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SOLUTIONS MANUAL - Engineering Tenth Edition BRAJA M. DAS © 2024 ...

Testbanks Dec 30, 2025 ★★★★☆ (4.0/5)
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SOLUTIONS MANUAL

For Principles of Foundation Engineering Tenth

Edition

BRAJA M. DAS

© 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.(Solutions Manual, All Chapters.100% Original Verified, A+ Grade) 1 / 4

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 2 / 4

1

© 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.Chapter 2 2.1 d.

(87.5)(9.81)

/

(1000)(0.05)



 3 17.17 kN m

  • 17.17
  • /

1 1 0.15

  

w   3 14.93 kN m a.Eq.

(2.12):1

  sw d G e 

 (2.68)(9.81)

14.93 1  e ; e = 0.76 b.Eq.

(2.6): 0.76

1 1 0.76

e n e

  



0.43 e.From Eq.

(2.14):(0.15)(2.68)

100 0.76



  





ws v V wG S Ve 53% 2.

  • a.From Eq s. (2.11) and (2.12), it can be seen that20.1

1 1 0.22

  



d w   16.48 kN/m 3

b.

3 (9.81)

16.48 kN/m 11

  



s w s d GG ee   Eq.

(2.15):(0.22)( )

ss e wG G . So 9.81

16.48 ;

1 0.22



 s s s G G G

2.67 3 / 4

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.

2.3 a.Eq.

(2.6):0.81

1 1 0.81

  



0.45 e n e b.Eqs.

(2.7), (2.14):(0.21)(2.68)

100 0.81



 





s wG S e 69.5% c.Eq.

(2.11):(1 )(2.68)(9.81)(1 0.21)

/

1 1 .0.81



 



sw Gw e   3 17.58 kN m d.Eq.

(2.12):(2.68)(9.81)

  • 3 /

1 1 .0.81

 



sw G e   3

  • .5 kN m
  • 2.4 a.Eq.

(2.12):1

  sw d G e  

Eq. (2.15):

s e G w So,1 w d e w e  







 ( )(9.81)

13.5 ;

(0.36)(1 )



 0.98 e e e

b. Eq. (2.6): 0.98

0.495

1 1 .0.98

   



e n e 0.5 c.Eq. (2.14

):0.98

0.36

  

s e G w 2.72 d

. Eq. (2.13):sat

(1 ) (13.5)(1 0.36) /    

d w



3 18.36 kN m

  • / 4

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