In the precise design of compression springs and tension springs, the calculation of shear stress is a central component of the strength analysis. It indicates the extent to which the spring material is stressed by the applied forces and thus forms an important basis for the spring’s mechanical load-bearing capacity, operational reliability, and service life. In compression springs, the axially acting compressive force in the spring wire is essentially converted into torsional stress.
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Shear stress for compression springs
In compression springs, the axially acting compressive force in the spring wire is essentially converted into a torsional stress. Under a static or quasi-static load on a compression spring, the resulting shear stress τ is calculated as follows.
Shear stress compression spring from force:
\Large\tau=\frac{8DF}{\pi d^{3}}Shear stress Compression spring from displacement:
\Large\tau=\frac{Gds}{\pi nD^{2}}
When a compression spring is subjected to dynamic loading, the corrected shear stressτk applies. The shear stress distribution across a spring’s wire cross-section is uneven; the highest stress occurs at the spring’s inner diameter. The maximum stress can be approximated using the stress correction factor k, which depends on the coil ratio (ratio of the mean diameter to the wire diameter) of the spring. For springs subjected to dynamic loading, the following applies:
Corrected shear stress compression spring:
\Large \tau_{{\kappa}}= \kappa \cdot \tauwhere the following holds for k (according to Bergsträsser): \Large \kappa=\frac{\frac{D}{d}+0.5}{\frac{D}{d}-0.75}
Now the comparison is made with the permissible voltage.
Permissible tension of compression spring:
\Large \tau_{{zul}}=0.5\cdot R_{{m}}or.
\Large \tau_{{czul}}= 0.56 \cdot R_{{m}}
The values for theMinimum tensile strength R m are dependent on the wire thickness and can be found in the standards of the corresponding materials.
As a rule, it must be possible to compress compression springs up to the block length, which is why the permissible stress for the block length is t czul to consider.
Under dynamic loading, the lower and upper stresses (tk1andtk2) of the corresponding stroke must be determined. The difference is the stroke stress. Neither the upper stress nor the stroke stress may exceed the corresponding allowable values. These values can be found in the fatigue strength diagrams of EN 13906-1:2002. If the stresses satisfy this comparison, the spring is fatigue-resistant with a limit load cycle of10⁷.
Shear stress for tension springs
When a tension spring is subjected to a static or quasi-static load, the existing shear stress τ is calculated as follows.
Shear stress:
\Large \tau=\frac{8DF}{\pi d^{3}}
When a tension spring is subjected to dynamic loading, no universally applicable fatigue strength values can be specified, as additional stresses may occur at the bends of the eyes, some of which may exceed the permissible stress levels. Tension springs should therefore be subjected to static loads whenever possible. If dynamic loading cannot be avoided, bent-on eyes should be avoided, and rolled or screwed-in end pieces should be used instead. It is advisable to conduct a service life test under the intended operating conditions. Surface hardening by shot peening is not feasible due to the closely spaced coils.
Corrected shear stress:
\Large \tau_{{\kappa}}=\kappa\cdot\tauAllowable stress:
\Large \tau_{{zul}}=0.45 \cdot R_{{m}}
The existing maximum stresstn at the maximum spring travelsn is set equal to the allowable stress. However, to avoid relaxation, only 80% of this spring travel should be utilized in practice.
\Large s_{{2}}= 0.8 \cdot s_{{n}}
Explanation of Formula Symbols
d = Wire diameter (mm)
D = Mean coil diameter (mm)
F = Spring force (N)
G = Shear modulus (N/mm²)
n = Number of active coils (units)
Rm = Minimum tensile strength (N/mm²)
s = Spring travel (mm)
τ = Shear stress (N/mm²)
τzul = Allowable shear stress (N/mm²)
τczul = Allowable shear stress at block length (N/mm²)
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