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SOLUTIONS MANUAL AN INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY

PRINCIPLES OF FOUNDATION ENGINEERING, TENTH EDITION BRAJA M. DAS

Principles of

FOUNDATION NGINEERING

Principles of

FOUNDATION ENGINEERING Tenth Edition

Tenth Edition

Braja M. Das

INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY

Principles of Foundation Engineering Tenth Edition

BRAJA M. DAS Dean Emeritus, California State University, Sacramento, California, USA

ISBN: 978-0-357-68466-5

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Contents Chapter 2 ............................................................................................................................ 1 Chapter 3 .......................................................................................................................... 11 Chapter 4 .......................................................................................................................... 19 Chapter 5 .......................................................................................................................... 25 Chapter 6 .......................................................................................................................... 37 Chapter 7 .......................................................................................................................... 49 Chapter 8 .......................................................................................................................... 55 Chapter 9 .......................................................................................................................... 69 Chapter 10 ........................................................................................................................ 75 Chapter 11 ........................................................................................................................ 79 Chapter 12 ........................................................................................................................ 95 Chapter 13 ...................................................................................................................... 107 Chapter 14 ...................................................................................................................... 113 Chapter 15 ...................................................................................................................... 125 Chapter 16 ...................................................................................................................... 137 Chapter 17 ...................................................................................................................... 151

Chapter 2

2.1

d.  

c.  

(87.5)(9.81)  17.17 kN /m 3 (1000)(0.05)

 1 w



17.17  14.93 kN /m 3 1  0.15

a. Eq. (2.12):  d 

14.93 

Gs w 1 e

(2.68)(9.81) ; e = 0.76 1 e

b. Eq. (2.6): n 

e 0.76   0.43 1  e 1  0.76

e. From Eq. (2.14): S

2.2

Vw wGs  (0.15)(2.68)     100  53% Vv e 0.76 

a. From Eqs. (2.11) and (2.12), it can be seen that

d 

 1 w



20.1  16.48 kN/m3 1  0.22

b.  d  16.48 kN/m3 

Gs w Gs (9.81)  1 e 1 e

Eq. (2.15): e  wGs  (0.22)(Gs ) . So 16.48 

9.81Gs ; Gs  2.67 1  0.22Gs

2.3

a. Eq. (2.6):

n

e 0.81   0.45 1  e 1  0.81

b. Eqs. (2.7), (2.14):

S

wGs  (0.21)(2.68)    100  69.5% e 0.81 

c. Eq. (2.11):



Gs w (1  w) (2.68)(9.81)(1  0.21)   17.58 kN /m3 1 e 1  .0.81

d. Eq. (2.12):



2.4

Gs w (2.68)(9.81)   14.53 kN /m3 1 e 1  .0.81

a. Eq. (2.12):  d 

Gs w 1 e

Eq. (2.15): Gs 

So,

e w

e   w w d    1 e 13.5 

b. Eq. (2.6): n 

(e)(9.81) ; e  0.98 (0.36)(1  e)

e 0.98   0.495  0.5 1  e 1  .0.98

c. Eq. (2.14): Gs 

e 0.98   2.72 w 0.36

d. Eq. (2.13):  sat   d (1  w)  (13.5)(1 0.36)  18.36 kN /m3 2 © 2024 Cengage Learning®. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.5

emax  e emax  emin

Eq. (2.23): Dr 

0.91  e ; e  0.63 0.91  0.48

0.65 

Gs w (2.67)(9.81)   16.07 kN /m3 1 e 1  .0.63

Eq. (2.12):  d 

Eq. (2.13):    d (1  w)  (16.07)(1  0.1)  17.68 kN/m3 2.6

Gs w (1  w) 1 e

Eq. (2.11):  

17 

(2.66)(9.81)(1  0.08) ; e  0.658 1 e emax  e emax  emin

Eq. (2.23): Dr 

emax  0.658 ; emax  1.045 emax  0.4

0.6 

Eq. (2.12):  d  2.7

Gs w (2.66)(9.81)   12.76 kN /m 3 1  emax 1  1.045

Soil A: A-1-a(0) Soil B: A-2-6(1) GI  0.01( F200  15)( PI  10)  0.01(33  15)(13  10)  0.54  1

Soil C: A-7-5(19) GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)  (72  35)[0.2  0.005(56  40)]  0.01(72  15)(25  10)  18.91  19

Soil D: A-4(5) GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)  (64  35)[0.2  0.005(35  40)]  0.01(64  15)(9  10)  4.585  5

Soil E: A-2-7(1) GI  0.01( F200  15)( PI  10)  0.01(30  15)(14  10)  0.6  1

Soil F: A-6(5) GI  ( F200  35)[0.2  0.005( LL  40)]  0.01( F200  15)( PI  10)  (55  35)[0.2  0.005(35  40)]  0.01(55  15)(14  10)  5.1  5

2.8

Soil A:

From Table 2.9 and Figure 2.9, symbol is SM Gravel portion = 100  92 = 8% < 15% From Figure 2.10 group name is silty sand

Soil B:

From Table 2.9 and Figure 2.9, symbol is SM Gravel portion = 100 100 = 0% < 15% From Figure 2.10 group name is silty sand

Soil C:

From Table 2.9 and Figure 2.9, symbol is MH Percent passing No. 200 sieve is 72 (> 30%) Percent sand = 100  72 = 28 Percent gravel = 100  100 = 0 From Figure 2.11, group name is sandy elastic silt

Soil D:

From Table 2.9 and Figure 2.9, symbol is ML Percent passing No. 200 sieve is 64 (> 30%) Percent gravel = 100 95 = 5 Percent sand = 9564 = 31 Group name is sandy silt (Figure 2.11)

Soil E:

From Table 2.9 and Figure 2.9, symbol is SM Gravel portion = 100  100 = 0 (< 15%) From Figure 2.10, group name is silty sand

Soil F:

From Table 2.9 and Figure 2.9, symbol is CL Percent passing No. 200 sieve is 55 (> 30%) 4

Percent sand = 100 55 = 45 Percent gravel = 100  100 = 0 From Figure 2.11, group name is sandy lean clay

2.9

Eq. (2.38):

k1 e13 (1  e2 )  k2 e23 (1  e1 ) 0.14 0.563 (1  0.79)  k2  0.343 cm /s k2 0.793 (1  0.56)

2.10

e e 0.92  0.72 log k  log ko  o  log(5.4  106 )  0.5eo (0.5)(0.92)

k  1.9×106 cm / s

2.11

At A:  = 0 u=0

 = 0 At B:  d (sand) 

Gs w (2.65)(9.81)   17.33 kN/m3 1 e 1  0.5

  (2)(17.33)  34.66 kN /m 2 u0      u  34.66  0  34.66 kN /m 2 At C:  sat(clay) 

(Gs  e) w (2.65  0.6)(9.81)   19.93 kN/m 2 1 e 1  0.6

  (2)(17.33)  (2)(19.93)  74.52 kN /m 2 u  (2)(9.81)  19.62 kN /m 2

   74.52  19.62  54.9 kN /m 2

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At D:  sat(clay) 

Gs w (1  w) (2.75)(9.81)(1  0.36)   18.44 kN/m3 1  wGs 1  (0.36)(2.75)

  (2)(17.33)  (2)(19.93)  (3)(18.44)  129.84 kN /m 2 u  (5)(9.81)  49.05 kN /m 2

   129.84  49.05  80.79 kN /m 2 2.12

Eq. (2.52): icr 

Gs  1 1 e

In densest state: icr 

In loosest state:

2.66  1  1.17 1  0.42

icr 

2.66  1  0.84 1  0.97

Range: 1.17 to 0.84 2.13

Eq. (2.56): Cc  0.009( LL  10)  0.009(41  10)  0.279 Eq. (2.67) Sc 

2.14

Cs 

       (2.6)(0.279) H c Cc  120  log  o log     0.052 m  52 mm 1  eo 1  1.3  82    o 

Cc 0.279   0.07 4 4

Eq. (2.71): Sc 

       Cs H c  C H log c  c c log  o  1  eo  o 1  eo   c 

(0.07)(2.6)  95  (0.279)(2.6)  120  log    log   1  1.3 1  1.3  82   95   0.037 m  37 mm 

2.15

3 a.  o  (2)[ d (sand) ]  (2)[ sat (sand)   w ]    [ sat (clay)   w ] 2  (2)(17.33)  (2)(19.93  9.81)  (1.5)(18.44  9.81)  67.85 kN /m2

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b. Eq. (2.62):

Cc 

c. Eq. (2.67): Sc 

e1  e2 0.905  0.815   0.299  2 200 log log 100  1

       H c Cc log  o  1  eo   o 

eo  wGs  (0.36)(2.75)  0.99

Sc  2.16

(3)(0.299)  115  log    0.1033 m  103.3 mm 1  0.99  67.85 

For 50% consolidation [Eq. (2.80)]: Tv 

  U (%) 



   (0.5)  0.197 4  100  4 2

Eq. (2.75):

Tv 

cv t H2

0.197 

2.17

(5.6 mm 2/ min)t  316, 607 min  219.9 days (3 1000 mm) 2

Eq. (2.75): Tv 

cv t H2

For 60% consolidation, Tv  0.287 (Figure 2.26). 0.287 

cv (6 min) ; cv  29.9 mm 2 /min (25) 2

For U = 50%, Tv = 0.197 (Figure 2.26)

0.197 

(29.9 mm 2 / min)t  2.45  1000    2  

; t  9887 min  6.87 days

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2.18

U

30  0.5 60

Tv (1) 

cv (1)t

Tv (2) 

cv 2t (2)(t )   8 106 t 2 2 H 2  11000    2  

2 1



(2)(t )  2  1000    2  

 2  106 t

So, Tv (1)  0.25Tv (2) . For the trial and error procedure, the following table can now be prepared.

Tv (1)

Tv (2)

U 1a

U 2a

U1 H1  U 2 H 2 U H1  H 2

0.05

0.2

0.26

0.51

0.34

0.1

0.4

0.36

0.7

0.473

0.125

0.5

0.4

0.76

0.52

0.385

0.73

0.5

0.1125 0.45 From Figure 2.26

So, Tv (1)  0.1125  2 106 t  56, 250 min  39.06 days 2.19

Normally consolidated clay; so c′ = 0. Eq. (2.93):

    1   3 tan 2  45   2     (115  230)  115 tan 2  45   ;    30° 2  2.20

Normally consolidated clay; soc′ = 0. Eq. (2.93)

 

 1   3 tan 2  45 

 

28  2 2   104 tan  45    288 kN /m 2 2  

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2.21

Eq. (2.93):

 

 1   3 tan 2  45 

 

     2c tan  45   2 2 

      Test 1: 329.2  82.8 tan 2  45    2c tan  45   2 2  

Eq. (a)

      Test 2: 558.6  165.6 tan 2  45    2c tan  45   2 2  

Eq. (b)

From Eqs. (a) and (b),    28°; c  30 kN /m2 2.22

Normally consolidated clay, so c   0; c   0 Refer to the figure below:

 65    24.8°  155 

  sin 1 

 3   3  u f  90  38  52 kN/m2 1  220  38  182 kN/m2  65    33.75°  117 

   sin 1 

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Chapter 3

3.1

Do2 - Di2 32 - 2.8742 Eq. (3.3): AR (%) = ´100 = ´100 = 8.96% Di2 2.8742

3.2 Depth from ground surface (m)

N60

cu (kN/m 2 ) [Eq. (3.8b)]

3.0

76.5

4.5

107.3

6.0

107.3

7.5

116.8

9.0

126.0

Average cu = 106.78 kN/m2

s o¢ (MN/m 2 )

OCR [Eq. (3.9)]

1 [(1.5)(16.5) + (1.5)(19 - 9.81)] 5.51 1000 = 0.03854 1 0.03854 + (1.5)(16.8 - 9.81) 1000 6.46 = 0.0490 1 0.0490 + (1.5)(16.8 - 9.81) 5.65 1000 = 0.0595 1 0.0595 + (1.5)(16.8 - 9.81) 1000 5.48 = 0.07 1 0.07 + (1.5)(16.8 - 9.81) 5.35 1000 = 0.0805 Average OCR = 5.69

3.3 Depth (m) 1.5 3.0 4.5 6.0 7.5 9.0

s o¢ (kN/m2) 18 × 1.5 = 27 18 × 3.0 = 54 18 × 4.5 = 81 18 × 6.0 = 108 108 + (1.5)(20.2 – 9.81) = 123.6 123.6 + (1.5)(20.2 – 9.81) = 139.2

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0.5

é ù ê ú 1 ú ê ; pa » 100 kN/m 2 Eq. (3.13): CN = ê æ s o¢ ö ú êç ÷ ú ëê è pa ø ûú Depth (m)

s o¢ (kN/m2)

N60

1.5 6 27 3.0 8 54 4.5 9 81 6.0 8 108 7.5 13 123.6 9.0 14 139.2 a Rounded off to nearest whole number 3.4

(N1)60a

1.92 1.36 1.11 0.96 0.9 0.85

12 11 10 8 12 12

From Problem 3.3, the average value of

1 ( N1 )60 = (12 + 11 + 10 + 8 + 12 + 12) = 10.83 » 11 6 Eq. (3.31b): f ¢ = 15.4( N1 )60 + 20 = (15.4)(11) + 20 = 33° 3.5

From Problem 3.3

f¢ (deg)

Depth (m)

s o¢ (kN/m )

pa (kN/m )

N60

[Eq. (3.30)]

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

6 8 9 8 13 14

34.7 34.9 34.0 31.4 34.9 34.9

Average ϕ' = 34.1° ≈ 34°

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3.6 Depth (m)

s o¢ (kN/m )

pa (kN/m )

1.5 3.0 4.5 6.0 7.5 9.0

27 54 81 108 123.6 139.2

100 100 100 100 100 100

Dr (%) [Eq. (3.22)]

N60

6 50.6 8 51.7 9 49.7 8 43.2 13 52.8 14 52.7 Average Dr = 50.12% ≈ 50%

3.7 1.7

s o¢

Depth (m)

(kN/m )

po (kN/m )

1.5 3.0 4.5 6.0 7.5 9.0

26.4 52.8 79.2 105.6 132.0 158.4

100 100 100 100 100 100

æ 0.06 ö ç 0.23 + ÷ D50 ø è 0.133 0.133 0.133 0.133 0.133 0.133

N60

Dr (%) [Eq. (3.23)]

5 52.9 11 55.5 14 51.1 18 50.2 16 42.3 21 44.3 Average Dr ≈ 49.4%

3.8 Depth (ft)

γ (lb/ft3)

s o¢

(lb/in.2) 10 106 7.36 15 106 11.04 20 106 14.72 25 118 18.82 30 118 22.92 35 118 27.02 40 118 31.12 a rounded to nearest whole number

pa(lb/in.2)

N60

f ¢ (deg)a [Eq. (3.30)]

14.7 14.7 14.7 14.7 14.7 14.7 14.7

7 9 11 16 18 20 22

34 34 35 37 36 36 36

1 Average f ¢ = (34 + 34 + 35 + 37 + 36 + 36 + 36) = 35.4° » 36° 7

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