Hydraulic Excavators (Classnote-2)

Hydraulic Excavators Excavators (II)

By

Er. Ash Ashok Shresth stha (DoR) IOE-BEM Construction Equipment(Elective) August 2010

Cycle time

• The The sum sum of time time requ requir ired ed to load load bu buck cket et,, swi swing ng loaded, dump and swing empty. • Typi Typica call cycle cycle elem elemen entt time times s un unde derr ave avera rage ge conditions, for 2 to 4 cum shovels will be – Lo Load ad bu buck cket et – Swing with load – Dump load – Return swing

16 August 2010

7-9 7-9 sec. se(depend c. on material type) 4-6 sec (depend on machine size) 2-4 sec (depend on dumping target) depend on machine size) 4-5 sec ( depend

IOE-BEM-CE(Electiv IOE-BEM-CE(Elective): e): Excavators: Excavators: By Ashok Shrestha Shrestha

Slide No. 2

Factor affecting production • Actu Actual al pro produ duct ctio ion n of of a shov shovel el is aff affec ecte ted d by the following factors: – Cl Class ass of ma mate teri rial al – He Heig ight ht of cut cut – Angl Angle e of of swi swing ng – Size Size of of haul haulin ing g unit units s – Oper Operat ator or ski skillll – Physica Physicall condit condition ion of the the shovel shovel Production efficiency efficiency ranges from 30 to 45 min per hour

16 August 2010


Slide No. 3

Effect of height of cut and Angle of swing • The The eff effec ectt of of hei heigh ghtt of of cut cut an and d Ang Angle le of of Swi Swing ng on 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. 4

Optimum height of cut • The opt optim imum um hei heigh ghtt of of cut cut ran range ges s ffro rom m 30 30 to to 50% of the maximum digging height. – 30 % for a easy to load load material materials s (i.e. (i.e. load sand, sand, gravel etc.) – 40 40% % for for comm common on ea eart rth h – 50 50% % for for poo poorl rly y bla blast sted ed rock rock,, or stic sticky ky clay clay

• The ide ideal al pro produ duct ctio ion n of shov shovel el is bas based ed on operating at a 90 0 swing and optimum height of cut. • The The iide deal al p pro rodu duct ctio ion n shou should ld b be e mul multi tipl plie ied d by by tthe he proper correction factor in order to correct the production for any given height and swing angle. 16 August 2010


Slide No. 5

Excavator Production • St Step eps s fo forr esti estima mati ting ng prod produc ucti tion on:: 1. Obta Obtain in the the hea heape ped d buck bucket et loa load d volu volume me (lc (lcm) m) fro from m the manufacturer performance data. 2. Apply Apply a b buck ucket et fill factor factor ba based sed on type type mat materi erial al being excavated. 3. Esti Estima mate te a pe peak ak cycl cycle e time time:: (Load bucket+ Swing with load+ Dump load+ Return swing)

4. Obtain Obtain the factor factor for an angle gle of swing swing and height height of cut from from the the ta table ble (% of optimum optimum depth depth vs angle angle of swing). 5. Apply ply a ef effi fic cien iency fa fact cto or (usuall (usually y 30 – 45 min min per per 60 min) min) 6. Con Confor form m the the produc production tion un units its to desire desired d volu volume me or weight (lcm to bcm) (lcm= bcm*(1+swell factor) 18 August 2010


Slide No. 6

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.

Typical cycle element times under average conditions, for 2 to 4 cum shovels Load bucket 7-9 sec. (Based on materials) Swin Sw ing g with ith load load 4-6 4-6 sec sec (Bas (Based ed on siz size of of m m/c /c)) Dump load 2-4 sec (Base on hauling unit) Return swing 4-5 sec (Based on size of of m/ m/c)

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) higher(6sec) value for bigger 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 specificat ion and multiply it by a factor within range of 0 0.3 .3 – 0.5. Take Take lower value (0.3) for for easy to load material and higher value (0.5) for very difficult material

Optimum depth of cut

=

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

=

50% of maximum degging height (for difficult blasted rock etc)

=

40% of maximum digging height (for common averge material)

18 August 2010

% of optimum depth

=

 Average height of cut    × 100  Optimum height of cut 


Slide No. 7

7. Com Comput pute e the the produc productio tion n rate rate,, usin using g ffoll ollowi owing ng formu formula la.  3600 × Q × F × ( AS : D )   E   ×  t    60   3600 × Q × F × ( AS : D )   E   1  P (bcm/hr) =  × ×  t    60   1 + S.F.  P (lcm/hr)

=

Where; P (lcm/hr) = Production Production in loose cubic cubic meter (volume) (volume) per hour P (bcm/hr) = Production Production in bank cubic meter meter (volume) per hour P (ton/hr) = Production Production in in tons (weight) per 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 Loose unit weight weight (N) (N)

18 August 2010


Slide No. 8

Example-1 •

A 3.8 3.8 cu.m cu.m.. shov shovel el hav havin ing g a ma maxi ximu mum m diggi digging ng hei height ght of of 10.4 10.4m m is bei being ng used to load poorly blasted rock. The face being worked is 3.7m high and the haul haul units can can be positioned positioned so that that the swing swing angle angle is only 600. What is the adjusted ideal production if the ideal cycle time is 21 sec. 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) (Given) Optimum height = 0.5 x 10.4 = 5.2m (Taken highest %, for poorly blasted rock) (30–50%)

% of optimum height

=

 Working height)  3.7   × 100 = × 100 = 71 . 15 % 5 .2  Optimum height 

Angle of Swing = 60 0 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)

 3600 × Q × F × ( AS : D )   E  Assuming Efficiency factor,   ×   E= 45/60 t    60   3600 × 3.8 × 0.9 × 1.08   45  P (lcm/hr) =   ×   = 475 (lcm / hr ) 21    60  P (lcm/hr)

18 August 2010

=


Slide No. 9

•

Example-2

A 2.3 2.3 cu.m cu.m.. shov shovel el hav havin ing g a maxi maximu mum m diggi digging ng hei heigh ghtt of 9 9.1 .1 m, m, will will be 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 fill factor (F)= 1 (taken from the table, for well-blasted well-blasted 100-110%) 100-110%) Ideal Cycle Cycle time (t) = (Load + Swing loaded loaded + Dump + Swing empty) empty) = 9 + 4+ 4 + 4 = 21 sec Assuming: Load = 9 sec (Taken maximum maximum value, value, as material is rock difficult 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)

Typical cycle element times under average conditions, for 2 to 4 cum shovels: Load bucket 7-9 sec. Swing with load 4-6 sec Dump load 2-4 sec Return swing 4-5 sec

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

18 August 2010

height

=

 Working   Optimum

height) 

 × 100 height 

=

6 .7 4 . 55


×

100

=

147 %

Slide No. 10

Example-2

(/contd…)

Angle of Swing = 120 0 Angle of Swing and depth (AS:D)= 0.79

(by interpolation)

(from the table 0.81- [{(0.81-0.75)/(1 [{(0.81-0.75)/(160-140 60-140)}*(147)}*(147-140)] 140)]

(Given) Efficiency Factor (E) = 30/60 (From the table for well blasted rock) % Swell = 60%

1  3600 × Q × F × ( AS : D )   E    P (bcm/hr) =  × ×  t    60   1 + swell 

 3600 × 2.3 × 1 × ( 0.79 )   30   1  P (bcm/hr) =  × ×  = 97 .3 21    60   1 + 0.6 

18 August 2010


Slide No. 11

Hydraulic Hoe: Production Estimating • The The same same ele eleme ment nts s tha thatt aff affec ectt sho shove vell prod product uctio ion n are applicable to hoe excavation operation. • Ho Hoe e c cycl ycle e tim times es are are app appro roxi xima mate tely ly 20 20% % long longer er than similar size shovel and work. • The The opt optim imum um de depth pth of cut cut for for ho hoe e iis s usua usualllly y in th the e range of 30 to 60%. • Standard da data fo for “C “Cycle titime” based sed on on bu bucket size and average conditions (30-60 0 swing angle, hauling unit at same level etc. is available). • No stan standar dard d dat data a and and fa fact ctor ors s base based d on on ang angle le of swing and depth of cut is available. 18 August 2010


Slide No. 12

Hydraulic Hoe: Production Estimating • St Ste ep-1: -1:

Bu Bucke cket

• Step-2:

Fill

si size

(lcm (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).

Fa Facto ctor:

(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. 13

Hydraulic Hoe: Production Estimating • Step-3: Cycle cle time (sec sec)

(Load + Swing load + Dump +

Swing empty). Typical excavation cycle times based on machine (bucket) size

The cycle times must be increased when loads are dumped into a smaller haul units. Small machine swing faster than large ones.

Depth of cut: 40 to 60% Swing Swing angle angle = 30 – 60 0 Loading haul units on the same level

10 August 2010


Slide No. 14

Hydraulic Hoe: Production Estimating • Step-4: Depth of cut

(Obtain m aximum aximum dig depth from manufacturer’s

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

18 August 2010


Slide No. 15

Hydraulic Hoe: Production Estimating • Ste tepp-5: 5: Effici ficien ency cy Fact actor: or: – 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.

– Operat Operator or eff effici iciency ency:: (Skil (Skilll of ope operat rator) or) – Equipment Equipment availability availability (Haul units availabili availability ty ‘x’% ‘x’% of the time) Machine’s working range based on size of machine (bucket) fitted with standard items (Boom, Arm etc.)

18 August 2010


Slide No. 16

Step-6: Compute production rate, using following formula f ormula.  3600 × Q × F   E  P (lcm/hr) =  ×  t    60   3600 × Q × F   E   1  P (bcm/hr) =  × ×  t    60   1 + S.F.  Where; P (lcm/hr) = Production Production in loose cubic cubic meter (volume) (volume) per hour P (bcm/hr) = Production Production in bank cubic meter meter (volume) per hour P (ton/hr) = Production Production in in tons (weight) per 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 Loose unit weight weight (N) (N)

18 August 2010


Slide No. 17

Example-3 A crawler hoe hoe having having a 2.8 cu.m cu.m bucket bucket is being conside considered red for use on a project to to excavate excavate dry clay clay from a borrow borrow pit. The The clay will 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 the table table maximum maximum depth depth of cut cut 7 – 8.2 m Depth of cut = 2.5 m  2.5    * 100  7   2.5    * 100  8.2  18 August 2010

=

35 .7 %

Checking for optimum depth of cut range 30% to 60%. =

30 . 4 %


Slide No. 18

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

Step-6: Production rate Calculation

 3600 × Q × F   E  P (lcm/hr) =  ×  t    60  P (lcm/hr)

 3600 × 2.8 × 0.85   50  =   ×   = 324 .5 (lcm/hr) 22    60 

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

 324 .5  P (bcm/hr) = =   = 240 .37 1 + Swell factor  1 + 0 .35  P (lcm/hr)

18 August 2010


Slide No. 19

PRODUCTIVITY OF HYDRAULIC EXCAVATOR: (Quick Method) The hourly production of hydraulic excavator can also be calculated as :-

 3600 × Q × E   3600 × ( q × k ) × E   P (lcm/hr) =   =  Ct (t × F )     Where; P (lcm/hr) = Production Production in loose cubic cubic meter (volume) (volume) per hour Q = Production per cycle (cu.m) Bucket Factor (k) = (q*K) Material Bucket facto q = bucket heaped capacity (cu.m) K = Bucket factor Moist loam or sandy clay 1.0 - 1. E= Job Efficiency factor Common soil 0.9 - 1 t = Standard Cycle time in seconds Sand and gravel 0.85 – 0 F = Time factor Hard tough clay 0.8 - 0 Ct = Cycle time in seconds =(t*F) Rock –well blasted

Rock – poorly blasted

0.6 - 0. 0.4 - 0

Standard Cycle time (t)= Actual cycle cycle time time = Actual

excavating time + swing time loaded+ dumping time +swing time empty OR standard standard cycle cycle time time * time tim time factor fact facto or Cycle Time (t)

Time Factor (F)

Standard cycle time based on bucket Capacity

Dumping Conditions Digging Conditions

EASY

NORMAL (Average)

RATHER DIFFICULT

DIFFICULT

BUCKET CAPACITY

Swing Angle /Time (Sec )

40 - 90 90

90 -180

BELOW 40 %

0.7

0.9

1.1

1.4

0.25

13 -15

15 -17

40 % - 75 75 %

0.8

1.0

1.3

1.6

0.4

13 -15

15 -17

0.45

14 -16

16 -18

0.7

16-18

1 8 – 21

0.9

1 8 - 20

2 0 – 23

1.2

2 0 - 22

22 - 25

OVER 75 %

0.9

1.1

1.5

1.8

Digging condition= (Digging depth/ Max. depth of cut)*100 Easy = Dump onto spoil pile Normal = Large dump target Rather difficult = Small dump target Difficult = Small dump target requiring maximum reach.

18 August 2010


Slide No. 21

Example-4 A 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 60 0 angle of swing) Job Efficiency (E) = (50/60)=0.83 P (lcm/hr)

=

P (bcm/hr)

18 August 2010

 3600 × 1.2 × 0 .85 × 0.83    = 106 .5 ( 22 1 . 3 ) ×   =

 106 .5    = 78 .9 1 . 35  

→

(For given 35% of swell)


Slide No. 22

Hydraulic Excavators (Classnote-2)

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