Parallel Stack

---# DIN 2093 Parallel Stacking (Same Direction Stacking)

DIN 2093 Parallel Stacking (Same Direction Stacking)

1. Definition and Arrangement

Parallel stacking (German Parallelschaltung, English parallel stacking) refers to stacking $n$ disc springs of identical dimensions in the same direction, i.e., all disc springs have their conical surfaces oriented identically. During loading, all disc springs are compressed simultaneously, jointly bearing the external load.

Schematic arrangement (example with $n=3$):

Load F
↓
╱▔▔╲ ← 1st disc (cone facing up)
╱ ╲
╱▔▔╲ ← 2nd disc (same direction)
╱ ╲
╱▔▔╲ ← 3rd disc (same direction)
╱ ╲
──────────

2. Force‑Deflection Relationship

The core mechanical characteristic of parallel stacking is: total deflection equals that of a single disc, total load equals the sum of loads of individual discs.

Let the axial force of a single disc spring at compression $s$ be $F_{single}(s)$, then the total force for $n$ discs in parallel is:

At the same time, the total compression (deflection) equals the compression of a single disc:

Important conclusion: Parallel stacking can multiply the load capacity, but cannot increase the elastic deflection.

Flat Force

When the assembly is compressed until each disc is completely flat, the total flat force is:

where $F_{flat,single}$ is given by the Almen‑Laszlo formula:

3. Stiffness and Energy Storage

3.1 Tangent Stiffness

Since the ordinate of the force‑deflection curve is scaled by $n$ while the abscissa remains unchanged, the tangent stiffness of the combined spring at any compression $s$ is also scaled by $n$:

where $k_{single}(s)$ is the tangent stiffness of a single disc spring (see DIN 2093 stiffness section).

3.2 Elastic Energy Storage

Energy storage equals the area under the force‑deflection curve, thus also multiplied by $n$:

For the fully flat state:

4. Effect of Friction on Load

In parallel stacking, friction exists between adjacent disc springs and between the disc springs and the guide element (mandrel or sleeve). During loading, frictional resistance opposes the external force, causing the actual required axial force to be higher than the theoretical value; during unloading, friction reverses direction, and the actual released force is lower than the theoretical value. This hysteresis effect causes the actual load to be 5 %~ 15 % higher (depending on lubrication, surface quality, number of discs).

Engineering recommendations:

• Preload setting: If tightening by torque method, use the measured or manufacturer-provided loading curve; if no data is available, multiply the theoretical force by 1.1 ~ 1.15 as an approximation.
• Precision applications: Determine the actual force‑deflection curve through loading tests and compensate for friction (e.g., adequate lubrication, use of guide elements).

5. Number of Discs Limit and Load Uniformity

DIN 2093 recommends a parallel disc count $n \le 4$, for the following reasons:

1. Uneven load distribution: Due to thickness tolerances, parallelism, and coaxiality, too many discs make it difficult to ensure uniform load sharing, and individual discs may become overloaded.
2. Friction accumulation: More discs increase the number of contact surfaces, increasing the proportion of friction, and the load‑deflection characteristic deviates more severely from theory.
3. Lateral stability: Tall parallel stacks may experience lateral instability (buckling) at larger compressions.
4. Heat dissipation: In dynamic applications, internal friction generates heat, and too many discs hinder heat dissipation.

If greater load capacity is required, priority should be given to increasing disc spring dimensions (outer diameter, thickness) or using mixed stacking (parallel and series combinations), rather than simply increasing the number of parallel discs.

6. Calculation Example

Given: Single disc spring specification (DIN 2093 Series A, 20×10.2×1.5):

• Single disc flat force $F_{flat,single} \approx 11\,000\ \text{N}$
• Single disc force at compression $s = 0.5\ \text{mm}$: $F_{single} \approx 6\,400\ \text{N}$
• Single disc stiffness $k_{single} \approx 12\,800\ \text{N/mm}$ (approximate)

Requirement: Bear a load of 18 000 N with unchanged deflection.

Solution: Use 3 discs in parallel ($n = 3$)

• Total flat force $F_{flat,total} = 3 \times 11\,000 = 33\,000\ \text{N} \gg 18\,000\ \text{N}$, satisfied.
• At $s = 0.5\ \text{mm}$, theoretical total force $F_{total} = 3 \times 6\,400 = 19\,200\ \text{N} > 18\,000\ \text{N}$, acceptable.
• Considering friction amplification (take 10%), actual required force is approximately $19\,200 \times 1.1 = 21\,120\ \text{N}$.
• Total stiffness $k_{total} \approx 3 \times 12\,800 = 38\,400\ \text{N/mm}$.

Space check: Total stack height $H = n \times t = 3 \times 1.5 = 4.5\ \text{mm}$, must match bolt length.

7. Comparison with Other Stacking Methods

Summary: DIN 2093 parallel stacking achieves multiplied load capacity through same-direction stacking while keeping deflection unchanged. Design must account for force increase due to friction (5 %~ 15 %) and strictly limit the number of parallel discs (≤4). This stacking method is suitable for applications requiring high load and small deflection.