Mechanical Engineering Design · Chapter 10
Mechanical Engineering Design · Part 3 — Design of Mechanical Elements

Mechanical Springs

Helical compression and extension springs from first principles: shear stress with curvature correction, spring rate by Castigliano, end styles and stability, material strength estimation, static and fatigue design procedures with the Zimmerli data, plus torsion, Belleville, and special springs.

10–1Stresses in Helical Springs

Learning objectives

  • Explain how axial load creates torsion and direct shear in helical spring wire.
  • Calculate spring rate, deflection, corrected stress, coil lengths, and closure margin.
  • Separate stiffness, static strength, fatigue, stability, surge, and manufacturability checks.
  • Select compression, extension, torsion, disk, or special spring behavior for a load path.
  • Iterate a spring design using traceable material, process, tolerance, and endurance data.

Load path in a close-coiled helix

An axial force \(F\) acts at mean radius \(D/2\), producing wire torque \(T=FD/2\) plus direct shear. Nominal torsional stress scales as \(FD/d^3\); therefore a small wire change strongly affects strength. The dimensionless spring index \(C=D/d\) describes curvature and formability.

Corrected body-coil shear
\[C=\frac{D}{d},\qquad K_B=\frac{4C+2}{4C-3},\qquad \tau_{\max}=K_B\frac{8FD}{\pi d^3}\]

Physical meaning

The correction is larger for tightly curved coils because inner fibers carry more shear. Stress is load demand; wire yield, ultimate, or endurance values are resistance and require a stated source and condition.

10–2The Curvature Effect

Choosing a stress factor

The direct-shear factor, Wahl factor, and Bergsträsser factor represent different approximations. Do not mix a factor with an equation calibrated for another convention. This chapter uses \(K_B\) consistently for corrected body stress. It is an elastic local model and does not include presetting residual stress, surface defects, or end-coil contact.

10–3Deflection of Helical Springs

Torsional compliance

Strain energy in the twisted wire leads to axial deflection. For a close-coiled round-wire spring with linear elastic modulus \(G\), constant mean diameter, and \(N_a\) active coils, rate is constant and deflection is proportional to force.

Rate and deflection
\[k=\frac{d^4G}{8D^3N_a},\qquad y=\frac{F}{k}=\frac{8FD^3N_a}{d^4G}\]

Rate scales with \(d^4\) but inversely with \(D^3N_a\). Consequently, changing wire size to fix stress also changes travel, so spring sizing is inherently iterative.

EXAMPLE 1Rate and stress of a compression spring
A round-wire compression spring has wire diameter \(d=4.0\ \text{mm}\), mean coil diameter \(D=32\ \text{mm}\), \(N_a=8\) active coils, and \(G=79\ \text{GPa}\). Under \(F=300\ \text{N}\), find spring rate, deflection, and curvature-corrected maximum shear stress.

Recognition strategy

The requested outputs separate stiffness from strength: rate and deflection come from torsional compliance, while maximum local shear uses a curvature correction. The spring index confirms whether the helical formulas are geometrically reasonable.

Given information

  • \(d=4.0\ \text{mm}\), \(D=32\ \text{mm}\), and \(N_a=8\).
  • \(G=79{,}000\ \text{MPa}\) and axial compressive load \(F=300\ \text{N}\).
  • Use the Bergsträsser factor \(K_B=(4C+2)/(4C-3)\) for the maximum shear estimate.

Model and interpretation

Model the body as a close-coiled, linear-elastic helix with round wire. Positive axial deflection is compression. Ends, seating, coil contact, residual stress, and dynamic surge are excluded from this body calculation.

Step-by-step solution
1. Compute the dimensionless spring index \(C=D/d=32/4=8.00\), which lies in the commonly manufacturable range and gives \(K_B=34/29=1.1724\).
2. Evaluate stiffness with consistent millimetre units: \(k=d^4G/(8D^3N_a)=4^4(79{,}000)/[8(32)^3(8)]=9.6436\ \text{N}/\text{mm}\).
3. Use the linear force-deflection law: \(y=F/k=300/9.6436=31.11\ \text{mm}\) of axial compression from the unloaded position.
4. Calculate local shear: \(\tau=K_B8FD/(\pi d^3)=1.1724[8(300)(32)/(\pi4^3)]=447.8\ \text{MPa}\).

Verification

Units: Using \(G\) in \(\text{N}/\text{mm}^2\) makes the rate \(\text{N}/\text{mm}\); the stress expression reduces from \(\text{N}\!\cdot\!\text{mm}/\text{mm}^3\) to \(\text{MPa}\).

Signs and direction: The force and deflection are reported as positive compressive magnitudes. Internal wire shear reverses around the section, while \(447.8\ \text{MPa}\) denotes the maximum magnitude at the inner fiber.

Plausibility: The uncorrected torsional value is \(382.0\ \text{MPa}\); multiplying by \(K_B>1\) must raise, not lower, the inner-fiber stress to \(447.8\ \text{MPa}\).

Final result: \(\boxed{k=9.64\ \text{N}/\text{mm},\quad y=31.1\ \text{mm},\quad \tau_{\max}=448\ \text{MPa}}\).

Engineering interpretation

The same geometry can be sufficiently compliant yet too highly stressed, demonstrating that stiffness and strength are separate requirements. Material shear strength, available travel, solid clearance, ends, and fatigue must be checked before use.

10–4Compression Springs

Ends, active coils, and closure

End style determines which coils deflect and how the spring seats. For squared-and-ground ends, a common idealization is \(N_t=N_a+2\) and \(L_s=dN_t\). Free length equals solid length plus available deflection only for the chosen nominal geometry; pitch and tolerances must prevent premature contact.

Squared-and-ground geometry
\[N_t=N_a+2,\qquad L_s=dN_t,\qquad L_0=L_s+y_s\]

10–5Stability

Buckling is a system problem

A long compression spring can bow before reaching intended travel. Stability depends on \(L_0/D\), end restraint, guide clearance, and deflection. Use the governing standard, supplier curve, or validated stability model for the actual seats. A guide rod can prevent global buckling but introduces rubbing, wear, and debris sensitivity.

10–6Spring Materials

Diameter-dependent, process-dependent properties

Wire tensile strength often follows \(S_{ut}=A/d^m\) over a specified diameter range. Constants \(A\) and \(m\), shear modulus, temperature derating, and torsional yield ratio must come from the applicable material specification or supplier certification. Coating, heat treatment, presetting, and shot peening alter performance.

Traceable strength correlation
\[S_{ut}=\frac{A}{d^m},\qquad S_{sy}=c_yS_{ut}\]

The coefficient \(c_y\) is not universal. Record alloy, wire size, condition, test basis, lot traceability, and whether minimum or typical values are being used.

10–7Helical Compression Spring Design for Static Service

Problem-solving strategy

  1. Define minimum and maximum forces, required travel, envelope, seats, temperature, life, and reliability.
  2. Select a traceable wire specification and trial standard wire diameter and spring index.
  3. Calculate corrected stress, rate, active coils, solid length, free length, and load at closure.
  4. Check operating and solid strength, buckling, clash allowance, tolerances, and manufacturable index and pitch.
  5. For cyclic service add fatigue and surge checks, then iterate geometry until every constraint converges.
Closure checks
\[F_s=k(L_0-L_s),\qquad \xi=\frac{F_s-F_{\max}}{F_{\max}},\qquad n_s=\frac{S_{sy}}{\tau_s}\]
EXAMPLE 2Static closure and solid-height audit
A squared-and-ground compression spring uses \(d=5.0\ \text{mm}\), \(D=40\ \text{mm}\), \(N_a=7\), and \(G=79\ \text{GPa}\). It must carry \(F_{\max}=450\ \text{N}\), with a specified 20 percent load overrun to solid. Given qualified material shear yield strength \(S_{sy}=720\ \text{MPa}\), determine free length and the factor of safety at solid.

Recognition strategy

The explicit overrun defines the solid load, connecting operating travel to geometry. This is an iterative static spring check: calculate rate, infer solid force and travel, establish lengths, then compare local stress with supplied shear yield strength.

Given information

  • Squared-and-ground ends give \(N_t=N_a+2=9\) and \(L_s=dN_t\).
  • \(F_{\max}=450\ \text{N}\), overrun \(\xi=0.20\), and \(S_{sy}=720\ \text{MPa}\).
  • \(d=5.0\ \text{mm}\), \(D=40\ \text{mm}\), \(N_a=7\), and \(G=79{,}000\ \text{MPa}\).

Model and interpretation

Treat the force-deflection relation as linear until nominal solid contact and define compression as positive. The closure margin is load based: \(F_s=(1+\xi)F_{\max}\). Manufacturing tolerances and contact nonlinearity require later allowance.

Step-by-step solution
1. Find \(C=40/5=8\), \(K_B=34/29=1.1724\), and \(k=5^4(79{,}000)/[8(40)^3(7)]=13.78\ \text{N}/\text{mm}\).
2. The specified overrun sets \(F_s=1.20(450)=540\ \text{N}\). Thus travel to nominal solid is \(y_s=F_s/k=540/13.78=39.19\ \text{mm}\).
3. For squared-and-ground ends, \(L_s=d(N_a+2)=5(9)=45.0\ \text{mm}\), and the nominal free length is \(L_0=L_s+y_s=84.19\ \text{mm}\).
4. At solid, \(\tau_s=K_B8F_sD/(\pi d^3)=515.9\ \text{MPa}\), so \(n_s=S_{sy}/\tau_s=720/515.9=1.396\).

Verification

Units: Rate times deflection returns newtons, and wire-stress substitution uses newtons and millimetres to produce \(\text{N}/\text{mm}^2=\text{MPa}\). All length additions are in millimetres.

Signs and direction: Compression, positive load, and positive shortening are used consistently. Solid height is a geometric length, whereas \(y_s\) is the positive shortening needed to reach it.

Plausibility: The operating deflection is \(450/13.78=32.66\ \text{mm}\), below the \(39.19\ \text{mm}\) closure travel, and their force ratio is the specified \(1.20\).

Final result: \(\boxed{L_0=84.2\ \text{mm},\quad \tau_s=516\ \text{MPa},\quad n_s=1.40}\).

Engineering interpretation

The nominal design survives the defined solid load against the supplied yield value, but free-length and coil-diameter tolerances can consume closure margin. Buckling, set removal, clash, seat flatness, and vendor stress-relaxation data remain release checks.

10–8Critical Frequency of Helical Springs

Surge and resonance

Stress waves travel along the wire. Periodic forcing near a spring mode can create surge, coil clash, noise, and fatigue far above a quasistatic prediction. Natural frequency depends on spring mass, attached masses, boundaries, and damping; use a validated dynamic model or test for the assembled system rather than a body-only formula when consequences are significant.

10–9Fatigue Loading of Helical Compression Springs

Alternating and mean shear

For a linear spring, split loads before calculating stress: \(F_a=(F_{\max}-F_{\min})/2\) and \(F_m=(F_{\max}+F_{\min})/2\). Apply the same geometric stress coefficient to each. Shot peening can improve endurance only when coverage, intensity, material condition, and quality are controlled.

Stress components and Gerber reference
\[\tau_a=K_B\frac{8F_aD}{\pi d^3},\qquad \tau_m=K_B\frac{8F_mD}{\pi d^3},\qquad S_{se}=\frac{S_{sa}}{1-(S_{sm}/S_{su})^2}\]

10–10Helical Compression Spring Design for Fatigue Loading

Two checks, not one

A fatigue-safe spring must also resist maximum-load yielding and solid-height overload. Conversely, a static pass says nothing about life. Choose the fatigue criterion required by project practice and use endurance coordinates with documented surface process, cycle definition, environment, temperature, and reliability.

EXAMPLE 3Fatigue and static checks for a peened spring
A shot-peened compression spring has \(d=4.5\ \text{mm}\), \(D=31.5\ \text{mm}\), and cycles from \(120\) to \(420\ \text{N}\). Supplied certified properties are \(S_{ut}=1500\ \text{MPa}\) and \(S_{sy}=675\ \text{MPa}\). Using supplied peened endurance coordinates \(S_{sa}=396.5\ \text{MPa}\), \(S_{sm}=534.3\ \text{MPa}\) and the Gerber construction, find fatigue and maximum-load static safety factors.

Recognition strategy

The load range requires alternating and mean stress, not a static-only comparison. Curvature-corrected stress is linear in load, and the Gerber load-line equation provides an auditable fatigue factor; maximum stress separately checks yielding.

Given information

  • \(F_{\min}=120\ \text{N}\), \(F_{\max}=420\ \text{N}\), \(d=4.5\ \text{mm}\), and \(D=31.5\ \text{mm}\).
  • Certified strengths \(S_{ut}=1500\ \text{MPa}\), \(S_{sy}=675\ \text{MPa}\), with \(S_{su}=0.67S_{ut}\) specified for this calculation.
  • Peened endurance coordinates are supplied data requiring confirmation for wire size, process coverage, temperature, cycle definition, and reliability target.

Model and interpretation

Use compression as positive, so \(F_a=(F_{\max}-F_{\min})/2\) and \(F_m=(F_{\max}+F_{\min})/2\). Assume linear elasticity, stable seating, no surge, and a body-coil failure location; ends are checked separately.

Step-by-step solution
1. Compute \(C=31.5/4.5=7.00\), hence \(K_B=(4C+2)/(4C-3)=30/25=1.200\). Loads are \(F_a=150\ \text{N}\) and \(F_m=270\ \text{N}\).
2. The stress coefficient is \(K_B8D/(\pi d^3)=1.0563\ \text{MPa}/\text{N}\), giving \(\tau_a=158.4\ \text{MPa}\), \(\tau_m=285.2\ \text{MPa}\), and \(\tau_{\max}=443.7\ \text{MPa}\).
3. With \(S_{su}=0.67(1500)=1005\ \text{MPa}\), derive \(S_{se}=S_{sa}/[1-(S_{sm}/S_{su})^2]=552.6\ \text{MPa}\).
4. Solve \(n\tau_a/S_{se}+(n\tau_m/S_{su})^2=1\): \(0.08055n^2+0.2867n-1=0\), whose positive root is \(n_f=2.17\). The static factor is \(n_s=675/443.7=1.52\).

Verification

Units: Every term in the Gerber equation is a stress ratio and dimensionless. The linear coefficient in \(\text{MPa}/\text{N}\) times each load component gives the corresponding stress component.

Signs and direction: Both loads are compressive, so mean shear is positive and alternating amplitude is nonnegative. Only the positive quadratic root represents a physically meaningful safety factor.

Plausibility: Recombining components gives \(\tau_m+\tau_a=443.6\ \text{MPa}\), matching maximum-load stress; \(\tau_m-\tau_a=126.8\ \text{MPa}\) matches minimum-load stress.

Final result: \(\boxed{n_f=2.17\ \text{ by Gerber},\qquad n_s=1.52\ \text{ at }420\ \text{N}}\).

Engineering interpretation

Both simplified criteria pass, with static margin smaller numerically. Release still depends on the required reliability, actual shot-peen specification and verification, surface quality, corrosion, relaxation, end stresses, tolerances, and avoidance of surge.

10–11Extension Springs

Initial tension and end risk

A close-wound extension spring does not extend until load exceeds initial tension \(F_i\), after which \(F=F_i+ky\). Hooks and loops introduce bending, torsion, curvature, and fretting and often govern before the body. Analyze each end geometry separately and inspect the transition radius.

Working line after initial tension
\[F=F_i+ky\qquad(F\ge F_i)\]

10–12Helical Coil Torsion Springs

The wire bends

A torsion spring is named for its applied output moment, but its wire is principally bent. Curvature raises inner-fiber normal stress. Coil diameter changes as the spring winds, so arbor clearance, leg motion, active end length, and moment direction must be modeled at maximum rotation.

Corrected inner-fiber bending stress
\[\sigma_i=K_i\frac{32M}{\pi d^3},\qquad K_i=\frac{4C^2-C-1}{4C(C-1)}\]

10–13Belleville Springs

Compact nonlinear compliance

A coned-disk spring provides high force in small axial space and may have a deliberately nonlinear load-deflection curve. Parallel nesting approximately multiplies load, while series stacking approximately multiplies travel; friction and uneven load sharing make actual stacks depart from ideal arithmetic. Use manufacturer curves and stack guidance.

10–14Miscellaneous Springs

Select by function and failure mode

Constant-force strip springs, volute springs, wave springs, elastomers, and gas springs serve different envelope, travel, damping, and force-law needs. Selection must include hysteresis, creep or relaxation, temperature, corrosion, wear, cycle life, mounting, and safe containment rather than comparing nominal rate alone.

10–15Summary

Chapter recap

  • Use mean coil diameter, active coils, and a consistent curvature factor for body calculations.
  • Keep stiffness, static strength, fatigue, closure, stability, and dynamics as distinct checks.
  • Iterate standard wire and manufacturable geometry using traceable properties and realistic tolerances.

Before moving on

You should be able to recognize the spring family, define its load and sign convention, calculate rate and corrected stress, audit static and fatigue margins, and identify the vendor, standard, tolerance, stability, surge, and test evidence still required.