SOLUTION MANUAL OF PRINCIPLES OF GEOTECHNICALL ENGINEERRIG_6ED

SOLUTION MANUAL OF PRINCIPLES OF GEOTECHNICALL ENGINEERRIG_6ED

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Unformatted text preview: INSTRUCTOR’S SOLUTIONS MANUAL % to accompany ISBN 0-495—07317—2 BRAJAM..DAS mam-195073172 THOMSON ENGINEERING?“ ' 9 80495 73113 mkmm.wmummwmwwm mlu‘mmmwfl» n..." -_. .. égis'zggeqtéad retrieggéIeSysgtEQaSL-Withqht the'_writ§§n peimgggpn-ofthé pfibl'i's "‘e Chapter ' Chapter Chapter Chapter -. .. 1 , . . . Chapter: Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter \0 on -.z oCZELn 4: at» N Ht—ns—nt—ar—tl—It—I \I-UI 4':- D) N l-I-l o .. a. 101 .. 105 .. 119 .t 125 CHAPTER 2 Percent retained Sieve No. Mass retained (g) on each sieve Percent finer 4 " 28 4.54 95.46 10 42 6.81 88.65 20 -- 48 7.78 80.87 40 128 20.75 60.12 60 221 35.82 243 ' 100 86 13.94 10.36 200 40 6.48 . 3.88 Pan 24 3.88 0 z 6 mg . .. . The graph for percent finer versus grain size is shown. 100 '80 Ch 0 A 0 Percent finer M O 10.0 5 .0 1.0 0.5 0.1 0.05 Grain size (mm) b. From the graph, Dlo = 0.14 mm, D30 = 0.27 mm, D60 = 0.42 mm. _ c” 13,0 0.14 2 2 d” C‘zfiLflflqM 2.2 _ b. 0.. d. Percent retained Sieve No. Mass retained (g) on each sieve Percent finer 4 0 0 100 10 44 7.99 92.01 - 20 ' 56 10.16 81.85 40 82 14.88 66.97 60 51 9.26 57.71 80 106 19.24 38.47 100 92 16.70 21.77 200 . 85 15.43 6.34 Pan 35 _ 6.34 0 2551 g The gram-size distribution curve is Shown- 100 80 Percent finer ON C 4; O t 20 0 . 10.0 5.0 1.0 0.5 0.1 0 05 Grain size (mm) From the giaph, D.50 = 0.3 mm, D30 = 0.17 mm, Dlo = 0.11 mm. Cu = 2:3— = 2.73 0.1 l (0.17)2 —'-——.....'—‘ “ — (0..11)(0.3) " ' Percent retained Sieve No. Mass retained (g) on each sieve Percent finer 4 0 0 100 10 40- 5.49 94.51 20 60 8.23 86.28 40 89 12.2 74.08 60 140 19.2 54.88 80 122 16.74 38.14 100 210 28.81 9.33 200 56 7.68 1.65 Pan 12 ' 1.65 - 0 2729 g The grain-size distribution curve is shown. 100 80 60 v 40 Percent finer 20 10.0 5.0 f 1.0 0.5 0.1 0.05 Grain size (mm) b. From the graph, D‘sD =. 0.27 mm, D30 = 0.17mm, Dm = 0.15-mm. c. C =£gl=L8 " 0.15 _ (0.17)2 _ “ ‘(0.15)(027) ' ' 2.4 Percent retained Sieve No. Mass retained (g) on each sieve Percent finer 4 0 0 100 6 30.0 6.0 94 10 48.7 9.74 84.26 20 127.3 25.46 58.8 40 96.8 19.36 39.44 60 76.6 15.32 24.12 100 55.2 11.04 13.03 200 43.4 8.68 4.40 Pan 22.0 4-4 0 2500 g The grain-size distribution curve is shown. ' 100 80' 60 40 Percent finer 20 10.0 3.0 1.0 0.3 0.1 0.05 Grain size (mm) b. From the graph, D60 ? 0.82 mm, D30 = 0.31 mm, Dlo = 0.12 mm. C. Cu = y = 6.83 0.12 (0.31)2 c = ——-——— = 0. (0.81)(0.12) 2.5 0.. Percent retained Sieve No. Mass retained (g) on each sieve Percent finer 4 0 0 100 '6 0 0 100 ID 0 0 100 20 9-1 1.82 98.18 40 249.4 49.88 48.3 - 60 179.8 35.96 12.34 100 22.7 4.54 7.8 200 15.5 3.10 47 Pan 23.5 4.70 0 2500 g The grain-size distribution curve is l'shcwn. 100 80 ON C) 4:. O Percent finer N _ O c I I I a I I I I I l I l l I 10.0- - 3.0 1.0 0.3 Grain size (mm) From the graph, D60 = 0.48 mm, D30 = 0.33 mm, D19 = 0.23 mm}; Cu = 9513- = 2.09 0.23 2 (033) . ._ 0 99 HM? 2.6 2.7 100 80 O\ C) 4:; 0 Percent finer N O 0 1 0 0.3 0.1 b. Percent passing 2 mm = 100 Percent passing 0.06 mm = 84 Percent passing 0.002 mm = 11 . Percent passing 2 mm = 100 Percent passing 0.05 mm = 80 Percent passing 0.002 mm = 11 . Percent passing 2 mm = 100 Percent passing 0.075 mm = 90 Percent passing 0.002 mm = 11 . Percent passing 2 mm = 100 - Percent passing 0.06 mm = 74 Percent passing 0.002 mm "—' 9 . Percent passing 2 mm = 100 Percent passing 0.05 mm = 70 Percent passing 0.002 mm = 9 0.03 Grain size (mm) a. The grain-size distribution curve is shown. 0.01 0.003 0.001 GRAVEL: 100 -— 100 = 0% SAND: 100 ~ 84 = 16% SILT: 84 —11 = 7.3% . CLAY: 11- 0=11% GRAVEL: 100 — 100 =0% ' SAND: 100 — so = 20% I SILT: 80 —11=69% CLAY: 11-- 0= 11% GRAVEL: 100 - 100=0% SAND: 100 - 90 = 10% SILT: 90 -II =79% CLAY: ll - O=ll% . The grain-size distribution is shown in the figure on the next page. GRAVEL: 100 — 100 = 0% .- SAND: 100 - 74 = 26% SILT: 74 -9 = 65% '- CLAY: 9 ~ o=9% _ GRAVEL: 100 —' 100 = 0% SAND: 100 - 7.0 =30% ' SILT: 7O -9 = 61% CLAY: 9 - 0 = 9% . Percent passing 2 mm = 100 100 80' Percent finer l .0 0.3 0.1 0.03 0 .01 0.003 0 .001 Grain size (mm) Figure 2.7a. GRAVEL: 100 — 100 = 0% SAND: '100 ~ 80 = 20%‘ SILT: 80 —9 = 71% CLAY: 9 — 0 =9% Percent passing 0.075 mm = 80 Percent passing 0.002 mm = 9 The grain-size distribution is shown. 100 4: ON 60 O O 9 Percent finer M O 1.0 0.3 0.1 0.03 0.01 Grain size (mm) 0.003 0 001 b.. Percent passing 2 mm = 100 Percent passing 0.06 mm = 30 Percent passing 0.002 mm = 6 c. Percent passing 2 mm = 100 Percent passing 0-05 mm = 26 Percent passing 0.002 mm = 9 d. Percent passing 2 mm = 100 Percent passing 0.075 mm = 34 Percent passing 0.002 mm = 6 2.9 sedimentation Equation (2.5): D (mm) = K L (cfn) V I (mm) GRAVEL: 100 - 100 = 0% SAND: 100 — 30 = 70% SILT: 30 —6 = 24% CLAY: 6 — o = 6%. GRAVEL: 100 - 100 = 0% SAND: 100 w 26 = 74% SILT: 26 v6 = 20% CLAY: 6 - 0 = 6% GRAVEL: 100 - 100 = 0% SAND: 100 - 34 = 66% SILT: 34 —6 = 28% CLAY: 6 - 0 = 6% G, = 2.7; temperature = 24°; L = cm; time, t = 60 minutes after the start of From Table 2.6, for G, = 2.7 and temperature = 24°, K = 0.01282. So 92 D = 0.01282 4 = 0.005 mm 60 2.10 For G, = 2.7 and temperature = 23°, K = 0.01279 (Table 2.6) D (m) = K L (cfn) = 0.012791 = 0.0046 mm 1 (mm) 100 .- CHAPTER 3 Gy +9}! Gr e e. 3”] =-—_£._w—.—.—w= S w+__.._._ : + 7““ 1+e 1+e 1-1-52:le 7" l+e 1’.” (1 +e)(ym- yawn erfi 7m - rd +e2’m" en em- rsax+ m= nar- 7d yam—74 e=—-——~— yd_?sat+YW , (1+w JG y : SE 3 W “at l+wsa,G, 7581+ wsale ysat= Gs yw+wsale yw Gs( yw + wsat 7w hwsat = ysat G: ya: ___ 7w 7w+w9t7wmwsatysat yw_ satU’sat—7w) 3.3 a“ y=E=££=122lblft3 V 0.1 b.. yd —L= 122 = 108.931blft3 1+w 1+3 100 Gsrw c“ yd = 1+e math; e=0.56 1+e e 0.56 d.‘ n=—= =036 l+e l+0.56 6.. f. w_G, = (0.12)(2.72) s: x100=58.3% 8 W511: 12;: =10.891b 1+w 1+_ 100 Ww= W— W5 =12.2 40.89 =1..311b Vw = fl: 0.021 {:3 624 y 19.2 1+w 1+_9_-§ 100 =17.5kN/m3 b. yd=17..5=Q—7£=-———; e=0.51 e 0.51 n: _ 1+e 1+0.51 = wG, _ (0.098)(2.69) e 0.51 = 0.338 S x 100 = 51.7% a. y=—=—-5—=123lblft3 C. d- e. f. y __ 123 1+w 1+0.098 _ G, yw l_ (2.66)(62.4) yd 112.02 e 0.482 n=—-—-—r= 1+e 1+0.482 = 112.02 lb I ft’ Yd: e -I = 0.482 = 0.325 we, _ (0.098)(2.66) e 0482 30.75 W“ _ 9.8 +...._ I 100 S: x100=54.1% =28 lb ._10_ 3.6 3.7 3..8 a. d. a.. b.. a. b.. W, = W - W3 = 30.75 —- 28 = 2.75 lb K, = Hi = E = 0.044 ft3 62.4 9’.» 3! 20.6 = = =17.67kN/m3 1+w l+0.l66 7d n=—f—'= 05-2 =0.34 1+e 1+o.52 we, u (0.166)(2.74) e 052 S = x 100 = 87.5% At 90% saturation, y _ Gm, +Se yw _ (Gs + Se) )3, _ (2.74 +0.9 x 0.52)(9.81) _ Water to be added = 20.7 - 20.6 = 0.] kN I m3 At 100% saturation, _ (G, + e) yw = (2.74 + 0.52)(9.81) y‘a‘ 1+e 1+0.52_ m 21.04 kN /m3 Water to be added = 21.04 - 20..6 = 0.44 kN I m3 y=Gayw+ngyw; Gsw=Se 1+ e w y“, +Se y“, [(O'Gxe) + (0..6)(e)](62.4) w - 0.17 _ y = --———-~———————-; 96 = —--~——-——-—-—--———- 1+e 1+e e = 0.59 G, 3 fi = W = 2,08 w 0.17 -11- 20.7 kN / m3 _ (G, +em _ (2.08+0.59)(62.4) . _ = 104.3 lb I 1:3 c r5“ l+e 1.59 3.9 a. e=wG, =(o.23)(2.67)=0.614 : (G, +e) yw 3: (2.67 +0.614)(62.4) 1'“ l+e 1+0..6]4 = 126.97 lb/ft3 7/ 126.97 13‘I = = yd 1+ w 11.0.23 _ (G, +Se) y“, _ (2.67 + 0.7 x 0.614)(62.4) 1+ e 1.614 = 103.2111/113 c. y = 119.8 113/113 - 3.10 yd =W=1735 kN/m3 1000 Gsyw _1= (2.68)(9.81) yd 17.85 _ M73 x 100 = 17.65% e=- — l = 0.473 e w=——- G, 268 3.11 a. e" 0'35 =0538 " 1— 0.35 =...._..<Gs+em 75“ 1+e 1.533 = 20.59 kNlm:i = G,yw(l + w) _ 18 = (2.69)(9.81)(1+ w). b. 7 1+ 8 1+ 0.538 w = 0.049 = 4.9% 3.12 a., y = M; 10573 = (62.4)(G, + 058) He ' 1+e G, = 1.694 + 1.194e -12- (a) (62.4)(6, + 0.7512) 1 + e 112.67 = From Equations (a) and (b), (62.4)(1.694 + 1.194e + 0.7513) _ 112.67 = , e = 0.81 1+e 11. From Equation(a), G, = 1.694 + (1 .194)(0.32) = 2.66 3-13 yd = M: W = 91.710/113 1+e 1+081 _ “(6, +e) _ (62.4)(2.66+ 0.81) _ y“. 119.6 lb/ 113 Wa1:e1'=(4..5)(}’sat — yd) = (4.5)(1196 - 91.7) = 125.5510 _ 3.14 e = em — 0.0,... ~ em) = 0.86 - (0..56)(0.86 — 0.43) = 0.619 = 109.7 Ila/1'13 1+e 1.619 1’ 3.15 e = 6m — D,(em ~ em) = 0.75 — (0.65)(0..75 ~ 0.52) = 0.6 _ G57“, _ (2.7)(62.4) ——-——————-——-—= 105.31b/ft3 yd 1+e 1.6 e *'e 3J6 a“ Dr = __HE‘__1 emax _ emin ' - e1 = emax — D,(em - em) = 0.9 — (0.4)(09 — 0.46) = 0.724 0,), (2.65)(9.81) 7'2 - =——~—-————— _ = 15.08 kN 1m3 1 + e, 1 + 0.724 b. _ e2 = em - 13,02max - em) = 0.9 — (0..75)(0..9 — 046) = 0.57 -13- (b) AH Ae el—ez AH_0.724—0.57 H 1+.»l He,” 2 1+0..724 AH= 0.1787 In = 178.7 mm *_ Gm 3.17 74'" wG, 1+ S 921: G5“ wGS 1+ 0.4 1137: 6’7" wGs 1+ 0.7 . 113.7 1+25wG F" E t' db, __=—'_"J— 10m qua Ions (3)311 0 92.1 1+1.429wG, W From Equations (a) and (c), [fig-yam) 92.1: w ' “(5211513) 3.18 a. Refer to the plot ; w= 0.12 = 12% of'w versus N. LL = 28.5 -14- (a) (b) (C) b“ PI= LL —PL = 28.5 - 12.2 = 16.3 w—PL “31-122 3‘.19 LI: — LL—PL 16.3 = 1.15 2 2 M —M V-V .02 = *L 2 _ ' f 3 0 SL [ M )(100) [ M ](Pw)(100) = 362—525](100) —(M](1)(100) .—.- 19.4% SR = &— ~ 31— — 1.85 rm ‘ (13.5)(1) ‘ M—M V-V 3.21 = _I_.2. ... ' f SL [ M )(100) [ M ](pw)(100) 2 2 {44-301 24.6—15.9 100 — —————— 1 100 =4618-2su =17.2s°/ 30”] )( ) [ 301 )(X ) 9 o M2 30.1 SR = = 1.89 1? pw (1 5~9)(1) -15- _ _15_ CHAPTER 4' 4.1 SOIL A: From Table 4.1, the soil is A-2-4. The GI for A—2-4 is zero, So, the classification is A-2-4(0). ' SOIL B: Table 4.1.. Soil is A—3. GI= 0. Classification: A-3(0) SOIL C: Table 4.1.. Soil is A»2-6. Equation (4.2): GI= 0.010?200 -15)(PI-10) = 001(12 —15)(13 * 10) = —0.09 = 0 Classification: A—2-6(0) SOIL D: Table 4.1.. Soil is A-2-7. Equation (4.2): GI= 0.0104"200 ~15)(PI — 1-0) = 0.01(30 -- 15)(18 - 10) = 12 z 1 Classification: A-2-7(1) ' SOIL B: Table 41.. Soil is A—l-b. The group index for A—l-b is 0. Classification: A-1-'b(0) I 42 SOIL A: Table 4.1.. Soil is A—7-5. Note: PI= 21, which is less than LL — .30 = 52? 30 = 22 GI = (F209 -— 35)[0..2- + 0.005(LL - 40)] + 0.010300 —15)(PI -10) = (72 — 35)[0..2 + 0.005(52 — 40)] + 0.01(72 ~- 15)(21 ~ 10) = 15.89 2 16 Classification: A—7—5(16) I I SOIL B: Table 4.._1_. Soil‘is A—6. _ GI (in00 -- 35)[0..2 + 0.005(LL — 40)] 1- 0.0107200 —'15)(PI —- 10) = (58 - 35)[0..2 + 0.005(38 -- 40)] + 0.01(58 - _15)(l2 — 10) = 5.23 z 5 Classification: A-I6(5) ' ' SOIL C: Table 4.1. Soil is A—7-6. Note: P1 = 14 is greater than LL - 30 f 11 GI = (F2.30 — 35)[0..2 + 0.005(LL - 40)] + 0.0107200 ~15)(PI -10) = (64 - 35)[0.2 + 0.005(41 — 40)] + 0..01(64 -- 15)(14 - 10) = 7.905 = 8 Classification: A—7—6(8) _ SOIL D: Table 41. Soil is A—6- _ G] = (F200 - 35)[O..2 + 0.005(LL — 49)] 4— 0.010%“, -15)(PI - 10) = (82 w 35)[0..2 + 0.005(32 — 40)] + 0..01(82 -— 15)(12 — 10) = 8.86 z 9 -17- Classification: A-6(9) SOIL B: Table 4.1.. Soil is A—6.. _ GI = (F20° - 35)[0.2 + 0.00501 - 40)] + 001(one —15)(PI-1_0)‘ = (48 — 35)[0.2 * 0.005(30 ~ 40)] + 0..01(48 — 15)(11 — 10) = 2.28 z 2 Classification: A-6(2) 4.3 SOIL 1: Finefi'action = % passing No. 200 sieve = 30% Coarse fi'action = 100 ~ 30 = 70% Gravel fi'action = 100 -- 70 = 30% Sand fiaction: 70 - 30 = 40% More than 50% of coarse fraction passing No. 4 sieve, so sandy soil. Table 4.2 and Figure 4.2: so _ _ _ Figure 4.3; more than 15% gravel. Clayey sand with gravel SOIL 2: Coarse fraction = 200 — 20 = 80% Gravel fraction = 100 — 48 = 52% Sand fiaction = 80 - 52 = 28% Table 4.2.. Gravelly soil. Table 4.2 and Figure 4.2: GC - Figure 4.3, 215% sand, so clayey gravel with sand SOIL 3: Coarse fi'action = 100 ~ 70 = 30% Gravel fiaction = 100 — 95 = 5% Sand fraction = 95 - 70 = 25% ' LL = 52; PI = 28.. From Table 4.2, it is a fine~grainedi soil. Table 4.2 and Figure 4.2: CH ' From Figure 4.4, 230% plus 200, % sand > % gravel, < 15% gravel, so sandy fat clay . . SOIL 4: Coarse fi'action = 100 -r 82 = 18% Gravel fraction = 100 — 100 = 0% Sand fi'action = 18 ~0 = 18% LL = 30; P1 = 19.. From Table 4.2 and Figure 42: CL Figure 4..4i lean clay with sand I SOIL 5: Coarse fi'action = 100 — 74 = 26% ' Gravel fraction = 100 - 100= 0% Sand fiaetion = 26% Fine-grained soil. L1. = 35; PI=21 -13- SOIL 6: SOIL 7: SOIL 8: SOIL 9: SOIL 10: SOIL 11: From Table 4.2 and Figure 42: CL Figure 4.4: lean clay with sand Coarse fraction = 100 ~ 26 —'= 74% Gravel fi'action = 100 -— 87 = 13% Sand fiaction = 74 — 13 = 61% Table 4.2: coarse-grained soil; LL = 38, P1 = 18 Table 4-2 and Figure 4.2: SC Figure 4.3: <15% gravel: clayey sand Coarse fraction == 100 - 78 = 22% Gravel fraction = 100 — 88 = 12% Sand fiaction = 22 -— 12 = 10% Table 41.2z'fine-grained soil. LL = 69; P1 = 33 Table 4.2 and Figure 42: CH _ ' _ Figure 4.4: <30% plus 200; %sand < % gravel: fat clay with gravel Coarse fraction = 100 - S7 = 43% Gravel fraction = 100 — 99 = 1% Sand fraction = 43 -- l = 42% LL : 54; PI= 26. Table 4.2 and Figure 4.2: CH Figure 4.4: 230% plus 200; % sand > % gravel: sandy fat clay Coarse fraction = 100 — 11 = 89% Gravel fraction = 100 71 = 29% Sand fraction = 89 —29 = 70% LL = 32; P1 = 16; C" = 48; CC = 2.9.. Table 4.2 and Figure 4.2: SP-SC Figure 4.3: poorly graded sand with clay and gravel Coarse fiaction = 100 — 2 = 98% Gravel fi'action = 100 - 100 = 0% Sand fi'action = 98 ~O = 98% C” = 7.2; Cc. = 2.2; Table 42: SW Figure 4.3: <15% gravel: well graded sand Coarse fraction = 100 — 65 = 35% Gravel fiaction = 100 -— 89 = 11% Sand fi’action = 35 - 11 = 24% LL = 44; P1 = 21 . Table 4.2 and Figure 4.2: CL Figure 4.4: sandy lean clay -19.. 4.4 SOIL 12: Coarse fiaction = 100 — 8 = 92% Gravel fi‘action = 100 — 90 = 10% Sand fi'action = 92 — 10 = 82% LL = .39; PI: 31; Cu = 3.9; Cc. = 2.1 Table 4.2 and Figure 4.2: SP-SC Figure 4.3: poorly graded sand with clay a. Percent passing No. 10 sieve = 90 Percent passing No. 40 sieve = 38 Percent passing No. 200 sieve = 13 P1 = 23 — 19 = 4 Referring to Table 4.1, the soil is A—1---b.. G1 = 0., So the soil is A-l-b(0). b. Coarse fi'action = 100 - 23 = 77% Gravel fiaction = 100 — 100 = 0% _ Sand fi'action = 77 --0 = 77% -. LL: 23; PI= 19 From Table 4.2 and Figure 4.2: SC From Figure 4.3: clayey sand -20- 5.1 From Equation (5.3): CHAPTER 5 5.2 2400 __ G,p,, _. (2.65)(1000) 9"" G,w‘. 2.65w 2300 1+ 1+ ~ - s 0" 'S=1009/ _ 2650 E 2100 ° 2.65w - on 1+ e5 «‘2‘ m g 1900 [email protected](%)mkg/m3 E . 1700 . 0 . 150 0 5 10 15 20 Moisture content (%) The plot between pd versus w is shown Equation (5.4): I 2400 w 1000 pm = p 1 = 1 2240 w + — w + —— G, 2.54 1000 ' “g 2080 _ w+ 0.3937 g; ‘ g 1920 1760 1600 0 5 10 15 20 w (%) ._21- 5.3 Refer to the following table, Weight of Moist unit Dry unit Volume, V wet soil, W weight, 3/“ content, w weight, yd” (1b / fi3) (lb / fi3) _I7'='i:; 73 ]+—"(96) ' 100 The plot of )4, vs. w is shown. From the plot, Maximum dry unit weight = 100.8 lb / ft’; Optimum moisture content = 15% 102 100 .8 lb_/_if_ * A 100 '3: E 33’ 98 Eb g 96 :s E 94 92 i 10 12 14 I6 18 20 22 Moisture content (%) yd L. 637w; e: 057W _I=W_1=0_66 l + e yd 1000.8 wG (0.15)(2.68) = ——’- = -——-—— = 60.97 9 0.66 ° -22.. 5,.4 Refer to the following table. Weight of wet soil Moist unit Dry unit in the mold, W weight, 7“ content, w weight, )1," (lb / 113) (1b / £13) a _HE:. b * y 9"}?! ydhll+w(%) 100 The plot of yd vs.. _w is shown, From the plot, Maximum dry unit weight = 117.3 lb / ft3 Optimiim moisture content = 11%' 120 117.31b/f‘t3 . 0 5 10 15 20 w (%) _ 6,7,, _1:(2.67)(62.4)- yd 117.3 we, (0.11)(2.67) S = —-——— = —— = 0.699 = 69.9% e 0.42 —1 = 0.42 -23.. 5.5 Weight of mold, V wet soil, W weight, 7/ (N) W=massinkg x 9.81; 314—- W/V The plot of yd vs.‘ w is shown. _ From the graph, mm, = 18.3 kN I m3 wo = 15.5? .5." . pt ° I E yd = 0.95 mm) = (0..95)(18...‘3) E =17..39kN/m3 _ i From the graph, w = 13% at 0.95 7.3.1,”) W (Va) 17.3 16 +___. 100 5-6 Vanna) = 17-3 kN / m3; ' Human) = a 1491 ml m3 _ 1 yflcompacred) =18~1 kN/ m3 Volume of soil to be excavated = (2000)[i%48'711) = 2427.9 :113 ;24.. G, x 9.31_ 1.7 Borrow Volume to be excavated Total pit yd at borrow pit fi'orn borrow pit '= Cost / m3 cost (kN I m3) [285853G5 / Ydamow pm] (3) (3) 5294. 1 In3 57 Dry unit weight of'solids required: 5000 yd kN = (5000)[ J: 23,8536, kN G: x 9.81 ——————=5.16G 1.7 3 1+0.9 , 559 111 GS X 9.31 Borrow pit B 5.8 Mass of'sand used to fill the hole and cone = 6.08 — 2.86 = 3.22 kg Sand used to fill the hole = 3.22 - 0.118 = 3.102 kg Volume of'the hole = = 0.00179 m3 3.34 = 6.00179 1865.9 p": (12.1) 1+ —— 100 or = Mm) 7d(m)_7d(min) 7d(fie1d) 0..72=' WI. .173 nus — I46 7dtfie1d) =1865..9 kg / m3 p = 1664.5 kg/m’ -25- R(%) = “m” x100: 16'77 17s yd(max) x 100 = 94.2% _ R = = m: I'm. yda-Ield) = I."3 . Dr = 7d(field) _. ydunin) 7dtm) = —‘ = 7d(max) " 7mm») Ydmetd) 16--98 ‘141-46 15-188 5.11 a,_ R=oflg4=M=l4fléb 18.6 , MM, : 17.43 kN Im-3 7d(mz_v<) b”. . Dr 2 rag-mid) - mm; mm) = [17.48 — 15.1 18.6 ]= 72.4% ydfl'nax) _ 74min) yd(fie]d) ] — c. y = no + w) = (17.41;)(1 + 0.08) = 18.88 kN / m3 5.12 Equation (5.17): 1 1' + + . 1.982 0.312 0.182 = 11.02 5.13 Equation (5.17): 3 1 1 ' S :17 + + =11.39 N [0.722 0.472 0.172 5.14 Refer to Figure 5.39.. For 1--m distance from the vibroflot, 1C 3 8.. There are 3 vibroflots. So, combined IC = 8 X 3 = 24.. With IC = 24, the probable relative density to be achieved is about 98%. '-26- "mm CHAPTER 6 6.1 k = 25‘— : w— = 0.0138 in./ sec Ah: (41.6)(35)(180 sec) 3 6.2 a“ k = 2%. = (580 cm )(35 cm) 5-——-——_ = 2.15 x 10'2 cm / sec Ah! (125 cm )(42 cm)(180 sec) b. v, = veg]; v = ki e 1+e v, = = (0.0215)( 3 (wig) 6.3 =25; 0.014 =—1°——_ Ah: (105)(h)(60) 42 cmXI +0.61 )= 0.068 cm / sec 35 cm 0.61 h=34cm v = ki = (0..014)[%) = 0.019 cm / sec _ g a ‘ (0.15)(15) _ .2 . . 6.4 a. k — 2..303[At)iogm[h2] (2.303)[ (3MB) Jlogwtzjg 6.88x 10 ml mm b. k = 2.303(3-é)10g10[%] At _ (0.15)(15) g _ g 0.0688 _ (2.303)[ mm ]log,o[h2 J... 0..4318log[h2 J I 0.1593 = @1063]; 1.433 = h2 = 17.3 in. -27- 6.5 6.6 6.7 6.8 aL h a x 38 650 k = 2.303 — 1 J— ; 0.175 = 2.303 1 —) [ J og‘°[h2] ( )[6.5x a) °g[300 a = 0.31 cm2 = 2.303(0'19g:850]10gw[~:15—3] = 0.234 cm / min = 0.0039 cm / see I? _ L __ (0.0039 cm/sec)(1.005 ><10'3 N «5/ m2) 3 3 =4x10'" m2 7.0 . 9.789x10 N/m . aL h b.. k=2..30. ——-- l -1- 3L“) ogwhz] 02.34 cm/ min = 2.30% 0'97 x 50.]1ogw[1§9] 16x6 h2 h2=478.3mm q=kiA 1...: 160+ 150 #008 125 r k=3mlday= 3 =0.00208m/min 24x60 A =2 x 500=1000m2 q = (0..00208)(008)(1000) = 0.166 m3 / min From the figure on the following page, i = M - m = sin a length — (00:6!) ' q = kiA = (k)(sina')(4..2cosa)( 1); k = 6.8 X 10" cm/ sec = 6.8 x 1045 ml sec -28- 6.9 6.10 Ground surface q = (6.8 x 10*)(sin10°)(4.,2oos10°)(3600) = 0.0176 m3 I hr / m \W_J ' to change to m" h ~17.6><10'3 m3lhrlm h l = [ L J ' 6050! q = kiA = k[h°°sa](H, 6030: x 1) '= [230035](3cos5) 1o 52 me = (0.0005)(0..0536)(2..99) = 8.01 x 10'5 1;? Is I m From Equation (6.3121): k, 1+el . 0.022 [1+0148)_ 0.0747 k2 e3 .’ k2 Q73 "02018 1+0..7 k2 = 0.059 cm I sec -29- W'- n, 0.3 I 61.1 1 n, = 0.131, e, = — 0.449 1—n, 1—01.31 _ 0.4 1-— 0.4 = 0667 122 = 0.4, 62 = Equation (6.3121): 3 3 3 k2 = k: e2 I +3el = k1 I + el 'e—2 = 0.1 z 0.37 ft 1' min 1 + e2 eI I + 92 (21 1.667 01.449 61.12 park”, = Rpflm) = (0118)(1720) = 1376 kg / In3 31 =.‘i&__1=W—1=0547 palm) 1720 as =§;&__1=Wfl—1=0933 pd(ficld) 1376 3 3 k2 = kg 82 1+3e] = 004(9-522] = 0.159 cm lsec 1+1:2 e, 0.547 1933 . 6.13 e, = em — (emax - emQD, = 0.17 n (0.7 _- 01.46)(0..8)= 0.508 92 = 0,7 — (0.7 — 0146x015) = 0.153 1: =1: 32— 3 “"1 =0..00 2 ' e, 1+1:2 0.508 Opening Percent Fraction between two (cm) passing consecutive sieves (%) E9}— = 8.52 x 10'3 cm 1' sec 158 6..14 100 - —-——20 80 ————12 68 28 ""40 ————23 0 -30- For fraction between f} _ 20 _ 40.7 98 sieve Nos” 30 and 40 D}; 404 x D; 59s — 0060 404 x 004250.595 "' ‘- For fraction between f4 __ 12 _ 44103 sieve Nos. 40 and 60 D3104 x D9595 _ 004250 404 x 0020595 " " ' For fraction between f; _ 40 __ 2362 8 sieve Nos. 60 and 100 D3404 x D9595 _ 0020 404 x 00150595 " - " For fiaction between f; _ 28 _ 2812 2 sieve Nos 100 and 200 DE 404 x 190.595 ‘ W ‘ ‘- 0 2 2100A: 2 100 3000166 09404 x 03,595 4044.98+44l..03+ 23623-423122 [1 0.53 l + 0.15 2 k = (1.99 x104)(0..0166)2[L] [ 65 ]= 0.0...
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