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 ]= 0.0108 cm / sec _ log 0.25 _ —0.602 _ _ =2..477 " log0.57l 41.243 er . Cm‘12x 10~6(1+0;8) H e] , 08m = 3.754 x 10“6 cm I see kl=C 0622"" .- . ‘ ' ' k3=[ '162 )(3.754x10"6)=0.709x10'6 cm/sec 6.16 logk =A’ loge+B’ _ logk, - 10ng H log(1.2 x104)—log(3.6 x 104) ._ 196 ' log eI — iog e2 log(0..8) — Iog(1..4)' f B’ = log k, - A’ log .2, = log(1.2 x 10*) - (1 .196)Iog(0.,8) = — 5.173 -31- log k3 =(1..96)log(0..62) - 5.73 = —6.,1369 k3 = 0.7 x 10“ cm / sec 61»,- 5= Em a" ' 17c2 1+ei 82 -6 " = M1? = ; 01667 = 096316" 0.91 x 10 2.2 1-9 "=10g(0.1667)= —0.77s '3 3398 Iog(0..6316) —0‘.l996 .5 c: M=W= 0.216x10‘6 eI n ' n e 0.93 898 I _6 4 ' k3 = C r: (0.216 x 10 ) = 0.075 x 10 cm [sec l+e L9 2.1..— 1.3) 0.5 x 2.1 6.18 logic = log k0 — :0; e = log(0.86 x 10*) —[ 90 = 0.1 x 10" cm [sec 1 6.19 km) = EUQH, +sz2 + ...,..) km) = flue-5x15) + (300 x 10-5)(2.5) + (3.5% 10*5)(3)] -5 = g-(15+750+10.5)=108.86x10'5 cm/sec _ H _ 7 _ .5 kneq) - E—W—T— 10 cm/sec —_—5'+'—“—"‘—-—:+ __s 1:] k2 10 300x10 3.5x10 kmeq) = 108.86x10"5 # 36.8 ' km) 2..96><10'5 -32- 6.20 k _ —H._~..___= 4 §l+_H_2+£1+H4 1 1 1 1 k 2 = 2124 1410-4 cm/sec 1 kH(eq) a §(H1k1+ H2152 + H3"; + H4k4) —:I(20><10*4 +2><10"‘t +l><10'4 +3><10’4)=615><10'4 cm/sec kmeq) _ 6.5)(10q4 _3.06 _ _4 - hm, 21.124x10 61.21 Equation (6.37): v = v1 = v2 = v3 = .. .1 4. keqi=k1il=k2i2=ksi3=n .4 .4 keq = 10-2 MA 450 150 300 Ah 01.1213 10-2 — = '2 A ; Ah = x [450] 10 [150) A 1213mm Similarly, ke fl = :3 x 10-3[-A-h—B] ‘1 450 150 300 Ah 1.21: 1 “3 — =3 10'3 ——3; Ah =4 ..4: 3x 0 [450) x [150) B 0 3mm he = h, 1- AhB = 287.87 - 40.43 = 247.44 mm -33- CHAPTER 7 7.1 Equation (7.14): hz : _hl_ki__ [ "‘ k2 ] H, —+w— H1 H2 6 cm = (18 cm)(0.002 cm/ sec) [0.002 cm/ se‘c k2 cm/ 560) (7 cm) w—————+———————— 7 cm 10 cm k2 = 0.0057 cm 1 sec 7.2 The flow net is shown. k=4 x 10" cm/sec H=H1" H2=6 " 111.80 _ 4x10”4 [45x4] q" 102 8 =9x10'6 m3/m/sec = 77.76 x 10'2 m3 lmlday 7.3 The flow net is shown. A5; 3;'Nd = 5 N Na «6 = [4 m/ sec](3 - 05)[%] =6x10'6 m3/m/sec = 0.518 1113 lm {day -35- 7.4 7.5 7.6 Based on the notations in Figure 7.12, H=(4—1.5)m=2.5m S=D=3.6m T’=D,=6m 3:39:06 7" 3 From the figure, i m 0.44 KH .4 q = (0.44)(2.5)[4 j g? »x60x60xi24 m/day]=0.38 m’lday/m The flow net is shown. q = “(fij = (0%? x 60 x 60 x 24 m /day)(10)[%] = 7.2 m3 I day I 111 Refer to the flow net shown in Problem 7.5 and the figure on the following page. The flow net has 12 potential drops. Also H = 10 111. So the head loss for each drop = (10/12)m. Thus”, _ Pressure head at D = (10 + 3.34) — 2(10/12) = 11.67 Similarly," ' Pressure head at 15‘ = (10 + 3.34) - 3(10/12) = 10.84 m Pressure head at F= (10 + 1.67) - 3.5(10/12) = 8.75 m -36- 7.7 E :1 67:1. "‘ 1.67:1.67' FmTH§+“‘_13-32 m—“m’hm’f‘m‘E Pressure head (m) 11 67 Pressure head at G = (10 + 1.67) ~ 8.5(10/12) = 4.586 m Pressure head at H = (1-0 + 3.34) 4 9( 10/12) = 5.84 m ” Pressure head at I'= (10 + 3.34) -'10(10/12)= 5 m " _ __ _. The pressure heads calculated above are shown in the above figure. The hydraulic uplifi force per unit length of the structure can now be calculated to be = 74., (area of the pressure head diagram)( 1) [w}1,67)+(m—4;¥§)(m7) +[8.75 :4586}18m32)' + [4.586;- 534 )(1‘6'7) + [ 5.8: + 5 = 9.81 )(157) = 9..81(1 8.8 + 16.36 + 122.16 + 8.71 + 9.05) = 1717.5 kN I ni' ' The flow net is shown on the next page. N f = 4; Nd = 14 _ NI _ 1.91 _ _4_ ; -s i q— 7d]_[102](1o 15)[I4]_(10 )(85)[14] = 2.429x10'5 m3 lm/sec z 2.1 m3 lm/day -37- 7.8 Problem 77 Refer to the flow net shown. for Problem 7. 7., At the bottom left-hand side of'the weir, the number of“ drops is about 1.5. At the bottom right-hand side of the weir, the number of drops is 10.. Ian—H, _10—15 N, 14 Ah: -= 0.507 m/drop- So, at the bottom left-hand side of'the weir, the uplift head is (10 + 3) — (Ah)(1..5) = 13 - (0..607)(1.5) = 12.09 In At the bottom of the right-hand side of the weir, the uplift head is (10 t 3) - (Ah)(10)= 1:3 - (0607)(10) = 6.93 m H——--————37 m -——--—-—-—-—H 6.93 m 12.09 111 So, uplift force as 9.81 x (37) (12.09-1-633] z 3452 mm -3 8- 7.9 F‘ortlu'scase,T’=8m;S=4m;H=Hl - H2=6m;B=8m;b——~B/2=4m a.. £=i=05 T' 8 ‘x‘=b—.x’=4-I=3m x’ 3 — = — = 0.75 b 4 i = i = 0.5 T' 8 From Figure 7J3, q/kH = 0.37 q = (0..37)( 0331 x 60 x 60 x 24](6) z 1.92 In: May lm 0.001 102 q = (0..4)[ xil60 x 60x 24)(6) a: 2.07 m3 / day l m 7.10 a] = 30°; 0;, = 45°; H= 8 In; and A = 8cot30 = 13.86 m. 0.315 = 4016 m. -39- d= H1 00th + Ll +(1H1 — H)cota'I + 0.3A = (l2)(cot45) + 4 + (12 - 8)cot30 + 4.16 = 27.09 m L_ d _ d2 _ H2 _27.09 [27.09]2_[ s ]2_171m c030;2 cos2 a2 sinza2 00545 00345 sin45 " 2x104 102 q = ithanaz sin a2 = )(1.71)][(tan45)(sin45)] = 2.418 x104 m3/sec/m a: 0.209 m3 [day/m 7.11 = HooterI = Scot28 = 15.05 m d= Hl com:2 + L. + (Hl — H)cotarl + 0.313 = 12cot35 + 7 + (12 — 8)cot28 + (0.3)(15.05) =17..14 + 7 + 7.52 + 4.52 = 36.18 In 2 2 2 2 L: d __ a; __ =36.18 [36.18) “(.8 )zzzém cosct2 cos a2 $111 :22 c0335 c0335 s1n35 q = kLtana2 sin a2 = [[£§(—)—129-:](226)][(t_an35)(sin 35)] =1...'36><10‘6 m3 Isec/mz 0.113 m3 [day/m 7112 From Problem 7.10, d = 27.09 m; H = 8 1n; 1:5 = 45" = 21-92 = 339 i H 8 m z 0.2 (Figure 7.17) mH (0.2)(8) sin or; sin45 L: =-- 2.26 m -40- 7.13 2x 10‘4 102 q = kLsinz a2 =[ ](2.26)(si1n2 45) = 2.26x10'6 m3 lsec/mz 0.195 m3 [day/m From Problem 7.11, d= 3618 m; H= 8 m; 65 ‘—‘ 35° Slnaz sm35 2 x10"4 102 q = kLtanaz sin (12 =[ )(226)(sin2 45) = 226x10"6 m3 lsec/mz 0.119 m3 lday/m .-41_ “mm:- -42, 8.1 8.2 (2)(115)= 230 230+(118)(4)= 702 The plot is shown. Depth (ft) cab/fifi Depth (9)000) = 900 + (21)(122) = 346 CHAPTER 8 u (1b I fiz) (ft) 0 . 0 2 -43- Depth (fl) 0' (1b / fl?) ——-—> 452.4 858 The plot is shown. 0(1b/ a?) u (lb / 112) 0" (lb I fiz) 0 30 66 8070 ‘ Depth (ft) - Depth (ft) . Depth (it) 8.3 Point A: a: 0; u = 0; 0’= 0 Point B: ' a= (4x162) = 64.8 mm:2 u=_0 0’= 64.8 - 0 = 64.8 kN I 1112 Point 0: a= 64.8 + (1.5)(13..4) = 92.4 kN I m.2 u = (1.5)(931) = 14.72 kN I m2 0': 92.4 -- 14.72 = 77.68 kN I 661 Point D: a: 92.4 +(9)(19..81)= 270.69 kN I m1 u = 14.72 + (9)(.9..81)..=. 10.3.01 kN. I m? a’= 270.69 — 103.01 = 167.68 kN I In2 The plot is shown on the next page. -44- 8.4 0(kN / m2) :4 (kN /m‘2) 0’ (kN / m2) Depth (In) Depth (In) Problem 8.3 6,9, (2.65)(9.81) 3 :——=—---= 17.33 kN/ 7““) 1+2 1+0.5 m (G +e)y (2.68+0.65)(9.81) ' - 3 :u—s MW) l+e 1+0..65 - m _ (G, +e)y,, _ (2.71 + 0.81)(9.81) y’m‘c'“) 1+ e 1 +0.81 =19..08 kN/m3 (4)(17..33)= 69.32 t __ 69.31 + (1.5)(193) = 99.02 (9..81)(1..5) = 14.72 343 99.02 + (3)(19.03)-= 156.26 (9.81)(1.5 + 3) = 44.15 112.11 The plot is Iéhown 021 the next page. -45- 0(kN /m2) u (kN / m2) a’(kN / m2) Depth (In) Depth (In) Depth (m) Problem 8.4 Gm, = (2.65)(9.81) =17..lkN/ 3 1+e 1+0.52 m 3-5 7d(sand) 3" : Gsyw +eyw = (2.68+0.52)(9.81) = kN / m3 1 + e 1 +1.52 ' 3 _ Gm. +671. ysaflclay) _ T 75mm) wG, =e; a, =3=fl=271 w 0.45 — =171-37kN/m? (2.71 + 122)(9.31)_ rsaxclay) "" _ (17.1)(3) = 51.3 51.3 + (20..46)(2.5) = 102.45 102.45 + (17.37)(2) = 137.19 (2..5)(9.81) = 24.53 77.92 (2.5 + 2)(9.81) = 44.15 93.04 cow> -45- The plot is shown, 0 (kN / m2) u (kN / m2) 0’ (kN / m2) Depth (m) Depth (m) Depth (m) 35 & yflfim)=5§sz=§§§9§2fil=11221b/fi3 1+e 1+0._49 many) = (—G’m = W = 119.9 lb / £13 1+e 1+09. 7 113/112 _ 0 (111‘.2)(15)= 1683 “1333 + (119.9)(12) = 3121.8 0 o .(62‘.4)£1_21_= 748.8 (G, +e)y,, (2.6s+0.49)(62.4) . 3 b.. =——-—-=——————=132..81b/fi Mm" 1+e 1+0‘.49 -47- 8.7 808 8.9 a=(15)(132.s) + (12)(119.9)= 3430.8 lb / 112 u = (27)(62..4) = 1684.8 lb / fi2 0’: 3430.3 —_ 1684.8 = 1746 lb / {12 Change of 0’: 2373 - 1746 = 6271b Ift2 0.. Let the height of'rise be 12. So, at any time, at the bottom of’the clay layer, a: (15 — h)(112..2) + (100328) + (12)(119..9) u = (12 + h)(62.4) Change of'a’: 300 -—- 2373 - [(1683 — 112.2}: + 132.3}: + 1433.8) - (748.8 + 62.4h)] h= 7.18ft . y' Gs—l 2.68—1 1.68 (cr=——--= = =—--—- yw 1+e 1+e l+e _ (1 + w)G, jg, = (l + 0.4)(2.7)(62.4) _ =113..41b/1=t2 7W1?) 1+wG, 1+(0..4)(.2«7) . Let the depth‘of the excavatiOn be H. So, (25 — H)(1 13.4) — ( 18)(62.4) = 0 = 0’ Hz 15.1 ft Let the depth ofexcavation-be H, and height of water be 12.. Given: H = 18 it. So, (25 ~ H)(I 13.4) + (h)(62.4) - (18)(62.4) = o _ (18)(62.4)—(25— 18X] 13.4) _ 62.4 h = 5.28 ft -48- 8.10 i=—=.-‘——=0.'.4 ' q .4 ki‘A = [gig—g x 60 m/ min](0..4)(0..45) = 0.013 m3 [min 8.11 a.. i=—=——=O‘.6 q = kiA = (0..1)(0..6)(0..5 X 1002 cm2) = 300 cm3 I see b. 7 G,—1_2.6s—1 1 =—-—=-—— -—— yw 1+3 "1+0.55— " Since I < i“, no boiling . c i=i =1L;108=£ H 2 2 h= 2.16 8.12 Refening to Figure 7.9, the block to_be considered on the downstream side is 7.5 fl (width) x 15 fl (depth) Ix 1- fl (length at right angle to the section), H1=15figH2=5fi H=H, —H2=15 - 5=10ft From the flow net, the hydraulic head can be estimated (see figure). -.49- Area of’the hydraulic head diaglam under the block of'soil shown in the figure is 0.5+ 0.333) A = H[ 0_.3_§3+o.233 (5.5)+H[ )(75—55) = 2.29H+0566H (2.29 + 0.566)(10) Average hydtaulic head = 7 5 z 3.81 ft 1'“ = = 0.254 Factor of safety: F3 = __2:'_ 2 M = 3,63 My, (olzs4x62u4) 8.1.3 Equation (8.19): W; = (5 mpg-)0; )+ (D[§}}’dm] I 2 r 5 I D F5: 371+ng = 5D 7 +§?F'D+"'2"7d(fi') U 51—1322;er 01' _ I . I . - y'+57F +--"""" (120—62..4)+~(—5X—12-7‘—63A—)+—1-9~5- w” (0.254)(624) a 5.43 8.14 ‘ Equation (8.25): 12 (mm) = C . eDm ' ' 2 10 C=10‘mm; h=——=85.5mm « _ (0‘.65)(0.18) C =50 mmz; h: 50 27.4 mm m——_' = 4 0.65)(0..18) Range: 85.5 mm to 427.4 mm -50- G,y,, (2.66)(62.4) .65 =—-—=—-———=110.66lb/fi3 81 MW) l+e 1+0.5 yw (Gs + Se) (62.4)[2.71+(0.5)(0.75)] 7(clay) =T=-——fm——= 110 Ib/ft3 w G, +8 62.4 2.72 +0.95) Yummy) = Iii-:7) = %é‘9—' = 1 / fl3 (1 10.66)(6) = 663.96 663.96 + (110m) = 1103.96 (~ 06.5)(62.4)(4) = e 124.8 The plot is given, aab / fiz) u (lb / f6) 0’ (lb / f6) 1103.96 1103 96 1599632 Depth(ft) Depthm) Depthm) (2.66 9.81 3-16 711mm) = “‘l—jgs—Z _ (9.81)[2.71 + (0.6)(0.‘75)] “1“? 1+ 0.75 = 171.4 mm3 =17.‘71 kN/m3 _ (9.81)(2.72 + 0.95) 7531mm "' =18..46 kN/m3 kN/m2 69..6+(171.71)(2‘.5)=113.88 . 113.88 113.38+(18.46)(3.5)=178.49 ' (3.5)(9.31)=34.34 144.15 6.5+.3.5=10 Ihe plot is given. 0(kN / m2) 2: (kN /m=) 0’ (kN / m2) -—-—-—. - 6‘5 "10 178.149 Depth (11:) Depth (111) Depth (m) -52- 8.17 From Problem 8.15, yam) = 110.66 lb / fi3; y“, (day) = 1 17.44 lb / {‘13 mm) = = 10133 15/113 I + 0.75 18+16=34 ' 1965.24+(16)(1 17.44)=3844.28 (I6)(62.4)=998.4 The plot is given. 0(lb I it?) u (lb / 112) 1306.28 Depth (fi) -53- 2845.88 0’ (lb/ fiz) 2845-88 Wm 8.18 From Problem 8.16, yam) = 17.4 kN / m3; y,”th = 18.46 kN / m3 _ (9.81)[2.7l 43(055)(0.75)] I + 075 Mm .. = 175 kN/m3 (17.4)(3) = 52.2 52.2 + (1.3)(17.5) = 83.7 . 0 . ' 83.7 +_(4)(18.46) = 157.54 (4x931) = 39.24 118.3 The plot is given. 0(kN / m2) u (kN / m2) 0’ (RN / m2) -9..‘71 39.24 Depth (m) Depth (m) Depth (m) -54- 9.1 9.2 a.. CHAPTER 9 0' + 2 1 0), a, [ ay — 0-,] 12 = _+ + :9, 0-3 2 2 0x=80kNlm2; q,=120kN/m2; rxy=+40kNlm2; 6: 145° 2 ) +(40)2 2 0: _120+30+ (120—80 0’3 — a, =_ 144.7_kN/m2; 0;, =55.3 kN/m2 X 2 003264-25 sin26 120+80 2 120 — 80 cos[(2)(145)] + 40 sin[(2)(145)] = 69.25 kN I m 3 + o", 0',‘ 7,, = sin29— 7,9, 003219 _120—so sin[(2)(145)] —40 cos[(2)(145)] _= -3_2.47"1;N1m-2 ¢=5001blfi2;o;=2501b/-fi2; rxy=---'801b/fi-2;t9=‘45° ' ' 0] _250+500i [250—500]2+(_30'52 (,3 2 2 a1 = 523.4 lblft’; 03 = 226.6 lb I {:2 __ 250+500 + 250—500 2 2 = M sin90- (-30) 00590 = "125 lb I ft: 0,, cos90—808in90=295 113m”l 2In -55- 9.3 a. The M'ohr’s circle is shown. (150,25) Shear stress (kN / m2) 0 Normal stress (kN / m2) C (100,—25) 0—01=l§02Lu3~9=~125kN/m2 013 =1f[l§—9;l—m) +(25)2 = 35.36 kN Im2 03 =63" =125 —- 35.36 = 89.64 MN] I m2 (+) a, = 51V =125 + 35336 =_160.36 kN I m2 (+) -- b3. 430,02 = tan“‘[35-] = 45° 25 0,, = 001 + 01 D‘cos65 = 125 + 353600365 = 139.9 kN / In:l (+) Tn = iD sin65 = 350365in6'5 .—_ 32.05“ I ma (_) . -_55- 9.4 a. The Mohr"s circle is shown. Shear stress (lb / fiz) Normal stress (lb / ftz) (125,—40) 56:: =10151b/ft2 =125“9° =1751b/fi2 52> 02 5:2} = 1{(17.5)2 + (40)2 = 43.66 lb [112 a, =(W = 107.5 + 43.66 = 151.16 lb I ft2 03 = 69‘ = 107.5 - 43.66 = 63.84 lb / rt2 (+) b. 470,02 = tan-{fl} = 66.37” 17.5 q, = "0"“0, + w“0, D sin(36.37') = 107.5 + 43.66sin'3637 = 13.3.4 lb / ft’ 1;, = 43..66cos.36..37 = 35.16 lb / ft2 -57- 9.5 a‘. The Mohr"s circle is shown, Shear stress (kN / m2) NOImal stress (kN / m2) (.50, - 30) q=5IV=95kNlmz; 03=27§ =30kN/m2 bu 0,, and 2;, are coordinates of D.. So q,= 94.2 kN/m’; 13,: 7.1 RN/m2(-') 91.6 a., The Mohr’s circle is shown on page 59‘. a,=_0_]V z 109.1kN/m2; og=6§ =25.9kN/m2 b‘. 0,, and 1;, are coordinates of D.‘ So q, = 29.1 kN-Iling 1;; 16,08 W I m2 -53- Shear stress (lb / fiz) (90,35) C Normal stress (lb / fiz) B (45, -35) Problem 96 P r z 11 2 (1b) (ft) (11) (Table 9.1) (lb '/ 'fi2)__ i Z (102+52)“: 11.18 1.12 0.0626 (102+52)°5= 11.18 1..12 0.0626 - 0.5 0.2733 16.4 An, _= 220.5113”? 9.7 9.8 Equation (9.16): q: [-flIons Equation (9.15): 2 2 (2000)(0.408) 1 2 IodA: ________...__....___.. =00771b/fi a 27:00? [014082 + (1.12)2 ] le LoadB: W[——l—] =1531b/fi2 2;:(10)2 (0.408)2+(1.12)2 3 3 Load c: W m—Tl—«z— =14.51b/ft2 27:00) (0.408) +(0.5) A0; = 0.77 +1.53 +145 = 16.81b/ft2 9.9 Equation (9.19): A0 = ____24223 2 _<2><100L<2..>’ +9.29% 2 . 2 It[(.x'l +x'2 )2 +22]2 +22)2 +22 £(52 +22): 3(22 +22) == 16.53 mm:2 3 3 9.10 Ac, = 2‘11” 2922 2 22+ 2 22 2 7t[(xl+.x2)+z] 7r(x2+z)+z 7’[(3+25)2 +(2.5)2]2 ,[asf +(25);]2 = 20.83 kNlm3 _ (2)(100)(25)3 +‘ (2)(260)(25)3 3 3 Adz =I 2qlz 2 + Zqzz 2 +x2 )2 +22] +22) +22 (2X750X3)3 2?: (3)3 35: 2 2 2 + 2 2 2 3(12 +3) 7r(4 +3) 911 = 0.55 + 0.027542 q: = 1252.7 lb / ft -60. 2qlz2 9.412 Aog atA due to q, = __..T :r(x2 +22) 01' (Aaz)l = 2W? = 71.96 kN Im2 24(2)? +(2)2] Vertical component 0qu = q2 sin45 2q2(sin45)23 ; (Adz) =04-0043q z:[(5)2 +(2)2]2 2 2 (AG—2}) = Horizontal component of q2 = q2 c0345 From Equation (94.21): 2q2xzz n.2q2(cos45)(5)(2)2 (A6213 2 2 2 75(x2 +22) 4(5): +(2)"’] % 0010742 Total vertical stress, 44=10kN/m =(Aa:.)1*+<442+(443 10 = 74496 + 040043;;2 + 0,0107%;2 __ 10—796 _ = 136 kN/ 92 0.015 m 9.13 B= 10 fl;q =2001b/ft2;x= 8'1t;z'=8 11 A0, 3£=£3Xfl=m 22% =0..24s" I, __ =1‘.6.,l From Table 94.4, 3 10 B 10 ‘1 A0: = (04.248)(200) = 49.6 lb I n1 9.14 ~23- : mm) = 1; 33 = 99 = 2.. From Table 9.4, A“: B 3 B 3 ‘ q = 0.409 Ac, = (60)(ot.409) = 24.54 kN / m2 -61- 9.15 Equation (9.24): A0: = L B 2 2 n‘flxz +22 u[—] ] + 3222} 2 9: (2)(1)(q)(15)(0.75)2 2 2 2r{[(15)2 + (0.75)2 J +(1)2(o.75)2} q = 119.4 kN / m2 9.16 Refer to the figure below. H—5m———n -h—————8{m'——“m'_—5m_+1 a; 5x18=90kNlm2 I I l I l I 5m: :Sm : 1 A6 ‘14 With the notations given in Figure 9.17, for the left side: B10 B _ = _ = 0; ~2— = E = 1. From Figure 9..18,13m-= 0.24 z 5 - - z ' 5 . ' - . For the right side, EL = § = 1.6; ii = 3 = 1. From Figme 9.18, 13”.) = 0.48 z 5 z _. Ac, = q[13(£) + [M] = (90)(0.24 + 0.48) = 64.8 kN / m2 _ -52- 9.17 At A: 75 ft 7.5 ft 14—60 ft—DH—H H—DH—60 fi—H GB ' I : 1 15ft: :let | l l I A 0 6A For the left side: i = 1'5 = 005 z 15 £3; ._. fl = 4 z 15 A0, =(30)(115)(0H463 + 0.468) = .3229 lb I ft” At B: H—GOfi—N [1—15 ft-—H¢———60 fi—-Di (:3ox115) Ib/ft2 -- A69 15ft l I I I I I I I E : 341 $3 15ft For the lefi side: i=3: 0 z 15 Fiji: 2 15 13 = 0142 A492 = (30)(115)(0‘.42 + 0.48) z 31051b/ft2 -63. For the right side: B "—1—: 0.5 2 §1=4 z 13 =0‘.468 For the right side: 5-2-1 2 15 i=6_0=4 z 15 13': 0.48 At C: “"6011",! “M75 fi_*—"60 fi—H For the left side: For the light side: i=0 35:15.: 2 z 15 fi=fl= "31459.: 2 15 z '15 13 =0.‘42 I3 =05 M2 = (30)(115)(0‘.5 f 0.42) z 276 lb I :11 9.18 Equation (9.30) and Table 9.6: q = 3500 lb / ft2 A0: (lb / 112) m.me 9.19 Equation (9.31) and Tables 9.7 and 9.8: q = 300 kN I In2 022795 030730 0.21662 0.28481 0.15101 014915 90.1 0.09192 004378 40.7 0.05260 0.00023 9.20 Refer to the Newmax'k’s chart. The plan is drawn-tic scale. 3—3—= 4 m. M z 65.. A0; = (IV)qM= (0.005)(300)(65) =97.5kN/m2 9.21 a. Equations (9.36) and (9.37): L n:— z E Z 10 5 =—-—-=2; — 5 m 5 Equation (9.34): A02 = 91;; I4 = 0.1999 A0; = (1800)(0..1999) = 359.8 lb I ft2 -65- =1 b“ Refer to the figure below“ For rectangle 1: m = 5:- = 006; n = g- = 12; I4 = 0.1431 2 6 For rectangle 2: m = g = 0.4; n = g = 12; I4 = 0.1063 ' . 3 4 ' For rectangle 3: m = ~5— = 0.6; n = E = 0.8; 14 = 0.1247 ‘ 2 4 ’ For rectangle 4: m = g = 0.4; n = ~5— = 0.8; 14 = 0.0931 1“ Ac, = q[14(,,+ 14(2) +146, + 14m] = (1800)(0..143l + 0.1063 + 0.1247 + 0.0931) =_ 841 lb/ft’ c.. Refer to the figure h"'."——.———10f't"—'——“—~—D|1-—3 fi—fl 7 5ft A I -65- A m stress at C due to rectanguiar area 13 f: x 5 ft 0-? — - stress at C due to rectangular area 3 ft x 5 ft 13 For rectangular area 13 ft x 5 ft: m = = 1; n = —5— = 2.6; 14 = 0202 5 For rectangular area 3 fl x 5 ft: m n = g = 1; I4 = 0.1361 fl Urn» UIHA H 5: 9‘ Ac, =q(0..202 — 0.1361) =(1800)(0..202 — 0.1.351) = 118.6lb/1’t2 9.22 Equations (9.41), (9.42), and (9.43): b=§=§=25fi 2 2 m=£=a=2 B 5 nI=£=—li:6 b 2.5 From Table 9.10, 15 = 0.095 A02 = qu = (1800)(0..095) = 171 kN I m2 -67- -68- CHAPTER 10 _ 2 10.1 Equati0n(10..1): Ls",=Aa13l E“ S IP B=3fi;L=6fl;m,=6/3=2 Table 10.1: Forml = 2, Ip= 1.2] 1—0.42 S, = 3000 3 ——————-_— ( X )(140x2000 113/112) (1.21) = 0.03267 11 = 039 in. 711 3x3 10.2 A0 = = 79 kN / m2; 1,, "—’ 0.88 (Table 10.1) 1—0322 S” = (mm 16 000 (0.88) = 0.0117 = 11.7 mm . . _ a (1-11?) 10.3 Equat1on(10.5). SeuAaBeIGIFIE E Aa= 100kN/m2 3, =11“? =1/(4X3f =3385m It 71' p, = 0.3; E0 = 16,000 kN / In2 E 16 000 -—= ° = —’——- = 11.82 ’6 k3, (400)(.3.385) i = 39.. = 5.91 3, 3.385 From Figure 10.4,16 z 0.89.. Prom-Equation (10.6): .69- = 0.815 4.5+10 3.385 3385 15x106 (2)(025)]3 16,000 + (7)0100) From Equation (10.7): IE=1— 1 33.5 exp(l.22p, w 0.4)[ 3‘ +1.6] D; :1— : 0.923 1 3.5 exp[(1.22)(0.;3) — 04}(£§§ + 1.6) 15 1-03.2 Se = (100)(3..385)(0..89)(0..815)(0.923)[ 1 6,000 = 0.01289 m z 13 mm 10.4 a. The plot is shown. L2 L1 1.0 0.09 0.08 0.7 20 50 100 200 500 1000 0’ (kN / m2) :10- b.‘ or": 120 kN/mz eI — e2 _ 0.985 - 0.85 0.. Co = -——-— = 0.448 10 9—5: log[fl) g a; 200 10.5 a‘. Height of’solids: — “EK— — ~——2§2g—— = 1.12 cm = 0.441 in. s _ AGSYW ‘- (40-91)(2-r54)2(2-58)(1) I 0‘ (lb/ flz) The e-log a’graph is plotted. 0.6 0.5 0.4 0.3 ‘ 400 1,000 10,000 20,000 0’- (lb / flz) b‘. From the graph, aj= 940 1b I ft‘I 0“ CC = _eI_—_eZ_ : z 0_133 o" lo 2 log[--—) g[ 6;] - 2000 ;71_ W; _ _—_1.17_gm_ : 1.356 cm = 0.539 in. 24vaw gens x 2.54)“ (2.72)(1) 10.6 H, Pressme, 0" (ton / fiz) 0.85 .—._ N “2" 0.80 E 0.75 / E 0 70 t‘ g .. 0-65 0.60 + 0.5 1.0 2.0 5.0 10.0 Pressure, a’(ton I ftz) 10.7 a. Sc. = CcH log[3~£+—AU—] 1 + e" a; Cc = 0-009(LL m 10) = 0.009(50 — 10) = 0.36 ' H; 0.. -— mm.) H. firm“) -62.4]1£ +[ Md“, - 62.4]7 = (110)(8)+(115 —62..4)(15) + (120— 62.4)[3-22) = 2158.6 lb / ft2 _ (0.36)(17 x 12) lo 2158.6+1000 Sc 1 + 0.9 21586 J = 6.39 in. -72- 10.8 I 0.9 I . b. so = C’H log[a‘]+ C‘H log[9-°—+'Ai] 1 + e0 0-,, l + e0 o". = (17)(12) gig-910% 2600 +0.36log 31586 = 3.79 in. 1 + 0.9 6 2158.6 2600 G, r“, = (2.68)(9.81) = =16..4. / 3 Mm“) 1+e 1+0.6 “(N m (Gs+e w 2.68+0.6 981 _ rsat(smd)=—_—_I+e)y =(—"‘i'_:('i%(—)-=20.“kN/m3 75a1(clay) = ———(.G’ + e) 7’" .= W =17..6 kN/m3 1+e . 1+1.2 a; = (16.43)(2) + (20.1 1 — 9..81)(1.5) + (17.6 — 9.81)??— = 58.06 kN / m2 Cc. = 0.009(LL - 10) -—- 0.009(45 # 10) = 0.315 _ Sc = (0315)(25 x 1000) Iog( 58.06 +140] = 190.8 mm 1 +12 5806 hm) = G, y... =(2.67)(981)=1658 RN [m3 1+ e 1+ 0.58 (G, - 1) y (2.67 — l)(931) 3 ' =-——-‘i=——~—=10.37kN/ rm) 1+ e 1.58 m 7(::Iay) = “(G3 — I) r” = ———(2'72 _ D031) = 8.03 kN / m3 1 + e 1 + 1.1 a; a (1.5)(1658) + (1.5)(10.37) + [36.03) = 48.46 kN / m2 Cc = 0.009(LL - 10) = 0.009(45 — 10) = 0-315 -73 - C,H a; CcH a; +Ao" 5,, = ' log + log —-— 1+e, ' ' 0'0 1 + co ac 9%](2) . I = — 5 log[ 16° )+ wlog(w) = 0.0524 m = 52.4 mm 2.1 480.46 2.1 1010 CC = fl = = 0‘33 10 “5 10g(@) g a; 110 Cc. = m; e; = e, - Cc log 3 = 12 - 0.83log[3—50] = 0.78 [0'5] a; 110 log ' 0'] ‘ 1011 cc 314319 =0419 433 = 1.1— 0.419 log[§-'§-) = 0.872 1 1012 I; = E”; U=50%; T, =0..197 2 dr (0.002 c1112 ls): .. 7 = ————-——. 019 [2.5x 100 cm] 2 t = 17.8 days -74- 104.13 The e-log 0’ plot is shown. 00’: tSf; e] = 0 ton f fi2 I 05+ Aa’= 2.5 tsf; e = 0.79.. ‘ 5 Ae = 0.96 — 0.79 = 0.17 _ HA9 _1+e0 = (75 x 12)(0.17) 1 + 0.96 = 7.8 in. 2.5 ton / f‘t2 so Ae 10.14 a. m,,= “" = M’ l+eav 1+ealv Ae = 0:1 - .e2 = 1.82 — 1.54 = 0.28 A0’= 02’ — af= 400 - 200 = 200 kN/m2 e _1.82+_1§i av_ 2 9&8.) _ 200 m‘, -- l + 1.68 = 1.68 = 0.000522 m2 IkN b. 6,, = k =0..003 em: lsec mv7w = 1: (0.000522 x 1002 cm2 ,kN)[ 9.81 100-3 cm3 I: = 1.53 X 10"" cm lsec '-75- 10.15 ’5 10.16 Equation (10.58): t: ‘50 = 2 Hl.(dr) cvtL C'th 2 H F '(dr) 2 _ HF‘(dr)tL 2 Hi (dr) _ (10 x 12 in.)2(140 sec) _ 2 8,064,000 sec = 93.3 days (0.5 m.) c,,t 2 (dr) :nxUz .932- 502 t2_U§’ t2 "3? t2 = 33.6 days 10.17 a. my = cv b. Sc (33] [0.81—0.73] 0,. = Aa' = 200-400 = l+eav l-i-eav l[ Ae a. [3.7) [ .—._—-.——.—_— 0.81 — 0.7 u—lz—Qffi = 0.000895 m2 lkN 0.81 + 0.7 1+eav 1+3“ 1+[ J 2 _ 3.1x10‘7 m/s _ my 7., “ (0..000895)(9.81 kN /m3) 4 2 _ 2 EH; (0.197)[2 m ] ‘50 = — _ k = 353.1)(10'7 1112 /'sec. __ ———-—-——.————_ “— — _, 0..00223x107sec#62 hrs 353.1x10 m2 / sec 6 If _ AeH _ (0.81 — 0.7)(4) l + «30 _ = 0.243 m 1 + 0.81 Sc at 50% = (0.5)(0..243) = 0.1215 m = 121.5 mm 0.4519x10'3 m2 lkN w») (00.197)[--~"--——-0'025 )2 2 = Tva =__3.__ = 0.,9x10“5 m2 min 30.4 t5!) cl? k= may“, = (0.4519 x 10-3 m2 /kN)(0.9 x 10'5 m2 lmin)(9081 kN I m3) = 3.99 X 10" m I min c t c t TED = v L = V F H g (dr) H 1% ‘(dr) .2...___’F . fli_ ‘F' Him Him’ J (3:102 2 (“A81 1020 a” my: a” M' Hem, 1+4?aw Ae = 10.21 — 0.96 = 0.25 Aa’=4 - 2=2tonlfi2 em, = 12142—096 =1n085 (9-23) 2 m, 5—“ 1+1..035 4, c, = k = ———-—1'8"-L———= 0.0962 ftz Iday "’W (0.06 1-12 lton)[ 62'4 ton/02] 2000 So = 0.06 ft: lton : Tij (0286)(9)2 : =—i=-——-=240.Sd 5" cv 00962 a” -77- 11. CC =—;—~=——4—-‘-‘=0.83 10% 02) mm 0': Sc = Q H log 92L“: = (0'83)(9)log(1)= 1.013 fl 1 + e0 0'; l + 1.21 2 St. at 60% = (0..6)(1‘.018) = 0.611 fl z 7.33 in. [35] [1.2—0.95] _ Aa' _ 220—110 _ .4 2 10.21 a.. m‘, — ——--—1+eav — —-—-—--——l + l2+095 —— 10.95x 10 m lkN 2 b.. cv = k mv 7w 2 k 0.0036 cm lsec = _4 2 2 3 [(10.95 x 10 )(100) cm lkN] 1002 kN lcm k = 3.87 X 10" cm / see 1022 a” :90 = (:30; 0,348 =W ‘1' [15— x 12 x 2.54) 2 c, = 4.27 x 10" cm2 lscc b Ilab _ tfidd . tlab _ 120 x 24 x 60 x 60 " H2 ‘ H2 5 _”'”_—2' “ 2 Wm Wield) [15 x2254) x 12 x 254] thb = 720 sec -73- Hum 30 10.23 . U°/ :— — a (0) [so )(100) = 37.5% b. T — cv’ - U=50%; I; =0..197 v'" 23 dr 0.003t 0.197 = ——-—2 (400 cm) ISO = 10,506,667 sec = 121.6 days = 0003: [fig-“)2 2 . 1'50 = 2,626,667 sec = 30.4 days c. _2; = 0.197 A6: +4A0'jn + AU; 10.24 Equation (10.66): Aagw = _ 6 Equation (9.39): Aal= qu Aa'av = _—___1 7‘48 + (4 x6803) +433 = 8.99 kN / m2 G, yw + wG, 7,, (2.7)(9.81)(1 + 0.35) 3 =m=“= 18.38 kN/ Mela” l+wG, 1+(o..35)(2‘.7) m o;=(1 x l4..5)+(l)(17.8 — 9.81) + (1)(18.38 — 9.81) = 31.015101/m2 Cc. = 0.009(LL — 10) = 0.00908 - 10) = 0.252 -79- Sc .__ (0252)(2) log[31.06 + 8.99 1+(0..35x 2.7) ' J = 00286 = 28.6 mm 3106 Is ' Table 9.10 Aa’= qI5 _ 18.65 + (4 x 9.49) + 3.6 "" 6 _ (0252)(2) logf 1.06 + 10.04 = 10.04 kN / m2 )= 0.0315 m = 31.5 mm 1 + 0.945 31.06 .80- CHAPTER 1 1 11.1 a. Normal stress, a’= 192 kN lmz; shear stress, I, = 120 kN / In2 -1 If -1 '= — =tan —— 232° ‘6 tan [0"] [192] b. I}- = a’tan 95' = 200tan32 = 124.97 kN / m2 50x50 Sh ‘f' ,8: —_ em me (r’)(1000x1000 J: (1:24.97)(0.05)2 = 0.312 kN = 312 N 11.2 Shear force, S = (2 x 2-in..-2)(a’tan¢’) = (4)(15)(tan_41) = 52.16 lb 11.3 Area of'specimen = [3(0032 = 0.0019634 m2 = A A graph of 2} versus a’wiil yield 115' = 29.19. 11.4 Area, A = 0.0019634 m2 A graph of' 2} versus U’will yield ¢’ x 21.9". -81- 11..5 Equation (1 L8)“ With 6’ = 0 a; = a; tan2(45 + = 105 tan2 (45 + gig] = 404.4 kN / m2 Aodw = 404.4 — 105 = 299.4 kN I m2 11.6 a. Equation (11A): 9=45+£~=45+£=63° 2 2 b. From Equations (9.8) and (9.9): a, : 0'1 +03 4-: a, ~a3 60526: 404.4+105+ 40444050050sz 2 2 2 = 350.3 kN/mz T: 01 _a3sinzg=wSfifi2x2fl=114.68kN/m2 For failure, 2; = a’tan¢’ = 35_0H8tan36 = 254.9 kN / m2 Since the developed shear stress. ris 114.68 kN / m2, which is less 254.9 kN / m2, the specimen did not fail. 11.7 ¢' = 25 + 0.181), = 25 + (0..18)(60) = 353° 0-; = agtan2[45+%)=18tan2(45+§:—'8)= 68.7 iii/inf 1108 0'1’=4:73:+Ac;§,r0.)=140-1-264=4041(N/Irf2 a; = agtan2[45+%) , 05 05 ' ¢'=2 tan" 9-3- —45 :2 tan-(19$) —45 =29.03° 0'3 140 -82- 11.9 0'; = 0'5 +1503“) = 0'3 tan2[45+%) A 0w“), = w——_...__.23 26 = 14.73 lb / in.2 mnz(45+i:—]——1 tan2(45+~é—)—l ’— 0':— 150' 11.10 0'3 =—“E’——=——i8£—= 122.96 kN/mz tan2[45+%]—1 tan2(45+%§]—1 111.11 a. a; = a; tan2[45 + (300+ 350) = 3oom2[4s + ¢' = 21.6° b. 9:45+—=45+~2——1—'§= 553° 2 2 c.. From Equations (9..8)-and (9.9); ' a, = 01w3 + 01—03 00520: 650+300 + 650-300cofl2x 553) 2 2 - . 2 . ' = 410.151.311.112 0-; —ag 650—300 z = c0326 = —2-——-cos(2 x 55.18) = 162.7 kN I m2 11.12 a; = a; tan2[45 + + 2c' tan[4s +%-] Specimen I: (15+ 31.4) = 46.4 = 15tan2[45 4.512,)... 26! tan[45 + (a) Specimen II: (25 + 47) = 72 = 25 tan2[45 + + 2c' tan(45 + (b) Equation (b) minus Equation (a): 72 —- 46.4 = 10tan2(45 +%); 915' = 26° I -83- From Equation (b): 72~25tan2[45+§ —_——-—26—3.1 = 2.49 lb / in.’ I C: 11.13 a; = 36tan2(45 +329] + (2)(2..49)tan[45+-2§6~] = 100.2 lb / in.2 ll..l4 6’ =0 Aa' c;3'§,=——-——--“rt—2;,—-——=—~~-1'9W-~—=0.59tonlft2 tan2[45+—m)--I tan2[45+———-)—1 2 2 ¢ 12+914 °5 - 11.15 alud3tan2[45+3); ¢=2[tan'l( 12‘ J —45:l=16° ' k - . , I II 0.5 I ai=agtan2[4s+iJ; ¢'=2 tan‘l[———~—m12+9'l4_6’83] —45 '= 28° 2 12-433 0.5 I 11.16 ¢ = 2[tan-‘ (iii-.23] _. 45] = 13° 05 ¢'=2[m—1[_.~_._14<;;;2;; 75) -45]=29.4o 1117 a. ot=a3tan2(45+§] 05 ¢ = 2[tan"l[10:;597) ~45] = 18.4° -34- bu a; =agtan2[45+%) Audw = kN I m2 11.18 a; = a; tan2[45+%—] = 105m2 [451?]: 2901.8 kN/ m2 11%, = 290.3 — 105 = 185.8 kN I m2 11.19 a, = 03 tan2[45 + = 15153112 (45 + = 32.97 lb / in.2 . A090) = 32.97 - 15 = 17.97 lb I 111.2 a; = a; 121111454?) Audm = I ill.2 11.20 g”: 0; ~ 03=2800 --1500= 1.:«100111/rtz 11.21 a; = a; tan2[45+-¢3) Audm = ‘68-5 kN I m2 -85.- 1 1.22 1 1.23 l 1.24 Audm = "' kN I m2 a“ q’==m+p’tana 23 = m + 50tana' 18 = m + 30tana: m = 10.51b/in3; a= 14° b4. 45' = sin"(tana) = sin'1(tan14) = 14.440 c: = m = 105 = 10.821b/iu.2 Cmv'sI) = 0‘1 1 + 0.0037131 0'0 a’= (3x17) + (3095 - 94.81) = 80.07 kN / m2 cm“ = [0.11 + (0..0037)(18)](80..07) = 14.14 kN I m2 -86... ' (a) (b) CHAPTER 12 12.110124 Ko=(1 - sin ¢')(OCR)sm¢-‘ '(dig) Pozl/ZKJHZ I I 2 0022-6: Ka Pa = 1A: Ka yHZ 32 0.307 (0.307)(110)(10) ('/2)(0..307)(110)(10)2 = 337.7 lb I ftz = 1688.5 lb I ft 0.361 (%)(0..361)(93)(12)2. (O..361)(98)(12) = 424.5 lb I ft’ = 254711411: 36 0.26 (0..26)(17..6)(3) ('/2)(0..26)(I7.6)(3)2 = 13.73 kN I m2 = 20.59 kN I m 40 0.217 (0..217)(18..2)(6) (1/2)(0..217)(18.,2)(6)2 = 23.7 kN I m2 12.5 to 12.3 K“ = tan2[45-.¢—) 12.6 :4 ft- =71.09kN/m Note: 1.. Pressure distribution is similar to that shown in Figure 12.1321; that is, Ug=0atz=0and 0Q=KayHatz=H. 2. E is the distance measured fiom the bo'ttotn of'the wall. ..s-7_ m 12.9 to 12.12 KP = tan2(45+%-] 12.9 34 3.537 (3.537)(110)(10) (1/2)(3.537)(110)(10)2 3.33:: =3890.7lblft’ =19,4541blft 36 3.352. (3.852)(105)(12) (1/2)(3.852)(105)(12)2 - = 4853.5 lb / n2 = 29,212 lb I ft 31 3.124 (3..124)(14.4)(5) (l/zxg.124)(14.4)(5)2 -- = 224.9 W l m2 = 562.3 1:le m 28 2.77 (2..77)(13.5)(4) (1/2)(2.77)(13.5)(4)2 = 149.6kN/m3 =299.2kN/m Notes 1. (7AM) = 0; ttiangular pressure distIibution 2. z“ is the distance measured fi'om the bottom of'the wall 12.13 Kg = tan2(45—i:-] = mi 0", (lb / fiz) Depth (ft) I IDepth (ft) z=0fi: 0;=0§,Ka=0; u=0 z=5 fi: a;=(105)(5)(‘/s)= I751b/ft2; 21:0 2 = 10 fi: 0; = [(105)(5) + (122 — 62.4)(5)](1/3) = 274.3 lb I ft2 u = (62.4)(5) = 312 lb / ft2 -88.. (V2)(5)(175) = 437.5 (173(5) = 875 (1/2)(5)(274‘.3 « 175) = 243.3 (I/z)(5)(312)- = 730 Pa = 22,340.23 lb / ft Resultant: Taking the moment about the bottom of'the wall, 5 5 5 5 [(437.3[5 + 5] + (879(3) + (243.3(3) + (730%)] - - 2340.8 __ 2916.7 + 21875 + 413.8 +1300 2340.8 E: = 2.91 ft 12.14 K“ = tan2 [45 - = 0.283.. Refer to the figure. 34,9 . 0:. (lb / ft?) ' u (lb I ftZ) 3 + 20 ——————————— -— Depth (ft) Depth (ft) 2' = 0 it: a; = 01,143, = (300)(o‘.2s3) = 84.9 lb / {8; u = o z = 6 ft: 0; = [300 + (6)(110)](0..283) = 271.68 1b / 112; u = o -39- z = 20 a: a; = [300 +(6)(110) + (126 — 62.4)(14)](0.283) = 523.66 lb/ 112 u = (62.4)(14) = 873.6 lb / fiz Area No. Area 1 (84.9)(20) = 1,698 2 (1/2)(6)(27l.68 - 84.9) = 560.34 3 (14)(271.68 - 84.9) = 2,614.92 4 (V2)(l4)(523.6 — 271.68) = 1,763.44 5 (V2)(14)(873.6) = 6,1 15.2 P, = 212,751.13 lb I ft Location of resultant: Taking the moment about the bottom of'the wall, 6 (1,698)[?) + (560..34)[14 + + (2,614.92)[-1;) +(1,763.44)[1;] +(6,115.2)[1—:~] z = 12,7519 _ 16,980 + 8,965.44 + 18,304.44 + 8,229.4 + 28,53 7.6 - . 12,7519 = 6.35 ft 12.15 Km) = tan2 [45 — = 03.33; Km) = tan2 [45 — = 026.. Refer to the figure. 0; (kN / m2) u (kN / m2) 29.43 Depth (m) -90- z = 0 m: a; = 01,112,“) = (15)(0.333) = 5 kN / ml; 11 = 0 z = 3m: 6:, = 61,19“): [(15.5)(3) + 15](0.333) = 20.48 kN / m2 a; = aLKam = [(15.5)(3) + 15](0..26) = 15.99 kN / m2 u = 0 z = 6 m: a; = 01,1642, = [15 + (15.5)(3) + (19 - 9.81)(3)](0.26) = 23.16 kN/m2 u = (9.81)(3) = 29.43 kN / 612 Area No. Area 1 . (6)(5) = 30 2 (%)(3)(20.4s - 5) = 23.22 3 (3)0599 - 5) = 32.97 4 (V:)(3)(23.16 — 15.99) = 10.76 - 5 (1/2)(3)(29.43)_ = 44.15 _ Pa = 2141.1. kN I In Location ofresultant: Taking the moment about the bottom 6 . 3 ' 3 i 3‘ 3 (30%) + (2.3.22)[3 + 3-] + (32.97)[-2—] + (10.70%) + (44.15)[.3—) f = . 141.1 90 + 92.38 + 49.46 + 10.76 + 44.15 28725 ' = —-——.—~—-——-—m = = 2.04 In 141.1 141.1 12.16 a. Equation (12.23): . 2 . ‘ OJ _ yzcosa1/1+sm ¢'—251n¢'cos1;ra a _ —__.__—_____ cosa +1/sm2 915' — sm2 0: - 01‘, = sin"' 5}“ — or +26 = sin-'(sfn 10]— 10 + (2)(5) = 20.32° sm 425' sm30 a, _ (15)(4)(cosIO)1/1 + sin2 30 — (2)(sin 30)(cos 20.321 " cole+Jsin2 30—51112 10 -91- =22.7kNIm2 Equation (12.25): fl=tan-.[ sin¢'sinyra J=m~,[ (sin30)(sin20.32) ]_18'1° 1— sin ¢' cos Va 1— (sin 30)(c0320.32) _ b.. Equation (12.27): cos(a— 6M} +sin2 ¢' —2sin¢'cosy/a cos2 9(cosa + x/sin2 ¢' — sin2 a) _ cos(10 ~ SN 1 + sin2 _30 — (2)(sin 30)(cos 20.32) c03210(cosIO-+ «1 sin2 .30 — sin2 10) Ka(R) = = 03 94 pa = ngZKM) = é(15)(4)2(o..394) = 47.28 kN I 111 Location and direction of resultant: — At at distance of H13 = 4/3 = 1.33 m above the bottom of the wall inclined at an angle ,8 = 18.1° to the normal drawn to the back face of the wall 12.17 TlfisisaRankjnecase since §’= 04. Pp =%7H2KP(R) cos(a — 6) 1+sin2 96' ~i~25in¢5'cosqgrjp cos2 0(cosa -— sin2 95" — sin2 at) Equation (12.33); K“) = a=0; t9= 10°; ¢'=36° . -1 sina . 4 sin0 = + —2t9= +0— 2 10 =—20° WP 8’“ [my] a m [sin36] _ (X ) ‘ cos(0 — 10)‘/fisin2 36 + (2)(sin 36) cos(-20) cos2 (_10)(cos0 — «lsin2 36 — sin2 0) l t . 1 Pp = 5(16‘5)(3)2(3.855) = 286.2 kN/ [11 KM): P = 3.855 -92- Equation (12,32): [3: my! sin.¢'sin 511,, 3 m4 sin365in(—20) $7.340 - l +- sm¢' cosy; p l + (sm 36)[cos(~20)] PP acts at a distance of H13 = 3/3 = 1 m from the bottom of the wall inclined at an angle fl = —7.34° to the normal drawn to the back face of the wall. 12.18 K p = tan2 [45 +329) = 3.‘ Refer to the figure. . ’ lb/ftz u lblft2 0,,( _) ( J 2469 312 Depth (ft) Depth (ft) z=0fl: ap’=0;u=0 z=5 ft: ap’= ylzKp=(105)(5)(3)=15751b/fi2; u=o z = 10 ft: 0; = [(105)(5) +- (122 - 62.,4)(5)](3) = 2469 113/ ft? u = (62.4)(5) = 312 lb / :12 (1/2)(5)(1575) = 3,937.5 (5)( 1575) = 7,875 ('/2)(5)(2469 — 1575) -—- 2,235 ('/z)(5)(312) = 730 PP = 214,828 lb / ft -93- Location of’the resultant: Taking the moment about the bottom of' the wall, 5 s 5 s (.393 7.5)[5 + a) + (7873(3) + (223 3(5) + (780)[-§] E = = 3.44 ft 14,828 12.19 a. H= 4.5 m; cu =19..3 kN/mz; y= 19.6 kN/ma; ¢'= 0 Kg 6a =fl_2cu At the top (2 = 0): on = ~26" = (—2)(19.3) = “38.6 kN/m2 At the bottom (2 = 4.5 m): 0‘, = (19..3)(4..5) - (2)(19..3) = 86.85 ~ 38.6 = 48.25 kN / m2 The pressure diagram is shown. n—~——~——..| 48.25 kN / m2 b. Equation (12.49): 2c” (2x193) = 1.97 m y 196 20 = c. Equation (12.51): Pa 2 %7H2 — 2ch = [%)(I9.6)(4.5)2 n (2)(19.3)(4.5) = 24.75 kN / m d. Equation (12.53): = (%)(19.6)(4.5)2 (2x1 9.3)(4-5) + 2c: 7 (2x193)2 19.6 P; = 31-sz —2c,,H+ = 62.76 kN/m -94- Resultant measured from the bottom: H“z° =M3034m 3 3 1220 a.. a", = 00K, —2c,,‘/Ka 00:7’Z‘HI; Ka=1 Atz=0: 00= 8 kN/rn2 q, = 8 — (2)093) = —30..6 kN / m2 Atz=4n5 m: T 2 94 00 = (19.6)(45) + s = 96.2 kN / m2 " on = a0 — @093) = 57.6 kN / m2 J The pressure diagram is shown. [4-———57r.6 kN/m2“*"'—'H_ ._ bu 0a=0u (WM?) - 26u=0v _ 2a,, ~-q _ 38.6—-8 " 1.56 m r 19.6 20 ct, Referring to the diagram in Part a: Pa = [—21-](294X57u6) -— [~2!-J(30t.6)(156) =_ 60.8 kN / m d‘. R, = [%)(2.94)(57.6) = 84.67 kN / In Location ofthe resultant from” the bottom of'the wall: = 0.98 -95- 12.21 K0 = tan2[45 —~ ?] = tan2[45 — 12—6] = 0563; Jig, = 0.754 N Equation (1252); Pa = ——KJHz —2 Kac’H+ -§-(0568)(19)(5)2 — (2)(0.754)(26)(5) + (2)5262 = 10.02 kN lm 12.22 Equation (12.63): 20 =2; lei—sing}. = (2X88) I+sln25 :25} fi 3/ 1—sm¢' 110 l—sm25 Atz=0fiz 0;:0 At z= 15 ft: ag= yngmcosa c' 88 — = —— = 0.053 )2 (110)(15) - ' - I ' . . . 1 For a = 10°, 0 = 25°, and c— = 0053, the value of'K;(R) e 0366 ‘ 72 a; = (110)(15)(o..366)(ces10) = 590.7 lb / 02 R, = %(15 — 2.51)(594..7) = 3714 lb I ft 1223 Use Equations (12.68) and (1269).. a:=0; 6= 10°; ¢'=36°; y= 18kNlm3;H=5 m 70.15kN/m 70.58kN/m P, is located at a vertical distance of 5/3 = 1.67 m above the bottom of the wall and is _, _ inclined at an‘ angle (5 ’to the normal drawn to the back face of the wall. 1 i -96- 12.24 a. _¢'=33°;;1;=90 ~19 —6’ =90 —- 5 - 20=65° Weight ofwedge ABC = —21—(1I..6)(7..l)(L23§) = 5721 lb / a = 5.271 kip / ft 2' The weight of each of'the wedges CBE, EBF, FBG, GBH = [%)(20)(4)(128) = 51201b/fi = 5.12 kip / fi Weight (kip / 11) 5.271 5.271 + 5.12 = 10.391 ABF 10391 + 5.12 = 15.511 ABG 15511 + 5.12 = 20.631 ABH 20.631 + 5.12 = 25.751 ABC ABE The graphical construction is shown. Pa = 5.2 kip I ft. m ‘—" Scale LIE—1 Load scale l4ki2/fi' -97- b. ¢'= 34°; 40:90 -0 H17 =73° Weight of'wedge ABC =%(16)(9)(116) = 8352 111/ 0 z 8.35 kip/ 0 The weight of each ofthe wedges CBE, EBF, FBG, GBH = [gamma 16) = 5104115/0 z 5.10 kip / ft Weight (kip / ft) 8.35 8.35 + 5.10 = 13.45 13.45 + 5.10 =18.55 18.55 + 5.10 = 23.65 23.65 + 5.10 = 28.75 ABC ABE ABF The graphical construction is shown. Pa = 6.3 kip / ft. Load scale fikm/fil -93- W*. 0.. = (1680)(9.81) 7 1000 Weight of'wedge ABC = %(5.25)(2.5)(16..48) = 108.15 kN / m The weight of'each of'the wedges CBE, EBF, FBG, GBH = l 2 Weight (kN/m)' 108.15 103.15 + 53.56 =161..71 ABF 161.71 + 53.56 = 215.27 ABG 215.27 + 53.56 = 2681.83 ABH 268.83 + 53.56 = 322.39 ABC ABE The graphical construction is shown. Pa= 125kN/m.. 5.5 m -99- [ )(1)(65)(I6..48) = 53.46 kN / m mum/m Scale 1.1L _ ' Load scale I SOkN/m =16..48 kN / m3; 0': 30°; 1/: 90 - 10 — 30 = 50° 6.5 m 12.25 Equation (12.74): PM =%-yH2(1 —k.)K;' 8 2 11:, = O; 0= 0; a= 0; ¢' = 35%;;- = g; k,, = 0.3.. From Table 12.9, Kf,’= 0.486 Pm, = (15)(6)2(1'— 0)(0..486) = 131.2 kN/m i 2 For 525’ = 35°, 3-;- FI'om Table 12.6,‘Ka = 0.2444. 1:, = észx‘, = %(15)(6)2(0.2444) = 66 RN / m Equation (12.83): H 6 Par AP“(0.6H) (66(3) + (13 1.2 — 66)(0.6 x 6) 'z‘ = —_.._._____ = ——————_ a 2.8 m pa: 131.2 12.26 Equation (12.84): P03 = KH ~ szQ, w c ’(H - zo)2N;c. Given: zo = 0; 6=-10°; 515’ = 15°; kh = 0.15 N {,0 = NM = 1.75 (Figure 12.3.3); Na, = 0.3 (Figuve 12.35); i = 1.3 (Figure 12.36); N;, = 1N”. So, P“ = (19)(6 — 0)2(1.3 x03) .- (20)(6 -- 0x175) = 56.76 kN lm I 12.27 20 = 0; n = 0; 6= 5°; {6’ = 20°; k,l = 0.25; N36 z 1.65; Na, 3 0.25; 21 = 1.65; N;,,= 2N”, Equation (12.84): Pa. = KH - ZOYN .1, - 61H - ZINE. = (100)(10 - 0)2(1.65 x 0.25) ~ (200)(10 -- __O)(-1..65) = 4125 — 3300 = 825 lb I ft -100— CHAPTER 13 13.1 Equation(l3..10): Pp: g—nyKP H _ 3 c036 00520 H‘ = = 3.19 m; y= 15.5 kN/m3. From Figure 13.4, with 19= +20° and ¢' = 35°, the value of'Kflem '~' 5.3 a _245_07 45' ¥_ "” From Table 13.1, R = 0.836. KP = (5..3)(0..836) = 4.431 Pp = %(15.5)(3.19)2(4.431) = 349.4 kN I m 13.2 %= Equation (1310): Pp = éyHfiKp 6= 0;H1= H: 15 fl; ¢’ = 30°; ar= 0.. From Figme 13.4, Kp(§4.¢j % 6.2 From Table 13.1, for ¢' = 30° and; = the value OH? is 0.35.. so Pp = %(100)(15)2(6..2 x 0.85) = 59,288 lb I ft 13.3 Equation(13..l4): Pp= WK, 6’ = %¢' = 20". From Table 1.3.3, for 45' = 30° and 6’= 20°, K; = 4.4 I; = %(100)(15)2(4..4) = 49,500 lb I ft 0512 13.4 $7 = .75.» = +0.4. From Figure 13.5, for; =+o..4 and ¢' = 30°, 1%.m e 9.8.. -'-101- From Iable13.2, R = 0.686 Pp = '33717276,, = (14.3)(2.5)2(9.8 x 0.686) = 311 kN / m l. 2 13.5 Equation (13.1.5): PW: észK; For 7" = 30°, — = E = 0.5, k" = 0, and kh = 0.3, the value of'K; = 3.17 (Figure 13.9) 6' 30 Pm = %(16)(5)2(3.7) = 740 kN / m 13.6 nu = 2—1“ =04. ¢'= 35°; 6’= 20°. Table 13.4: P“ = 0.248 5 m 05sz Pa = (0.243)(05)(16)(5)2 = 49.6 kN I m 13.7 nu =fl: 0.3. 15.6m [ c_’ 3 i = 01.1 )4“! (18)(15.6) Pa = 0.122 0.574?2 From Table 1.3.5, for {15' = 20°; 6’= 15°, Pa = (0.122)(0.5)(18x156)2 x 267.2 kN 13.8 Refer to Figure 13.15.. 6., = 0..65;’Htan2[45 — % (0..65)(16){9)ten2(45 - 32-9] = 31.2 kN / mi Refer to the diagram on the next page. 2A = (317-2)(3)(% ) A =' 70-2 kN /m 4 ' 1 Hence, \ 1 -102- Bl = (31.2)(3) ~ 702 = 234.4 kN / m (31.2)(2) BZ=C1= =31.‘2kN/m Again, taking the moment about C2, we have 213 = (31 -2)(4)( % ) D = 124.8 kN / m So, C2 = (31.2)(4) -- 124.8 = 0 The strut loads are as follow: At level A: (A)(s) = (701.2)(2) = 140.4 kN At level B: (B1 + Bz)(s') = (23.4 + 31H2)(2) = 109.2 kN At level C: (0(8) = (C: + C2)(s) = (31.2 + 0)(2) = 62.4 kN At level D: (D)(s) = (124.8)(2) = 249.6 kN -IO3- '~ 1 04- CHAPTER 14 14.1 Equation (14.15): F = .—“'—._ + “W ’ yHcoszfl tanfi tanfl l4 tan25 2.5 = ——-—2-——-_ + (18)(H)(cos 20)(ta.n‘20) tan20 2.5 = a°£+1..28 H H= 1.98 m 14.2 Equation (14.16): Her = i.__2____1..___.. = .—_ 26.8 ft y cos fl(tanfi—tan¢’) 110 cos 20(tan20—tan15) = (1850x931): 14‘s 75“ 1000 18.15 kN/In3 7’ =‘18..15 — 9.81:8..34kN/m3 Equation (14.. 18): s ysat HCOSZIB tanfl ysat = 25 '+fl_‘an_20:1.31 . fisxsxcgsz 15)(tan_15) 18..15tan15 _ 14.4 m = = W = 128.7 lb / 03 e .. y’ = 128.7 — 62.4 = 66.3 lb / {13 -105- - ___c' + .211. W“ rchoszfltanfl r53: tanfi 500 66.3 tan20 = _----~———-—— + _ (128.7)(2oxcos2 20)(tan 20) 128.7 tan 20 F; = 0.604 + 0.515 z 1.12 14.5 Equation (14.42): Her = :13: sinfi cos¢' = (4)(9.6) sin60cole z 5.83 m y l— cos(fi— 425') 15.72 1- cos(60— 10) _ 145 H =55. M; =.(25)(4)[(sin45)(coszo) I. a 7 l-cosofi—sé’) 18 l—cos(45—20) ' ' 25 -1 tan20] 4..7 = w; I=c—=._—= 2' '=tan _...I....._. : I. O 1 Fs 25 0,, FS 25 10kN/m, ¢d [ 2'5 828 From Equation (14.40): — ——--——- -784_m 7 I—cos(fi—¢;) 18 H=fl m = (4x10) (sin45)(coss.2s) lmcos(45—8428) " ’ 14.8 H=531[—S‘-‘19L°£”—]; y=1151blft3 7 1‘005(13—¢3) 5 5 (1b I 112) (deg) 14.9 p= 1700 kg / m3 --106- 1000 c’=18kN/m2; ¢'=20° F52;c;=~£—=—1—§=91<1\I/m2 F, 2 115:: = tan"[ta;¢ )ztan'l “320]: 1031° H = 4c; _ (41(9).[W] ... 23.74 m 7 _ 16.68 1~—cos(26.57'—10.31) 1-“ 008(fi- 115;) 14.10 m = 0.185 (fi'om Figure 14.9). Equation (14.47): H ca 500 = = —-——-— = 24.56 ft — Toe fail ct ym (l 10)(0..185) are 14.11 m = 0.135 for [3: 56° (Figure 14.9): c 500 2 =—"=—=2001b/ft 6" F, 2.5 ed 200 =——-=———-—=9.83ft 7m (100)(0.185) 14--12 fl = tan" —i— = 63..43°. For fl: 63.43°,-m' = 0.196 (Figure 14.9). (a) cd =c—"=£§-§=16.275kN/m2 F, 2 _ c_d _ .._.L6_'275_ ___. 4.39 m _ ym " (18..9)(0..196) ..1 07.. ym (18.9)(0.196) Since fl > 53°, it is a toe circle Refer to Figure 14.14 and Example Problem 14.5, H r. = CI . 6 2 sm a sin--— 2 From Figure 14.10, a «- 37°; Bx 69°. So 3.78 8.78 . r=___"'_'—=—“—m= 12.93 m (2 sin 37)[sin (2X005X0566) I 14'” a" D z 31%: 1‘41; 7'5” =1“ RN lms» P0rfi= 40°, D = 1.41; m = 0.175 Her = cu ym cu = (8.5)(18..5)(0.175) = 27.5 kN / m2 1).. From Figure 14.9, mid point circle 0. Hum Figure 14.11, 71 = 0.7 Distance = nH= (0.7)(8.5) = 5.95 m a.H-= (33(4) =24 14.15 Equation (14.56): cR= 023:0) 5 C nation 14.55 : = "‘“m E4 ( ) m WE Table 14.1, m = 0.0478 0.0478 —-—§—-— F ‘l.4l '2 Osmium); ‘“ -108— cu 14.16 F; 2 7H M.,6=30°;1’-1=12m;c,,=40kN/m2 From Figure 14.16, for k, = 0.4 and D = 1, M = 4 40 (1 8X12) F, = (4) = 0.74 14.17 1:; =-%M. fl=60°; cu=1000 113/112; y=1151blfi2;H=50 fi;k,,'=-0'..3 7 From Figure 14.17, M = 3.3 F _ 1000 . — ,mm); 0.57 14.18 a. fl=tm“l[%]=26.57° F3 =—I—=2.75 tangy tan20 From Figure 14.22, C “mu: 4. 0.05 71‘!“ um 01’ 700 05 = ————-—‘—— = 349.7 ft (1 10)(Hcr )(tan 20) b. F’ = ul— = 2.14 . ,6: tan"‘[i) = 337° tan ¢' tan25 1.5 Figure 14.22, _._c_.._ z 0,035; —-—fl—' = 0.035 I 7H“ tan¢' (110)(Hcr )(tan 25) He, = 417.8 ft —109— c‘. 16 = tan"[%]=18u4° —F’ = —~—l = 3.73 tan¢’ tan 20 Figure 14.22: c' _ 30 “005 mummy (15.8)(Hcrxtanls) “ Hag 142m _ 1 , 14.19 )6: tan '[3]=26.57°; ¢ = 10°; y =1201b/ft3;H==40ft;c’=6001b/fl2;c;=myH The plot och ,versus F 45. is shown. From this 126.: F¢,=F,=l.46 "-110- 14.20 5:49; ¢’=20°;c" =6001b/fi2; y =1151b/fi3;H=25fl F‘, = tan¢’ 6;: El. = with From the graph, given Fax: qu: F5 = 2.0 4 14.21 p= tan"[§1§] = 218°; ¢' = 12°; y =17..5 kN/mfl; H= 10 m;'¢' = 18 kN=/ m2; c; = From the graph on the next page, Fc._-,=- F¢.= Fs = 1.6 -lll-- Problem 14.21 14.22 fl = MING] = 45° _€._ = —_.E__ = 0.786.. Figure 14.22, E z 6.9 _ K 7Htan¢' (17.1)(5xtan15) tans!“ - " F, = (6.9)(tan15) = 1.85 14.23 a.. Refer to the figure. 0 f2 = sin30°; x = 80 fl x :40 = sin35° ladlUS, R R = 40 = 69.73 fl sin35 With radius R = 69.73 ft, the trial failure circle has been drawn. Weight of slice, Area of slices, A ' _W,, =A x y (112) . (kip / fl) ——— 59.8 ., 32 (20x32 +20) z 20 £127.77 £61.98 -113- F _ -R6c'+(ZW,,cosa,,)tan¢' " _ 2W; sinan (69..73)[[—1—§6)(70)](04) + (12‘7..77)(tan 20) 61.98 = 1.3 (Note: Accuracy can be increased by increasing the number of'slices‘.) b, AsinParta, Ezsina; .x= _H = ,5 =10m .x srna srn30 (i) i=s' 2), or ,5 =R=7‘.78m R 2 sm40 With radius R = 7.78 m, the trial surface has been drawn. _ -114- The following table can now be prepared. Weight of slice, Area ofsliees, A W" = A X y a n W" 005a,, (ml) (kN / In) (deg) (kN / In) (—3.92% = 53 116.28 ' 91.63 ___..._(2)(4“: + 2'8) = 7.0 119.7 112.48 40.94 71.32 I 71.43 (3)(2.8) 2 £295.14 £150.02 4.2 2 F _ .R6c'+(ZW,,cosa,,)tan¢' ‘ _ ZW,,sina,, (773)[[—”—)(80)](18) + (295.14)(tan 15) 18° = 1.8.3 1 5002 (Note: Accuracy will improve with smaller slices.) 14.24 ¢' = 25°; ,6: 26.57°; r, = 0.5;"; = “5 —— = 0.05. 7H (1 15x20) Using Table 14.1, the following table can be prepared. -1 14.25 ¢'=20°;,B=18..43°;r;,=0.5;—£—-- 6 —- = 0.05 7H (20)(5) Using Table 14.. 1, the following table can be prepared. 14.26 r, = 0.25; H= 25 11; fl= 30°; ¢"= 20°; c '= 100 1b / 112; y= 115 lb / 113 _ 7Htan¢' _ (115)(25)(tan20) ' _ c' — 100 A“ = 1046 From Figure 14..29(b), for toe circle: M = 25, Dz 1105 From Figure 14..30(b), N, z 25, D 3 1., So N, = 25 _ 1\f..c',_(2.s)(100)_087 ‘_ yH "(115x25)" ' 14.27 )1 . . — — 3.52 c“ c' 20 r" = 0.5; fl = 20° Figure I4..29(c): N, z 14, D x 1.25 Figure 1430(6): N, z 15, D 2 1.25.. SON, z 25;D z 1.25 F: = M = (145)(20) = M ‘ 7H (17.5)(15) -116- 14..28 fl= 20°; ¢’= 15°;r,,=0..5; y= 17‘.5kN/m3;c’ =20kN/m2;H=15m F s (caiculatcd) Jrrchssnma-d) 14.29 p: tan"[%)= 18..43°; 96' = 25°; c" = 12 kN / m3; y= 19 kN lmz; r" = 0‘.25;H=12‘.63 m -117- Fflcalculatcd) From the gxaph, F, x 1.72 1.9 3'" be 1.37 1.6 l .5 1 .6 I .7 Fdassumed) 1.8 -118- 1.9 CHAPTER 15 15.1 Equation(15.11): qu= c'M-t- qu + 53/3111; ¢'=23°;M= 31..6I;Nq = 17.31; N7: 13.7 (Table 15.1) qafl = 37" z %[(14)(31.61) +(o..7 x 16.8)(17..81) + %(16.8)(0.8)(13.7):l = 243 kN / m 2 S 15.2 ¢’= 20°. From Table 15.1, M. = 17.69; N4 = 7.44; N, = 3.64 q; = :4 = %[(14.2)(17.69) + (0.5 x 13.2)(7.44) + %(13..2)(1.2)(3.64)] = 119.6 kN / m 2 15-3 ¢ = 0.. From Iable15..1, NC = 5.7; M, =1;N,,= 0 1 41111" s—? 5' (cu M +q) = %[(2600)(5.7) + (3 x 110)] = 373311;)”:2 7 15.4 ¢’= 20°. From Table 15.2, NC = 11.85; N4 = 3.88; N, = 1.12 Equation (15.16): g; = go’N; + qN; + éyBN’, F 5 gm, = fl = x 14.2](1135) + (0.5 x 132)(3.33) + %(13.2)(12)(1.12)] =53.2kN/m2 15.5 For continuous footing, Equation (15.39): la, = 11;], = 11,, = 1 Also, for vertical load: A“. = 2?,- = )1”. = 1 So, q. = com... + qzlqu. + £731,111, ¢'= 28°. From Tables 15.3 and 15.4, M. = 25.8; N] = 14.72; Ny= 1119. From Table 15.5 -119- D I ,1“, = 1 + 0.2 -—f 12:.[45 + 9i] = 1 + 02(fl] 12:1[45 + E] = 1.29 B 2 0.8 2 D r 2“ = 22d = 1+ 0.1 —f tan[45+ 3] = 1+ 01(3) 10(45 +23] = 1.146 B 2 0.8 2 q" 1 [(14)(25.8)(1.29) + (0.7 x 16.8)(14..72)(1.146) _ _ .. 2 gal] — F - 3 +%(16.8)(0..8)(1..l46)(11.9) ]— 252.1 kN l m S 15.6 ¢'= 20°; N6 = 14.83; M, = 6.4; Ny= 2.871 (Tables 15.3 and 15.4) A“, = 1 + 0.269% tan[45 + = 1.1 19 1.2 2 Aqd = 2,, = 1+ 018—3] m[45+ 329] = 1.06 q 1 (14.2)(14.33)(1-119)+(0.5 x 18.2)(6..4)(l..06) 2 £10: _= F = 3 +%(132)(1.2)(237.1)(1.06) = “0'2 “N1 m S 15.7 ¢ = 0; NC = 5.14; N? = 1;N,,= 0 (Tables 15.3 and 15.4). Shape and inclination factms are equal to 1.. . u I gall = q? 3 FR)»: Nclcd + qlcd] S 5' 7 D! 3 Table 15.5: ,1 d =1+02 —+- =1+02 — =1.171; 2 4 =1 6' B 5 ‘1 gm = %[(2600)(5.14)(I.171) + (3 x 110)(1)] =39951b/ft’ 15.8 Equation (15.12): 1],, = qu + 0..4y’BN,,., ¢'= 35°; Nq = 41.44; Ny= 4541 q = yh + wa- h)=(105 x 2) + (118 — 62..4)(4 ~ 2) = 321-2 113/112 -120— "32 32 Qal1=q =—P:—(qu+0‘-42’BN,) F l — 52 [(3212)(41.44) + (0.4)(118 — 62..4)(5)(45..41)]-I—= 153 kip 9“" - F 1000 15.9 (25' = 259. From Table 1511,1116 = 25.13; M, = 12.72; N, = 8134 y: (l800)(9.81) = 1000 17.66 kN / m3 q" = 1130’Nc + qu + 0.14yavBNr q = ny= (11.2)(17..66) = 21.19 kN I 1112 ya” = 71371 yD + y’(B -— 12)] ~— [Equation (15124)] D== h—Df=2—1..2-—~0..sm = (1980)(9.8l)= 191.42 RN / m3 1000 sat y“ = 1i8'[(17..60)(0.8) + (19.142 — 91'.81)(1.s — 0.13)] '=" 131.19 kN/m3 «1,, ~—— (1 ..3)(23..94)(25;;1'3) 4r (21 ..19)(12..72)- + (0.4)(13.,19)(1.,3)(8.E34) = 11:30.3 kN / m2 Qfl1_;0122§¢92=W=-1221m 5' 15.10 05’ = 40°.. From Table 15.1, .qu7 = 81.27; N}, = 1151.31 qafl = %_ z {Tqu +0n4yBN71= §.[(0.92 x18..1)(81..27) +(0..4)(18_I)(B)(1151-313] 3 S - ' (a) = 4510.1 + 278.33 670 1 I ’ qau = 732— . _ _ _ 0 (b) '4121- B = 0.98 m 15.11 4 = 35°. From Table 151,111,, = 41.44; N,= 45.41 qau = gm x 115)(41.44) + (0.4)(115)(B)(45.41)] = 3177 + 696.38 = 925x 1000 B2 = 3177 4- 696.33 9411 B = 4 ft 1512 ¢' = 35°. From Tables 15.3 and 15.4, M, = 33.3; Nr= 37.152 q" = qu AM” + % “Rwy/1,11,. q = 321.2 kN / m2 (see Problem 15.8). Table 15.5: ' D 1 21¢, = ,1“, = 1+ 0..1[—Ef—]tan[45 + = 1+ 0.1%) tan[45__+ = 1.154 11¢, = 51,, =. 1-+-_0..1[%]tan2[45 + = _1 +06] tanz (45 4+ = 1.369 q" = (321..2)(33..3)(1.154)(1.369) +%(55..6)(5)(37.152)(1..154)(1.369) = 25,056 lb 102 _ (26,056)(5)2 3" (3)0000) ' p _ 15.13 a. For vertical load, Equation (15.4.3): 1 1 qu = (12qu zlquq+-2—,1,,, 1rd yB N}, c' = 0; ¢'= 30°. Tables 15.3 and 15.4: Nq = 18.4; Ny= 15.668 --122— B' = B — 2x = 4.5 — (2)(0.5) = 3.5 11; L’ = 4.5 11 Table 15.5: 2.4, = 2,, = 1+ 0.1[%] tan2[45 J?) = 1+ 018—?)ng (45 +15)1=1.233 D I W, = ,1”, -_~ 1 + 0.1[Ff] tan[45 + = 1 + 0.16%] tan(45 + 15) = 1.173 ,, = (105)(3..5)(I..233)(1.173)(18.4) + (0.5)(1.233)(1.173)(1'05)(3.5)(15.668) =13,9441b/fi2 tu’L’ (13,944)(35)(45) Qau = —— = ——---——-— = 54.9 kip F, (4)0000) b. 'B'=6-—2>'<o.5=5.0fi;L'=6fl ¢' = 25°. Tables 15.3 and 15.4: NC = 20.72; Nq = 10.66; Ny= 6.765 I it“ = 1+ 0.2[3 )tan2[45 + = 1+ 02[%)tan2 (45 +1215)=1.,411 Ag, = 2,, = 1+ 0.1(%)tan2[45 + a 1 + 0.1g)ng (57.5) = 1.205 . D r A“, = 1 + 0.2[Ff] tan[45 + = 1 + 02(45—5)tan(57.5) = 1.283 11 D W, = ,1”, = 1 + 0.1[ Bf )tan[45 = 1 + 0.16% tan(57.5) = 1141 q" = c’Nc 232a, '+ 4114, 11w,sz + % 'yB’erzflxrg So qu = (400)(20.72)(1.411)(1.283) + (120)(4..5)(10.66)(l..205)(l. 141) + (0.5)(120)(5)(6..765)(l.205)(1.141) = 25,709 lb / 1’12: 25.71 kip / fi2 -123- _ (25.7l)(5 x 6) Qall _ = 192.8 kip . l , 0.. q“ = gigs/14d Nq + -2— yB zip/1,31%, B’ = 2.5 — (2 X 0.2) = 2.1 In; L’ = 2.5 m = (2000)(9.s 1) = 19.62 kN/m3 1000 7 W = 42°. Tables 15.3 and 15.4: M, = 85.38; Nr= 139.316 B' ¢' . 2.1 42 119: = 1115 2'” l + 133112 + = 1 + tanz + = D . 24,, = A", = 1+ 0.1[-B—{] mega = 1+ 0,1[%]m{45 +-423)_=1.135 q" = (I9.62)(1.5)(85.38)(I..424)(1.135) +£-(19-62)(2.l)(l39.316)(1..424)(1..135) = 8,700 kN / m2 98“ = “tu L I: (8700)(2.1)(2.5) = 11,419 F, 4 15.14 Equation(IS..55): BF 6 I ' 2 qua.) =qu(P) -—. =(3850) — =23,1001b/fi BP . 1 Qall = A9210?) = [23,100 4 4 J(6)2 =207,9001b =. 207.9 kip 15.15 gm = 248.9 kN / m2. Equation 15.54: gm = gm; gum = 248.9 kN / m2 7! 2 - Aqum [law (2489) _ __ = ___..._ =«260.6s kN 981! F; 3 -124~ CHAPTER 17 17.] Equation (17.4): 2 _ _2 2 _ 2 Do Dl x 100 = (35) (3.375) Area ratio, A, (9%) = 2 2 x 100= 7.54% D,- (3.375) 2._ 2 . 2 h 2 17.2 A, = D° 2D' x100= Wx100=&88% D. (2.875) 17.3 (No.50 = CNN60 = [9.78 (Jrjmw) .. Given: y= 14.5 kN / m3. do Now the following table can prepared. ’ Rounded-011' .Nw 122 +' 20.3[91] 17.4 ¢'=tan-1 -; pa'zloom/m2 'Pa. Average 45 ’ = 36° -125- .n. .n 1 17.5 Equation (l‘7..l I): 17' N60 [023+ng [ J D,(%) = ——-——— ——-- (100) Given 7! = 176 (Nl)60 = CN N60 =[ Depth, 2 (fl) 5 10 15 20 25 .30 05 0.06 98 I 0 9 15.5 kN / 1113.. The following table can now be prepared. 4 73.3 x 73% 68.6 x 69% 73.3 = 73% 68.6 = 69% 71.5 x 72% 2 1+0; JNw ton ' ' _ . 5x 110' 2000 I 9I‘42 z 9 10x11" — 0.55 I 10.32 = 10 2000 ' lellO =0..825 1.1 - - . z 1 17.5 110 2.5120—-62..4 __= [( x )+ ( )12000 1.035 I 0.983 8.845 % 9 1“035+5020-624)=14.” 1235 = 13 5020—624) - . 1....179 —~—————-—-—=1.323 1 .3. =10 + 2000 - 0 32 ~126v 17.7 a. Average value of (NI )60 z 1.3 b.. Equation (15.50): 3.283 +1 2 S, gamma) 2 11198(N1)50(“§28—B) Fd D E. = 1 + 0.3%?) = 1 + 0%?) = 1.248 . 2 (1.11....) =(11-98)(13)[£2§)£)—fi] (1.241%?) = 258.1 kN / m2 (3.28)(2) [qc ] 17.8 Equation (17.25): p“ = 7.6429133;26 . Use pa z 100 kN / m2. 17.9 Equation(17.23): cu = ‘10]; “a .. N. x 18.3 c = 920—(8)(18) u = 42.4 mm2 18.3 17.10 From Equations (17.20) and (17.21): 11,: 2 to 34.. = 2 to 30.05) = 410 to 615 kN I m2 17.1 1 From Equation (17.29): Recover ratio, R = (%)( 100) = 75% -127- ...
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This note was uploaded on 11/25/2010 for the course CIVIL CE333 taught by Professor Dr.ghamdi during the Spring '10 term at Alabama State University.

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SOLUTION MANUAL OF PRINCIPLES OF GEOTECHNICALL ENGINEERRIG_6ED

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