Solution manual for soils and foundations 8th edition by liu

Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Instructor’s Manual to accompany Soils and Foundations Eighth Edition Cheng Liu Jack B Evett Upper Saddle River, New Jersey Columbus, Ohio Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Copyright © 2014, 2008, 2004, 2001, 1998, 1992 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458 Pearson Prentice Hall All rights reserved Printed in the United States of America 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 Prentice Hall™ is a trademark of Pearson Education, Inc Pearson® is a registered trademark of Pearson plc Prentice Hall® is a registered trademark of Pearson Education, Inc Instructors of classes using Fennimore, Sustainable Facility Management: Operational Strategies for Today, 1/e may reproduce material from the instructor’s manual for classroom use 10 ISBN-13: 978-0-13-511393-6 ISBN-10: 0-13-511393-8 Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L CHAPTER2 (2-1) Sieve No Size Opening 3/8 in No.4 Mass Retained (mm) 9.50 (g) 4.75 42 Percent Retained 2.00 146 No 40 No 100 0.425 0.150 No 200 0.075 458 218 73 Pan Percent Passing Retained No 10 Cumulative Percent 63 1000 4.2 4.2 14.6 45.8 21.8 64.6 86.4 35.4 7.3 6.3 93.7 100.0 6.3 18.8 100 95.8 81.2 13.6 100.0 Gradation curve is given on page (2-2) (2-3) (1) Eq (2-3): PI = 29 - 19 = 10% AASHTO classification: A-2-4 (0) (2) USCSclassification: SC (a) (b) (c) (d) Eq (2-10): Eq (2-11): Eq (2-12): Eq (2-15): w = [(62 - 50)/50)(100) = 24.0% 'y = 62/0.56 = 110.7Ib/ft3 'Yd = 50/0.56 = 89.3 Ib/ft" 2.64 = 50/(62.4Vs); = 0.304 ft3 v, Vw = (62 - 50)/62.4 = 0.192 ft3 v; = 0.56 - 0.304 = 0.256 ft3 (2-4) Eq (2-7): e = 0.256/0.304 = 0.84 (e) Eq (2-8): n = [(0.064 + 0.192)/0.56)(100) (f) Eq (2-9): S = [0.192/(0.064 + 0.192))(100) = 75.0% (a) Eq (2-10): (b) Eq (2-15): 2.66 = 122.7 /(lVs); Vv= 72 - 46.1 = 25.9 em" Eq (2-7): e = 25.9/46.1 = 0.56 (c) (d) Eq (2-8): = (141.5 Eq (2-9): Eq (2-11): Eq (2-12): (e) (f) (2-5) v; = 45.7% w = [(141.5 -122.7)/122.7](100) v, = 46.1 = 15.3% em" n = [(72 - 46.1)/72)(100) = 36.0% -122.7)/1 = 18.8 cm3 S = (18.8/25.9)(100) = 72.6% 'y = (141.5/72)(62.4) = 122.6 lb/ft" 'Yd = (122.7/72)(62.4) = 106.3lb/ft3 Work with ft3 of specimen = = (a) Eq (2-10): 18 (Ww/Ws)(100); Ww 0.18Ws Ww+ Ws= 118.5; 0.18Ws+ Ws= 118.5; Ws= 100.4lb Eq (2-12): 'Yd = 100.4/1 = 100.4lb/ft3 (b) 2.72 = 100.4/(62.4Vs); ft3 Eq (2-7): e = 0.41/0.59 = 0.69 Ww = (0.18)(100.4) = 18.1 ib Vw = 18.1/62.4 = 0.29 ft3 Eq (2-15): v, = 0.59 ft3 v; = - 0.59 = 0.41 (c) Eq (2-9): (2-6) S = (0.29/0.41)(100) = 70.7% Work with ft3 of specimen Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L N Gradation Curve for Problem 2••1 D50 II 0.68 II1Ift 010 • 0.105 C • u III!II D6D 010 0.90 • 0.105 • 8.57 o~ ~ ~ ~ 100 10.0 1.0 0.1 Grain Diameter (mm) Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L (a) (b) (c) {d} Eq (2-7): 0.56 = Vv/Vs; v; = 0.S6Vs v; + v, = 1; 0.S6Vs + Vs = 1; Vs = 0.64 ft3 Eq (2-1S): 2.64 = WJ[(O.64)(62.4)J; WS= 10S.41b Eq (2-1O): 15 = (Ww/l0S.4)(100); Ww = 15.8 Ib W = 15.8 + 105.4 = 121.2 Ib Eq (2-11): 'Y = 121.2/1 = 121.2 Ib/ft3 Eq (2-12): 'Yd = 105.4/1 == 105.4Ib/ft3 v; = 1- 0.64 = 0.36 ft3 Eq (2-8): n = {0.36/1)(100} = 36.0% Vw= 15.8/62.4 = 0.253 ft~ Eq {2-9}: S = {0.253/0.36){100} = 70.3% (2-7) Work with ft3 of specimen (a) Eq (2-10): 42 = (Ww/Ws)(100); Ww = 0.42Ws Ww+Ws = 112.8; 0.42Ws + Ws == 112.8; Ws == 79.4lb Ww = 112.8 - 79.4 == 33.4 Ib Vw= 33.4/62.4 = 0.54 ft3 Vs == 1- 0.54 == 0.46 ft3 Eq (2-7): e = 0.54/0.46 == 1.17 {Because S = 100%, v; == Vw.} (b) Eq (2-15): G, = 79.4/[{0.46){62.4)] = 2.77 (2-8) Work with ft3 of specimen (a) Eq (2-10): 26 = (Ww/1S.80)(100); Ww = 4.11 kN Vw== 4.11/9.81 == 0.419 m3 Vs= - 0.419 == 0.581 m 'Ysat == (4.11 + 15.80)/1 == 19.91 kN/m (b) Eq (2-7): e == 0.419/0.581 == 0.72 (Because = 100%, v, = Vw') (c) Eq (2-15): Gs == 15.80/[(0.581){9.81)j == 2.77 (2-9) Work with ft3 of specimen (a) Eq (2-7): 1.33 == Vv/Vs; v; = 1.33 v, v; + v, = 1; 1.33Vs + v, == 1; v, == 0.43 ft3 Vw = 1- 0.43 == 0.57 ft3 Ww = (62.4}{0.57) = 35.6 Ib Eq (2-10): 48 = (35.6/Ws){100); Ws == 74.2 Ib W == 74.2 +35.6 = 109.8 Ib Eq (2-11): 'y == 109.8/1 = 109.8 Ib/ft (b) Eq (2-15): Gs = 74.2/[{0.43)(62.4)J = 2.77 (2-10) Work with ft3 of specimen (a) Eq (2-10): 35 == (Ww/Ws}(100); Ww = 0.35Ws Vv= Ww/62.4 = 0.3SWJ62.4 == 0.005609Ws Eq (2-15): 2.70 == WJ[(Vs){62.4)) Vs = WJ[(2.70)(62.4)] = O.00593SWs Eq (2-7): e == (0.005609Ws)/(0.005935Ws) == 0.945 (b) From (a), v; = 0.94SVs v; + Vs = 1; 0.94SVs + v, = 1; v, == 0.514 ft3 v; == 1- 0.514 == 0.486 ft3 Ww = (62.4}{0.486) == 30.3 Ib Eq (2-15): 2.70 = W,/[{0.514)(62.4)]; Ws == 86.6 Ib Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L W = 86.6 + 30.3 = 116.9 Ib Eq (2-11): 'Y = 116.9/1 = 116.9 lb/ft" (2-11) Work with ft3 of specimen (a) Eq (2-7): 0.85 = Vv/Vs; v = 0.85Vs v; + Vs = 1; 0.85Vs + Vs = 1; Vs = 0.54 ft3 v; = 1- 0.54 = 0.46 ft3 Eq (2-9): 42 = (Vw/0.46)(100); Vw= 0.193 ft3 Ww = (62.4)(0.193) = 12.04 Ib Eq (2-15): 2.74 = Ws/[(0.54)(62.4)]; Ws = 92.331b Eq (2-10): w = (12.04/92.33)(100) = 13.04% W = 92.33 + 12.04 = 104.37 Ib/ft3 (b) Eq (2-11): 'Y = 104.37/1 = 104.4lb/ft3 (2-12) (a) Ww = 1.445 -1.301 = 0.144 kN Eq (2-10): w = (0.144/1.301)(100) = 11.1% (b) Eq (2-15): 2.65 = 1.301/[(Vs)(9.81)]; v, = 0.050 m3 v; = 0.082 - 0.050 = 0.032 m Eq (2-7): e = 0.032/0.050 = 0.64 (c) Eq (2-8): n = (0.032/0.082)(100) = 39.0% (d) v; = (1.445 -1.301)/9.81 = 0.015 m Eq (2-9): S = (0.015/0.032)(100) = 46.9% (e) Eq (2-11): 'Y = 1.445/0.082 = 17.62 kN/m3 (f) Eq (2-12): 'r, = 1.301/0.082 = 15.87 kN/m (2-13) Work with m3 of specimen (a) Ww + Ws = 18.55 kN Eq (2-10): 12.3 = Ww/Ws)(100); Ww = 0.123Ws 0.123Ws + vs, = 18.55; Ws = 16.52 kN Eq (2-12): 'yd = 16.52/1 = 16.52 kN/m3 (b) Eq (2-15): 2.72 = 16.52/[(Vs)(9.81)]; v, = 0.619 m3 v; = 1- 0.619 = 0.381 m Eq (2-7): e = 0.381/0.619 = 0.62 (c) Ww = (0.123)(16.52) = 2.03 kN Vw= 2.03/9.81 = 0.207 m3 Eq (2-9): S = (0.207/0.381)(100) = 54.3% (2-14) Work with m3 of specimen Ww+ Ws = 18.85 kN Eq (2-10): 5.2 = (Ww/Ws)(100); Ww = 0.052Ws 0.052Ws + Ws = 18.85; Ws = 17.92 kN Ww = (0.052)(17.92) = 0.93 kN Vw= 0.93/9.81 = 0.095 m Eq (2-15): 2.66 = 17.92/[(Vs)(9.91)]; v, = 0.687 m3 v; = - 0.687 = 0.313 m Eq (2-7): eo = 0.313/0.687 = 0.46 Eq (2-18): Dr = [(0.92 - 0.46)/(0.92 - 0.38)](100) = 85.2% (2-15) Vs = V - v; Eq (2-7): e = Vv/(V- Vv)= (Vv/V)/(V/V - Vv/V) Because Vv/V = n, e = n/(l- n) Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L (2-16): V = v, + Vs Eq (2-8): n = Vv/(Vv + Vs) = (Vv/Vs)/(VvlVs + VJVs) Because VvlVs e, n e/(e + 1) = (2-17) = Work with ft3 of specimen (a) Eq (2-8): 38 = (Vv/V)(100) Vv = 0.38V = (0.38)(l) = 0.38 ft3 v, = 1= 1- 0.38 = 0.62 ft3 v; Eq (2-7): e = 0.38/0.62 = 0.61 (b) Ws = (0.62)(2.66)(62.4) = 102.9 Ib Eq (2-9): 35 (Vw/0.38)(100); Vw = = 0.133 ft3 Ww = (0.133)(62.4) = 8.30 Ib W = 8.30 + 102.9 = 111.2 Ib Eq (2-11): 'y = 111.2/1 = 111.2 Ib/ft3 (2-18) Let subscript "b" denote soil in borrow pit and subscript "d" denote soil in the dam Eq (2-7): 1.12 {V.)b + {Vs)b = Vb; = (Vv)b/(Vs)b; {Vv)b = 1.12{Vs)b; 1.12{Vs)b + {Vs)b= Vb (Vb is the total volume of soil from the borrow pit.) {Vs)b= 0.4717Vb Eq (2-7): 0.78 = {Vv)i{Vs)d; {Vv)d = 0.78{Vs)d {Vv)d+ {Vs)d= 5,000,000 0.78{Vs)d +{Vs)d = 5,000,000; {Vs)d= 2,808,989 m Because {Vs)b (Vs)d, 0.4717Vb 2,808,989 v; = 5,955,000 m3 = (2-19) = Work with ft3 of specimen Ww + Ws = 128.21b Eq (2-10): 14.5 = (Ww/Ws)(100); Ww = 0.145Ws 0.145Ws + Ws = 128.2; Ws = 112.0 Ib Ww = 128.2 - 112.0 = 16.2 Ib (initially) Drying reduces the weight of water by 128.2 - 118.8, or 9.4 lb Hence, the weight of water after drying is 16.2 - 9.4, or 6.8 lb Eq (2-10): (2-20) wnew = (6.8/112.0)(100) = 6.1% Work with m3 of specimen (a) Ww + Ws = 17.98 Eq (2-10): 7.6 = (Ww/Ws)(100); Ww = 0.076Ws 0.076Ws+ Ws= 17.98; Ws=16.71 kN Ww = 17.98 -16.71 = 1.27 kN Eq (2-15): 2.65 = 16.71/[(Vs)(9.81)]; v; = 1- 0.643 = 0.357 m3 Eq (2-7): Eq (2-18): (b) eo = 0.357/0.643 = 0.56 62 = [(emax - 0.56)/(emax - v, = 0.643 0.35)](100); m emax = 0.90 v: + Vs = Eq (2-7): 0.90 = VvIVs; v; = 0.90Vs 0.90Vs + Vs = 1; Vs = 0.526 m v; = - 0.526 = 0.474 m3 = = Eq (2-15): 2.65 WJ[(0.526)(9.81)]; Ws 13.67 kN Eq (2-10): 7.6 (Ww/13.67)(100); Ww 1.04 kN W = 13.67 + 1.04 = 14.71 kN Eq (2-11): 'ym;n = 14.71/1 = 14.71 kN/m = = Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L (2-21) Work with ft3 of specimen v; + Vs = Eq {2-7}: 0.85 = vsv: v, = 0.85 v, 0.85 v, + v, = 1; v, = 0.541 ft3 v; = - 0.541 = 0.459 ft3 Eq {2-9}: 30.4 = (Vw/0.459)(100); v; = 0.140 ft3 For the degree of saturation to be 100%, Vwmust be equal to Vv' Hence, a volume of water of 0.459 0.140, or 0.319 ft3, must be added to each cubic foot This is {0.319)(62.4}, or 19.9 Ib of water {2-22} (a) Work with ft3 of specimen v; + Vs = Eq (2-18): 47 = [(0.95 - eo}/{0.95 - 0.38)](100); eo = 0.68 Eq (2-7): 0.68 = VvlVs; v, = 0.68Vs 0.68Vs + v, = 1; v, = 0.595 ft3 v; = 1- 0.595 = 0.405 ft3 Eq (2-15): 2.65 = WJ[{0.595)(62.4}]; WS= 98.41b Eq {2-12}: 'r, = 98.4/1 = 98.4Ib/ft Ww = {0.405)(62.4} = 25.3 Ib (if saturated) W = 25.3 + 98.4 = 123.7 Ib Eq (2-11): 'y= 123.7/1 = 123.7Ib/ft3 {b} Work with m3 of specimen v; + Vs = Eq {2-18}: 47 = [(0.95 - eo)/{0.95 - 0.38)]{100}; eo = 0.68 Eq {2-7}: 0.68 = vJv; v; = 0.68Vs 0.68Vs + Vs = 1; Vs = 0.595 m v- = 1- 0.595 = 0.405 m Eq {2-15}: 2.65 = WJ[(0.595)(9.81)]; WS= 15.47 kN Eq (2-12): 'Yd = 15.47/1 = 15.47 kN/m3 Ww = {0.405)(9.81} = 3.97 kN (if saturated) W = 15.47 + 3.97 = 19.44 kN Eq (2-11): 'y = 19.44/1 = 19.44 kN/m Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L CHAPTER (3-1) (a) = 0.600 ton/ft'' CN = 0.77log10 (20/0.600) = 1.17 Ncorrected = (1.17)(26) = 30 10 ft = 3.048 m and 120 Ib/ft = 18.85 kN/m3 Po = (10)(120)/2000 Eq (3-1): (b) Po = (3.048}{18.85) = 57.45 (3-2) (a) Po = [(8)(120) Eq (3-1): (b) (3-4) + (2)(120 - 62.4)]/2000 = 0.538 CN = 0.77log10 (20/0.538) = 1.21 ton/ft'' Ncorrected = (1.21}{26) = 31 = 1.08 klps/ft'' = 51 71 kN/m2 Ncorrected = (26)(100/51.71)1/2 = = 36 = (a) Po (7)(20.40) 142.8 kN/m2 Eq (3-2): CN = 0.77 IOg10(1915/142.8) = 0.868 (b) Ncorrected = (0.868)(22) = 19 Eq (3-3): Ncorrected = (22)(100/142.8)1/2 = 18 (a) Po = (2)(20.40) + (5)(20.40 - 9.81) = 93.75 kN/m2 (b) Ncorrected = (1.01)(22) = 22 Eq (3-3): Ncorrected = (22)(100/93.75)1/2 Eq (3-2): (3-5) = 34 Po Eq (3-3): (3-3) kN/m2 Ncorrected = (26)(100/57.45)1/2 Eq (3-3): Eq, (3-4): c eN = O.77log lO (1915/93.75) = 61/{{rr)[{4/12)2(8/12}{1/2) = 1.01 = 23 + (4/12)3(1/6)]} = 449 Ib/ft2 From Fig 3-17 with PI = 40%, Il = 0.85 Hence, ccorrected = (0.85)(449) = 3821b/ft (3-6) A plot of time versus distance is given on page slopeline = 0.083/80 = 0.001038 slopeline = (0.093 - 0.08675)/(180 -100) = 0.00007813 VI = reciprocal of slopeline = 1/0.001038 = 963 ft/sec V2 = reciprocal of slopeline = 1/0.00007813 = 12,800 ft/sec L = 82 ft (from plot on page 8) Eq (3-6): HI = (82/2)[(12,800 - 963)/(12,800 + 963)]1/2 = 38 ft With VI = 963 ft/sec, according to Table 3-3, the subsurface material in the first layer is estimated to be normal sand or loose sand above the water table With V2= 12,800 ft/sec, according to Table 3-3, the subsurface material in the second layer is estimated to be hard limestone, basalt, granite, or unweathered gneiss (3-7) Electrode Resistance Resistivity (ft) (ohms) (ohm-ft) 10 20 12.73 800 351 800 1151 275 289 1426 1715 Spacing Cumulative Resistivity (ohm-ft) 30 2.79 1.46 40 1.15 50 60 330 317 2045 2362 70 80 1.05 0.84 1.21 1.00 532 503 2894 3397 90 100 0.97 0.95 549 597 3946 4543 Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L + ,r ~ l·t :r- I , • t H:or it L ~ f P.-t • J ~ ~! 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Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 39 (10-23) For to 20 ft depth, because the N-value (10) < 15, Eq (10-21): 13 :: {10/15)[1.5 - (0.135)(20/2)°.5] = 0.715 Eq (10-19): fSl = (0.715)[(0.116)(20/2)] = 0.83 kip/ft2 Asurfacel= (4){n)(20) = 251.3 ft2 For 20 to 28 ft depth, N-value > 15 Eq (10-20): 13 = 1.5 - (0.135)(20 + 8/2)°·5 = 0.839 fs2 = (0.839)[(0.116)(20) + (0.128)(8/2)] = 2.38 kip/ft2 Asurface2= (4)(n){8) = 100.5 ft2 Because N-value (35) < 50, Eq (10-24): qb = (1.20)(35) = 42 kips/ft2 Abase = (n)(4)2/4 = 12.57 ft2 Eq (10-14): Qultimate = (0.83){251.3) + (2.38){100.5) + (42)(12.57) = 976 kips Oaliowable= 976/2.5 = 390 kips Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L CHAPTER11 (11-1) Eq {11-5}: Ko == 1- sin 30°:: 0.5 (a) Latera1 earth pressure l T +1 -\ III = 16.50 klt/m2 1m L L ~16.S0 + (0.5) (16.50 - 9.81) (1) == 19.85 klil/m2 (b) :Lateral water P%eJlsureJ T • + -L ~ ~ (c) '1'otal lateral T -+ (9.81) (1) = 9.81 kH/.2 pha.~ J I \ 16 SO ):)il/m2 1• -L ~ ~19.85 Eq (11-9): (11-2) :II: 29~66 kN/.;J Po = (16.50)(2}/2 + [(16.50 + 29.66)/2](1) == 39.58 kN per meter of wall Eq (11-14): Ka = (1- sin 35°)/(1 + sin 35°) = 0.271 Eq (11-10) P, (1/2){120)(2S)2(O.271) 10,160 Ib per foot of wall Point of application == H/3 = 25/3 == 8.33 ft above base of wall = {11-3} + '.81 = Eq (11-11): Ka = (cos 10o}{[cos 100 - (cos210o- cos2 3S0)}0.5/[cos10° + (cos2100 Eq (11-10): Pa:: (l/2)(18.85)(7.62)2(O.2818) == 154.2 kN per meter of wall Point of application = H/3 = 7.62/3 = 2.54 m above base of wall - cos' 3So)]0.5}::0.2818 40 Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 41 (11-4) T lS I h IZ 0.11 ~t -F:o~ ~t : ~90o -L~: -' _ tan 10° = AB/15 AB = 2.64 ft tan 15° = Be/AB = h/2.64 h=0.71ft Eq (11-10): Pa' = (1/2)(125)(15.71}2(0.373) = 57541b perfaat of wall W = (y)(AB){H)/2 = (125}(2.64)(15 + 0.71)/2 = 25921b per foot of wall Ph = Pa' cas ~ = (5754){cos 15°) = 5558 Ib per foot of wall P, = Po' sin f3 = (5754){sin 15°) = 14891b perfoot of wall ~v = w + Pv = 2592 + 1489 = 40811b per foot of wall I H = Ph = 5558 Ib per foot of wall Pa = [(~ V)2 + (I H)2]0.S = (40812 + 55582)°.5 (11-5) = 6900 Ib per foot of wall See Fig 11-1S(a) r 350 lb.lft2 f 25 - 19.29 25 f!t x = r' 19.29 J ft 1182 lb/ft2 = 5.71 ft + ~ y = 6.43 ft .1 2c tan (45° - 4>/2) = (2)(250) tan (45° - 20°/2) = 350 Ib/ft2 'YHtan2 (45° - cp/2) - 2c tan (45° - cp/2) = (125)(25) tan2 (45° - 20°/2) - (2)(250) tan (45° - 20°/2) = 1182 Ib/ft2 2c/fYtan {45° - /2}] = (2){250)/((125) Or tan (45 - 20 /2)] = 5.71 ft Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 42 350/1182 = (25 - xlix 350x = (1182)(25 - x) x = 19.29 ft Resultant = (1/2){1182)(19.29) = 11,400 Ib/ft Y:=x/3 :=19.29/3 :=6.43 ft above base of wall (11-6) Eq (11-18): Bracket e + {[sin (35° + 20°) sin (35° - OO)]/[sin(90° - 20°) sin (90° + OO)]}O.5:= 1.707 Ka = sin2 {90° + 35°)/[sin2 90° sin {90° - 200)(1.707}2] = 0.245 Eq (1l-10): Pa = (1/2)(120){25}2(0.245) = 9190 Ib perfoot of wall (11-7) Eq (11-18): Bracket= + {[sin (30° + 25°) sin (30° -15°)]/[sin (80° - 25°) sin (80° + 15°)]}0.5 1.510 Ka :=sin2 (80° + 300}/[sin2 800 sin (800 - 25°){1.510}2] = 0.487 Eq (1l-10): p = (1/2)(125}(15}2(0.487) = 6850 Ib per foot of wall (11-8) Eq (11-14): Ka = (1- sin 30°)/(1 + sin 30°) = 0.3333 Eq (1l-10): Pa = (1/2)(115)(25)2(0.3333) = 11,980 Ib per foot of wall Point of application for Pa = H/3 = 25/3 :=8.33 ft above base of wall Eq (11-20): P' = (500)(25)(0.3333) = 4170 Ib per foot of wall Point of application for P' = H/2 = 25/2 = 12.50 ft above base of wall (a) Total active earth pressure = 11,980 + 4170 = 16,150 Ib per foot of wall (b) let the point of application of the total active earth pressure be "h" ft above base of wall Take moments of forces at base of wall 16,150h = (11,980)(8.33) + (4170)(12.S0} h = 9.41 ft Thus total active earth pressure acts at 9.41 ft above base of wall = (11-9) Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 43 AWl = (1/2)(7.9)(18.9)(125) = 9330 Ib/ft AW2 = (1/2)(10.4}(22.8)(125) = 14,820 Ib/ft AW3 = (1/2)(12.0){28.2)(125) = 21,150 Ib/ft From figure, Pa = 6900 Ib/ft Note: Original scales of figure were in = ft and in = 5000 lb/ft However, the figure was reduced in the production process (11-10) f h: iQQ :4.35 ft 115 I / I / / J I I- I / / I I y• 9.41 AWl = (1/2){S}(2S}(11S) + (115)(4.35)(S) = 9690 Ib/ft AW2 = (1/2){10)(25)(115) + (115){4.35)(10) = 19,380 Ib/ft Aw3 = (1/2)(15)(25)(115) + (115)(4.35}{15} = 29,070 Ib/ft AW4 = (1/2)(20)(25)(115) + (115}(4.35)(20) 38,760 !b/ft From figure, P, = 16,100 Ib/ft Y = (H/3)[(H + 3h)/(H + 2h)J = {25/3){[25 + (3)(4.35)]/[25 + (2)(4.35)]) 9.41 ft above base of wall Note: Original scales of figure were in = ft and in = 10,000 Ib/ft However, the figure was reduced = = in the production process Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 44 (11-11) From Fig 11-33(a), P = 0.65H'Y tan2 (450 - 1.5 OK (b) Without passive pressure Eq {12-1}: (F.5')sliding (0.58)(19.71}/6.68 1.71 > 1.5 OK (a) Eq (12-2): = = With passive pressure Eq (11-15): Kp:: (I + sin 35°)/{1sin 35°):: 3.690 Eq (11-12): Pp = (1/2)(O.115)(5)2{3.690) == 5.3 kips/ft Eq (12-1): (F,S'}sliding = [(O.58}(19.71) + 5.31/6.68 == 2.5 > 2.0 OK (c) Base pressure calculations location of resultant R (= I V) if R acts at x ft from the toe Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 48 x = r Mtoe/r V = {r Mr - r Mo)/r v = {125.97 - 44.15)/19.71 = 4.15 ft e = 10/2 - 4.15 = 0.85 ft < L/6 = 10/6 = 1.67 ft OK (in middle third) Eq (9-11): q = 19.71/(10 x I) ± (19.71 x 0.85)(10/2)/[(1)(10)3/12] ±0 q = 1.97 ± 1.01 (12-4) ql = 0.96 kips/ft qR = 0.96 kips/ft Eq (12-11): T= 9.464 kN Eq (11-14): Eq (12-4): 0.000762 = T/[(0.090)(138,000)] Ka= (1- sin 36°)/{1 + sin 36°) = 0.260 9.464 = (17.0)(8)(0.260)(5)(0.4) = 0.669 m Assume a F.5 of 1.5 0= /2= 36°/2 = 18° Eq (12-7): Lmin= (1.5)(0.260){0.669)(0.4)f[{2){0.090)(tan Eq (12-9): Eq (12-8): LRankine = tan (45° -36°/2) 18°)] = 1.78 m = 4.08 m Ltotat=4.08 + 1.78 = 5.86 m Eq (12-10): (Ltotal)minimum = (0.80)(8) = 6.40 m > 5.86 m Therefore, use Ltotal= 6.40 m and a horizontal spacing of 0.669 m (12-5) 3.4 • 2.0 • 5.1 Eq (11-14): at I(a = (1- sin 33°}J{l + sin 33°) = 0.295 Component 1: P = (1/2)(17.6)(3.4)2(0.295) = 30.0 kN/m M = (30.0)[(3.4)(2/3) - 1.2J = 32.0 kN-m/m {P = active earth pressure or water pressure, and M = clockwise moment about tie point.} Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L 49 Component 2: P = (3.4)(17.6)(5.1)(0.295) = 90.0 kN/m M = (90.0)[(5.1)(1/2) + 2.2] = 427.5 kN-m/m Component 3: P = (1/2)(17.6 - 9.81)(5.1)2(0.295) = 29.9 kN/m M = (29.9)[(5.1)(2/3) + 2.2] = 167.4 kN-m/m Component 4: P = (1/2)(9.81)(5.1)2 = 127.6 kN/m M = (127.6)[(5.1)(2/3) + 2.2] = 714.6 kN-m/m l P = 277.5 kN/m l M = 1341.5 kN-m/m Rwp (i.e., Ps) = (1/2)(9.81)(5.7)2/1.5 = 106.2 kN/m l Momentsj, point= [(P6)min][(3.7)(2/3) + + 1.6] + (106.2)[(5.7)(2/3) + 1.6]-1341.5 =0 (P6)min= 126.6 kN/m Eq (11-15): Kp= (1 + sin 34°)/(1- sin 34°) = 3.537 (P6)mm(i.e., maximum mobilizable P6) = [(1/2)(18.86 - 9.81)(3.7)2(3.537)]/1.5 = 146.1 kN/m > 126.6 kN/m OK T + Ps + (P6)mm= PI + P2 + P3 + P4 T + 106.2 + 126.6 = 277.5 T= 44.7 kN/m (12-6) Eq (11-15): Kp= (1 + sin 30°)/(1- sin 30°) = 3.00 Eq (11-14): Ka= (1- sin 30°)/(1 + sin 30°) = 0.333 Capacity of deadman = tie-back tension = [(1/2)(Y)(H)2(Kp- Ka)]/F.s 79 = [{1/2)(18.39)(H)2(3.00 - 0.333)]/1.5 H = 2.20 m Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L CHAPTER 13 (13-1) f 20 ft From Fig 13-3, L = D/sin a ~ = arc tan (1/1.5) = lS/sin 15° = 58.0 ft = 33.69° = (D/sin fj) sin (fj - a) = (lS/sin 33.69°) sin (33.69° _15°) = 8.67 ft Eq (13-1): W = (58.0){8.67}(110)/2 :: 27,660 Ib/ft, or 27.66 kips/ft Eq (13-4): F.S = [(0.220)(58.0) + (27.66)(cos lSo)(tan 12°))/(27.66 sin lSo) h (13-2) Cd = c/F.S.c = 350/1.5 = 233.33Ib/ft2 = = = 2.58 > 1.5 OK = tan cPd (tan d= (tan 10°)/1.25 :=;0.14106 d:=;8.03° 9.82 From Fig 13-10 with Ns:=;9.82 and d= 8.03°, (13-8) 13 :=;37° ~ W cos a :=;306 + 1410 + 2380 + 3050 + 3480 + 3540 + 3210 + 2190 + 600 = 20,166 Ib ~ W sin a = -38 - 74 + 124 + 429 + 934 + 1570 + 2000 + 2040 + 766 :=;77511b Eq (13-13): F.S :=;[(225)(40) + {20,166)(tan 15°)]/7751 = 1.86 Full file at https://TestbankDirect.eu/Solution-Manual-for-Soils-and-Foundations-8th-Edition-by-L ... https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L... https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L... https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L Solution Manual for Soils and Foundations 8th Edition by Liu Full file at https://TestbankDirect.eu /Solution- Manual- for- Soils- and- Foundations- 8th- Edition- by- L