1Faculty Of Engineering

Civil Engineering Department

Construction Technology-I CIVL 596

barrage construction projectSubmitted to:

PROF. DR. TAHİR ÇELİK

Project Number 02 ------- Group Number 04

Students’ Name Students’ Number

BILAL MOHAMMED PIROT 145292

SEVAR DILKHAZ SALAHADDIN 145233

HUSSEIN GHANDOUR 135919

SPRING 2014-2015

2BARRAGE

CONSTRUCTION

A careful survey will be the first requirement in order to determine the optimum site for construction of a barrage

Particularly where a large barrage is required to dam a deep estuary, presents the greatest technical challenge

A brief Introduction

The purpose of the Dam/ Barrage studies is to provide a control structure at source of river that will maintain the predetermined water level

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Calculated Data in accordance to, G04 (Even Number) Required Compacted soil = 4*109 m3

Factor for converting the compacted soil to loose soil is (1.41) from Table Loose soil = 5.64109 Lm3

Softy clay soil, and the clay soil quarry is above the barrage level Job condition: good, and Management condition: good Finish in 8 years, working with two shafts, 52 weeks/year, 6 days/week, and 16

hours/day (each shaft 8 hours)

The site condition of the projectConversion of soil type

4EQUIPMENTS

Excavation Equipment

Name = hydraulic shovel 6090

Capacity = 52 m3, bottom dump

No. Of cycles =145cycles/hour from (table 2)

Swing factor =1.05 from (table 2)

Bucket fill factor = 0.8 from (table 3)

Job efficiency = 0.75 from (table 4)

Production = 4,750.20 lm3/hr. = 76,003.20 lm3/ day

The soil that should be excavated in one hour

= 141.23 * 103 Lm3/hr.

Number of Hydraulic shovels = 141.23 * 103 / 4,750.20 = 29.73 ≈ 30 shovels

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Bucket fill factors for excavators

Job efficiency

Standard cycles per hour

Swing factor

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TRUCK

Name = Truck 793D

Capacity = 176 m3

Travel Time = Filling + Dumping + Hauling+ Returning

FILLING TIME

The number of passes of shovel = 176/52 = 3.385 ≈ 4 passes

Time for 4 passes = Filling time = (60×4)/145 = 1.65 minutes

DUMPING TIME

Assume = 2.0 minutes

HAULING EQUIPMENT

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Time = Distance / (Average Speed Factor Max. Speed)

HAULING

TIME

Average speed factor 0.928 (from the front Table)

Distance = 1.2 km

Section 03, HaulingSection 03, Returning

Section 02, HaulingSection 02, Returning

Section 01, Hauling As an Example

Average speed factors

Section 01, HaulingSection 01, Returning

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TIME for section 1, hauling

=

1.2 60/ (0.928 39)

=

1.989 Minutes

Performance and Retarder Curve for Section 01, Hauling

MAX. SPEED

• Weight of truck = 383,673 kg, and assume that the penetration = 3 cm

• Rolling resistance factor (kg/ton) = 20 + 6×(penetration in cm) = 20+6×3 = 38 kg/ton

• Effective grade = grade% + rolling resistance factor/10 = 0+ (38/10) = 3.8 %From the

• Performance and Retarder Curve (Figure 4) ….. Speed = 39 km/hr.

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Time = Distance / (Average Speed Factor Max. Speed)

RETURNINGTI

ME

Average speed factor 0.928 (from the front Table)

Distance = 1.2 km

Effective grade = grade% + rolling resistance factor/10 Rolling resistance factor (kg/ton) = 20 + 6(penetration in cm)Assume that the penetration = 0.0Rolling resistance factor = 20+60= 20 kg/ton

Section 01, Returning As an Example

Average speed factors

TIME = 1.2 60/ (0.928 17) = 4.564 Minutes

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In one working day we have 960 Minutes

In one day each truck has = 960/103.433 = 9.28 9 cycles

Volume of Soil will transport by one truck per day = 176*9 = 1,584 Lm3/day

 = 2.28 106 /1,584 = 1,439.4

𝐍𝐮𝐦𝐛𝐞𝐫 𝐨𝐟 𝐭𝐫𝐮𝐜𝐤𝐬 1440 TRUCKS

TOTAL HAULING TIME = 22.814 Minutes

TOTAL RETURNING TIME = 76.969 Minutes

Travel Cycle Time = 103.433 Minutes

GRADING EQUIPMENT 11

Name = Komatsu D575A-3 Super Dozer

Typical dozer operating speed = 4 km/hr.Blade width = 7.4 m from

Assume the thickness of layers of soil = 0.4 m

Production of dozer /hour = 0.4 * 7.4 * (4 * 1000) = 11,840 m3/hr.

Production of dozer per day = 11,840 × 16 = 189,440 m3/day

Total volume of soil that will be transported by 1440 trucks = 2,280,960 m3 Number of Dozers = 12.041 ≈ 13 dozers/day

DOZER

12COMPACTING EQUIPMENT

COMPACTOR

Name = Soil Compactor 825H

Width compacted per pass = 4.39 m

Assume the thickness of layers of soil = 0.4 m

Total time required = 0.271 hr. = 16.26 minutes

Production of one dozer in 16.26 min. = 400m * 4.39m * 0.4m = 702.40 m3

Production of one dozer/day = (16 * 702.40) / (0.271) = 41,470.11 m3/day Total volume of soil that will be transported by all trucks to the site = 2,280,960 m3/day

The number of compactors = 2,280,960 / 41,470.11 = 55.003 56 compactors

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PROJECT COST

TOTAL COST FOR EACH EQUIPMENT=

INITIAL COST + REPAIR COST + FUEL COST + SERVICE COST + OPERATOR COST + STORAGE COST

Assume Initial cost of one shovel = 10,000,000 $

Cost of Hydraulic Shovel 6090 As an example

Hourly repair cost = ×

Yearly repair cost = Hourly repair cost ×16 × 6 × 52

Year digit = 1 (for first year) and = 2 (for second year)

Sum of years’ digits = 36 (we have 8 years)

Total working hours per year = 16*6*52 = 4,992 hr.

Life time repair cost = Repair cost factor * Price of equipment = 0.90 * 10,000,000 = 9,000,000 $

YearYear digit

Sum of years’ digits

Life time repair cost/total working hours ($/hr.)

Hourly repair cost-$-

Daily repair cost-$-

Weekly repair cost-$-

Yearly repair cost-$-

1st Year 1 36 1,802.88 50.08 801.28 4,807.68 249,999.36

2nd Year 2 36 1,802.88 100.16 1,602.56 9,615.36 499,998.72

3rd Year 3 36 1,802.88 150.24 2,403.84 14,423.04 749,998.08

4th Year 4 36 1,802.88 200.32 3,205.12 19,230.72 999,997.44

5th Year 5 36 1,802.88 250.40 4,006.40 24,038.40 1,249,996.80

6th Year 6 36 1,802.88 300.48 4,807.68 28,846.08 1,499,996.16

7th Year 7 36 1,802.88 350.56 5,608.96 33,653.76 1,749,995.52

8th Year 8 36 1,802.88 400.64 6,410.24 38,461.44 1,999,994.88

8,999,976.96Total repair cost for one shovel

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Storage Cost: Assume 350 $/month 8 years = 350*12 *8 = 33,600 $ Total for 30 Shovels = 33,600*30 = 1,008,000 $

Fuel cost: Horse power 4,500 hpSevere load conditions = 0.048 (gal/h/hp.) (Table 15) Consumption/hour = 0.048 * 4,500 = 216.0 $Assume cost of 1 gal of fuel = 0.40/gal $Fuel cost per hour = 216.0 * 0.40 = 86.40 $/hr.8 years working = 86.40 * 16 *6*52*8 = 3,450,470.4 $Total fuel cost for 30 Shovels = 3,450,470.4* 30 = 103,514,112 $

Service Cost:Service cost factor for severe conditions = 50% (Table 16)Service cost= factor * fuel cost/hour = 0.50 * 86.40 = 43.20 $/hr.8 years = 43.20 * 16 *6*52*8 = 1,725,235.20 $Total fuel cost for 30 Shovels = 1,725,235.20 * 30 = 51,757,056 $

 Operator Cost: Assume 10 $/hr.8 years operator cost = 10*16*6*52*8 = 399,360 $Total for 30 Shovels = 30 * 399,360 = 11,980,800 $

TOTAL COST OF THE PROJECT = 10,795,412,913.92 US DOLLARS

15REFERENCES

1. Course Lectutes of Construction Technology-I / auth. Prof. Dr. Tahir Çelik // Construction Economics. - Famagusta : [s.n.], 2015.

2. Dozers [Online] / auth. Komatsu Company for Dozers // komatsuamerica.com. - 2015. - http://www.komatsuamerica.com/equipment/dozers.

3. Hydraulic Shovel 6090 [Online] / auth. CAT Company for Hydraulic Shovels // cat.com. - 2015. - http://www.cat.com/en_US/products/new/equipment/hydraulic-mining-shovels/hydraulic-mining-shovels/18490663.html.

4. Soil Compactor 825H [Online] / auth. CAT Company for Compactors // cat.com. - 2015. - http://www.cat.com/en_ZA/products/new/equipment/compactors/soil-compactors/18593477.html.

5. Truck 793D [Online] / auth. CAT Company for Trucks // cat.com. - 2015. - http://www.cat.com/en_US/products/new/equipment/off-highway-trucks/mining-trucks/13894258.html.

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