Extension Spring Calculator: Rate, Extension and Stress
The spring rate, the extension under a force and the body shear stress of a helical extension spring.
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How this calculator works
An extension spring is wound with its coils pressed together, so it carries an initial tension before the coils start to separate. Beyond that the spring behaves like a compression spring with the same rate, k = G d⁴ / (8 D³ Na). The extension is the force above the initial tension divided by the rate.
The shear stress in the body uses the Wahl correction. The stresses in the hooks are higher than in the body and often govern, so check the hook design with the spring maker. Springs that cycle need a fatigue check.
Formulas used
- k = \frac{G d^4}{8 D^3 N_a}
- y = \frac{F - F_i}{k}
- \tau = K_W\frac{8 F D}{\pi d^3}
Worked example
With the default numbers in the calculator above:
- Spring index: 8
- Spring rate: 1.929 N/mm
- Extension beyond the closed length: 23.33 mm
- Shear stress in the body at the working force: 361.8 MPa
- Shear stress at the initial tension: 90.45 MPa
Change the numbers in the calculator to see the full working, every step, for your own case.
Common questions
What is initial tension?
The force needed to start opening the coils of a close-wound extension spring.
Why do extension springs break at the hooks?
The hook bend has a high stress concentration. Fatigue cracks usually start there.
Is the rate the same as a compression spring?
Yes, for the active coils. The difference is the initial tension and the end hooks.
Related calculators
Further reading: Shigley, Mechanical Engineering Design (mechanical springs).
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