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

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A Reference Answer

The shear strength of the ring itself is based on the shear area of the material. The formula is $P_r = π · D · t · τ_{ult} / S_f$, where $D$ is the shaft/bore diameter, $t$ is the ring thickness, and $τ_{ult}$ is the ultimate shear strength of the material (typically taken as $0.6 · σ_{tensile}$ for steel). It is important to note that this calculation assumes the load is applied uniformly and that the groove and retained component have minimal radii. If the retained component has a large corner radius, it will apply the load further out on the ring, creating a moment arm that leads to 'dishing' and reduces the effective shear capacity.

A Reference Answer

The axial load capacity of a retaining ring assembly is often limited by the groove material's yield strength, not the ring itself. The maximum thrust load $P_g$ based on groove deformation is $P_g = \frac{D · d · π · σ_y}{S_f}$, where $D$ is the shaft/bore diameter, $d$ is the groove depth, $σ_y$ is the yield strength of the groove material, and $S_f$ is the safety factor (typically 2). If the groove material is soft (e.g., aluminum), the groove wall will 'dish' or deform at loads much lower than the ring's shear strength. Engineers must ensure the edge margin (the distance from the groove to the end of the shaft) is at least 3 times the groove depth to prevent 'blowout' of the groove wall.

A Reference Answer

External spiral retaining rings are limited by their ability to 'cling' to the groove at high speeds. Centrifugal force causes the ring to expand radially. The speed $N$ at which the ring will lose its grip is given by $N = ± π \sqrt{\frac{E · g · (D_g - D_i)}{4 · ρ · R_m^3 · (1 + ν)}}$, where $E$ is the modulus, $g$ is gravity, $D_g$ is the groove diameter, $D_i$ is the free-ring ID, $ρ$ is the material density, $R_m$ is the mean radius, and $ν$ is Poisson's ratio. For high-speed applications, 'Self-Locking' features are used, where a tab on one turn locks into a slot on the other, mechanically preventing the ring from expanding. This allows the ring to operate at RPMs far exceeding the theoretical limit of a standard ring.

A Reference Answer

As a wave spring is compressed, the waves flatten, causing the mean diameter $D_m$ to expand. This expansion is defined by $\Delta D = α · f$, where $α$ is the expansion coefficient and $f$ is the deflection. For a spring operating in a bore, the initial clearance must be sufficient to prevent the spring from binding. If binding occurs, the spring rate $k$ increases exponentially, leading to unpredictable load behavior and potential fatigue failure at the wave crests. The calculation for the minimum bore diameter $D_b$ should be $D_b > D_{outer} + \frac{0.02 · (H_f - H_o) · n^2}{D_m}$, where $H_f$ is the free height and $H_o$ is the operating height. This ensures the spring remains 'free-floating' throughout its entire stroke.

A Reference Answer

The solid height $H_s$ is the physical limit of the spring's axial dimension under maximum load. For a crest-to-crest wave spring with $N$ turns and shim ends, the formula is $H_s = (N + 1) · t$, where $t$ is the material thickness. However, manufacturing tolerances on the flat wire thickness (typically $\pm 0.0005$ inches) and the wave height must be considered. In reality, the 'theoretical' solid height is rarely achieved due to wave nesting imperfections; thus, engineers use a 'measured' solid height which is often 1.1 times the theoretical value. If the application requires a precise hard stop, the shim ends must be ground to a specific parallelism tolerance, often within $0.002$ inches TIR (Total Indicator Reading).

A Reference Answer

Nested wave springs are produced from a single continuous filament of flat wire coiled in parallel. This configuration multiplies the load capacity by the number of turns $N$ while maintaining a low profile. The load equation is $P = \frac{E · b · t^3 · f · n^4}{D_m^3 · L}$, where $L$ is a constant related to the nesting friction. Unlike crest-to-crest springs which increase deflection, nested springs increase force. A critical design consideration is the inter-turn friction; as the spring deflections, the layers slide against each other, creating a slight hysteresis in the load-deflection curve. This is beneficial for damping in high-vibration aerospace actuators but requires careful lubrication with molybdenum disulfide (MoS2) to prevent galling.

A Reference Answer

In advanced wave spring design, the linear approximation of load $P = k · f$ fails as deflection $f$ exceeds 80% of the available work stroke. The L-factor, or the non-linearity correction, accounts for the change in the wave geometry and the shifting contact points. As the spring is compressed, the radius of curvature at the crests changes, effectively shortening the moment arm. The revised stress formula becomes $\sigma = \frac{3 · π · P · D_m}{4 · b · t^2 · n^2}$. If the spring is designed near its elastic limit, this non-linearity can lead to permanent set, especially in materials like SAE 1070 carbon steel. Designers must use Finite Element Analysis (FEA) to map the stress distribution across the wave peak and valley to ensure the maximum fiber stress does not exceed the minimum yield strength $\sigma_y$ of the material.

A Reference Answer

The spring rate $k$ for a crest-to-crest wave spring is governed by the number of turns $N$, the number of waves per turn $n$, and the material properties. The standard formula for the rate is $k = \frac{E \cdot b \cdot t^3 \cdot n^4}{I_d \cdot D_m^3 \cdot N}$, where $E$ is the Modulus of Elasticity, $b$ is the radial wall, $t$ is the material thickness, and $D_m$ is the mean diameter. The factor $I_d$ is a correction factor for the curvature. In high-precision applications, the addition of shim ends increases the rate and provides a 360-degree contact surface, reducing the effective active number of waves. The designer must account for the transition from the inactive shim to the active wave, which typically involves a linearity correction of 5-10% in the load-deflection curve as the spring approaches its solid height.

A Reference Answer

The edge margin ($E_m$) is the distance from the edge of the groove to the end of the shaft or bore. If $E_m$ is too small, the material between the groove and the face of the part may shear or 'blow out' under axial load. The required $E_m$ is roughly $3 imes d$ (three times the groove depth). The stress in this region is modeled as a shear tear-out: $\tau = \frac{P}{\pi imes D imes E_m}$. In aerospace applications with lightweight aluminum housings, the edge margin must be carefully calculated and often verified by FEA to prevent catastrophic failure of the retention shoulder under impact loading.

A Reference Answer

Spiral rings are installed by winding them into a groove. The stress during expansion (for external rings) or contraction (for internal rings) must remain below the material's yield strength to avoid permanent deformation. The maximum fiber stress during installation is $\sigma_{inst} = \frac{E imes t imes (D_g - D_f)}{D_g imes D_f}$, where $D_g$ is the groove diameter and $D_f$ is the free diameter. If $\sigma_{inst} > S_y$, the ring will not return to its original shape, resulting in a loose fit. For 302 Stainless Steel rings, the maximum allowable expansion is typically limited to $1\%$ to $2\%$ of the diameter to maintain a 'snug' fit.

A Reference Answer

Ring shear capacity ($P_r$) is the axial load required to physically shear the ring's cross-section, calculated as $P_r = \frac{D imes t imes ext{\pi} imes S_s}{K}$, where $S_s$ is the shear strength of the ring material ($S_s \approx 0.577 imes S_u$). Usually, $P_r$ is significantly higher than the groove deformation capacity ($P_g$). For a ring to fail in shear, the groove must be deep and the housing material must be extremely hard. In most engineering failures, the groove wall yields first, leading to an angular deflection of the ring (dishing), which causes it to climb out of the groove before shear stress reaches the ultimate limit.

A Reference Answer

The thrust capacity is often limited by the shear strength of the groove material rather than the ring itself. The allowable thrust load $P_g$ is given by $P_g = \frac{D imes d imes ext{\pi} imes S_y}{K}$, where $D$ is the shaft/bore diameter, $d$ is the groove depth, $S_y$ is the yield strength of the groove material, and $K$ is a safety factor (typically 2). If the groove material is soft (e.g., aluminum), the groove wall will yield and deform ('dish'), causing the ring to 'pop out' under load. In such cases, increasing the groove depth or using a harder housing material is necessary to meet the axial load requirements.

A Reference Answer

For external rings, centrifugal force causes the ring to expand radially, which can eventually lead to the ring lifting out of the groove. The maximum RPM ($V$) is calculated using the formula $V = \frac{715}{D_o} \sqrt{\frac{E imes I imes ext{cl}}{_x000d_ho imes A imes R_m^4}}$, where $D_o$ is the shaft diameter, $E$ is the modulus, $I$ is the moment of inertia, 'cl' is the centrifugal clearance (groove depth - ring radial wall), $\rho$ is the density, $A$ is the cross-sectional area, and $R_m$ is the mean radius. Engineers must ensure the operating RPM is at least 20% below this theoretical limit. If higher speeds are required, a 'self-locking' feature (a tab and slot) is integrated to mechanically prevent expansion.

A Reference Answer

The Work Height ($H_w$) is the axial space the spring occupies under a specific load. To avoid plastic deformation, the stress at $H_w$ must not exceed the yield strength ($S_y$) of the material. For nested springs, the stress is $S = \frac{3 \pi \cdot P \cdot D_m}{4 \cdot b \cdot t^2 imes N^2}$. The safety factor is $SF = S_y / S$. If $H_w$ is too close to the solid height ($H_s = n \cdot t$), the spring enters the non-linear range where the actual load exceeds the calculated linear load due to contact between turns. Designers must ensure that $H_w > H_s + (0.1 \times f_{total})$ to maintain linearity and prevent early fatigue failure caused by local stress concentrations at the solid height.

A Reference Answer

Friction in multi-turn wave springs occurs at the contact points between waves (crests) and at the interface between the spring and the bore or shaft. This leads to a hysteresis loop in the load-deflection curve where the loading force $P_{load} = P_{theoretical} + F_{friction}$ and the unloading force $P_{unload} = P_{theoretical} - F_{friction}$. The magnitude of $F_{friction}$ is a function of the coefficient of friction $\mu$ and the normal force at the contact points. In high-frequency applications, this friction generates heat, which can lead to localized thermal expansion and a reduction in the fatigue life. For precision applications, lubrication or PEEK coating is applied to reduce $\mu$ and flatten the hysteresis loop.

A Reference Answer

17-7PH (Type 631) stainless steel achieves its properties through a combination of cold reduction and precipitation hardening. When designing for high-temperature environments, the relaxation or 'set' is influenced by the heat treatment condition (e.g., CH900). Relaxation is modeled by the Arrhenius equation where the rate of creep is $\epsilon = A \sigma^n e^{-Q/RT}$. For a wave spring, stress $\sigma$ must be kept below 75% of the minimum tensile strength of the CH900 condition ($210-240$ ksi) to minimize permanent set. If the operating temperature exceeds $650^{\circ}F$ ($343^{\circ}C$), the material undergoes significant loss of elastic modulus, and the design must be derated by approximately 5-10% to account for thermal relaxation.

A Reference Answer

The radial wall $b$ is a critical factor in determining both the load capacity and the outer-diameter expansion of the spring during compression. For a nested wave spring, the load $P$ follows $P = \frac{E \cdot b \cdot t^3 imes N^4 imes f}{1.59 \cdot D_m^3} imes n_{nested}$. As the spring is compressed, the radial wall tends to expand due to the flattening of the waves. If the radial wall is too large relative to the mean diameter, the spring may bind in its housing or experience non-linear stress peaks at the inner diameter. Stress is calculated as $S = \frac{3 \pi \cdot P \cdot D_m}{4 \cdot b \cdot t^2 imes N^2}$, showing that stress decreases linearly as the radial wall $b$ increases for a constant load.

A Reference Answer

In Crest-to-Crest wave springs, the spring rate $K$ is inversely proportional to the number of turns $n$ and directly proportional to the fourth power of the number of waves $N$ per turn. The governing equation for the load $P$ is $P = \frac{E \cdot b \cdot t^3 imes N^4 imes f}{1.59 \cdot D_m^3 imes n}$, where $E$ is the Modulus of Elasticity, $b$ is the radial wall, $t$ is the material thickness, $f$ is the deflection, and $D_m$ is the mean diameter. Increasing the number of turns $n$ effectively adds springs in series, reducing the overall rate, while increasing the wave count $N$ stiffens the spring exponentially. Engineers must balance $N$ to avoid 'bottoming out' or exceeding the material's elastic limit at the wave peaks.

A Reference Answer

The installation stress $S_i$ is a function of the expansion distance. For a spiral ring, $S_i = \frac{E t (D_s - D_g)}{D_g (D_g - t)}$, where $D_s$ is the shaft diameter and $D_g$ is the free diameter of the ring. It is essential that $S_i$ remains below the yield strength of the material to prevent permanent deformation (setting). If the stress exceeds yield, the ring will not snap back into the groove tightly, resulting in a loose fit and reduced thrust capacity. For materials like 302 Stainless Steel, the yield is lower than Carbon Steel, necessitating wider grooves or specialized installation tools.

A Reference Answer

The edge margin is the distance from the groove to the end of the shaft or bore. If this margin is too small, the groove wall may shear off under load. The required margin $z$ is calculated using $z = \frac{3 F}{D \pi \sigma_y}$, where $F$ is the thrust load. As a rule of thumb, for steel housings, $z$ should be at least $3 \cdot d$ (three times the groove depth). For lighter alloys, this should be increased to $5 \cdot d$. Insufficient edge margin leads to 'Groove Wall Blowout', a catastrophic failure mode in hydraulic and pneumatic cylinders.

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