Tote dumper design

Transcription

Tote dumper design

Tote dumper design
Tutorial
This example shows the steps necessary to design a tote dumper.
Tote dumper requirements:
Tote dimensions: 120 cm x 120 cm x 120 cm
Tote weight, Wt: 400 Kg
Cycle time, ts: 60-80 sec.
Warning – This tutorial is to
serve as a guide covering most
design areas. User is to consult
and respect local government
regulations where the unit is to
operate.
Dumping height: 140 cm
Dumping inclination: 30° from horizontal
Operating temperature: 5°C to 40°C
TOTE
Step 1: Dumper layout & geometry
Step 2: Hydraulic design
Step 3: Dumper structure design & stress calculation
1
DESIGN THROUGH ITERATIONS
Because the dumper geometry, its structure and the hydraulic components
used to put it in motion are so intricately connected, a global design method is
required.
STRUCTURE
Several iterations on the geometry, hydraulic components, and structure
are required to arrive at a functional solution.
This tutorial will minimize the number of iterations required. It will also
allow you to optimize your solution.
2
1.0 Dumper layout & geometry
The tote dumper shown below is a steel structure made largely of
rectangular hollow tubes. A tote is loaded in the lifting bin. Two hydraulic
actuators raise and tilt the bin with its tote. It is emptied and is lowered back
to the floor level.
The lifting bin is to be strong enough to withstand the stresses through the
complete lift-and-tilt cycle. Its weight will be substantial and comparable to
the tote weight.
TOTE
LIFTING BIN
HYDRAULIC
ACTUATOR
3
Two layouts are to be made where the following items are shown:
• The tote.
• The bin and its pivot position (A).
• The actuators and their anchoring positions (B and 4).
Layout A is for tote loading.
Layout B is for tote dumping.
4
The hydraulic actuators are to be sized for force and stroke for the complete
lifting cycle. Both force and stroke are affected by the actuators anchoring
positions. Along with the tote and the dumping height, they impose a
dimensional constraint on the design.
Actuator length [B4] between retracted and extended conditions is calculated
by:
[B4] = ( [AB]2 + [A4]2 – 2[AB][A4]cos(v) )0.5
5
To simplify the layout process, open the MS Excel file
PackexOneToteDumperDesign.xls.
In worksheet DUMPER, input data is to be entered only in the white cells.
Tote weight, Wt, is the weight of the tote with its content.
Bin weight, Wd, is the weight of the bin. It is unknown for the first iteration
and is to be approximated equal to the tote weight.
Units Metric Imperial checkboxes. Choose Metric if your enter your data in
Kg and cm. Choose Imperial if your enter your data in lbs and inch. Switching
from metric to imperial or to imperial to metric will not convert your data.
6
PackexOneToteDumperDesign.xls
BIN
NODES 1,2,3,4,0,A,B. Enter in columns X and Y the coordinates of the
points on the bin and the base when the bin is in loading position. Nodes
1,2,3,4 represents the intersections of two or more trusses in the bin. Node 0
represents the tote center of gravity (usually at mid-height). Node A
represents the pivot position of the bin. Node B represents the actuator
bottom anchor location.
TRUSSES 1,2,3,4,5. Enter in columns “height”, “width” and “wall thickness”
the data for each truss of the bin. Bin truss is defined as the rectangular
hollow steel tube located between the two nodes indicated.
HEIGHT
WALL THICKNESS
WEIGHT
WIDTH
7
BIN OUTPUT DATA TABLE
Each table row represents a series of information for a specific bin angle,
from 0° (loading) to 120° (dumping).
Column “B-4” shows the hydraulic actuator length, from the bottom clevis to
the top clevis. The selected actuator must be able to operate within these
limits. User can change the locations of nodes 4 and B, the retracted
actuator extremities, and see the impact on the actuator length.
Column “Actuator Force F4” shows the minimum required force from the
hydraulic actuator. The data is also plotted under the table.
Data in others columns is used for the mechanical stress analysis (step 3).
8
By adjusting the bin pivot position (node A), the actuator anchor positions
(nodes B and 4), the tote location through its center of gravity (node 0), a
compromise can be reached that will satisfy:
•The dumper overall dimensions.
•The loading and dumping dimensions requirements.
•The hydraulic actuators extended and retracted lengths.
•The hydraulic force.
9
2.0 Hydraulic Design - Actuators
The dumper operates through two linear actuators which are powered by an
hydraulic unit; basically a pump, a electric motor, an oil tank, and a pressure
regulator.
For each actuator, from the file PackexOneToteDumperDesign.xls:
Required maximum force, F4 = 957 Kg (2109 lbs)
Stroke, s = 77 cm (30.3 in), difference between extended and retracted
actuator lengths
The actuator has a pivot connection at
both extremities, the cap clevis and the
rod end. Using overall length, stroke and
manufacturer’s data, the minimum
required rod diameter to prevent buckling
is Dr = 34.9 mm (1.375 in).
Rod end
The actuator minimum bore size when
operating at a pressure p = 10342 kPa
(1500 psi) is:
Dbm = {4 F4 / (π p) } 0.5 = 34 mm (1.34 in)
The selected bore size must be above and
is Db = 50.8 mm (2 in)
If the actuator selected reaches its end of
stroke in operation, it is important that it
includes a cushioning device (a piston
attachment that reduces the oil flow near
the stroke end). Consult manufacturer.
Cap clevis
10
2.1 Hydraulic Design - Actuators
The actuators selected have the following characteristics:
Bore diameter: 50.8 mm (2 in)
Rod diameter: 34.9 mm (1.375 in)
Maximum stroke: 86.4 cm (34 in)
Piston seal, near zero leakage
Hydraulic fluid:
Petroleum based oil
density: 900 Kg/m3
viscosity: 50 cSt (50 x 10-6 m2/s) at 40°C
Actuator Friction
From manufacturer’s data at 10342 kPa (1500 psi):
Running friction, Ff = 65 Kg (143.3 lbs)
Breakaway friction Fb = 86 Kg (189.6 lbs)
The breakaway friction is the highest value and will be used for pressure
calculation.
11
2.2 Hydraulic Design - Circuit
Hydraulic circuit diagram
12
In the diagram shown on the previous page, the goal is to reach the best
compromise between manufacturing cost and operating cost.
The hydraulic unit is made of a fixed-displacement pump powered by an
electric motor. The pump is mounted on top an oil tank which includes an oil
strainer, an oil return filter, a pressure gauge and a pressure regulator.
The pump supplies the manual 4-way / 3-position valve. The supply oil
pressure is regulated by the unit pressure regulator.
OPERATION
When the valve spool is shifted to the left,
the flow is sent to the flow divider/combiner.
The flow is divided in two equal flows and
this division is unaffected by pressure.
Each flow goes through the open channel of
a counterbalance valve and enters the cap
end of the actuator. The rod extends and lifts
the bin.
The oil on the other side of the actuator
piston flows into the hydraulic unit tank.
13
When the manual valve spool is shifted to
the right, the flow is sent to the head end of
both actuators.
The rod retracts because of the hydraulic
pressure and the bin weight.
The oil on the other side of the actuator
piston flows into the counterbalance valve.
The pressure setting of this valve is 30%
higher than the pressure created by the bin
weight inside the actuator. The oil flow is
regulated preventing the bin to fall down.
Both flows then enter the flow
combiner/divider which synchronizes the
actuators retraction.
When the manual valve spool is in neutral
position, the oil flow is blocked and returned
to the tank through the pressure regulator.
The bin is held in place because the oil
inside the actuators is blocked by the
counterbalance valve.
14
2.5 Hydraulic Design - Cycle
Working Cycle
The tote dumper operates in 3 phases cumulating in a cycle time (ts)
between 60 and 80 seconds.
Phase A. Bin Lift. Both actuators extend at a slow and steady rate (ta = 40
sec).
Phase B. Dumping. Both actuators rest in their fully-extended position (tb =
5 sec).
Phase C. Bin Descent. Both actuators retract at a faster rate. The duration
time, tc , depends on the pump selected to meet the lifting rate.
15
Phase A – Bin lift, actuator extension
Each actuator will support a force F4 between 818 Kg and 957 Kg (ref. F4 ,
PackexOneToteDesign.xls) during their stroke of s = 77 cm.
Since the acceleration and deceleration periods are very short, the average
rod speed can be approximated by:
Va = 60 s / (100 ta) = 1.16 m/min
Oil flow entering the actuator is:
Qa(in) = Va (π Db2 / 4000000) = 0.0023 m3/min (2.3 L/min) (0.62 US
gal/min)
Oil flow exiting the actuator is:
Qa(out) = Va (π (Db2 – Dr2) / 4000000) = 0.0012 m3/min (1.2 L/min) (0.33 US
gal/min)
Oil exiting the actuators will go back to the tank through port B of the
manual 4/3 valve. For this valve, pressure drop for a combined flow of 2.4
L/min is Pd6 = 150 kPa (22 psi)
16
Phase A – Bin lift, actuator extension
The oil pressure on the supply side will vary between:
Pa(min)
= 9.81 (F4min + Fb) / (π Db2 / 4000) + Pd6 { (Db2 – Dr2) / Db2 }
= 4454 kPa (646 psi)
Pa(max)
= 9.81 (F4max + Fb) / (π Db2 / 4000) + Pd6 { (Db2 – Dr2) / Db2 }
= 5127 kPa (743 psi)
Where Db = 50.8 mm (bore diameter), Dr = 34.9 mm (rod diameter).
The pump rated flow is 2 Qa(in) = 4.6 L/min (1.24 US gal/min)
17
Phase B – Dumping, actuator stopped and fully extended
Each actuator will support a force of 831 Kg (ref. F4 ,
PackexOneToteDesign.xls). The oil pressure will be:
Pb(max) = 9.81 (F4) / (π Db2 / 4000) = 4022 kPa (583 psi)
Where Db = 50.8 mm (bore diameter)
The oil will be trapped between the actuator piston and the counterbalance
valve. The counterbalance valve pressure setting should be:
Pv = 1.3 Pb(max) = 5229 kPa (758 psi)
18
Phase C – Bin descent, actuator retraction
During the descent, the tote is empty or may still be full. The empty tote
weight is 100 Kg. Each actuator will support a force between 477 Kg and
957 Kg (ref. F4 , PackexOneToteDesign.xls) during their stroke of s = 77
cm. The oil pressure on the return side will vary between:
Pc min = 9.81 (F4min - Fb) / (π Db2 / 4000) = 1892 kPa (274 psi)
Pc max = 9.81 (F4max - Fb) / (π Db2 / 4000) = 4215 kPa (611 psi)
Where Db = 50.8 mm (bore diameter)
But the pressure setting of the counterbalance valve (5229 kPa, 758 psi)
needs to be overcome by the pressure on the pilot port. To have a flow of
oil, the pressure on the supply side will be:
Pc in = {π Db2 Pv – 9.81 F4 / 1000} / (π Db2 R + π Dr2 )
= 172 kPa (25 psi)
Where R = 3, counterbalance valve pilot ratio.
Pc out = Pcmax + Pc in { (Db2 – Dr2) / Db2 } = 4306 kPa (625 psi)
19
Phase C – Bin descent, actuator retraction
Oil flow entering the actuator is equal to the flow calculated in phase A. This
is because the pump is a fixed-displacement type.
Qc(in) = Qa(in) = 2.3 L/min
Since the acceleration and deceleration periods are very short, the average
rod speed can be approximated by:
Vc = Qc(in) / (π (Db2 – Dr2) / 4000000) = 2.19 m/min
tc = 60 s / (100 Vc ) = 21 sec.
Oil flow exiting the actuator is:
Qc(out) = Vc (π Db2 / 4000000) = 0.004 m3/min (4.4 L/min) (1.2 US gal/min)
20
2.11 Hydraulic Design - Piping
Piping pressure loss
Pressure loss, ∂P = f L ρ vf2 / (2 d) (units Pa)
Where
L : piping length (units meter)
ρ = 900 Kg/m3 : oil density
vf = Q / (π d2 / 4): oil speed (units meter/s)
d : piping internal diameter (units meter)
f = 64 / Re (for Re < 2000, laminar flow)
Re = vf d / v : Reynolds number
v = 50 x 10-6 m2/s (50 cSt) : oil viscosity
Q : maximum oil flow (units m3 / s), during the bin descent
Hose
A
B
C
0.0191
0.0159
0.0159
0.0159
0.0159
0.0159
2
4
0.5
7
4
7
0.0015
0.0007
0.0007
0.0004
0.0007
0.0004
speed, vf
0.56
0.37
0.37
0.20
0.37
0.20
Re
197
118
118
62
118
62
f
0.325
0.541
0.541
1.024
0.541
1.024
∂P
4072
8479
1060
7835
8479
7835
diameter,
d
length, L
Flow, Q
D
E
F
21
2.12 Hydraulic Design - Piping
The maximum pressure losses, 8.5 kPa (1.2 psi) are in hoses B and E .
Theses losses are very low because the bin motion is slow.
Hose A has an internal diameter of 19.1 mm (3/4”) and a pressure rating
of 105 Bar (1500 psi).
All other hoses have an internal diameter of 15.9 mm (5/8”) and a
pressure rating of 130 Bar (1885 psi).
22
2.13 Hydraulic Design - Lift diagram
This diagram was built by selecting valves with the required flow and
pressure settings.
23
2.14 Hydraulic Design - Descent diagram
This diagram was built by selecting valves with the required flow and
pressure settings.
24
2.15 Hydraulic Design - Valve selection
Counterbalance valve
To be located as close as possible to the actuator.
BIN LIFT
Outlet pressure: Pa = 5127 kPa (743 psi)
Maximum free flow: (2.34 L/min) (0.62 US gal/min)
Pressure drop: Pd2 = 300 kPa (44 psi) (manufacturer’s data)
BIN DESCENT
Operating pressure: Pv = 5229 kPa (758 psi)
Inlet pressure: Pc out = 4306 kPa (625 kPa)
Maximum regulated flow: 4.4 L/min (1.2 US gal/min),
Pilot ratio: 3 (for variable load)
25
Flow combiner/divider valve
BIN LIFT
Outlet pressure: (5127 kPa) + Pd2 = 5427 kPa (787 psi)
Maximum regulated flow: (4.68 L/min) (1.24 US gal/min)
BIN DESCENT
Maximum regulated flow: 8.8 L/min (2.4 US gal/min),
Inlet pressure: (1600 kPa) + Pd3 = 2800 kPa (406 psi)
26
Manual 4/3 valve
BIN LIFT
(A) pressure: (5427 kPa) + Pd4 = 5527 kPa
(802 psi)
(P-A) flow (P): (4.68 L/min) (1.24 US gal/min)
(P-A) Pressure drop: Pd6 = 300 kPa (44 psi)
(manufacturer’s data)
(P) pressure: (5527 kPa) + Pd6 = 5827 kPa
(845 psi)
(B-T) flow (P): (2.4 L/min) (0.66 US gal/min)
(B-T) Pressure drop: Pd6 = 150 kPa (22 psi)
(B) pressure: 150 kPa (22 psi)
BIN DESCENT
(A-T) flow: 8.8 L/min (2.4 US gal/min),
(A-T) Pressure drop: Pd5 = 1600 kPa (232 psi)
(A) pressure: 1600 kPa (232 psi)
(P-B) flow: 4.68 L/min (1.24 US gal/min),
(P-B) Pressure drop: Pd5 = 700 kPa (102 psi)
(P) pressure: Pc in + Pd5 = 872 kPa (126 psi)
27
2.18 Hydraulic Design - Hydraulic unit
A typical hydraulic power unit
has the following components:
Pump.
Electric motor.
Oil tank.
Pressure relief valve.
Suction strainer.
Pressure gage.
Oil level gage.
Breather and fill cap.
Drain plug.
Return filter (option).
Heat exchanger (option).
…
28
Pump and Motor
The pump has a displacement of D = 2.8 x 10-6 m3 (0.17 in3) per revolution
and an efficiency of n = 0.73 (manufacturer’s data). The driving motor is
turning at 1750 rpm.
This displacement at that speed gives the required flow:
Q = 1750 D = 0.0048 m3/min (4.8 L/min)
Its operation cycle has 3 phases:
1. LIFT
•
•
•
•
Duration, t1 = 40 sec.
Flow, Q1 = 4.68 L/min (1.24 gal/min)
Maximum pressure, P1 = 5827 kPa (845 psi)
Power, H1 = (Q1/ 60000)(1000 P1) / n = 623 Watts (0.84 Hp)
2. DUMPING (all pump flow goes back to the tank through the relief valve)
•
•
•
•
•
Relief valve pressure, P2 = 6000 kPa (870 psi)
Input torque, T2 = (1000 P2 D)/ (2 π n) = 3.64 N*m (32.2 lbf-in)
3. DESCENT
•
•
•
•
Maximum pressure, P3 = 872 kPa (126 psi)
29
Pump and Motor
Root mean square power, Hrms, is the average power consumed through the
complete operation
Hrms = { (H12 t1 + H22 t2 + H32 t3) / (t1 + t2 + t3) }0.5 = 518 Watts (0.7 Hp)
The selected motor is a 0.75 kW (1 Hp) totally-enclosed fan-cooled AC
electric motor. Its starting torque must be above T2.
30
Tank
Besides its function as an hydraulic oil reservoir, the tank provides:
•Enough volume to let returning fluid slow down. This lets heavier
contaminants settle and entrained air escape.
•Surface area to transfer heat from the fluid to the environment.
•Air space above the fluid to accept air that bubbles out of the fluid.
•A baffle that separates fluid entering the reservoir from fluid entering the
pump suction line.
Tank minimum volume, Vtmin = Q1 * (3 minutes) = 14 L (3.72 US Gal)
Tank actual volume, Vt = 19 L (5 US Gal)
In our application, the oil velocity is too low to require the installation of a
heat exchanger.
31
3.0 Dumper structure design & stress calculation
Overview
The dumper structure has two subassemblies: the lifting bin and the fixed
base. Both subassemblies are made of welded hollow rectangular steel
tubes. The anchors for the hydraulic actuators are made of thick steel plate
welded to the tubes.
In section 3, the material stresses will be calculated in all critical
components. Using finite elements, the calculation will be done by the MS
Excel file PackexOneToteDumperDesign.xls.
ACCESS TO THE STRESS CALCULATION IS GRANTED ONLY TO
SUBSCRIBERS.
LIFTING BIN
BASE
32
Lifting Bin
The lifting is made of hollow rectangular tubes welded together. The tote
rests on a steel plate floor and leans on a steel plate wall during the lift.
wall
tubes
floor
anchor
anchor
33
Lifting Bin
The components which support the greatest stresses are:
1.The tubes (trusses) forming the sides of the bin.
2.The floor plate.
Stress calculation for the tubes are done using finite element method.
Although even a brief explanation of this method is beyond the scope of this
tutorial, the following figure gives an overview.
Finite element method for trusses.
A truss is defined as an element bound by two nodes. A node contains the
coordinates of the element extremity and an identification number.
Element #1
Node #2
Force F
F1, x1
F2, x2
Node #1
F1, F2 are the internal forces and x1, x2 are the displacements (strains)
associated with nodes 1 and 2 and caused by external force F.
In this example, x1 = 0 and F2 = F. The strain x2 is proportional to the
element rigidity and F1 = -F2. Solving complex truss assemblies is done by
finding the stiffness (rigidity) matrix of the assembly, applying known loads
and displacements to resolve the unknown forces and displacements.
Stresses are calculated with the internal forces in each element.
34
Lifting Bin - Trusses
Each side of the bin is a weld assembly of 4 tubes called trusses, it supports
half of the combined tote and bin weight. The forces acting on this assembly
are shown below.
In above figure, nodes are tagged 1 to 4. Trusses are identified as:
Truss #1: node 1 and node 2
35
Using file PackexOneToteDumperDesign.xls, user can enter the tote weight,
the bin approximated weight, the nodes coordinates, the truss section
dimensions.
36
By clicking on the blue arrow, all internal forces of each truss are calculated
for different inclinations of the lifting bin (0° tote loading, 120° tote dumping).
The internal forces are the axial load, the shear load, the bending moment.
The maximum stress is calculated from the combination of all theses forces.
Shear load
Axial load
Bending moment
37
Lifting Bin – Floor plate
The steel floor plate supports completely the tote weight when the bin is in
loading position. The maximum stresses are on the perimeter at the
extremities the rectangular tubes.
Maximum
stress
Floor plate
Maximum stress, σ, for a square plate with an uniform surface load:
Surface load, Pp = (Wt + Wp ) / (a2) = 3.2 kPa (0.46 psi)
Where Wt = 400 Kg (tote weight), Wp = 126 Kg (plate weight), a = 1270 mm
(plate length and width)
Stress, σ = -0.511 Pp a2 / t2 = -26376 kPa (-3825 psi), compressive stress
Where t = 10 mm (plate thickness)
38
Base - Trusses
Each side of the base is a weld assembly of 3 tubes called trusses, it
supports half of the combined tote and bin weight. The forces acting on this
assembly are shown below.
F5y
F5x
F8y
F8x
In above figure, nodes are tagged 5 to 8. Trusses are identified as:
39
Base - Trusses
Using file PackexOneToteDumperDesign.xls, user can enter the nodes
coordinates, the truss section dimensions for the base. By clicking on the
blue arrow, all internal forces of each truss are calculated for different
inclinations of the lifting bin (0° tote loading, 120° tote dumping). The internal
forces are the axial load, the shear load, the bending moment. The
maximum stress is calculated from the combination of all theses forces.
40

Tote dumper design

Transcription

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