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. 1610 (3549 ) 19 84.980 (187360 ) 513 580 . *

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*( 1 N = 0.1 kg approx)1 newton per square metre = 1 pascal (Pa)

1 kilo pascal = 1 000 Pa

1 mega pascal = 1 000 000 Pa

1 bar = 100 000 Pa

1 bar = 1 kg / cm2 (approx)UNITS OF PRESSURE*

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Non Hydraulic Excavator*

Hydraulic Excavator*

Hydraulic Operation*

Hydraulic Operation*

Hydraulic Operation*

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Hydraulic Operation*

Hydraulic Operation

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Hydraulic Operation*

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*1. Straight and Angling Blade Capacity:

2. Semi-U and U-Blade Capacity:

*Example 1 : find Straight and Angling Blade Capacity if Blade width equal (2.9 m) and Effective blade height equal (1.55 m) ?

SolutionW = 2.9 m ; H' = 1.55 m

*Example 2 : find Semi-U Blade Capacity if Blade width equal (6.321 m) , Effective blade height equal (2.179 m) and angle blade equal (30) ? W = 6.321 m ; H' = 2.179 m ; = 30 ; Z = 0.4 W

Solution

*Production, LCY/h = C x S x V x B x E

C = Cycles/h (Table 2)S = Swing-Depth Factor (Table 3)V = Heaped Volume , LCYB = Bucket Fill Factor (Table 1)E = Job Efficiency

** ProblemFind the expected production in cubic meters per hour of a hydraulic excavator. Heaped bucket capacity is 3/4 m3. The material is sand and gravel Job efficiency is 50 min/h. Average depth of cut is 4 m. Maximum depth of cut is 6 m. Average swing is 90. = 0.9144 = 1.0936133 3.281

*Production, LCY/h = C x S x V x B x EC = Cycles/h (Table 2)S = Swing-Depth Factor (Table 3)V = Heaped Volume , m3B = Bucket Fill Factor (Table 1)E = Job EfficiencySolution Job efficiency is 50 min/h = 50 / 60 = 0.83

*Bucket Fill Factors for ExcavatorsTab 1Solution (B)

**Standard Cycles per Hour for Hydraulic Backhoes (Tab 2)

Solution(C)

*Swing-Depth Factor (Tab 3)

Solution (S)

*Cyclic Output = 250 cycles/60min (Table 2)Swing-Depth Factor = 1.1 (Table 3)Bucket Fill Factor = 0.95 Table 1Job Efficiency = 50 /60 = 0.83V = Heaped Volume = 0.75 m3Production = 250 cycles x 1.1 (swing-depth) x 0.75 m3 x 0.95 (bucket fill factor) x 0.83 (job eff.)

= 162.6 m3 /hrSolutionProduction, LCY/h = C x S x V x B x EC = Cycles/h (Table 2)S = Swing-Depth Factor (Table 3)V = Heaped Volume , m3B = Bucket Fill Factor (Table 1)E = Job Efficiency

**Shovel Production ExampleProblemFind the expected production in m3 per hour of 1.5 m3 hydraulic shovel equipped with a front-dump bucket. The material is common earth averageThe average angle of swing is 75 degrees.The job efficiency is 45 min / hr.

*Standard Cycles per Hour for Hydraulic Scavator(Table 4)(Table 5)

**Standard Cycles = 150/60 min (Table 4)Swing factor = 1.05 (Table 5 ) Bucket Volume = 1.5 m3Bucket Fill Factor = 0.9 (Table 1 ) Job Efficiency = 45 / 60 = o.75Q Production = 150 cycles x 1.05 (swing factor) x 1.5 m3 x 0.9 (bucket fill factor) x 0.75

= 159.5 m3 /hrSolution

*ExampleWhat is the production capacity to Bulldoze has the following specifications and works under the following conditions:Clay soilTransport distance (M) : 50 meters on the flat land

Blade dimension (D): 3 meters wide and 1 meter high

Coefficient time (C) : 50 minutes in an hour Speed Push (SP) : 3 km / h Speed Back (SB) : 7 km / h

constant time (HT) : a half minutes for each session

*SolutionFrom the table (1) Correction factor Bulge equal ( 0.83)

NO.Soil typeCorrection factor BulgeMass per cubic meter (kg)1Dry clay0.8514002Wet clay0.818003Clay0.8316004Sandy clay mixed with gravel0.918505Dry sandy0.8919506Wet sand0.882150

**V = The size of the soil at full load in cubic meters,W = Width Blade Bulldozer,X = (1.5 H) for Sand soil and (1.67 H) for Clay soil,H= Height Blade Bulldozer,Y = Inclination angle of the soil , equal 30 degree for the sand soil and 33 degree of the clay soil.

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