1  Structure and Characteristics of Ring Springs

(1) Structure of ring springs

A ring spring consists of outer rings with internal conical surfaces and inner rings with external conical surfaces, fitted together (Fig. 18-1a). The dimensions and number of pairs of inner and outer rings are determined according to the magnitude of the applied load and the required deformation.

Fig. 18-1  Structure of a ring spring

a) Free state   b) Loaded state

D₁ — inner diameter of the spring;  D₂ — outer diameter of the spring;  H₀ — free height of the spring;  δ₀ — axial spacing between the outer rings;  β — cone half-angle of the spring ring;  δ — axial spacing between the outer rings after loading;  H — height of the spring after loading.

When an axial load F acts on the end face of the rings (Fig. 18-1b), normal pressure acts on the conical contact surfaces between the outer and inner rings. This causes the diameter of the outer ring to increase and places it in tension, while the diameter of the inner ring decreases and places it in compression. The rings move relative to each other along the conical surfaces and are forced into one another, shortening the axial dimension of the spring and producing an axial spring deformation f.

Ring springs are generally installed on a guide sleeve or mandrel. Because the outer diameter of the outer ring increases and the inner diameter of the inner ring decreases after loading, an appropriate clearance should be provided between the outer ring and the guide sleeve and between the inner ring and the mandrel (generally about 2% of the diameter).

(2) Characteristics of ring springs

Because the outer and inner rings slide relative to each other along the mating conical surfaces, a large friction force acts at the contact surfaces. During loading, the axial force F is balanced by the surface pressure and friction force; therefore, the effective axial load is reduced, which is equivalent to increasing the spring stiffness. During unloading, friction hinders recovery of the elastic deformation, which is equivalent to reducing the spring force. As shown in Fig. 18-2, the characteristic curve of a ring spring during one loading and unloading cycle is OABO. Without friction, it would be OC. During unloading, the characteristic curve starts from point B rather than point E because friction causes elastic hysteresis.

The characteristic curve clearly shows that the area OABO represents the work consumed and converted into heat by friction during the loading and unloading cycle. Its magnitude can reach approximately 60%–70% of the work performed during loading (area OADO). Therefore, ring springs have a high buffering and vibration-damping capability, and their energy-storage capacity per unit volume of material is greater than that of other types of springs.

Fig. 18-2  Characteristic curve of a ring spring

(3) Applications of ring springs

Ring springs are commonly used where space is limited but strong buffering is required, for example, in hangers for large pipelines (Fig. 18-3a), supports for vibrating machinery (Fig. 18-3b), and connecting parts of heavy railway vehicles.

Ring springs can also be used to form composite springs, either by installing two sets of ring springs with different diameters concentrically (Fig. 18-4), or by combining a ring spring with another cylindrical helical spring.

Fig. 18-3  Examples of ring spring applications

a) Large pipeline hanger   b) Support for vibrating machinery

Fig. 18-4  Composite ring spring

A ring spring consists of many pairs of inner and outer rings. Therefore, when damage or wear occurs, only individual rings need to be replaced, making repair relatively easy and economical.

To prevent scoring and adhesion between the conical surfaces of the ring spring, reduce wear, eliminate noise, and cool the working surfaces, lubricant should be applied between the contact surfaces. Common solid lubricants include graphite, and grease may also be used.

2  Design Calculations for Ring Springs

2.1  Force Analysis of Ring Springs

As shown in Fig. 18-5, the end face of the ring spring is subjected to an axial force F, while the conical surfaces of the inner and outer rings are subjected to normal pressure N. The friction force generated by relative sliding of the surfaces is Nfμ (fμ is the coefficient of friction). Therefore, during loading, the force equilibrium equation is:

F = N sinβ + N fμ cosβ = N(sinβ + fμ cosβ)

N = F / (sinβ + fμ cosβ)     (18-1)

Fig. 18-5  Force analysis of a ring spring

2.2  Stress Calculation for the Outer Ring

(1) Tensile stress in the outer ring

As shown in Fig. 18-5, the radial component of force acting on the outer ring is:

Fᵣ = 2(N cosβ − N fμ sinβ)     (18-2)

The radial component per unit length on the center circumference of the outer-ring cross-section is:

p₂ = Fᵣ/(πD₀₂) = 2N(cosβ − fμ sinβ)/(πD₀₂)     (18-3)

where D₀₂ is the center diameter of the outer-ring cross-section, as shown in Fig. 18-6.

Fig. 18-5

Under the radial component of force, the cross-section of the outer ring is subjected to tension, and the tensile force is F₂. The radial force acting on an infinitesimal circumferential length ds is p₂ds. Its projection on the y-axis is:

p₂ ds sinα = p₂ (D₀₂/2) dα sinα

Therefore, the equilibrium equation with F₂ is:

2F₂ − ∫₀^π p₂ sinα (D₀₂/2) dα = 0

F₂ = p₂D₀₂/2     (18-4)

Because the wall thickness of a ring spring is small, generally 1/15–1/30 of the diameter, the tensile stress in the outer-ring cross-section can be calculated approximately as:

σ′₂ = F₂/A₂ = p₂D₀₂/(2A₂) = N(cosβ − fμ sinβ)/(πA₂)     (18-5)

Substituting Eq. (18-1) gives:

σ′₂ = [F/(πA₂)] [(cosβ − fμ sinβ)/(sinβ + fμ cosβ)] = F/(πA₂γ)     (18-6)

A₂ = h(b + s/2) = hb + h²tanβ/4     (18-7)

γ = (sinβ + fμ cosβ)/(cosβ − fμ sinβ) = (tanβ + tanρ)/(1 − tanρ tanβ) = tan(β + ρ)     (18-8)

where A₂ is the cross-sectional area of the outer ring, as shown in Fig. 18-7; γ is a coefficient; and ρ is the friction angle, tanρ = fμ.

Fig. 18-7

(2) Compressive stress in the outer ring

In addition to the tensile stress in the cross-section of the outer ring, compressive stress σc also acts on the conical contact surface:

σc = N/(πDl)     (18-9)

From Eq. (18-5):

σc = 2σ′₂A₂ / [D(h − δ₀)(1 − fμ tanβ)]     (18-10)

Due to lateral deformation, the circumferential tensile stress on the inner surface of the outer ring is:

σ″₂ = (1/μ)σc     (18-11)

Therefore, the maximum stress on the inner surface of the outer ring is:

σ₂ = σ′₂ + σ″₂ = σ′₂ + 2σ′₂A₂/[μD(h − δ₀)(1 − fμ tanβ)]

= F/(πA₂γ) [1 + 2A₂/{μD(h − δ₀)(1 − fμ tanβ)}]     (18-12)

D = 1/2[(D₂ − 2b) + (D₁ + 2b₁)]     (18-13)

l = (h − δ₀)/(2 cosβ)     (18-14)

where D is the mean diameter of the conical contact surface, as shown in Fig. 18-7; l is the contact-surface width, as shown in Fig. 18-7; μ is Poisson's ratio of the material.

2.3  Stress Calculation for the Inner Ring

The radial component of force acting on the inner ring is the same as that acting on the outer ring, namely Fᵣ. The radial component per unit length on the center circumference of the inner-ring cross-section is:

p₁ = Fᵣ/(πD₀₁) = 2N(cosβ − fμ sinβ)/(πD₀₁)     (18-15)

As shown in Fig. 18-6b, under the radial component of force, the cross-section of the inner ring is compressed. The compressive force is:

F₁ = p₁D₀₁/2     (18-16)

The compressive stress in the cross-section is:

σ₁ = F₁/A₁ = p₁D₀₁/(2A₁) = N(cosβ − fμ sinβ)/(πA₁)     (18-17)

Substituting Eq. (18-1) gives the maximum stress in the inner ring:

σ₁ = F/(πA₁) [(cosβ − fμ sinβ)/(sinβ + fμ cosβ)] = F/(πA₁γ)     (18-18)

A₁ = h(b₁ + s/2) = hb₁ + h²tanβ/4     (18-19)

where D₀₁ is the center diameter of the inner-ring cross-section; A₁ is the cross-sectional area of the inner ring, as shown in Fig. 18-7.

The compressive stress σc on the contact surface of the inner ring is the same as that of the outer ring and can be calculated using Eq. (18-10). Due to lateral deformation, the circumferential tensile stress on the inner-ring surface can also be calculated using Eq. (18-11). Therefore, the stress on the outer surface of the inner ring is σ₁ − σ″₂. Obviously, the maximum compressive stress in the cross-section of the inner ring does not occur at the contact surface but within the cross-section, and its magnitude can be calculated using Eq. (18-18), namely σ₁.

To satisfy the strength requirements, the maximum stresses of the outer and inner rings calculated from Eqs. (18-12) and (18-18), respectively, should both be less than the allowable stress.

2.4  Deformation Calculation of Ring Springs

During loading and unloading, the relationship between load and deformation is different because the direction of the friction force on the contact surface is different.

(1) During loading

After the ring spring is subjected to an axial force (Fig. 18-8), the outer-ring diameter increases and the inner-ring diameter decreases due to the radial component of force. The relationship between the radial deformation amounts Δr₁ and Δr₂ and the cross-sectional stresses is approximately:

Fig. 18-8

Inner ring:

Δr₁ = σ₁D₀₁/(2E)

Outer ring:

Δr₂ = σ₂D₀₂/(2E)

The total radial deformation is:

Δr = Δr₁ + Δr₂ = (σ₁D₀₁ + σ₂D₀₂)/(2E)     (18-20)

As shown in Fig. 18-8, before loading, point K on the conical surface of the outer ring contacts point M on the conical surface of the inner ring. After loading, because of the change in diameter, the inner and outer rings slide along the conical surface, and K and M no longer contact each other. The change in their axial position is the axial deformation. From the figure, the axial deformation of a ring spring for one pair of conical contact surfaces is:

f/n = Δr/tanβ = (σ₁D₀₁ + σ₂D₀₂)/(2E tanβ)     (18-21)

Substituting Eqs. (18-6) and (18-18):

f/n = F/(2πEγ tanβ) (D₀₁/A₁ + D₀₂/A₂)

Thus, the total axial deformation of the ring spring is:

f = nF/(2πEγ tanβ) (D₀₁/A₁ + D₀₂/A₂)     (18-22)

where n is the number of pairs of conical contact surfaces.

(2) During unloading

As shown in Fig. 18-9, the direction of friction changes during unloading. Therefore, the relationship between the normal pressure on the conical surface and the axial force F is:

N = F/(sinβ − fμ cosβ)     (18-23)

The radial component of force acting on the outer and inner rings is:

Fᵣ = 2(N cosβ + N fμ sinβ)

= 2F(cosβ + fμ sinβ)/(sinβ − fμ cosβ) = 2F/γ′     (18-24)

γ′ = (sinβ − fμ cosβ)/(cosβ + fμ sinβ) = (tanβ − tanρ)/(1 + tanβ tanρ) = tan(β − ρ)     (18-25)

Fig. 18-9  Force analysis of a ring spring during unloading

Similarly, from Eqs. (18-3), (18-6), (18-15), (18-18), and (18-22), the relationship between deformation and load Fᵣ during unloading is:

f = nFᵣ/(2πEγ′ tanβ) (D₀₁/A₁ + D₀₂/A₂)     (18-26)

During unloading, when the elastic deformation of the ring spring begins to recover, its deformation is equal to the final deformation f during loading. Therefore, from Eqs. (18-22) and (18-26):

Fᵣ/γ′ = F/γ

Thus, the load Fᵣ when elastic deformation begins to recover is:

Fᵣ = F γ′/γ     (18-27)

where γ′ is a coefficient.

2.5  Deformation Energy of Ring Springs

From Eqs. (18-22), (18-26), and Fig. 18-2, the characteristic curve of a ring spring is linear. Therefore, during loading, the deformation energy absorbed by the ring spring is:

U = 1/2 Ff     (18-28)

During unloading, the deformation energy released by the spring is:

Uᵣ = 1/2 Fᵣf = 1/2(F γ′/γ)f = U γ′/γ     (18-29)

Therefore, during one loading cycle, the energy consumed by friction in the ring spring is:

U₀ = U − Uᵣ = (1 − γ′/γ)U = ψU     (18-30)

ψ = 1 − γ′/γ

where ψ is the damping coefficient.

2.6  Test Load and Deformation under the Test Load

Let the deformation under the test load Fₛ be fₛ. According to Eq. (18-22), the test load of the spring is:

Fₛ = [2πEγ tanβ fₛ/n] / (D₀₁/A₁ + D₀₂/A₂)     (18-31)

fₛ = n/2 (δ₀ − δmin)     (18-32)

where n is the number of pairs of conical contact surfaces, calculated from Eq. (18-22); δmin is the minimum spacing that should be retained between spring rings at the working limit position.

2.7  Calculation of Structural Parameters of Ring Springs

With reference to the dimensional symbols shown in Fig. 18-7, the following principal geometric relationships can be obtained:

(1) Ring diameters

Outer diameter of the outer ring:

D₂ = D₁ + 2(b + b₁) + (h − δ₀)tanβ     (18-33)

Inner diameter of the outer ring:

D′₂ = D₂ − 2(b + 1/2 h tanβ)     (18-34)

Center diameter of the outer-ring cross-section:

D₀₂ ≈ D₂ − 1.3b     (18-35)

Outer diameter of the inner ring:

D′₁ = D₁ + 2(b₁ + 1/2 h tanβ)     (18-36)

Center diameter of the inner-ring cross-section:

D₀₁ ≈ D₁ + 1.3b     (18-37)

(2) Ring height h

Generally, h = (1/5.6–1/3)D. If h is too small, the cross-sectional area of the ring is small and the stress is high; at the same time, the conical contact area is small and the surface stress increases. If h is too large, manufacturing becomes difficult because the thickness of the general ring cross-section is relatively small.

(3) Ring thicknesses b and b₁

The ring thickness is related to the strength requirements. For initial selection, b = b₁ ≥ (1/6–1/2)h may generally be used. Because the outer-ring cross-section is subjected to tensile stress while the inner-ring cross-section is subjected to compressive stress, and the tensile fatigue strength of the material is lower than its compressive fatigue strength, for the same height h and material, b should be greater than b₁ to obtain approximately similar strength. Generally, b = 1.3b₁ may be used.

(4) Number of rings n₀

Including the single-cone rings at both ends, the total number of inner and outer rings n₀ is:

n₀ = n + 1     (18-38)

Generally, the inner rings at both ends are made as single-cone rings; therefore, the number of outer rings is:

n₁ = n/2     (18-39)

where n is the number of pairs of conical contact surfaces, determined from Eq. (18-22) according to the required total deformation f under load F.

(5) Free height H₀ and solid height Hᵦ

As shown in Fig. 18-7, the free height of the ring spring is:

H₀ = 1/2 n(h + δ₀)     (18-40)

where δ₀ is the spacing between adjacent outer rings (or inner rings) in the free state; generally δ₀ = (1/4)h.

When the ring spring is compressed until the end faces of adjacent outer rings (or inner rings) contact each other, its solid height is:

Hᵦ = 1/2 nh     (18-41)

To ensure spring stability, Hᵦ ≥ 4f.

After the ring spring reaches solid height, the spring stiffness tends toward infinity and the spring loses its spring function. To avoid this condition, a minimum spacing δmin ≥ 1 mm should be retained at the spring working limit position. For larger diameters or lower machining accuracy, the minimum spacing should be larger. Generally, for lower accuracy:

δmin ≈ D/50     (18-42)

For higher accuracy:

δmin ≈ D/100     (18-43)

(6) Cone half-angle β

From Eqs. (18-8) and (18-22), when the half-angle β is small, the spring stiffness is small. If β < ρ, self-locking will occur during unloading, i.e. the spring cannot rebound. If β is too large, under the same friction angle ρ, γ′/γ is larger. According to Eq. (18-27), the load Fᵣ at which elastic deformation recovers during unloading is larger, meaning that the hysteretic attenuation caused by friction is smaller. The area OABO in the characteristic curve of Fig. 18-2 is smaller, and the buffering and vibration-absorption capability of the ring spring is reduced.

For design, β = 12°–20° may be selected. When the contact surface machining accuracy is high, β = 12° may be selected. Under normal machining accuracy, a slope of 1:4 is often used, corresponding to β = 14°3′. When lubrication conditions are poor and the coefficient of friction is large, β should be selected somewhat larger to avoid self-locking.

(7) Coefficient of friction fμ and friction angle ρ

For ring springs under good lubrication conditions, the coefficient of friction and friction angle of the conical contact surfaces may be selected according to the following conditions:

Heavy-duty service with contact surfaces not precision-finished:  ρ ≈ 9°   fμ ≈ 0.16

Heavy-duty service with precision-finished contact surfaces:  ρ ≈ 8°30′   fμ ≈ 0.15

Light-duty service with precision-finished contact surfaces:  ρ ≈ 7°   fμ ≈ 0.12

When designing a ring spring, its overall dimensions are generally subject to structural limitations. Therefore, the outer diameter D₂, inner diameter D₁, free height H₀, solid height Hᵦ, and other geometric dimensions should be checked to ensure that they satisfy the structural dimensional restrictions.

3  Materials and Allowable Stresses of Ring Springs

Common materials for ring springs include 60Si2MnA, 65Si2MnWA, 55CrSiA, and 50CrMn spring steels.

The allowable stresses for ring springs are shown in Table 18-1.

Table 18-1  Allowable stresses of ring springs (MPa)

① The mean allowable stress [σm] is the average of the tensile stress [σ₂] in the outer-ring cross-section and the compressive stress [σ₁] in the inner-ring cross-section.

For ring springs made of any material, it must be ensured that when the spring is compressed to solid height, its stress does not exceed the elastic limit of the material.

4  Manufacturing and Technical Requirements

Ring spring blanks may be produced by open-die forging or rolling. For small production quantities, open-die forging may be used, followed by machining to obtain the finished shape and dimensions. For mass production, seamless steel tube may be cut into blanks and rolled to the finished shape and dimensions using a dedicated ring rolling machine. After forming, the rings are heat-treated after passing inspection. When necessary, the contact surfaces are ground after heat treatment.

The general surface roughness of the conical contact surfaces is Ra = 1.6–0.4 μm, and the surface hardness after heat treatment is 40–60 HRC.

Because the ring thickness is small, particular attention should be paid during manufacturing to preventing distortion of the rings. To ensure interchangeability of the rings during assembly, the cone angle and height dimensions of each ring must be within the specified tolerances.

The working drawing of the ring spring parts should specify the test load and corresponding deformation for each conical contact surface, so that finished-product quality inspection can be carried out.

5  Recommended Structural Parameters of Ring Springs

Table 18-2 lists recommended values and characteristics of ring spring structural parameters and may be used as a reference when designing springs.

Table 18-2  Recommended structural parameters and characteristics of ring springs

① When calculating the axial deformation f/n of one pair of contact surfaces (n is the number of pairs of contact surfaces), take the modulus of elasticity E = 206 × 10³ MPa.

6  Slotted Ring Springs

Slotted ring springs (Fig. 18-10) are manufactured from round steel or steel tube. They are equivalent to ring springs formed as an integral unit. When subjected to an axial load, each ring deforms. For fixing purposes, threads may be provided at the ends. This type of spring can be used as either a compression spring or a tension spring. Because of its precise stiffness, it is particularly suitable for precision instruments.

Fig. 18-11 shows the force analysis of one ring of a slotted ring spring under axial load. From this force condition, the stiffness equation is:

F′ = 256Ebh³/(nπD³)     (18-44)

The maximum stress equation is:

σmax = 3πFD/(16bh²)     (18-45)

where D is the mean diameter of the spring ring; b is the radial width of the spring ring; h is the axial thickness of the spring ring; n is the number of turns of the spring ring; E is the elastic modulus of the material.

Fig. 18-10  Slotted ring spring

Fig. 18-11  Force analysis of a slotted ring spring

Example 18-1  Design of a Ring Spring Buffer

Design a ring spring buffer device subjected to a load of 1.2 × 10⁶ N, with an axial deformation of 26 mm and an installation axial deformation of 10 mm.

Solution

(1) Selection of material

The material selected is 60Si2MnA. From Table 2-7, the yield point σs (RₑL) = 1375 MPa and the modulus of elasticity E = 206 × 10³ MPa. From Table 18-1, the allowable stress of the outer ring [σ₁] = 900 MPa and that of the inner ring [σ₂] = 1200 MPa.

(2) Selection of structural dimensions

According to Table 18-2 and the applied load, the following structural dimensions are selected:

Outer ring:

Outer diameter: D₂ = 489 mm

Thickness: a = 24.5 mm

Edge thickness: b = 13 mm

D′₂ = D₂ − 2a = (489 − 2 × 24.5) mm = 440 mm

D₀₂ = D₂ − a = (489 − 24.5) mm = 464.5 mm

Inner ring:

Inner diameter: D₁ = 428.5 mm

Thickness: a₁ = 21 mm

Edge thickness: b₁ = 9.5 mm

D′₁ = D₁ + 2a₁ = (428.5 + 2 × 21) mm = 470.5 mm

D₀₁ = D₁ + a₁ = (428.5 + 21) mm = 449.5 mm

Ring height: h = 78 mm

β = arctan[(a − b)/(h/2)] = arctan[(24.5 − 13)/(78/2)] = 16.4°

Spring pitch: t = 102 mm

Take the coefficient of friction fμ = 0.16, corresponding to a friction angle ρ = 9°.

δ₀ = t − h = (102 − 78) mm = 24 mm

From Eq. (18-13), the mean diameter of the spring is:

D = 1/2[(D₂ − 2b) + (D₁ + 2b₁)]

= 1/2[(489 − 2 × 13) + (428.5 + 2 × 9.5)] mm = 455.3 mm

From Eqs. (18-7) and (18-19), the cross-sectional areas of the outer and inner rings are:

A₂ = hb + h²tanβ/4 = 78 × 13 + 78² × tan16.4°/4 mm² = 1462 mm²

A₁ = hb₁ + h²tanβ/4 = 78 × 9.5 + 78² × tan16.4°/4 mm² = 1189 mm²

γ = tan(β + ρ) = tan(16.4 + 9) = 0.475

(3) Number of pairs of conical contact surfaces

From Eq. (18-22), the number of pairs of conical contact surfaces is:

n = [2πEγ tanβ f/F] / (D₀₁/A₁ + D₀₂/A₂)

= [2π × 206 × 10³ × 0.475 × tan16.4° × 26] / [1.2 × 10⁶ × (449.5/1189 + 464.5/1462)] = 5.6

Take 6 pairs of conical contact surfaces.

(4) Stress of the rings

From Eq. (18-12), the maximum stress on the inner surface of the outer ring is:

σ₂ = F/(πA₂γ) [1 + 2A₂/{μD(h − δ₀)(1 − fμtanβ)}]

= [1.2 × 10⁶/(π × 1462 × 0.475)] [1 + 2 × 1462/{0.3 × 455.3 × (78 − 24) × (1 − 0.16tan16.4°)}] MPa

= 779 MPa

This is less than the allowable value [σ₂] = 900 MPa.

From Eq. (18-18), the compressive stress in the inner ring is:

σ₁ = F/(πA₁γ) = [1.2 × 10⁶/(π × 1189 × 0.475)] MPa = 677 MPa

This is less than the allowable value [σ₁] = 1200 MPa.

(5) Spring deformation

From Eq. (18-22), the actual axial deformation of the spring is:

f = nF/(2πEγ tanβ) (D₀₁/A₁ + D₀₂/A₂)

= [6 × 1.2 × 10⁶/(2π × 206 × 10³ × 0.475 × tan16.4°)] (449.5/1189 + 464.5/1462) mm = 28 mm

(6) Test load and deformation under the test load

Because the ring diameter in this design is relatively large, the minimum spacing of the spring is calculated from Eq. (18-42):

δmin = D/50 = 455.3/50 mm ≈ 9 mm

Then, from Eq. (18-32), the test deformation is:

fₛ = n/2(δ₀ − δmin) = 6/2(24 − 9) mm = 45 mm

The corresponding test load is obtained from Eq. (18-31):

Fₛ = [2πEγ tanβ fₛ/n] / (D₀₁/A₁ + D₀₂/A₂)

= [2π × 206 × 10⁷ × 0.475 × tan16.4° × 45] / [6 × (449.5/1189 + 464.5/1462)] N

= 1.95 × 10⁶ N

(7) Strength check under the test load

The test stresses corresponding to the test load Fₛ are obtained from Eqs. (18-12) and (18-18):

σ₂ₛ = Fₛ/(πA₂γ) [1 + 2A₂/{μD(h − δ₀)(1 − fμtanβ)}]

= [1.95 × 10⁶/(π × 1462 × 0.475)] [1 + 2 × 1462/{0.3 × 455.3 × (178 − 24) × (1 − 0.16tan16.4°)}] MPa

= 1270 MPa

σ₁ₛ = Fₛ/(πA₁γ) = [1.95 × 10⁶/(π × 1189 × 0.475)] MPa = 1099 MPa

Both are less than the yield point σs = 1375 MPa.

(8) Determination of related structural parameters of the spring

Fig. 18-12  Assembly drawing of the ring spring in Example 18-1

Technical requirements

1. Before assembly, the ring components shall be cleaned and shall be free of rust and oil contamination.

2. Apply an appropriate amount of lubricating oil to the contact surfaces and rust-preventive oil to the remaining surfaces.

3. Check the contact spots on the contact surfaces; the contact area shall be greater than 60%.

4. Compress the spring three times under the test load; the residual deformation of its height shall not exceed 3 mm.

1) Number of ring components:

Total number of ring components: n₀ = n + 1 = 6 + 1 = 7

Number of outer-ring components: n₁ = n/2 = 6/2 = 3

2) Spring height:

Free height, according to Eq. (18-40):

H₀ = 1/2 n(h + δ₀) = 1/2 × 6 × (78 + 24) mm = 306 mm

Height under the working load: H = H₀ − f = (306 − 28) mm = 278 mm

Height under the test load: Hₛ = H₀ − fₛ = (306 − 45) mm = 261 mm

Installation height: H₁ = H₀ − f₁ = (306 − 10) mm = 296 mm

(9) Spring working drawing

Fig. 18-12 is the spring assembly drawing, and Fig. 18-13 is the working drawing of the spring rings.

Fig. 18-13  Working drawing of the ring components in Example 18-1

a) Outer ring   b) Inner ring   c) End ring

Technical requirements

1. Material: 60Si2MnA; hardness after heat treatment: 45–50 HRC.

2. Chamfer and remove burrs.

3. The surface shall be free of rust and cracks.

4. Oxidation treatment shall be applied to the surface.

Editorial note: The symbols [σ₁] and [σ₂] in the original Chinese source are used inconsistently in part of Example 18-1. This English edition preserves the source notation and numerical results rather than silently changing the original engineering data.