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CE403 Construction Methodology

CE403 Construction Methodology. Excavators Shovels, Draglines, Hoes, and Clamshells. http://www.youtube.com/watch?v=DL8Oymr5vqE. Back Hoe Production Estimating. Production, LCY/h = C x S x V x B x E C = Cycles/h (Table 3-3) S = Swing-Depth Factor (Table 3-4)

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CE403 Construction Methodology

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  1. CE403Construction Methodology Excavators Shovels, Draglines, Hoes, and Clamshells

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

  3. Back Hoe Production Estimating • Production, LCY/h = C x S x V x B x E C = Cycles/h (Table 3-3) S = Swing-Depth Factor (Table 3-4) V = Heaped Volume , LCY B = Bucket Fill Factor (Table 3-2) E = Job Efficiency

  4. Back Hoe Production Estimating • Cycle = Load Bucket + Swing with Load + Dump load + Return Swing • Swing-depth factor – accounts for angle of swing and depth of cut

  5. Standard Cycles per Hour for Hydraulic Backhoes

  6. Swing-Depth Factor

  7. Bucket Fill Factors for Excavators

  8. Back Hoe Production Estimating • Efficiency Dependent on: • Management Conditions • Skill, Training & Motivation of Workers • Selection, Operation & Maintenance of Equipment • Planning, Job Layout, Supervision & Coordination of Work • Job Conditions • Topography & Work Dimensions • Surface & Weather Conditions • Specification Requirements for Work Methods or Work Sequence Required

  9. Back Hoe Production Estimating

  10. Back Hoe Production Estimating • 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…

  11. Hoe Production Example Problem 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.

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

  13. Example 3-4 Solution Cyclic 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/h

  14. Swing-Depth Factor (Tab 3-4)

  15. Example 3-4 Solution Cyclic 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/h

  16. Job Management • Major Factor Controlling Hydraulic Excavator • Maximum depth • Working radius • Dumping Height • Density of Material

  17. Shovels

  18. Crowding & Breakout Forces

  19. Shovel ~ Production Estimating Production, LCY/h = C x S x V x B x E C = Cycles/hour (Table 3-6) S = Swing Factor (Table 3-6) V = Heaped Bucket Volume, LCY B = Bucket Fill Factor (Table 3-2) E = Job Efficiency

  20. Standard Cycles per Hour for Hydraulic Shovels

  21. Shovel Production Example Problem Find 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.

  22. Standard Cycles per Hour for Hydraulic Shovels < 5 yd)

  23. Shovel Production Example Solution Standard Cycles = 150/60 min (Table 3-6) Swing factor = 1.05 (Table 3-6) Bucket Volume = 3.0 LCY Bucket Fill Factor = 1.0 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/h

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

  25. Dragline

  26. Optimal Digging Area

  27. Expected Production = Ideal Output x Swing-Depth Factor x Efficiency Dragline Production

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

  29. Swing-Depth Factor (Tab 3-9)

  30. 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%

  31. Ideal Dragline Output, in BCY/hr

  32. 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

  33. 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

  34. Swing-Depth Factor

  35. 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

  36. CLAMSHELL

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