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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
16 August 2010 Slide No. 3IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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.
16 August 2010 Slide No. 4IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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.
16 August 2010 Slide No. 5IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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)
18 August 2010 Slide No. 6IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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
=
=
=
18 August 2010 Slide No. 7IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
+
×
×
×××=
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
18 August 2010 Slide No. 8IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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
18 August 2010 Slide No. 9IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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%)
18 August 2010 Slide No. 10IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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 =
+
×
×
×××=
18 August 2010 Slide No. 11IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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.
18 August 2010 Slide No. 12IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
Hydraulic Hoe: Production Estimating
• 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]
18 August 2010 Slide No. 13IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
Hydraulic Hoe: Production Estimating
• 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.
10 August 2010 Slide No. 14IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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%.)
18 August 2010 Slide No. 15IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
Hydraulic Hoe: Production Estimating
• 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)
18 August 2010 Slide No. 16IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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 (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
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
18 August 2010 Slide No. 17IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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%.
18 August 2010 Slide No. 18IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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 =
+
=
+
=
18 August 2010 Slide No. 19IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
PRODUCTIVITY OF HYDRAULIC EXCAVATOR:
(Quick Method)
The hourly production of hydraulic excavator can also be calculated as :-
Where;
P (lcm/hr) = Production in loose cubic meter (volume) per hour
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
18 August 2010 Slide No. 20IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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.
18 August 2010 Slide No. 21IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha
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)
18 August 2010 Slide No. 22IOE-BEM-CE(Elective): Excavators: By Ashok Shrestha