SOLUTIONS MANUAL for Principles of Foundation Engineering 10th Edition by Braja M. Das-stamped by StudyGuide

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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3.9

1 Between depths 20 ft and 30 ft, average N 60 = (11 + 16 + 18) = 15. 3 Eq. (3.32): Es = paa N 60 = (2000)(10)(15) = 300, 000 lb/ft 2 » 2083 lb /in.2

3.10

Eq. (3.21)

é ù æs¢ ö Dr (%) = 12.2 + 0.75 ê 222 N 60 + 2311 - 711(OCR) - 779 ç o ÷ - 50Cu2 ú è pa ø ë û

0.5

At a depth of 3.0 m:

é æ 55 ö 2ù Dr (%) = 12.2 + 0.75 ê 222(9) + 2311 - (711)(2) - (779) ç ÷ - (50)(2.8) ú è 100 ø ë û = 46.3%

0.5

At a depth of 4.5 m:

é æ 82 ö 2ù Dr (%) = 12.2 + 0.75 ê 222(11) + 2311 - (711)(2) - (779) ç ÷ - (50)(2.8) ú è 100 ø ë û = 48.2%

0.5

At a depth of 6.0 m:

é æ 98 ö 2ù Dr (%) = 12.2 + 0.75 ê 222(12) + 2311 - (711)(2) - (779) ç ÷ - (50)(2.8) ú è 100 ø ë û = 48.9% 1 Average Dr = (46.3 + 48.2 + 48.9) = 48.7% » 48% 3 3.11

Eq. (3.37): cu =

0.5

T K

7p d 3 (7)(p )(0.0635)3 = Eq. (3.39): K = 6 6

= 93.8 ´10-5 cu (VST) =

0.051 kN × m = 54.4 kN/m2 -5 3 93.8 ´ 10 m

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Eqs. (3.37) and (3.43a) cu (corrected) = l cu (VST) = [1.7 - 0.54 log(PI)](54.4) = [1.7 - 0.54 log(46 - 21)](54.4) = 51.4 kN /m 2

3.12

Eq. (3.45): OCR = b

cu (field)

s o¢

s o¢ = 59.5 kN/m2 (from Problem 3.2) PI = 46 – 21= 25

b = 22(PI)-0.48 = (22)(25)-0.48 = 4.69

æ 54.4 ö OCR = (4.69) ç ÷ = 4.29 è 59.5 ø 3.13

7p d 3 (7)(p )(2)3 From Eq. (3.39): K = = = 29.32 in.3 6 6 = 0.017 ft 3 cu (VST) =

T 23 = = 1352.9 lb /ft 2 K 0.017

From Eqs. (3.42) and (3.43a):

cu (corrected) = [1.7 - 0.54log(PI)](1352.9)] = [1.7 - 0.54log(58 - 29)](1352.9) = 1231.6 lb /ft 2 3.14

Eq. (3.48): b =

1 1 = = 4.18 0.08 + 0.0055(PI) 0.08 + 0.0055(29)

Eq. (3.45): OCR = b

cu (field) æ 1352.9 ö = (4.18) ç ÷ = 4.22 s o¢ è 1340 ø

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3.15 Depth (m)

s o¢ (MN/m2)

qc (MN/m2)

f ¢ (deg)

1.5

0.024

2.06

42.05

3.0

0.048

4.23

42.15

4.5

0.072

6.01

41.98

6.0

0.096

8.18

42.03

7.5

0.120

9.97

41.93

9.0

0.144

12.42

42.07

f¢ (average) ≈ 42° 3.16

Note OCR = 1, Qc = 1 Dr (%)

3.17

Depth (m)

qc (kN/m2)

s o¢ (kN/m2)

[Eq. (3.54)]

1.5

2060

37.1

3.0

4230

44.7

4.5

6010

48.2

6.0

8180

52.3

7.5

9970

120

54.6

9.0

12420

144

58.3

σo = (2)(18) + (20)(4) = 116 kN/m2, qc = 0.8 × 1000 = 800 kN/m2 Eq. (3.59): cu =

qc - s o 800 - 116 = » 45.6 kN /m2 Nk 15

s o¢ = (2)(18) + (20 - 9.81)(4) = 76.76 kN/m 2 1.01

æ q -so ö Eq. (3.63): OCR = 0.37 ç c ÷ è s o¢ ø

1.01

æ 800 - 116 ö = (0.37) ç ÷ è 76.76 ø

» 3.37

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3.18.

Eq. (3.64):

46 + 180 öæ 326.5 - 42.4 ö æ Dp ö æ E p = 2(1 + µs )(Vo + vm ) ç ÷ = (2)(1 + 0.5) ç 535 + ÷ç ÷ 2 è Dv ø è øè 180 - 46 ø = 4121.6 kN /m 2 3.19

KD =

po - uo 280 - (9.81)(8 - 3) = = 2.43 s o¢ 95

æK ö Eq. (3.71): K o = ç D ÷ è 1.5 ø

0.47

æ 2.43 ö - 0.6 = ç ÷ è 1.5 ø

0.47

- 0.6 = 0.65

Eq. (3.72): OCR = (0.5K D )16 = (0.5 ´ 2.43)16 = 1.37

Eq. (3.75): Es = (1 - µ s2 ) ED = (1 - µ s2 )(34.7)( p1 - po )

= (1 - 0.352 )(34.7)(350 - 280) = 2131 kN/m2

3.20

KD =

po - uo 260 - (4)(9.81) = = 3.2 s o¢ (2)(14.5) + (4)(19.8 - 9.81)

Eq. (3.77): f ¢ = 31 +

3.21

Eq. (3.83): v =

105 =

KD 3.2 = 31 + = 38.2° 0.236 + 0.066 K D 0.236 + (0.066)(3.2)

1 - µs ; µ = 0.32 g (1 - 2µ s )(1 + µ s ) s g

Es 0.68 ´ ; E = 14,136 kN/m 2 18 (0.36)(1.32) s 9.81

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3.22

A time-distance plot is shown.

1 15.24 ´ 10-3 Slope of 0a = = v1 7.5

7.5 ´ 103 v1 = = 492 m /s (top layer) 15.24 v2 =slope of ab =

10 ´ 103 » 1390 m /s 7.19

v3 = slope of bc =

20 ´ 103 = 3390 m /s 5.9

xc = 7.5 m

Z1 =

1 v2 - v1 1 1390 - 492 xc = (7.5) = 2.6 m 2 v2 + v1 2 1390 + 492

Eq. (3.85): Z 2 =

2Z1 v32 - v12

2Z v 2 - v 2 ù é v v ù 1é êTi 2 - 1 3 1 ú ê 3 2 ú ; Ti 2 » 20 ´10-3 s 2ê (v3 )(v1 ) ú ê v32 - v22 ú ë ûë û

(v3 )(v1 )

(2)(2.6) (3390)2 - (492)2 = = 0.0105 (3390)(492)

v3v2

(3390)(1390)

v -v 2 3

2 2

So, Z 2 =

(3390) 2 - (1390) 2

= 1524

1 (0.02 - 0.0105)(1524) = 7.24 m 2 18

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Chapter 4

4.1

a. Eq. (4.2): RC =

g d (field) æ 16.35 ö = (100) = 90.4% g d (max) çè 18.08 ÷ø

b. Eq. (4.3): RC =

A 1 - Dr (1 - A)

γ d (min) γ d (max)

14.46 = 0.8 18.05

Hence, 0.904 =

0.8 1 - Dr (1 - 0.8)

Dr = 0.575 = 57.5% 4.2

a. Eq. (4.13): For standard Proctor test, E = 600 kN·m/m3 wopt = (1.99 - 0.165ln E ) PI = [1.99 - 0.165ln(600)](14) = 13.08%

Eqs. (4.14), (4.15) and (4.16):

γ d (max) = L - Mwopt = (14.34 + 1.195ln E ) - (-0.19 + 0.073ln E ) wopt = [14.34 + 1.195ln(600)] - [-0.19 + 0.073ln(600)](13.08) = 21.98 - 3.62 = 18.36 kN /m 3 b. Eq. (4.13): For modified Proctor test, E=2700 kN·m/m3 wopt = (1.99-0.165lnE)PI=[1.99-(0.165) ln(2700)](14)=9.61% Eqs. (4.14),(4.15) and(4.16):

g d (max) = (14.34 + 1.195ln E ) - (-0.19 + 0.073ln E ) wopt = [14.34 + (1.195) ln(2700)] - [ -0.19 + (0.073) ln(2700)](9.631) = 23.78 - 3.72 = 20.06 kN /m 3

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a. Eq. (4.11): wopt = (1.95 - 0.38log E)PL=(1.95-0.38log600)(18) = 16.09%

4.3

g d (max) = 22.68e

-0.0183 wopt (%)

= 22.68e( -0.0183)(16.09) = 16.9 kN /m 3

b. wopt = (1.95- 0.38log E)PL = (1.95 - 0.38log2700)(18) = 11.63%

g d (max) = 22.68e( -0.0183)(11.63) = 18.33 kN/m3 14 ö æ 3 Compacted g d = 18.4 kN/m3 . g = 18.4(1 + w) = (18.4) ç1 + ÷ = 20.98 kN/m . è 100 ø

4.4

Total weight of moist soil needed = (20.98)(20,000) = 419,600 kN Total volume of soil to be excavated = For the compacted fill, g d =

4.5

419, 600 = 23, 573 m 3 17.8

Gsg w Gsg w = kN/m3 1 + 0.6 1.6

The total dry weight of fill needed, Ws =

8, 000Gsg w = 5000Gsg w kN/m3 1.6

Borrow pit

Void ratio

Dry unit Weight, γd (kN/m3)

Volume of soil to be excavated = Ws/γd (m3)

Cost ($)

0.82

Gsγw/1.82

9,100

9,100 × 9 = 81,900

0.91

Gsγw/1.91

9,550

9,550 × 7 = 66,850

0.95

Gsγw/1.95

9,750

9,750 × 8 = 78,000

0.75

Gsγw/1.75

8,750

8,750 × 11 = 96,250

Borrow Pit B is the cheapest. 4.6

Eq. (4.19):

é ù é 3 ù 3 1 1 1 1 S N = 1.7 ê + + + = (1.7) + + + ú ê ú 2 2 2 2 2 2 ë (2) (0.7) (0.65) û ë ( D50 ) ( D20 ) ( D10 ) û = 3.86 Rating¾Excellent (SN between 0 and 10)

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4.7

é ù 3 1 1 Eq. (4.19): S N = (1.7) ê + + + ú = 3.15 ¾ Excellent 2 2 2 ë (3.2) (0.91) (0.72) û

4.8

a. Eq. (4.21): S( p ) =

b. Tv =

s o¢ + Ds (¢p ) (0.27)(8) 110 + 75 Cc H c log = log = 0.241 m 1 + e0 s o¢ 2.02 110

cv t . For 80% consolidation, Tv = 0.567 (Chapter 2). H2

0.567 =

(0.52)(t ) æ8ö ç ÷ è2ø

; t = 17.45 months

c. For t2 = 12 months [Eq. (4.25)], Tv =

cvt2 (0.52)(12) = = 0.39 2 H2 æ8ö ç ÷ è2ø

From Figure 4.19 for Tv = 0.39, U » 52%

Ds (¢p ) 75 = = 0.682. Refer to Eq. (4.24) s o¢ 110

0.52 =

Ds (¢ f ) = 1.445 ìï é Ds (¢ f ) ù üï Ds (¢p ) log í1 + 0.682 ê1 + úý êë Ds (¢p ) úû þï îï log(1 + 0.682)

;

Ds (¢ f ) = (1.445)(75) = 108.4 kN/m2 4.9

a. S( p ) =

s o¢ + Ds (¢p ) (0.3)(5) Cc H c 72 + 58 log = log = 0.192 m » 192 mm. 1 + eo s o¢ 2 72

b. Tv = 0.567 =

(0.1 cm2 / min)t æ 5 ´100 cm ö ç ÷ 2 è ø

t = 354,375 min = 246.09 days = 8.2 months

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c. For t2 = 12 months, Tv =

cvt2 (0.1 cm2 / min)(12 ´ 30 ´1440) = = 0.829 2 H2 æ 500 cm ö ç ÷ è 2 ø

From Figure 4.19, U » 83% From Figure 4.17, for U = 83% and

Ds (¢p )

s o¢

58 = 0.806, the value of 72

Ds (¢ f ) » 0.3, Ds (¢p )

Ds (¢ f ) = (0.3)(58) = 17.4 kN/m2 4.10

a. Eq. (4.30): n =

de 4.5 = =9 2rw (2)(0.25)

r 0.35 Eq. (4.31): S = s = = 1.4 rw 0.25 Eq. (4.32): Tr =

cvr t2 (0.3)(6) = = 0.089 de2 (4.5)2

Eq. (4.29): 2 kh æ n 2 - S 2 ö n2 ænö 3 S m= 2 ln ç ÷ - + + ç ÷ ln S n - S 2 è S ø 4 4n 2 k s è n 2 ø

æ 92 - 1.42 ö 92 1.42 æ 9 ö 3 ln + + (2) ç ÷ ln(1.4) = 1.82 ç ÷ 2 92 - 1.42 è 1.4 ø 4 (4)(92 ) 9 è ø

æ -8Tr ö æ -8 ´ 0.089 ö Eq. (4.28): U r = 1 - exp ç ÷ = 0.347 = 32.4% ÷ = 1 - exp ç è 1.82 ø è m ø b. Tv =

cvt2 (0.3)(6) = = 0.089 2 H dr2 æ9ö ç ÷ è2ø

Tv = 0.089 =

p æ Uv % ö

ç ÷ ; U v = 33.7% 4 è 100 ø 22

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Eq. (4.27): U v ,r = 1 - (1 - U r )(1 - U v ) = 1 - (1 - 0.324)(1 - 0.337) = 0.552 = 55.2%

4.11

cv = cvr = 0.042 ft 2 /day = 15.33 ft 2 /year

Vertical drainage:

cv t2 15.33t2 = = 0.613t2 H 2 æ 10 ö2 ç ÷ è 2ø

Eq. (a)

Radial drainage: Eq. (4.32):

cvr t2 15.33t2 = = 0.426t2 de2 62

Eq. (b)

de (6)(12) = = 4.5 2rw (2)(8)

t2 (yr)

Tv [Eq. (a)]

Uv (Table 2.12)

Tr [Eq. (b)]

Ur [ Eqs. (4.28) and (4.34)]

Ur,v [Eq. (4.27)]

0.2

0.123

0.395

0.085

0.553

0.73

0.4

0.245

0.555

0.170

0.80

0.91

0.8

0.49

0.76

0.341

0.96

0.99

1.0

0.613

0.82

0.426

0.98

0.997

4.12

Eq. (4.41): Tc =

Tv =

cv tc (0.015)(30) = = 0.015 H2 (5.5) 2

cv t2 (0.015)(50) = = 0.0248 H2 (5.5)2

From Figure 4.23, U v » 15% For the sand drain, n =

de 2.5 = = 17.86 2rw (2)(0.07)

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Eq. (4.37):

Tr =

Trc =

cvr tc (0.015)(30) = = 0.072 de2 (2.5)2

cvr t2 (0.015)(50) = = 0.12 de2 (2.5)2

Eq. (4.36): U r = 1 -

Eq. (4.38): A =

1 [exp( ATrc ) - 1]exp(- ATrc ) ATrc

2 m

n2 3n2 - 1 17.862 (3)(17.86)2 - 1 ln( n ) = ln(17.86) = 2.14 n2 - 1 4n2 17.862 - 1 (4)(17.86) 2

2 = 0.935 2.14

ATrc = (0.935)(0.072) = 0.067 Ur = 1-

1 [exp(0.067) - 1]exp(-0.067) = 0.033 = 3.3% 0.067

U v ,r = 1 - (1 - U r )(1 - U v ) = 1 - (1 - 0.033)(1 - 0.15) = 0.178 = 17.8%

S( p ) =

Cc H s ¢ + Ds ¢ (0.3)(5.5) æ 80 + 70 ö log o = log ç ÷ = 0.256 m = 256 mm 1 + eo s o¢ 1 + 0.76 è 80 ø

Settlement at 50 days = (256)(0.178) » 45.6 mm

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Chapter 5 5.1

1 a. Eq. (5.10): qu = c¢N c + qN q + g BNg 2

f′ = 28°. Table 5.1: Nc = 31.61; Nq = 17.81; Nγ = 13.7 qall =

qu 1 é 1 ù = ê(400)(31.61) + (110)(3)(17.81) + (110)(3)(13.7) ú FS 4 ë 2 û

= 5195 lb /ft 2 b. f′ = 35°. Table 5.1: Nc = 57.75; Nq= 41.44; Nγ = 45.41

qall =

qu 1 é 1 ù = ê(0)(57.75) + (17.8)(1.2)(41.44) + (17.8)(1.5)(45.41) ú FS 4 ë 2 û

= 372.8 kN /m 2 c. Table 5.1: f′ = 30°; Nq = 22.46; Nγ = 19.13 Eq. (5.19), with c′ = 0:

qu 1 = (qN q + 0.4g BNg ) FS FS 1 = [(2 ´ 16.5)(22.46) + (0.4)(16.5)(3)(19.13)] = 280 kN /m 2 4

qall =

5.2

Eq. (5.19); c′ = 0: qall =

qall =

qu 1 = (qN q + 0.4g BNg ) FS FS

Qall 1805 = 2 B2 B

From Table 5.1 for f′ = 34°, Nq = 36.5; Nγ = 38.04

1805 1 = [(1.5)(15.9)(36.5) + (0.4)(15.9)( B)(38.40)] B2 3 By trial and error, B = 2 m 25 ®

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5.3

1 a. Eq. (5.30): qu = c¢N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BNg Fg s Fg d Fg i 2 Vertical load, so Fci, Fqi, and Fγi are equal to 1. Continuous foundation, so Fcs, Fqs, and Fγs are equal to 1.

f′ = 28°. Table 5.3: N c = 25.8; N q = 14.72; Ng = 16.72. Table 5.4:

Fqd = 1 + 2 tan f ¢(1 - sin f ¢) 2

Fcd = Fqd -

1 - Fqd N c tan f ¢

æ 3ö = 1 + 2 tan 28(1 - sin 28) 2 ç ÷ = 1.299 B è 3ø

= 1.299 -

1 - 1.299 = 1.321 (25.8)(tan 28)

Fg d = 1 qall = =

qu 1 æ 1 ö = ç c¢N c Fcd + qN q Fqd + g BNg Fg d ÷ FS FS è 2 ø 1é 1 ù (400)(25.8)(1.321) + (110 ´ 3)(14.72)(1.299) + (110)(3)(16.72)(1) ú ê 4ë 2 û

= 5675 lb /ft 2 b. c′ = 0

qall =

1 æ 1 ö ç qN q Fqd + g BNg Fg d ÷ FS è 2 ø

f′ = 35°. Table 5.3: Nq = 33.3; Nγ = 48.03. Table 5.4:

Fqd = 1 + 2 tan f ¢(1 - sin f ¢)2

æ 1.2 ö = 1 + 2 tan 35(1 - sin 35) 2 ç ÷ = 1.2 B è 1.5 ø

Fg d = 1 1é 1 ù qall = ê(1.2 ´17.8)(33.3)(1.2) + (17.8)(1.5)(48.03)(1) ú = 373.7 kN /m 2 4ë 2 û 26 ®

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1 c. qu = qN q Fqd Fqs + g BNg Fg d Fg s 2

f′ = 30°. From Table 5.3: Nq = 18.4; Nγ = 22.4. Table 5.3: Fγs = 0.6; Fγd = 1; Fqs = 1 + tanf′ = 1.577

æ2ö = 1 + 0.289 ç ÷ = 1.193 B è3ø

Fqd = 1 + 2 tan f ¢(1 - sin f ¢) 2 qall =

1é 1 ù (2)(16.5)(18.4)(1.193)(1.577) + (16.5)(3)(22.4)(0.6)(1) ú ê 4ë 2 û

= 368.8 kN /m 2 5.4

1 æ 1 ö 2é ¢ ùæ 1 ö Qall = ç ÷ B êc N c Fcs Fcd Fci + qN q Fqs Fqd Fqi + g BN g Fg s Fg d Fg i ú ç ÷ 2 è FS ø ë û è cos 20° ø Note Qall = inclined allowable load.

f′ = 25°. Table 5.3: Nc = 20.72; Nq = 10.66; Nγ = 10.88. Fcs = 1 +

B Nq æ 5 öæ 10.66 ö = 1 + ç ÷ç ÷ = 1.514 L Nc è 5 øè 20.72 ø

Fqs = 1 +

B æ5ö tan f ¢ = 1 + ç ÷ (tan 25) = 1.466 L è5ø

Fg s = 1 - 0.4

B = 1 - 0.4(1) = 0.6 L

Fqd = 1 + 2 tan f ¢(1 - sin f ¢)2

Fcd = Fqd -

1 - Fqd N c tan f ¢

æ3ö = 1 + 0.311ç ÷ = 1.187 B è5ø

= 1.187 -

1 - 1.187 = 1.206 (20.72)(tan 25)

Fg d = 1 2

æ 20 ö Fci = Fqi = ç1 - ÷ = 0.605 è 90 ø 27 ®

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æ b ö æ 20 ö Fg i = ç1 - ÷ = ç1 - ÷ = 0.04 è f ¢ ø è 25 ø

é (600)(20.72)(1.514)(1.206)(0.605) ù 2 æ 1 öæ 5 ö ê ú æ 1 ö = 156.8 kip qall = ç ÷ ç ÷ ê + (115 ´ 3)(10.66)(1.466)(1.187)(0.605) ú ç ÷ è 3 ø è 1000 ø è cos 20° ø êë+ (0.5)(115)(5)(10.88)(0.6)(1)(0.04) úû 5.5

From Eqs. (5.28) and (5.30):

qall(net) =

1 é 1 ù c¢N c Fcs Fcd + qN q Fqs Fqd + g BN g Fg s Fg d - q ú ê FS ë 2 û

Note: ( Fci = Fqi = Fg i = 1)

Qall(net) = qall(net) BL

f′ = 25°. Table 5.3: Nc = 20.72; Nq = 10.66; Nγ = 10.88. Table 5.4:

æ B ö æ Nq ö æ 2 öæ 10.66 ö Fcs = 1 + ç ÷ ç ÷ = 1 + ç ÷ç ÷ = 1.343 è L ø è Nc ø è 3 øè 20.72 ø

æBö æ2ö Fqs = 1 + ç ÷ tan f ¢ = 1 + ç ÷ (tan 25) = 1.31 èLø è3ø æBö æ2ö Fg s = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.73 èLø è3ø Fqd = 1 + 2 tan f ¢(1 - sin f ¢) 2

Fcd = Fqd -

1 - Fqd N c tan f ¢

æ 1.5 ö = 1 + 2 tan 25(1 - sin 25) 2 ç ÷ = 1.233 B è 2 ø

= 1.233 -

1 - 1.233 = 1.257 (20.72)(tan 25)

Fg d = 1

g ¢ = (19.5 - 9.81) = 9.69 kN/m3

28 ®

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q = (1 ´ 17) + (9.69)(0.5) = 21.85 kN/m3 Qall(net) =

(2)(3) é(70)(20.72)(1.343)(1.257) + (21.85)(10.66)(1.31)(1.233) ù ú 3 êë + (0.5)(9.69)(2)(10.88)(0.73)(1) - 21.85 û

= 5760 kN 5.6

1 cʹ = 0. Eq. (5.30): qu = qN q Fqs Fqd + g BNg Fg s Fg d 2

qu B 2 Qall = qall B = FS 2

2 1 æ öB Qall = ç qNq Fqs Fqd + g BNg Fg s Fg d ÷ 2 è ø FS

Eq. (a)

f′ = 30°. Table 5.3: Nq = 18.4; Nγ = 22.4. Table 5.4:

æBö Fqs = 1 + ç ÷ tan f ¢ = 1 + tan 30 = 1.577 èLø æBö Fg s = 1 - 0.4 ç ÷ = 0.6 èLø Fqd = 1 + 2 tan f ¢(1 - sin f ¢) 2

0.577 æ2ö = 1 + 2 tan 30(1 - sin 30) 2 ç ÷ = 1 + B B èBø

Fg d = 1 From Eq. (a):

3330 =

B2 é ù æ 0.577 ö 1 (2)(16.5)(18.4)(1.577) ç1 + ÷ + (16.5)( B)(22.4)(0.6)(1) ú ê 4 ë B ø 2 è û

B3 + 8.64 B 2 + 4.98B = 120.13 B»3m

29 ®

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5.7

From Eq. (5.48):

Ir =

Es 2(1 + µ s )(c¢ + q¢ tan f ¢)

Bö æ æ 1ö q¢ = g ç D f + ÷ = (17) ç1 + ÷ = 25.5 kN/m 2 2ø è è 2ø

Ir =

1400 = 6.38 2(1 + 0.35)(72 + 25.5 tan 20)

From Eq. (5.49):

I r (cr) = =

1ì éæ Bö æ f ¢ öù ü íexp êç 3.3 - 0.45 ÷ cot ç 45 - ÷ ú ý 2î Lø è 2 øû þ ëè 1ì éæ 1ö æ 20 ö ù ü íexp êç 3.3 - 0.45 ÷ cot ç 45 - ÷ ú ý = 40.36 2î 2ø è 2 øû þ ëè

Since Ir(cr) > Ir, use Eqs. (5.50) and (5.52)

éæ (3.07 sin f ¢) log(2 I r ) ö ù ïü Bö ïìæ Fg c = Fqc = exp íç -4.4 + 0.6 ÷ tan f ¢ + êç ÷ú ý Lø 1 + sin f ¢ ø û ïþ ïîè ëè ìæ 1ö éæ (3.07 sin 20) log(2 ´ 6.38) ö ù ü = exp íç -4.4 + 0.6 ÷ tan 20 + êç ÷ ú ý = 0.534 2 1 + sin 20 è ø è øû þ ë î

Fcc = Fqc -

1 - Fqc N c tan f ¢

Fcc = 0.534 -

. For ϕʹ = 20°, Nc = 14.83.

1 - 0.534 = 0.448 14.83tan 20

Now from Eq. (5.47):

1 qu = c¢N c Fcs Fcd Fcc + qN q Fqs Fqd Fqc + g BNg Fg s Fg d Fg c 2 From Table 5.3, for f′ = 20°, Nc = 14.83; Nq = 6.4; Nγ = 5.39.

30 ®

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Table 5.4:

æ Nq ö æ B ö æ 6.4 öæ 1 ö Fcs = 1 + ç ÷ ç ÷ = 1+ ç ÷ç ÷ = 1.216 è 14.83 øè 2 ø è Nc ø è L ø

æBö æ1ö Fqs = 1 + ç ÷ tan f ¢ = 1 + ç ÷ (tan 20) = 1.182 èLø è2ø æBö æ1ö Fg s = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.8 èLø è2ø Fqd = 1 + 2 tan f ¢(1 - sin f ¢)2

Fcd = Fqd -

1 - Fqd N c tan f ¢

æ1ö = 1 + 2 tan 20(1 - sin 20)2 ç ÷ = 1.315 B è1ø

= 1.315 -

1 - 1.315 = 1.373 (14.83)(tan 20)

Fg d = 1 Thus,

qu = (72)(14.83)(1.216)(1.373)(0.448) + (1 ´ 17)(6.4)(1.182)(1.315)(0.534) 1 + (17)(1.0)(5.39)(0.8)(1)(0.534) 2 = 909 kN /m 2 5.8

B′ = B - 2e = 1.5 - (2)(0.15) = 1.2 m; L′ = 1.5 m

1 Eq. (5.60): qu¢ = qN q Fqs Fqd + g BNg Fg s Fg d 2

f′ = 36°. Table 5.3: Nq = 37.75; Nγ = 56.31 æ B¢ ö æ 1.2 ö Fqs = 1 + ç ÷ tan f ¢ = 1 + ç ÷ (tan 36) = 1.58 è L¢ ø è 1.5 ø Fqd = 1 + 2 + tan f ¢(1 - sin f ¢)2

æ 1 ö = 1 + 2 tan 36(1 - sin 36)2 ç ÷ = 1.165 B è 1.5 ø

æ B¢ ö æ 1.2 ö Fg s = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.68 è L¢ ø è 1.5 ø

Fg d = 1 31 ®

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1 qu¢ = (1 ´ 17)(37.75)(1.58)(1.165) + (17)(1.2)(56.31)(0.68)(1) 2 2 = 1571.9 kN/m Qall =

5.9

qu¢ B ¢L ¢ (1571.9)(1.2)(1.5) = = 707.3 kN FS 4

Eq. (5.64): Qall =

f ¢ = 36°.

Qult BL é 1 ù = qN q ( e ) Fqs ( e ) + g BNg ( e ) Fg s ( e ) ú ê 4 4 ë 2 û

e 0.15 = = 0.1. From Figures 5.26 and 5.27, N q ( e ) » 38; Ng ( e ) » 32 B 1.5

Eqs. (5.66) and (5.67):

Fqs ( e ) = 1 æ 2e öB é æ 3 öæ e ö ù æ B ö Fg s ( e ) = 1 + ç - 0.68 ÷ + ê0.43 - ç ÷ç ÷ ú ç ÷ èB øL ë è 2 øè B ø û è L ø

ù æ 1.5 ö æ 1.5 ö é æ3ö = 1 + [(2)(0.1) - 0.68] ç ÷ + ê0.43 - ç ÷ (0.1) ú ç ÷ è 1.5 ø ë è2ø û è 1.5 ø = 1 - 0.48 + 0.28 = 0.8

Qall = 5.10

(1.5)(1.5) é 1 ù (1)(17)(38)(1) + (17)(1.5)(32)(0.8) ú = 546.98 kN » 547 kN ê 4 2 ë û

c ¢ = 0; f ¢ = 35°

1 qu = qN q Fqs Fqd + g BNg Fg s Fg d 2 Table 5.3: Nq = 33.3; Nγ = 48.03.

Fqs = Fg s = 1 Fg d = 1 Fqd = 1 + 2 tan f ¢(1 - sin f ¢)2

æ 0.9 ö = 1 + 2 tan 35(1 - sin 35) 2 ç ÷ = 1.127 B è 1.8 ø

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1 qu = (16)(0.9)(33.3)(1)(1.127) + (16)(1.8)(48.03)(1)(1) 2 = 540.4 + 691.6 = 1232 kN/m 2 Eq. (5.68): RK = 1 -

qu ( e ) qu

k D f 0.9 æeö = = 0.5, a = 1.754; k = 0.8 Eq. (5.69): RK = a ç ÷ . For B 1.8 èBø

1.754(0.12)0.8 = 1 -

qu ( e ) qu

= 1-

qu ( e ) 1232

. qu (eccentric) = 835.8 kN/m 2

Qu = Bqu ( e ) = (1.8)(835.8) = 1504.4 kN /m 5.11

1 qu = qN q Fqs Fqd + g BNg Fg s Fg d 2

q = (2)(105) + (2)(118 - 62.4) = 321.1 lb/ft 2 e=

5 - 2 = 0.5 ft. 2

Table 5.3: f ¢ = 35°; N q = 33.3; Ng = 48.03. Table 5.4: Fqs = Fg s = Fg d = 1

æ4ö Fqd = 1 + 2 tan 35(1 - sin 35) 2 ç ÷ = 1.204 è5ø 1 qu = (321.2)(33.3)(1)(1.204) + (118 - 62.4)(5)(48.03)(1)(1) 2 2 » 19,554 lb/ft

eö é æ æ 0.5 ö ù 2 Eq. (5.72): qu ( e ) = qu ç1 - 2 ÷ = 19,554 ê1 - (2) ç ÷ ú = 15, 643 lb/ft Bø è è 5 øû ë

Qu = Bqu ( e ) = (5)(15, 643) = 78, 215 lb /ft » 78.2 kip /ft

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5.12

70 = 0.156 m; c¢ = 0 450

1 é ù Eq. (5.64): Qult = B 2 ê qN q ( e ) Fqs ( e ) + g ¢BNg ( e ) Fg s ( e ) ú 2 ë û

f ¢ = 30°;

e 0.156 = B B

Qult = (450)(FS)= (450)(6)= 2700 kN

Fqs ( e ) = 1 B=L

é æ 2 ´ 0.156 ö æ 0.156 ö ù 2 Fg s ( e ) = 1 + ç - 0.68 ÷ (1) + ê0.43 - (1.5) ç ÷ ú (1) B è ø è B øû ë 0.234 ö æ 0.312 ö æ = 1+ ç - 0.68 ÷ + ç 0.43 ÷ B ø è B ø è 0.312 0.234 = 0.75 + B B Hence,

ì 1 0.312 0.234 ö ü æ 2700 = B 2 í(1.2 ´16)[ N q ( e ) ](1) + (19 - 9.81)( B)[ Ng ( e ) ] ç 0.75 + ÷ý 2 B B øþ è î é 0.312 0.234 ö ù æ 2700 = B 2 ê19.2 N q ( e ) + (4.595)( B) Ng ( e ) ç 0.75 + ÷ B B ø úû è ë TRIAL AND ERROR: Let B = 2 m;

Eq. (a)

e 0.156 = = 0.078; f ¢ = 30° B B

From Figures 5.26 and 5.27, N q ( e ) » 20; Ng ( e ) » 14 Right-hand size of Eq. (a):

é 0.312 0.234 ö ù æ (2) 2 ê(19.2)(20) + (4.595)(2)(14) ç 0.75 + ÷ 2 2 ø úû è ë = 1942 kN < 2700 kN 34 ®

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Let B = 2.5 m;

e 0.156 = = 0.0624; f ¢ = 30° B 2.5

From Figures 5.26 and 5.27, N q ( e ) » 17; Ng ( e ) » 16 Right-hand side of Eq. (a):

é 0.312 0.234 ö ù æ (2.5) 2 ê(19.2)(17) + (4.595)(2.5)(16) ç 0.75 + ÷ 2 2 ø úû è ë = 2946 kN > 2700 kN So, B » 2.4 m 5.13

eB 0.12 1ö e 0.36 æ = = 0.08 ç between zero and ÷ ; L = = 0.16 (< 0.5). B 1.5 6 ø L 2.25 è So, Case II; Figure 5.31 applies. From Figure 5.31,

L1 L = 0.92; 2 = 0.33 L L

L1 = (0.92)(2.25) = 2.07 m; L2 = (0.33)(1.5) = 0.495 m 1 1 Eq. (5.80): A¢ = ( L1 + L2 ) B = (2.07 + 0.495)(1.5) = 1.923 m2 2 2 Eq. (5.81): B ¢ =

A¢ 1.923 = = 0.929 m L1 2.07

Eq. (5.82): L¢ = 2.07 m

1 qu¢ = qN q Fqs Fqd + g B ¢Ng Fg s Fg d 2 Table 5.3: For f ¢ = 35°; N q = 33.3; Ng = 48.03

æ B¢ ö æ 0.929 ö Fqs = 1 + ç ÷ tan f ¢ = 1 + ç ÷ (tan 35) = 1.314 è L¢ ø è 2.07 ø æ B¢ ö æ 0.929 ö Fg s = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.82 è L¢ ø è 2.07 ø 35 ®

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æ 0.8 ö Fqd = 1 + 2 tan 35(1 - sin 35) 2 ç ÷ = 1.316 è 1.5 ø

Fg d = 1 1 qu¢ = (17 ´ 0.8)(33.3)(1.314)(1.316) + (17)(0.929)(48.03)(0.82)(1) 2 2 = 1094.13 kN/m Qall =

5.14

qu¢ B ¢L ¢ (1094.13)(0.929)(2.07) = = 526 kN FS 4

e öæ b° ö æ a. From Eq. (5.95): Qu ( ei ) = qu B ç1 - 2 ÷ç1 - ÷ B ø è f¢ ø è

Df B

1 qu = qN q Fqd + g BNg Fg d 2

f ¢ = 38°; N q = 48.93; Ng = 78.03 (Table 5.3) æ 1 ö Fqd = 1 + 2 tan 38(1 - sin 38) 2 ç ÷ = 1.165 è 1.4 ø

Fg d = 1 1 qu = (17.5 ´ 1)(48.93)(1.165) + (17.5)(1.4)(78.03)(1) = 1953.43 kN/m 2 2 1 ö æ ç 2- ÷ 1.4 ø

é æ 0.15 ö ù é æ 18 ö ù è Qu ( ei ) = (1953.43)(1.4) ê1 - (2) ç ÷ ú ê1 - ç ÷ ú è 1.4 ø û ë è 38 ø û ë 1.5- 0.7

e öæ b° ö æ b. Eq. (5.96): Qu ( ei ) = qu B ç1 - 2 ÷ç1 ÷ B ø è f¢ ø è

» 941.4 kN /m

Df B

æ 1 ö 1.5- 0.7ç ÷ è 1.4 ø

é æ 0.15 ö ù é æ 18 ö ù = (1953.43)(1.4) ê1 - (2) ç ÷ ú ê1 - ç ÷ ú è 1.4 ø û ë è 38 ø û ë = 1130.9 kN /m

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Chapter 6

6.1

Friction ratio, m =

f2¢ 35 = = 0.875 f1¢ 40

(2 + m)f1¢ (2 + 0.875)(40) = 3 3 = 38.33°

¢ = f(eq)

¢ = 38.33° , From Table 5.3, for f(eq) N c (eq) » 78 N q (eq) » 62 Ng (eq) » 79

é æBö ù 1 é æ B öù qu = qN q (eq) ê1 + ç ÷ tan feq¢ ú + g BNg (eq) ê1 - 0.4 ç ÷ ú è L øû ë èLø û 2 ë é æ5ö ù qu = (105 ´ 3.5)(62) ê1 + ç ÷ tan 38.33ú ë è5ø û é æ1ö æ 5 öù + ç ÷ (105)(5)(79) ê1 - (0.4) ç ÷ ú è2ø è 5 øû ë

= 43, 274 lb/ft 2 6.2

From Eq. (6.6)

1 qu = g BNg q 2 From Figure 6.4, for

Df B

= 1,

a = 20°, and f ¢ = 30°, the value of Ng q = 40 . 37 ®

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æ1ö So, qu = ç ÷ (110)(4)(40) è2ø = 8800 lb/ft2

6.3

1 2

Eq. (6.10): qu = qN q* Fqs* + g BNg* Fg*s

H 0.9 m = = 0.6; f ¢ = 40° . Figures 6.9 and 6.10: Nq* » 380; Ng* » 200 B 1.5 Eqs. (6.11) and (6.12) and Figure 6.11:

æBö æ 1.5 ö Fqs* = 1 - m1 ç ÷ = 1 - (0.46) ç ÷ = 0.724 èLø è 2.5 ø æBö æ 1.5 ö Fg*s = 1 - m2 ç ÷ = 1 - (0.52) ç ÷ = 0.688 èLø è 2.5 ø

Qall =

qu BL (1.5 ´ 2.5) é 1 ù = (1.2 ´ 17)(380)(0.724) + (17)(1.5)(200)(0.688) ú ê FS 3 2 ë û

= 9208.6 kN » 9209 kN

6.4

H 0.6 = = 0.4; f ¢ = 35°. N q* » 340; Ng* » 100 B 1.5 Eqs. (6.11) and (6.12) and Figure 6.11:

æBö æ 1.5 ö Fqs* = 1 - m1 ç ÷ = 1 - (0.55) ç ÷ = 0.45 èLø è 1.5 ø æBö æ 1.5 ö Fg*s = 1 - m2 ç ÷ = 1 - (0.58) ç ÷ = 0.42 èLø è 1.5 ø

Qall =

qu BL (1.5 ´ 1.5) é 1 ù = (15 ´ 1)(340)(0.45) + (15)(1.5)(100)(0.42) ú ê FS 3 2 ë û

= 2075.6 kN

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6.5

Eq. (6.13): qu = qNq* + 0.4g BNg*

f ¢ = 30°;

H 1.75 = = 1. B 1.75

From Figure 6.9, Nq* » 22; and, from Figure 6.12, Ng* » 78. qu = (1 ´ 17)(22) + (0.4)(17)(1.75)(78) = 374 + 928.2 = 1302.2 kN/m 2

qall =

1302.2 = 325.55 kN/m2 4

Qall = qall ( B ´ L) = (325.55)(1.75 ´ 1.75) » 997 kN 6.6

Eq. (6.21): é ù æBö ê 0.5 ç H ÷ - 0.707 ú è ø ú cu + q qu = 5.14 ê1 + 5.14 ê ú êë úû é ù æ 1.5 ö ê 0.5 ç 0.7 ÷ - 0.707 ú è ø ú (115) + (18.5)(1) = 5.14 ê1 + 5.14 ê ú êë úû

= 651.5 kN/m2

6.7

æ è

Eq. (6.43): qu = ç1 + 0.2

Bö æ B ö æ 2ca H ö ÷ cu (2) N c + ç1 + ÷ç ÷ + g 1D f Lø è L øè B ø

cu (2) B 585 = 0; = = 0.585 L cu (1) 1000

From Figure 6.16 and Eq. (6.45):

qu = (585)(5.14) +

ca » 0.975; ca = (0.975)(1000) = 975 lb/ft 2 cu (1)

(2)(975)(1.65) + (121)(1.65) = 4279 lb/ft 2 3

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CHECK ¾ Eq. (6.44):

Bö æ qu = qt = ç1 + 0.2 ÷ cu (1) N c + g 1D f = (1000)(5.14) + (121)(1.65) = 5339.7 lb/ft 2 Lø è So, qu = 4279 lb/ft 2 .

qall =

6.8

qu 4279 = = 1426 lb /ft 2 FS 3

cu (2) 43 B 0.92 = = 0.754; = = 0.597 L 1.22 cu (1) 72

From Figure 6.16:

ca » 0.975; ca = (0.975)(72) = 70.2 kN/m 2 cu (1)

Eq. (6.43):

Bö æ æ B ö æ 2c H ö qu = ç1 + 0.2 ÷ cu (2) N c + ç1 + ÷ç a ÷ + g 1 D f Lø è è L øè B ø é (2)(70.2)(0.76) ù = [1 + (0.2)(0.754)](43)(5.14) + (1 + 0.754) ê úû + (17)(0.92) 0.92 ë = 254.3 + 203.43 + 15.64 = 473.4 kN/m 2 CHECK ¾ Eq. (6.44): Bö æ qu = qt = ç1 + 0.2 ÷ cu (1) N c + g 1D f = [1 + (0.2)(0.754)](72)(5.14) + (17)(0.92) Lø è = 441.5 kN/m 2 (USE) Qu = (441.5)(0.92)(1.22) = 495.5 kN

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6.9

Eq. (6.22): qu = cu (1) N c Fcs Fcd + q cu (2) 43 H 0.76 = = 0.826; = = 0.597 B 0.92 cu (1) 72

From Figure 6.13, N c » 5. From Table 5.4: (f = 0) .

æ B ö æ Nq ö æ 0.92 öæ 1 ö Fcs = 1 + ç ÷ ç ÷ = 1+ ç ÷ç ÷ = 1.151 è L ø è Nc ø è 1.22 øè 5 ø æ Df ö æ 0.92 ö Fcd = 1 + 0.4 ç ÷ = 1 + 0.4 ç ÷ = 1.4 è 0.92 ø è B ø So, qu = (72)(5)(1.151)(1.4) + (17)(0.92) = 597.74 kN/m 2

Qu = (597.74)(0.92)(1.22) = 670.9 kN 6.10

Eq. (6.40)

1 é ù qu = êg 1 ( D f + H ) N q (2) Fqs (2) + g 2 BNg (2) Fg s (2) ú 2 ë û

B ö æ 2 D f ö K s tan f1¢ æ + g 1H 2 ç 1 + ÷ ç 1 + - g 1H ÷ Løè H ø B è

f2¢ = 32°. Table 5.3: N q (2) = 23.18; Ng (2) = 30.22 Table 5.4:

æBö æ 1.5 ö Fqs (2) = 1 + ç ÷ tan f2¢ = 1 + ç ÷ (tan 32) = 1.625 èLø è 1.5 ø æBö æ 1.5 ö Fg s (2) = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.6 èLø è 1.5 ø Eq. (6.42):

q2 g 2 N g (2) = q1 g 1 N g (1)

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f1¢ = 40°. Ng (1) = 109.41 (Table 5.3) q2 (16.7)(30.22) = = 0.256 q1 (18)(109.4) Figure 6.15: f1¢ = 40°;

q2 = 0.256; K s » 5 q1

qu = [(18)(1.5 + 1)(23.18)(1.625) + (0.5)(16.7)(1.5)(30.22)(0.6)] æ 1.5 öæ (2)(1.5) ö 5 tan 40 + (18)(1) 2 ç1 + - (18)(1) ÷ç1 + ÷ 1 ø 1.5 è 1.5 øè = (1695 + 227.1) + 402.8 - 18 = 2306.9 kN/m 2

1 2

CHECK ¾ Eq. (6.41): qt = g 1D f N q (1) Fqs (1) + g 1BNg (1) Fg s (1)

f1¢ = 40°. Table 5.3: N q (1) = 64.2; Ng (1) = 109.41 æBö æ 1.5 ö Fqs (1) = 1 + ç ÷ tan f1¢ = 1 + ç ÷ (tan 40) = 1.839 èLø è 1.5 ø æBö æ 1.5 ö Fg s (1) = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.6 èLø è 1.5 ø

1 qt = (18)(1.5)(64.2)(1.839) + (18)(1.5)(109.41)(0.4) 2 = 3187.7 + 886.2 = 4073.9 kN/m 2 So, qu = 2306.9 kN/m 2

qu (net) = 2306.9 - (1.5 ´18) » 2280 kN/m2

Qall(net)

qu (net) B 2 FS

(2280)(1.5)2 = 1282.5 kN 4

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6.11

1 c1¢ = 0. Eq. (6.47): qt = g 1D f N q (1) Fqs (1) + g 1BNg (1) Fg s (1) 2 Table 5.3: For f1¢ = 30°, N q (1) = 18.4; Ng (1) = 22.4 Table 5.4:

æBö æ4ö Fqs (1) = 1 + ç ÷ tan f1¢ = 1 + ç ÷ (tan 30) = 1.385 èLø è6ø æBö æ4ö Fg s (1) = 1 - 0.4 ç ÷ = 1 - 0.4 ç ÷ = 0.733 èLø è6ø

1 qt = (98)(3)(18.4)(1.385) + (98)(4)(22.4)(0.733) 2 = 7492.3 + 3218.2 = 10, 711.5 lb/ft 2 » 10.7 kip/ft 2

1 2

Eq. (6.48): qb = g 2 D f N q (2) Fqs (2) + g 2 BNg (2) Fg s (2) Table 5.3: f2¢ = 38°; N q (2) = 48.93; Ng (2) = 78.03 Table 5.4:

æ4ö Fqs (2) = 1 + ç ÷ (tan 38) = 1.52 è6ø æ4ö Fg s (2) = 1 - 0.4 ç ÷ = 0.733 è6ø

1 qb = (108)(3)(48.93)(1.521) + (108)(4)(78.03)(0.733) 2 2 = 36, 467.2 lb/ft » 36.47 kip/ft 2

æ Hö Eq. (6.46): qu = qt + (qb - qt ) ç1 - ÷ è Dø

D » B (top¾loose sand) 2

æ 2ö qu = 10.7 + (36.47 - 10.7) ç1 - ÷ = 17.14 kip/ft 2 è 4ø Qall =

qu BL (17.14)(4)(6) = = 102.84 kip FS 4

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6.12

f1 = f2 = 0. From Eq. (6.47), qt = cu (1) N c (1) Fcs (1) + g 1 D f N q (1) Fqs (1)

B = 0. Note, for f1 = 0 : N c (1) = 5.14; Ng (1) = 0; N q (1) = 1; Fcs (1) = 1; Fqs (1) = 1 L (Tables 5.3 and 5.4). So, qt = (48)(5.14) + (16.5)(1) = 263.22 kN/m 2

Similarly, qb = (96)(5.14) + (17.5)(1) = 510.94 kN/m 2

Eq. (6.46): Clay; D » B 2

æ Hö æ 0.8 ö qua = qt + (qb - qt ) ç1 - ÷ = 263.22 + (510.94 - 263.22) ç1 ÷ è Dø è 1.5 ø

= 317.2 kN /m 2 6.13

Eq. (6.23): qu = cu (1) mN c Fcs Fcd + q

cu (2) cu (1)

48 = 2 (> 1) 96

H 0.8 = = 0.533 (> 0.5) B 1.5 Table 6.2: m = 1

æ B ö æ Nq ö Fcs = 1 + ç ÷ ç ÷ = 1+ 0 = 1 è L ø è Nc ø æ Df ö æ 1 ö Fcd = 1 + 0.4 ç ÷ = 1 + (0.4) ç ÷ = 1.267 è 1.5 ø è B ø qu = (48)(1)(5.14)(1)(1.267) + (16.5)(1) = 329.1 kN /m 2

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6.14

1 2

Eq. (6.51): qu = qN qV q + g BNg V g

x 2 = = 1.67. B 1.2

f ¢ = 35°. Table 5.1: N q = 41.44; Ng = 45.41.

For f ¢ = 35°, from Figure 6.23, V q = 1.38; V g = 2.39

1 qu = (1 ´ 16.8)(41.44)(1.38) + (16.8)(1.2)(45.41)(2.39) = 2054.72 kN/m2 2 q(net) = qu - g D f = 2054.72 - (1´16.8) = 2037.92 kN/m2 Net allowable bearing capacity =

6.15

2037.92 » 509.5 kN /m 2 4

æg Bö From Eq. (6.49): qu = cu (2) N c (T ) + D f g 2 N q (T ) + ç 2 ÷ Ng (T ) è 2 ø

W = 1.5 m; B = 1 m;

W 1.5 = = 1.5 B 1

From Figures 6.19-6.21: N c (T ) » 15; N c (T ) » 7.2; Ng (T ) » 12

æ 17 ´1 ö 2 qu = (40)(15) + (1)(17)(7.2) + ç ÷ (12) = 824.4 kN /m è 2 ø 6.16

a. B = 1 m; H = 4 m. Since B < H , N s = 0.

1 = = 1.0. B 1

Eq. (6.54): qu = cu N cq Figure 6.26:

qall =

cu Ncq FS

b 2 = = 2; b = 60°; Ncq » 6.5 B 1 =

(68)(6.5) = 147.3 kN /m2 3

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b. Ncq

qu = cu N cq

b (m)

B (m)

b/B

[Figure 6.26]

(kN/m2)

4.2

285.6

5.5

374

6.5

442

7.0

476

7.0

476

7.0

476

7.0

476

Note: D f /B = 1 6.17

1 2

Eq. (6.53): qu = g BNg q

Df B

Df b 4 b 6 = 1; = = 1.5. From Figure 6.25 for = 1 and = 1.5, f ¢ = 40° B 4 B 4 B

and b = 30°, the value of N g q » 135.

qu = 0.5g BNg q = (0.5)(110)(4)(135) = 29,700 ld/ft 2 6.18

é æ q öù Eq. (3.55): f ¢ = tan -1 ê0.1 + 0.38log ç c ÷ ú è s o¢ ø û ë Depth (m)

qc (MN/m 2 )

s o¢ (MN/m 2 )

f ¢(deg)[Eq. (3.55)]

1.73

0.033

3.6

0.066

37.2

4.9

0.099

36.6

6.8

0.132

36.9

8.7

0.165

0.248

37 Average f ¢ » 37°

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From Figure 6.33, for f = 37°, N q » 43; Ng » 60

tan q =

kh 0.2 = = 0.2 1 - kv 1 - 0

From Figure 6.34, for tan q = 0.2 and f ¢ = 37° ,

Ng E Ng

= 0.39 and

N qE Nq

= 0.58

Ng E = (0.39)(60) = 23.4; N qE = (0.58)(43) = 24.9 Eq. (6.59):

1 1 quE = qN qE + g BNg E = (16.5 ´1)(24.9) + (16.5)(1.5)(23.4) = 700.5 kN /m 2 2 2 6.19

Df B

1 = 0.67. For f ¢ = 37°, FS = 4. Using Figure 6.35 and interpolating, 1.5

kh* » 0.32. From Figure 6.36 for kh* » 0.32 and f ¢ = 37° , the value of

tan a AE » 1.1. Eq. (6.60): -4

æV2 ö 0.32 tan a AE ç ÷ = 0.174 0.30 è Ag ø = 0.0062 m = 6.2 mm

k* S Eq = 0.174 h A

-4

æ 0.352 ö (1.1) ç ÷ è 0.3 ´ 9.81 ø

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Chapter 7 7.1

The following table can now be prepared. Note: B = 2 m (diameter); z = 3 m;

qo = 150 kN/m 2 z æBö ç ÷ è2ø

r (m)

r æBö ç ÷ è2ø

Ds a qo

Ds (kN/m2 )

0 0.4 0.8 1

0.146 0.141 0.127 0.118

21.9 21.15 19.05 17.7

3 3 3 3 a Table 7.2

7.2

See Eq. (7.1)

Load P @ (kN)

r (m)

z (m)

r z

1.0 1.41

0.0844 0.0309

0.234 0.172

1.118

0.0629

0.699

B C

100 200

(6 + 6 )

= 8.48

6 6

400

(62 + 32 )0.5 = 6.708

ì ü ï ï ï ïï P 3 ï 3 Ds z = 2 í kN/m 2 2.5 ý 5/2 2 z ï é ærö ù ï é æ r ö2 ù ï 2p ê1 + ç ÷ ú ï 2p ê1 + ç ÷ ú îï êë è z ø úû þï êë è z ø úû

2 0.5

Ds z = S1.105 kN /m 2

7.3

Eq. (7.6)

Ds z =

2q1 z 3 2q2 z 3 (2)(90)(3)3 (2)(325)(3)3 + = + p [( x1 + x2 )2 + z 2 ]2 p [ x22 + z 2 ]2 p[(6.5)2 + (3) 2 ]2 p[2.52 + 32 ]2

= 24.6 kN /m 2

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7.4

B = 36ft; q = 900 lb/ft 2 ; x = 27ft; z = 15 ft 2 x (2)(27) 2 z (2)(15) Ds z = = 1.5; = = 0.833 . From Table 7.4, = 0.18 B 36 B 36 q Ds z = (0.18)(900) = 162 lb /ft 2

7.5

Eq. (7.17): m =

Area 1: m =

B L ; Eq. (7.18): n = . Refer to areas is Figure 7.13. z z

4 8 = 0.4 n = = 0.8 I1 » 0.09314 (Table 7.7) 10 10

Area 2: m =

6 8 = 0.6 n = = 0.8 I 2 » 0.12474 (Table 7.7) 10 10

Area 3: m =

4 10 = 0.4 n = = 1.0 I 3 » 0.10129 (Table 7.7) 10 10

Area 4: m =

6 10 = 0.6 n = = 1.0 I 4 » 0.13605 (Table 7.7) 10 10

Ds = qo ( I1 + I 2 + I 3 + I 4 ) = (3000)(0.09314 + 0.124749 + 0.10129 + 0.13605) = 1365.66 lb /ft 2 7.6

Refer to Figure 7.13 Area

z (ft)

B (ft)

L (ft)

1 2 3 4

20 20 20 20

4 10 4 10

8 8 12 12

B z 0.2 0.5 0.2 0.5

L z 0.4 0.4 0.6 0.6

Iw (Table 7.10) 0.0214 0.0459 0.0282 0.061

S 0.1565

Ds = (2500)(0.1565) » 391.25 lb /ft 2

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7.7

Referring to the figure shown and Eq. (7.31)

é H 2 I a ( H 2 ) - H1I a ( H1 ) ù Ds av = 4qo ê ú H 2 - H1 ë û

qo =

50 ´ 2000 lb = 4000 lb/ft 2 5´5

H1 = 3 ft; H 2 = 13ft æ5ö B çè 2 ÷ø For I a ( H 2 ) : m2 = = = 0.192 H2 13 æ5ö L çè 2 ÷ø n2 = = = 0.192 H2 13 Figure 7.16: I a ( H 2 ) » 0.09

æ5ö B çè 2 ÷ø For I a ( H1 ) : m2 = = = 0.83 H1 3 æ5ö L çè 2 ÷ø n2 = = = 0.83 H1 3 Figure 7.16: I a ( H1 ) » 0.212

é (13)(0.09) - (3)(0.212) ù Ds av = (4)(4000) ê » 854 lb /ft 2 ú 13 - 3 ë û

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7.8

Point A:

B1 3 B 20 = = 0.6; 2 = = 4. Figure 7.23: I ¢ » 0.472 z 5 z 5

Ds = (2)(10 ´ 17)(0.472) = 160.5 kN /m2 Point B:

For

B1 0 B 20 = = 0; 2 = = 4; I ¢ » 0.415 z 5 z 5

For

B1 6 B 20 = = 1.2; 2 = = 4; I ¢ » 0.485 z 5 z 5

Ds = (10 ´17)(0.415 + 0.485) = 153 kN/m2

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Point C:

For

B1 26 B 20 = = 5.2; 2 = = 4; I ¢ » 0.5 z 5 z 5

For

B1 0 B 20 = = 0; 2 = = 4; I ¢ » 0.415 z 5 z 5

Ds = (10 ´ 17)(0.5 - 0.415) = 14.45 kN/m2

7.9

Eq. (7.5):

ì ü ï ï ïï ïï 1 Ds = qo í1 . qo = 150 kN/m 2 ; B = 2 m ý 3/2 2 ï é æ Bö ù ï 1 + ï ê ç ÷ ú ï îï êë è 2 z ø úû þï

ì ü ï ï ïï ïï 1 2 Ds t (i.e., z = 3 m) = 150 í1 ý = 21.93 kN/m 2 3/2 ï é æ2ö ù ï ï ê1 + ç ÷ ú ï ïî êë è 6 ø úû ïþ ì ü ï ï ïï ïï 1 2 Ds m (i.e., z = 4.5 m) = 150 í1 ý = 10.46 kN/m 2 3/2 ï é æ2ö ù ï ï ê1 + ç ÷ ú ï ïî êë è 9 ø úû ïþ

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ì ü ï ï 1 ïï ïï 2 Ds b (i.e, z = 6 m) = 150 í1 ý = 6.04 kN/m 3/2 2 ï é æ 2ö ù ï ï ê1 + ç ÷ ú ï îï ëê è 12 ø ûú þï

Ds av = 7.10

21.93 + (4)(10.46) + 6.04 = 11.64 kN /m 2 6

Referring to Figure 7.20:

H1 H 3 6 = = 3; 2 = =6 æ Bö æ2ö æ Bö æ 2ö ç ÷ ç ÷ ç ÷ ç ÷ è 2 ø è2ø è 2 ø è2ø From Figure 7.20:

Ds av » 0.082 qo Ds av = (0.082)(150) = 12.3 kN /m 2

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Chapter 8

8.1

Eq. (8.1): Se = Ai A2

qo B Es

D f = 4 ft; H = 40 ft; B = 5ft; L = 10 ft

Df 4 H 40 L 10 = = 8; = = 2; = = 0.8 B 5 B 5 B 6 Figure 8.1: A1 = 0.85; A2 = 0.93

Se = (0.85)(0.93) 8.2

3000(5) = 0.0439 ft » 0.53 in. 1875 ´144

Eqs. (8.16) and (8.17): Se = 0.93qo (a B ¢)

B¢ =

1 - µs2 Is I f Es

6.25 L 10 H ¢ 32 - 2.5 = 3.125 ft; m¢ = = = 1.6; n¢ = = = 9.44 2 B 6.25 æ B ö æ 6.25 ö ç ÷ ç ÷ è2ø è 2 ø

Tables 8.2 and 8.3: F1 = 0.597; F2 = 0.025

I s = F1 +

1 - 2µs 1 - (2)(0.3) F2 = 0.597 + (0.025) = 0.611 1 - µs 1 - 0.3

D f = 2.5 ft;

Df B

2.5 L = 0.6; = 1.6. Table 8.4: I f » 0.77 6.25 B

2 æ 6.25 ö æ 1 - 0.3 ö Se = (0.93)(3000) ç 4 ´ ÷ (0.611)(0.77) = 0.0324ft = 0.389 in. ÷ç 2 ø è 3200 ´144 ø è

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8.3

Eq. (8.16): Se = qo (a B ¢ )

B¢ = m¢ =

1 - µ s2 Is I f Es

B 2 = = 1m; qo = 210 kN/m2 ; µs = 0.3; a = 4 2 2 L 3.2 H¢ ¥ = = 1.6; n¢ = = =¥ B 2 æBö æBö ç ÷ ç ÷ è2ø è2ø

Eq. (8.5) and Tables 8.2 and 8.3:

I s = F1 +

Df B

1 - 2µ s 1 - 2µ s F2 = 0.697 + (0) = 0.697 1 - µs 1 - µs

1.6 B = 0.8; = 0.625. Table 8.4: I f » 0.73 2 L

æ 1 - 0.32 ö Se = (210)(4 ´1) ç ÷ (0.697)(0.73) = 0.0458 m = 45.8 mm è 8500 ø

8.4

m¢ =

L 3.2 H 4 = = 1.6; n¢ = = =4 B 2 æ Bö æ2ö ç ÷ ç ÷ è 2 ø è2ø

Tables 8.2 and 8.3: F1 = 0.46; F2 = 0.057

I s = F1 +

Df B

1 - 2µs 1 - (2)(0.3) F2 = 0.46 + (0.057) = 0.493 1 - µs 1 - 0.3

1.2 B 2 = 0.6; = = 0.625. Table 8.4: I f » 0.785 2 L 3.2

Se = qo (a B¢)

æ 1 - 0.32 ö 1 - µs2 I s I f = (210)(4 ´1) ç ÷ (0.493)(0.785) Es è 8500 ø

= 0.0348 m = 34.8 mm

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8.5

Eqs. (8.16) and (8.17): Se = 0.93qo (a B ¢)

1 - µs2 Is I f Es

1.8 1.8 15 - 1 = 0.9 m; m¢ = = 1; n¢ = = 15.56 2 1.8 æ 1.8 ö ç ÷ è 2 ø

B¢ =

Tables 8.2 and 8.3: F1 » 0.52; F2 » 0.01

I s = F1 +

1 - 2µs 1 - (2)(0.04) F2 = 0.52 + (0.01) = 0.523 1 - µs 1 - 0.4

µs = 0.4;

Df B

1 B = 0.46; = 1. Table 8.4: I f » 0.83 1.8 L

æ 1 - 0.42 ö Se = (0.93)(190)(4 ´ 0.9) ç ÷ (0.523)(0.83) = 0.0151 m = 15.1 mm è 15, 400 ø 8.6

From Eq. (8.19) the equivalent diameter,

Be =

4BL

(4)(2)(1)

= 1.56 m

qo = 190 kN/m 2

Eo 9000 = = 11.54 kBe (500)(1.56)

H 2 = = 1.28 Be 1.56 From Figure 8.6 for b = 11.54 and H / Be = 1.28 , the value I G = 0.68.

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Eq. (8.23):

IF =

æ ö 3 ç ÷ æ 2t ö Ef 4.6 + 10 ç B ÷ç ÷ ç Eo + e k ÷ è Be ø è 2 ø p 1 = + = 0.788 4 é ù ê ú é (2)(0.23) ù 3 15 ´106 úê 4.6 + 10 ê ú ê 9000 + æ 1.56 ö (500) ú ë 1.56 û ç ÷ êë úû è 2 ø 4

Eq. (8.24):

IE = 1-

æB ö 3.5exp(1.22µ s - 0.4) ç e + 1.6 ÷ ç Df ÷ è ø 1 = 1= 0.917 æ 1.56 ö 3.5exp[(1.22)(0.4) - 0.4] ç + 1.6 ÷ è 1 ø

Eq. (8.22):

Se =

qo Be I G I F I E (190)(1.56)(0.68)(0.788)(0.917) 1 - µ s2 = (1 - 0.42 ) Eo 9000

(

)

= 0.0136 m » 13.6 mm

8.7

Eq. (8.22): Se =

qo Be I G I F I E 1 - µ s2 Eo

(

)

qo = 150 kN/m 2

Be =

4B2

(4)(3) 2

= 3.385 m

µ s = 0.3; Eo = 16, 000 kN/m 2

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Eo 16, 000 = = 11.82 kBe (400)(3.385)

H 20 = = 5.91 Be 3.385 From Figure 8.6, I G » 0.89. From Eq. (8.23):

IF =

æ ö 3 ç ÷ æ 2t ö Ef 4.6 + 10 ç B ÷ç ÷ ç Eo + e k ÷ è Be ø è 2 ø p 1 = + = 0.815 4 é ù ê ú é (2)(0.25) ù 3 15 ´106 úê 4.6 + 10 ê ú ê16, 000 + æ 3.385 ö (400) ú ë 3.385 û ç ÷ êë úû è 2 ø 4

Eq. (8.24):

IE = 1-

æB ö 3.5exp(1.22 µ s - 0.4) ç e + 1.6 ÷ ç Df ÷ è ø 1 = 1= 0.923 æ 3.385 ö 3.5exp[(1.22)(0.3) - 0.4] ç + 1.6 ÷ è 1.5 ø

æ 1 - 0.32 ö Se = (150)(3.385)(0.89)(0.815)(0.923) ç ÷ = 0.0193 m = 19.3 mm è 16,000 ø 8.8

Eq. (8.27): At z = 0:

éæ 10 ö ù æL ö I z = 0.1 + 0.0111ç - 1÷ = 0.1 + 0.0111 êç ÷ - 1ú = 0.107 èB ø ëè 6.25 ø û 59

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Eq. (8.28):

z1 éæ 10 ö ù æL ö = 0.5 + 0.0555 ç - 1÷ = 0.5 + 0.0555 êç ÷ - 1ú = 0.53 B èB ø ëè 6.25 ø û

Eq. (8.26): I z ( m ) = 0.5 + 0.1

qo = q - q =

q -q qz¢(1)

3000 = 20.83 lb/in.2 144

z1 = (0.53)(6.25) = 3.31 ft

qz¢(1) = ( D f + z1 )g = (2.5 + 3.31)(115) = 668.15 lb/ft 2 = 4.64 lb/in.2 I z ( m) = 0.5 + 0.1

Eq. (8.29):

20.83 = 0.71 4.64

z2 éæ 10 ö ù æL ö = 2 + 0.222 ç - 1÷ = 2 + 0.222 êç ÷ - 1ú = 2.13 B èB ø ëè 6.25 ø û

z2 = (2.13)(6.25) = 13.31 ft

Iz ( Dz ) Es

Depth (ft)

Dz (in.)

Es (lb/in.2)

0 - z1 = 0 - 3.31

39.72

3200

0.107 + 0.71 = 0.409 2

0.00508

z1 - z2 = 3.31 - 13.31

120

3200

0.71 + 0 = 0.355 2

0.01331

S 0.01839 Eq. (8.25): Se = C1C2 (q - q )S

1z (Dz ) Es

q = g D f = (115)(2.5) = 287.5 lb/ft 2 = 1.997 lb/in.2 qo = q - q =

3000 = 20.83 lb/in.2 144

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æ q ö æ 1.997 ö æ 5 years ö C1 = 1 - 0.5 ç ÷ = 1 - 0.5 ç ÷ = 0.952; C2 = 1 + 0.2 log ç ÷ = 1.34 è 20.83 ø è 0.1 ø èq -qø

Se = (0.952)(1.34)(20.83)(0.01839) = 0.489 in. 8.9

Eq. (8.26): I z ( m ) = 0.5 + 0.1

q -q qz¢(1)

q = g D f = (18)(1.5) = 27 kN/m2 qo = q - q = 195 - 27 = 168 kN/m 2

qz¢(1) = ( D f + z1 )g = (1.5 + B)g = (4)(18) = 72 kN/m2

I z ( m) = 0.5 + 0.1

168 = 0.652 72

See the figure below for the strain influence factor diagram.

Layer No.

0.2 0 ��-----+

£5 (kN/m 2 ) Qr----.------+

2.0 _______ 0:.5_?_ _ _ 2.0 ____ 6, 000

0 2.5

0. 65 2

2• 5 - - - - - - - -

® - - - 8.0

12, 000

8.0 - - - - - -

1 0 , 000 - - 10.0 - - - - - - - - - - - - 10.0 - - - - - Depth below foundation /ml

Depth below foundation /ml

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Depth (m)

Dz (m)

Es (kN/m2)

Iz ( Dz ) Es

0-2

2.0

6,000

0.38

0.000127

2-2.5

0.5

12,000

0.606

0.0000253

2.5-8

5.5

12,000

0.413

0.000189

8-10

2.0

10,000

0.087

0.000017

S 0.0003583

q = g D f = (18)(1.5) = 27 kN/m2 æ q ö æ 27 ö æ 10 ö C1 = 1 - 0.5 ç ÷ = 1 - 0.5 ç ÷ = 0.92; C2 = 1 + 0.2log ç ÷ = 1.4 è 195 - 27 ø è 0.1 ø èq -qø

Se = C1C2 (q - q )S

Iz (Dz ) = (0.92)(1.4)(195 - 27)(0.0003583) Es

= 0.0775 m = 77.5 mm 8.10 Depth (ft)

N60

Average N60 = 11

N 60 æ B + 1 ö2 From Table 8.9: Allowable qnet = ç ÷ Fd Se 4 è B ø B = 6 ft; Se = 1 in. æ Df ö æ3ö Fd = 1 + 0.33 ç ÷ = 1 + (0.33) ç ÷ = 1.165 è6ø è B ø 2

11 æ 6 + 1 ö 2 qnet = ç ÷ (1.165)(1) » 4.36 kip /ft 4è 6 ø 62

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8.11

æ Bö z¢ Eq. (8.41): = 1.4 ç ÷ BR è BR ø æ 1.2 ö z ¢ = (1.4)(0.3) ç ÷ è 0.3 ø

0.75

= 1.188 2

é æ L öù ê1.25 ç B ÷ ú æ B ö0.7 æ q¢ ö Se è øú Eq. 8.42: = a1a 2a 3 ê ç ÷ ç ÷ BR ê 0.25 + L ú è BR ø è pa ø êë B úû

Normally consolidated sand:

a1 = 0.14

a2 =

1.71 1.71 = = 0.06 1.4 ( N 60 ) (11)1.4

From Table 8.11, a 3 = 1. q¢ = 210 kN/m 2 é æ 4.2 ö ù 1.25 ç 0.7 ÷ ê Se 1.2 ø ú æ 1.2 ö æ 210 ö è úç = (0.14)(0.06)(1) ê ÷ ç ÷ 0.3 ê 0.25 + 4.2 ú è 0.3 ø è 100 ø êë 1.2 úû

Se » 0.019 m = 19 mm 8.12

B = 5 ft; L = 10 ft; d = 5 ft; e = 0.5 ft

æBö æ5ö Eq. (8.47): f L / B = 0.8 + 0.2 ç ÷ = 0.8 + (0.2) ç ÷ = 0.9 èLø è 10 ø æeö æ 0.5 ö Eq. (8.48): f e (center) = 1 - 0.33 ç ÷ = 1 - (0.33) ç ÷ = 0.967 èBø è 5 ø 2

æd ö æ 10 ö Eq. (8.50): fd (center) = 1 - ç ÷ = 1 - ç ÷ = 0.988 è 90 ø è 90 ø 63

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b = 26.6°, so it is 2H:1V slope. æ dö Eq. (8.53): f b ,d = 0.7 ç1 + ÷ è Bø

0.15

æ 5ö = 0.7 ç1 + ÷ è 5ø

0.15

= 0.777

f = f L / B f e fd f b ,d = (0.9)(0.967)(0.988)(0.777) = 0.668

Now the following table can be prepared.

p p ( m)

æ DR ö ç ÷ è Ro ø 0.002

(lb/in.2 )

Se1 B

7.2

0.004

0.00048

Se 2 (in.) 0.0288

2.27

qo 4 (lb/in.2 ) 10.92

Qo 5 (kip) 78.62

24.2

0.00096

0.0576

2.17

35.08

252.58

0.008

32.6

0.00192

0.1152

2.0

43.55

313.56

0.012

42.4

0.00288

0.1728

1.86

52.68

379.3

0.024

68.9

0.00576

0.3456

1.7

78.24

563.33

0.05

126.1

0.012

0.72

1.35

113.72

818.78

0.08

177.65

0.0192

1.152

1.1

130.54

939.89

0.10

210.5

0.024

1.44

1.07

150.46

1083.31

0.20

369.6

0.048

2.88

0.9

222.2

1599.84

Se æ DR ö =ç ÷ (0.24) B è Ro ø

æS ö æS ö Se = ç e ÷ ( B) = ç e ÷ (60) in. èBø èBø

Figure 8.17

f Gp p ( m) = 0.668Gp p ( m) = qo

Qo (kip) = qo BL =

(qo lb/in.2 )(60 in. ´120 in.) 1000

The plot of Se versus Qo is shown. From the plot, Qo at 1 in. settlement is 900 kip.

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8.13

Referring to the figure and Eqs. (7.22) and (7.26):

é H 2 I a ( H 2 ) - H1I a ( H1 ) ù Ds av = 4qo ê ú H 2 - H1 ë û qo =

900 = 268.7 kN/m2 1.83 ´1.83

H1 = 1.22 m; H 2 = 4.27 m æ 1.83 ö B çè 2 ÷ø For I a ( H ) : m2 = = = 0.214 H2 4.27 2

æ 1.83 ö L çè 2 ÷ø n2 = = = 0.214 H2 4.27 Figure 7.16: I a ( H 2 ) » 0.105 65

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æ 1.83 ö B çè 2 ÷ø For I a ( H ) : m1 = = = 0.75 H1 1.22 1

æ 1.83 ö L çè 2 ÷ø n1 = = = 0.75 H1 1.22 Figure 7.16: I a ( H1 ) » 0.205

é (4.27)(0.105) - (1.22)(0.205) ù Ds av = (4)(268.7) ê = 69.9 kN /m 2 ú 4.27 - 1.22 ë û 8.14

Ds i =

900 = 96.75 kN/m2 2 (1.83 + 1.22)

Ds n =

900 = 43 kN/m2 2 (1.83 + 2.745)

Ds b =

900 = 24.19 kN/m2 2 (1.83 + 4.27)

1 Ds av = [96.75 + (4)(43) + 24.19] = 48.82 kN /m 2 6 8.15

s o¢ = (1.52)(15.7) + (1.22)(19.24 - 9.81) +

3.05 (19.24 - 9.81) 2

= 23.8 + 11.5 + 14.4 = 49.7 kN/m2 s o¢ + Ds av ¢ = 49.7 + 69.9 = 119.6 kN/m 2 Se ( p ) =

Cs H c s¢ C H s ¢ + Ds ¢ log c + s c log o 1 + eo s o¢ 1 + eo s c¢

(0.06)(3.05) 100 (0.25)(3.05) 119.6 log + log 1.68 49.7 1.68 100 = 0.0684 m = 68.4 mm =

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8.16

From Problems 8.14 and 8.15,

s o¢ = 49.7 kN/m 2

s o¢ + Ds av¢ = 49.7 + 48.82 = 98.52 kN/m 2 Se ( p ) =

Cs H c s ¢ + Ds ¢ (0.06)(3.05) 98.52 log o = log 1 + eo s c¢ 1.68 49.7

= 0.0324 m = 32.4 mm

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Chapter 9

9.1

æ D f öù é æ 0.195B öù é Eq. (9.12): qnet(u) = 5.14cu ê1 + ç ÷ú ÷ú ê1 + 0.4 ç ë è L øû ë è B øû é (0.195)(8) ù é æ 3 öù = (5.14)(120) ê1 + 1 + (0.4) ç ÷ ú ê ú 18 ë ûë è 8 øû = 771 kN /m 2

9.2 Depth (m)

N60

1.5

3.0

4.5

6.0

7.5

9.0

10.5

Average N60 = 10.86 » 11 Eq. (9.14): qnet(all) =

N60 æ 0.33D f ö æ Se ö ç1 + ÷ç ÷ £ 0.67 N60 Se (mm) 0.08 è B ø è 25 ø

N 60 é (0.33)(1.5) ù æ Se ö æ Se ö 1+ ç ÷ = 13.74 N 60 ç ÷ ê ú 0.08 ë 5 û è 25 ø è 25 ø

æ 50 ö qnet(all) = (13.74)(11) ç ÷ = 302.3 kN /m 2 è 25 ø

69 ®

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9.3

æ 30 ö æ 30 ö qnet(all) = 13.74 N 60 ç ÷ = (13.75)(11) ç ÷ = 181.4 kN /m 2 è 25 ø è 25 ø

9.4

a. B = 20 m; L = 20 m; cu = 30 kN/m2; γ = 18.5 kN/m3 Eq. (9.21): D f =

Q 48 ´ 1000 kN = = 6.49 m Ag (20 ´ 20)(18.5)

æ 0.195B ö æ 0.4 D f ö 5.14cu ç1 + ÷ç1 + ÷ L øè B ø è b. Eq. (9.23): FS = Q - g Df A é (0.195)(20) ù æ 0.4 D f ö (5.14)(30) ê1 + úû ç1 + B ÷ 20 ë è ø 2= 3 æ 48 ´10 ö ç ÷ - 18.5D f è 20 ´ 20 ø 240 - 37 D f = 184.27 + 3.69 D f

Df = 1.37 m

9.5

æ 0.195B ö æ 0.4 D f ö 5.14cu ç1 + ÷ç1 + ÷ L øè B ø è Eq. (9.23): FS = Q - g Df A FS = 2; cu = 20 kN/m 2 ; B = L = 20 m; Q = 48 ´ 103 kN; g = 18.5 kN/m 3

é (0.195)(20) ù æ 0.4 D f ö (5.14)(20) ê1 + úû ç1 + B ÷ 20 ë è ø 2= 3 æ 48 ´10 ö ç ÷ - 18.5D f è 20 ´ 20 ø 240 - 37 D f = 122.85 - 2.48 D f

Df = 3.39 m

70 ®

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9.6

B = 10 m; L = 12 m; Q = 30,000 kN

1 ¢ = (Ds l¢ + 4Ds m¢ + Ds b¢ ) Eq. (7.32): Ds av 6

Eq. (8.61): ml =

L 12 = = 1.2 B 10

z (m)

z æBö ç ÷ è2ø

Ic (Table 8.13)

¢ = qo I c (kN/m 2 ) Ds av

4.0

0.8

1.2

» 0.81

æ 30 ´1000 ö 0.81ç ÷ = 202.5 è 10 ´12 ø

6.6

1.32

1.2

» 0.64

æ 30 ´1000 ö 0.64 ç ÷ = 160.0 è 10 ´12 ø

9.2

1.84

1.2

» 0.45

æ 30 ´1000 ö 0.45 ç ÷ = 112.5 è 10 ´12 ø

nl =

1 Ds av¢ = [202.5 + (4 ´ 160) + 112.5] = 159.2 kN/m2 6 Sc =

Cc H c s ¢ + Ds av¢ log o 1 + eo s o¢

s o¢ = (16)(4.2) + (2)(18 - 9.81) + (2.6)(17.5 - 9.81) = 103.58 kN/m 2 s c¢ = 105 kN/m 2 . So s o¢ » s c¢ (normally consolidated). Sc =

(0.38)(5.2) æ 103.58 + 159.2 ö log ç ÷ = 0.425 m 1 + 0.88 103.58 è ø

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9.7

We will need to use Table 7.7 At the top of the clay layer: m =

B 10 = = 2.5 z 4 L 12 = =3 z 4

Ds t¢ = 0.242qo At the middle of the clay layer: m =

B 10 = = 1.515 z 6.6

L 12 = = 1.818 z 6.6

Ds m¢ = 0.22qo At the bottom of the clay layer: m =

B 10 = = 1.09 z 9.2

L 12 = = 1.30 z 9.2

Ds b¢ = 0.189qo 1 Ds av ¢ = (Ds t¢ + 4Ds m¢ + Ds b¢ ) 6 1 æ 30, 000 ö 2 = ç ÷ [0.242 + (4)(0.22) + 0.180] = 54.63 kN/m 6 è 10 ´12 ø

Sc =

Cc H c s ¢ + Ds av¢ (0.38)(5.2) 103.58 + 54.63 log o = log = 0.193 m 1 + eo s o¢ 1 + 0.88 103.58 2

9.8

æ B +1 ö æ 25 + 1 ö 3 Eq. (9.42): k = k1 ç ÷ = (55) ç ÷ = 14.9 lb /in. è 2B ø è 50 ø

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9.9

B = 30 ft; L = 70 ft

Bö æ k( B´B ) ç1 + 0.5 ÷ Lø è Eq. (9.44): k = 1.5 From Problem 9.8, k(B×B) = 14.9 lb/in.3

é æ 30 ö ù (14.9) ê1 + (0.5) ç ÷ ú è 70 ø û ë k= = 12.1 lb /in.3 1.5 9.10

From Eq.(9.44): é æ 0.7 ö ù k( B´ B ) ê1 + (0.5) ç ÷ú k(1´0.7) è 1 ø û 1.35 ë = = =1 k(5´3.5) é æ 3.5 ö ù 1.35 k( B´ B ) ê1 + (0.5) ç ÷ú è 5 øû ë

k(5´3.5) =

k(1´0.7) 1

18 = 18 MN /m3 1

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Chapter 10

10.1

Df B

æD ö 5 = 1.25; f ¢ = 35°. Table 10.1: ç f ÷ » 5. So it is a shallow foundation. 4 è B øcr

m = 0.25; Ku = 0.936 Eq. (10.3):

é æ D f öù æ D f ö Fq = 1 + 2 ê1 + m ç ÷ú ç ÷ K u tan f ¢ B B è øû è ø ë = 1 + 2[1 + (0.25)(1.25)](1.25)(0.936)(tan 35) = 3.15 Qu = Fq Ag D f = (3.15)(4 ´ 4)(112)(5) = 28.224 lb » 28.2 kip 10.2

2 = 1.67 B 1.2 æD ö Eq. (10.7): ç f ÷ = 0.107cu + 2.6 = (0.107)(74) + 2.5 = 10.4 > 7 è B øcr-square =

æD ö So, ç f ÷ = 7. è B øsq æD ö æ Df ö é æ L öù Eq. (10.8): ç f ÷ =ç 0.73 + 0.27 ç ÷ú ÷ ê è B øû è B øcr-rectangular è B øcr-square ë æ Df ö é æ 2.4 ö ù = 7 ê0.73 + 0.27 ç ÷ ÷ ú < 1.55 ç è 1.2 ø û ë è B øcr-square

æ Df ö ç ÷ = 8.89. Hence, this is a shallow foundation. è B øcr Df B = 1.57 = 0.188 Eq. (10.9): a ¢ = 8.89 æ Df ö ç ÷ è B øcr æBö æ 1.2 ö * Eq. (10.10): Fc-rectangular = 7.56 + 1.44 ç ÷ = 7.56 + 1.44 ç ÷ = 8.28 èLø è 2.4 ø 75 ®

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From Figure 10.3, with a ¢ = 0.188 , the value of b ¢ = 0.275. Eq. (10.13):

Qu = A(b ¢ Fc6cu + g D f ) = (1.2 ´ 2.4)[(0.275)(8.28)(74) + (18)(2)] » 589 kN 10.3

H1 2 = =2 D1 1

f ¢ = 30° æH ö Table 10.2: For f ¢ = 30° , the value of ç 1 ÷ = 4 . So this is a shallow anchor è D1 øcr

From Eq. (10.24):

Qu = Fqg AH1 +

p Dag ( H n2 - H12 ) Ku (cr) tan f ¢ 2

From Figure 5.6, for f ¢ = 30° and

H1 = 2 , the magnitude of Fq » 6 . Also from D1

Table 10.2, K µ¢ (cr) = 0.9

p æ 12 + 7 ö æp ö 2 2 Qu = (6)(108) ç ´12 ÷ (2) + ç ÷ (108)(6 - 2 )(0.9) tan 30 2 è 2 ´12 ø è4 ø = 3251 lb 10.4

H1 = 2 m; D1 = 0.45 m H1 2 = = 4.44 D1 0.45 From Eq. (10.31): æ H1 ö ç ÷ = 0.107 cu + 2.5 = (0.107)(45) + 2.5 = 7.3 è D1 øcr æH ö æH ö Since ç 1 ÷ = 7.3 ismore than 7, the actual ç 1 ÷ = 7 . è D1 øcr è D1 øcr

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For this case,

æH ö H1 = 4.4 and is less than ç 1 ÷ = 7.So it is a shallow anchor. D1 è D1 øcr

From Eq. (10.33):

æ D + Dn ö Qu = A(cu Fc + g H1 ) + p ç 1 ÷ ( H n - H1 )cu è 2 ø æ H1 ö ç ÷ è D1 ø = 4.44 = 0.63; here 7 æ H1 ö ç ÷ è D1 øcr

Fc » 7.9 (Figure 10.9) æp ö æ 0.45 + 0.3 ö Qu = ç ´ 0.452 ÷ [(45)(7.9) + (17.8)(2)] + p ç ÷ (6 - 2)(45) 2 è4 ø è ø = 274.3 kN

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Chapter 11

11.1

a. Eq. (11.18):

Qp = Ap q¢Nq* £ Ap ql

Given f2¢ = 42! . From Table 11.5, Nq* = 525

Ap q¢Nq* = (0.356 ´ 0.356)(17.5 ´12)(525) = 13,973 kN Ap ql = Ap (0.5 pa Nq*tan f2¢ ) = (0.356 ´ 0.356)[(0.5)(100)(525)(tan 42°)] = 2995.5 kN Hence, Qp = 2995.5 kN b. Eq. (11.31):

Es = mpa = (600)(100) = 60, 000 kN/m 2

æ f ¢ - 25 ö æ 42 - 25 ö ÷ = 0.1 + 0.3 ç ÷ = 0.355 è 20 ø è 20 ø

Eq. (11.32):

µs = 0.1 + 0.3 ç

Eq. (11.33):

æ f ¢ - 25 ö æ q¢ ö D = 0.005 ç1 ÷ç ÷ 20 ø è pa ø è

Eq. (11.30):

Eq. (11.29):

æ 42 - 25 öæ 17.5 ´12 ö D = 0.005 ç1 ÷ç ÷ = 0.00158 20 øè 100 ø è Es Ir = 2(1 + µs )q¢tan f2¢ Ir =

60, 000 = 117.09 2(1 + 0.355)(17.5 ´12)(tan 42)

I rr =

Ir 117.09 = = 98.8 1 + I r D 1 + (117.09)(0.00158)

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Qp = Aps o¢ Ns*

Eq. (11.24):

é1 + 2(1 - sin 42) ù (17.5 ´12) = 116.3 kN/m 2 ú 3 ë û

s o¢ = ê

Ns* » 160 (Table 11.7). Q p = (0.356 ´ 0.356)(116.3)(160) = 2358 kN

c. Eq. (11.41):

f2¢ = 42°,

Qp = q¢Nq* Ap

L 12 = = 33.7. From Figure 11.20, N q* » 100 D 0.356

Q p = (17.5 ´12)(100)(0.356 ´ 0.356) = 2661.5 kN

11.2

a. Eq. (11.45):

L¢ = 15D = (15)(0.356) = 5.34 m

For z = 0 to z = 5.34 m: f = Ks o¢ tan d ¢ At z = 5.34 m : f = (1.3)(17.5 ´ 5.34)[tan(0.8 ´ 30)] = 54.1 kN/m2

( f z =0 + f z =5.34 m )

pL¢ + f z =5.34 m p(12 - L¢) 2 (0 + 54.1) = (4 ´ 0.356)(5.34) + (54.1)(4 ´ 0.356)(12 - 5.34) 2

Qs =

= 205.69+513.08 = 718.77 kN b. Eq. (11.50): Qs = K (s o¢ ) tan(0.8f1¢)pL 17.5 ´ 12 s o¢ = = 105 kN/m 2 2 L 12 = = 33.7 D 0.356

f1¢ = 30°. From Figure 11.23, K » 0.35 Qs = (0.35)(105) tan (24)(4 ´ 0.356)(12) = 279.6 kN

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11.3

From Problem 11.1 Qp =

2995.5 + 2358 + 2661.5 » 2672 kN 3

Qs =

718.77 + 279.6 = 499.2 kN 2

Qall = 11.4

Qp + Qs FS

2672 + 499.2 = 792.8 kN » 793 kN 4

a. Eq. (11.19) will apply: q p = ql = 0.5 pa Nq* tan f

f ¢ = 40! ; Nq* = 346 (Table 11.5). So, q p = (0.5)(2000)(346)(tan 40) = 290,328 lb/ft 2 » 290 kip/ft 2 2

æ p öæ 15 ö Q p = Ap q p = ç ÷ç ÷ (290) » 356 kip è 4 øè 12 ø b. f ¢ = 40! ; I rr = 50; Ns* » 93 (Table 11.7). Qp = Aps o¢ Ns*

æ 1 + 2Ko ö ÷ q¢; K o = 1 - sin 40 = 0.357 è 3 ø

s o¢ = ç

q¢ = (10)(100) + (10)(116 - 62.4) + (50)(122.4 - 62.4) = 4536 lb/ft 2 é1 + (2)(0.357) ù (4536) = 2591.6 lb/ft 2 » 2.59 kip/ft 2 ú 3 ë û

s o¢ = ê

éæ p öæ 15 ö2 ù Qp = êç ÷ç ÷ ú (2.59)(93) » 296 kip ëêè 4 øè 12 ø ûú c. Qp =

356 + 296 » 326 kip (estimated) 2

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æ 15 ö d. L¢ = 15D = (15) ç ÷ = 18.75 ft. Let us assume L¢ » 20 ft. Qs = pLf av è 12 ø æ 0.1´10 ö (tan 0.6 ´32) = 9.57 kip Qs ( z =0 to 10 ft) = (p ´1.25)(10) (1.4) ! çè 2 ÷ø "#$# % K

s o¢ ( av)

é (10 ´ 0.1) + (10)(0.116 - 0.0624) ù s o¢ (between z = 10 to 20 ft) = ê = 0.768 kip/ft 2 ú 2 ë û Qs (z =10 to 20 ft) = (p ´1.25)(10)(1.4)(0.768)(tan 0.6 ´ 32) = 14.7 kip Qs (z = 20 to 70 ft) = (p ´1.25)(50)(1.4)[(10 ´ 0.1) + (10)(0.116 - 0.0624)]

´ (tan 0.6 ´32) = 188 kip

Qs = 9.57 + 14.7 + 188 » 212 kip 11.5

The pile tip is 9 m below the ground surface. The average value of N60 for a distance of 10D above and 5D below the pile tip is: N 60 =

5 + 16 + 18 + 21 = 15 4

é L ù Eq. (11.42): Qp = Ap ê0.4 pa N 60 æç ö÷ ú £ Ap (4 pa N 60 ) è D øû ë

é é æ L öù æ 9 öù Ap ê0.4 pa N 60 ç ÷ ú = (0.305)2 ê(0.4)(100)(15) ç ÷ ú = 1647 kN è D øû è 0.305 ø û ë ë Ap (4 pa N60 ) = (0.305)2[(0.4)(100)(15)] = 558.15 kN N 60 (9 m below ground surface) =

4 + 8 + 7 + 5 + 16 + 18 = 9.7 » 10 6

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From Eq. (11.52):

Qs = pLf av = pL[0.02 pa ( N 60 )]

= (4 ´ 0.305)(9)[(0.02)(100)(10)] = 219.6 kN Qall = 11.6

Qp + Qs FS

558.15 + 219.6 » 194 kN 4

Qp = Ap [19.7 pa ( N60 )0.36 ] = (0.305)2[(19.7)(100)(15)0.36 ] = 485.8 kN Qs = pL[0.224 pa ( N 60 )0.26 ] = (4 ´ 0.305)(9)[(0.224)(100)(10)0.27 ] = 479.6 kN Qall =

11.7

485.8 + 479.6 = 241.4 kN 4

æc ö æ 1600 ö Eqs. (11.40) and (11.38): I rr = I r = 347 ç u ÷ - 33 = (347) ç ÷ - 33 = 244.6 è 2000 ø è pa ø 4 p 4 Eqs. (11.37): N c* = (ln I rr + 1) + + 1 = (ln 244.6 + 1) + 2.57 = 11.24 3 2 3 2

æ 15 ö æ 1600 ö Q p = Ap cu N = ç ÷ ç kip/ft 2 ÷ (11.24) = 28.1 kip è 12 ø è 1000 ø * c

Eqs. (11.68): Qs = a cu pL

cu 1600 = = 0.8. Table 11.12: a = 0.54. pa 2000 æ 1600 öæ 15 ö Qs = (0.54) ç kip/ft 2 ÷ç 4 ´ ÷ (50) = 216 kip è 1000 øè 12 ø

Qall =

Qp + Qs FS

28.1 + 216 » 81.4 kip 3

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11.8

æ 1600 öæ 15 ö Eq. (11.20): Q p = 9cu Ap = (9) ç ÷ç ÷ = 22.5 kip è 1000 øè 12 ø

Qs = pLf av fav = l (s o¢ + 2cu )

Eq. (11.58):

s o¢ =

(121)(50) = 3025 lb/ft 2 = 3.025 kip/ft 2 2

L = 50 ft =15.24 m. Table 11.11: l = 0.2 æ 15 ö Qs = ç 4 ´ ÷ (50)(0.2)[3.025 + (2)(1.6)] = 311.25 kip è 12 ø Qall =

11.9

22.5 + 311.25 » 111.25 kip 3

L = 15 m; l = 0.2 s o¢ =

(15 ´ 18) = 135 kN/m2 2

Qs = pLl (s o¢ + 2cu ) = (4 ´ 0.38)(15)(0.2))(135 + 2 ´ 80) = 1345.2 kN Qall =

1345.2 » 448.4 kN 3

16 11.10 a. Qs = p[ L1a1cu (1) + L2a 2cu (2) ]. p = (4) æç ö÷ = 5.33 ft è 12 ø

cu (1) pa cu (2) pa

700 = 0.35. Table 11.12: a1 » 0.78 2000

1500 = 0.75. Table 11.12: a 2 » 0.56 2000

Qs = 5.33[(20)(0.78)(700) + (40)(0.56)(1500)] = 237,292 lb » 237 kip

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cu (av) =

(700)(20) + (1500)(40) = 1233.3 lb/ft 2 60

1 1 (20)(2360) + (40)(2360 + 4760) 2 s o¢ (av) = 2 = 2766.7 lb/ft 2 60

Length of pile = 60 ft » 18.3 m; l » 0.18

Qs = pLl[s ¢o (av) + 2cu (av) ] =

1 æ 16 ö ç 4 ´ ÷ (60)(0.18)[2766.7 + (2)(1233.3)] 1000 è 12 ø

= 301.4 kip » 301 kip c.

f av(1) = (1 - sin fR¢ ) tan fR¢ s ¢o (av) æ1 ö = (1 - sin 20)(tan 20) ç ´118 ´ 20 ÷ = 282.6 lb/ft 2 è2 ø æ 2360 + 4760 ö 2 f av(2) = (1 - sin 20)(tan 20) ç ÷ = 852.6 lb/ft 2 è ø

Qs = p[ L1 f av(1) + L2 f av(2) ] æ 1 ö æ 16 ö =ç ÷ ç 4 ´ ÷ [(20)(282.6) + (40)(852.6)] è 1000 ø è 12 ø = 212 kip

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11.11

Eq. (11.87): qu (design) =

qu (lab)

11, 400 = 2280 lb/in.2 5

f¢ ö 36 ö æ æ Nf = tan 2 ç 45 + ÷ = tan 2 ç 45 + ÷ = 3.852 2ø 2 ø è è Eq. (11.88): Qp (all) =

qu (design) ( Nf + 1) Ap FS

For HP 14 ´ 102, Ap = 30 in.2

Qp (all) = 11.12

(2280)(3.852 + 1)(30) = 110.6 kip (3)(1000)

Eq. (11.96):

se (1) =

(Qwp + x Qws ) L Ap E p

[300 + (0.57)(600)](18) = 0.00334 m = 3.34 mm (0.406 ´ 0.406)(21´106 kN/m 2 )

Eq. (11.97):

se (2) =

qwp D Es

éæ 300 öæ 0.406 ö ù (1 - µs2 ) I wp = êç (1 - 0.382 )(0.85) ÷ç 3 ÷ú ëè 0.406 ´ 0.406 øè 30 ´10 ø û

= 0.0179 m = 17.9 mm

æQ ö D Eq. (11.99): se (3) = ç ws ÷ (1 - µs2 ) I ws è pL ø Es I ws = 2 + 0.35

18 = 4.33 0.406

é ù æ 0.406 ö 600 se (3) = ê (1 - 0.382 )(4.33) = 0.001 m = 1 mm ç ú 3 ÷ ë (4 ´ 0.406)(18) û è 30 ´10 ø

se = 3.34 + 17.9 + 1.0 = 22.24 mm

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11.13

Eq. (11.96): se (1) =

(Qwp + x Qws ) L

Eq. (11.97): se (2) =

qwp =

Qwp Ap

Ap E p

qwp D Es

[58 + (0.62)(280)](15) = 0.0018 m = 1.8 mm (0.305) 2 (21´106 )

(1 - µ s2 ) I wp

58 = 623.5 kN/m 2 (0.305) 2

é (623.5)(0.305) ù 2 ú (1 - 0.3 )(0.85) = 0.00049 m = 4.9 mm 30,000 ë û

So, se (2) = ê

æQ ö D Again, from Eq. (11.99): se (3) = ç ws ÷ (1 - µs2 ) I ws è pL ø Es I ws = 2 + 0.35

L 15 = 2 + 0.35 = 4.45 D 0.305

é ù æ 0.305 ö 280 So, se (3) = ê (1 - 0.32 )(4.45) = 0.00063 m = 0.63 mm ú ç ÷ ë (4 ´ 0.305)(15) û è 30, 000 ø

se = 1.8 + 4.9 + 0.63 = 7.33 mm 11.14

1 1 bh3 = (0.305) 4 = 0.0007 m 4 ; E p = 21 ´ 106 kN/m 2 12 12

Eq. (11.113): T = 5

Ep I p nh

(21´106 )(0.0007) = 1.098 m 9200

L 30 = > 5 — So, long pile. M g = 0. From Eq. (11.108): T 1.098

xz ( z ) = Ax

QgT 3 Ep I p

; Qg =

xz ( z ) E p I p AxT 3

At z = 0, xz = 12 mm = 0.012 m ; Ax = 2.435 (Table 11.17)

(0.12)(21´106 )(0.0007) Qg = = 54.7 kN (2.435)(1.098)3

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Check for moment capacity: M z (max) = Fg

Qg =

Ip = AmQgT d 2

2 Fg I p dAmT

From Table 11.17, the maximum value of Am is 0.772. So

Qg =

(2)(21,000)(0.0007) = 113.7 kN (0.305)(0.772)(1.098)

Use Qg = 54.7 kN 11.15

Check for bending failure: æ 0.0007 ö Ip ÷ = 96.39 kN × m M y = Fg = (21, 000) ç d ç 0.305 ÷ 2 è 2ø

My D g Kp 4

f¢ ö æ D 4g tan 2 ç 45 + ÷ 2ø è

96.39 = 232 (0.305) (16) tan 2 (45 + 15) 4

Qu ( g ) e = 0, » 50 D K p D 3g

From Figure 11.44a, with

Qu ( g ) = 50 tan 2 (60)(0.305)3 (16) = 68 kN Check for pile head deflection: Eq. (11.125): h = 5

nh 9200 =5 = 0.91 m -1 6 Ep I p (21 ´ 10 )(0.0007)

h L = (0.91)(30) = 27.3. From Figure 11.45a for h L = 27.3, e = 0. L

xz ( z =0) ( E p I p ) 5 (nh ) 5Qg L

» 0.15

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Qg =

3 5

xz ( z =0) ( E p I p ) (nh ) (0.15) L

2 5

(0.012)[(21´106 )(0.0007)]5 (9200) 5 = = 32.5 kN (0.15)(30)

Use Qg = 32.5 kN 11.16

Eq. (11.129): Qu =

EH E S +C

H E = 40 kip-ft = 40 ´ 12 kip-in.; E = 0.85 Qu =

(0.85)(40 ´12) = 2040 kip 1 + 0.1 10

Qall =

2040 = 340 kip 6

Qu =

EWR h WR + n Wp S + C WR + W p 2

11.17

W p = weight of (pile + cap) =

90 ´ 100 + 2.4 = 11.4 kip 1000

é ù ê (0.85)(40 ´12) ú é12 + (0.35) 2 (11.4) ù Qu = ê úê ú = 1167.9 kip 1 12 + 11.4 ë û ê ú + 0.1 ë 10 û Qall =

1167 » 292 kip 4

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11.18

11.19

Qu =

EH E EH E L (0.85)(40 ´12)(90 ´12) ; = = 0.5 2 Ap E p (2)(29.4)(30 ´106 kip/in 2 ) EH E L S+ 2 Ap E p

Qu =

(0.85)(40 ´12) = 680 kip 0.1 + 0.5

Qall =

680 » 227 kip 3

Eq. (11.137b): Qn =

¢ )g ¢f H 2f tan d ¢ p(1 - sin ffill 2

(p ´ 0.450)(1 - sin 25)(17.5)(42 )tan(0.5 ´ 25) = 2 = 25.3 kN 11.20

Qn =

¢ )g ¢f H 2f tan d ¢ p(1 - sin ffill 2

(p ´ 0.450)(1 - sin 25)(19.8 - 9.81)(42 ) tan(0.5 ´ 25) 2 = 14.5 kN =

11.21

Eq. (11.138): L1 =

L1 =

L - H f é L - H f g ¢f H f ù 2g ¢f H f + ê L1 ë 2 g ¢ úû g¢ (50 - 10) é 50 - 10 (110)(10) ù (2)(110)(10) êë 2 + 124 - 62.4 úû - 124 - 62.4 L1 1514.3 - 35.7; L1 » 25 ft L1

Eq. (11.140):

Qn = p(1 - sin fclay )g ¢f H f tan d ¢L1 +

L12 p (1 - sin fclay )g ¢ tan d ¢ 2

16 ö æ = ç p ´ ÷ (1 - sin 20)(110)(10) tan(0.5 ´ 20)(25) 12 ø è 90 ®

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æ 16 ö (25)2 ç p ´ ÷ (1 - sin 20)(124 - 62.4) tan(0.5 ´ 20) è 12 ø + 1

= 22719 lb » 22.7 kip 2

11.22

æ 16 ö Eq. (11.44): Q p = Ap qc = ç ÷ (168.1) = 298.8 kip è 12 ø

L 60 æ 16 ö = = 45; p = (4) ç ÷ = 5.33 ft 16 D è 12 ø 12 Depth from Ground surface (ft) 0-20 20-45 45-60

(ft)

Fr (%)

fc = Frqc (kip/ft2)

20 25 15

2.3 2.7 2.8

1.345 2.03 4.71

a¢ (Figure 11.25) 0.44 0.44 0.44

pDLa ¢f c (kip) 63.09 119.3 165.69 S348.08 kip

Qall = 11.23

Qp + Qs FS

298.8 + 348.08 = 161.72 kip » 162 kip 4

Piles acting individually: Eq. (11.144): SQu = n1n2 [9 Ap cu( p) + a pcu L]

cu 86 = = 0.86; Table 11.12: a » 0.52 pa 100

é æp ù ö SQu = (3)(3) ê(9) ç ´ 0.3152 ÷ (86) + (0.52)(p ´ 0.315)(86)(20) ú ø ë è4 û = (9)(60.3 + 885.1) = 8508.6 kN Qall =

Qu 8508.6 = » 2836 kN FS 3

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Piles acting as a group: Eq. (11.145): SQu = Lg Bg cu ( p ) Nc* + 2S( Lg + Bg )cu DL

(2)(0.315) æDö Lg = Bg = (n - 1)d + 2 ç ÷ = (2)(0.6) + = 1.515 m 2 è2ø

Lg Bg

= 1;

L 20 = = 13.2. From Figure 11.57, N c* = 9. So Bg 1.515

SQu = (1.515)(1.515)(86)(9) + (2)(1.515 + 1.515)(86)(20) = 1776.5 + 10,423.2 = 12,199.7kN Qall =

12,199.7 = 4066.6 kN 3

So, Qall = 2836 kN 11.24

Piles acting individually: Eq. (11.144): SQu = n1n2 [9 Ap cu ( p ) + a pcu L]

cu 860 = = 0.43; Table 11.12 : a » 0.72 pa 2000 ìï éæ p öæ 12 ö 2 ù üï 12 ö æ SQu = (3)(3) í(9) êç ÷ç ÷ ú (860) + (0.72) ç p ´ ÷ (860)(45) ý 12 ø è îï êëè 4 øè 12 ø úû þï = (9)(6078 + 87,537) lb » 842.5 kip Piles acting as a group: Eq. (11.145): SQu = Lg Bg cu ( p ) Nc* + 2S( Lg + Bg )cu DL

æDö Lg = Bg = (n - 1)d + 2 ç ÷ = (2)(2.25 ft) + 1ft = 6 ft è2ø

Lg Bg

= 1;

L 45 = = 7.5. From Figure 11.57, N c* = 9. So Bg 6 92 ®

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11.25

SQu =

(6)(6)(860)(9) + (2)(6 + 6)(860)(45) = 1207.4 kip 1000

Qall =

842.5 = 280.8 » 281 kip 3

Piles acting individually: SQu = n1n2 [9 Ap cu ( p ) + a1 pcu (1) L1 + a 2 pcu (2) L2 + a 3 pcu (3) L3 ]

Depth (m)

cu (kN/m2)

0-5

0.25

0.87

5-11

0.45

0.72

11-17

0.6

0.62

(Table 11.12)

é (9)(0.356)2 (60) + (0.87)(4 ´ 0.356)(25)(5) ù SQu = (4 ´ 4) ê ú = 13,087 kN ë + (0.72)(4 ´ 0.356)(45)(6) + (0.62)(4 ´ 0.356)(60)(6) û Piles acting as a group: Eq. (11.145): SQu = Lg Bg cu ( p ) Nc* + 2S( Lg + Bg )cu DL Lg = 3 + 0.356 = 3.356 m; Bg = 3.356 m

Lg Bg

3.356 L 17 = 1; = = 5.07 3.356 Bg 3.356

From Figure 11.57, N c* = 9. So

SQu = (3.356)2 (60)(9) + (2)(3.356 + 3.356) ´ [(25)(5) + (45)(6) + (60)(6)] = 16,217 kN Qall =

13, 087 » 4362 kN 3

93 ®

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11.26

Layer 1

Ds ¢ (kN/m2)

s o¢ + Ds ¢

1335 = 22.23 (2.75 + 5)2

217.42

s o¢ (kN/m2) (15.72)(3) + (18.55 – 9.81)(3)+ (13)(19.18 – 9.81) = 47.16 + 26.22 + 121.81 = 195.19

(kN/m2)

195.19 + (5)(19.18 – 9.81) + (2.5)(18.08 – 9.81) = 262.72

1335 = 5.74 (2.75 + 12.5)2

268.46

262.72 + (2,5)(18.08 – 9.81) + (1.5)(19.5 – 9.81) = 297.93

1335 = 3.6 (2.75 + 16.5)2

301.53

Dsc (1) =

(0.8)(10) æ 217.42 ö log ç ÷ = 0.208 m 1 + 0.8 è 195.19 ø

Dsc (2) =

(0.31)(5) æ 268.46 ö log ç ÷ = 0.0073 m 1+1 è 262.72 ø

Dsc (3) =

(0.26)(3) æ 301.53 ö log ç ÷ = 0.0024 m 1 + 0.7 è 297.93 ø

SDsc » 217.7 mm

94 ®

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Chapter 12

12.1

Eq. (12.18): Qp (net )  Ap q ( N q*  1)

q  L1 c  L2 s  (6)(15.6)  (3)(17.6)  146.4 kN/m2 Ap 

 4

Db2 

 4

(2) 2  3.14 m 2

Eq. (12.19): Nq*  0.21e0.17   0.21e(0.17)(35)  80.59 L 63   4.5. Assume approximately 5. Db 2

Figure 12.10:    Qp (net)  (3.14)(146.4)[(0.815  80.59) 1]  29,734 kN

Qall  12.2

Qp (net ) FS



29, 734  9911 kN 3

Eq. (12.5): Qp (net )  Ap [q ( N q*  1) Fqs Fqd Fqc ]

   35. Table 12.1: N q  33.3 Fqs  1.7 (Table 12.1)  L  1  9  Fqd  1  2 tan  1  sin  2 tan 1    1  0.255 tan    1.344 2  Db 

Table 12.1: I cr  119.3 Eq. (12.10): I rr 

Ir 1  Ir 

95 ®

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Es 2(1   s )q  tan  

Ir 

Es  600 pa  (600)(100)  60,000 kN/m2 Table 12.1:  s  0.25

Ir 

60, 000  234.1 2(1  0.25)(146.4)(tan 35)

Eq. (12.14):   n

q pa

n = 0.0025 (Table 12.1)  146.4    (0.0025)    0.00366  100 

I rr 

234,1  126.08 1  (234.1 0.00366)

So I rr  I cr . Hence Fqc  1. Qp (net)  (3.14)[(146.4)(33.1  1)(1.7)(1.344)(1)]  33,925 kN

Qall(net) 

12.3

33,925  11, 308 kN 3

Eqs. (12.40) and (12.42):

Qs   * cu pL1  (0.4)(35)( 1.2)(6)  316.7 kN 12.4

q   (6.25)(17.8)  (2.5)(18.2)  156.75 kN/m2 Ap 

 4

(1.75) 2  2.405 m 2

Nq*  0.21e0.17   0.21e(0.17)(32)  48.39 L 8.75  5 Db 1.75

96 ®

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Figure 12.10:   0.79 Qp (net)  Ap q ( N q*  1)  (2.405)(156.75)[(0.79)(48.39)  1]  14, 034 kN Qall(net) 

12.5

14, 034  3509 kN 4

   32. Table 12.1: N q  23.18

Fqs  1.625  8.75  Fqd  (0.276) tan 1    1.379  1.75 

I cr  85.49 I rr 

Ir 1  Ir 

Ir 

Es 2(1  s )q  tan  

Es  400 pa  (400)(100)  40,000 kN/m2

 s  0.205 (Table 12.1) Ir 

40, 000  169.45 2(1  0.205)(156.75) (tan 32)

Eq. (12.14):   n

I rr 

q  156.75   (0.00325)    0.00509 pa  100 

169.45  90.98 1  (169.45  0.00509)

I rr  I cr , so Fqc  1. Q p (net )  Ap [q( N q*  1) Fqs Fqd Fqc ]  (2.405)[(156.75)(23.18  1)(1.625)(1.379)(1)]  18, 737 kN Qall(net) 

18, 737  4684 kN 4

97 ®

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12.6

a. Qs   *cu pL1  (0.4)(32)(  1)(6.25)  251.3 kN b. Qs   i*cu (i ) p( L1  1.5)  (0.55)(32)(  1)(6.25  1.5)  262.6 kN

12.7

a. Qp  Ap cu (2) N c*

Nc*  1.33[ln( I r )  1] cu (2) pa



75  0.75 100

From Eq. (12.38) and Figure 12.17: I r 

Fs  219 3cu (2)

Nc*  1.33[ln  219   1]  8.5 Qp 

 4

(1.5) 2 (75)(8.5)  1126.6 kN

b. Qs   *cu (1) pL1   *cu (1) pL1  (0.4)(  1.5)[(50)(6)  (75)(7)]  1555 kN c. Qw  12.8

1126.6  1555  894 kN 3

a. Eq. (12.44):  L    30   2 q p  6cub 1  0.2   (6)(2400) 1  (0.2)     43, 200 lb/ft Db   3   

Check: 9cu  (9)(2400)  21.600 lb/ft 2 . So use q p  21, 600 lb/ft 2   (21, 600)   (3) 2 4 Q p (net )  q p Ap   152.68 kip 1000

98 ®

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b. Qs   i*cu (1) p( L1  5)   i*cu (2) p( L1  Ds )  [(0.55)(  3)(1400)(20  5)  (0.55)(  3)(2400)(10  3)]  195,941 lb  195.14 kip

c. Qw 

12.9

152.68  195.94  116.2 kip 3

a. Eq. (12.31): q p  57.5 N60  (57.5)(19)  1092.5 kN/m 2 Since Db  1.27 m, q pr 

1.27  1.27  2 qp    (1092.5)  693.7 kN/m Db  2 

[Eq. (12.32a)] settlement of base 25   1.25% Db 2  1000

Figure 12.12: Using the trend line,

end bearing  0.41 Ap q pr

  Qall(net)  (0.41)   2 2  (693.7)  893.5 kN 4  

b. Eqs. (12.29) and (12.30):

Qs  fi pL1 fi  1 ozi  L  2

 11  2  (18)  99 kN/m 2

   1     ozi

 11   2

1  3   4 zi0.5  1.5  (0.244) 

0.5

 0.928

fi  (0.928)(99)  91.9 kN/m2 99 ®

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Qs  (91.9)(  1)(11)  3175.8 kN settlement 25   2.5% Ds 1 1000

Figure 12.13: Using the trend line,

side load transfer  0.9 f1 pL

Qs (net)  (0.9)(3175.8)  2858.2 kN c. Qw  893.5  2858.2  3752 kN 12.10 a. Eq. (12.33): With 35% gravel, 6  

1  7  8 zi0.75  2  0.15 zi0.75  2   0.15   2

0.75

 1.658

L   (18)(6)(1.658)  fpL1   1 1  ( Ds )( L1 )    [( )(0.75)(6)  1265.7 kN 2   2    Eq. (12.31): q p Ap   5 N 60 Ap  (57.5)(29)  (1.25) 2   2046.3 kN 4 

Qu  1265.7  2046.3  3312 kN

allowable settlement 12   0.016  1.6% Ds (0.75)(1000) From Figure 12.14a: Side load = (0.92)(ultimate side load) = (0.92)(1265.7) = 1164.4 kN

allowable settlement 12   0.0096  0.96% Db (1.25)(1000)

100 ®

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From Figure 12.12: end bearing = (0.3)(2046.3) = 613.9 kN Hence, Q  1164.4  613.9  1778 kN

12.11 a. L1  6  1.5  4.5 m; cu (1)  50 kN/m2 L2  7  1.5  5.5 m; cu (2)  75 kN/m2

fi pL1  (0.55)[(50)( 1.5)(4.5)  (75)( 1.5)(5.5)]  1652.3 kN  L    13   2 q p  6cub 1  0.2   (6)(75) 1  (0.2)     1230 kN/m Db   1.5    

Check: q p  9cub  (9)(75)  675 kN/m2 Use qb  675 kN/m2   Qu  1652.3  (675)  1.52   2845 kN 4 

allowable settlement 25   1.66% Ds (1.5)(1000) From Figure 12.20: side load  (0.87)(ultimate side load)

allowable settlement 25   1.66% Db (1.5)(1000) From Figure 12.21: end bearing  (0.77)(ultimate end load)   Q  (0.87)(1652.3)  (0.77)(675)  1.52   2356 kN 4  

101 ®

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12.12 From Problem 12.7(c), Qw  894 kN

Qws  0.8Qw  (0.8)(894)  715.2 kN

Qwp  894  715.2  178.8 kN Eq. (11.96): se (1) 

(Qwp   Qws ) L Ap E p

[178.8  (0.65)(715.2)](13)  2 6   1.5  (20 10 ) 4   236.8 106 m  0.24 mm



Eq. (11.98): se (2) 

Qwp C p Db q p

q p  cu (2) Nc*  (75)(9)  675 kN/m2 ; Cp  0.03

se (2) 

(178.8)(0.03)  0.0053 m  5.3 mm (1.5)(675)

Eq. (11.100): I ws  2  0.35

L 13  2  0.35  3.03 D 1.5

Assume  s  0.3; Es  12, 000 kN/ m 2 Q  D Eq. (11.99): se (3)   ws  (1   s2 ) I ws  pL  Es  715.2   1.5   (1  0.32 )(3.03)  0.004 m  4 mm     ( 1.5)(13)   12, 000 

Total settlement se  se(1)  se(2)  se(3)  0.24  5.3  4  9.54 mm 12.13 From Problem 12.8(c), Qw  116.2 kip

Qws  0.83Qw  (0.83)(116.2)  96.45 kip

Qwp  116.2  96.45  19.75 kip 102 ®

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Eq. (11.96): se (1) 



(Qwp   Qws ) L Ap E p [19.75  (0.65)(96.45)](30) 6   2   3 10 144   3    103 4  

 0.00081 ft  0.0097 in.

Eq. (11.98): se (2) 

Qwp C p Db q p

q p  cu (2) Nc*  (2.4 kip/ft 2 )(9)  21.6 kip/ft 2 ; C p  0.03

se (2) 

(19.75)(0.03)  0.0091 ft  0.11 in. (3)(21.6)

Q  D Eq. (11.99) se (3)   ws  (1   s2 ) I ws  pL  Es

I ws  2  0.35

L 30  2  0.35  3.107 D 3

Assume  s  0.3; Es  2, 000 lb/in.2 = 288 kip/ft2. So  96.45   3  2 se (3)     (1  0.3 )(3.107)  0.01 ft  0.12 in.   (  3)(30)   288 

Total settlement se  se(1)  se(2)  se(3)  0.0097  0.11  0.12  0.24 in.   12.14 Eq. (12.69): Q p  Ap [4.83(qu ) 0.51 ]   (1.5) 2  [(4.83)(1.81) 0.51 ]  11.55 MN 4 

Eq. (12.71): Qs  0.8(qu )0.5 ( DL)  (0.8)(1.81)0.5 [( )(1.5)(8)]  40.58 MN Qall 

11.55  40.58  13.03 MN 4

103 ®

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  Ds K p  12.15 a. Eq. (12.48): Qc  1.57 D ( E p RI )   E R  p I  

0.57

2 s

E p  22 106 kN/m2

RI = 1

Ip 

 D4 64



( )(1.25) 4  0.1198 m4 64

  35    K p  tan 2  45    tan 2  45    3.69 2 2     (17.5)(1.25)(35)(3.69)  Qc  (1.57)(1.25) (22  10 )(1)   (22  106 )(1)   2

Qg Qc



0.57

 326, 630 kN

260  0.0008 326, 630

From Figure 12.25, xo Ds  0.0025

xo  (0.0025)(1.25)  0.00313 m  3.13 mm b. Eq. (12.50):

  Ds K p  M c  1.33Ds3 ( E p RI )   E R  p I  

0.4

 (17.5)(1.25)(3.5)(3.69)   (1.33)(1.25) (22 10 )(1)   (22 106 )(1)    1,586,544 kN  m 3

0.4

From Figure 12.27, for

Qg Qc

 0.0008,

M max  0.000375 Mc

M max  (0.000375)(1,586,544)  594.9 kN-m

104 ®

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D   1.25  M max  s   594.9    2  2    c.  tensile    3104 kN /m 2 Ip 0.1198

E p RI (22 106 )(1)   7787.2  Ds K p (17.5)(1.25)(35)(3.69)  L     6 ; Lmin  (6)(1.25)  7.5 m  Ds  min

18, 000  1.81 m 28, 000

12.16 Ds  2.257

L1 

18, 000  8.79 m  (1.81)(360)

 max 

18, 000



 6995.6 kN/m 2

(1.81) 2

Try L2  6.5 m Eq. (12.54):    2c tan rc 2z   z   max exp      1    (1   ) Ec Db  c r   Er    (2)(0.35)(tan 34) (2)(6.5)   6995.6 exp     1.35 1.81   1  0.35  0.6  

= 2174 kN/m 2  2200 kN/m 2 (O.K.) Eq. (12.59): 

  1  

 z  L  Qw 2174  18, 000   1    max   Db L2  6995.6  ( )(1.81)(6.5) 2

= 335.6 kN/m2  360 kN/m 2 (O.K.) So, use length of 6.5 m. 105 ®

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12.17 Let us use L2 = 5.5 m    (2)(0.35)(tan 34) (2)(5.5)   z   max exp     1.35 1.81   1  0.35  0.6  

 (6995.6)(0.371)  2595 kN/m2  3000 kN/m2 

  1  

(O.K.)

 z  L  Qw 2595  18, 000   1     max   Db L2  6995.6  ( )(1.82)(5.5) 2

= 362kN/m2  360 kN/m2 Use length 5.5 m.

106 ®

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Chapter 13

13.1

Gs = 2.96; g w = 9.81 kN/m3 From Eq. (13.5): Gsg w (2.69)(9.81) gd £ = æ LL ö æ LL ö 1+ ç ÷ (Gs ) 1 + 2.69 ç ÷ è 100 ø è 100 ø 26.39 = 1 + 0.0269 LL

gq below which LL

collapse will occur (kN/m3)

20.8

18.8

17.16

15.78

14.60

13.59

12.71

The plot is shown below.

107 ®

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13.2

LL = 0.31; Gs = 2.64. For collapse, Eq. (13.5):

gd £

Gsg w (2.64)(9.81) = = 14.24 kN/m3 1 + (LL)(Gs ) 1 + (0.31)(2.64)

Since field gd = 13.2 kN/m3 is less than 14.24 kN/m3, collapse is likely to occur. 13.3

s c¢ = 1700 lb/ft 2 ; s o¢ = 1200 lb/ft 2 s c¢ 1700 = = 1.42 (within the range of 0.8 to 1.5) s o¢ 1200 The soil is normally consolidated.

13.4

LL = 50; w = 20%; Z = 8 m. From Figure 13.13, sw (free) = 3% Eq. (13.10):

DS F = 0.0033Zsw(free) = (0.0033)(8)(3) = 0.0792 m = 79.2 mm 13.5

LL = 55; w = 20%; Z = 8 m. Figure 13.13: sw(free) » 4.5%

DSF = (0.0033)(8)(4.5) = 0.1188 m » 119 mm 13.6

DS =

1 é1 1 1 1 ù (0.6 + 0)(2) + (0.6 + 1.5)(2) + (1.5 + 2.75)(2) + (2.75 + 4.75)(2) ú ê 100 ë 2 2 2 2 û

= 0.1445 ft = 1.73 in.

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13.7

Refer to the figure in Problem 13.6.

Total swell, DS (in.)

Depth (ft) 12

1 æ 1 ö 0 + (0.6 + 0)(2) ç ÷ (12) = 0.072 2 è 100 ø

1 æ 1 ö 0.072 + (1.5 + 0.6)(2) ç ÷ (12) = 0.324 2 è 100 ø

1 æ 1 ö 0.324 + (1.5 + 2.75)(2) ç ÷ (12) = 0.834 2 è 100 ø

1 æ 1 ö 0.834 + (2.75 + 4.75)(2) ç ÷ (12) = 1.734 2 è 100 ø

The plot of DS versus depth is shown below. From the figure, the depth of undercut is: 5.40 – 4 = 1.4 ft below the bottom of the foundation.

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13.8

1 1 é1 ù (0.4 + 0)(1) + (0.4 + 0.75)(1) + (0.75 + 1.5)(1) ú ê 1 2 2 2 DS = ê ú 1 1 100 ê ú + (1.5 + 3.1)(1) + (3.1 + 5.5)(0.5) ëê 2 2 ûú = 0.0635 m = 63.5 mm

13.9

Depth (m)

Total swell, DS (m)

1 æ 1 ö 0 + (0.4 + 0)(1) ç ÷ = 0.002 2 è 100 ø

1 æ 1 ö 0.002 + (0.75 + 0.4)(1) ç ÷ = 0.00775 2 è 100 ø

1 æ 1 ö 0.00775 + (0.75 + 1.5)(1) ç ÷ = 0.019 2 è 100 ø

1 æ 1 ö 0.019 + (1.5 + 3.1)(1) ç ÷ = 0.042 2 è 100 ø

1.5

1 æ 1 ö 0.042 + (3.1 + 5.5)(0.5) ç ÷ = 0.0635 2 è 100 ø 110 ®

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The plot of DS versus depth is shown. From the figure, the depth of undercut is: 2.5 – 1.5 = 1 m below the bottom of the foundation.

13.10 Eq. (13.14): U = p Ds Z s sw ¢ tan f ¢ps

Ds = 1 m; Z = 9 m U = (p ) (1)(9)(600)(tan 20) = 6174.6 kN Eq. (13.17): U =

cu N c p ( Db2 - Ds2 ) FS 4

é (150)(6.14) ù æ p ö 2 2 2 6174.6 = ê ç ÷ ( Db - 1 ) = 241.1( Db - 1) ú 3 ë ûè 4 ø Or, 25.61 + 1 = Db2 Db = 5.16 m

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æp ö (cu Nc ) ç ÷ ( Db2 - Ds2 ) è4ø 13.11 Eq. (13.17): FS = U -D æp ö (150)(6.14) ç ÷ ( Db2 - 12 ) 723.35( Db2 - 1) 4ø è 4= = (6174.6) - (1500) 4674.6 Or, 25.85 = Db2 - 1 Db = 5.2 m

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Chapter 14

14.1

Ko = 1 - sin j ¢ = 1 - sin 30 = 0.5 At z = 0 m:

s h¢ = 0; u = 0

At z = 3 m:

s h¢ = (0.5)[(16.5)(3)] = 24.75 kN/m 2 ; u = 0

At z = 6 m:

s h¢ = (0.5)[(16.5)(3) + (3)(19 - 9.81)] = 38.54 kN/m 2

u = (3)(9.81) = 29.43 kN/m2 The pressure diagram is shown.

1 1 1 Po = (3)(24.75) + (24.75)(3) + (3)(38.54 - 24.75) + (3)(29.43) 2 2 2 = 37.125 + 74.25 + 20.685 + 44.145 = 176.21 kN /m

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Taking moment about the bottom of the wall. é 3ö æ æ3ö æ 3ö æ 3 öù ê(37.125) ç 3 + 3 ÷ + (74.25) ç 2 ÷ + (20.685) ç 3 ÷ + (44.145) ç 3 ÷ ú è ø è ø è ø è øû z=ë 176.21 =

14.2

148.5 + 111.375 + 20.685 + 44.145 = 1.84 m 176.21

Given, PI = 26 Eq. (14.3):

Ko = 0.44 + 0.0042(PI) = 0.549

Eq. (14.4):

Ko (OC) = Ko(NC) OCR = (0.549)( 2) = 0.776

At z = 0 m:

s h¢ = 0

At z = 5 m:

s h¢ = (5)[(18)(0.776)] = 69.84 kN/m 2

Po = z= 14.3

1 (5)(69.84) = 174.6 kN /m 2

5 = 1.67 m 3

Ko = 1 - sin 30 = 0.5 At z = 0 ft:

s h¢ = K os o¢ = (0.5)(300) = 150 lb/ft 2

At z = 4 ft:

s h¢ = (0.5)[300 + 4(105)] = 360 lb/ft 2 ; u = 0

At z = 10 ft: s h¢ = (0.5)[300 + 4(105) + (6)(122 - 62.4)] = 538.8 lb/ft 2

u = (6)(62.4) = 374.4 lb/ft 2

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P = A1 + A2 + A3 + A4 + A5 = 600 + 420 + 2160 + 536.4 + 1123.2 = 4839.6 lb/ft z=

14.4

(600)(8) + (420)(7.34) + (2160)(3) + (536.4)(2) + (1123.3)(2) = 3.65 ft 4839.6

1æ 2c¢ ö Eq. (14.15): Pa = ç H ÷ g HK a - 2c¢ K a 2 çè g K a ÷ø

(

)

j¢ ö 24 ö æ æ K a = tan 2 ç 45 - ÷ = tan 2 ç 45 - ÷ = 0.422; K a = 0.649 2ø 2 ø è è Pa =

1é (2)(10) ù 5.5 [(18.7)(5.5)(0.422) - (2)(10)(0.649)] ê 2ë (18.7)(0.649) úû

1 = (3.852)(43.4 - 12.98) = 58.6 kN /m 2

14.5

jö æ (a) K a = tan 2 ç 45 - ÷ 2ø è

j =0 Ka = 1 The pressure distribution diagram is shown.

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(b) Eq. (14.12):

zc =

2c (2)(630) = = 11.15 ft g K a (113)(1)

1æ 2c ö Pa = ç H ÷ g HK a - 2c K a 2 çè g K a ÷ø 1é (2)(630) ù = ê 21 [(113)(21)(1) - (2)(630)(1)] = 5481 lb /ft 2ë 113 úû

(

14.6

)

j¢

j1¢ = 34°; K a (1) = tan 2 æç 45 - 1 ö÷ = tan 2 æç 45 2ø

34 ö ÷ = 0.283 2ø

j¢ 2ø

j 2¢ = 25°; K a (2) = tan 2 æç 45 - 2 ö÷ = tan 2 (45 - 12.5) = 0.406; K a (2) = 0.637 è

At z = 0 ft:

s a¢ = 0

At z = 8.2 ft (top layer):

s a¢ = g 1 zKa (1) = (107)(8.2)(0.283) = 248.3 lb/ft 2

At z = 8.2 ft (bottom layer):

s a¢ = g 1zKa (2) - 2c2¢ Ka(2) = (107)(8.2)(0.406) - (2)(209)(0.637) = 356.22 - 266.27 = 89.95 lb/ft 2

s a¢ = [(107)(8.2) + (140 - 62.4)(14.8)](0.406) - (2)(209)(0.637)

At z = 23 ft:

= 822.5 - 266.27 = 556.23 lb/ft 2 u = (62.4)(14.8) = 923.5 lb/ft 2

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The pressure diagram follows:

Pa = area of pressure diagram 1 + 2 + 3 + 4 = 1018 + 1331 + 3450 + 6834 = 12, 633 lb/ft = 12.6 kip/ft 14.7

a. j ¢ = 34°; a = 10°; Ka = 0.294 (Table 14.3); s a¢ = g zKa At z = 2 m:

s a¢ = (2)(17)(0.294) = 10 kN /m 2

At z = 4 m:

s a¢ = 19.99 kN /m 2

At z = 6 m:

s a¢ = 29.99 kN /m 2

1 1 b. Pa = g H 2 K a = (17)(6)2 (0.294) = 89.96 kN /m 2 2

14.8

6 2 m (from the bottom of the wall) inclined = 3 at an angle of 10° with the horizontal

Eq. (14.27): zc =

2c¢ 1 + sin j ¢ (2)(250) 1 + sin 25 = = 6.82 ft g 1 - sin j ¢ 115 1 - sin 25

c¢ 250 = = 0.099 » 0.1 g z (115)(22)

At z = 22 ft:

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For j ¢ = 25°, a = 10°. From Table 14.4, K a¢ = 0.296. So

s a¢ = g zK a¢ cos a = (115)(22)(0.296) cos10 = 737.5 lb/ft 2

At z = 22 ft:

1 Pa = (737.5)(22 - 6.82) » 5598 lb /ft 2 14.9

2 Note: d ¢ = 20°; j ¢ = 30°. So d ¢ = j ¢; a = 10°; b = 80°. 3 From Table 14.6, Ka = 0.4376. 1 1 Pa = g H 2 K a = (16.5)(6)2 (0.4376) = 129.97 kN /m 2 2 Acts a distance of 2 m from the bottom of the wall inclined at an angle of 20° to the normal drawn to the back face of the wall.

14.10 a. Eq. (14.28): Pa =

1 d ¢ 20 2 K ag H 2 . = = ;a = 10°; b = 85°; j ¢ = 30° 2 j ¢ 30 3

Table 14.6: K a = 0.3857

Pa =

1 (0.3857)(105)(12) 2 = 2916 lb /ft 2

Acts a distance of

12 = 4 ft from the bottom of the wall inclined at an angle of 3

20° to the normal drawn to the back face of the wall.

d 15 1 = = ; a = 20°; b = 85°. Table 14.7: K a = 0.4708 j ¢ 30 2 1 Pa = (0.4708)(105)(12)2 = 3559 lb /ft 2 Acts a distance of 4 ft from the bottom of the wall inclined at an angle of 15°to the normal drawn to the back face of the wall.

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jö æ 14.11 K p = tan 2 ç 45 + ÷ ; j = 0; K p = 1 2ø è

s p = s o K p + 2c K p

At z = 0 ft:

= 0 + (2)(630)(1) = 1260 lb/ft 2 At z = 21 ft: s p = (113)(21)(1) + (2)(630)(1)

= 2373 + 1260 = 3633 lb/ft 2

The pressure distribution diagram is shown.

Pp = Areas of 1 + 2 = 26, 460 + 24,916.5 = 51, 376.5 lb /ft z=

(26, 460)(10.5) + (24,916.5)(7) = 8.8 ft from the bottom of the wall 51,376.5

1 14.12 Eq. (14.70): Pp = g H 2 K p + 2c ¢H K p 2

j¢ ö 24 ö æ æ K p = tan 2 ç 45 + ÷ = tan 2 ç 45 + ÷ = 2.371. 2ø 2 ø è è

K p = 1.54

1 Pp = (18.7)(5.5)2 (2.371) + (2)(10)(5.5)(1.54) = 840 kN /m 2 119 ®

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j¢ ö æ 14.13 K p (1) = tan 2 ç 45 + 1 ÷ = tan 2 (45 + 17) = 3.537 2ø è K p (2) = tan 2 (45 + 12.5) = 2.463; K p(2) = 1.57 At z = 0 ft:

s ¢p = 0

At z = 8.2 ft (first layer):

s ¢p = g 1 zK p (1) = (107)(8.2)(3.537) = 3103.36 lb/ft 2

= 3.1 kip/ft 2 At z = 8.2 ft (bottom layer): s ¢p = g 1 zK p (2) + 2c¢ K p(2)

= (107)(8.2)(2.463) + (2)(209)(1.57) = 2161 + 656 = 2817 lb/ft 2 » 2.8 kip/ft 2 u=0

s ¢p = [(107)(8.2) + (140 - 62.4)(14.8)](2.463) + (2)(209)(1.57)

At z = 23 ft:

= 4989.7 + 656 = 5645.7 lb/ft 2 » 5.65 kip/ft 2 u = (14.8)(62.4) = 923.5 lb/ft 2 » 0.92 kip/ft 2 The pressure distribution diagram follows. σ ρ' (kip/ ft 2 ) 1

3.1

2.8

8.2

u (kip / ft 2 )

5.65

23 z (ft)

0.92 z (ft)

Pp = area of pressure diagrams 1 + 2 + 3 + 4 = 12.71 + 41.44 + 21.09 + 6.81 = 82.05 kip /ft

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14.14 j ¢ = 30°; d ¢ = 20°; a = 10°; b = 80° Eq. (14.80): K p = 7.16 (Table 14.17)

1 1 Pp = g H 2 K p = (16.5)(6)2 (7.16) = 2126.5 kN /m 2 2 Acting at a distance of 2m from the bottom of the wall inclined at an angle of 20° to the normal drawn to the back face of the wall. 14.15 f ¢ = 36°;

d¢ = 0.5.Table 14.7: for a = 0° and b = 85°, Ka = 0.2718. f¢

Eq. (14.31):

é sin 85 ù é (2)(30) ù sin b ù é 2q ù = 17 + ê = 32 kN/m3 ú ú ê ú ê ú ë sin( b + a ) û ë H û ë sin(85 + 0) û ë 4 û é

g eq = g + ê

1 1 Pa = ( K ag eq H 2 ) = (0.2718)(32)(4) 2 = 69.58 kN /m 2 2 14.16 Eq. (14.45): P =

q [ H (q 2 - q1 )] 90

æ b¢ ö -1 æ 1.5 ö ÷ = tan ç ÷ = 24.4° èHø è 3.3 ø

q1 = tan -1 ç

æ a¢ + b¢ ö -1 æ 2.5 ö ÷ = tan ç ÷ = 37.1° è H ø è 3.3 ø

q 2 = tan -1 ç P=

25 [(3.3)(37.1 - 24.4)] = 11.64 kN /m 90

1 14.17 Eq. (14.53): Pae = g H 2 (1 - kv ) K ae 2 For kh = 0.2,

d ¢ 20 2 = = ; f ¢ = 30°; a = 10°; b = 85° f ¢ 30 3

é kh ù -1 æ 0.2 ö Eq. (14.55): q ¢ = tan -1 ê ú = tan ç ÷ = 11.3° è1- 0 ø ë (1 - kv ) û 121 ®

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Eqs. (14.57) and (14.58):

b ¢ = b - q ¢ = 85 - 11.3 = 73.7° a ¢ = q ¢ + a = 11.3 + 10 = 21.3° æ sin 2 b ¢ ö 1 2 ¢ ¢ Eq. (14.56): Pae = g H (1 - kv )[ K a ( b , a )] ç ÷ 2 2 è cos q ¢ × sin b ø From Eq. (14.29) for b′= 73.7°; a′ = 21.3°; d′ = 20°; and f′ = 30°, the value of

Ka (b ¢, a ¢) = 0.6676. æ sin 2 73.7 ö æ1ö 2 Pae = ç ÷ (18.2)(5) (1 - 0)(0.6676) ç ÷ = 143.77 kN /m 2 è2ø è cos11.3 × sin 85 ø Eq. (12.28): Pa = For

1 ( K ag H 2 ) 2

d¢ 2 = : f ¢ = 30°; a = 10°; b = 85°; Ka = 0.3857 (Table 14.6) f¢ 3

1 Pa = (0.3857)(18.2)(5) 2 = 87.75 kN /m 2

DPae = 143.77 - 87.75 = 56.02 kN/m æHö æ5ö 0.6 H (DPae ) + ç ÷ Pa (0.6)(5)(56.02) + ç ÷ (87.75) è3ø = è 3ø Eq. (14.60): z = = 2.19 m Pae 143.77 14.18 Eq. (14.64): c* =

c¢ 113 = = 0.098 » 0.1 g H (115)(10)

f ¢ = 25°; kh = 0.2 Figure14.23(b): Pae* = 0.167

Pae = Pae* g H 2 = (0.167)(115)(10) 2 = 1920.5 lb /ft

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14.19 a = 17.5° and f ¢ = 35°;

a = 0.5 f¢

K p = ( R¢)[ K p (d ¢/f ¢=1) ] From Figure 14.30(a), K p (d ¢ /f ¢ =1) = 23.

d ¢ 10 = = 0.29. Table 14.19: R¢ » 0.53. f ¢ 35

Hence, K p = (23)(0.53) = 12.19

1 1 Pp = g H 2 K p = (16.5)(4)2 (12.19) = 1609.1 kN /m 2 2

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Chapter 15 15.1

Refer to the diagram. f1¢ = 32°; a = 10°. From Table 14.3, K a = 0.321.

1 1 Pa = ( H ¢)2 g 1K a = (9.577) 2 (16.8)(0.321) = 247.31 kN/m 2 2

Pv = Pa sin10° = 42.92 kN/m; Ph = Pa cos10° = 243.55 kN/m Refer to the table. Section

Weight (kN/m)

Moment arm from C (m)

Moment about C (k/N-m/m)

(8)(0.4)(g c ) = 75.46

1.9

143.4

(0.5)(0.2)(8)(g c ) = 18.86

1.63

30.7

(5.6)(0.96)(g c ) = 126.77

2.8

354.96

8 + 8.617 (3.5)(16.8) = 488.6 2

1.5 + 0.6 + 1.75 = 3.85

1880.7

5.6

240.5

Pv = 42.94 S752.53

S 2650.26

Note: g c = 23.58 kN/m3

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M O = Ph

H ¢ (243.55)(9.577) = = 777 kN-m 3 3

FS(overturning) =

2650.26 = 3.41 777

æ2ö æ2ö SV tan ç ÷ f2¢ + B ç ÷ c2¢ è3ø è3ø FS(sliding) = Pa cos a éæ 2 ö ù æ2ö (752.53) tan êç ÷ (28) ú + (5.6) ç ÷ (30) è3ø ëè 3 ø û = = 1.5 243.55

B SM R - S M O 2560.26 - 777 = 2.8 = 0.31 m 2 SV 752.33

qtoe =

SV æ 6e ö 752.53 é (6)(0.31) ù 1+ = 179 kN/m 2 ç1 + ÷ = ê ú B è Bø 5.6 ë 5.6 û

B ¢ = B - 2e = 5.6 - (2)(0.31) = 4.98 m 1 Eq. (15.21): qu = c2¢ N c Fcd Fci + qN q Fqd Fqi + g 2 B¢Ng Fg d Fg i 2 From Table 5.3 for f2¢ = 28° : N c = 25.8; N q = 14.72; Ng = 16.72

æ 1.75 ö Fqd = 1 + 0.299 ç ÷ = 1.105 è 4.98 ø Fcd = Fqd -

1 - Fqd N c tan f2¢

æ Yö Fqi = Fci = ç 1 - ÷ è 90 ø

= 1.105 -

1 - 1.105 = 1.11 25.8 tan 28

æ P cos a ö æ 243.55 ö Y = tan -1 ç a = tan -1 ç ÷ » 18° ÷ è 752.53 ø è åV ø

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æ 18 ö Fqi = Fci = ç1 - ÷ = 0.64 è 90 ø

q = 1.75g 2 = (1.75)(17.6) = 30.8 kN/m 2 Fg d = 1 2

æ Y ö æ 18 ö Fg i = ç1 - ÷ = ç1 - ÷ » 0.13 è f2¢ ø è 28 ø

So, qu = (30)(25.8)(1.11)(0.64) + (30.8)(14.72)(1.105)(0.64) 1 + (17.6)(4.98)(16.72)(1)(0.13) 2 = 549.8 + 321 + 95.3 » 966 kN/m 2

FS(bearing) = 15.2

qu 966 = » 5.4 qtoe 179

Ka = 0.26; H ¢ = 7.3 m 1 1 Pa = g ( H ¢) 2 K a = (18.08)(7.3)2 (0.26) = 125.25 kN/m 2 2

Ph = 125.25 kN/m; Pv = 0

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Weight (kN/m)

Moment arm from C (m)

Moment about C (kN-m/m)

(0.3)(6.5)(g c ) = 45.98

0.8 + 0.3 + 0.15 = 1.25

57.48

(0.5)(0.3)(6.5)(g c ) = 22.99

0.8 + 0.2 = 1.00

22.9

(3.4)(0.8)(g c ) = 64.14

1.70

109.04

(6.5)(2)(18.08) = 235.4

2.40

564.1

Section

S368.15

S753.52

gc = 23.58 kN/m2 æ 7.3 ö M O = (125.25) ç ÷ = 304.78 kN-m è 3 ø FS(overturning) =

753.52 = 2.47 304.78

éæ 2 ö ù æ2ö (3.68.15) tan êç ÷ (15) ú + (3.4) ç ÷ (30) è3ø ëè 3 ø û FS(sliding) = = 1.06 125.25 e=

3.4 753.52 - 304.78 = 0.481 m 2 368.15

qtoe =

368.15 é (6)(0.481) ù 1+ = 200.19 kN/m 2 ê ú 3.4 ë 3.4 û

B¢ = B - 2e = 3.4 - (2)(0.481) = 2.438 m 1 qu = c2¢ N c Fcd Fci + qN q Fqd Fqi + g 2 B ¢Ng Fg d Fg i 2

æ 1.5 ö Fqd = 1 + 0.294 ç ÷ = 1.181 è 2.438 ø 1 - Fqd 1 - 1.181 Fcd = Fqd = 1.181 = 1.243 Nc tan f2¢ 10.98 tan15 æ 125.25 ö Y = tan -1 ç ÷ = 18.79° è 368.15 ø 2

æ 18.79 ö Fci = Fqi = ç 1 ÷ = 0.626 90 ø è 128

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æ 18.79 ö Fg i = ç1 ÷ = 0.064 15 ø è

qu = (30)(10.98)(1.243)(0.626) + (1.5)(19.65)(1.181)(0.626) 1 + (19.65)(2.438)(2.65)(1)(0.064) 2 = 346.85 kN/m 2

FS(bearing) = 15.3

346.85 346.85 = = 1.73 qtoe 200.19

Refer to the figure.

f1¢ = 32°; H ¢ = 6.8 m 32 ö æ K a = tan 2 ç 45 - ÷ 2 ø è = 0.307 1 Pa = Ph = g ( H ¢ ) 2 K a 2 1 = (16.5)(6.8)2 (0.307) 2 = 117.1 kN/m Refer to the table. Section

Weight (kN/m)

Moment arm from C (m)

Moment about C (kN-m/m)

(0.5)(2)(6)(g c ) = 141.48

3.77

533.38

(0.6)(6)(g c ) = 84.89

2.8

237.69

141.48

1.83

258.9

(5.85)(0.8)(g c ) = 110.35

2.925

322.8

(0.5)(2)(6)(16.5) = 99.00

4.43

438.57

(0.75)(6)(16.5) = 74.25

5.475

405.52

S 2196.86

S651.45

gc = 23.58 kN/m3 129

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MO =

H ¢Pa æ 6.8 ö =ç ÷ (117.1) = 265.4 kN-m 3 è 3 ø

FS(overturning) =

SM R 2196.86 = = 8.28 SM O 265.4

æ2 ö æ2ö æ2ö SV tan ç f2¢ ÷ + ç ÷ c2¢ B (651.45) tan(14.66) + ç ÷ (40)(5.85) è3 ø è3ø è3ø FS(sliding) = = Pa 117.1 =

15.4

170.42 + 156 = 2.79 117.1

d¢ 2 = . From Table 14.6, for f1¢ = 32°, a = 0, b = 71.57°, Ka = 0.45, d ¢ = 21.33°. f1¢ 3 1 Pa = (16.5)(6.8)2 (0.45) = 171.67 kN/m 2

Ph = 171.67cos(21.33 + 18.43) = 131.97 kN/m Pv = 171.67sin(21.33 + 18.43) = 109.8 kN/m

Refer to sections in the figure shown for problem 15.3.

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Section

Weight (kN/m)

Moment arm from C (m)

Moment about C (kN-m/m)

141.48

3.77

533.38

84.89

2.8

237.69

141.48

1.83

258.9

110.35

2.925

322.8

109.8

4.611

506.29

S588

M O = Ph

S1859.06

H¢ æ 6.8 ö = (131.97) ç ÷ = 299.13 kN-m 3 è 3 ø

FS(overturning) =

1859.06 = 6.2 299.13

æ2 ö æ2ö æ2ö SV tan ç f2¢ ÷ + ç ÷ c2¢ B (588) tan(14.66) + ç ÷ (40)(5.85) è3 ø è3ø è3ø FS(sliding) = = = 2.35 Ph 131.97 15.5

a. Eq. (15.33): CIE =

kv = 0; tan q ¢ =

CIE =

sin(b - d ¢ ) - cos(b - d ¢ ) tan f2¢ (1 - kv )(tan f2¢ - tan q ¢)

kh = 0.3 1 - kv

sin(90 - 15) - cos(90 - 15) tan 36 = 1.82 (1 - 0)(tan 36 - 0.3)

For kh = 0.3, f1¢ = 30° and d1¢ =

f1¢ 2

, the value of Kae = 0.563 (Table 14.8)

1 1 Ww = g 1H 2 (1 - kv ) K aeC AE = (18)(7)2 (1 - 0)(0.563)(1.82) = 451.9 kN 2 2 With a factor of safety of 2, Ww = 903.8 kN /m

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æ 0.2 Av2 ö b. kh = Aa ç ÷ è Aa D ø tan q ¢ =

CIE =

0.25

é (0.2)(0.15) 2 ù = (0.25) ê ú ë (0.25)(2 in.) û

0.25

= 0.077

0.077 = 0.077 1- 0

sin(90 - 15) - cos(90 - 15) tan 36 = 1.198 (1 - 0)(tan 36 - 0.077)

f¢ 2

æ è

ö ø

From Table 14.8, K ae » 0.35 ç for f1¢ = 30° and d1¢ = 1 , a = 0, kh = 0.077 ÷ . 1 1 Ww = (FS) g 1H 2 (1 - kv ) K aeC AE = (2) (18)(7) 2 (1 - 0)(0.35)(1.198) 2 2 = 369.8 kN /m

15.6

b¢ = 2 m; a¢ = 1.5 m; q = 70 kN/m2 s o¢ (1) = g z

z (m)

(kN/m2) 34 68 102 136

2 4 6 8 †

15.7

z 2b ¢ 0.5 1.0 1.5 2.0

s o¢ (2)

s o¢ = s o¢ (1) + s o¢ (2)

(kN/m2)

(kN/m2) 64.0 87.1 118.15 150.0

30† 19.1† 16.15†† 14††

Eq. (15.38); ††Eq. (15.39)

Eq. (15.42): M = 1.4 -

0.4b¢ (0.4)(2) = 1.4 = 0.686. So use M = 1.0. 0.14 H (0.14)(8)

Refer to Figure 15.36.

æ b¢ + a¢ ö -1 æ b¢ ö -1 æ 3.5 ö -1 æ 2 ö b = tan -1 ç ÷ - tan ç ÷ = tan ç ÷ - tan ç ÷ è z ø èzø è z ø èzø æ b¢ + a¢ ö b 1 é -1 æ b¢ + a¢ ö -1 æ b¢ ö ù ÷ - = ê tan ç ÷ + tan ç ÷ ú è z ø 2 2ë è z ø è z øû

a = tan -1 ç =

1 é -1 æ 3.5 ö -1 æ 2 ö ù tan ç ÷ + tan ç ÷ ú ê 2ë è z ø è z øû

K a = tan 2 (45 - 17.5) = 0.271; g 1 = 17 kN/m3 ; q = 70 kN/m 2 132

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s a¢ (1) = K ag z

z (m) 2 4 6 8 †

15.8

(kN/m2) 9.21 18.43 27.64 36.86

α (deg) 52.63 33.9 24.35 18.84

β (deg) 15.26 14.62 11.83 9.59

æ è

34 ö ÷ = 0.2827 2ø

s a¢ (2)† (kN/m2) 14.94 7.11 3.15 1.57

s a¢ = s a¢ (1) + s a¢ (2) (kN/m2) 24.15 25.54 30.79 38.43

Eq. (15.41)

f¢ ö 2ø

æ è

a. K a = tan 2 ç 45 - 1 ÷ = tan 2 ç 45 Eq. (15.52) t=

(g HK a SV S H )[ FS( B ) ] wf y

(16)(10)(0.2827)(1)(1.25)(3) (0.12)(260, 000)

= 0.005437 m » 5.44 mm

b. Eq. (15.51). At z = 0,

f ¢ ö FS( P ) SV S H ( K ag z ) æ L = H tan ç 45 - 1 ÷ + 2 ø (2w tan fµ¢ )(g z ) è = 10 tan(45 - 17) +

(3)(0.2827)(1)(1.25) (2)(0.12)(tan 25)

= 14.79 m 15.9

a. Check for overturning:

1 1 Pa = g 1H 2 K a = (16)(10) 2 (0.2827) = 226.16 kN/m 2 2

æ 10 ö M O = Pa z ¢ = (226.16) ç ÷ = 753.87 kN-m/m è3ø

L = 14.79 m

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Eq. (15.54):

æ L2 ö (10)(16)(14.79)2 æLö M R = ( HL)(g 1 ) ç ÷ = H g 1 ç ÷ = » 17,500 kN-m/m 2 è2ø è2ø FS(overturning) =

17,500 = 23.2 753.87

b. Check for sliding: Eq. (15.56):

FS(sliding) =

ö (16)(10)(14.79) tan éæ 2 (34) ö ù ÷ú êç øû ø= ëè 3 = 4.37 226.16

æ2 è

g 1HL tan ç f1¢ ÷ 3 Pa

c. Check for bearing capacity:

f2¢ = 25°. From Table 5.3, N c = 20.72; Ng = 10.88 1 qult = c2¢ N c + g 2 LNg 2 1 = (30)(20.72) + (15.5)(14.79)(10.88) 2 = 621.6 + 1247.1 » 1869 kN/m 2

s v ( h) = g 1h = (16)(10) = 160 kN/m2 FS(bearing) =

1869 = 11.68 160

æ è

15.10 f1¢ = 30°. K a = tan 2 ç 45 -

Eq. (15.62): SV =

30 ö ÷ = 0.333. 2ø

Tall g 1 zK a [FS( B ) ]

At z = H = 6 m: SV =

16 = 0.336 m (15.9)(6)(0.333)(1.5) 134

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SV K a [FS( B ) ] H -z + f¢ ö 2 tan fF¢ æ tan ç 45 + 1 ÷ 2ø è

At z = 0 (maximum length):

6 (0.336)(0.333)(1.5) + = 3.47 + 0.23 = 3.7 m tan(45 + 15) éæ 2 ö ù 2 tan êç ÷ (30) ú ëè 3 ø û

Eq. (15.66): ll =

SV K a [FS( P ) ] (0.336)(0.333)(1.5) = = 0.115 m --- Use 1 m 4 tan fF¢ éæ 2 ö ù 4 tan êç ÷ (30) ú ëè 3 ø û

15.11 Check for overturning:

1 Pa = (15.9)(6)2 (0.333) = 95.3 kN/m 2 M O = Pa

6 = 190.6 kN-m/m 3

Use L = 3.7 for all depths.

MR =

H g 1L2 (6)(15.9)(3.7) 2 = = 653 kN-m/m 2 2

FS(overturning) =

653 = 3.43 190.63

æ2 ö g 1HL tan ç f ¢ ÷ è 3 ø = (15.9)(6)(3.7)(tan 20) = 1.35 Check for sliding: FS(sliding) = Pa 95.3 1

Check for bearing capacity: For f2¢ = 20°, from Table 5.3, N c = 14.83; Ng = 5.39

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1 1 qult = c2¢ Nc + g 2 LNg = (55)(14.83) + (16.8)(3.7)(5.39) = 983.17 kN/m2 2 2

FS(bearing) =

983.17 = 10.31 (15.9)(6)

Note: FS against sliding is below acceptable values, so increase L to about 8 m.

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Chapter 16

16.1

Refer to Figure 16.11 in the text. L1 = 3 m; L2 = 8 m.

g = 16 kN/m3 ; g sat = 18 kN/m3 ; j ¢ = 30°

g ¢ = 18 - 9.81 = 8.19 kN/m3 30 ö 1 30 ö æ æ K a = tan 2 ç 45 - ÷ = ; K p = tan 2 ç 45 + ÷ = 3 2ø 3 2ø è è æ1ö s 1¢ = g L1 K a = (16)(3) ç ÷ = 16 kN/m 2 è3ø æ1ö s 2¢ = (g L1 + g ¢L2 ) K a = [(16)(3) + (8.19)(8)] ç ÷ = 37.84 kN/m 2 è3ø

L3 =

37.84 = 1.73 m (8.19)(3 - 0.333)

1 1 1 P = (3)(16) + (8)(16) + (8)(37.84 - 16) + (1.73)(37.84) 2 2 2 = 24 + 128 + 87.36 + 32.73 = 272.09 kN/m

3ö 8 ö æ 2 ´1.73 ö æ æ P( z ) = (24) ç 9.73 + ÷ + (128)(5.73) + (87.36) ç1.73 + ÷ + ç ÷ (32.73) 3ø 3ø è 3 ø è è z=

1412.8 = 5.19 m 272.09

s 5¢ = (g L1 + g ¢L2 ) K p + g ¢L3 ( K p - K a ) = [(16)(3) + (8.19)(8)](3) + (8.19)(1.73)(2.667) = 378.35 kN/m2

A1 =

378.35 = 17.32 (8.19)(2.667) 137 ®

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A2 =

(8)(272.09) = 99.65 (8.19)(2.667)

A3 =

(6)(272.09)[(2)(5.19)(8.19)(2.667) + 378.35] = 2070.4 (8.19)2 (2.667)2

A4 =

(272.09)[(6)(5.19)(378.35) + (4)(272.09)] = 7339.8 (8.19)2 (2.667)2

L44 + 17.32 L34 - 99.65L24 - 2070.4 L4 - 7339.8 = 0; L4 = 11.7 m

D = L3 + L4 = 1.73 + 11.7 = 13.43 m b. Total length = 3 + 8 + (1.3)(13.43) » 28.5 m c.

z¢ =

2P (2)(272.09) = = 4.99 m » 5 m ( K p - K a )g ¢ (8.19)(2.667)

1 1 M max = P( z + z ¢ ) - g ¢ z ¢ 3 ( K p - K a ) = (272.09)(5.19 + 5) - (8.19)(5)3 (2.667) 6 6 = 2317.5 kN-m /m 16.2

a. Refer to Figure 16.11 in the text.

L1 = 4 m; L2 = 8 m; f ¢ = 32° g = 16.1 kN/m3 ; g sat = 18.2 kN/m3 ; g ¢ = 18.2 - 9.81 = 8.39 kN/m 3 32 ö 32 ö æ æ K a = tan 2 ç 45 - ÷ = 0.307; K p = tan 2 ç 45 + ÷ = 3.255 2ø 2ø è è

s 1¢ = g L1 K a = (16.1)(4)(0.307) = 19.77 kN/m 2 s 2¢ = (g L1 + g ¢L2 ) K a = [(16.1)(4) + (8.39)(8)](0.307) = 40.38 kN/m 2 L3 =

40.38 = 1.63 m (8.39)(3.255 - 0.307)

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1 1 1 (4)(19.77) + (8)(19.77) + (8)(40.38 - 19.77) + (1.63)(40.38) 2 2 2

= 39.54 + 158.16 + 82.44 + 32.91 = 313.05 kN/m

4ö 8ö æ æ P( z ) = (39.54) ç 9.63 + ÷ + (158.16)(5.63) + (82.44) ç1.63 + ÷ 3ø 3ø è è é (2)(1.63) ù + (32.91) ê úû 3 ë

1713.91 = 5.47 m 313.05

s 5¢ = (g L1 + g ¢ L2 ) K p + g ¢ L3 ( K p - K a ) = [(16.1)(4) + (8.39)(8)](3.255) + (8.39)(1.63)(2.948) = 468.4 kN/m2 Eq. (16.17): A1 =

468.4 = 18.94 (8.39)(2.948)

Eq. (16.18): A2 =

(8)(313.05) = 101.25 (8.39)(2.948)

Eq. (16.19): A3 =

(6)(313.05)[(2)(5.47)(8.39)(2.948) + 468.4] = 2268.94 (8.39)2 (2.948)2

Eq. (16.20): A4 =

(313.05)[(6)(5.47)(468.4) + (4)(313.05)] = 8507.44 (8.39)2 (2.948)2

Eq. (16.16): L44 + 18.94 L34 - 101.25L24 - 2268.94 L4 - 8507.44 = 0 L4 = 11.68 m. Thus,

D = L3 + L4 = 1.63 + 11.68 = 13.31 m b. Total length: 4 + 8 + (1.3)(13.31) = 29.3 m

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c. Eq. (16.21): z ¢ =

2P (2)(313.05) = =5m g ¢( K p - K a ) (8.39)(2.948)

1 Eq. (16.22): M max = P( z + z¢) - g ¢z¢3 ( K p - K a ) 6 1 = (313.05)(5.47 + 5) - (8.39)(5)3 (2.948) 6 = 2762 kN -m /m 16.3

Ka = tan 2 (45 -18) = 0.26; K p = tan 2 (45 + 18) = 3.85 Eq. (16.24): s 2¢ = g LK a = (15)(5)(0.26) = 19.5 kN/m2 Eq. (16.28): L3 =

LK a (5)(0.26) = = 0.36 m K p - K a 3.85 - 0.26

1 1 1 1 Eq. (16.29): P = s 2¢ L + s 2¢ L3 = (19.5)(5) + (19.5)(0.36) 2 2 2 2

= 48.75 + 3.51 = 52.26 kN/m Eq. (16.30): z =

L(2 K a + K p ) 3( K p - K a )

(5)[(2)(0.26) + 3.85] = 2.03 m (3)(3.85 - 0.26)

Eq. (16.27): s 5¢ = g LK p + g L3 ( K p - K a ) = (15)(5)(3.85) + (15)(0.36)(3.85 - 0.26)

= 288.75 + 19.39 = 308.14 kN/m2 Eqs. (16.32) to (16.35):

A1¢ =

s 5¢ 308.14 = = 5.72 g ( K p - Ka ) (15)(3.85 - 0.26)

A2¢ =

8P (8)(52.26) = = 7.76 g ( K p - Ka ) (15)(3.85 - 0.26)

A3¢ =

A4¢ =

6 P [2 z g ( K p - K a ) + s 5¢ ]

g (K p - Ka ) 2

(6)(52.26)[(2)(2.03)(15)(3.85 - 0.26) + 308.14] = 56.96 (15) 2 (3.85 - 0.26) 2

P(6 z s 5¢ + 4P) (52.26)[(6)(2.03)(308.14) + (4)(52.26)] = = 71.4 g 2 ( K p - Ka )2 (15)2 (3.85 - 0.26)2

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Eq. (16.31): L44 + 5.72 L34 - 7.76 L24 - 56.96 L4 - 71.4 = 0; L4 » 3.5 m

Dtheory = L3 + L4 = 0.36 + 3.5 = 3.86 m Maximum Moment: Similar to Eq. (16.21). 2P (2)(52.26) = = 1.39 m ( K p - K a )g (15)(3.85 - 0.26)

z¢ =

1 1 M max = P ( z + z ¢) - g ¢z ¢3 ( K p - K a ) = (52.26)(2.03 + 1.39) - (15)(1.39)3 (3.85 - 0.26) 6 6 = 178.73 - 24.1 = 154.63 kN-m /m

16.4

j¢ ö æ K a tan 2 ç 45 - ÷ = 0.307; K p = 3.255 2ø è s 2¢ = g LK a = (16)(4)(0.307) = 19.65 kN/m 2 LK a (4)(0.307) = = 0.417 m K p - Ka 2.948

L3 =

1 1 1 P = s 2¢ L + s 2¢ L3 = (19.65)(4 + 0.417) = 43.4 kN/m 2 2 2

L(2 K a + K p ) 3( K p - K a )

(4)(0.614 + 3.255) = 1.75 m (3)(2.948)

s 5¢ = g LK p + g L3 ( K p - Ka ) = (16)(4)(3.255) + (16)(0.417)(2.948) = 208.32 + 19.67 = 227.99 kN/m2

A1¢ =

s 5¢ 227.99 = = 4.83 g ( K p - Ka ) (16)(2.948)

A2¢ =

8P (8)(43.4) = = 7.36 g ( K p - Ka ) (16)(2.948)

A3¢ =

6 P [2 z g ( K p - K a ) + s 5¢ ]

g (K p - Ka ) 2

(6)(43.5)[(2)(1.75)(16)(2.948) + 227.99] = 46.1 [(16)(2.948)]2

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A4¢ =

P(6 z s 5¢ + 4P) (43.5)[(6)(1.75)(227.99) + (4)(43.4)] = = 50.2 g 2 ( K p - Ka )2 [(16)(2.948)]2

L44 + 4.83L34 - 7.36 L24 - 46.1L4 - 50.2 = 0; L4 » 3.26 m Dtheory = L3 + L4 = 0.417 + 3.26 » 3.7 m z¢ =

2P (2)(43.4) = » 1.36 m ( K p - K a )g (2.948)(16)

1 1 M max = P ( z + z ¢) - g ¢z ¢3 ( K p - K a ) = (43.4)(1.75 + 1.36) - (16)(1.36)3 (2.948) 6 6 = 134.97 - 19.77 = 115.2 kN-m /m

16.5

j ¢ = 36°; Ka = tan 2 (45 -18) = 0.26; K p = tan 2 (45 + 18) = 3.85; K p - Ka = 3.59

8P (8)(20) = = 2.476 g ( K p - K a ) (18)(3.59) 12 PL (12)(20)(5) = = 18.57 g ( K p - Ka ) (18)(3.59) 2

é ù é (2)(20) ù 2P ê ú =ê ú = 0.383 ëê g ( K p - K a ) ûú ë (18)(3.59) û Eq. (9.36): D4 - 2.476D2 - 18.57 D - 0.383 = 0; D » 3 m z¢ =

2P (2)(20) = = 0.787 m ( K p - K a )g (18)(3.59)

Eq. (16.38): M max = P( L + z ¢) -

g z¢3 ( K p - K a ) 6

= (20)(5.787) -

(18)(0.787)3 (3.59) 6

= 115.74 - 5.25 = 110.49 kN-m /m

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16.6

a. Refer to the figure below.

j¢ ö 1 æ K a tan 2 ç 45 - ÷ = tan 2 (45 - 15) = 2ø 3 è

j¢ ö æ K p tan 2 ç 45 + ÷ = tan 2 (45 + 15) = 3 2ø è 1 s 1¢ = g L1K a = (16)(3) = 16 kN/m2 3 1 s 2¢ = (g L1 + g ¢L2 ) K a = [(16)(3) + (18 - 9.81)(5)] = 29.65 kN/m2 3

P1 = Areas of 1 + 2+ 3 =24 + 80 + 34.13 = 138.13 kN/m

æ5ö (24)(6) + (80)(2.5) + (34.13) ç ÷ è 3 ø = 2.9 m z1 = 138.13

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Eq. (16.48):

D 2 [4c - (g L1 + g ¢L2 )] - 2 DP1 -

P1 ( P1 + 12cz1 ) =0 (g L1 + g ¢L2 ) + 2c

D 2 [(4)(35) - 88.95] - 2 D(138.13) -

138.13[138.13 + (12)(35)(2.9)] =0 88.95 + (2)(35)

51.05D2 – 276.52D – 1178.5 = 0; D = 8.25 m b. Length of sheet pile = 3 + 5 + (1.4)(8.25) = 19.55 m c. Eq. (16.42): s 6 = 4c - (g L1 + g ¢L2 ) = (4)(35) - [(16)(3) + (18 - 9.81)(5)]

= 51.05 kN/m2 Eq. (16.49): z ¢ =

138.13 = 2.71 m 51.05

Eq. (16.50):

s 6 z ¢2

(51.05)(2.71) 2 2 2 = 774.9 - 187.46 = 587.44 kN -m /m of the wall

M max = P1 ( z ¢ + z1 ) -

16.7

= (138.13)(2.71 + 2.9) -

Eq. (16.61):

4 D 2c - 2 PD -

P( P + 12cL) =0 2c

(4)(45) D 2 - (2)(8) D -

(8)[8 + (12)(45)(4)] =0 (2)(45)

180D2 - 16D - 192.7 = 0; D » 1.1 m Eq. (16.63): M max = P( L + z¢) -

z¢ =

4cz¢2 2

P 8 = = 0.044 m 4c (4)(45)

M max = (8)(4.044) - (2)(45)(0.044) 2 = 32.18 kN-m /m

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16.8

a. Refer to the figure. j ¢ = 34°.

j¢ ö j¢ ö æ æ K a tan 2 ç 45 - ÷ = 0.283; K p tan 2 ç 45 + ÷ = 3.537 2ø 2ø è è s 1¢ = g L1 K a = (17)(4)(0.283) = 19.244 kN/m 2

s 2¢ = (g L1 + g ¢ L2 ) K a = [(17)(4) + (19 - 9.81)(9)](0.283) = 42.65 kN/m 2 L3 =

42.65 42.65 = » 1.43 m g ¢( K p - Ka ) (9.19)(3.254)

P = Area of 1 + 2 + 3 + 4 = 38.49 + 173.2 + 105.33 + 30.49 = 347.51 kN/m z=

1 [(38.49)(11.76) + (173.2)(5.93) + (105.33)(4.43) + (30.49)(0.95)] = 5.68 m 347.51

Eq. (16.67):

L34 + 1.5L24 (l2 + L2 + L3 ) -

3P[( L1 + L2 + L3 ) - ( z + l1 )] =0 g ¢( K p - Ka )

L34 + 1.5L24 (2 + 9 + 1.43) -

(3)(347.51)[(4 + 9 + 1.43) - (5.68 + 2)] =0 (9.19)(3.254) 145

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L34 + 18.645 L24 - 235.32 = 0; L4 » 3.3 m

D = 3.3 + L3 = 3.3 + 1.43 = 4.73 b. Eq. (16.65): s 8¢ = g ¢( K p - Ka ) L4 = (9.19)(3.254)(3.3) » 98.68 kN/m2 The pressure distribution diagram is shown in the figure on the previous page. c. Eq. (16.66):

1 1 F = P - [g ¢( K p - K a ) L24 ] = 347.51 - [(9.19)(3.254)(3.3)2 ] = 184.68 kN /m 2 2 16.9

a. Dactual = (1.3)( Dtheory ) = (1.3)(4.73) » 6.15 m For zero shear, use Eq. (16.69):

1 1 s 1¢L1 - F + s 1¢( z - L1 ) + K ag ¢( z - L1 ) 2 = 0 2 2

Let z–L1 = x

1 1 s 1¢L1 - F + s 1¢x + K ag ¢x 2 = 0 2 2 1 1 (19.244)(4) - 148.68 + 19.244 x + (0.283)(9.19) x 2 = 0 2 2

x 2 + 14.8x - 112.46 = 0; x » 5.7 m z = x + L1 = 5.7 + 4 = 9.7 m Taking the moment about the point of zero shear

æ x2 ö 1 1 4ö æ æ xö ¢ ¢ M max = - s 1L1 ç x + ÷ + F ( x + 2) - s 1 ç ÷ - K ag ¢x 2 ç ÷ 2 3ø è è3ø è 2ø 2 With s1¢ = 19.244 kN/m, L1 = 4, x = 5.7 m, F = 184.68 kN/m,

M max » 759 kN-m/m

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b. H ¢ = L1 + L2 + Dactual = 4 + 9 + 6.15 = 19.15 m » 20 m

Hʹ Section

I(m /m)

(m)

PZ-35

504 ´10–6

PZ-27

255.9 ´10–6

log r

S(m /m)

M d = Ss all

Md M max

16.47 ´ 10–4

–2.783

262.9´ 10–5

552

0.727

32.6´ 10–4

–2.481

166.66´ 10–5

350

0.46

10.91´10-7 H ¢4 EI

If logr and Md/Mmax are plotted in Figure 16.27 with the curve for loose sand as the separating line, it can be seen that PZ-35 will be sufficient, however, PZ-27 is not suitable. 16.10 g ¢ = g sat - g w = 19 - 9.81 = 9.19 kN/m3

g av¢ =

g L1 + g ¢ L2 (17)(4) + (9.19)(8) = = 11.79 kN/m3 L1 + L2 4+8

j¢ ö æ K a = tan 2 ç 45 - ÷ = tan 2 (45 - 17.5) = 0.271 2ø è Eq. (16.75):

s a¢ = CK ag av L = (0.68)(0.271)(11.79)(12) = 26.07 kN/m 2 Eq. (16.76):

s ¢p = Rs a¢ = (0.6)(26.07) = 15.64 kN/m2 Eq. (16.79): 2 2 æ l ö L é æ l öù æ 1.5 ö 12 é æ 1.5 ö ù D 2 + 2 DL ç1 - 1 ÷ - ê1 - 2 ç 1 ÷ ú = D 2 + 2 D(12) ç1 1 - (2) ç ÷ ú = 0 ÷ ê è 12 ø 0.6 ë è 12 ø û è Lø R ë è L øû

D2 + 21D - 180 = 0; D » 6.6 m Check for R: R =

L( L - 2l1 ) (12)(12 - 3) = = 0.592 - - O.K. D(2 L + D - 2l1 ) (6.6)(24 + 6.6 - 3)

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Eq. (16.80):

F = s a¢ ( L - RD) = (26.07)[12 - (0.6)(6.6)] = 209.6 kN/m Eq. (16.81):

éæ RD ö2 æ 2l1 ö æ RD ö ù M max = 0.5s a¢ L2 êç1 ÷ - ç ÷ ç1 ÷ú L ø è L øè L ø úû êëè 2 é 0.6 ´ 6.6 ö æ 2 ´1.5 öæ 0.6 ´ 6.6 ö ù 2 æ = (0.5)(26.07)(12) êç1 ÷ -ç ÷ç1 ÷ú 12 ø è 12 øè 12 ø úû êëè = 528.2 kN-m /m 16.11 a. Refer to the figure.

j¢ ö j¢ ö æ æ K a = tan 2 ç 45 - ÷ = 0.26; K p = tan 2 ç 45 + ÷ = 3.85; K p - K a = 3.59 2ø 2ø è è s 1¢ = g L1 K a = (17)(3)(0.26) = 13.26 kN/m 2

s 2¢ = (g L1 + g ¢L2 ) = [(17)(3) + (19.5 - 9.81)(8)](0.26) » 33.4 kN/m 2

P1 = Areas of 1 + 2 + 3 = 19.89 + 106.08 + 80.56 = 206.53 kN/m

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(19.89)(9) + (106.08)(4) + (80.56)(2.67) 179.01 + 424.32 + 215.1 = = 3.96 m 206.53 206.53 Eq. (16.83): s 6 D 2 + 2s 6 D( L1 + L2 - l1 ) - 2 P1 ( L1 + L2 - l1 - z1 ) = 0 z1 =

s 6 = 4c - (g L1 + g ¢L2 ) = (4)(40) - 128.52 = 108 - 86.3 = 31.48 kN/m 2 So, 31.48D2 + (2)(31.48)(9.5) D - (2)(206.53)(3 + 8 - 1.5 - 3.96) = 0 or

D2 + 19D - 72.69 = 0; D » 3.3 m b. P1 - s 6 D = F [Eq. (16.82)] 206.53 – (31.48)(3.3)= 102.6 kN/m 6.12

Eq. (16.93):

5.4 æ H 2 ö Pu = ç ÷ tan f ¢ è Bh ø

0.28

5.4 æ 0.92 ö = ç ÷ tan 32 è 0.3B ø

g BhH 0.28

(17)( B)(0.3)(0.9)

= 52.39 B 0.72 kN B (m)

Pu = 52.39B0.72 (kN)

0.3

0.6

36.3

0.9

48.6

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16.13 Pu = M g q (g h2 ) BFs × H = 0.9 m: h = 0.3 m; g = 17 kN/m3 ; f ¢ = 32°

B (m)

H (m)

h (m)

B h

H h

Mg q

(Figure 16.43)

(Figure 16.44)

(kN)

0.3

0.9

0.3

3.0

»18

1.86

15.37

0.6

0.9

0.3

3.0

'18

1.3

21.48

0.9

0.3

3.0

1.13

28.0

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Chapter 17

17.1

Eq. (17.8)

f¢ ö 35 ö æ æ s a = 0.65g HK a = 0.65g H tan 2 ç 45 - ÷ = (0.65)(17)(6.5) tan 2 ç 45 - ÷ 2ø 2ø è è = 19.6 kN/m 2

19.46 kN/m2

19.46 kN/m2 1m

2m A

2m B1

1.5 m

SM B1 = 0

æ3ö (19.46)(3) ç ÷ è 2 ø = 43.79 kN/m A= 2

B1 = (19.46)(3) - 43.79 = 14.59 kN/m SM B2 = 0

æ 3.5 ö (19.46)(3.5) ç ÷ 2 ø è C= = 59.6 kN/m 2

B2 = (19.46)(3.5) - 59.6 = 8.51 kN/m Strut load at A = (43.79)(spacing) = (43.79)(3) = 131.4 kN Strut load at B = (B1 + B2)(spacing) = (14.59 + 8.51)(3) = 69.3kN Strut load at C = (59.6)(3) = 178.8 kN

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17.2

a. For the sheet pile, refer to shear force diagram. 24.33 kN

0.75 m

B′

1.25 m

14.49 kN

19.46 kN

1 M A = (1)(19.46) = 9.73kN × m/m 2 29.18 kN 8.51 kN B′′

0.437 m

1.563 m

1.5 m

30.42 kN

1 M B¢¢ = (0.437)(8.51) = 1.86 kN × m/m 2 1 M C = (1.5)(29.18) » 21.9 kN/m 2 S=

21.9 kN × m/m = 0.129×10-3 m3 /m of wall 3 170 ´ 10 kN × m/m

b. For wales, M max =

Bs 2 . 8

Bs 2 (69.38)(32 ) S= = = 0.459×10-3 m 3 /m 3 8s all (8)(170 ´ 10 )

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17.3

40 ö æ K a = ç 45 - ÷ = 0.217 2 ø è

s a = 0.65g HK a = (0.65)(18)(6.5)(0.217) = 16.5kN/m 2 SM B1 = 0 æ3ö (16.5)(3) ç ÷ è 2 ø = 37.13kN/m A= 2

16.5 kN/m2

16.5 kN/m2 1m

2m A

2m B1

1.5 m

B1 = (16.5)(3) - 37.13 = 12.37 kN/m SM B2 = 0 æ 3.5 ö (16.5)(3.5) ç ÷ 2 ø è C= = 50.53 kN/m 2

B2 = (16.5)(3.5) - 50.53 = 7.22 kN/m Strut load at A = (37.13)(4) = 148.5 kN Strut load at B = (12.37 + 7.22)(4) = 78.4 kN Strut load at C = (50.53)(4) = 202.12 kN

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17.4

Refer to the pressure diagram in Problem 17.3. The shear force diagram is given next. 1m 20.63 kN

2m 0.75 m

A B′

B1 12.37 kN

16.5 kN

25.78 kN

B′′

C 24.75 kN

7.22 kN 0.438 m

1.562 m

1.5 m

It can be seen that M C will be maximum.

1 M C = (1.5)(24.75) = 18.56 kN/m 2

17.5

18.56 kN × m/m

s all

18.56 = 0.109×10-3 m 3 /m of wall 3 170 ´10

a. H = 8 m; H s = 3 m; H c = 5. Eq. (17.12):

1 1 [g s H s + ( H - H s )g c ] = [(17.5)(3) + (5)(18.2)] H 8 3 = 17.94 kN /m

g av =

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Eq. (17.11):

1 [g s K s H s2 tan fs¢ + ( H - H s )n¢qu ] 2H 1 = [(17.5)(1)(32 )(tan 34) + (5)(0.75)(55)] (2)(6)

cav =

= 19.53 kN /m 2

g av H (17.94)(8) = = 7.35. The pressure diagram will be like Figure 17.9a. cav 19.53

s a = g av H ç1 è

4cav ö 4 ö æ 2 ÷ = (17.94)(8) ç1 ÷ = 65.4 kN/m g av H ø è 7.35 ø

Also check: sa = 0.3gavH = (0.3)(17.94)(8) = 43.06 kN/m2 Use sa = 65.4 kN/m2

17.6

a. Eq. (17.13): cav =

Eq. (17.14): g av =

1 [(5)(2125) + (10)(1565) + (10)(1670)] = 1719 lb /ft 2 25 1 [(5)(111) + (10)(107) + (10)(109)] = 108.6 lb /ft 2 25

g av H (108.6)(25) = = 1.58. Use Figure 17.9b. cav 1719

sa = 0.3gavH = (0.3)(108.6)(25) = 814.5 lb/ft2 17.7

gH c

(17.5)(7) = 4.08. Use Figure 17.9a. 30

s a = g H ç1 è

4c ö 4 ö æ ÷ (low), or ÷ = (17.5)(7) ç1 gH ø è 4.08 ø

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sa = 0.3gavH = (0.3)(17.5)(7) = 36.75 kN/m2 - Use

1.75 m

3m 36.75 kN/m2

B 2m C

5.25 m

1m 36.75 21 kN/m2 kN/m2

36.75 kN/m2

1m A

1m C

SM B1 = 0 æ 36.75 ´ 2.252 ö æ 1 1.75 ö öæ ç ÷ + ç ´ 36.75 ´1.75 ÷ç 2.25 + ÷ 2 3 ø 93 + 91 è2 øè è ø A= = = 61.3 kN/m 3 3

1 é ù B1 = ê(36.75 ´ 2.25) + (1.75)(36.75) ú - 61.3 = 53.55 kN/m 2 ë û

SM B2 = 0 æ3ö (36.75 ´ 3) ç ÷ è 2 ø = 82.69 kN/m C= 2

B2 = (36.75 ´ 3) - 82.69 = 27.56 kN/m Strut load at A = (61.3)(5) = 306.5 kN Strut load at B = ( B1 + B2 )(5) = (53.55 + 27.5)(5) = 405.55 kN Strut load at C = (82.69)(5) = 413.45 kN

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17.8

Refer to the load diagrams in Problem 17.7. The shear diagrams are given below (not to scale). 50.8

B′ 10.5 kN/m

53.55 kN/m

C B′′

27.56 kN/m 36.75 kN/m 2m

X1 =

53.55 kN/m

53.55 = 1.46 m 36.75

é 1 æ 36.75 ö ù 1 M A = ê (1) ç ´1÷ ú = 3.5 kN × m/m øû 3 ë 2 è 1.75

æ X2 ö æ 1.462 ö M B¢ = B1 X 1 - 36.75 ç 1 ÷ = (53.55)(1.46) - (36.75) ç ÷ è 2 ø è 2 ø = 78.18 - 39.17 = 39.01 kN × m/m » 39 kN × m /m X2 =

27.56 kN/m

27.56 = 0.75 m 36.75

1 M C = (1)(36.75) = 18.38 kN × m/m 2 1 M B¢¢ = (27.56)(0.75) = 10.34 kN × m/m 2

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Maximum moment at B¢: M max = 39 kN × m/m

17.9

gH c

M max

s all

39 kN × m/m = 22.94×10-5 m 3 /m 170 ´103 kN/m3

(17.5)(7) = 2.04 < 4 60

s a = 0.3g H = (0.3)(17.5)(6) = 36.75 kN/m 2 1m

1.75 m

3m 36.75 kN/m2

3.5 m

2m C

1.75 m

1m 1.75 m

1.25 m 1.75 m

2.25 m 36.75 kN/m2

36.75 kN/m2

3m A

2m B1

1m C

Same reactions at A and B1 as in Problem 17.7. A = 61.3 kN/m; B1 = 53.55 kN/m. Taking the moment about B2, SM B2 = 0.

æ 1.252 ö 1 1.75 ö æ (36.75) ç ÷ + (36.75)(1.75) ç1.25 + ÷ 2 ø 2 3 ø è è C= = 43.78 kN/m 2 So B2 = 34.32 kN/m.

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Strut load at A = (61.3)(5) = 306.5 kN Strut load at B = ( B1 + B2 )(5) = (53.55 + 34.32)(5) = 439.35 kN Strut load at C = (43.78)(5) = 218.9 kN

æ 0.2 B¢¢ ö cH 5.14c ç1 + ÷+ L ø B¢ è 17.10 Eq. (17.23): FS = gH +q Since T >

B B 6 , B¢ = = = 4.24 m 2 2 2

B¢¢ = 2B¢ = 2 ´ 4.24 = 6 m q=0 é (0.2)(6) ù (30)(7) (5.14)(30) ê1 + + 18 úû 4.24 ë FS = = 1.75 (17.5)(7) æ 0.2 B¢ ö cN c ç1 + ÷ L ø è 17.11 Eq. (17.22): FS = q c ö æ çg + - ¢ ÷ H H B ø è B¢ =

B 6 = = 4.24 m 2 2

é (0.2)(4.24) ù (60)(5.7) ê1 + (12.5) úû ë FS = = 1.65 60 ö æ ç17.5 + 0 ÷ (7) 4.24 ø è

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