04- Soil Improvement - Amazon S3s3.amazonaws.com/zanran_storage/web.eng.fiu.edu/ContentPages/... ·...

04- Soil Improvement 01 Can a contractor attain 105 of Standard Proctor 02 Find the optimum moisture content (OMC) 03 Find the maximum dry unit weight in SI units 04 What is the saturation S at the OMC 05 Number of truck loads required 06 What is the saturation S at the OMC 07 Definition of the relative compaction (RC) 08 The relative compaction (RC) of a soil 09 Converting volumes from borrow pits and truck loads 10 Ranges of water and fill required for a road 11 Soil improvement techniques 12 Find the family of saturation curves for compaction 13 Water needed to reach maximum dry unit weight in the field 14 Multiple choice for fill volumes and truck load requirements for a levee 15 Multiple choice compaction of a levee Symbols for Soil Improvement e rarr Voids ratio GSrarr Specific gravity of the solids of a soil n rarr Porosity of the soil OMCrarr Optimum moisture content S rarr Degree of saturation V rarr ( 31 9441030 ft mminusequiv 4 3 Standard Proctor mold ASTM D-698)

Vararr Volume of air VSrarr Volume of solids VV rarr Volume of voids (water + air) VW rarr Volume of water wrarr Water content VSrarr Volume of soil sample γrarr Unit weight of the soil γd rarr Dry unit weight γb rarr Buoyant unit weight of the soil γSAT rarr Saturated unit weight of the soil γS rarr Unit weight of the solid γW rarr Unit weight of water γdfield rarr Dry unit weight in the field

90

Compactionndash01 Can a contractor attain 105 of Standard Proctor (Revision Aug-09) A contractor has reported several compaction tests that have exceeded the 98 Standard Proctor required by the Specifications of a paving project In fact several of the field tests reported Relative Compactions of 105 of Standard Proctor Is this possible Solution Definitely yes Remember that the Proctor test series (Standard and Modified) prescribes a specific amount of energy into a standard mold (130th of 1 ft3) Therefore the contractor could impart more energy into the field compaction than was performed in the laboratory to attain the 100 of Standard (or Modified) Proctor Standard Proctor follows the procedure of ASTM D-698 or AASHTO T-99 Modified Proctor follows the procedure of ASTM D-1557 or AASHTO T-180

91

Compactionndash02 Find the optimum moisture content (OMC) (Revision Aug-09) A Standard Proctor test has yielded the values shown below Determine

a) The maximum dry unit weight γd and its OMC remember V = 130 ft3 b) The moisture range for 93 of maximum dry unit weight

No Weight of wet soil (lb) Moisture 1 326 824 2 415 1020 3 467 1230 4 402 1460 5 336 1680

Solution

The formulas used are and 1

= =+d

WV w

γγ γ

W (lb) w () γ (lbft3) γd (lbft3) 326 824 978 9035 415 1020 1245 1130 467 1230 1401 1248 402 1460 1206 1052 336 1680 1008 8630

92

Compactionndash03 Find the maximum dry unit weight in SI units (Revision Aug-09) Using the table shown below a) Estimate the maximum dry weight of a sample of road base material tested under Standard Proctor ASTM D-698 (all weights shown are in Newton)

Note that the volume 3

3 43

1 1 944 1030 3532

mV ftft

minus⎛ ⎞= =⎜ ⎟

⎝ ⎠3mtimes

b) Find the OMC c) What is the appropriate moisture range when attaining 95 of Standard Proctor

Trial No 1 2 3 4 5

W(Newton) 145 156 163 164 161 ω () 20 24 28 33 37

SolutionTrial No 1 2 3 4 5

( )3mkN

VW

=γ 154 165 173 174 171

( )31 mkN

d ωγγ+

= 128 133 135 131 125

93

Compaction-04 What is the saturation S at the OMC (Revision Aug-09) The results of a Standard Compaction test are shown in the table below

ω () 62 81 98 115 123 132 γ (kNm3) 169 187 195 205 204 201

ωγγ+

=1d 159 173 178 184 182 178

a) Determine the maximum dry unit weight and the OMC b) What is the dry unit weight and moisture range at 95 RC (Relative Compaction) c) Determine the degree of saturation at the maximum dry density if Gs = 270

Solution a) γd max = 184 kNm3 OMC = 115

b) γd at 95 = (095)(184) = 175 kNm3

The moisture range w for 95 RC is from 875 to 1375

c)

( )( )( )( )

( )

max

max

0115 270 184981

071184270981

= = =⎛ ⎞minus minus ⎜ ⎟⎝ ⎠

s d

w

ds

w

wG

SG

γγγγ

Saturation S = 71

94

Compaction-05 Number of truck loads required (Revision Feb-09) The in-situ moisture content of a soil is 18 and its moist unit weight is 105 pcf The specific gravity of the soil solids is 275 This soil is to be excavated and transported to a construction site and then compacted to a minimum dry weight of 1035 pcf at a moisture content of 20 a) How many cubic yards of excavated soil are needed for 10000 yd3 of compacted fill b) How many truckloads are needed to transport the excavated soil if each truck can carry 20 tons

Solution

( ) ( )

but therefore

105 89 versus 10351 1 018

) Volume to be excavated 1

= =

=

= = = =+ +

=

borrow site construction site

borrow site borrow site construction site construction site

d construction sited borrow site

W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089

27 10511630) Number of truck loads

20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95

Compaction-06 What is the saturation S at the OMC (Revision Aug-09) The Standard Proctor test results of a clayey gravel soil are shown below Find the degree of saturation S at the optimum condition assume that Gs = 260

Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205

γd kNm3 194 201 206 209 208 206 205 Use γw = 10 kNm3 for simplicity

Solution

( )( ) ( )( )

It is known that 1

therefore at the OMC th

The degree of saturation at th

e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus

⎛ ⎞ minus= there4 = = minus =⎜ ⎟

⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96

Compaction-07 Definition of the relative compaction (RC) (Revision Aug-09) The relative compaction (RC) of a sandy road base in the field is 90 The maximum and minimum dry unit weights of the sand are γd(max) = 204 kNm3 and γd(min) = 139 kNm3 Determine the field values of

a) The dry unit weight in the field b) Relative density (of compaction) Dr c) The moist unit weight γ when its moisture content is 15

Solution The relative compaction RC is the dry unit weight obtained in the field as compared to the Standard Proctor obtained in the laboratory

( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1

) The relative compaction is

090204

090 204

) The relative density

84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184

) The moist unit weight is

768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97

Compaction-08 The relative compaction (RC) of a soil (Revision Aug-09) A Standard Proctor compaction test performed on a crushed limestone (Gs = 270) obtained a maximum dry unit weight of 90 pcf at OMC A field compacted sample showed a moisture of 28 and a unit weight of 1037 pcf

a) Find the relative compaction (RC) b) Find the degree of saturation S of the field soil sample

Solution

( )

( )

( ) ( )( )

( )( )( )

max

Relative Compaction =

1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+

⎛ ⎞there4 = minus = minus =⎜ ⎟

⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98

Compaction-09 Converting volumes from borrow pits and truck loads (Revision Aug-09) The sandy base course of a highway embankment is 30 m wide and 15 m thick The sandy soil is trucked in from a borrow pit with a water content of 15 and a voids ratio of 069 Specifications require the base to be compacted to a dry unit weight of 18 kNm3 Determine for 1 km of length of embankment the following a) The dry unit weight of sandy soil from the borrow pit assuming that GS = 270 b) The number of 20 m3 truckloads of sandy soil required to construct the base course c) The weight of water per truck load of sandy soil and d) The degree of saturation S of the in-situ sandy soil Solution

( )

( )( )( ) 3 3

(

(

3

)

27 (98)a) The borrow pit soil dry unit weight

1 1 069

b) The volume of the finished embankment 30 15 1 = 45 10

Volume of borrow pit soil require

157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157

18 45 10 Number of truck trips 157 20

c) W eight of dry soil in 1 truck-load 20 157 314

W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN

( )( )015 270d) Degree of saturation 059

47 per truck load

59069

= = = =SwGSe

kN

99

Compaction-10 Ranges of water and fill required for a road (Revision Aug-09) From the Standard Proctor compaction curve shown below a) Give two possible reasons that may cause a Proctor test curve to cross the ZAV curve b) What is the water content range (in gallons) needed to build a 24 ft wide street 1000

feet long with a 16rdquo thick compacted base at 98 Standard Proctor

Solution a) The two relationships that describe the zero-air-voids unit weight are

1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠

The only way these curves could cross are (1) a wrong value for Gs and (2) wrong values for the moisture w the saturation S and the moist unit weight

100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max

b ) The total volume V for the street isftV thick ft wide ft long ftin

The maximum dry unit weight pcf a t an OMC The moisture range at 98 of Standard Proctor is 83

γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10

to 143The dry unit weight at 98 SP is = (098)(1195pcf) = 1171pcf

The total weight of the soil in the base W is

W V pcf ft lbs

To find the amount of water calculate the amount

γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w

of water at the low endversus the high end of the moisture range at 98 SP

x lb gallons ftWLow end at 83 V gal

lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft

The spread is 21700 gallons so that the answer can be expressed as anaverage with a rangeThe volume of water required is 4535

0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠

gallons 10850 gallonsplusmn

101

Compaction ndash 11 Soil improvement techniques (Revision Aug-09) What types of soil improvements would you consider for a clay soil using the liquefiable methods shown below

Notice the portion of the chart that includes clays The soil improvement methods include (1) compaction grouting (2) vibro-replacement (3) drains (4) compaction piles (5) jet grouting (6) admixtures (7) deep soil mixing (8) soil reinforcement (9) surcharging fills (10) electro-kinetic injection and (11) pre-compression

102

Compaction-12 Find the family of saturation curves for compaction (Revision Aug-09) This problem expands Compaction-06 where a Modified Proctor compaction test is performed on a clayey gravel road base The solids have a specific gravity of 265 The compaction data yielded the following binomial values for γdγw versus w

w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205

a) Find the γd max and the OMC from the compaction curve b) Find the degree of saturation S at the above conditions c) Calculate the percentage of air for a given porosity n and the saturation S d) Find the equation that describes the points of equal saturation e) Determine the equation for S = 100 f) Discuss the characteristics of this last curve and equation

Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se

γγ γ⎛ ⎞ ⎛ ⎞= there4 = = =⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠⎝ ⎠

c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =

= minus minus minus = minus minus +

= minus =

n

That is the percentage of air a is equal to the porosity n times the factor (1 - S) where S is the degree of saturation

d) Consider a family of curves of equal saturation

volume of water volume of watervolume of air+volume of water 1 volume of soil grains

1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus

⎛ ⎞⎜ ⎟ ⎛ ⎞⎝ ⎠= there4 + minus there4⎜ ⎟⎛ ⎞ ⎝ ⎠minus ⎜ ⎟⎝ ⎠

Curves can be plotted for varying values of a and saturation S

104

If w = 0 all curves pass thru γs

Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d

These are hyperbolas with the w-axis as an asymptote

105

Compaction-13 Water needed to reach maximum dry unit weight in the field (Revision Aug-09) A Standard Proctor test yields the values listed below for a soil with Gs = 271 Find (a) The plot of the dry unit weight versus the water content (b) The maximum dry unit weight (c) The optimum moisture content (d) The dry unit weight at 90 of Standard Proctor (e) The moisture range for the 90 value (f) The volume of water (in gallons) that must be added to obtain 1 cubic yard of soil at the maximum density if the soil was originally at 10 water content

w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d

OMC (20)Moisture Range (13 - 26)

=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115

a) The plot of γd (dry unit weight) versus w (water content) (b) From the plot the maximum dry unit weight is γdmax = 1075 pcf (c) From the plot the optimum moisture content is OMC = 20 (d) The 90 value of the maximum dry unit weight γdmax = (09)(1075) = 968 pcf

106

(e) The moisture range for the 90 value is approximately from 13 to 26 (f) Soil at 10 moisture

( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW

but W lb and W W lb W W W lb

= = there4 =

= = = = there4 = + = + = Soil at 20 moisture

( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =

Therefore need to add the following water (215 lb) ndash (9 lb) = 125 lbft3

there4

33

3

3 3 3

125 748 27 40

Add 40 gallons of water per cubic yard of compacted soil

gallons dy 624 1

= =lb ft gallons ftAdded water

ft lb ft yd

107

Compaction-14 Fill volumes and truck load requirements for a levee (Revision Aug-09) Your company has won a contract to provide and compact the fill material for an earth levee with the dimensions shown below The levee fill is a silty clay soil to be compacted to at least 95 of maximum Standard Proctor of γd-max = 106 pcf at an OMC of 18 Your borrow pit has a silty clay with an in-situ moist density of 1121 pcf at 18 and a Gs = 268 When the soil is excavated and loaded on to your trucks the voids ratio of the material is e = 147 Your trucks can haul 20 cubic yards of material per trip a) Determine the volume of fill required for the levee b) Determine the volume required from the borrow pit c) Determine the required number of truckloads

Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27

b) To find the volume required from the borrow pit consider that the weight of solids is the samein both

ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee

sd s d d borrow borrow d levee levee

d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )

Number of truck-loads required is based on

268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee

d hauled hauled d levee levee

d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677

Number of truck-loads 13470 7350 =15

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf

V ydtruck capacity yd truck load

γγ

109

Compaction-15 Multiple-choice compaction of a levee (Revision Aug-09) A townrsquos new reservoir will impound its fresh water with a small earth levee rectangular in plan The perimeter of the levee wall will be 2200 feet long by 1750 feet wide and its cross-section is shown below The reservoir is to be built with a silty clay soil with a specific gravity of 270 available from three different local sources Specifications call for a compacted soil at the levee with a dry unit weight of 97 pcf at an OMC of 31 percent Assume all voids are totally devoid of any gas (or air) The borrow suppliers quoted the following Pit Price ($yd3) Gs S () w () A 105 269 65 22 B 091 271 49 22 C 078 266 41 18

Water supply reservoirrsquos top view

Levee cross-section

110

Questions

1 What is the leveersquos cross-sectional area a) 675 ft2

b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2

2 What is the approximate total volume V of soil required for the levees

a) 2550 ft3 b) 300000 yd3 c) 900000 yd3 d) 122 Myd3 e) 075 Myd3

3 What is the approximate volume of water impounded (stored) in the reservoir a) 6300000 ft3 b) 134 acre-ft c) 7 Mft3 d) 154 acre-ft e) 45 106 gallons

4 What is the unit weight γ of soil of the compacted earth levee

a) 127 pcf b) 97 pcf c) 86 pcf d) 98 pcf e) 128 pcf

5 What is the designed voids ratio e of the compacted soil in the levee

a) 11 b) 092 c) 084 d) 079 e) 12

6 What is the voids ratio e of the material in pit A

a) 098 b) 091 c) 102 d) 11 e) 072

7 What is the voids ratio e of the material in pit B

a) 093 b) 122 c) 081

111

d) 101 e) 100

8 What is the voids ratio e of the material in pit C

a) 110 b) 112 c) 108 d) 105 e) 117

9 Assume that the voids ratio e of pit A is 098 What is the equivalent volume required of

pit A to place 12 million cubic yards of compacted soil in the levees a) 129 Myd3 b) 120 Myd3 c) 095 Myd3 d) 097 Myd3 e) 096 Myd3

10 Assume that the voids ratio e of pit B is 081 what is the equivalent volume required of

pit B to place 12 million cubic yards of compacted soil in the levees a) 100 Myd3 b) 102 Myd3 c) 118 Myd3 d) 105 Myd3 e) 107 Myd3

11 Assume that the voids ratio e of pit C is 110 what is the equivalent volume required of pit C to place 12 million cubic yards of compacted soil in the levees

a) 134 Myd3 b) 137 Myd3 c) 125 Myd3 d) 123 Myd3 e) 121 Myd3

12 Which pit offers the cheapest fill

a) Pit A b) Pit B c) Pit C d) Both A and C e) Both B and C

112

Solutions 1 Answer is e

Cross-sectional area

A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2

2 Answer is e Perimeter (approx) p = (2)(2200) ft + (2)(1750) ft = (4400 + 3500) = 7900 ft = 2633 yd Volume (approx) V = p A = (283 yd2)(2633 yd) = 746000 yd3 = 075 Myd3

3 Answer is d

Perimeter of inner bottom of reservoir P = 2 x 2010 + 2 x 1560 = 4020 + 3120 = 7140ft = 2380 yd

113

Cross-sectional area A = 12

(25rsquo x 75rsquo) = 9375 ft2 = 104 yd3

Volume of water (approx) V = P x A = 7140 ft x 938 ft3 = 67 Mft3

= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +

27w)d d pw cfγγ γ γ= there4 = + = + =

5 Answer is c In general Se = Gsw but S = 1 (from statement Va = 0) there4 e = Gsw = (270)(031) = 084 6 Answer is b

For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117

9 Answer is a The leveersquos voids ratio e = 084 = Vv Vs therefore for 1yd3 of solids we will need 184 yd3 of soil since pit A has an e = 098 need 198 yd3 of soil for each 1 yd3 of solids or

VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3

10 Answer is c Pit B has an e = 081

VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3

11 Answer is b Pit C has an e = 110

VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114

The cost of Pit A = (129 x 106 yd3) ($105 yd3) = $135 106

The cost of Pit B = (118 x 106 yd3) ($091 yd3) = $107 106

The cost of Pit C = (137 x 106 yd3) ($078 yd3) = $107 106

Both pits B and C are lowest

115

Compactionndash01 Can a contractor attain 105 of Standard Proctor (Revision Aug-09) A contractor has reported several compaction tests that have exceeded the 98 Standard Proctor required by the Specifications of a paving project In fact several of the field tests reported Relative Compactions of 105 of Standard Proctor Is this possible Solution Definitely yes Remember that the Proctor test series (Standard and Modified) prescribes a specific amount of energy into a standard mold (130th of 1 ft3) Therefore the contractor could impart more energy into the field compaction than was performed in the laboratory to attain the 100 of Standard (or Modified) Proctor Standard Proctor follows the procedure of ASTM D-698 or AASHTO T-99 Modified Proctor follows the procedure of ASTM D-1557 or AASHTO T-180

91




Solution


= =+d

WV w

γγ γ


92



3 43

1 1 944 1030 3532

mV ftft

minus⎛ ⎞= =⎜ ⎟

⎝ ⎠3mtimes


Trial No 1 2 3 4 5

W(Newton) 145 156 163 164 161 ω () 20 24 28 33 37


( )3mkN

VW

=γ 154 165 173 174 171

( )31 mkN

d ωγγ+

= 128 133 135 131 125

93


ω () 62 81 98 115 123 132 γ (kNm3) 169 187 195 205 204 201

ωγγ+

=1d 159 173 178 184 182 178



b) γd at 95 = (095)(184) = 175 kNm3


c)

( )( )( )( )

( )

max

max

0115 270 184981

071184270981

= = =⎛ ⎞minus minus ⎜ ⎟⎝ ⎠

s d

w

ds

w

wG

SG

γγγγ

Saturation S = 71

94


Solution

( ) ( )

but therefore

105 89 versus 10351 1 018


= =

=

= = = =+ +

=




W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089


20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115




Solution


= =+d

WV w

γγ γ


92



3 43

1 1 944 1030 3532

mV ftft

minus⎛ ⎞= =⎜ ⎟

⎝ ⎠3mtimes


Trial No 1 2 3 4 5

W(Newton) 145 156 163 164 161 ω () 20 24 28 33 37


( )3mkN

VW

=γ 154 165 173 174 171

( )31 mkN

d ωγγ+

= 128 133 135 131 125

93


ω () 62 81 98 115 123 132 γ (kNm3) 169 187 195 205 204 201

ωγγ+

=1d 159 173 178 184 182 178



b) γd at 95 = (095)(184) = 175 kNm3


c)

( )( )( )( )

( )

max

max

0115 270 184981

071184270981

= = =⎛ ⎞minus minus ⎜ ⎟⎝ ⎠

s d

w

ds

w

wG

SG

γγγγ

Saturation S = 71

94


Solution

( ) ( )

but therefore

105 89 versus 10351 1 018


= =

=

= = = =+ +

=




W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089


20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115



3 43

1 1 944 1030 3532

mV ftft

minus⎛ ⎞= =⎜ ⎟

⎝ ⎠3mtimes


Trial No 1 2 3 4 5

W(Newton) 145 156 163 164 161 ω () 20 24 28 33 37


( )3mkN

VW

=γ 154 165 173 174 171

( )31 mkN

d ωγγ+

= 128 133 135 131 125

93


ω () 62 81 98 115 123 132 γ (kNm3) 169 187 195 205 204 201

ωγγ+

=1d 159 173 178 184 182 178



b) γd at 95 = (095)(184) = 175 kNm3


c)

( )( )( )( )

( )

max

max

0115 270 184981

071184270981

= = =⎛ ⎞minus minus ⎜ ⎟⎝ ⎠

s d

w

ds

w

wG

SG

γγγγ

Saturation S = 71

94


Solution

( ) ( )

but therefore

105 89 versus 10351 1 018


= =

=

= = = =+ +

=




W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089


20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


ω () 62 81 98 115 123 132 γ (kNm3) 169 187 195 205 204 201

ωγγ+

=1d 159 173 178 184 182 178



b) γd at 95 = (095)(184) = 175 kNm3


c)

( )( )( )( )

( )

max

max

0115 270 184981

071184270981

= = =⎛ ⎞minus minus ⎜ ⎟⎝ ⎠

s d

w

ds

w

wG

SG

γγγγ

Saturation S = 71

94


Solution

( ) ( )

but therefore

105 89 versus 10351 1 018


= =

=

= = = =+ +

=




W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089


20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


Solution

( ) ( )

but therefore

105 89 versus 10351 1 018


= =

=

= = = =+ +

=




W W W V

V V

pcf pcf pcfw

a

γ

γ γ

γγ γ

( )

3 3

33

3 3

10350000 1163089


20 2000824

⎛ ⎞=⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠= =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝

minus

⎠

pcfyd ydpcf

feet lbydyd feet

bton lb

truc

truck l

k ton

oads

95


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


Test 1 2 3 4 5 6 7 w 300 445 585 695 805 946 990 d

w

γγ 194 201 206 209 208 206 205


Solution

( )( ) ( )( )

It is known that 1



e saturation is

20900695 260 07426

e OMC is

0 10 209

s s s ds

d d

d OMCOMC OMC s

s d

wGSe wG S but ee

S w G

γ γ γγ γ

γγ γ

minus


⎝ ⎠

⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟minus minus⎝ ⎠ ⎝ ⎠

74

96




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115




( )( )( )

( ) ( )3

(max)

3( )

( ) (min) (max)

(max) (min)

3

( )

1


090204

090 204


84

i

s

= = =

there4 = =

⎛ ⎞⎛minus= ⎜ ⎟⎜⎜ ⎟minus⎝ ⎠⎝

d field d field

d

d field

r

d field d dr

d d d field

a RC

RCkN m

kN m

b D

D

kN m

γ γγ

γ

γ γ γγ γ γ

( )( )

( ) ( )( )3 3

184 139 204 0768204 139 184


768

212 1 184 1 015

⎞ minus ⎛ ⎞= =⎟ ⎜ ⎟⎜ ⎟ minus ⎝ ⎠⎠

= + = + =d kN m

c

w kN m

γ

γ γ

=

97



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115



Solution

( )

( )

( ) ( )( )

( )( )( )

max


1037) 8101 1 028

810 090900

) 11

270 6241 1 108

810

028 270070

10

90

Satu8

= = =+ +

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

= there4 + =+


⎝ ⎠=

there4 = = =

moistd field

d field

d

s w s wd

d

s w

d

s

s

a pcfw

pcfRCpcf

G Gb ee

Ge

Se wG

wGSe

γγ

γ

γ

γ γγγ

γγ

ration S=70

98


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


( )

( )( )( ) 3 3

(

(

3

)


1 1 069



157

d

= = =+ +

=

=

S wd

d base

d borrow pi

Ge

V m m km lon

k

x m

Nm

g

γγ

γγ

( ) ( )

( )

3 3

)

3 3

3

33

18 45 10157



W eight of water

2 58

= (

0 truck-loads

015)(314 )

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

t

d

d

V x m

x mm

kNW m kNm

wW kN


47 per truck load

59069

= = = =SwGSe

kN

99




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115




1 11

s w s wzav zav

s

G G andwGe w

S

γ γ γγ γ= = =+ +⎛ ⎞+ ⎜ ⎟

⎝ ⎠


100

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115

( ) ( )( ) 3116 24 1 000 32 00012

119 5 11d-max



γ

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

( )( ) ( )63117 1 32 000 3 75 10



W V pcf ft lbs


γ= = =

( )6

3

3

3 75 10 7 480 0831 0 083 1

34 50062 4

0 14

ww

w

w



lbft

High end at 143 V

γ

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠= = asymp

⎛ ⎞⎜ ⎟⎝ ⎠

=( )

6

3

3

3 75 10 7 4831 0 143 1

56 20062 4

x lb gallons ft

gallbft


0

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠ asymp

⎛ ⎞⎜ ⎟⎝ ⎠


101



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115



102


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


w() 300 445 585 695 805 946 990 γd γw 194 201 206 209 208 206 205


Solution

(kN

m3)

w ()

d

OMC (77)= 209 kNm3

d max

2 3 4 5 6 7 8 9 10 1118

185

19

195

20

205

21

215

22

103

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115

b) ( )( )max

max

20977 265 76265 209

s ds

w d

wGSe wG Se


c)

1

n

1-n

a

nS

A

W

S

( )

11 1 1 1

(1 )

V W W WW

V

a

V V V Vns VV V V

a nS n nSa n S V

= = = =


= minus =

n




1

d

wd

w sd

s

d s

w s wSS

wγ γγ γ γ

=+

S

ww SS S

γγ

γγ

γγγ

= = =minus



104


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


Asymptotes a = 0

a1

a2

a3

0

2

4

6

8

10

12

0 3 6 9 12 15 18 21 24 27

w

γ d

γd = (1-a2)γs

S = 100

S1

S2

0

02

04

06

08

1

12

0 10 20 30 40

w

γ d


105


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


w () 10 13 16 18 20 22 25 γ (pcf) 98 106 119 125 129 128 123

γd (pcf) 89 94 1026 1059 1075 1049 984

Solution

(pcf

)

w ()

d


=(090)(1075) = 968 pcfd field

= 1075 pcfd max

0 10 20 3085

90

95

100

105

100

105

110

115


106


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


( )( )

010 010

89 010 01 89 9 89 9 98

ww s

s

s w s s w

Ww W WW


= = there4 =


( )( )

020 020

1075 020 02 1075 215 129

ww s

s

s w s s w

Ww W WW


= = there4 =

= = = = there4 = + =


there4

33

3

3 3 3

125 748 27 40


gallons dy 624 1


ft lb ft yd

107


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115


Solution

( )( ) ( )( ) ( )( ) ( )23

1 1 450a) Volume of levee = 20 40 20 20 20 60 233002 2 27


ft longft cyft cy

⎛ ⎞⎡ ⎤⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠

( )( )

( )

( ) ( )

( ) ( )

( ) ( )

( )

( )

But or or

1121where 95 095 106 10071 1 018

10072330095

s borrow s levee


d borrow d levee

d leveeborrow levee

d borrow

W W

W W V V VV

pcf and pcfw

pcfV V cypcf

γ γ γ γ

γγ γ

γγ

=

= = =

= = = = =+ +

there4 = = 24700cy⎡ ⎤=⎢ ⎥

⎣ ⎦

108

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115

( )( )

( ) ( )

( ) ( )

( )

( )

( ) ( )


268 6241007 =

1 1 147

=

=

⎛ ⎞there4 = ⎜ ⎟⎜ ⎟

⎝ ⎠

= =+ +

s hauled s levee


d leveehauled levee

d hauled

S wd levee d hauled

W W

V V

V V

Gbut pcf ande

γ γ

γγ

γγ γ

( )( ) 3 3

( )

3

3

= 677

100723300 34700677


⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟ minus⎝ ⎠ ⎝ ⎠

d leveehauled levee

d hauled

hauled

pcf

pcfV V yd ydpcf


γγ

109



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115



Levee cross-section

110

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115

Questions


b) 2100 ft2

c) 2350 ft2

d) 2700 ft2

e) 2550 ft2







a) 11 b) 092 c) 084 d) 079 e) 12


a) 098 b) 091 c) 102 d) 11 e) 072


a) 093 b) 122 c) 081

111

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115

d) 101 e) 100


a) 110 b) 112 c) 108 d) 105 e) 117









112



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115



A = 12

(30)(90) + (30)(10) + 12

(30)(60) = 1350 + 300 + 900 = 2550 ft2 = 283 yd2


3 Answer is d


113




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





115




= (67)(106 ft3) 3748 gallons

ft⎛⎜ ⎟⎝ ⎠

⎞= 501x106 gal = ( )( )6 32

167 10 ft -acre 243560 ft

⎛⎜⎝ ⎠

⎞⎟ = 154 acre-ft

4 Answer is a

From ( )( )(1 1) 97 1 031(l +



For Pit A e = SG wS

= (269) 022 065

⎛⎜⎝ ⎠

⎞⎟= 091

7 Answer is b

For Pit B e = SG wS

= (271) 022 049

⎛⎜⎝ ⎠

⎞⎟= 122

8 Answer is e

For Pit C e = SG wS

= (266) 018 041

⎛⎜⎝ ⎠

⎞⎟= 117


VA = 12 x 106 yd3 198 184

⎛⎜⎝ ⎠

⎞⎟= 129 Myd3


VB = 12 x 106 yd3 181 184

⎛⎜⎝ ⎠

⎞⎟= 118 Myd3


VC = 12 x 106 yd3 210 184

⎛⎜⎝ ⎠

⎞⎟= 137 Myd3

12 Answer is e

114





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