Knowledge Center

Design & beräkningar

Practical answers for wave spring and retaining ring selection, installation, materials and troubleshooting.

298 Published questions
2 Core topics
1:1 Question intake

If the published answers do not match your application, send us your question and our team will review it.

Questions & Answers

Engineer-reviewed questions

Loading questions...

A Reference Answer

To prevent rolling failure, the groove depth $d$ must be sufficient to ensure the ring remains seated even when subjected to the maximum expected thrust load. The rule of thumb is that the ring should be seated at least $1/3$ to $1/2$ of its radial wall thickness $b$ into the groove. Mathematically, the stability is governed by the ratio of the groove depth to the ring's thickness. For a standard safety factor of $2.0$, the minimum groove depth $d_{min}$ is calculated based on the groove material's yield strength $\sigma_y$ and the thrust load $P$: $d_{min} = (P \cdot S) / (D \cdot \pi \cdot \sigma_y)$. In soft materials like plastics or aluminum, the groove must be significantly deeper than in hardened steel to prevent the material from shearing or 'plowing' under the ring's edge.

A Reference Answer

The radial wall $b$ is the width of the flat wire used to coil the spring. In the stress equation $\sigma = (6 \cdot P \cdot D_m) / (n^2 \cdot b \cdot t^2)$, the stress is inversely proportional to $b$. A wider radial wall reduces the bending stress for a given load $P$, thereby increasing the fatigue life. However, a wider $b$ also increases the spring rate $k$ and the total solid height of the spring. Furthermore, if $b$ is too large, the 'radial expansion' during compression becomes more pronounced, increasing the risk of interference with the housing. Designers must balance $b$ to achieve the desired $k$ while keeping stresses within the allowable limits of the Goodman diagram for the selected material.

A Reference Answer

The shear strength of a spiral retaining ring is the maximum axial thrust load it can withstand before the material is physically sheared through its cross-section. It is calculated as $P_s = \pi \cdot D \cdot t \cdot \tau_{ult}$, where $D$ is the shaft/bore diameter, $t$ is the ring thickness, and $\tau_{ult}$ is the ultimate shear strength. $\tau_{ult}$ is typically estimated as $0.6$ to $0.75$ times the ultimate tensile strength $S_{ut}$ for most steels. Unlike tensile strength, which measures the resistance to pulling apart, shear strength measures the resistance to adjacent planes of the material sliding past one another. In ring design, the shear capacity of the ring usually exceeds the yield capacity of the groove, making the groove the critical design limit.

A Reference Answer

In the spring rate formula $k = (E b t^3 n^4) / (D_m^3)$, the mean diameter $D_m$ is raised to the third power. This means that $k$ is inversely proportional to $D_m^3$. If the tolerance on the diameter is large, the resulting spring rate can vary significantly. For instance, a $2\%$ increase in $D_m$ results in approximately a $6\%$ decrease in $k$. In precision medical devices, where constant force is required, $D_m$ must be tightly controlled through specialized coiling techniques. Furthermore, $D_m$ changes as the spring is compressed; as the waves flatten, the spring expands radially. If the design does not account for this expansion (the 'breathing' of the spring), the spring will bind against the housing, causing the rate $k$ to become infinite.

A Reference Answer

Axial play is the total possible movement of the retained components along the shaft or housing axis. It is calculated as $Play = G_w - (T_{ring} + \sum T_{comp})$, where $G_w$ is the groove width, $T_{ring}$ is the ring thickness (including its dish/flatness tolerance), and $\sum T_{comp}$ is the sum of the tolerances of all retained components. In high-precision optical or aerospace assemblies, axial play must be minimized to prevent vibration-induced wear. This is often managed by using a wave spring in tandem with the spiral ring to provide a constant axial preload, or by selecting a 'balanced' spiral ring which has a more uniform thickness across its circumference, reducing the 'snaking' effect within the groove.

A Reference Answer

Wave springs are manufactured from cold-rolled flat wire rather than round wire. The cold-rolling process increases the dislocation density within the crystalline lattice, significantly raising the yield strength $\sigma_y$ and hardness, but reducing ductility. This work-hardening is critical for achieving high spring rates in small envelopes. The final spring rate $k$ is sensitive to the resulting thickness $t$ due to the cubic relationship $k \propto t^3$. A variation of just $0.01$ mm in thickness can result in a $10-15\%$ change in load. For fatigue life, the surface finish from rolling is superior to drawn wire, reducing crack initiation sites, provided that the subsequent heat treatment (stress relieving) is performed to stabilize the structure without excessive decarburization.

A Reference Answer

The maximum RPM $N$ for a spiral retaining ring is limited by the point where centrifugal forces overcome the ring's radial cling on the groove bottom. The formula is $N = \sqrt{(448 \cdot E \cdot I \cdot g) / (\rho \cdot A \cdot R_m^3 \cdot (R_o - R_i))}$ where $E$ is the modulus, $I$ is the moment of inertia, $g$ is gravity, $\rho$ is material density, and $R_m$ is the mean radius. As the ring rotates, the center of mass generates a radial force $F_c = m \cdot \omega^2 \cdot r$. For external rings, this force acts to expand the ring. Engineering designs must ensure a safety factor of at least $2.0$ relative to the operating RPM. If the calculated limit is exceeded, designers should specify a 'Self-Locking' feature, which utilizes a tab-and-slot mechanism to mechanically prevent diameter expansion under high angular velocity.

A Reference Answer

The theoretical spring rate $k$ for a multi-turn Crest-to-Crest wave spring is derived from the formula $P/f = (E b t^3 n^4) / (D_m^3 N Z)$, where $E$ is the Modulus of Elasticity, $b$ is the radial width, $t$ is the material thickness, $n$ is the number of waves per turn, $D_m$ is the mean diameter, and $N$ is the number of turns. The inclusion of shim ends introduces a correction factor to the active number of turns. Non-linearity typically occurs as the spring approaches its solid height, usually around $80\%$ of the total available travel. This is caused by the 'bottoming out' effect where the wave peaks begin to flatten against the contact surfaces, effectively reducing the active length of the beam and increasing the spring rate exponentially. In high-precision applications, designers must account for the expansion of the radial wall $b$ during compression using $D_{outer-max} = D_{outer} + (0.015 \cdot f \cdot n^2 / D_m)$ to prevent binding in the housing.

A Reference Answer

The total axial displacement $\delta_{total}$ is the sum of the initial clearance, the ring deflection, and the groove deformation. $\delta_{total} = (W - t) + \frac{P}{k_{ring}} + \frac{P}{k_{groove}}$, where $W$ is the groove width and $t$ is the ring thickness. If the groove is too wide, the ring can 'tilt', which increases the effective clearance and reduces the contact area. This tilting leads to a non-uniform pressure distribution $p(x)$, which can exceed the yield strength of the housing at the outer edge. For precision assemblies, 'end-play' is minimized by using thicker rings or shims to achieve a 'snug' fit.

A Reference Answer

As a wave spring approaches its solid height $L_s = n \times t$, the load-deflection curve deviates from the linear $P = kf$. This is due to 'wave flattening' where the contact area between waves increases, effectively shortening the active beam length. The spring rate $k$ increases exponentially. Designers must account for this 'stop' mechanism. If the system requires a hard stop, the spring must be designed such that the working load is reached before $80\%$ deflection. Exceeding this causes excessive stress $\sigma$ and can lead to 'coining' of the material, where the wave peaks are permanently flattened.

A Reference Answer

For a spiral retaining ring, the shear area is not simply the circumference times thickness. Because it is a continuous spiral, the shear area $A_s = \pi D t \times (n)$ where $n$ is the number of turns. For a standard 2-turn ring, the shear area is $2\pi D t$. The axial load capacity $P_r$ is then $P_r = \frac{A_s S_s}{K}$. Interestingly, while a 2-turn ring has twice the shear area of a 1-turn ring, the groove yield $P_g$ often remains the bottleneck. Therefore, adding turns increases the ring's strength but does not help if the housing material is the weak link.

A Reference Answer

Edge-winding involves bending flat wire on its edge, which significantly work-hardens the material before it is even formed into waves. This increases the initial tensile strength but reduces the remaining ductility. For materials like 302 SS, the modulus $E$ can shift slightly, and the internal residual stresses $\sigma_{res}$ must be relieved through a stress-relieving heat treatment (e.g., $315^\circ C$ for 30 minutes). If not relieved, the 'Springback' is unpredictable, leading to inconsistent free heights $H$ and loads $P$. The design must assume the 'as-heat-treated' properties for accurate load prediction.

A Reference Answer

The 'Grip' or seating force of a retaining ring is determined by its radial wall $b$ and the amount of interference between the ring's free ID and the groove diameter. The radial pressure $q$ is given by $q = \frac{2 E I (D_g - D_f)}{D_m^2 b}$. Since $I = \frac{t b^3}{12}$, the grip force is proportional to $b^3$. A larger radial wall significantly increases the force required to expand the ring, which improves the maximum RPM limit but makes installation more difficult and increases the risk of over-stressing the material during assembly.

A Reference Answer

The spring rate $k$ for a wave spring is proportional to the cube of the number of waves $N$. From the formula $P/f = k = \frac{E b t^3 N^4 K}{D_m^3}$, we see that doubling the number of waves from $N=3$ to $N=6$ increases the stiffness by a factor of 16 ($2^4$), assuming all other parameters are constant. This extreme sensitivity allows designers to fine-tune loads in very small increments. However, increasing $N$ reduces the maximum possible deflection $f$ before the waves interfere with each other, necessitating a balance between load capacity and stroke.

A Reference Answer

The axial thrust capacity based on ring shear is calculated as $P_r = \frac{D t \pi S_s}{K}$ where $D$ is the shaft/bore diameter, $t$ is the ring thickness, $S_s$ is the shear strength of the ring material, and $K$ is the safety factor (usually 3). For most spring steels, $S_s \approx 0.6 \times S_{ut}$ (ultimate tensile strength). It is critical to compare $P_r$ (ring shear) with $P_g$ (groove yield). The lower value determines the system's limit. In high-impact applications, $K$ should be increased to 5 or higher to account for dynamic loading and potential fatigue at the groove interface.

A Reference Answer

Nested wave springs consist of multiple turns wound in parallel. The total load $P_{total}$ is the product of the number of turns $n$ and the load of a single turn $P_s$, such that $P_{total} = n \times \frac{E b t^3 N f K}{D_m^3}$. However, friction between the layers must be accounted for, typically introducing a hysteresis loop in the load-deflection curve. The thickness $t$ in the stress equation $\sigma = \frac{3 \pi P D_m}{2 N^2 b t^2}$ refers to the individual layer thickness. This configuration is ideal for applications like heavy-duty valve seals where space is restricted but $P$ must be very high ($> 5000$ N).

A Reference Answer

Centrifugal lifting occurs when the centrifugal force exceeds the ring's grip on the groove. The maximum RPM $N$ is calculated using $N = \sqrt{\frac{0.48 C_1 E I g}{\mu R^3 (1+S) V_g}}$ where $E$ is the modulus, $I$ is the moment of inertia, $\mu$ is the mass per unit length, and $V_g$ is the volume. For an external ring in SAE 1070 carbon steel, once the speed reaches the point where the radial expansion $\Delta D = \frac{12 \rho \omega^2 R^4}{E g}$ exceeds the groove depth, the ring loses its axial retention capability. High-speed applications often require 'self-locking' features where a tab engages a slot to mechanically prevent expansion.

A Reference Answer

The theoretical load $P$ for a Crest-to-Crest wave spring is derived from the beam deflection formula adjusted for the sinusoidal geometry. The standard equation is $P = \frac{E b t^3 N f K}{D_m^3 n}$ where $E$ is the Modulus of Elasticity, $b$ is the radial wall, $t$ is the material thickness, $N$ is the number of waves per turn, $f$ is the deflection, $D_m$ is the mean diameter, and $n$ is the number of turns. The Modulus of Elasticity $E$ is critical as it defines the stiffness; for instance, using 17-7PH CH900 ($E \approx 200$ GPa) versus Inconel X-750 ($E \approx 213$ GPa) significantly alters the spring rate $k = P/f$. The linear range is typically maintained between $20\%$ and $80\%$ of the available deflection before bottoming out or entering the non-linear high-stress zone where $f > 0.8(h-t)$.

A Reference Answer

Static thrust formulas do not account for 'impact' or 'shock' loads where the kinetic energy $E_k = \frac{1}{2} m v^2$ must be dissipated. For impact loading, the effective capacity of a spiral ring is reduced by $50\%$ or more. The designer must ensure that the ring does not undergo dynamic dishing. The impact capacity is often tested by drop-weight methods. To improve impact resistance, a 'heavy-duty' spiral ring with increased material thickness $t$ and a self-locking feature is used to prevent the ring from momentarily expanding and 'jumping' the groove during a high-G shock event, common in downhole drilling jars.

A Reference Answer

Under thrust load, all spiral rings exhibit some degree of 'dishing' as the radial wall $w$ twists. The allowable dish angle $\theta$ is typically limited to $6-10$ degrees. If $\theta$ exceeds this, the ring may lose its grip on the groove. The dish angle can be estimated by $\theta \approx \frac{M}{EI} L$, where $M$ is the moment caused by the thrust load offset. For heavy-duty applications, using a 'multiple-turn' ring (e.g., 3-turn vs 2-turn) increases the torsional stiffness and reduces the dish angle, thereby increasing the effective thrust capacity and safety margin against roll-out.

No matching questions

TOP