Principles of Geotechnical Engineering 9th Edition Das Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/principles-of-geotechnical-engineering-9th-edition-das-solutions-manual/
81 © 2018 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 10 10.1 a. 2 2 3 1 2 2 xy x y x y τ σ σ σ σ σ σ + ± + = xσ =175kN/m2; y σ =145kN/m2; τxy =+40kN/m2 2 2 3 1 (40) 2 175 145 2 175 145 + ± + = σ σ xσ 1σ = 202.72kN/m2; 3σ = 117.28kN/m2 b. ° = + + + = 22 sin2; cos2 2 2 θ θ τ θ σ σ σ σ σ xy x y x y n xyτ 2kN/m 177 = + + + = )]22)(2sin[(40 )]22)(2cos[( 2 175 145 2 175 145 nσ 239.2kN/m = = = 40cos[(2)(22)] sin[(2)(22)] 2 175 145 cos2 sin2 2 θ xy x y n τ θ σ σ τ 10.2 a. x σ =2193lb/ft2; y σ =3906lb/ft2; xyτ =919lb/ft2; θ=168º 2 2 3 1 (919) 2 2193 3906 2 2193 3906 + ± + = σ σ 1σ = 4305.75lb/ft2; 3σ = 1793.25lb/ft2
b. 22cos[(2)(168)]919sin[(2)(168)] n σ +− =++ = 2 3458.16lb/ft
3906219339062193
82 © 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 39062193sin[(2)(168)](919)cos[(2)(168)] 2 n τ =−=− 2 1187.91lb/ft
2 2 1 2 1 kN/m 27 63 90 ; kN/m63 2 36 90 = = = + = O O OO 2 2 2 1 34.2kN/m 21) ( 2 90 36 = + = B O ) ( kN/m28.8 2 + = = = 34 63 3 OS σ ) ( kN/m 97.2 2 + = + = = 34 63 1 ON σ ° = = ∠ 37.87 27 21 tan 1 2 1O BO b. ) ( kN/m42 2 + = = = cos(52.12) 34 63 cos(52.12) 1 1 D O OO σn ) ( sin(52.12) 1 + = = kN/m27 2 D O τn
10.3 a. TheMohr’scircleisshown.
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2 2 1 12 630450 540 lb/ft 54045090 lb/ft 2 OOOO + ===−= ; 222 1 (90)(110)142.1 lb/ft OB =+= 1 540142.1 σON==+= 2 682.1lb/ft 3 540142.1 σOS==−= 2 397.1lb/ft ° = = ∠ 50.71 90 110 tan 1 2 1O BO b. 2lb/ft 517.06 = = + = cos(80.71) 142 540 30) cos(50.71 1 1 D O OO σn 2lb/ft 140.24 = = sin(80.71) 142 nτ
10.4 a. TheMohr’scircleisshown.
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P isthepole. Fromthegraph, ; 1 2kN/m 178 ≈ = ON σ 2kN/m62 ≈ = OS σ3 b. nσ and nτ arecoordinatesof D.So nσ ≈ 176kN/m2; nτ ≈ 8kN/m2 ( )
10.5 a. TheMohr’scircleisshown.
P isthepole.Fromthegraph, ; 1 2kN/m 27.8 ≈ = ON σ 2kN/m12.3 ≈ = OS σ3 b. nσ and nτ arecoordinatesof D. So, nσ ≈ 12.6kN/m2; nτ ≈ 2.1kN/m2
10.6 a. TheMohr’scircleisshown.
10.7
10.8 Eq.(10.15):
10.9 Eq.(10.15):Inthiscase,
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Load @ P (kN) r (m) z (m) z r I1 (Table10.1) 1 2 I z P σz = ∆ (kN/m2) B C D 125 250 500 8 (82 +82)0.5 =11.31 (82 +42)0.5 =8.944 10 10 10 0.8 1.13 0.89 0.1386 0.0628 0.1111 0.173 0.157 0.555 ∆σz =∑ ∑∑0.885kN/m2
229.16kN/m = + + + = + + + + = ∆ 2 2 2 3 2 2 2 3 2 2 2 2 3 2 2 2 2 2 1 3 1 ] 4 )(440)( ] )4( [( )(110)( ] [ 2 ] ) [( 2 π π π π z x z q z x x z q σz
x2 =0 271.68kN/m = + + + + = + + + + = ∆ 2 2 2 3 2 2 2 3 2 2 2 2 3 2 2 2 2 2 1 3 1 ] 4 )(440)( ] )4( [(6 )(110)( ] [ 2 ] ) [( 2 π π z x π z q z x x π z q σz 10.10 2 2 2 2 3 2 2 2 2 2 1 3 1 ] [ 2 ] ) [( 2 z x π z q z x x π z q σz + + + + = ∆ 1 2 2 2 3 2 2 2 3 1 000415 98 64 ] 7 3800)( ] 7 [26 )( )( 77 q q + = + + + = π π q1 = 28,964lb/ft 10.11 2 2 2 3 1 1 ] [ 2 to due at z x π z q q A σ z + = ∆ ,or 2 2 2 2 3 1 11.94kN/m ] )5( [( )(375)( ) ( = + = ∆ π σz
2 = q2 sin63°
Verticalcomponentof q
10.14 Refertothefigureonthefollowingpage.
Fortheleftside(withthenotationsgiveninFigure10.18):
86 © 2018 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 2 2 2 2 3 2 2 00215 ) ];( )5( )5. [(12 (sin63)( 2 ) ( q q z z = ∆ + = ∆ σ π σ Horizontalcomponentof q2 = q2 cos63° FromEq.(10.17): 2 2 2 2 2 2 2 2 2 2 2 3 002749 ] 5 [12 )( (cos63)(12 2 ) ( 2 ) ( q q z x xz q σz = + = + = ∆ π π
∆ z σ =58kN/m2 =(∆ z σ )1 +(∆ z σ )2 +(∆ z σ )3 58=11.94+0.00215q2 +0.002749q2 kN/m 9402 = 2q 10.12 B =48ft; q =1450lb/ft2; x =28.8ft; z =21ft 347 FromTable10.4, 875. 48 )(21) 2 ;2.1 48 )8. )(28 2 = ∆ = = = = q B z B x zσ ∆ z σ =(0.347)(1450)= 503.15lb/ft2
818
8 )( 2 8 )( 2 = ∆ = = = = q B z B x zσ ∆ z σ =(700)(0.818)= 572.6kN/m2
Totalverticalstressincrease,
10.13
FromTable10.4,
39 10.19,Figure From 20 60 20 0 ) 2( 2 1 = = = = = L I z B z B Fortherightside: 495 10.19,Figure From 20 60 ; 75 20 55 ) 2( 2 1 = = = = = R I z B z B
87 © 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ∆ z σ = q[I2(L) + I2(R)]=(3332)(0.39+0.495)= 2948.8lb/ft2 10.15 At A: Fortheleftside: 416 12 5 1 = = z B ; 21 12 50 2 = = z B ; I2(L) =0.47 Fortherightside: 416 1 = z B ; 21 2 = z B ; I2(R) =0.47 ∆ z σ =(382.5)(0.47+0.47)= 359.55kN/m2 At B: Fortheleftside: 0 12 0 1 = = z B ; 21 12 50 2 = = z B ; I2(L) =0.425
88 © 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Fortherightside: 833 12 10 1 = = z B ; 21 12 50 2 = = z B ; I2(R) =0.485 ∆ z σ =(382.5)(0.425+0.485)= 348kN/m2 At C: Fortheleftside: 0 1 = z B ; 21 12 50 2 = = z B ; I2(L) =0.425 Fortherightside: 05 12 60 1 = = z B ; 21 12 50 2 = = z B ; I2(R) =0.485 ∆ z σ =(382.5)(0.5–0.425)= 28.68kN/m2
10.16 RefertoNewmark’schart.
Theplanisdrawntoscale.
= AB 6m. M ≈ 65.
∆ z σ =(IV) q M =(0.005)(565)(65)
= 183.62kN/m2
10.17 Point A:
Eqs.(10.33)and(10.34):
Eq.(10.31): ∆ z σ = q I3;Table10.10: I3 =0.2279
∆ z σ =(330)(0.2279)= 75.2kN/m2
Point B: Refertothefigure.
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5.1 6 9 6 18 = = = = = = z B m z L n
90 © 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. × × = ∆ 9m 5m area dueto atB stress 9m 23m area dueto atB stress zσ Forrectangulararea23m × 9m: 0.2288 ;5.1 6 9 ; 83 6 23 3 = = = = = I n m Forrectangulararea5m × 9m: 0.1788 ;5.1 6 9 ; 833 6 5 3 = = = = = I n m ∆ z σ = q(0.2288–0.1788)=(330)(0.05)= 16.5kN/m2 Point C: Refertothefigure. ForArea1: 0.1777 ;9.0 6 4.5 ;2.1 6 2.7 3 = = = = = I n m ForArea2: 0.1899 ;9.0 6 4.5 ;8.1 6 10 3 = = = = = I n m ForArea3: 0.1431 ;6.0 6 6.3 ;2.1 6 2.7 3 = = = = = I n m ForArea4: 0.1521 ;6.0 6 6.3 ;8.1 6 10 3 = = = = = I n m ∆ z σ = q[I3(1) + I3(2) + I3(3) + I3(4)]=(330)(0.1777+0.1899+0.1431+0.1521) = 218.72kN/m2
10.18 Eqs.(10.36),(10.38),(10.39),and(10.40):
10.19 Eq.(10.27)andTable10.7:
10.20 UsingWestergaard’ssolution(Table10.14):
AcomparisonofBoussinesq’ssolution(fromProblem10.19)andWestergaard solutionispresentedinthefollowingfigure.
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m 5.4 2 9 2 = = = B b ; 2 9 18 1 = = = B L m z (m) 3 6 9 12 15 b z n = 1 0.67 1.33 2 2.66 3.33 I4 (Table10.11) 0.91 0.561 0.481 0.357 0.259 4qI z = ∆σ (kN/m2) 300.3 185.1 158.7 117.8 85.5
q
2 R (m) z (m) R z q σz ∆ ∆ z σ (kN/m2) 2.5 0 0 1 39.24 2.5 2 0.8 0.7562 29.53 2.5 4 1.6 0.3961 15.54 2.5 8 3.2 0.1322 5.18 2.5 10 4 0.0869 3.41
=(4)(9.81)=39.24kN/m
R (m) z (m) R z q zσ∆ ∆ z σ (kN/m2) 2.5 0 0 1 39.24 2.5 2 0.8 0.5076 19.92 2.5 4 1.6 0.2487 9.76 2.5 8 3.2 0.0858 3.37 2.5 10 4 0.0572 2.24
10.21 Eq.(10.28)andTables10.8and10.9:
10.22 Refertothefigurebelow.
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q =(5)(9.81)=49.05kN/m2 z (m) r (m) R (m) R z R r A′ B′ ∆ z σ (kN/m2) 2 0 4 0.5 0 0.55279 0.35777 44.66 2 2 4 0.5 0.5 0.49075 0.34214 40.85 2 4 4 0.5 1.0 0.28156 0.13591 20.48 2 6 4 0.5 1.5 0.09499 -0.03455 2.964 2 8 4 0.5 2.0 0.03701 -0.02651 0.51
FromEq.(10.21):
CRITICALTHINKINGPROBLEMS
10.C.1Fromthegivenstressconditionsonplanes EB and FG,thepoints I (25,10)and J (10,−5)arelocated,andtheMohr’scircleisdrawnasshownbelow.
93 © 2018 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. ° = = 18.43 6 2 tan 1 1α ° = = 63.43 1 2 tan 1 2α ° = ° ° = = 45 18.43 63.43 1 2 α α α ° = = 26.56 2 1 tan 1 δ ; ° = 53.13 2δ
1726 sin53.13 45 180 5 6 2 2 1 sin2 2 2 1 = ° × × = = ∆ π π δ α B x π q σz 2kN/m 69.04 = = ∆ 1726) (400)( zσ
a. Fromthegraph, 28.5kN/m ≈ 3σ and 2kN/m 31.25 ≈ 1σ
b. Thepole P islocatedbydrawingaline, IP,inclinedatanangle25° withthe horizontal.Theline PJ determinestheinclinationoftheplane FG inspace. Therefore,theanglebetweenplanes EB and FG isgivenby ∠ IPJ ≈ 108° °
c. Bydrawinghorizontalandverticallinesfromthepole P,thepoints L (31,2.5) and K (9.2,−2.5)arelocated.
Therefore,thestressesontheplanes AB and BC are:
σAB = 31kN/m2; τAB = +2.5kN/m2
σBC = 9.2kN/m2; τBC =‒2.5kN/m2
10.C.2a. Verticalstressincreaseduetowheelload:
y =0.305m; R =0.15m; q =565kN/m2
Overburdenpressureatthemiddleofthelayer=0.305
Totalverticalpressure, ∆σy:
Thesevaluesareenteredintothefollowingtable
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Element r (m) R y R r A′ B′ ∆σy (kN/m2) A 0.457 2.03 3.05 0.02221 0.00028 12.7 B 0.267 2.03 1.78 0.05278 0.04391 54.63 C 0 2.03 0 0.10557 0.17889 160.71
× 19.4=5.92kN/m2
At A: σy-A =12.7+5.92= 18.62kN/m2 At B: σy-B =54.63+5.92= 60.55kN/m2
C: σy-C =160.71+5.92= 166.63kN/m2
At
Element at Horizontal stress, σx (kN/m2) Shear stress, τ (kN/m2) Vertical stress, σy (kN/m2) σ1 (kN/m2) σ3 (kN/m2) αi (deg) A 25 17 18.62 39 4.5 55 B 32 45 60.55 93 1 48 C 7 0 166.63 167 7 0 b. Elementat A:
≈
σ3 ≈ 4.5kN/m2
TheMohr’scircleisshown. σ1
39kN/m2;