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Principles Of Foundation Engineering, 8th Edition Solution Manual

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    An ’s Solutions Manual to Accompany PRINCIPLES OF FOUNDATION ENGINEERING, 8 TH EDITION, SI EDITION BRAJA M. DAS

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    Contents Chapter Page 2.................................................................................................................................1 3...............................................................................................................................13 4...............................................................................................................................25 5...............................................................................................................................37 6...............................................................................................................................49 7...............................................................................................................................55 8...............................................................................................................................69 9...............................................................................................................................75 10..............................................................................................................................95 11............................................................................................................................107 12............................................................................................................................113 13............................................................................................................................121 14............................................................................................................................133 15............................................................................................................................147 16............................................................................................................................157 Note that there are no end of chapter questions for Chapter 1.

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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 2.3 a. 0.55          e e e w G w s ; 1 ) 12 . 0 1 )( 81 . 9 )( 65 . 2 ( 18.79 . 1 ) 1 (

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    2 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 kN/m 16.78        12 . 0 1 79 . 18 1 w d 2.4 a. w e G s  . 0.97                    e e e e w e w d ; 1 ) 81 . 9 ( 36 . 0 18.43 . 1 ) ( b. 0.49      97 . 0 1 97 . 0 1 e e n c. 2.69    36 . 0 97 . 0 w e G s d. 3 kN/m 18.23          97 . 0 1 ) 81 . 9 )( 97 . 0 69 . 2 ( 1 ) ( sat e e G w s 2.5 From Eqs. (2.11) and (2.12): 3 kN/m 98 . 16 08 . 0 1 34 . 18     d Eq. (2.12): 53 . 0 ; 1 ) 81 . 9 )( 65 . 2 ( 16.98 ; 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 D r 3 kN/m 13.4        94 . 0 1 ) 81 . 9 )( 65 . 2 ( 1 max (min) e G w s d

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    3 2.6 Refer to Table 2.7 for classification. Soil A: A-7-6(9) ( Note : PI is greater than LL  30.) GI = (F 200 – 35)[0.2 + 0.005(LL – 40)] + 0.01(F 200 – 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 = = (6 4 – 35)[0.2 + 0.005(35 – 40)] + 0.01(64 – 15)(9 – 10) 4. 5 8 5 ≈ 5 Soil E: A-2-6(1) GI = 0.01(F 200 – 15)(PI – 10) = 0.01(3 3 – 15)(1 3 – 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)( 2 4 – 10) 19.2 ≈ 19 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

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    4 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 2.8 43 . 0 1 39 . 18 ) 81 . 9 )( 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 ; k 2 = 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

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    5 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 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 m ax 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 

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    6 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/m 2 u = 0 σ  = 50.52 kN/m 2 At C : σ = σ B + (20.81)(1.5) = 50.52 + 31.22 = 81.74 kN/m 2 u = (9.81)(1.5) = 14.72 kN/m 2 σ  = 81.74 – 14.72 = 67.02 kN/m 2 At D : σ = σ C + (18.55)(5) = 81.74 + 92.75 = 174.49 kN/m 2 u = (9.81)(6.5) = 63.77 kN/m 2 σ  = 174.49 – 63.77 = 110.72 kN/m 2 2.13 Eq. (2.54): C c = 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

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    7 2.15 a. Eq. (2.53): 0.392 150 300 log 792 . 0 91 . 0 log 1 2 2 1                      e e C c 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%, T v = 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 C c b. 0.736                      2 2 1 2 2 1 ; 120 200 log 82 . 0 0.377 ; log e e e e C c 2.17 Eq. (2.73): 2 H t C T v v  . For 60% consolidation, T v = 0.286 (Table 2.11). Lab time: min 6 49 min 8 6 1   t

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    8 /min cm 5057 . 0 ; ) cm 8 . 3 ( 6 49 286 . 0 2 2         v v C C Field: U = 50%; T v = 0.197 days 6.29 min 9060 ; cm 2 100 05 . 3 ) 5057 . 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, T v (1) = 0.25 T v (2) . The following table can be prepared for trial and error procedure. T v (1) T v (2) U 1 U 2 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, T v (1) = 0.1125 = 2  10 -6 t ; t = 56,250 min = 39.06 days 2.19 Eq. (2.84): . 2 H t C T c v c  t c = 60 days = 60  24  60  60 sec; 2 2  H m = 1000 mm.

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    9 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 T v = 0.0207 and T c = 0.0415, U = 5%. So S c = (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 T v = 0.069 and T c = 0.0415, U  23%. So S c = (0.23)(120) = 27.6 mm 2.20          N S 1 tan Normal force, N (N) Shear force, S (N)          N S 1 tan (deg) 222.4 489.3 667.2 193.5 424.8 587.1 41.02 40.96 41.35 From the graph,   41  2.21 Normally consolidated clay; c  = 0.            2 45 tan 2 3 1 ;  38             ; 2 45 tan 207 4 . 662 207 2 2.22            2 45 tan 2 3 1 ;  30             ; 2 45 tan 138 276 138 2

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    10 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/m 2 2.25  25                             7 . 89 8 . 220 7 . 89 8 . 220 sin sin 1 3 1 3 1 1                  3 1 3 1 1 sin 2 3 2 3 kN/m 75 . 51 95 . 37 7 . 89 ; kN/m 85 . 182 95 . 37 8 . 220          34             75 . 51 85 . 182 75 . 51 85 . 182 sin 1 Normally consolidated clay; c = 0 and c  = 0 2.26            2 45 tan 2 3 1 . 2 2 1 kN/m 305.9 2 20 45 tan 150           2 kN/m 61.9                       u u u ; 2 28 45 tan 150 9 . 305 ; 2 45 tan 2 2 3 1

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    11 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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    12

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    13 Chapter 3 3.1 Eq. (3.3): 8.96%        100 73 73 2 . 76 100 (%) 2 2 2 2 2 2 i i o R D D D A 3.2 Depth from ground surface (m) N 60 c u (kN/m 2 ) [Eq. (3.8)] o  (MN/m 2 ) OCR [Eq. 3.9)] 3.0 5 92.4 03854 . 0 )] 81 . 9 19 )( 5 . 1 ( ) 5 . 16 )( 5 . 1 [( 1000 1    5.51 4.5 8 129.6 0490 . 0 ) 81 . 9 8 . 16 )( 5 . 1 ( 1000 1 03854 . 0    6.46 6.0 8 92.4 0595 . 0 ) 81 . 9 8 . 16 )( 5 . 1 ( 1000 1 0490 . 0    5.65 7.5 9 141.1 07 . 0 ) 81 . 9 8 . 16 )( 5 . 1 ( 1000 1 0595 . 0    5.48 9.0 10 152.2 0805 . 0 ) 81 . 9 8 . 16 )( 5 . 1 ( 1000 1 07 . 0    5.35 Average c u = 121.5 kN/m 2 Average OCR = 5.69

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