05 excavators 3

CE403Construction Methodology

Excavators

Shovels, Draglines, Hoes, and Clamshells

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http://www.youtube.com/watch?v=DL8Oymr5vqE

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Back Hoe Production Estimating

Production, LCY/h = C C x SS x VV x BB x EE

CC = Cycles/h (Table 3-3)

SS = Swing-Depth Factor (Table 3-4)

VV = Heaped Volume , LCY

BB = Bucket Fill Factor (Table 3-2)

EE = Job Efficiency

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Cycle = Load Bucket + Swing with Load + Dump load + Return Swing

Swing-depth factor – accounts for angle of swing and depth of cut

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Standard Cycles per Hour for Hydraulic Backhoes

Small Medium Large1.0 YD 1¼ to 2¼ YD 2½ YD

sand, gravel,

loam

common earth,

soft clay

tough clay,

rock

120

Hard 110 160 130 100

Average 135 200 160

Type of MaterialWheel Tractor

Excavator Machine Size

Soft 170 250 200 150

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Swing-Depth Factor

Depth of Cut

(% of Max)

45º 60º 75º 90º 120º 180º

30 1.33 1.26 1.21 1.15 1.08 0.9550 1.28 1.21 1.16 1.10 1.03 0.9170 1.16 1.10 1.05 1.00 0.94 0.8390 1.04 1.00 0.95 0.90 0.85 0.75

Angle of Swing, deg

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Bucket Fill Factors for Excavators

Material Bucket Fill Factor Common Earth, Loam 0.80-1.10 Sand & Gravel 0.90-1.00 Hard Clay 0.65-0.95 Wet Clay 0.50-0.90 Rock, Well-Blasted 0.70-0.90 Rock, Poorly-Blasted 0.40-0.70

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Efficiency Dependent on:– Management ConditionsManagement Conditions

Skill, Training & Motivation of Workers Selection, Operation & Maintenance of Equipment Planning, Job Layout, Supervision & Coordination of Work

– Job ConditionsJob Conditions Topography & Work Dimensions Surface & Weather Conditions Specification Requirements for Work Methods or Work Sequence

Required

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Management Conditions Job-Site Conditions Excellent Good Fair

Excellent 0.84 0.81 0.76 Good 0.78 0.75 0.71

Fair 0.72 0.69 0.65 Poor 0.63 0.61 0.57

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Can also estimate efficiency through number of effective working minutes per hour.

Eg., 50-min/h – actual work is done 50 minutes per hour…the other ten minutes spent on breaks, smoke break, bath room, thinking…

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Hoe Production Example

ProblemProblem

Find the expected production in loose cubic yards per hour of a small hydraulic excavator.

Heaped bucket capacity is 3/4 CY.

The material is sand and gravel with a bucket fill factor of 0.95.

Job efficiency is 50 min/h.

Average depth of cut is 14 ft.

Maximum depth of cut is 20 ft.

Average swing is 90.

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Standard Cycles per Hour for Hydraulic Backhoes (Tab 3-3)

Small Medium Large1.0 YD 1¼ to 2¼ YD 2½ YD

sand, gravel,

loam

common earth,

soft clay

tough clay,

rock

120

Hard 110 160 130 100

Average 135 200 160

Type of MaterialWheel Tractor

Excavator Machine Size

Soft 170 250 200 150

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Example 3-4

SolutionSolutionCyclic Output = 250 cycles/60min (Table 3-3)

Swing-Depth Factor = 1.00 (Table 3-4)

Bucket Fill Factor = 0.95

Job Efficiency = 50 /60 = 0.833

Production

= 250 cycles x 1.00 (swing-depth) x 0.75 CY x 0.95 (bucket fill factor) x 0.833 (job eff.)= 148 LCY/h148 LCY/h

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Swing-Depth Factor (Tab 3-4)

Depth of Cut

(% of Max)

45º 60º 75º 90º 120º 180º

30 1.33 1.26 1.21 1.15 1.08 0.9550 1.28 1.21 1.16 1.10 1.03 0.9170 1.16 1.10 1.05 1.00 0.94 0.8390 1.04 1.00 0.95 0.90 0.85 0.75

Angle of Swing, deg

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Example 3-4

SolutionSolutionCyclic Output = 250 cycles/60min (Table 3-3)

Swing-Depth Factor = 1.00 (Table 3-4)


Job Efficiency = 50 /60 = 0.833

Production

= 250 cycles x 1.00 (swing-depth) x 0.75 CY x 0.95 (bucket fill factor) x 0.833 (job eff.)= 148 LCY/h148 LCY/h

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Job Management

Major Factor Controlling Hydraulic Excavator– Maximum depth– Working radius– Dumping Height– Density of Material

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Shovels

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Crowding & Breakout Forces

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Shovel ~ Production Estimating

ProductionProduction, LCY/h = CC x SS x VV x B B x EE

CC = Cycles/hour (Table 3-6)

SS = Swing Factor (Table 3-6)

VV = Heaped Bucket Volume, LCY

BB = Bucket Fill Factor (Table 3-2)

EE = Job Efficiency

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Standard Cycles per Hour for Hydraulic Shovels

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Shovel Production Example

ProblemProblemFind the expected production in LCY per hour of 3 CY

hydraulic shovel equipped with a front-dump bucket.

The material is common earth with a bucket fill factor of 1.0.

The average angle of swing is 75 degrees.

The job efficiency is 0.80.

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Standard Cycles per Hour for Hydraulic Shovels

< 5 yd)

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Shovel Production Example

SolutionSolution Standard Cycles = 150/60 min (Table 3-6)

Swing factor = 1.05 (Table 3-6)

Bucket Volume = 3.0 LCY


Job Efficiency = 0.80

QProduction = 150 cycles x 1.05 (swing factor) x 3.0 cy x 1.0 (bucket fill factor) x 0.80 (eff) = 378 LCY/h378 LCY/h

Draglines

Longest reach for digging and dumping of any member of the crane-shovel family.

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Dragline

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Optimal Digging Area

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Dragline Production

Expected Production

= Ideal Output

x Swing-Depth Factor

x Efficiency

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Ideal Dragline Output, in BCY/hr (Tab 3-7)

¾ 1 1¼ 1½ 1¾ 2 2½ 3 3½ 4 5Light moist clay or loam 130 160 195 220 245 265 305 350 390 465 540Sand and gravel 125 155 185 20 235 255 295 340 380 455 530Common earth 105 135 165 190 210 230 265 305 340 375 445Tough clay 90 110 135 160 180 195 230 270 305 340 410Wet, sticky clay 55 75 95 110 130 145 175 210 240 270 330

Type of MaterialBucket Size, CY

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Swing-Depth Factor (Tab 3-9)

Depth of Cut% of

Optimum30º 45º 60º 75º 90º 120º 150º 180º

20 1.06 0.99 0.94 0.90 0.87 0.81 0.75 0.7040 1.17 1.08 1.02 0.97 0.93 0.85 0.78 0.7260 1.25 1.13 1.06 1.01 0.97 0.88 0.78 0.7280 1.29 1.17 1.09 1.04 0.99 0.90 0.82 0.76100 1.32 1.19 1.11 1.05 1.00 0.91 0.83 0.77120 1.29 1.17 1.09 1.03 0.98 0.90 0.82 0.77140 1.25 1.14 1.06 1.00 0.96 0.88 0.81 0.75180 1.15 1.05 0.98 0.94 0.90 0.82 0.76 0.71200 1.10 1.09 0.94 0.90 0.87 0.79 0.73 0.69

Angle of Swing, deg

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Dragline Example

Determine the expected dragline production in LCY per hour based on the following information:– Dragline Size: 2 cyd– Swing Angle: 120 degrees– Average Depth of Cut: 7.9 ft– Material: Common Earth– Job Efficiency: 50min/h– Soil Swell = 25%

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Ideal Dragline Output, in BCY/hr

¾ 1 1¼ 1½ 1¾ 2 2½ 3 3½ 4 5Light moist clay or loam 130 160 195 220 245 265 305 350 390 465 540Sand and gravel 125 155 185 20 235 255 295 340 380 455 530Common earth 105 135 165 190 210 230 265 305 340 375 445Tough clay 90 110 135 160 180 195 230 270 305 340 410Wet, sticky clay 55 75 95 110 130 145 175 210 240 270 330

Type of MaterialBucket Size, CY

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Dragline Example

Solution: – Ideal Output: 230 BCY/h– Optimum Depth of Cut: 9.9 ft– Actual Depth/Optimum Depth: 7.9/9.9 x 100 = 80%– Swing Depth Factor: 0.90– Efficiency factor: 50/60 = 0.833– Volume Change Factor = 1+0.25 = 1.25– Estimated Production = 230 x 0.90 x 0.833 x 1.25 =

216 LCY/h

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Dragline Example

Solution: – Ideal Output: 230 BCY/h– Optimum Depth of Cut: 9.9 ft (Tab 3-8)– Actual Depth/Optimum Depth: 7.9/9.9 x 100 = 80%– Swing Depth Factor: 0.90– Efficiency factor: 50/60 = 0.833– Volume Change Factor = 1+0.25 = 1.25– Estimated Production = 230 x 0.90 x 0.833 x 1.25 =

216 LCY/h

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Swing-Depth Factor

Depth of Cut% of

Optimum30º 45º 60º 75º 90º 120º 150º 180º

20 1.06 0.99 0.94 0.90 0.87 0.81 0.75 0.7040 1.17 1.08 1.02 0.97 0.93 0.85 0.78 0.7260 1.25 1.13 1.06 1.01 0.97 0.88 0.78 0.7280 1.29 1.17 1.09 1.04 0.99 0.90 0.82 0.76100 1.32 1.19 1.11 1.05 1.00 0.91 0.83 0.77120 1.29 1.17 1.09 1.03 0.98 0.90 0.82 0.77140 1.25 1.14 1.06 1.00 0.96 0.88 0.81 0.75180 1.15 1.05 0.98 0.94 0.90 0.82 0.76 0.71200 1.10 1.09 0.94 0.90 0.87 0.79 0.73 0.69

Angle of Swing, deg

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Dragline Example

Solution: – Ideal Output: 230 BCY/h– Optimum Depth of Cut: 9.9 ft– Actual Depth/Optimum Depth: 7.9/9.9 x 100 = 80%– Swing Depth Factor: 0.90– Efficiency factor: 50/60 = 0.833– Volume Change Factor = 1+0.25 = 1.25– Estimated Production = 230 x 0.90 x 0.833 x 1.25 =

216 LCY/h

CLAMSHELL

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CLAMSHELL

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Clamshell Bucket

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Clamshell Production

Production = C x V x B x E

C – Cycles per hour

V – Bucket Capacity

B – Bucket Fill Factor

E – Job Efficiency Factor

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Clamshell Production Example

ProblemProblem

Estimate production in LCY per hour for a medium-weight clamshell excavating loose earth.

Heaped bucket capacity is 1.0 CY.

The soil is common earth with a bucket fill factor of 0.95.

Estimated cycle time is 40s.

Job efficiency is estimated at 50 min/h.

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Clamshell Production Example

SolutionSolution

Production = C x V x B x E

Production = (3,600(sec/h)/40 sec) x 1.0 CY x 0.95 (bucket fill factor) x 50/60 (job eff.)

= 71 LCY/h

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Trenching Machines

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Trenchless Technologies

Deterioration of: Water Lines Sewer Lines Gas Lines

Urban Areas Inaccessible Areas:

Beneath Buildings Under Roadways

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Trenchless Technology: Cast In Place Pipe

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Trenchless Technology - Cast In Place Pipe

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Trenchless Technology - Cast In Place Pipe

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Pipe BurstingA process that…

Breaks an Existing Pipe

Expands Broken Shards into Surrounding Soil

Pulls in the New Carrier Line Simultaneously

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Pneumatic Pipe Bursting

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Pneumatic Components

Bursting Tool

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Upsizing Considerations

0-25% 25-50% 50-125%

Class A Routine and

generally considered favorable

Class B Challenging to

moderately difficult

Class C Very challenging

to extremely difficult

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Trenchless Technology:Pipe Bursting

CE403Construction Methodology

End of Lecture

05 excavators 3

Documents

factor x

swingdepth factor table

swingdepth x

load bucket swing

bucket sides

hoe production estimatingcycle

hoe production estimatingcan

heaped bucket capacity