60 80 100 200 Pan Mass of soil retained on each sieve (g) 40 60 89 140 122 210 56 12 ∑ 729 g Percent retained on each sieve 0.0 5.49 8.23 12.21 19.20 16.74 28.81 7.68 1.65 Percent finer 100.00 94.51 86.28 74.07 54.87 38.13 9.33 1.65 0.00 b D10 = 0.17 mm; D30 = 0.18 mm; D60 = 0.28 mm c Cu = D60 0.28 = = 1.647 ≈ 1.65 D10 0.17 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part D302 0.18 d Cc = = = 0.68 ( D60 )( D10 ) (0.28)(0.17) 2.6 a Sieve no 10 20 40 60 100 200 Pan Mass of soil retained on each sieve (g) 0 9.1 249.4 179.8 22.7 15.5 23.5 ∑ 500 g Percent retained on each sieve 0.0 0.0 0.0 1.82 49.88 35.96 4.54 3.1 4.7 Percent finer 100.00 100.00 100.00 98.18 48.3 12.34 7.8 4.7 0.00 b D10 = 0.21 mm; D30 = 0.39 mm; D60 = 0.45 mm c Cu = D60 0.45 = = 2.142 ≈ 2.14 D10 0.21 d Cc = 0.39 D302 = = 1.609 ≈ 1.61 ( D60 )( D10 ) (0.45)(0.21) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2.7 a b Percent passing mm = 100 GRAVEL: 100 – 100 = 0% SAND: 100 – 73 = 27% SILT: 73 – = 64% CLAY: – = 9% Percent passing 0.06 mm = 73 Percent passing 0.002 mm = c Percent passing mm = 100 Percent passing 0.05 mm = 68 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 68 = 32% SILT: 68 – = 59% CLAY: – = 9% d Percent passing mm = 100 Percent passing 0.075 mm = 80 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 80 = 20% SILT: 80 – = 71% CLAY: – = 9% © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2.8 a b Percent passing mm = 100 Percent passing 0.06 mm = 30 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 30 = 70% SILT: 70 – = 65% CLAY: – = 5% c Percent passing mm = 100 Percent passing 0.05 mm = 28 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 28 = 72% SILT: 72 – = 67% CLAY: – = 5% d Percent passing mm = 100 Percent passing 0.075 mm = 34 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 34 = 66% SILT: 66 – = 61% CLAY: – = 5% © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part λqd = + tan φ ′(1 − sin φ ′)2 2⎛ ⎞ = + tan 32(1 − sin 32) ⎜ ⎟ = 1.184 B ⎝ 1.5 ⎠ Df λ γd = q all 16.5 ⎡(21)(35.49)(1.266) + (17.5)(1)(23.18)(1.184)⎤ 1⎢ ⎥ = 606.8 kN/m = ⎥ 3⎢ + (17.5)(1.5)(30.22)(1) ⎣ ⎦ φ′ = 24º; Nc = 19.32; Nq = 9.60; Nγ = 9.44 ⎛4⎞ ⎝6⎠ λcd = + 0.4⎜ ⎟ = 1.266 λqd = + tan φ ′(1 − sin φ ′)2 2⎛ 4⎞ = + tan 24(1 − sin 24) ⎜ ⎟ = 1.209 B ⎝6⎠ Df λ γd = qall 16.6 ⎡(1500)(19.32)(1.266) + ( 4)(118)(9.60)(1.209)⎤ 1⎢ ⎥ = ⎢ = 11,377 lb/ft ⎥ + (118)(6)(9.44)(1) ⎢⎣ ⎥⎦ q all = ⎛ ⎞ ⎜ c ′N c λcd + qN q λ qd + γBN γ λγd ⎟ Fs ⎝ ⎠ φ′ = 0º; Nc = 5.14; Nq = 1.0; Nγ = ⎛ Df ⎝ B λcd = + 0.4⎜⎜ ⎞ 0.75 ⎞ ⎟⎟ = + 0.4⎛⎜ ⎟ = 1.12 ⎝ ⎠ ⎠ λqd = + tan φ ′(1 − sin φ ′)2 Df B =1 λ γd = 1 q all = [(37)(5.14)(1.12) + (0.75)(19.5) + 0] = 37.94 kN/m 142 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 16.7 Eq (16.12): qu = 1.3c ′N c + qN q + 0.4γ ′BN γ φ′ = 32º; Nc = 44.04; Nq = 28.52; Nγ = 26.87 q = γ h + γ′ (Df – h) = (16 × 0.9) + (18.9 – 9.81)(1.2 – 0.9) = 17.12 kN/m2 Qall = qu B B = (1.3c ′N c + qN q + 0.4γ ′BN γ ) Fs Fs 1.752 [(1.3)(17)(44.04) + (17.12)(28.52) + (0.4)(18.9 − 9.81)(1.75)(26.87)] Qall = 3.5 = 1428 kN 16.8 q = γ Df = (16 × 1.2) = 19.2 kN/m2 D = h – Df = (1.2 + 0.5) – 1.2 = 0.5 m Eq (16.7): γ av = [γD + γ ′( B − D)] = [(16)(0.5) + (18.9 − 9.81)(1.75 − 0.5)] = 11.06 kN/m 1.75 B qu B B2 Fs = = (1.3c ′N c + qN q + 0.4γ ′BN γ ) Qall Qall Fs = 16.9 1.75 [(1.3)(17)(44.04) + (17.12)(28.52) + (0.4)(11.06)(1.75)(26.87)] = 3.58 1428 φ′ = 22º From Table 16.1, Nc = 20.27; Nq = 9.19; Nγ = 5.09 γ = (1750)(9.81) = 17.16 kN/m 1000 q u = 1.3c ′N c + qN q + 0.4γ av BN γ q = γ Df = (1.5)(17.16) = 25.74 kN/m2 γ av = [γD + γ ′( B − D )] B D = h – Df = 2.5 – 1.5 = m 143 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part γ sat = (1950)(9.81) = 19.13 kN/m 1000 γ av = [(17.16)(1) + (19.13 − 9.81)( − 1)] = 13.24 kN/m qu = (1.3)(28)(20.27) + (25.74)(9.19) + (0.4)(13.24)(2)(5.09) = 1028.3 kN/m2 Qall = (qu ) B (1028.3)(2) = = 1175 kN Fs 3.5 16.10 From Eq (16.12): q all = (1.3c ′N c + qN q + 0.4γBN γ ) Fs φ′ = 29º; Nc = 34.24; Nq = 19.98; Nγ = 16.18 (Table 16.1) qall = [(1.3)(900)(34.24) + ( 4.5)(116)(19.98) + (0.4)(116)( B )(16.18)] = 12622.6 + 187.7 B q all = 250000 B2 From Eqs (a) and (b), (a) (b) 250,000 = 12,622.6 + 187.7 B By trial and error, B2 B ≈ 4.31 ft 16.11 φ′ = 25º From Table 16.1, Nq = 12.72; Nγ = 8.34 q all = [( 2.1 × 19)(12.72) + (0.4)(19)( B )(8.34)] = 203 + 25.35 B q all = 550 = 203 + 25.35 B B2 B ≈ 1.5 m 144 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 16.12 q all = qu = Fs Fs ⎛ ′ ⎞ ⎜ c N c λcs λcd + qN q λ qs λ qd + γBN γ λγs λγd ⎟ ⎝ ⎠ φ′ = 32º; Nc = 35.49; Nq = 23.18; Nγ = 30.22 (Table 16.2) ⎛ B ⎞⎛ N q ⎝ L ⎠⎝ N c λcs = + ⎜ ⎟⎜⎜ ⎛ Df ⎝ B λcd = + 0.4⎜⎜ ⎞ 23.18 ⎞ ⎟⎟ = + ⎛⎜ ⎟ = 1.653 35 49 ⎝ ⎠ ⎠ ⎞ 1.2 ⎞ ⎟⎟ = + 0.4⎛⎜ ⎟ = 1.274 ⎝ 1.75 ⎠ ⎠ ⎛B⎞ ⎝L⎠ λ qs = + ⎜ ⎟ tan φ ′ = + tan 32 = 1.624 λqd = + tan φ ′(1 − sin φ ′)2 2⎛ ⎞ = + tan 32(1 − sin 32) ⎜ ⎟ = 1.157 B ⎝ 1.75 ⎠ Df ⎛B⎞ ⎝L⎠ λγs = − 0.4⎜ ⎟ = 0.6 λ γd = q = γ h + γ′ (Df – h) = (16 × 0.9) + (18.9 – 9.81)(1.2 – 0.9) = 17.12 kN/m2 Qall = (1.75) ⎡(17)( 35.49)(1.653)(1.274 ) + (17.12)( 23.18)(1.624 )(1.157 )⎤ ⎢ ⎥ + (18.9 − 9.81)(1.75)( 30.22)( 0.6)(1) ⎢ ⎥⎦ ⎣ ≈ 1890 kN/m 16.13 a For vertical load, Eq (16.44): q u = q N q λ qd λ qs + γB ′Nγ λγd λγs c′ = 0, φ′ = 31º Table 16.2: Nq = 20.63; Nγ = 25.99 B′ = B – 2x = 2.5 – (2)(0.2) = 2.1 m; L′ = 2.5 m ⎛ B′ ⎞ ⎛ 2.1 ⎞ ⎟ tan φ ′ = + ⎜ ⎟ tan 31 = 1.504 ⎝ L′ ⎠ ⎝ 2.5 ⎠ λqs = + ⎜ ⎛ B′ ⎞ ⎛ ⎞ ⎟ = − 0.4⎜ ⎟ = 0.664 ⎝ L′ ⎠ ⎝ ⎠ λγs = − 0.4⎜ 145 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part λqd = + tan φ ′(1 − sin φ ′)2 2⎛ ⎞ = + tan 31(1 − sin 31) ⎜ ⎟ = 1.134 B′ ⎝ 2.1 ⎠ Df λ γd = 1 qu = (19)(1)(1.504)(1.134)(20.63) + (19)(2.1)(25.99)(1)(0.664) 2 = 1012.8 kN/m Qall = qu B ′L′ (1012.8)(2.1)(2.5) = = 1063.4 kN Fs (5) b B′ = – (2)(0.9) = 4.2 ft; L′ = ft φ′ = 26º Nc = 22.25; Nq = 11.85; Nγ = 12.54 qu = c ′N c λcs λcd + qN q λqs λqd + γB ′N γ λγs λγd ⎛ B ′ ⎞⎛⎜ N q ⎟ ⎝ L ′ ⎠⎜⎝ N c λcs = + ⎜ ⎞ 4.2 ⎞⎛ 11.85 ⎞ ⎟⎟ = + ⎛⎜ ⎟⎜ ⎟ = 1.373 ⎝ ⎠⎝ 22.25 ⎠ ⎠ ⎛ B′ ⎞ ⎛ 4.2 ⎞ ⎟ tan 26 = 1.341 ⎟ tan φ ′ = + ⎜ ⎝ L′ ⎠ ⎝ ⎠ λqs = + ⎜ ⎛ B′ ⎞ ⎛ ⎞ ⎟ = − 0.4⎜ ⎟ = 0.72 ⎝ L′ ⎠ ⎝ ⎠ λγs = − 0.4⎜ ⎛ Df ⎝ B′ λcd = + 0.4 tan −1 ⎜⎜ ⎞ ⎞ ⎟⎟ = + 0.4 tan −1 ⎛⎜ ⎟ = 1.006 ⎝ ⎠ ⎠ λqd = + tan φ ′(1 − sin φ ′)2 2⎛ ⎞ = + tan 26(1 − sin 26) ⎜ ⎟ = 1.293 B′ ⎝ 4.2 ⎠ Df λ γd = qu = (900)( 22.25)(1.373)(1.006) + (115)( 4)(11.85)(1.341)(1.293) +(0.5)(115)( 4.2)(12.54)(0.72)(1) = 39,291 lb/ft = 39.29 kip/ft 146 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Qall = (39.29)( 4.2)(6) = 198 kip c φ′ = 38º; c′ = Nq = 48.93; Nγ = 78.03 B′ = 1.5 – (2)(0.1) = 1.3 m; L′ = 1.5 m γ = (1800)(9.81) = 17.66 kN/m 1000 For vertical load, Eq (16.44): q u = q N q λ qd λ qs + γB ′Nγ λγd λγs ⎛ B′ ⎞ ⎛ 1.3 ⎞ ⎟ tan φ ′ = + ⎜ ⎟ tan 38 = 1.677 ⎝ L′ ⎠ ⎝ 1.5 ⎠ λqs = + ⎜ ⎛ B′ ⎞ ⎛ 1.3 ⎞ ⎟ = − 0.4⎜ ⎟ = 0.653 ⎝ L′ ⎠ ⎝ ⎠ λγs = − 0.4⎜ λqd = + tan φ ′(1 − sin φ ′)2 ⎛ 1.5 ⎞ = + tan 38(1 − sin 38) ⎜ ⎟ = 1.266 B′ ⎝ 1.3 ⎠ Df λ γd = 1 qu = (17.66)(1.5)(1.266)(1.677)( 48.93) + (17.66)(1.3)(78.03)(1)(0.653) = 3336.7 kN/m2 Qall = qu B′L′ (3336.7)(1.3)(1.5) = = 1301.3 kN (5) Fs ⎛B 16.14 Eq (16.57): qu ( F ) = qu ( P ) ⎜⎜ F ⎝ BP Qall = Aqu ( F ) ⎞ ⎛5⎞ ⎟⎟ = (6800)⎜ ⎟ = 170,000 lb/ft ⎝2⎠ ⎠ ⎛ 170,000 ⎞ =⎜ ⎟(5) = 85,000 lb = 85 kip ⎝ ⎠ 16.15 qu(P) = 320 kN/m2 Eq (16.56): qu ( F ) = qu ( P ) ; qu ( F ) = 320 kN/m2 147 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ⎛π⎞ ( 2.5) (320) Aqu ( F ) ⎜⎝ ⎟⎠ = 392.7 kN Qall = = Fs CRITICAL THINKING PROBLEM C.16.1 The footing is placed at a depth of 1.5 m Part (a) B=1m Eq (16.49): qnet = N 60 ⎡ S ( mm) ⎤ Fd ⎢ e 0.05 ⎣ 25 ⎥⎦ 2×B=2×1=2m N60 should be averaged up to a distance of m below the foundation or up to a depth = 1.5 + = 3.5 m Therefore, N60-avg = (12 + 7)/2 ≈ 10; Se = 20 mm ⎛D Fd = + 0.33⎜⎜ f ⎝ B qnet = ⎞ ⎛ 1.5 ⎞ ⎟⎟ = + 0.33⎜ ⎟ = 1.495 ≤ 1.33 So, Fd = 1.33 ⎝ ⎠ ⎠ 10 ⎡ 20 ⎤ (1.33) ⎢ ⎥ = 212.8 kN/m 0.05 ⎣ 25 ⎦ Qall-net = (qnet )( A) (212.8)(1) = ≈ 71 kN 3 B = 1.5 m Eq (16.50): qnet = N 60 ⎛ B + 0.3 ⎞ ⎡ S ( mm ) ⎤ ⎜ ⎟ Fd ⎢ e 0.08 ⎝ B ⎠ ⎣ 25 ⎥⎦ × B = × 1.5 = m 148 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part N60 should be averaged up to a distance of m below the foundation or up to a depth = 1.5 + = 4.5 m Therefore, N60-avg = (12 + 7)/2 ≈ 10; Se = 20 mm ⎛D Fd = + 0.33⎜⎜ f ⎝ B ⎞ ⎛ 1.5 ⎞ ⎟⎟ = + 0.33⎜ ⎟ ≈ 1.33 ⎝ 1.5 ⎠ ⎠ qnet = ⎛ + ⎞ ⎡ 20 ⎤ ⎜ ⎟ (1.33) ⎢ ⎥ = 153.2 kN/m 0.08 ⎝ 1.5 ⎠ ⎣ 25 ⎦ Qall-net = (qnet )( A) (153.2)(1.5) = ≈ 115 kN 3 B=2m Eq (16.50): qnet N ⎛ B + ⎞ ⎡ S ( mm ) ⎤ = 60 ⎜ ⎟ Fd ⎢ e 0.08 ⎝ B ⎠ ⎣ 25 ⎥⎦ 2×B=2×2=4m N60 should be averaged up to a distance of m below the foundation or up to a depth = 1.5 + = 5.5 m Therefore, N60-avg = (12 + + 8)/3 = 9; Se = 20 mm ⎛D Fd = + 0.33⎜⎜ f ⎝ B ⎞ ⎛ 1.5 ⎞ ⎟⎟ = + 0.33⎜ ⎟ = 1.247 ⎝ ⎠ ⎠ qnet ⎛ + ⎞ ⎡ 20 ⎤ = ⎜ ⎟ (1.247 ) ⎢ ⎥ = 114.7 kN/m 0.08 ⎝ ⎠ 25 ⎣ ⎦ Qall-net = (qnet )( A) (114.7)(2) = ≈ 229 kN 3 B=3m Eq (16.50): qnet N ⎛ B + ⎞ ⎡ S ( mm ) ⎤ = 60 ⎜ ⎟ Fd ⎢ e 0.08 ⎝ B ⎠ ⎣ 25 ⎥⎦ 149 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 2×B=2×3=6m N60 should be averaged up to a distance of m below the foundation or up to a depth = 1.5 + = 7.5 m Therefore, N60-avg = (12 + + + 19)/4 ≈ 12; Se = 20 mm ⎛D Fd = + 0.33⎜⎜ f ⎝ B ⎞ ⎛ 1.5 ⎞ ⎟⎟ = + 0.33⎜ ⎟ = 1.165 ⎝ ⎠ ⎠ qnet ⎛ + ⎞ ⎡ 20 ⎤ = ⎟ (1.165) ⎢ ⎥ = 112.77 kN/m ⎜ 0.08 ⎝ ⎠ 25 ⎣ ⎦ Qall-net = (qnet )( A) (112.77)(3) = ≈ 338 kN 3 The design chart is shown below 400 Qall-net (kN) 300 200 100 0.5 1.5 2.5 B (m) Part (b) For any design footing size B, Qall−net ≥ Qapplied ……………………………………………………… (1) Qapplied = 250 kN From the chart it is found that condition (1) is satisfied when B ≈ 2.25 m 150 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Part (c) φ′ = 33º; c′ = 0; Nq = 32.33; Nγ = 31.94 (Table 16.1); B = 2.25 m qall-net = = qu −net qu − q = = (1.3c′N c + qN q + 0.4 γBN γ − q ) Fs 3 [(1.5)(17)(32.33) + 0.4(17)(2.25)(31.94) − (1.5)(17)] = 429.2 kN/m2 Qall-net = qall − net (B ) = ( 429 )( 2.25) ≈ 2173 kN Part (d) Net allowable column load calculated by the Terzaghi’s bearing capacity equation (Part c) is significantly higher than that calculated by the method based on N60 and limting settlement value (Part b) In actual design of foundations, both bearing capacity (based on shear strength) and settlement criteria need to be satisfied It is possible that the allowable column load calculated in Part (c) will change (decrease) if a settlement limit is imposed Considering the uncertainty in sampling and testing of granular materials for the determination of shear strength, the method based on N60 and settlement criteria may be preferable for these conditions 151 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 152 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 17 17.1 Shelby tube A: Do2 − Di2 (76.2) − (73) Eq (17.6): AR (%) = (100) = × 100 = 8.96% ≤ 10% (73) Di2 Samples can be considered undisturbed Suitable for grain size distribution, Atterberg limits, consolidation and unconfined compression tests Shelby tube B: Do2 − Di2 (3.5) − (3.375) Eq (17.6): AR (%) = (100) = × 100 = 7.54% ≤ 10% Di2 (3.375) Samples can be considered undisturbed Suitable for grain size distribution, Atterberg limits, consolidation and unconfined compression tests Split spoon sampler: Do2 − Di2 (50.8) − (35) Eq (17.6): AR (%) = (100) = × 100 = 110% ≥ 10% Di2 (35) Samples can be considered as highly disturbed Suitable for grain size distribution and Atterberg limits tests, but not for consolidation and unconfined compression tests 17.2 Depth (m) 10 17.3 σ 0′ (kN/m2) 34 68 102 136 170 ⎡σ ′ ⎤ CN = ⎢ ⎥ ⎣ pa ⎦ 1.71 1.21 0.99 0.857 0.767 −0.5 N60 (N1)60 = CN N60 10 11 14 ≈12 ≈12 ≈11 ≈12 ≈7 ⎡ ⎤ ⎢ ⎥ N 60 ⎥ ; pa ≈ 100 kN/m2 φ ′ = tan −1 ⎢ ⎢ ⎛ σ o′ ⎞ ⎥ ⎢12.2 + 20.3⎜⎜ p ⎟⎟ ⎥ ⎝ a ⎠ ⎦⎥ ⎣⎢ 153 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Depth (m) 10 17.4 σ o′ (kN/m ) 34 68 102 136 170 po φ′ (deg) N60 (kN/m ) [Eq (17.20)] 100 20.1 100 10 21.0 100 11 18.48 100 14 19.37 100 10.9 Average φ′ ≈ 18º ⎡ ⎛ 0.06 ⎞ ⎟ ⎢ N 60 ⎜⎜ 0.23 + D50 ⎟⎠ ⎢ ⎝ Eq (17.18): Dr (%) = ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎛ 98 ⎞⎥ ⎜⎜ ⎟⎟⎥ (100) ⎝ σ o′ ⎠⎥ ⎥ ⎦ Given γ = 15.7 kN/m3 The following table can now be prepared Depth z (m) 1.5 3.0 4.5 6.0 7.5 17.5 σ o′ = γ z (kN/m2) 23.55 47.1 70.65 94.2 117.75 ⎛ σ′ ⎞ ( N1 )60 = CN N 60 = ⎜⎜ ⎟⎟ ⎝ pa ⎠ Depth (m) 1.5 4.5 7.5 D50 (mm) 0.3 0.3 0.3 0.3 0.3 Dr (%) 99.5 ≈ 100 74.2 ≈ 74 71.7 ≈ 72 70.4 ≈ 70 66.4 ≈ 66 N60 10 14 18 20 −0.5 N 60 ; φ ′ = 27.1 + 0.3( N1 ) 60 − 0.00054( N ) 60 σ o′ (kN/m2) 1.5 × 18 = 27 × 18 = 54 4.5 × 18 = 81 ⎡(5.3 × 18) ⎤ ⎢⎣ + 0.7(18.8 − 9.8)⎥⎦ = 101.7 101.7 + 1.5(18.8 − 9.8) = 115.2 115.2 + 1.5(18.8 − 9.8) = 128.7 N60 CN (N1)60 φ′ 11 1.924 1.36 1.11 15.4 ≈ 15 12.24 ≈ 12 12.2 ≈ 12 (deg) 31.4 30.6 30.6 12 0.99 11.89 ≈ 12 30.6 15 17 0.931 0.881 13.96 ≈ 14 31.2 14.97 ≈ 15 31.4 Average φ′ ≈ 31° 154 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 17.6 a The properties should be averaged over a distance of 2B or m below the footing, i.e up to a depth of 5.5 m Therefore, design N60 = (8 + + 11) / = 9.33 ≈ and design φ ′ (deg) = (31.4 + 30.6 + 30.6) / = 30.8 ⎛ Df b Fd = + 0.33⎜⎜ ⎝ B ⎞ ⎟⎟ = + (0.33)⎛⎜ ⎞⎟ = 1.247 ⎝ ⎠ ⎠ N ⎛ B + ⎞ ⎛ + ⎞ ⎛S ⎞ ⎛ 25 ⎞ = 60 ⎜ ⎜ ⎟ (1.247)⎜ ⎟ = 185.53 kN/m ⎟ Fd ⎜ e ⎟ = 0.08 ⎝ B ⎠ 25 08 25 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ q net Qnet = qnet × B = 185.53 × = 742 kN 17.7 ⎛ qc ⎞ ⎜⎜ ⎟⎟ p 0.26 Use pa ≈ 100 kN/m2 Eq (17.39): ⎝ a ⎠ = 7.64 D50 N 60 Depth (m) 1.5 4.5 7.5 17.8 N60 11 12 15 17 D50 (mm) 0.28 0.28 0.28 0.28 0.28 0.28 qc (kN/m2) 4,390 4,938 6,036 6,585 8,231 9,328 Eq (17.33): E s = 3qc Using qc from Problem 17.7: Depth (m) 1.5 4.5 7.5 qc (kN/m2) 4,390 4,938 6,036 6,585 8,231 9,328 Es (kN/m2) 13,170 14,814 18,108 19,755 24,693 27,984 155 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 17.9 Eq (17.35): cu = cu = qc − σ c ; Nk ≈ 18.3 Nk 26000 − (22)(118) = 1278.9 lb/ft 18.3 17.10 From Eq (17.43): ⎛ 4.5 ⎞ Recovery ratio, R = ⎜ ⎟(100) = 56.25% ⎝ ⎠ 156 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... accessible website, in whole or in part 2.7 a b Percent passing mm = 100 GRAVEL: 100 – 100 = 0% SAND: 100 – 73 = 27% SILT: 73 – = 64% CLAY: – = 9% Percent passing 0.06 mm = 73 Percent passing... Percent passing mm = 100 Percent passing 0.05 mm = 68 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 68 = 32% SILT: 68 – = 59% CLAY: – = 9% d Percent passing mm = 100 Percent passing... publicly accessible website, in whole or in part 2.8 a b Percent passing mm = 100 Percent passing 0.06 mm = 30 Percent passing 0.002 mm = GRAVEL: 100 – 100 = 0% SAND: 100 – 30 = 70% SILT: 70 –