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Solutions Manual For Principles of Foundation Engineering 10th Edition By Braja M. Das (All Chapters, 100% original verified, A+ Grade) CA$22.85
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Solutions Manual For Principles of Foundation Engineering 10th Edition By Braja M. Das (All Chapters, 100% original verified, A+ Grade)

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Solutions Manual For Principles of Foundation Engineering 10th Edition By Braja M. Das (All Chapters, 100% original verified, A+ Grade) Solutions Manual For Principles of Foundation Engineering 10th Edition By Braja M. Das (All Chapters, 100% original verified, A+ Grade)

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  • December 4, 2023
  • 162
  • 2023/2024
  • Exam (elaborations)
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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 who le or in part. (Solutions Manual, All Chapters. 100% Original Verified, A+ Grade) 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 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)317.17 kN m
c. 17.17/1 1 0.15  w314.93 kN m
a.Eq. (2.12):
1sw
dG
e
(2.68)(9.81)14.931e
; e = 0.76 b.Eq. (2.6): 0.76
1 1 0.76ene  0.43
e.From Eq. (2.14):
(0.15)(2.68)1000.76   ws
vV wGSVe53%
2.2 a.From Eq s. (2.11) and (2.12), it can be seen that
20.1
1 1 0.22  dw
16.48 kN/m3 b. 3 (9.81)16.48 kN/m11  s w s
dGG
ee
Eq. (2.15):
(0.22)( )ss e wG G . So 9.8116.48 ;1 0.22s
s
sGGG2.67 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  0.45ene
b.Eqs. (2.7), (2.14):
(0.21)(2.68)1000.81  swGSe69.5%
c.Eq. (2.11):
(1 ) (2.68)(9.81)(1 0.21)/1 1 .0.81   swGw
e317.58 kN m
d.Eq. (2.12):
(2.68)(9.81)4 3 /1 1 .0.81  swG
e31 .5 kN m
2.4 a.Eq. (2.12):
1sw
dG
e
Eq. (2.15): seGw
So,
1w
de
w
e

( )(9.81)13.5 ;(0.36)(1 )0.98eee
b. Eq. (2.6): 0.980.4951 1 .0.98   ene0.5
c.Eq. (2.14 ):
0.98
0.36  seGw2.72
d. Eq. (2.13):
sat (1 ) (13.5)(1 0.36) /    dw 318.36 kN m

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