Knowledge Center

การออกแบบและการคำนวณ

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

The shear capacity of the ring $F_r$ is determined by the formula $F_r = D t \pi \tau_{ult} / K$, where $t$ is the ring thickness, $D$ is the groove diameter, and $\tau_{ult}$ is the ultimate shear strength of the material. For most spring steels, $\tau_{ult} \approx 0.6 \cdot \sigma_{uts}$. In applications involving high-frequency impact or shock loads, the safety factor $K$ should be increased to at least 3. If $F_r$ is less than the required service load, the ring thickness must be increased, or a multi-turn ring must be specified to distribute the shear over a larger area.

A Reference Answer

Thrust capacity is often limited by the groove material rather than the ring itself. The allowable thrust load $F_g$ is given by $F_g = \frac{D %d \pi \sigma_y}{K}$, where $D$ is the shaft/bore diameter, $d$ is the groove depth, $\sigma_y$ is the yield strength of the groove material, and $K$ is a safety factor (typically 2). For soft materials like Aluminum 6061-T6, the groove will often deform plastically at the edge, causing the ring to 'dish' or tilt. This tilting reduces the effective shear area, leading to premature ejection. Engineers must ensure the groove depth $d$ is sufficient to keep the compressive stress below the yield point.

A Reference Answer

For external rings, centrifugal force causes the ring to expand, potentially losing its grip on the groove. The maximum RPM is calculated as $N_{max} = \sqrt{\frac{4.48 \cdot 10^{11} E I}{w \rho R_m^3 (R_o - R_i)}}$, where $w$ is the weight of the material per unit length and $\rho$ is the density. If the application speed exceeds $N_{max}$, the ring must be designed with a 'Self-Locking' feature. This involves a tab-and-slot mechanism that mechanically prevents the ring from expanding beyond the groove diameter, allowing it to withstand speeds that would otherwise cause a standard spiral ring to fail.

A Reference Answer

Shim ends, also known as flat ends, provide a $360^{\circ}$ contact surface compared to the point contact of plain ends. This feature significantly reduces the localized contact stress on mating components, which is vital when the spring interfaces with soft materials like Aluminum or plastics. Mathematically, the shim end acts as a rigid boundary condition, improving the stability of the stack and ensuring that the load $P$ is distributed uniformly across the circumference. However, the addition of shim ends increases the solid height $H_s = (Z+2)t$, which must be accounted for in the axial space claim.

A Reference Answer

When a wave spring is compressed, the waves flatten, causing a slight increase in the outside diameter ($OD$). The expansion can be approximated by the formula $\Delta OD = 0.02 \cdot \frac{f \cdot (OD + ID)}{N^2}$, where $f$ is the deflection and $N$ is the number of waves. For springs operating in a bore, this expansion is critical; if the clearance between the $OD$ and the bore is insufficient, the spring will bind, causing a sudden non-linear increase in spring rate and potential fatigue failure. Engineers should specify a bore diameter that accounts for both the manufacturing tolerance and this functional expansion.

A Reference Answer

Nested wave springs are produced by coiling multiple layers of wire in parallel rather than end-to-end. The load $P$ produced by a nested spring is proportional to the number of turns $n$ such that $P_{nested} = n \cdot P_{single}$. This configuration is used when high forces are required in extremely tight radial and axial envelopes. Unlike Crest-to-Crest springs, where turns act in series to increase deflection and decrease rate, Nested turns act in parallel to increase load for a given deflection. The total spring rate $k_{total} = \frac{E b t^3 N^4 n}{4 D_m^3}$, making them ideal for preloading high-capacity bearings in aerospace actuators.

A Reference Answer

Stress at work height is calculated using the formula $S = \frac{3 \pi P D_m}{4 b t^2 N^2}$, where $P$ is the load at work height. For standard applications, the calculated operating stress should not exceed the minimum tensile strength of the material to avoid permanent set. In high-performance alloys like 17-7PH CH900, the allowable stress can reach up to $80\%$ of the tensile strength. However, if the spring is compressed to 'solid height', the stress often exceeds the yield point, leading to plastic deformation. Engineers must use the formula for solid height stress $S_s = \frac{E t N^2 f}{D_m^2}$ to determine if a stop is required in the assembly to prevent over-compression.

A Reference Answer

The spring rate $k$ for a multi-turn Crest-to-Crest wave spring is derived using a modified beam deflection formula for curved segments. The fundamental equation is $k = \frac{E b t^3 N^4}{4 D_m^3 Z}$, where $E$ is the Young's Modulus, $b$ is the radial wall width, $t$ is the material thickness, $N$ is the number of waves per turn, and $Z$ is the number of active turns. It is critical to note that the rate is inversely proportional to the number of turns $Z$, meaning doubling the turns halves the rate, while the rate increases with the fourth power of the number of waves $N$. Engineers must also apply a correction factor $K$ for curvature when the ratio of mean diameter to radial wall is low, typically $D_m/b < 8$.

A Reference Answer

Unlike external rings, internal rings are 'pushed' into the groove by centrifugal force, which actually increases their security. However, at extreme speeds, the centrifugal force can cause the ring to expand so much that the hoop stress $ ext{σ}_h = _x000d_ho imes v^2$ exceeds the material's yield strength. If the ring plastically expands, it will lose its 'set' and may not contract back into the groove when the rotation stops, leading to failure during the next startup. The design limit is typically set so that the centrifugal stress remains below $70\%$ of the material's yield strength $S_y$.

A Reference Answer

A spiral retaining ring is made from coiled flat wire, which means it has no 'ears' or lugs, providing a full $360°$ contact surface. The shear capacity $P_s$ is calculated as $P_s = _x000c_rac{D imes T imes ext{π} imes ext{τ}_s}{FOS}$, where $T$ is the total ring thickness and $ ext{τ}_s$ is the shear strength of the material (approx. $0.6 imes UTS$). Unlike stamped circlips, which can have stress concentrations at the lug holes, spiral rings distribute the shear load uniformly. However, for multi-turn rings, the calculation must ensure the load is shared across all turns, which requires the groove to be deep enough to support the entire radial wall $b$ of the ring.

A Reference Answer

Dishing occurs when a spiral retaining ring is subjected to high axial loads that cause it to tilt within the groove. This creates a moment $M$ on the ring's cross-section. The resulting stress is a combination of the initial installation stress and the applied bending stress. The dish angle $ heta$ is proportional to the applied load and inversely proportional to the ring's stiffness $E I$. As the ring dishes, the load is no longer applied uniformly, concentrating the stress on the inner edge of the ring. This can lead to permanent 'cupping' of the ring. Engineers use a safety factor, typically $2$ or $3$, to ensure that the ring stays within the elastic range to prevent this geometry-induced failure.

A Reference Answer

The axial load capacity of a retaining ring assembly is often limited by the shear strength of the groove material rather than the ring itself. The maximum thrust load $P_g$ based on groove yield is $P_g = _x000c_rac{D imes d imes ext{π} imes ext{σ}_y}{K imes FOS}$, where $D$ is the shaft/bore diameter, $d$ is the groove depth, and $ ext{σ}_y$ is the yield strength of the housing material. If the axial load exceeds $P_g$, the groove 'lips' will deform plastically, causing the ring to 'dish' (tilt). This dishing reduces the effective contact area and leads to the ring being 'cammed' out of the groove, a common failure in aluminum housings.

A Reference Answer

The maximum RPM for an external ring is limited by centrifugal force, which causes the ring to expand radially. The speed $N$ at which the ring begins to lift off the groove is given by $N = _x000c_rac{1}{ ext{rad/sec conversion}} imes _x000c_rac{1}{D_m} imes _x000c_rac{ ext{Groove Depth}}{ ext{Expansion Constant}}$. More formally, the critical velocity $V_c$ is derived by balancing the centrifugal force $F_c = m imes _x000c_rac{v^2}{R}$ against the ring's internal elastic grip. The formula used by engineers is $V = _x000c_rac{1}{ ext{π} D_i} imes ext{sqrt} _x000c_rac{48 E I g riangle}{w _x000d_ho A R_m^4}$, where $ riangle$ is the radial clearance between the ring and the groove, $_x000d_ho$ is the density, and $I$ is the moment of inertia. Above this speed, the ring loses its 'cling' and can be ejected.

A Reference Answer

17-7PH (UNS S17700) is semi-austenitic and undergoes significant work hardening during the cold-rolling and coiling phases. The transformation from austenite to martensite is strain-induced. During the 'on-edge' coiling of a wave spring, the outer fibers of the flat wire experience higher strain, leading to a gradient of hardness across the cross-section. This results in a non-uniform residual stress state that must be relieved during the CH900 heat treatment process. If the coiling tension is not precisely controlled, the resulting wave heights will vary, leading to a load tolerance deviation. Precision springs often require a 'preset' operation where they are compressed to solid height to stabilize the metallurgical structure and reduce subsequent load loss in service.

A Reference Answer

The stability of a wave spring is inversely proportional to the wave period. A higher number of waves $n$ increases the radial stiffness and reduces the likelihood of lateral buckling. However, as $n$ increases, the spring rate $k$ increases by a power of 4 ($n^4$), making the spring much stiffer. For multi-turn springs, the ratio of free height $H_f$ to mean diameter $D_m$ is critical; if $H_f/D_m > 1.5$, internal or external guidance (a rod or a bore) is mandatory. The buckling limit is reached when the axial load induces a tangential stress that exceeds the critical buckling load $P_{cr} = _x000c_rac{_x0008_eta imes E imes I}{L^2}$, modified for the periodic geometry of the waves.

A Reference Answer

Nested wave springs involve multiple layers of flat wire coiled in parallel. The total load $P_{total}$ is the sum of the loads of each individual turn $n$, effectively $P_{total} = n imes P_{single}$. However, the stress calculation must account for the friction between layers. The primary bending stress is $ au = _x000c_rac{3 imes _x000d_ho imes P imes D_m}{2 imes b imes t^2 imes n^2}$, but in nested configurations, a correction factor $K$ for curvature and inter-layer contact is applied. The nested design allows for extremely high loads in a compact radial space, often used in automotive torque converters. Stress relaxation must be evaluated carefully, as the inner layers experience slightly higher compressive loads due to the radius of curvature variations during the winding process.

A Reference Answer

As a wave spring is compressed from its free height $H_f$ toward its solid height $H_s$, the waves flatten, causing the mean diameter $D_m$ to expand. This radial expansion can be approximated by $ riangle OD = 0.045 imes _x000c_rac{w^2 imes f}{D_m}$, where $w$ is the wave height and $f$ is the deflection per wave. In high-precision aerospace valves, ignoring this expansion leads to 'bore binding,' where the spring OD interferes with the housing ID. This creates parasitic friction, hysteresis in the load-deflection curve, and potential mechanical galling. Engineers must specify a clearance ratio, typically ensuring the housing $ID > OD_{max}$ at the full work height.

A Reference Answer

The spring rate $k$ for a crest-to-crest wave spring is fundamentally derived from the deflection of a curved beam. For a spring with $N$ turns and $n$ waves per turn, the load $P$ is expressed as $P = _x000c_rac{E imes b imes t^3 imes n^4 imes f}{C imes D_m^3 imes N}$, where $E$ is the Young's Modulus, $b$ is the radial wall, $t$ is the thickness, $f$ is the deflection, and $D_m$ is the mean diameter. The constant $C$ varies based on end conditions. Shim ends provide a flat surface for load distribution, which effectively adds two non-functional half-waves to the stack. While this increases the solid height $H_s = (N imes t) + (2 imes t_{shim})$, it significantly improves the linearity of the spring rate at the beginning and end of the stroke by preventing the 'point-loading' of wave peaks against the mating surfaces.

A Reference Answer

In a perfect theoretical model, grooves have $90^\circ$ sharp corners. In reality, tools have a radius $R$, and the mating component often has a chamfer $C$. These features reduce the effective contact area between the ring and the groove wall. The thrust capacity must be derated using a factor $K$. If the chamfer on the retained part is too large, it creates a 'wedge' effect that tries to expand the ring (bore type) or contract it (shaft type) out of the groove. As a rule of thumb, the maximum allowable chamfer or radius is $0.5 \times d$ (groove depth). If the chamfer exceeds this, a backup washer must be used to provide a square face for the ring.

A Reference Answer

Dish or coning refers to the axial deflection of the ring when subjected to a thrust load. Because a spiral ring is essentially a wound flat wire, it behaves like a very stiff truncated cone under load. The 'Moment of Inertia' of the wire cross-section $I = _x000c_rac{b t^3}{12}$ resists this twisting. If the load is eccentric or the groove is not square, the ring will 'dish' more severely. Excessive dishing reduces the effective shear area and can cause the ring to 'walk' out of the groove. We calculate the maximum allowable load before the ring 'dishes' past a critical angle (usually $7^\circ$) using the formula $M = E I \theta / R$, where $M$ is the applied moment from the thrust load.

No matching questions

TOP