Methods Of Construction: 4. Excavation Equipment

Methods of Construction 4. Excavation Equipment

Civil Engineering Fourth Year (2019 – 2020)

2 Hours / Week, 3 Units

Recommended Reading • Chapter 9, Robert Peurifoy et. al. - Construction Planning, Equipment, and Methods

Excavating Equipment The selection of the appropriate type and size of construction equipment often affects the required amount of time and effort and thus the job-site productivity of a project. It is therefore important for site managers and construction planners to be familiar with the characteristics of the major types of equipment most commonly used in construction.

Excavating Equipment An excavator is defined as a power-driven digging machine. The major types of excavators used in earthmoving operations include hydraulic excavators and the members of the cable-operated crane-shovel family (shovels, draglines, hoes and clamshells). Dozers, loaders and scrapers can also serve as excavators.

BACKHOE EXCAVATOR

Hydraulic Excavators Backhoe A backhoe is an excavator designed primarily for excavation below grade. It digs by pulling the dipper back toward the machine. This machine is also called hydraulic hoe or hydraulic excavator-backhoe.

Production Estimating Production (LCY/H) = C * S * V * B * E Where ,

C = Cycles/h (Table 3-3) S = Swing-depth factor (Table 3-4) V = heaped bucket Volume (LCY or LCM, Loose Cubic Yard or metre) B = Bucket fill factor (Table 3-2) , E = job Efficiency

Example: Find the expected production in loose cubic yards per hours of a small hydraulic excavator. Heaped bucket capacity is 0.75 CY (0.57 CM). 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 (4.3 m). Maximum depth of cut is 20 ft (6.1 m) and average swing is 90.

Solution: Cycle output = 250 cycles/60 min (Table 3-3) Swing-depth factor = 1.00 (Table 3-4),

Bucket Volume= 0.75 LCY

Bucket fill factor = 0.95,

Job efficiency= 50/60 = 0.833

Production = 250 * 1.00 * 0.75 * 0.95 * 0.833 = 148 LCY/h Or

= 250 * 1.00 * 0.57 * 0.95 * 0.833 = 113 LCM/h

Power shovel Power shovel is used to excavate & load the earth into trucks or tractor-pulled wagons or conveyor belts. Capable of excavating all classes of earth except solid rock. Can be operated on soft ground. May be mounted on rubber- tired wheels. It is useful for small jobs where considerable traveling is necessary & where the road surfaces

and ground are firm.

POWER SHOVEL

Size of a power shovel: Size of power shovel indicated by the size of dipper (bucket) in m3. Due to the swelling of soil the bank measure volume of a dipper will be less than the loose volume.

If a dipper of 2 m3 is used for excavating a soil with 25 % swelling,

therefore the bank measure volume is equal 2.0/1.25 = 1.6 m3

Power shovel are commonly available in the following sizes

(0.3-2.0) m3

Output of power shovels The out put of power shovel is affected by the following factors 1- Class of material.

2- Depth of cut. 3- Angle of swing 4- Job condition.

5- Management conditions.

‫نوع المادة‬

‫ارتفاع القطع‬/‫عمق‬ ‫زاوية الدوران‬ ‫ظروف العمل‬

‫ادارة المشروع‬

6- Size of hauling units.

‫حجم المكائن الناقلة‬

7- Skill of the operators.

‫مهارة المشغلين‬

8- Physical condition of the shovel. 9- handling of oversize material . 10- cleanup of loading area. 11- hauling -unit exchange

‫حالة وهيئة المجرفة‬ ‫ازالة كتل المادة الكبيرة‬ ‫تنظيف مكان التحميل‬ ‫تبديل وحدات النقل او السحب‬

Out put of a power shovel 60 Out put (m3/hr) = cycle time * Time factor * Capacity Cycle time= Load + Swing loaded + Dump + Swing empty

Actual depth of cut Optimum depth % = Optimum depth of cut * 100

Example – 1 Determine the out put of power shovel if; Size of shovel (dipper) =1.6m3 Type of soil is good common earth, Depth of cut (actual) = 3.6m Average angle of swing of 60˚. Solution:

from Table 8.2 Optimum out put =229 Optimum depth = 3.1

The percent of optimum depth = (3.6/3.1) * 100 = 116% From table 8.3 the depth swing factor is 1.13 by interpolation The probable out put = 229 * 1.13 = 259 m3/hr.

Example -2: Determine the probable out put in m3 for a power shovel which have the following properties Size of the shovel (Dipper)= 0.8 m3, Type of soil is Stiff & hard soil, Depth of cut (Actual) = 2.25m, Average angle of swing is 75˚ Job condition is fair & Management is good. Efficiency = 50/60

Solution: Optimum out put = 111m3/hr and Optimum depth is = 2.7m Percent of optimum depth = 2.25/2.7 * 100 = 83.3% Depth- swing factor = 1.05 (by interpolation) Job management factor = 0.69

Probable out put = 111*1.05*0.69 = 80.4 m3/hr If consider the working hours (Efficiency =50/60) The probable out put = 80.4*(50/60) = 67 m3

Example-3: Determine the number of power shovels to excavate 400,000 m3 of sand & gravel material at 4 months. If : Size of power shovel = 1.2 m3 Depth of cut ( actual ) = 3 m

working minutes /hour = 55 min. Angle of swing = 110˚ working hours /one day = 8hr.

Percent of stopping = 15% Job conditions is Good. Management conditions is Fair

Solution Time factor = 55/60 = 0.92 Job-management factor = 0.71

Optimum depth factor = 2.1 m and Optimum out put = 206 m3/hr Optimum depth % = 3/2.1 * 100 = 142.86 % Depth –swing factor = 0.84 Actual out put = 206*0.92*0.71*0.84= 111.82 m3/hr = 111.82* 8 = 894.56 m3/day Actual working day = ( Duration * ( 1-stoping ) ) = (4*30)*(1-0.15 ) = 102 days

Out put /4month = 894.56*102 = 91245.12 m3/4month No. of power shovel = Quantity of soil/output of shovel = 400000/91245.12 = 4.4 shovels

Apply 4 shovel for 102 days & 1 shovel for 41 days (0.4*102 = 41)

LOADERS

• Loaders Loaders are used in construction work to handle and transport bulk material, such as earth and rock, to load trucks and to excavate earth .It is designed to excavate at or above wheel/track level. Buckets: The most common buckets are:1. The one-piece conventional type (general purpose) 2. The hinged-jaw (multipurpose) 3. The rock bucket (heavy duty production having a Vshaped cutting edge)

• Loader Buckets

Operating Loads:To position the bucket to dump, a loader must maneuver and travel with load, so a check must be made of payload weight once the bucket volumetric load is determined. A wheel loader is limited to an operating load by weight that is less than 50% of rated full-turn static tipping load considering the combined weight of the bucket and load measured from the center of gravity of the extended bucket at its maximum reach.

Production rate for Wheel Loaders:The critical factors to be considered in choosing a loader are:1. The type of material 2. The volume of material to be handled

The production rate for a wheel loader will depend on: 1. Fixed time required to load the bucket, maneuver with four reversals of direction, and dump the load

2. Time required to travel from loading to dumping position 3. Time required to return to the loading position 4. The actual volume of material handled each trip

When travel distance is more than minimum it will be necessary to add a travel time to the fixed cycle time. For travel distances of less than 100 feet, a wheel loader should be able to travel with a loaded bucket at about 80% of its maximum speed in low gear and return empty at about 60% of its maximum speed in second gear. In the case of distances over 100 ft, return travel distance should be at about 80% of its maximum speed in second gear. If the haul surface is not well maintained or is rough these speeds should be reduced accordingly.

•

Loader Loading Cycle

DRAGLINE EXCAVATORS

DRAGLINES The dragline is a versatile machine capable of wide range of operations. It can handle materials that range from soft to medium hard.

The greatest advantage of a dragline over other machines is its long reach

for digging and dumping.

Example:

DOZERS

Clamshell

CHAIN TRENCHER

CHAIN TRENCHER

Excavator and Loader

Grader

Scraper