Solutions to Supplementary Problems

Transcription

1 Solutions to Supplementry Problems Chpter 8 Solution 8.1 Step 1: Clculte the line of ction ( x ) of the totl weight ( W ).67 m W = 5 kn W 1 = 16 kn 3.5 m m W 3 = 144 kn Q 4m Figure 8.10 Tking moments bout the toe Q : + M Q = = knm But, Equting, M Q = Wx = 61 x 61x = x = =.81 m 61 Introduction to Soil Mechnics, First Edition. Bél Bodó nd Colin Jones. 013 John Wiley & Sons, Ltd. Published 013 by John Wiley & Sons, Ltd INDD 1 6//013 4:50:0 PM

2 Introduction to Soil Mechnics Step : Drw the forces cting on the wll. P = 305 kn P p = kn 0.67 x =.81 m W = 61 kn.6 m Q F = 383 kn Figure Step 3: Check for overturning by tking moments bout the toe. Overturning moment: M o = = 793 knm esisting moment: M = = 1794 knm 1794 Therefore, F s = =.6 > 793 The wll is stisfctory ginst overturning. Step 4: Check ginst sliding by considering the horizontl forces. Horizontl disturbing force: H D = P = 305 kn Sum of horizontl resisting forces: + H = F + P p = = kn esultnt horizontl force: + H = = kn And the fctor of sfety ginst sliding: H Fs = = = 1.6 < H 305 D The unstisfctory fctor of sfety cn be incresed by: ) eplcing the snd below the bse with dense grvel, thus incresing the vlue of m = tn f, hence tht of F. b) Incresing the vlue of P p, by depositing mteril in front of the wll INDD 6//013 4:50:04 PM

3 Solutions to Supplementry Problems: Chpter 8 3 Step 5: Determine the position ( x ) of the resultnt verticl force ( ) reltive to the toe, due to ll verticl nd horizontl forces cting on the wll. W = 61 kn 305 kn.81 m kn x.6 m 0.67 m Q 383 kn Figure Tking moments bout the toe: + æm Q = = 1001 knm Note: F = 383 kn hs no moment bout Q. But, M Q = x nd = W = 61 kn M = 61x Q 1001 Equting, MQ = 61x = 1001 x = = 1.64 m 61 Lc of wll 61 kn 1.64 e 0.36 m Eccentricity e = 1.64 = 0.36 m b = 1m m d = 4m 61 kn 1.33 m Middle third 4 Middle third = = 1.33m 3 Therefore, = 61 kn is locted within the middle third, hence there is no tension t the bse. Figure INDD 3 6//013 4:50:07 PM

4 4 Introduction to Soil Mechnics Step 6: The mximum/minimum pressures under the bse re clculted from (8.57): f f f mx min min mx 6e = 1± bd d = 1 = 70kN/m = 1+ = 35kN/m > 00kN/m 4 4 Therefore, the soil is overstressed. The mximum pressure cn be lowered below 00 kn/m by widening the bse, sy, to d = 4. m nd crry out the checks gin. Solution 8. Erth pressure coefficients 1 sin3 K = = m 1+ sin3 1 Kp = = 3.6 K Depth of tie rod From (8.79): T b = Kp γ K h F s Where, b = m T Expressing depth of the tie: h = Kp γb K F s 10 = =.67 m 17 ( ) INDD 4 6//013 4:50:1 PM

5 Solutions to Supplementry Problems: Chpter 8 5 From (8.67): γ KP z T = K( H + z) Fs = (9 + z) z = z + z 1.63z ( ) 14.1 = z z 1.63z errnging, 1.33 z 5.53z = 0 Solving the qudrtic: z = 5.63 m Drw the section to scle nd position the nchor. Mesure the length of the tid rod. h =.67 m L = 15. m 9 GL m nchor 6.33 m GL 3 z = 5.63 m 61 (Scle: 1 cm to m) Figure φ = 45 + = 61 φ 45 = = 9 esults: Depth of penetrtion: z = 5.63 m Depth of tie: h =.67 m Length of tie: L = 15.0 m INDD 5 6//013 4:50:16 PM

6 6 Introduction to Soil Mechnics Solution 8.3 From the figure, σ v sinφ = σ v c cotφ + Or Substituting, Simplifying, σv σv φ sinφ c cotφ+ = From trigonometry: cotφ = cos sinφ cosφ σ v σv sinφ c + = sinφ Verticl overburden pressure t crck depth: σ = z γ σv σv c cosφ+ sinφ = v 0 c Substituting, z 0 γ = K σ v c cosφ = ( 1 sinφ) φ σ v = cos c From the text: 1 sin φ c Hence, z0 =, which is formul (8.3). γk cosφ φ = 1 1 sin K Solution nkine theory 4m Figure f = 3 g = 17 kn/m 3 P 1 sin3 K = = sin3 From (8.30): Kγ H P = = = kn INDD 6 6//013 4:50:19 PM

7 Solutions to Supplementry Problems: Chpter 8 7. Grphicl method (Grph 8.8) ) Drw five tril slip surfces nd clculte the weight of ech wedge per metre length of wll. Tble 8.1 Wedge Angle ( ) Wedge re (m ) Weight (kn) x.9 = = 1.46 = = = = = = = = = b) Drw the W nd forces, cting on ech wedge. As their direction is importnt only, plce them, for clrity of the digrm, where the ground nd the slip plne intersect. c) Drw the force polygons, bering in mind tht in the bsence of wll friction, the ctive force is horizontl. d) Mesure the mgnitude of the mximum vlue of the ctive force s P = 4 kn per metre length of wll, induced by wedge. This result is prcticlly the sme s the one clculted by nkine s theory. 3. Trigonometric method Becuse the ngle of wll friction is ignored, the polygon of ll forces cting on wedge is simplified. It is now right-ngled tringle, which cn be solved by bsic trigonometry. A formul for P cn now be derived in these terms nd pplied to ech wedge in succession INDD 7 6//013 4:50:3 PM

8 Wll H = 4m GL P Weight W 49.5 kn 78.5 kn kn 136 kn 16.8 kn Compct snd f = 3 g = 17 kn/m 3 d = Coulomb s method See lso Problem 8.9 for Culmnn s method Scle 4 cm = 1m Scle 1 cm = 10 kn Ws weight (kn) Clmnn line P Criticl slip surfce Grph INDD 8 6//013 4:50:33 PM

9 Solutions to Supplementry Problems: Chpter 8 9 The digrm below shows wedge in generl, the forces cting on it nd the ngles relevnt to the derivtion. H g f x P p Slip surfce W 90 f f Norml GL f 90 + f Unknown length x : H tnα = x H x = tnα Are of wedge: xh H A = = tnα Weight of wedge: γh W = γ A= tnα Figure Force tringle: P 90 + f f W P tn nd tn W Substituting W : ( α φ) = P = W ( α φ) ( α φ ) γh tn P = tnα (8.89) Figure The ctive force cn now be clculted for s mny slip plnes s desired nd the lrgest chosen. For this problem: Tbulting the clcultions: ( α ) ( α ) 17 4 tn tn 3 P = = kn tnα tnα Tble 8.13 Wedge P (kn) Bold fce = mximum vlue s before INDD 9 6//013 4:50:33 PM

10 10 Introduction to Soil Mechnics Notes 1. Formul (8.89) pplies only to f -soil bove the wter tble, horizontl ground surfce nd smooth wll.. The slip surfce, corresponding to the mximum ctive force, cn be found in this cse directly by formul (8.90), derived below. Formul (8.30): Formul (8.89): From which, γh P = K Equting these ( α φ) γh tn P = tnα ( α φ ) tn K = tnα Expressing the criticl ngle for mximum P : K tn α = tn( α φ ) tnα tnφ Ktnα = (From trigonometry) 1 + tnαtnφ ( K ) K tnα + tnφ tn α = tnα tnφ ( K tn φ) tn α + ( K 1) tnα + tnφ = 0 Dividing by tn f results in qudrtic eqution K 1 K α α tnφ tn + tn + 1 = 0 1 K But K < 1 Ktn α + tnα + 1= 0 tnφ Solving by the stndrd qudrtic formul: K ± 1 K 1 tnφ tnφ tnα = K ( ) ( φ) 4K The terms under the squre root re equl to zero however. Proof: 1 sinφ 1 + sinφ 1 + sinφ sinφ 1 K = 1 = = 1+ sinφ 1+ sinφ 1+ sinφ 1 K φ φ φ φ sin cos cos 1 sin = = = tnϕ 1 + sinφ sinφ 1 + sinφ 1 + sinφ ( )( ) ( φ) 1 K 4 1 sin φ 4 1 sinφ 1 + sinφ 1 sinφ = = = = φ 4 4K tn 1+ sin 1+ sin 1 + sinφ INDD 10 6//013 4:50:40 PM

11 Solutions to Supplementry Problems: Chpter 8 11 Therefore, 1 K tnφ 4K = 0 1 K nd tnα = K tnφ The criticl ngle is given by: α 1 1 K = tn φ K tn (8.90) Thus, only f of the cohesive soil hs to be known for the determintion of the criticl ngle in problems of similr type. In this cse: 1 sin3 f = 3 hence, K = = sin From (8.90): α = tn = tn3 ( ) 17.4 tn 61 3 From (8.89): P = = 41.78kN tn61 There is no need, therefore, to employ grphicl construction in order to solve this problem. Solution 8.5 Force tringle: W b + f Figure P p xh Are of wedge: A = where x = h tn β γh Weight of wedge: W = tnβ tn P + = p W ( β φ) Pp = Wtn( β+ φ) ( ) γh tn β+ φ Therefore, the pssive fore: P p = (8.91) tnβ INDD 11 6//013 4:50:5 PM

12 1 Introduction to Soil Mechnics For ngle of inclintion β : ( β+ φ) tn Kp = tnβ After the substitution of tn( β φ) tnβ+ tnφ + =, the derivtion for b is simi- 1 tnβ tnφ lr to tht for, in the ctive cse, yielding: β K p 1 1 = tn K tnφ p (8.9) Solution 8.6 ) Active cse: (8.90): α = tn = Pssive cse: (8.9): β = tn = b) Active cse: (8.89): ( ) 17 9 tn 61 3 P = = 11.5 kn tn61 ( + ) 17 tn 9 3 Pssive cse: (8.91): P p = = kn tn9 These results re the sme s shown on Figures 8.4 nd Note: Both nd b re functions of the known single vrible f. They cn, therefore be represented grphiclly (Grph 8.10) for ese of reference. Being stright lines, the formule my be simplified further to: ) Active cse: = = 61 As before b) Pssive cse: b = x 3 = 9 α = φ (8.93) β = φ Conclusion: Provided the conditions of vlidity, given on Grph 8.10, re stisfied, grphicl solution is unnecessry s formule (8.93), hence (8.89) nd (8.91) yield direct nswers INDD 1 6//013 4:51:06 PM

13 Solutions to Supplementry Problems: Chpter 8 13 Solution 8.7 Active cse () Forces cting on the wedge (b) Force tringle: h Wll P d H W f f g W = gh tn (90 + f + d ) P d 90 d f W Figure P sin( α φ) By the sine rule: = W sin( 90 + φ+ δ α) ( ) γh sin α φ And the ctive force: P = (8.94) tnα sin(90+ φ+ δ α Pssive cse () f W b Wll P p d (b) W d b + f P p (90 f d b) Figure αh W = tn β Applying the sine rule: Pp sin( β φ) = W sin( 90 + φ+ δ β) INDD 13 6//013 4:51:15 PM

14 14 Introduction to Soil Mechnics ( ) γh sin β φ And the pssive force: P p = (8.95) tnβ sin(90+ φ+ δ β Note: The criticl ngle nd b which yield the mximum nd minimum vlues of P nd P p respectively depend on f nd d. They cnnot be expressed by simple formule. Although the Culmnn digrm would indicte the ngles, it is esier to clculte series of vlues for P nd P p nd plot the figures on grph pper to locte the criticl vlues e.g. () P Active (b) P p Pssive mx. P min. P p Criticl ngle Criticl ngle b Figure Solution 8.8 Active cse: 9m Figure P f = 3 g = 17 kn/m 3 d = 17.6 ( ) sin( α 3) ( α) ( α ) ( α) : P = tnα sin sin 3 = tnα sin Tbulting vlues of P for sy Tble P (kn) After plotting these figures on Grph 8.11, the results re found to be: P = kn α = 58 o INDD 14 6//013 4:51:0 PM

15 Solutions to Supplementry Problems: Chpter Active cse GL f = 3 g = 17 kn/m 3 P = ctive force (kn) 180 Wll 9m kn 17.6 Slip plne = ngle of slip surfce Grph 8.11 Wll GL φ=3 γ=17kn/m Slip plne 17.6 m P p = pssive force (kn) 300 Pssive cse kn b =ngle of slip surfce Grph INDD 15 6//013 4:51:4 PM

16 16 Introduction to Soil Mechnics Pssive cse Wll f = 3 P p m d = 17.6 g = 17 kn/m 3 b Figure 8.10 (8.95) : ( β ) ( β) ( β+ φ) 17 sin Pp = tnβsin( β 34 sin + 3 = tnβsin 40.4 The pssive force is clculted for 5 b 5 nd tbulted. The ngles re within lower rnge, s the slip surfces re fltter thn in the ctive cse. Tble 8.15 b P p (kn) These figures re plotted on Grph 8.1 nd the results locted s: Pp = 10kN β = 18 Notes: 1. The effect of wll friction is decrese in the ctive force nd n increse in the pssive one. Compring the vlues clculted in Problems 8.6 nd 8.8. Tble 8.16 d = 0 d > 0 P (kn) P p (kn) b Problem For more complicted cses, the grphicl solution is esier INDD 16 6//013 4:51:4 PM

17 Solutions to Supplementry Problems: Chpter 8 17 Solution 8.9 The execution of the bove procedure is self-explntory for wll with verticl fce nd smooth surfce ( d = 0). GL Prllel to bseline Direction of P f = 3 4m Wll b Compct snd f = 3 g = 17 kn/m 3 d = Bseline Culmnn line P = 41.7 See lso Problem 8.4 for Coulomb s method W = weight of wedge Scle: 1cm = 10 kn Grph 8.13 f = 3 Culmnn s method INDD 17 6//013 4:51:7 PM

18 18 Introduction to Soil Mechnics Solution 8.10 Given: e mx = 6% e min = 53% G s =.66 p = 30 D r = 69% Step 1: Clculte the in-situ voids rtio (e) required, by formul (1.47), in order to determine the unit weight of the snd. ( e e) 100 mx Dr = emx emin Dr 69 e = emx ( emx emin) = 6 (6 53) = 55.8% Step : As the ground wter tble is ner the surfce of the fine snd, it is resonble to ssume, tht it is sturted, due to cpillry ction, between GL nd GWL. The sturted density is given by (1.4): γ st Gs + e = γ w = 9.81 = = 0.3 kn/m 1 + e Also, below the wter tble, the submerged density is: 3 γ = γ st γ w = kn/m Step 3: In order to clculte erth pressures, reference hs to be mde to section 8.4, relting to ctive expnsion. It is explined there, tht horizontl ctive pressure cn occur only fter the wll moves forwrd sufficiently to llow the soil to expnd from the K o -stte. During the expnsion process, the coefficient of erth pressure decreses from K o to K. If however the soil is not llowed to expnd, becuse the wll does not move, thn the coefficient remins K o nd the horizontl pressure is given by s H = K o s v, insted of s H =K s v. In this problem, the wll of the buried box section cnnot move forwrd, hence the ctive force hs to be estimted by pplying K o nd not K. 3 Pressures: σ = = 36.5 kn/m 1 σ = 36.5 kn/m σ = = 1.6 kn/m 3 σ = q = 70kN/m 4 σ = = 11.77kN/m ( wterpressure) w INDD 18 6//013 4:51:7 PM

19 Solutions to Supplementry Problems: Chpter 8 19 Note, tht both wlls re subjected to the sme pressures. As consequence of symmetric ground conditions nd the stiffness of the structure, there is neither ctive expnsion nor pssive compression. The section is subjected to pressure in K o -stte, hence K o must be used in the clcultions. Snd q = 70 kn/m g st 1.8 m Underpth P 4 P 1 g 1. m 3m 1.5 m 1.8 m P 0.6 m P 3 P w 0.4 m Q Figure 8.13 Coefficient of pressure t rest: K o = 1 sin f = 1 sin30 = Forces: P1 = = kn cting t y1 = 1.8 m P = = 1.90 kn cting t y = 0.6 m 1.6 P3 = = 3.78 kn cting t y3 = 0.4 m P4 = = kn cting t y4 = 1.5 m Pw = 1. = 7.06 kn cting t yw = 0.4 m esultnt force = 154. kn (sy 154 kn) Summing the moments of forces bout corner Q: M Q = ( ) = = 04.6 knm Equting the moment of the resultnt with the sum of moments yields its line of ction g INDD 19 6//013 4:51:3 PM

20 0 Introduction to Soil Mechnics 70 kn/m 154 kn 154 kn 1.33 m = 86 kn/m Figure γ = γ = = 1.33 m 154 Assuming the weight of the underpth, in mximum use, imprts pressure of 16 kn/m to the ground, the chrcteristic lods on the box section cn be indicted s shown. Solution 8.11 Pln view Apply formul (8.57): f mx min 6e = 1 ± bd d m 00 kn/m e 3m O Pressure digrm where b = m d = 3m min f = 0kN/m mx f = 00 kn/m Figure 8.15 Substituting: 6e 100 Mx : 00 = 1 + = e 100 Min : 0 = ( 1 e) e = 0.5m hence = = d 3 Check: Middle third = = = 0.5 m. This is why f 6 6 min = INDD 0 6//013 4:51:33 PM