SOLUTIONSMANUAL - Engineering  Ninth Edition  BRAJA M. DAS ...

SOLUTIONS MANUAL 

For Principles of Foundation  Engineering  Ninth Edition 

BRAJA M. DAS 

NAGARATNAM  SIVAKUGAN 

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Contents Chapter 2 ..........................................

................................................................................... 1

Chapter 3 ........................................................................................................................... 11 Chapter 5 ........................................................................................................................... 23 Chapter 6 ........................................................................................................................... 31 Chapter 7 ........................................................................................................................... 49 Chapter 8 ........................................................................................................................... 59 Chapter 9 ........................................................................................................................... 69 Chapter 10 ......................................................................................................................... 83 Chapter 11 ......................................................................................................................... 89 Chapter 12 ......................................................................................................................... 91 Chapter 13 ....................................................................................................................... 113 Chapter 14 ....................................................................................................................... 123 Chapter 15 ....................................................................................................................... 125 Chapter 16 ....................................................................................................................... 131 Chapter 17 ....................................................................................................................... 141 Chapter 18 ....................................................................................................................... 153 Chapter 19 ....................................................................................................................... 171 2 / 4

1 © 2019 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 From Eq. (2.18), e ρG ρ ws d + = 1e+ × == 1

100080.2

925.0 2450 ; e = 0.0571 From Eq. (2.6), Porosity, = + = + =

0571.01

0571.0

1e e n

0.054 

2.2 From Eq. (2.13), the dry density 3 kg/m 6.1786

153.01

2060 1 = + = + = w ρ ρ d From Eq. (2.18), e ρG ρ ws d + = 1

511.01

6.1786

100070.2

1=− × =−= d ws ρ ρG e

Once saturated, from Eq. (2.19), 3 kg/m 2125.1=× + + = + + =1000

)511.01(

)511.070.2(

)1( )( satw s ρ e eG ρ 2.3 Let’s consider a 16m 2 area in plan. The initial volume of this soil is V = 1 × 0.5 = 0.5 m 3 . Volume of the solids is Vs.s s V V e − == 5.0 9.0 ; Vs = 0.2632 m 3

W s = 0.2632 × 2.68 × 9.81 = 6.919 kN The new volume after compaction = 1 × 0.455 = 0.455 m 3

The dry unit weight,

36.919

15.21 kN/m 0.455 d γ= = From Eq. (2.12), e γG γ ws d + =

1 3 / 4

2 © 2019 Cengage Learning ® . All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

0.729=−

×

=−=1

21.15

81.968.2

1 d ws γ γG e   From Eq. (2.13), 3 kN/m 18.25=+= )20.01)(21.15(γ

2.4 At the compacted road base, the weight of solids, W s = 120,000 × 19.5 = 2,340,000 kN At the borrow pit, the dry unit weight, 3 kN/m 16.13

085.01

5.17 1 = + = + = w γ γ d

Volume of the pit, 3 m 145,080== 13.16

000,340,2

V

The moisture content has to be increased from 8.5% (at the borrow pit) to 14.0% (at the road base). The quantity of water to add,

2,340,000

× (0.14 ‒ 0.085) = 128,700 kN Volume of water to be added = 3 m 13,119.3= 81.9

700,128

2.5  3 lb/ft 99.9

105.01

4.110 1 = + = + = w γ γ d

e γG γ ws d + = 1

; 655.01

9.99

4.6265.2

=− × =e From Eq. (2.23), 60.6%=× − − = 100

515.0870.0

655.0870.0

r D

  2.6 a. A616a c. A63

• A616b d. A6766

2.7 Soil A: % of gravel = 50, % of sand = 13, % of fines = 37

D 10 = 0.035 mm, D30 = 0.061 mm, D60 = 9.8 mm 1 Cu = 280; C c = 0.02 LL = 58, PL = 34, PI = 24 1 plots below the A6line; hence, silt The soil can be described as poorly (gap) graded sandy silty gravel        with a group symbol of GM.

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