Calculation Pusat Pelupusan Sampah

8/8/2019 Calculation Pusat Pelupusan Sampah

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Example 4.5

Solution :

Where

1 - 0.5 = 0.5

Substituting these calculated values into Equation 4.31 yields

=

= (110)(2)(0.364) + (2)(220 + 660)(0.364)/[0.949 - (0.316)(0.364)]

= (80.08 + 640.64)/0.834

= 720.72/0.834

= 864 lb/ft (12.61 kN/m)

So,

Ultimate Strenght >Anchor Resistance Capacity > Allowable Strength

Calculate the resistant capacity of a given geomembrane in a L-shaped rectangular anchor trench ofdimensions. The geomembrane is 60-mil (1.5-mm) HDPE with an ultimate strength (at yield) 2,100 l14,500 kN/m) and an allowable strength 840 lb/in (5,800 kN/m).The runout length is 3 feet (0.9m

cover soil is 1 foot (0.3m). The anchor trench is 2 feet (0.6m) wide and 2 feet (0.6m) deep. The sideangle is 18.4 degrees [ 3(H) : 1 (V) ]. The unit weight of soil is 110 lb/ft (17.3 kN/m). the soil friction

30 degrees. The friction angle between the soil and the geomembrane is 20 degrees.

110 X 1 = 100 lb/ft2 (5.27 kN/m2)

110 X (1 + 0.5 x 2) = 110 x 2 = 220 lb/ft 2 (10.53 kN/m2)

110 X (1 + 2) = 330 lb/ft2 (15.80 kN/m2)

(110)(2)(tan 20) + 2(220)(2) + (330)(2)(tan 20)/[cos 18.4 - (sin 18.4)(tan 20

to the following inequalities :

2,100 lb/in2 > 1,200 lb/in2 > 840 lb/in2

(14,500 kN/m2 > 8,270 kN/m2 >5,800 kN/m2)

The results of the calculation indicate the design achor resistance capacity falls between the yield stallowable stress of a geosynthetic membrane liner. Therefore, the achor trench dimensions are acce


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Example 8.4

Assume that

Then

and

where and

Thus,

and

A landfill cell is constructed on a pipe slope of 1% and a bottom liner grade perpendicular to the pipThe horizontal distance from upstream to pipe is 100 ft (30 m). The granular soil drainage layer is 20.3 m) thick. The hydraulic conductivity of the drainage material is 0.01 cm/sec. The amount of leacnflow rate is estimated to be 3,000 gallons/acre/day (1,024 mm/year). Estimate the maximum leachover the primary liner.

SolutionThe actual landfill base grade is shown as Figure 8.14. Liquids always flow along the maxigrade. The largest grade in Figure 8.14 is not perpendicular to the leachate collection pipes. Thus, thmaximum leachate flow distance from upstream boundary to the leachate collection pipe should beargerthan 100 feet (30 m). The actual leachate flow grade, S, and the maximum leachate flow distahe upstream boundary to the leachate collection pip, L, can be calculated as follows:

a. the slope of the bottom liner grade perpendicular to the leachate collectionpipe is S1;

b. the slope of the leachate collection pipe is S2.

n figure 8.14, it can be seen that the landfill base grade varies with change of x. The landfill base grS(x), can be expressed as follows:


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The landfill base grade is a maximum in which

Because can never be equal to 0,

and

and

Thus,

or

where maximum landfill base grade (i.e., actual leachate flow grade);slope of the bottom liner grade perpendicular to the leachate collection pipe; an

slope of the leachate collection pipe.

where

b

slope of the bottom liner grade perpendicular to the leachate collection pipe; an

slope of the leachate collection pipe

For the maximum landfill base grade, Smax

,

Smax

=

S1=

S2=

The horizontal distance from upstream boundary to the leachate collection pipe along themaximum drainage grade is

Lmax

=horizontal distance from upstream boundary to the leachate collection pipealong the maximum drainage grade, ft or m;

horizontal distance from upstream boundary to the leachate collection pipe,which is perpendicular to the leachate collection pipe, ft or m;

S1=

S2=

For the example problem in which S1

= 2% and S2

= 1%, the maximum base grade (i.e., actual leach

grade) is


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=

= 0.02

=

= 112 feet = 1,341 inches (33.6 m)

Based on a given inflow rate,

Moore's 1980 Method:

(1,341.6)(0.0180)[1.545 + 1 - (68.956)(0.0288)]

(1,341.6)(0.0180)(0.559)

13.4 inches > 12.0 inches (336 mm > 300 mm)


(1,341.6)(0.0288 - 0.0224)

(1,341.6)(0.0064)

8.6 inches , 12.0 inches (216 mm < 300 mm)

Giroud's 1992 Method:

1 - 0.12 exp{-[log(1.0358646)^(58) ]^2 }

1 - 0.12 exp[-(log 1.0358646)^(58) ]^2 }1 - 0.12 exp(-0.00091477)

0.88

Smax

= (S1

2+S2

2)0.5

[(0.02)2+(0.01)2]0.5

The horizontal distance from the upstream boundary to the leachate collection pipe alonmaximum drainage grade is

Lmax =

100 [(0.01/0.02)2]0.5

r = 3,000 gallons/acre/day = 40.33 inches/year = 0.115 inch/day (2.81 10-3 m/day)

ymax =(1,341.6)(0.1105/340.16)1/2[340.16 0.02242/0.1105 + 1

- (340.16 0.0224/0.1105)(0.02242 + 0.1105/340.16)1/2]

ymax

=

(1,341.6)[(0.1105/340.16 + 0.02242

)1/2

- 0.0224](1,341.6)[(0.000325 + 0.00052)1/2 - 0.0224]

j =

1 - 0.12 exp{-[log(1.6 0.1105(340.160.0224^2 ))^(58) ]^2 }


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11.8 inches < 12.0 inches (299.8 mm < 300 mm)

McEnroe's 1993 Method:

R =

0.11505/(340.16 x 0.0224^2) = 0.6474 > 1/4

B =

(4 x 0.6474 -1)^(12) = 1.2608

(11.2608) tan^(-1)[(2 x 0.6474 -1)/1.2608]}

(1,341.6)(0.0224)(0.7957)exp (- 0.7027)

(1,341.6)(0.0224)(0.7959)(0.4952)

11.8 inches < 12.0 inches (299.8 mm < 300 mm)

ymax

=

(0.8800)(1,341.6)[(4 0.1105/340.16 + 0.0224^2)^1/2 - 0.0224](2

(0.8800)(1,341.6)(0.0424 - 0.0224)/(2 0.9997)

(0.8800)(1,341.6)(0.02004)/(2 0.9997)

Because R > 1/4, we use Equation 8.36 to calculate ymax

:

ymax

=

, . . . - . x . . .exp{(11.2608)tan^(-1)[(2 x 06474 x 0.0224 - 1)1.2608] -


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Example 8.5

Solution: k 0.01 cm/sec = 340.16 in/day (0.1 mm/day);

L 450 ft = 5,400 in (135 m);

k 30 in/year = 0.08(2.1 mm/day);22 in/day

k = 0.1;

k = 0.0995, cos = 0.9950


= (5400)(0.0822/340.16)^1/2 [340.16 X 0.1^2/0.08

- (340.16 X 0.1/0.0822) (0.1^2 + 0.0822/340.16) (5400)(0.0155)[41.382 + 1 - (413.820)(

(S400)(0.0155)(0.586)

49.0 inches (1,245 mm)


(5400)(0.0822/340.16 + 0.1^2)^1/2 - 0.

(5400)(0.000242 + 0.01)^1/2 - 0.1]

(5400)(0.00120)

6.5 inches (165 mm

Giround's 1992 Method:

j =

1 - 0.12 exp {- [log(1.6 X 0.0822/340.16 + 0.1^2 )^(58) ]

1 - 0.12 exp {- [log(0.0386642 )^(58)

1 - 0.12 exp [- (log 0.13094)^2 ]

1 - 0.12 exp(- 0.77956)

0.95

(0.9450)(5,400)[4 X 0.0822/340.16 + 0.1^2)^1/2 - 0.1] / (2

(0.9450)(5,400)(0.10472 - 0.1) / (2 X 0.9950)

(0.9450)(5,400)(0.00472) / (2 X 0.9950)

12.1 inches (307 mm)

known/in Thelopeangle is

A final cover for a municipal solid waste landfill is constructed on a slope of 10% with a madistance of 450 feet (135 m). The granular soil drainage layer is 24 inches (0.3 m) thick. Trainfall percolating though the cover is istemated to be 30 inches/year (762 mm/year). Thconductivity of the drainage material is 0.01 cm/sec. Estimate the maximum. saturated deliner.

tan

5.7 , sin

ymax

=

)] ymax =

ess andtable. y

max=


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McEnroe's 1993 Method :

R =

0.0822/(340.16 X 0.0995^2) = 0.0244 < 1/4

A = (1 - 4 X 0.0244)^1/2 = 0.0902^1/2 = 0.950

(5400 X 0.1)(0.0244 -0.0244 X 0.1 + 0.0244^2 X 0.1^2)^1/2

{[(1 - 0.950 - 2 X 0.0244)(1 + 0.950 - 2 X 0.0244 X 0.1)]

/[(1 + 0.950 - 2 X 0.0244)(1 - 0.950 - 2 X 0.0244 X 0.1)]}^1/(2X0.950)

(5400 X 0.1)(0.0244 -0.0244 X 0.1 + 0.0244^2 X 0.1^2)^1/2

{[(1 - 0.950 - 2 X 0.0244)(1 + 0.950 - 2 X 0.0244 X 0.1)]

/[1 + 0.950 - 2 X 0.0244)(1 - 0.950 - 2 X 0.0244 X 0.1)]}^1/(2X0.950) (5400 X 0.1)(0.0220)^1/2{[0.0012)(1.945)]/[(1.901)(0.0451)]}^0.526

5400 X 0.1 X 0.148 X 0.150

12.0 Inches (305 mm)

of 2%.inches

hateate head

Because R < 1/4, use Equation 8.34 to calculate y

ume

ce from

ymax

=

ade ,


8/25

Example 9.1

Solution:

1) Determine Amount of Leachate Generation

Area of the upgradient portion of the landfill cell:

1,000 X (3 X 6.5)

Maximum leachate flow generated for the upgradient portion of the cell:

0.0786 X 195,000

Area of the downgradient portion of the landfill cell:

1,000 X 100 =

Maximum leachate flow generated at the downgradient portion of the cell:

0.0355 X 100,000

Maximum leachate flow generated for the entire cell:

2) Selection of Pipe Size

Select 6-in (150-mm) SDR 11 HDPE pipe:

SDR

So t = 6/11 0.55 in (0.014 m)

= 6 -2 X 0.55 = 4.9 in = 0.408 ft (0.124 m)

= 0.408/4 = 0.102 ft (0.031 m)

A =

A leachate collection pipe is placed in the middle of a rectangular landfill cell. Upgradientcollection pipe is adjacent to a 3(H):1(V) side slope with a height of 65ft (19.5 m). The dowthe leachate collection pipe rests on a bottom floor sloping at 2% with a width of 100 ft (3length of this leachate collection pipe is 1,000 ft (300 m). The peak leachate generation ra

area is 0.0786 ft2/day/ft2 (0.0240 m3/day/m2), and the peak leachate generation rate at theis 0.0355 ft3/day/ft2 (0.0108 m3/day/m2). Design a perforated solid wall pipe to meet both dperforation requirements.

Au= 195,000 ft2 (18,000 m3)

(Qu)

max= (q

u)

maxX A

u

15,327 ft2/day = 0.1774 ft3/sec (5.023 X 10-3 m3/sec

Ad= 100,000 ft2 (9,000 m2)

3,550ft3/day = 0.0411 ft3/sec (1.164 X 10^-3 m3/sec)

0.1774 + 0.0411 = 0.2185ft3/sec (6.187 X 10^-3 m3/sec)

Do/t

Do/SDR

Di=

ate flow rh

= Di/4

= 3.1416 X (0.408/2 = 0.131 ft2 (1.217 X 10^-2 m2)

For HDPE pipe, n 0.011


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Manning's Equation:

Q =

(1.49/0.011) X 0.131 X (0.102)^2/3 X (0.01)^1/2)

135.455 X 0.131 X 0.218 X 0.1

3) Pipe Perforation

Maximum Leachate Inflow per Unit Length of Pipe

0.0786 X 195 + 0.0355 X 100

15.327 + 3.55

Also,

Discharge coefficient C = 0.62 ;

Bernoulii Equation:

0.62 X 0.000341 X 0.1

Number of Perforation Holes:

N =

0.0002184/0.00002114

10.35 holes/ft (34 holes/m)

So, use 12 holes/ft (40 holes/m); that is 6 holes per foot (20 holes per meter) each side as

g the

0.387 ft3/sec (10.96 X 10^-3 m3/sec) > Qmax

= 0.2185 ft3/sec (6.187 X

Qin

=

18.877 ft3/day/ft = 0.0002184 ft3/sec/ft (2.209 X 10^-5 m3/sec/m)

Assume that the diameter of a perforation hole, d, in the pipe is 0.25 inch (6 mm). Then tharea of a hole on the perforated pipe is

Ab

=

3.1416 X (0.25/12/2)^2 = 0.0003341 ft2 (0.0000316 ft3)

Limiting leachate entrance velocity vent

= 0.1 ft/sec (0.03 m/sec)

Qb

=

0.00002114 ft3/sec (5.986 x 10^-7 m3/sec)


10/25

Example 9.2

Solution:The maximum load applied on the pipe is given by

([(115)(2)+(60)(100)+(115)(1)+(110)(3.5)+(90)(0.5) ]812)/((1-40.((230+6,000+115+385+45)812)/(0.917)

(6,775812)/0.917

4,925 lb/ft = 410 lb/in (72 kN/m)

The maximum pressure applied on the pipe can be obtained from

410/8

From Table 9.5,

and

,

For

The soil reaction modulus is given by

E' = 2 X 4,439

The thickness of pipe given by

=8/11 0.73 in (0.1847 m)

An 8-inch (200-mm) SDR 11 HDPE perforated pipe with 8, 0.25-inch (6-mm) holes per footside per foot) is selected as a primary leachate collection pipe. The maximum load actinga 2-ft (0.6-m) protective sand layer (Y

sand= 115 lb/ft3 or 18 kN/m3), 100-ft (30-m) solid was

or 9.4 kN/m3), 12-inch (0.3-m) gas venting layer (ysand

= 115 lb/ft3 0r 18 kN/m3), 18-inch (0.

clay layer (yclay

= 110lb/ft3 or 17.3 kN/m3), 24-inch (0.6-m) drainage and protective layer (y

17.3 kN/m3

), and 6-inch (0.15-m) topsoil (ytop = 90 lb/ft3

or 14 kN/m3

). Assume Bedding andeflection lag factor D

L=1.0, elastic modulus of the pipe material for 50 years at 73F (23

= 28,200 lb/in2, (194, 000 kN/m). Poisson's ration of pipe material = 0.3. The bedding mpoorly graded gravel (GP) with 85% standard density. What will be the deflection ratio (%)buckling pressure of the project?

WC

=

Ptp

= 51.3 lb/in2 (354 kN/m2)

Ptp = 40 lb/in2 , ES = 4,100 lb/in2

Ptp

= 60 lb/in2 , ES

= 4,700 lb/in2

Ptp

= 51.3 lb/in2,

ES

= 4,100 + (51.3 - 40)(4,700 - 4,400)/20 = 4,100 + 339 = 4,439 lb/in 2

8,878 lb/in2 (61,200 kN/m2)


11/25

Also,

Mean radius of the pipe, r = 3.635 in (0.0923 m);

and

Modified Lowa Formula:

((1.0)(0.11)(410)(48.03))/((28,200)(0.0324)+(0.06

2,166/(914+26,011)

0.08 in (2.0 mm)

Deflection Ratio:

Deflection ratio =

= (0.08/7.27) X 100%

= 1.1% < 2.7% (ok, as shown inTable 9.4)

Wall Buckling of Pipe:

Poisson's ratio of pipe material, = 0.3;

Mean radius of the pipe, r = 3.635 in (0.0923 m).

Thus,

2[9,756(913.6848.03) ]^(12)

2 X (185,589)^1/2

2 X 431

The factor of safety for pipe wall buckling is, then,

0.9950)

FS = 862/51.3 = 16.8 > 2 (OK)

e ect on ag actor,L

= . ;

ximum horizontale moment ofhydraulic

pth over the cover

Bedding angle = 0, K = 0.11;

Elastic modulus of the pipe material, E =28,200 lb/in2 (194,000 kN/m3);

Soil reaction modulus, E' = 8,878 lb/in2 (61,200 kN/m3);

Inertia moment of the pipe wall per unit length, in

4

/in = in

3

, given by

= (0.73)3/12 = 0.389/12 = 0.0324 in3 (5.276 X 10

((1.0)(0.11)(410)(3.635)^3)/((400,000)(0.032

Modulus of soil reaction , E' = 8,878 lb/in2, (61,200 kN/m2);

Modulus of elasticity of the pipe material, E = 28,200 lb/in2 (194,000 kN/m2);

Momentof inertia of the pipe wall per unit lengt, I = 0.0324 in3 (5.276 X 10^-7 m

Pcr

=

2{[8,878((1-0.3^2 ) )][((28,2000.034))(

862 lb/in2 (5,943 kN/m2)


12/25

Example 9.3

Solution:

p =(3.75) 11.78 ft (3.5 m)

= 0.445

= (0.5)(11.78)(0.455)(70) (210)^2 (0.445)= 3,681,500 lb (16,400 kN)

=

3,681,500/3.98

FS =

5,000/6,424

0.78 < 1.0 (No Good)

Refer to Figure 9.9(a). The primary leachate removal HDPE manhole riser is circinside diameter of 3 ft (0.9 m) and an outside diameter of 3.75 ft (1.13 m). Thesurrounding the manhole riser is given by Hf = 210 ft (63 m); the unit weight ofywaste = 70 lb/ft3 (11 kN/m3); the internal friction angle of solid waste is givenfriction angle between the concrete manhole riser wall and solid waste is = 24downdrag force, stress on the concrete, and factor of safety if the compressive s

lb/in2 (34,500 kN/m2).

KO

= = 1 - sin 33 = 1 - 0.545 = 0.455

tan = tan 24

a.The downdrag force is

Qn=

b.The downdrag stress is

3,681,500[(4)(3.75^2-3.0^2 ) ]

925,000 lb/ft2

6,424 lb/in2 (44,300 kN/m2)

c. Finally, the factor of safety against concrete overstress is


13/25

Example 12.1

The filling procedure of a new minicipal solid waste landfill is listed in Table 12.3.

Table 12.3 Solid waste Filling Record

Time PeriodHeight of Solid Waste Filled

Feet Meter

1st Month 12 3.62nd Month 18 5.4

3rd Month 16 4.8

4th Month 10 3

5th Month 14 4.2

Assume the following:

Original applied pressure for solid waste

= 0.26;

Modified secondary compression index, 0.07; and

Calculate the total settlement of the top of the landfill at the end of the 5th mont

Solution:

Calculate the solid waste depth over the mid-level of each waste layer:

(0.5)(12) + 18 + 16 + 10 + 14 = 64 ft (19.2 m)

(0.5)(18) + 18 + 10 + 14 = 49 ft (14.7 m)

(0.5)(16) + 10 + 14 = 32 ft (9.6 m)

(0.5)(10) + 14 = 19 ft (5.7 m)

(0.5)(14) = 7 ft (2.1 m)

Calculate the total overburden pressure acting on the mid-level of each waste la

f the leachatengradient side ofm). The totale at the side slope

bottom floor arearainage and

Unit weight of solid waste, ywaste

= 70 lb/ft3 (11 kN/m3);

1000 lb/ft2; (48 kN/m2);

Modified primary compression index, C'c

Secondary settlement starting time, t1

= 1 month

H1=

H2=

H3=

H4=

H5=

70 X 64 = 4480 lb/ft2 (215 kN/m2)

70 X 49 = 3430 lb/ft2 (164 kN/m2)

70 X 32 = 2240 lb/ft2 (107 kN/m2)

70 X 19 = 1330 lb/ft2 (64 kN/m2)

70 X 7 = 490 lb/ft (23 kN/m) < o= 1000 lb/ft2 (48


14/25

Calculate the settlement of each waste layer:

First Layer:

0.26 X 12 X log (4480/1000) = 0.26 X 12 X 0.651 = 2.03 ft (

0.07 X 12 X log (4.5/1) = 0.07 X 12 X 0.653 = 0.55 ft (0.17

2.03 + 0.55 = 2.58 ft (0.79 m)

Second Layer:

0.26 X 18 X 1og (3430/1000) = 0.26 X 18 X 0.535 = 2.51 ft (

0.07 X 18 X log (3.5/1) = 0.07 X 18 X 0.544 = 0.69 ft (0.212.51 + 0.69 = 3.20 ft (0.98 m)

Third Layer:

0.26 X 16 X log (2240/1000) = 0.26 X 12 X 0.350 = 1.46 ft (

0.07 X 16 X log (2.5/1) = 0.07 X 16 X 0.398 X = 0.45 ft (0.21

1.46 + 0.45 = 1.91 ft (0.58 m)

Fourth Layer:

0.26 X 10 X log (1330/1000) = 0.26 X 10 X 0.124 = 0.32 ft (

0.07 X 10 X log (1.5/1) = 0.07 X 18 X -.176 = 0.22 ft (0.0670.32 + 0.22 = 0.54 ft (0.16 m)

Fifth Layer:

0 [becaus

0 + 0 = 0

Calculate the total settlement of the landfill at the end of the 5th month:

hown in Figure 9.3.

2.58 + 3.20 + 1.91 + 0.54 + 08.23 ft (2.51 m)

8.23/70 = 11.8%

0^-3 m3/sec) (OK)

e cross-sectional

= 490 lb/ft2 (23 kN/m2) < = 1000 lb/ft2 (48 kN/m2)]

0 (because t2

= 0.5 month , t1= 1 month)


15/25

Example 13.1

Solution Using Equation 13.9 to solve for the FS-value rasults in a value of 1.25,

agreement woth the curves of Figure 13.4:

a = 14.6 kN/m

b = - 21.3 kN/m

c = 3.5 kN/m

Thus FS = 1.25

512) )This value can be confirmed using Figure 13.4.

Commen In general, this is too low of a value for a final cover soil factor-of-safet

i.e., 4 holes pern the pipe includese (Y

waste= 60 lb/ft3

45-m) compacted

ilt= 110lb/ft3 or

e = 0,C) temperature E

terial of the pipe isand critical

The following are given: a 30-m slope with a uniformly thick 300-mm-deep coveweight of 18 kN/m3. The soil has a friction angle of 30 and zero cohesion (i.e., itcover soil is placed directly on a geomembrane as shown in FIgure 13.3. Direct sresulted in an interface friction angle between the cover soil and geomembraneadhesion. What is the FS-value at a slope angle of 3(H)-to-1(V) (i.e., 18.4)?

There are many possible options to increase the value (e.g., changing te geomethe use of toe berms, taperred cover soil thickness, and veneer reinforcement, sSoong, 1998). Nevertheless, this general problem will be used throughout this scomparison with other cover soil slope stability situations.


16/25

Example 13.3

Solution Using the design curves of Figure 13.10 along with Equation

* From Figure 13.9, at 20 km/hr and 3.0 seconds, the bulldozer's qccel

* From Equation 13.13

a = 88.8 kN/m

)(8,878)(48.03)) b = - 107.3 kN/m

c = 17.0 kN/m

Thus FS = 1.03This value can be confirmed using Figure 13.10.

CommenThis problem solution can now be compared with those of th

Example 13.1 Cover soil along with no bulldozer loading:

Example 13.2 Cover soil plus bulldozer moving up slope:

Example 13.3 Cover soil plus bulldozer moving down slope:

The following are given: a 30-m-long slope with uniform cover soil of 3kN/m3. The soil has a friction angle of 30 and zero cohesion (i.e., it isa bulldozer moving from the crest of the slope down to the toe. The buof 30 kN/m and tracks that are 3.0 m long and 0.6 m wide. The estimathe time to reach this speed is 3.0 seconds. The cover soil to geomemzero adhesion. What is the FS-value at a slope angle of 3(H)-to-1(V) (i.

-7 m3)

)+(0.061)(1,000)(3.635)^3 )

The inherent danger of a bulldozer moving down the slope is readily acomes about by the bulldozer decelerating instead of accelerating. This arguably the more severe condition, due to the extremely short timmotion. Clearly, only in unavoidable situations should the cover soil pldown the slope. If it is unavoidable, an analysis should be made of theconstruction specifications should reflect the precise conditions madeand ground contact pressure of the equipment shoud be stated alongthe cover soil placement operations. Truck traffic on the slopes can althan illustrated here and shoud be avoided in all circumstances.

)

635)^3 ]}^(12)


17/25

Example 13.4

Solution

(62.4)(0.5)[44 - (0.5)(0.5)(0.949)]/(0.333) = 4,100.3 lb/ft (5

(0.5){(110)(2 - 0.5) [(2)(44)(0.949) - 2 - 0.5] + (115)(0.5)[(2)

(0.5)(13,366.98 + 4,773.19)/[(0.316)(0.949)] = 30,245.3 lb/f

Using Equation 13.36

a =

b =

- (23.4)(0.625) + (7.8)(0.316)(0.949)(0.625 - 0.404)}

- (459.8 + 12,892.1 - 1,572.0 - 14.6 + 0.5) = -11,766 (-166

c =

[(30,245.3)(0.949) - 4,100.3 + (7.8)(0.316)](0.316)(0.625)(0.

ar in cross section with aneight of solid wasteolid waste is given byy w = 33; and interface. Determine the totalrength of concrete is 5,000

A 44-ft (13.2-m) high and 3(H):1(V) slope has cover sand with a uniforweight og 110 lb/ft3 (17.3 kN/m3). The cover sand has a friction angleSeepage occurs parallel to the slope and the seepage water head in thsaturated unit weight of sand is 115 lb/ft (18 kN/m). The interface frictand geomembrane is 22 degrees and zero adhesion. What is the facto

The side slope angle is at 18.4 for a 3(H):1(V) slope. Hence,

sin = sin (18.4) = 0.316, cos = cos (18.4) = 0.949, tan = tan (

H = 44 ft (13.2 m), h = 2 ft (0.6 m), hw

= 0.5 ft (0.15 m), = 110 lb/ft3

sat

= 115 lb/ft3 (18 kN/m3), w

= 62.4 lb/ft3 (9.81 kN/m3), = 32, =

tan = tan (32) = 0.625, tan = tan (22) = 0.404

(0.5)(62.4)(0.5)2 = 7.8 lb/ft (0.11 kN/m)

(0.5)(62.4)(0.5)2/(0.333) = 23.4 lb/ft (0.33 kN/m)

(0.5){(110)[(2)2-(0.5)2] + (115)(0.5)2}/[(0.316)(0.949)] = 73

(30,245.3)(0.316)(0.949) + (7.8) [1- - (0.949)2] = 9,071 (128

- {(735.7)(0.625) + (30,245.3)[(0.316)

2

(0.625) + (0.949)

2

(0.


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11,766 + [(-11,766^2 - (4)(9,071)(1,963

Commen

h.

er:

The seriousness of seepage forces in a slope of this type is ibeen 100% of the drainage layer thickness, the FS-value wothe result using a horizontal assumption of saturated cover

give essentially identical low FS-values. Clearly, the teachinadequate long-term drainage above the barrier layer in covseepage forces from occuring. Figure 13.15 shows a sand laseepage force.

kN/m2)


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62 m)

)

0.77 m)

)

45 m)

m)

098 m)

)


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which in seen to be in

y and a redesign is necessary.

soil at a units a sand). The

hear testing hasof 22 with zero

ry of the situation,ee Koerner andction for


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13.13, the solution can be obtained.

ration is 0.19g.

e previous two examples:

FS = 1.25

FS = 1.24

FS = 1.03

00-mm thickness at a unit weight of 18 sand). It is placed on the slope usingldozer has a ground contact pressureed equipment speed is 20 km/hr, andrane friction angle is 22 degrees with., 18.4)?

parent. Note, that the same resultsharp breaking action of the bulldozer

s involved when stopping forwardcement equipment be allowed to workspecific stability situation and then the design. The maximum weight

with suggested operator movement ofo give stresses as high or even higher


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8.02 kN/m)

44)(0.949) - 0.5]}/[(0.316)(0.949)]

(427.6 kN/m)

or SI units)

404) = 1,963 (28 for SI units)

thickness of 2 ft (0.6 m) at a unitf 32 degrees and zero cohesion.e sand layer is 6 inches (0.15 m). Theon angle between sand drainage layerof safety at a slope of 3(H):1(V)?

8.4) = 0.333.

(17.3 kN/m3),

2

.7 lb/ft (10.4 kN/m)

for SI units)

04)] - (4,100.3)(0.949)(0.404)


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]^0.5)/(2)(9,071)

mediately obvious. Had the saturationld have been still lower. Futhermore,oil with the same saturation ratio will

of this example problem is thatr soil slopes must be provided to avoider sliding failure sideslope caused by


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Calculation Pusat Pelupusan Sampah

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