Hydraulic Excavators (Classnote-2)

Hydraulic Excavators (II)

By

Er. Ashok Shrestha (DoR)

IOE-BEM

Construction Equipment(Elective) August 2010

Cycle time • The sum of time required to load bucket, swing

loaded, dump and swing empty.

• Typical cycle element times under average

conditions, for 2 to 4 cum shovels will be

– Load bucket 7-9 sec. (depend on material type)

– Swing with load 4-6 sec (depend on machine size)

– Dump load 2-4 sec (depend on dumping target)

– Return swing 4-5 sec (depend on machine size)

16 August 2010 Slide No. 2IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha

Factor affecting production

• Actual production of a shovel is affected by the following factors:

– Class of material

– Height of cut

– Angle of swing

– Size of hauling units

– Operator skill

– Physical condition of the shovel

Production efficiency ranges from 30 to 45 min per hour


Effect of height of cut and Angle of swing

• The effect of height of cut and Angle of Swing on

Shovel production published by PCSA from field

study can be used:

The percent of optimum height of cut, in the table, is

obtained by dividing the actual height of cut by the

optimum height for the given material and bucket, and

then multiplying the result by 100.


Optimum height of cut• The optimum height of cut ranges from 30 to

50% of the maximum digging height.

– 30 % for a easy to load materials (i.e. load sand,

gravel etc.)

– 40% for common earth

– 50% for poorly blasted rock, or sticky clay

• The ideal production of shovel is based on

operating at a 900 swing and optimum height of

cut.

• The ideal production should be multiplied by the

proper correction factor in order to correct the

production for any given height and swing angle.


Excavator Production• Steps for estimating production:

1. Obtain the heaped bucket load volume (lcm) from

the manufacturer performance data.

2. Apply a bucket fill factor based on type material

being excavated.

3. Estimate a peak cycle time:

(Load bucket+ Swing with load+ Dump load+ Return swing)

4. Obtain the factor for angle of swing and height of

cut from the table (% of optimum depth vs angle of swing).

5. Apply a efficiency factor (usually 30 – 45 min per 60 min)

6. Conform the production units to desired volume or

weight (lcm to bcm) (lcm= bcm*(1+swell factor)


Typical cycle element times under average conditions, for

2 to 4 cum shovels

Load bucket 7-9 sec. (Based on materials)

Swing with load 4-6 sec (Based on size of m/c)Dump load 2-4 sec (Base on hauling unit)

Return swing 4-5 sec (Based on size of m/c)

Obtain heaped cubic capacity of bucket from manufacturer’s specification for given model

Select the suitable Fill factor for type of material to be excavated.

Estimate the pick cycle time base on machine size, materials type and dumping condition.-Load time, for easy loading material take lower value (7sec) and higher value for difficult (9 sec) -Swing time, take lower value(4sec)for smaller m/c and for higher(6sec) value for bigger size m/c.--Dump time, for dumping in a hauling unit take higher value and lower value for free dumping.

Obtain AS:D from the table.maximum depth of cut from manufacturer’s specification and multiply it by a factor within range of 0.3 – 0.5. Take lower value (0.3) for easy to load material and higher value (0.5) for very difficult material

100cut ofheight Optimum

cut ofheight Averagedepth optimum of % ×

=

material) avergecommon (for height digging maximum of 40%

etc)rock blasteddifficult (for height degging maximum of 50%

material) load easy to(for height digging maximum of 30%cut ofdepth Optimum

=

=

=


+

×

×

×××=

S.F.1

1

60

):(3600(bcm/hr) P

E

t

DASFQ

Where;

P (lcm/hr) = Production in loose cubic meter (volume) per hour

P (bcm/hr) = Production in bank cubic meter (volume) per hour

P (ton/hr) = Production in tons (weight) per hour

Q = Heaped bucket capacity (lcm)

F = Bucket fill factor

AS:D = Angle of swing and depth (height) of cut correction factor

t = Cycle time in seconds

E = Efficiency minutes per hour (take 30-45 if not given)

S.F. = Swell Factor

Luw = Loose unit weight (N)

7. Compute the production rate, using following formula.

×

×××=

60

):(3600(lcm/hr) P

E

t

DASFQ


Example-1• A 3.8 cu.m. shovel having a maximum digging height of 10.4m is being

used to load poorly blasted rock. The face being worked is 3.7m high and the haul units can be positioned so that the swing angle is only 600. What is the adjusted ideal production if the ideal cycle time is 21 sec.

×

×××=

60

):(3600(lcm/hr) P

E

t

DASFQ

Bucket size (Q)= 3.8 m3 (Given)Bucket fill factor = 0.9 (taken from the table, for poorly blasted 85-100%) Ideal Cycle time (t) = 21 sec. (Given)

Optimum height = 0.5 x 10.4 = 5.2m (Taken highest %, for poorly blasted rock) (30–50%)

%15.711002.5

7.3100

height Optimum

height) Workingheight optimum of % =×=×

=

Angle of Swing = 600

Angle of Swing and depth (AS:D) =1.08 (by interpolation)

(from the table 1.03+(71.15-60)*(1.12-1.03)/(80-60)

)/( 47560

45

21

08.19.08.33600(lcm/hr) P hrlcm=

×

×××=

Assuming Efficiency factor,

E= 45/60


Example-2• A 2.3 cu.m. shovel having a maximum digging height of 9.1 m, will be

used on a highway project to excavate well-blasted rock. The average face height is expected to be 6.7 m. Most of the cut will require an average 1200 swing of the shovel in order to load the haul unit. Determine the estimated production in cubic meter bank measure. Take efficiency 30 min in 60 minute.

Bucket size (Q)= 2.3 m3 (Given)

Bucket fill factor (F)= 1 (taken from the table, for well-blasted 100-110%)

Ideal Cycle time (t) = (Load + Swing loaded + Dump + Swing empty)

= 9 + 4+ 4 + 4 = 21 secTypical cycle element times under average conditions, for 2 to 4 cum

shovels:Load bucket 7-9 sec.

Swing with load 4-6 secDump load 2-4 sec

Return swing 4-5 sec

%14710055.4

7.6100

height Optimum

height) Workingheight optimum of % =×=×

=

Assuming:Load = 9 sec (Taken maximum value, as material is rock difficult to load)Swing loaded = 4 sec (Taken smaller valve, as being smaller sized m/c)Dump = 4 sec (Taken maximum value, as it is to be loaded into haul units)Swing empty = 4 sec (Taken smaller valve, as being smaller sized m/c)

Optimum height = 0.5 x 10.4 = 4.55m (Taken highest %, for poorly blasted rock) (30–50%)


Example-2 (/contd…)

+

×

×

×××=

swell

E

t

DASFQ

1

1

60

):(3600(bcm/hr) P

Angle of Swing = 1200

Angle of Swing and depth (AS:D)= 0.79 (by interpolation)(from the table 0.81- [{(0.81-0.75)/(160-140)}*(147-140)]

3.976.01

1

60

30

21

)79.0(13.23600(bcm/hr) P =

+

×

×

×××=


Efficiency Factor (E) = 30/60 (Given)

% Swell = 60% (From the table for well blasted rock)

Hydraulic Hoe: Production Estimating

• The same elements that affect shovel production

are applicable to hoe excavation operation.

• Hoe cycle times are approximately 20% longer

than similar size shovel and work.

• The optimum depth of cut for hoe is usually in the

range of 30 to 60%.

• Standard data for “Cycle time” based on bucket

size and average conditions (30-600 swing angle,

hauling unit at same level etc. is available).

• No standard data and factors based on angle of

swing and depth of cut is available.



• Step-1: Bucket size (lcm) (From the

manufacturer specification for the size of bucket to be used. Many different size buckets will fit the same

machine. Interested in heaped capacity).

• Step-2: Fill Factor: (From the table for

corresponding type of material. Heaped capacity is base on 1:1 material angle of repose. It must be adjusted

based on the characteristics of material being handled).

[Bucket volumetric capacity (lcm) = Heaped capacity *Fill Factor]



• Step-3: Cycle time (sec) (Load + Swing load + Dump +

Swing empty).

Typical excavation cycle times based on machine (bucket) size

Depth of cut: 40 to 60%

Swing angle = 30 – 600

Loading haul units on the same level

The cycle times

must be increased

when loads are

dumped into a

smaller haul units.

Small machine

swing faster than

large ones.


Hydraulic Hoe: Production Estimating• Step-4: Depth of cut (Obtain maximum dig depth from manufacturer’s

data and check for optimum depth of cut within the range of 30% to 60%.)



• Step-5: Efficiency Factor: – Bunching (In actual operation cycle time is never constant. When loading haul

unit they will sometime bunch. The effect of bunching is a function of the no. of

haul units.

– Operator efficiency: (Skill of operator)

– Equipment availability (Haul units availability ‘x’% of the time)


Machine’s working range based on size of machine

(bucket) fitted with standard items (Boom, Arm etc.)

+

×

×

××=

S.F.1

1

60

3600(bcm/hr) P

E

t

FQ

Where;


P (bcm/hr) = Production in bank cubic meter (volume) per hour

P (ton/hr) = Production in tons (weight) per hour

Q = Heaped bucket capacity (lcm)

F = Bucket fill factor

t = Cycle time in seconds

E = Efficiency minutes per hour (take 30-45 if not given)

S.F. = Swell Factor

Luw = Loose unit weight (N)

Step-6: Compute production rate, using following formula.

×

××=

60

3600(lcm/hr) P

E

t

FQ


Example-3A crawler hoe having a 2.8 cu.m bucket is being considered for use on a project to excavate dry clay from a borrow pit. The clay will be loaded in trucks having a loading height of 3m. Soil-boring information indicates that below, 2.5 m, the material changes to an unacceptable silt material. What is the estimated production of the hoe in cubic meter bank measure, if the efficiency factor is equal to a 50-min hour.?

Step-1: Size of Bucket (Q)= 2.8 cu.m

Step-2: Bucket fill factor (F)= 85% (taken average of 80-90 from the table,

. for hard clay)

Step-3: Cycle times (t) = 22 sec (from the table, for nearest bucket size 3 cum)

Step-4: Optimum depth of cut to be within 30% to 60%

From the table maximum depth of cut 7 – 8.2 m

Depth of cut = 2.5 m

%4.30100*8.2

2.5=

%7.35100*7

2.5=

Checking for optimum depth of

cut range 30% to 60%.


Example-3 /….contdStep-5: Efficiency factor (E)= 50 min per hour (given)

×

××=

60

3600(lcm/hr) P

E

t

FQ

Step-6: Production rate Calculation

(lcm/hr) 5.32460

50

22

85.08.23600(lcm/hr) P =

×

××=

Swell factor = 35% for type of materials from the table

37.24035.01

5.324

factor Swell1

(lcm/hr) P(bcm/hr) P =

+

=

+

=


PRODUCTIVITY OF HYDRAULIC EXCAVATOR:

(Quick Method)

The hourly production of hydraulic excavator can also be calculated as :-

Where;


Q = Production per cycle (cu.m)

= (q*K)

q = bucket heaped capacity (cu.m)

K = Bucket factor

E= Job Efficiency factor

t = Standard Cycle time in seconds

F = Time factor

Ct = Cycle time in seconds =(t*F)

×

×××=

××=

)(

)(36003600(lcm/hr) P

Ft

Ekq

Ct

EQ

Material Bucket factor (K)

Moist loam or sandy clay 1.0 - 1.1

Common soil 0.9 - 1.0

Sand and gravel 0.85 – 0.95

Hard tough clay 0.8 - 0.9

Rock –well blasted 0.6 - 0.75

Rock – poorly blasted 0.4 - 0.5


Bucket Factor (k)

Standard Cycle time (t)= Standard Cycle time (t)= Standard Cycle time (t)= Standard Cycle time (t)= excavating time + swing time excavating time + swing time excavating time + swing time excavating time + swing time loaded+ dumping time +swing time empty loaded+ dumping time +swing time empty loaded+ dumping time +swing time empty loaded+ dumping time +swing time empty

ORORORORActual cycle time =Actual cycle time =Actual cycle time =Actual cycle time = standard cycle time * standard cycle time * standard cycle time * standard cycle time * timetimetimetime factorfactorfactorfactor

BUCKET

CAPACITY

Swing Angle /Time (SecSwing Angle /Time (SecSwing Angle /Time (SecSwing Angle /Time (Sec)

40 - 90 90 -180

0.25 13 -15 15 -17

0.4 13 -15 15 -17

0.45 14 -16 16 -18

0.7 16-18 18 – 21

0.9 18 - 20 20 – 23

1.2 20 - 22 22 - 25

Digging Digging Digging Digging ConditionsConditionsConditionsConditions

Dumping ConditionsDumping ConditionsDumping ConditionsDumping Conditions

EASYNORMAL (Average)

RATHER DIFFICULT

DIFFICULT

BELOW 40 %BELOW 40 %BELOW 40 %BELOW 40 % 0.7 0.9 1.1 1.4

40 % 40 % 40 % 40 % ---- 75 %75 %75 %75 % 0.8 1.0 1.3 1.6

OVER 75 %OVER 75 %OVER 75 %OVER 75 % 0.9 1.1 1.5 1.8

Digging condition= (Digging depth/ Max. depth of cut)*100Easy = Dump onto spoil pileNormal = Large dump targetRather difficult = Small dump targetDifficult = Small dump target requiring maximum reach.


Time Factor (F)Standard cycle time based on bucket

Capacity

Cycle Time (t)

Example-4A contractor has a project to construct a large office building with an underground parking

garage. He has decide to use a Hydraulic excavator to excavate for the parking garage

and load the excavated material into a dump trucks. The maximum digging depth of the

excavator is 6m and it is equipped with a 1.2 cum bucket size. The material to be

excavated is a tough clay at average depth of cut 3m and job condition are considered to

be average. Angle of swing 600 and work an average of 50 min per hour. What is the

estimated productivity in bank Cum per hour if the swell of the excavated material is 35%.

Bucket size (q) = 1.2 cum

Bucket factor (k) = 0.85 (taken average value of 0.8 and 0.9)

Digging condition = (3/6)*100 = 50%

Time Factor (F) = 1.3 (for digging and rather difficult dumping condition (dump truck))

Standard time (t) = 22 sec (1.2cum bucket size and 600 angle of swing)

Job Efficiency (E) = (50/60)=0.83

5.106)3.122(

83.085.02.13600(lcm/hr) P =

×

×××=

9.7835.1

5.106(bcm/hr) P =

= → (For given 35% of swell)


Documents

Hydraulic Excavators (Classnote-2)