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Answers to Problems for Principles of Foundation Engineering, 8th Edition – Braja M. Das

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The document contains various equations and calculations related to soil mechanics, including parameters such as soil classifications, consistency, and properties derived from laboratory tests. It discusses the behavior and physical characteristics of different soil types, and presents formulas for determining soil properties based on empirical models. Additional resources are referenced for accessing complete data and contact information is provided for further inquiries.

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1 Chapter 2 2.1 d. 3 kN/m 17.17    ) 05 . 0 )( 1000 ( ) 81 . 9 )( 5 . 87 ( c. 3 kN/m 14.93        15 . 0 1 17 . 17 1 w d a. Eq. (2.12): 0.76        e e e G w s ; 1 ) 81 . 9 )( 68 . 2 ( 14.93 . 1 b. Eq. (2.6): 0.43      76 . 0 1 76 . 0 1 e e n e. Eq. (2.14): 53%           ) 100 ( 76 . 0 ) 68 . 2 )( 15 . 0 ( e G w V V S s v w 2.2 a. From Eqs. (2.11) and (2.12), it can be seen that, 3 kN/m 16.48        22 . 0 1 1 . 20 1 w d b. e G e G s w s        1 ) 81 . 9 ( 1 kN/m 16.48 3 Eq. (2.14): ). )( 22 . 0 ( s s G wG e   So, 2.67    s s s G G G ; 22 . 0 1 81 . 9 48 . 16 s m t b 9 8 @ g m a i l . c o m You can access complete document on following URL. Contact me if site not loaded https://unihelp.xyz/ Contact me in order to access the whole complete document - Email: smtb98@gmail.com WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 - Telegram: https://t.me/solutionmanual
Chapter 2 2 2 2.3 a. 0.55          e e e w G w s ; 1 ) 12 . 0 1 )( 4 . 62 )( 65 . 2 ( 119.5 . 1 ) 1 ( b. 0.355    55 . 0 1 55 . 0 n c. 57.8%     100 55 . 0 ) 65 . 2 )( 12 . 0 ( e G w S s d. 3 lb/ft 106.7        12 . 0 1 5 . 119 1 w d 2.4 a. w e Gs  . 0.97                    e e e e w e w d ; 1 ) 4 . 62 ( 36 . 0 85.43 . 1 ) ( b. 0.49      97 . 0 1 97 . 0 1 e e n c. 2.69    36 . 0 97 . 0 w e Gs d. 3 lb/ft 115.9          97 . 0 1 ) 4 . 62 )( 97 . 0 69 . 2 ( 1 ) ( sat e e G w s
Chapter 2 3 2.5 From Eqs. (2.11) and (2.12): 3 lb/ft 108 08 . 0 1 64 . 116    d Eq. (2.12): 53 . 0 ; 1 ) 4 . 62 )( 65 . 2 ( 108 ; 1        e e e G w s d Eq. (2.23): 0.94         max max max min max max ; 44 . 0 53 . 0 82 . 0 e e e e e e e Dr 3 lb/ft 85.2        94 . 0 1 ) 4 . 62 )( 65 . 2 ( 1 max (min) e G w s d 2.6 Refer to Table 2.7 for classification. Soil A: A-7-6(9) (Note: PI is greater than LL 30.) GI = (F200 – 35)[0.2 + 0.005(LL – 40)] + 0.01(F200 – 15)(PI – 10) = = (65 – 35)[0.2 + 0.005(42 – 40)] + 0.01(65 – 15)(16 – 10) 9.3  9 Soil B: A-6(5) GI = = (55 – 35)[0.2 + 0.005(38 – 40)] + 0.01(55 – 15)(13 – 10) 5.4  5 Soil C: A-3-(0) Soil D: A-4(5) GI = = (64– 35)[0.2 + 0.005(35 – 40)] + 0.01(64 – 15)(9 – 10) 4.585 ≈ 5 Soil E: A-2-6(1) GI = 0.01(F200 – 15)(PI – 10) = 0.01(33 – 15)(13 – 10) = 0.54 ≈ 1 Soil F: A-7-6(19) (PI is greater than LL  30.) GI = = (76– 35)[0.2 + 0.005(52 – 40)] + 0.01(76 – 15)(24 – 10) 19.2 ≈ 19
Chapter 2 4 4 2.7 Soil A: Table 2.8: 65% passing No. 200 sieve. Fine grained soil; LL = 42; PI = 16 Figure 2.5: ML Figure 2.7: Plus No. 200 > 30%; Plus No. 4 = 0 % sand > % gravel – sandy silt Soil B: Table 2.8: 55% passing No. 200 sieve. Fine grained soil; LL = 38; PI = 13 Figure 2.5: Plots below A-line – ML Figure 2.7: Plus No. 200 > 30% % sand > % gravel – sandy silt Soil C: Table 2.8: 8% passing No. 200 sieve. % sand > % gravel – sandy soil – SP Figure 2.6: % gravel = 100 – 95 = 5% < 15% – poorly graded sand Soil D: Table 2.8: 64% passing No. 200 sieve Fine grained soil; LL = 35, PI = 9 Figure 2.5 – ML Figure 2.7: % sand (31%) > % gravel (5%) – sandy silt Soil E: Table 2.8: 33% passing No. 200 sieve; 100% passing No. 4 sieve. Sandy soil; LL = 38; PI = 13 Figure 2.5: Plots below A-line – SM Figure 2.6: % gravel (0%) < 15% – silty sand Soil F: Table 2.8: 76% passing No. 200 sieve; LL = 52; PI = 24 Figure 2.5: CH Figure 2.7: Plus No. 200 is 100 – 76 = 24% % gravel > % gravel – fat clay with sand
Chapter 2 5 2.8 43 . 0 1 117 ) 4 . 62 )( 68 . 2 ( 1 e ; 1            d w s w s d G e G Eq. (2.37):                                       43 . 0 1 43 . 0 63 . 0 1 63 . 0 22 . 0 ; 1 1 3 3 2 2 3 2 1 3 1 2 1 k e e e e k k ; k2 = 0.08 cm/s 2.9 From Eq. (2.41): n n n e e e e k k ) 6316 . 0 ( 1667 . 0 ; 9 . 1 2 . 1 2 . 2 9 . 2 10 91 . 0 10 2 . 0 or ; 1 1 6 6 2 1 1 2 2 1                                      898 . 3 1996 . 0 778 . 0 ) 6316 . 0 log( ) 1667 . 0 log(      n cm/s 10 0.075 6 -                                ) 10 216 . 0 ( 9 . 1 9 . 0 1 10 216 . 0 2 . 1 ) 2 . 2 )( 10 2 . 0 ( ) 1 ( 6 998 . 3 3 6 998 . 3 6 1 1 1 e e C k e e k C n n
Chapter 2 6 6 2.10 The flow net is shown. /m/s m 10 17.06 3 6 -                            8 ) 4 )( 25 . 5 ( 10 10 5 . 6 So, m. 25 . 5 75 . 1 7 cm/s; 10 5 . 6 2 4 2 1 max 4 q H H h k 2.11 a. 7825 . 0 3 2 7825 . 0 3 2 10 6 . 0 1 6 . 0 ) 2 . 0 ( 4622 . 2 ) 1 ( 4622 . 2                           e e D k cm/s 0.041  b. 32 . 2 6 . 0 3 32 . 2 10 6 . 0 3 ) 2 . 0 ( 2 . 0 4 . 0 6 . 0 1 6 . 0 ) 35 ( ) ( 1 35                           D C e e k u cm/s 0.171 
Chapter 2 7 2.12 3 dry(sand) kN/m 16.84 55 . 0 1 ) 81 . 9 )( 66 . 2 ( 1        e G w s   3 sat(sand) kN/m 81 . 0 2 48 . 0 1 ) 48 . 0 66 . 2 ( 81 . 9 1           e e G w w s 3 sat(clay ) kN/m 18.55 ) 74 . 2 )( 3478 . 0 ( 1 ) 3478 . 0 1 )( 81 . 9 )( 74 . 2 ( 1 ) 1 (          s w s wG w G At A: σ = 0; u = 0; σ = 0 At B: σ = (16.84)(3) = 50.52 kN/m2 u = 0 σ = 50.52 kN/m2 At C: σ = σB + (20.81)(1.5) = 50.52 + 31.22 = 81.74 kN/m2 u = (9.81)(1.5) = 14.72 kN/m2 σ = 81.74 – 14.72 = 67.02 kN/m2 At D: σ = σC + (18.55)(5) = 81.74 + 92.75 = 174.49 kN/m2 u = (9.81)(6.5) = 63.77 kN/m2 σ = 174.49 – 63.77 = 110.72 kN/m2 2.13 Eq. (2.54): Cc = 0.009(LL – 10) = 0.009(42 – 10) = 0.288 Eq. (2.65): mm 87.2                  110 155 log 82 . 0 1 mm) 1000 7 . 3 )( 288 . 0 ( log 1 o o o c c c e H C S 2.14 Eq. (2.69): mm 56.69                                           128 155 log 82 . 0 1 ) 3700 )( 288 . 0 ( 110 128 log 75 . 0 1 mm) 3700 ( 5 288 . 0 log 1 log 1 c o o c c o c o c s c e H C e H C S
Chapter 2 8 8 2.15 a. Eq. (2.53): 0.392 150 300 log 792 . 0 91 . 0 log 1 2 2 1                      e e Cc From Eq. (2.65): o o o c c c e H C S        log 1 Using the results of Problem 2.12, 2 kN/m 87 . 88 ) 81 . 9 55 . 18 ( 2 5 ) 81 . 9 81 . 20 ( 5 . 1 ) 84 . 16 )( 3 (        o 953 . 0 ) 74 . 2 )( 3478 . 0 (    s o wG e mm 194.54           87 . 88 50 87 . 88 log 953 . 0 1 mm) 5000 )( 392 . 0 ( c S b. Eq. (2.73): 2 H t C T v v  . For U = 50%, Tv = 0.197 (Table 2.11). So, days 609 sec 10 5262 ; cm) 500 ( 10 36 . 9 197 . 0 4 2 4       t t 2.16 a. Eq. (2.53): 0.377                      120 360 log 64 . 0 82 . 0 log 1 2 2 1 e e Cc b. 0.736                      2 2 1 2 2 1 ; 120 200 log 82 . 0 0.377 ; log e e e e Cc
Chapter 2 9 2.17 Eq. (2.73): 2 H t C T v v  . For 60% consolidation, Tv = 0.286 (Table 2.11). Lab time: min 6 49 min 8 6 1   t /min in. 0788 . 0 ; ) 5 . 1 ( 6 49 286 . 0 2 2         v v C C Field: U = 50%; Tv = 0.197 days 6.25 min 9000 ; 2 12 10 ) 0788 . 0 ( 197 . 0 2           t t 2.18 5 . 0 60 30   U t t H t C T v v 6 2 2 1 ) 1 ( ) 1 ( 10 2 2 1000 2 ) )( 2 (             t t H t C T v v 6 2 2 2 ) 2 ( ) 2 ( 10 8 2 1000 1 ) )( 2 (             So, Tv(1) = 0.25Tv(2). The following table can be prepared for trial and error procedure. Tv(1) Tv(2) U1 U2 U H H H U H U    2 1 2 2 1 1 (Figure 2.22) 0.05 0.10 0.125 0.1125 0.2 0.4 0.5 0.45 0.26 0.36 0.40 0.385 0.51 0.70 0.76 0.73 0.34 0.473 0.52 0.50 So, Tv(1) = 0.1125 = 2  10-6 t; t = 56,250 min = 39.06 days
Chapter 2 10 10 2.19 Eq. (2.84): . 2 H t C T c v c  tc = 60 days = 60  24  60  60 sec; 2 2  H m = 1000 mm. 0415 . 0 ) 1000 ( ) 60 60 24 60 )( 10 8 ( 2 3        c T After 30 days: 0207 . 0 ) 1000 ( ) 60 60 24 30 )( 10 8 ( 2 3 2         H t C T v v From Figure 2.24 for Tv = 0.0207 and Tc = 0.0415, U = 5%. So Sc = (0.05)(120) = 6 mm After 100 days: 069 . 0 ) 1000 ( ) 60 60 24 100 )( 10 8 ( 2 3 2         H t C T v v From Figure 2.24 for Tv = 0.069 and Tc = 0.0415, U  23%. So Sc = (0.23)(120) = 27.6 mm 2.20          N S 1 tan Normal force, N (lb) Shear force, S (lb)          N S 1 tan (deg) 50 110 150 43.5 95.5 132.0 41.02 40.96 41.35 From the graph,   41 2.21 Normally consolidated clay; c = 0.            2 45 tan2 3 1 ;  38             ; 2 45 tan 30 96 30 2
Chapter 2 11 2.22            2 45 tan2 3 1 ;  30             ; 2 45 tan 20 40 20 2 2.23 c = 0. Eq. (2.91): 2 kN/m 387.8                     2 28 45 tan 140 2 45 tan 2 2 3 1 2.24 Eq. (2.91):                      2 45 tan 2 2 45 3 1 c                    2 45 tan 2 2 45 tan 140 368 2 c (a)                    2 45 tan 2 2 45 tan 280 701 2 c (b) Solving Eqs. (a) and (b),  = 24; c = 12 kN/m2 2.25  25                             13 32 13 32 sin sin 1 3 1 3 1 1                  3 1 3 1 1 sin 2 3 2 3 lb/in. 5 . 7 5 . 5 13 ; lb/in. 5 . 26 5 . 5 32          34             5 . 7 5 . 26 5 . 7 5 . 26 sin 1 Normally consolidated clay; c = 0 and c = 0 2.26            2 45 tan2 3 1 . 2 2 1 kN/m 305.9 2 20 45 tan 150          
Chapter 2 12 12 2 kN/m 61.9                       u u u ; 2 28 45 tan 150 9 . 305 ; 2 45 tan 2 2 3 1 2.27 a. ) log( 6 . 1 4 . 0 10 26 50 D C D u r              30.7 )] 13 . 0 )[log( 6 . 1 ( ) 1 . 2 )( 4 . 0 ( ) 53 . 0 )( 10 ( 26 b. b ae    1 327 . 2 09 . 0 21 . 0 097 . 0 101 . 2 097 . 0 101 . 2 15 85                    D D a 081 . 0 ) 327 . 2 )( 398 . 0 ( 845 . 0 398 . 0 845 . 0       a b  33.67     081 . 0 ) 68 . 0 )( 327 . 2 ( 1

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Answers to Problems for Principles of Foundation Engineering, 8th Edition – Braja M. Das

  • 1. 1 Chapter 2 2.1 d. 3 kN/m 17.17    ) 05 . 0 )( 1000 ( ) 81 . 9 )( 5 . 87 ( c. 3 kN/m 14.93        15 . 0 1 17 . 17 1 w d a. Eq. (2.12): 0.76        e e e G w s ; 1 ) 81 . 9 )( 68 . 2 ( 14.93 . 1 b. Eq. (2.6): 0.43      76 . 0 1 76 . 0 1 e e n e. Eq. (2.14): 53%           ) 100 ( 76 . 0 ) 68 . 2 )( 15 . 0 ( e G w V V S s v w 2.2 a. From Eqs. (2.11) and (2.12), it can be seen that, 3 kN/m 16.48        22 . 0 1 1 . 20 1 w d b. e G e G s w s        1 ) 81 . 9 ( 1 kN/m 16.48 3 Eq. (2.14): ). )( 22 . 0 ( s s G wG e   So, 2.67    s s s G G G ; 22 . 0 1 81 . 9 48 . 16 s m t b 9 8 @ g m a i l . c o m You can access complete document on following URL. Contact me if site not loaded https://unihelp.xyz/ Contact me in order to access the whole complete document - Email: smtb98@gmail.com WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 - Telegram: https://t.me/solutionmanual
  • 2. Chapter 2 2 2 2.3 a. 0.55          e e e w G w s ; 1 ) 12 . 0 1 )( 4 . 62 )( 65 . 2 ( 119.5 . 1 ) 1 ( b. 0.355    55 . 0 1 55 . 0 n c. 57.8%     100 55 . 0 ) 65 . 2 )( 12 . 0 ( e G w S s d. 3 lb/ft 106.7        12 . 0 1 5 . 119 1 w d 2.4 a. w e Gs  . 0.97                    e e e e w e w d ; 1 ) 4 . 62 ( 36 . 0 85.43 . 1 ) ( b. 0.49      97 . 0 1 97 . 0 1 e e n c. 2.69    36 . 0 97 . 0 w e Gs d. 3 lb/ft 115.9          97 . 0 1 ) 4 . 62 )( 97 . 0 69 . 2 ( 1 ) ( sat e e G w s
  • 3. Chapter 2 3 2.5 From Eqs. (2.11) and (2.12): 3 lb/ft 108 08 . 0 1 64 . 116    d Eq. (2.12): 53 . 0 ; 1 ) 4 . 62 )( 65 . 2 ( 108 ; 1        e e e G w s d Eq. (2.23): 0.94         max max max min max max ; 44 . 0 53 . 0 82 . 0 e e e e e e e Dr 3 lb/ft 85.2        94 . 0 1 ) 4 . 62 )( 65 . 2 ( 1 max (min) e G w s d 2.6 Refer to Table 2.7 for classification. Soil A: A-7-6(9) (Note: PI is greater than LL 30.) GI = (F200 – 35)[0.2 + 0.005(LL – 40)] + 0.01(F200 – 15)(PI – 10) = = (65 – 35)[0.2 + 0.005(42 – 40)] + 0.01(65 – 15)(16 – 10) 9.3  9 Soil B: A-6(5) GI = = (55 – 35)[0.2 + 0.005(38 – 40)] + 0.01(55 – 15)(13 – 10) 5.4  5 Soil C: A-3-(0) Soil D: A-4(5) GI = = (64– 35)[0.2 + 0.005(35 – 40)] + 0.01(64 – 15)(9 – 10) 4.585 ≈ 5 Soil E: A-2-6(1) GI = 0.01(F200 – 15)(PI – 10) = 0.01(33 – 15)(13 – 10) = 0.54 ≈ 1 Soil F: A-7-6(19) (PI is greater than LL  30.) GI = = (76– 35)[0.2 + 0.005(52 – 40)] + 0.01(76 – 15)(24 – 10) 19.2 ≈ 19
  • 4. Chapter 2 4 4 2.7 Soil A: Table 2.8: 65% passing No. 200 sieve. Fine grained soil; LL = 42; PI = 16 Figure 2.5: ML Figure 2.7: Plus No. 200 > 30%; Plus No. 4 = 0 % sand > % gravel – sandy silt Soil B: Table 2.8: 55% passing No. 200 sieve. Fine grained soil; LL = 38; PI = 13 Figure 2.5: Plots below A-line – ML Figure 2.7: Plus No. 200 > 30% % sand > % gravel – sandy silt Soil C: Table 2.8: 8% passing No. 200 sieve. % sand > % gravel – sandy soil – SP Figure 2.6: % gravel = 100 – 95 = 5% < 15% – poorly graded sand Soil D: Table 2.8: 64% passing No. 200 sieve Fine grained soil; LL = 35, PI = 9 Figure 2.5 – ML Figure 2.7: % sand (31%) > % gravel (5%) – sandy silt Soil E: Table 2.8: 33% passing No. 200 sieve; 100% passing No. 4 sieve. Sandy soil; LL = 38; PI = 13 Figure 2.5: Plots below A-line – SM Figure 2.6: % gravel (0%) < 15% – silty sand Soil F: Table 2.8: 76% passing No. 200 sieve; LL = 52; PI = 24 Figure 2.5: CH Figure 2.7: Plus No. 200 is 100 – 76 = 24% % gravel > % gravel – fat clay with sand
  • 5. Chapter 2 5 2.8 43 . 0 1 117 ) 4 . 62 )( 68 . 2 ( 1 e ; 1            d w s w s d G e G Eq. (2.37):                                       43 . 0 1 43 . 0 63 . 0 1 63 . 0 22 . 0 ; 1 1 3 3 2 2 3 2 1 3 1 2 1 k e e e e k k ; k2 = 0.08 cm/s 2.9 From Eq. (2.41): n n n e e e e k k ) 6316 . 0 ( 1667 . 0 ; 9 . 1 2 . 1 2 . 2 9 . 2 10 91 . 0 10 2 . 0 or ; 1 1 6 6 2 1 1 2 2 1                                      898 . 3 1996 . 0 778 . 0 ) 6316 . 0 log( ) 1667 . 0 log(      n cm/s 10 0.075 6 -                                ) 10 216 . 0 ( 9 . 1 9 . 0 1 10 216 . 0 2 . 1 ) 2 . 2 )( 10 2 . 0 ( ) 1 ( 6 998 . 3 3 6 998 . 3 6 1 1 1 e e C k e e k C n n
  • 6. Chapter 2 6 6 2.10 The flow net is shown. /m/s m 10 17.06 3 6 -                            8 ) 4 )( 25 . 5 ( 10 10 5 . 6 So, m. 25 . 5 75 . 1 7 cm/s; 10 5 . 6 2 4 2 1 max 4 q H H h k 2.11 a. 7825 . 0 3 2 7825 . 0 3 2 10 6 . 0 1 6 . 0 ) 2 . 0 ( 4622 . 2 ) 1 ( 4622 . 2                           e e D k cm/s 0.041  b. 32 . 2 6 . 0 3 32 . 2 10 6 . 0 3 ) 2 . 0 ( 2 . 0 4 . 0 6 . 0 1 6 . 0 ) 35 ( ) ( 1 35                           D C e e k u cm/s 0.171 
  • 7. Chapter 2 7 2.12 3 dry(sand) kN/m 16.84 55 . 0 1 ) 81 . 9 )( 66 . 2 ( 1        e G w s   3 sat(sand) kN/m 81 . 0 2 48 . 0 1 ) 48 . 0 66 . 2 ( 81 . 9 1           e e G w w s 3 sat(clay ) kN/m 18.55 ) 74 . 2 )( 3478 . 0 ( 1 ) 3478 . 0 1 )( 81 . 9 )( 74 . 2 ( 1 ) 1 (          s w s wG w G At A: σ = 0; u = 0; σ = 0 At B: σ = (16.84)(3) = 50.52 kN/m2 u = 0 σ = 50.52 kN/m2 At C: σ = σB + (20.81)(1.5) = 50.52 + 31.22 = 81.74 kN/m2 u = (9.81)(1.5) = 14.72 kN/m2 σ = 81.74 – 14.72 = 67.02 kN/m2 At D: σ = σC + (18.55)(5) = 81.74 + 92.75 = 174.49 kN/m2 u = (9.81)(6.5) = 63.77 kN/m2 σ = 174.49 – 63.77 = 110.72 kN/m2 2.13 Eq. (2.54): Cc = 0.009(LL – 10) = 0.009(42 – 10) = 0.288 Eq. (2.65): mm 87.2                  110 155 log 82 . 0 1 mm) 1000 7 . 3 )( 288 . 0 ( log 1 o o o c c c e H C S 2.14 Eq. (2.69): mm 56.69                                           128 155 log 82 . 0 1 ) 3700 )( 288 . 0 ( 110 128 log 75 . 0 1 mm) 3700 ( 5 288 . 0 log 1 log 1 c o o c c o c o c s c e H C e H C S
  • 8. Chapter 2 8 8 2.15 a. Eq. (2.53): 0.392 150 300 log 792 . 0 91 . 0 log 1 2 2 1                      e e Cc From Eq. (2.65): o o o c c c e H C S        log 1 Using the results of Problem 2.12, 2 kN/m 87 . 88 ) 81 . 9 55 . 18 ( 2 5 ) 81 . 9 81 . 20 ( 5 . 1 ) 84 . 16 )( 3 (        o 953 . 0 ) 74 . 2 )( 3478 . 0 (    s o wG e mm 194.54           87 . 88 50 87 . 88 log 953 . 0 1 mm) 5000 )( 392 . 0 ( c S b. Eq. (2.73): 2 H t C T v v  . For U = 50%, Tv = 0.197 (Table 2.11). So, days 609 sec 10 5262 ; cm) 500 ( 10 36 . 9 197 . 0 4 2 4       t t 2.16 a. Eq. (2.53): 0.377                      120 360 log 64 . 0 82 . 0 log 1 2 2 1 e e Cc b. 0.736                      2 2 1 2 2 1 ; 120 200 log 82 . 0 0.377 ; log e e e e Cc
  • 9. Chapter 2 9 2.17 Eq. (2.73): 2 H t C T v v  . For 60% consolidation, Tv = 0.286 (Table 2.11). Lab time: min 6 49 min 8 6 1   t /min in. 0788 . 0 ; ) 5 . 1 ( 6 49 286 . 0 2 2         v v C C Field: U = 50%; Tv = 0.197 days 6.25 min 9000 ; 2 12 10 ) 0788 . 0 ( 197 . 0 2           t t 2.18 5 . 0 60 30   U t t H t C T v v 6 2 2 1 ) 1 ( ) 1 ( 10 2 2 1000 2 ) )( 2 (             t t H t C T v v 6 2 2 2 ) 2 ( ) 2 ( 10 8 2 1000 1 ) )( 2 (             So, Tv(1) = 0.25Tv(2). The following table can be prepared for trial and error procedure. Tv(1) Tv(2) U1 U2 U H H H U H U    2 1 2 2 1 1 (Figure 2.22) 0.05 0.10 0.125 0.1125 0.2 0.4 0.5 0.45 0.26 0.36 0.40 0.385 0.51 0.70 0.76 0.73 0.34 0.473 0.52 0.50 So, Tv(1) = 0.1125 = 2  10-6 t; t = 56,250 min = 39.06 days
  • 10. Chapter 2 10 10 2.19 Eq. (2.84): . 2 H t C T c v c  tc = 60 days = 60  24  60  60 sec; 2 2  H m = 1000 mm. 0415 . 0 ) 1000 ( ) 60 60 24 60 )( 10 8 ( 2 3        c T After 30 days: 0207 . 0 ) 1000 ( ) 60 60 24 30 )( 10 8 ( 2 3 2         H t C T v v From Figure 2.24 for Tv = 0.0207 and Tc = 0.0415, U = 5%. So Sc = (0.05)(120) = 6 mm After 100 days: 069 . 0 ) 1000 ( ) 60 60 24 100 )( 10 8 ( 2 3 2         H t C T v v From Figure 2.24 for Tv = 0.069 and Tc = 0.0415, U  23%. So Sc = (0.23)(120) = 27.6 mm 2.20          N S 1 tan Normal force, N (lb) Shear force, S (lb)          N S 1 tan (deg) 50 110 150 43.5 95.5 132.0 41.02 40.96 41.35 From the graph,   41 2.21 Normally consolidated clay; c = 0.            2 45 tan2 3 1 ;  38             ; 2 45 tan 30 96 30 2
  • 11. Chapter 2 11 2.22            2 45 tan2 3 1 ;  30             ; 2 45 tan 20 40 20 2 2.23 c = 0. Eq. (2.91): 2 kN/m 387.8                     2 28 45 tan 140 2 45 tan 2 2 3 1 2.24 Eq. (2.91):                      2 45 tan 2 2 45 3 1 c                    2 45 tan 2 2 45 tan 140 368 2 c (a)                    2 45 tan 2 2 45 tan 280 701 2 c (b) Solving Eqs. (a) and (b),  = 24; c = 12 kN/m2 2.25  25                             13 32 13 32 sin sin 1 3 1 3 1 1                  3 1 3 1 1 sin 2 3 2 3 lb/in. 5 . 7 5 . 5 13 ; lb/in. 5 . 26 5 . 5 32          34             5 . 7 5 . 26 5 . 7 5 . 26 sin 1 Normally consolidated clay; c = 0 and c = 0 2.26            2 45 tan2 3 1 . 2 2 1 kN/m 305.9 2 20 45 tan 150          
  • 12. Chapter 2 12 12 2 kN/m 61.9                       u u u ; 2 28 45 tan 150 9 . 305 ; 2 45 tan 2 2 3 1 2.27 a. ) log( 6 . 1 4 . 0 10 26 50 D C D u r              30.7 )] 13 . 0 )[log( 6 . 1 ( ) 1 . 2 )( 4 . 0 ( ) 53 . 0 )( 10 ( 26 b. b ae    1 327 . 2 09 . 0 21 . 0 097 . 0 101 . 2 097 . 0 101 . 2 15 85                    D D a 081 . 0 ) 327 . 2 )( 398 . 0 ( 845 . 0 398 . 0 845 . 0       a b  33.67     081 . 0 ) 68 . 0 )( 327 . 2 ( 1