lecture # 3 - universitas brawijaya · example 1: determine the hourly average production of...

EARTHMOVING VEHICLES

CIVIL ENGINEERING DEPARTMENT

UNIVERSITY OF BRAWIJAYA

LECTURE # 3

Topics :

Bulldozer

Ripper

Backhoe

Power shovel

Dragline

Clamshell

Loader

LAND CLEARING VEHICLE DOZER

This vehicle can be used for:

Land clearing

Dozing (pushing material)

Ripping

Assisting Scrapper

…dozing…

…ripping

…

Assisting Scrapper

TYPE OF DOZER

BLADE ADJUSTMENT

Tiltin

g

Angle

BLADE ADJUSTMENT

Pitch

BLADE ADJUSTMENT

TYPE OF BLADE :

A. Straight Blade (S-Blade)

For scrapping and stockpiling the soil. It can be used for hard soil as well.

B. Universal Blade (U-Blade)

Used for:

- Soil Reclamation

- Material preparation activity, etc.

C. Angling Blade (A-Blade)

Used for:

To dump the unwanted material aside

To open a driveway

To dig a channel

D. Cushion Blade (C-Blade)

to push the scrapper

Steps to calculate Bulldozer Productivity

A. By Using The Company Chart / Brosure

1. Calculate all correction factors (E) :

1. Correction for Work Condition (tabel E.01 [2] )

2. Correction for Soil Slope (chart E.02 [2] )

3. Correction for Soil Compaction to Standard (p. 38 [2] )

etc…

2. total of correction factor is the result from multiplying all

related correction factor.

3. Get the value of bulldozer max production from tabel E.1.01 [2]

according to vehicle type and span of moving.

4. Actual production = Max Production Total of correction factor

n

i

iEE1

MAXIMUM PRODUCTION OF BULLDOZER BY TYPES

MAXIMUM PRODUCTION OF BULLDOZER BY TYPES

1. Efficiency 100%

2. Steady time for changing gear

0.05 mnt

3. Soil Density 2300 lb/LCY or

1790 kg/m3 (BM) or 1370

kg/m3 (LM)

4. Swell 30%

5. Traction Coefficien :

a. Whell : 0.4 atau lbh

b. Track : 0.5 atau lbh

6. Hydraulic controlled

1. Calculate all correction factors (E)

(table E.01 [2] )

2. Correction for soil slope (graphic E.02 [2] )

Ideal production values are based on a soil density of 2,300 lb/LCY.

tunit weigh material Actual

(standard) Y2,300lb/LC correctweightMaterial

3. Correction for soil compaction (page. 38 [2] )

EXAMPLE 1:

Determine the hourly average production of Crawler Bulldozer D8 S-type Cat. (with a tilt cylinder) that being used for hard clay soil with an average thrust distance of 150 ft (45 m), the road slope 15%. Please use the method of 'slot dozing' (one way). Material density 2650 lb / LCY, the ability of the operator is MODERATE, the efficiency of the working time of 50 minutes / hour.

1. Calculate all correction factors (E)

(table E.01 [2] )

(table E.01 [2] )

2. Correction of land slope (graphic E.02 [2] )

Ideal production values are based on a soil density of 2,300 lb/LCY.

tunit weigh material Actual

(standard) Y2,300lb/LC correctweightMaterial

Correction of soil campation (page. 38 [2] )

= 2300/2650

= 0,87

Answer :

Correction of slope : 1,19 (graphic E.02 [2] )

Hard Clay : 0,80 (table E.01 [2] )

Slot dozing : 1,20 „

Operator ability (moderate) : 0,75 „

Time efficiency (50 mnt/hour) : 0,84 „

Weight Correction = 2300/2650 : 0,87

Answer :







Correction Factor = 1,19 x 0,8x 1,2x 0,75x 0,84 x 0,87 = 0,626

MAX PRODUCTION OF BULLDOZER BASE ON TYPE

Answer :








Max Production of bulldozer = 550 LCY/hour

Answer :








Max Production of bulldozer = 550 LCY/hour

Actual Production = 550 x 0,626 = 344 LCY/hour

B. Theoritical Method:

1. Calculate Blade Capacity per siklus : q = LH2

2. Calculate Wight Load

3. Compare with critical traction whether the vehicle can be used or not

4. If possible to be used, then :

- Determine speed when pushing forward (va) and move backword (vb)

- Calculate Circle Time : (minute)

5. Calculate all correction :

1. Correction of work condition (tabel E.01 [2] )

2. Correction of land slope (grafik E.02 [2] )

Total Correction Factor :

6. Actual Production :

tv

d

v

dW

ba

s

n

i

iEE1

EW

qQs

60

EXAMPLE

Calculate the estimated production of bulldozers using the following data:

Topsoil unit weight : 2300 lb/BCY with Swell = 43%

Coeffien of Traction : 0,9

Pushing Distance : 60 m = 0,0373 mil

Bulldozer Power : 105 HP

S-Blade : length 3,15 m and height 0,96 m

Total weight of bulldozer : 11.700 kg (25.800 lb)

Steady time (gear changing) : 0,1 mnt

Answer :

1. Blade capacity per siklus :

q = LH2 = 3,15 (0,96)2 = 2,9 Lm3 (1 yar kub = 0.76455 m3)

= 3,79 LCY

= 3,79/1,43 = 2,65 BCY

2. Weight Load = 2300 lb/BCY x 2,65 BCY = 6095 lb

3. Pushing Power = 6095 lb

Critical Traction = 0,9 x 25.800 lb = 23.220 lb > 6095 lb …

- bulldozer can be used..

Answer :

4. Cycle time:

DBP = 6095 lb

See table E.2.01 (2)

Answer :

Speed to push forward and backward

Push Speed = 4,92 mph

(interpolating from table E.2.01 [2] )

Backward Speed, use 3rd gear = 4,4 mph

Cycle Time

Push until 60 m, v = 4,92 mph :

= 0,455 mnt

Backward 60 m, v = 4,4 mph, :

= 0,51 mnt

Steady Time = 0,1 menit

Siclus time = 0,455 + 0,51 + 0,1 = 1,065

mnt 6092,4

0373,0at

mnt 604,4

0373,0bt

5. Correction Factor :



Operator Capability (moderate) : 0,75 „

Time Efficiency (50 mnt/hour) : 0,84 „

Correction Factor = 0,8x1,2x0,75x0,84 = 0,605

6. Actual Production of Bulldozer :

= = 90,3 BCY/hour

605,0065,1

6065,2

60 E

WqQ

s

If hard soil is founded in land clearing, then ripper will be used

Ripper :

- It can be pulled by a tractor

- As an attachment in the tractor

RIPPER PRODUCTION

1. Calculate production capacity of ripping device per trip :

q = length distance depth

2. Calculate Circular Time :

(t = steady time = backward time)

3. Calculate total correction: working time efficiency and work condition factor and procedure (table IV.4.01.05 page. 64 [2] )

4. Calculate Production per hour :

5. Actual Production = 80 % s/d 90 % of hourly production (Q).

tv

dWs

EW

qQs

60

EXAMPLE

Calculate ripper single shank production that being pulled by tractor type D9H CAT. with data as follow:

•Space of ripping : 0,915 m

•Depth of ripping : 0,610 m

•Length of ripping : 91 m

•Speed of ripping : 1,6 km/jam = 26,6 m/mnt

•Comeback time : 0,25 mnt

•Work time efficiency : 60 mnt/hour

1. Calculate production per trip

Production per trip:

q = 91x0,915x0,61 = 50,8 Bm3/trip

2. Calculate cycle time :

Ws = = 3,67 mnt

Total trip per hour = 60/3,67 = 16,35 trip/hour

25,06,26

91

tv

dWs

Ws = = 3,67 mnt

Total trip per hour = 60/3,67 = 16,35 trip/hour

3. Production per hour :

Q = 50,8x16,35 = 830,6 Bm3/hour

25,06,26

91

The result is still 10% - 20% higher.

So, actual production = 80% x 830,6 = 664,5 Bm3/hour or = 90% x 830,6 = 747,5 Bm3/hour

So, the real production between 664.5 to 747.5 Bm3/jam (not yet being multiply with correction factor)

BACKHOE

hoe

back shovel,

pull shovel

GERAKAN-GERAKAN BACKHOE

1. Put pressure on the boom to force the bucket teeth or

cutting edge into the ground

2. Raise the

stick and roll

the bucket

until it is full

3. Dump

the

excavated

material

4. Swing

the

bucket

back to

the

starting

position

BACKHOE

Factors that affect hoe production are the :

-Width of the excavation

-Depth of the cut

-Material type

-Working radius for digging and duming

-Required bucket dumping height

Steps to calculate hourly production of backhoe

1. Calculate Bucket Capacity

2. Determine cycle time

3 Calculate correction factor: E = Ei

4 Calculation backhoe production: Q = q 60/Ws E

PRODUCTION CALCULATION

Correction factors(Ei) :

1. Working time efficiency

2. Working condition efficiency (table IV.4.01.05 [2] pp. 64)

3. Bucket fill factor (table IV.4.01.06 [2] pp. 64)

4. Check Swing factor and depth of cut ;

- Determine optimum depth of cut for a hoe (depends on the type of material being excavated and bucket size, see table IV.4.01.04 [2] pp. 63)

- Determine the ratio of the actual depth of cut and the optimum depth of cut, expressed as a percent

- Determine swing factor and depth of cut (table IV.4.1.03 [2] pp. 63).

%100cut ofdepth optimum

cut ofdepth actual(%)Optimum depth

EXAMPLE

Determine the hourly output for 1 ¾ -cu-yd backhoe.

Notes :

-Material : sand

-swell 43%

-depth of cut: 6 feet, -Angle of swing : 90

-Job Condition : average

Jawab :

1. Bucket Capacity (q)

Bucket 1,75 cu.yd = 2 cu yd (heaped bucket volume)

Swell : 43%

Bucket Capacity : q = 2/1,43 = 1,399 BCY

2. Cycle Time

Bucket fill : 7 detik

Loading haul and swing : 10 detik

Dumping : 5 detik

Back to the starting position : 5 detik

Fix time : 4 detik

Total Cycle Time : 31 detik = 0,517 menit

3. Calculate correction factor: E = Ei

Working hour efficiency= 50/60 : 0,84

Working condition average : 0,65

Bucket fill factor (soil) : 0,85

Optimum depth of cut= 9,7 feet

b. Percentage of Optimum depth of cut:

c. Determine swing factor and depth of cut (table IV.4.1.03 [2] pp 63).

: 0,91

%60%8,61%1009,7

6cut ofdepth Optimum

%100cut ofdepth optimum

cut ofdepth actual(%)Optimum depth

Total correction factor : E = 0,840,650,850,91 = 0,42

6.Hourly Production

42,0517,0

60399,1

60 E

WqQ

s

= 68,19 BCY/jam

LOADERS

WHEEL & TRACK LOADERS

LOADERS PRODUCTION

STEP 1: BUCKET SIZE

LOADERS can

usually be

equipped with

several

different size

and type

buckets.

STEP 1: BUCKET SIZE

Buckets are rated in both

Heaped and Struck

capacities.

HEAPED CAPACITY is

the rating of interest for PRODUCTION ESTIMATING.

STEP 2: FILL FACTOR

HEAPED CAPACITY is based

on the SAE standard of a 2:1

material repose angle.

The FILL FACTOR adjusts

HEAPED CAPACITY in LCY

based on the type of material

being handled.

STEP 2: FILL FACTOR

The FILL FACTORS for

wheel and track loaders

are different because of

differences in the

breakout force that can be

developed.

STEP 2: FILL FACTOR

HEAPED BUCKET CAPACITY

X FILL FACTOR

= VOLUME

(LCY)

STEP 3: MATERIAL CONVERSION

The bucket load was in

LCY. It should be

converted to TONS in order

to check tipping load. We

may require the load in

BCY.

LOADER CAPACITY

STEP 4: CHECK TIPPING

A loader must maneuver

and travel with the load.

WHEEL LOADER

OPERATING WEIGHT is

limited to 50% of full-turn

static tipping load.


A loader must maneuver

and travel with the load.

TRACK LOADER

OPERATING WEIGHT is

limited to 35% of static

tipping load.


WHEEL LOADER:

Volume X Material Unit Wt

< 0.50 Static Tipping Load full-

turn


TRACK LOADER:


< 0.35 Static Tipping Load

STEP 5: CYCLE TIME

LOADER CYCLE TIME has

three components :

(1) Fixed Time to load,

Maneuver with four reversals of

direction (min. travel), and

Dump.

(2) Travel Time

(3) Return Time

LOADER PRODUCTION

STEP 5: FIXED CYCLE TIME

(1) Fixed Time

Time to load

Maneuver with four reversals

of direction (min. travel)

Dump

LOADER PRODUCTION

STEP 5: CYCLE TIME

(2) Travel Speed

(3) Return Speed

Check

Manufactures’ data

for gear speeds.

STEP 5: CYCLE TIME

Travel Speed with a loaded bucket -

Average speed should be about 80%

of maximum in low range.

STEP 5: CYCLE TIME

Return Speed with an empty bucket -

Average speed should be about 60% of

maximum in intermediate range.

STEP 5: CYCLE TIME

(2) Travel Time TRAVEL DISTANCE (FT)

88 TRAVEL SPEED (MPH)TRAVEL TIME (MIN)

(3) Return Time RETURN DISTANCE (FT)

88 RETURN SPEED (MPH)RETURN TIME (MIN)

STEP 5: CYCLE TIME

TOTAL CYCLE TIME =

Fixed Time

+ Travel Time

+ Return Time

STEP 6: EFFICIENCY FACTOR

When selecting a minutes per

hour efficiency factor visualize

the work site and consider:

• Materials being handled, bank or

broken

•Size of the loader’s dumping

target,

(large or small)

STEP 7: PRODUCTION

Working mins / hr

Total cycle time (min)

X Heaped Capacity (LCY)

X Fill Factor

= Production (LCY/hr)

PRACTICAL EXERCISE

You are using a track loader to

load uniform 1½ in. aggregates

from a stockpile. The loader is

equipped with a 2.6 cy bucket

(heaped). What is the minimum

loose volume production that can

be expected per loader cycle?

PE NO. 1

STEP 1: BUCKET SIZE:

• The loader is equipped with a

2.6 cy bucket (heaped).

HEAPED CAPACITY is the rating

of interest for PRODUCTION

ESTIMATING.

PE NO. 1

STEP 2: FILL FACTOR

The FILL FACTORS for wheel and track

loaders are different because of

differences in the breakout force that can

be developed.

Fill Factor: 90 to 110

PE NO. 1

STEP 2: FILL FACTOR


What is the minimum loose volume production?

Therefore use the lower value:

Fill Factor: 90

PE NO. 1


X FILL FACTOR

= VOLUME (LCY)

2.6 LCY X 0.9 = 2.34 LCY

PE NO. 2

Estimate the best production in tn/hr for

a CAT 938F loader using the largest

possible general purpose bucket w/

bolted on cutting edges. The loader is

being used to load trucks from a

stockpile of mixed dry aggregates

(gravel). Assume a 55 min-hr efficiency.

PE NO. 2

STEP 1: BUCKET SIZE:

• CAT 938G - largest possible

general purpose bucket w/

bolted on cutting edges

3.65 cy heaped capacity

PE NO. 2

STEP 2: FILL FACTOR

• Mixed dry aggregates

(gravel).

• CAT 938G is a wheel loader


PE NO. 2

STEP 2: FILL FACTOR


• Estimate the best production

Therefore use the higher

value:

Fill Factor: 100

PE NO. 2


X FILL FACTOR

= VOLUME (LCY)

3.65 LCY X 1.0 = 3.65 LCY

PE NO. 2

STEP 3: MATERIAL CONVERSION

Material unit weight

• mixed dry aggregates (gravel).

2,490 lb/cy Table 4.3

3.65 cy X 2,490 lb/cy = 9,089

lb

PE NO. 2


WHEEL LOADER:


< 0.50 full-turn Static Tipping

Load

20,260 lb

PE NO. 2


WHEEL LOADER:


< 0.50 Static Tipping Load full-

turn

9,089 lb < 0.5 X 20,130 lb

9,089 lb < 10,065 lb ok

PE NO. 2

STEP 5: CYCLE TIME (1) Fixed Time to load, maneuver with four reversals of

direction and dump.

27 to 30 sec

• best production

Therefore use 0.45 min

PE NO. 2

STEP 6: EFFICIENCY FACTOR

•55 min-hr efficiency.

STEP 7: PRODUCTION

Working mins / hr

Total cycle time (min)

X Heaped Capacity (LCY)

X Fill Factor

= Production (LCY/hr)

PE No. 2

STEP 7: PRODUCTION

55 mins / hr

0.45 min X 3.65 LCY X 1.0

= 397 LCY/hr

PE No. 2

STEP 7: PRODUCTION

lb/tn 2,000

lb/LCY 2,490LCY/hr 397

= 494 tn/hr

PE No. 2

THE END

lecture # 3 - universitas brawijaya · example 1: determine the hourly average production of...

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