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- > No.000307 Transfer with Automatic Tension Mechanism
No.000307 Transfer with Automatic Tension Mechanism
89
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The slack of the belt is absorbed by an automatic tension mechanism.
Relevant category
Cylinder with Guides
| Product name | Cylinder with Guides |
|---|---|
| Part number | MGCLF50-100 |
* Orange colored cells in the table below indicate the part numbers used in this example.
Selection criteria
Use this item to apply steady tension to chains.
Available sizes
■Cylinder with Guides
| Cyl. Bore | Bearing | Sliding Bearing | Linear Bushing | Cylinder Width (mm) | Overall Length (Listed below + St) | Thickness (mm) | ||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| St | 10 | 20 | 25 | 30 | 40 | 50 | 75 | 100 | 10 | 20 | 25 | 30 | 40 | 50 | 75 | 100 | ||||
| φ12 | ○ | ○ | ― | ○ | ○ | ○ | ○ | ○ | ○ | ○ | ― | ○ | ○ | ○ | ○ | ○ | 56 | 39 | 22 | |
| φ16 | ○ | ○ | ― | ○ | ○ | ○ | ○ | ○ | ○ | ○ | ― | ○ | ○ | ○ | ○ | ○ | 62 | 43 | 25 | |
| φ20 | ― | ○ | ― | ○ | ○ | ○ | ○ | ○ | ― | ○ | ― | ○ | ○ | ○ | ○ | ○ | 72 | 47 | 30 | |
| φ25 | ― | ○ | ― | ○ | ○ | ○ | ○ | ○ | ― | ○ | ― | ○ | ○ | ○ | ○ | ○ | 86 | 47.5 | 38 | |
| φ32 | ― | ― | ○ | ― | ― | ○ | ○ | ○ | ― | ― | ○ | ― | ― | ○ | ○ | ○ | 112 | 48 | ||
| φ50 | ― | ― | ○ | ― | ― | ○ | ― | ○ | ― | ― | ― | ― | ― | ― | ― | ― | 146 | 72 | 60 | |
Selection steps
■Cylinder Selection Steps
1. Define various conditions of the load.
↓
2. Calculate cylinder output (In case of double acting type)
↓
3. Determine the cylinder bore
↓
4. Determine the theoretical reference speed.
↓
5. Verify the cylinder's cushion mechanism.
↓
6. Verify the lateral load applied to the cylinder.
Accuracy Info
■Non-rotating Accuracy of Cylinder with Guides
Non-rotating accuracy: A value of runout due to clearances between the guide rod and bearings represented in an angle of rotation about the piston rod.
| Cyl. Bore | Non-rotating Accuracy of Link Bar Tip | |
|---|---|---|
| Sliding Bearing | Linear Bushing | |
| φ12 | ±0.12° | ±0.06° |
| φ16 | ±0.10° | ±0.06° |
| φ20 | ±0.09° | ±0.05° |
| φ25 | ±0.08° | ±0.05° |
| φ32 | ±0.06° | ±0.04° |
| φ50 | ±0.05° | － |
Performance info.
■Cylinder with Guides Operating Pressure Range
Min. Operating Pressure (MPa): 0.1
Max. Operating Pressure (MPa): 1.0
Pressure Resistance (MPa): 1.5
■Theoretical Thrust
（N)
| Cyl. Bore (mm) | Operating Direction | Operating Pressure (MPa) | |
|---|---|---|---|
| 0.4 | 0.5 | ||
| 12 | Outstroke | 45 | 57 |
| Instroke | 34 | 42 | |
| 16 | Outstroke | 80 | 101 |
| Instroke | 60 | 75 | |
| 20 | Outstroke | 126 | 157 |
| Instroke | 94 | 118 | |
| 25 | Outstroke | 196 | 245 |
| Instroke | 151 | 189 | |
| 32 | Outstroke | 322 | 402 |
| Instroke | 241 | 302 | |
| 50 | Outstroke | 785 | 982 |
| Instroke | 660 | 825 | |
■Allowable Lateral Load
（N・m)
| Cyl. Bore (mm) | Stroke (mm) | |||||||
|---|---|---|---|---|---|---|---|---|
| 10 | 20 | 25 | 30 | 40 | 50 | 75 | 100 | |
| 12 | 24 | 19 | 16 | 14 | 12 | 37 | 31 | |
| 16 | 40 | 33 | 28 | 24 | 21 | 55 | 46 | |
| 20 | 52 | 45 | 39 | 35 | 55 | 46 | ||
| 25 | 69 | 60 | 52 | 47 | 73 | 62 | ||
| 32 | 166 | 131 | 107 | 91 | ||||
| 50 | 296 | 245 | 241 | |||||
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Application Overview
Purpose
- Use a cylinder attached to a sprocket as a take up. The force should be applied to add tension on the belt.
Target workpiece
- Shape: aluminum block
- Size: W230 x D60 x H60mm
- Weight: 1kg
Design Specifications
Operating Conditions or Design Requirements
- Outer dimensions: W979 x D380 x H683mm
- Conveyor feed speed: 100mm/s
Required Performance
- Load: 90N
Selection Criteria for Main Components
- A motor with a torque of 3.2N·m and a moment of inertia of 3.2 x 10⁻²kgm² or higher is selected.
Design Evaluation
Verification of main components
- [Selection of motor]
Mass of chain and workpiece: M = 9kg
Sprocket diameter: D = 98mm
Sprocket weight: m = 1.4kg, number of sprockets: 6, efficiency of chain and sprocket: η = 0.8, friction coefficient of sliding surface: μ = 0.3
Friction force of sliding surface: F
F = Mgμ = 9 x 9.8 x 0.3 = 26.5N
Load torque: T
T = FD/2η = 26.5 x 98 x 10⁻³/2 x 0.8 = 1.6N
If we use 2 as the safety factor, T = 1.6 x 2 = 3.2N
Moment of inertia: J
Chain and workpiece: J 1
J1 = M{πD/2π｝² = 9{π x 98 x 10⁻³/2π}² = 2.2 x 10⁻²kgm²
Sprocket: J 2
J2 = 1/8 x mD2 = 1/8 x 1.4(98 x 10⁻³)2 = 0.17 x 10⁻²kgm²
As there are 6 sprockets,
J 3 = 0.17 x 10⁻² x 6 = 1 x 10⁻²kgm²
Total: J = J 1 +J 3 = 3.2 x 10⁻²kgm²
Other Design Consideration
- The cylinder for belt tension always applies force in the belt straining direction.
- The bracket of the idler is welded with a shaft and a plate.
Explore Similar Application Examples
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