How do extension spring ends differ from compression spring ends?

They need hooks or loops, with stress concentrated at the hook bend

They are squared and ground to seat properly

They need no special end detail

What is the effect of initial tension F_i in an extension spring?

Spring does not extend until applied load exceeds F_i, then F = F_i + k y

Spring extends from the first newton, with F = k y

It reduces the spring stiffness k

It makes the coils spaced apart

Practice · 6 marks

Extension springs, ends and spring combinations

Short answer

Compression spring ends are plain, ground, squared, or squared & ground; extension springs have hooks or loops and coils wound tight with initial tension. Initial tension provides a preload; no extension until load exceeds it. Equivalent stiffness: series 12 N/mm, parallel 50 N/mm. Series extension under 300 N is 25 mm.

Differentiate between compression and extension springs with respect to end details. Explain the initial tension in an extension spring. Two springs of stiffness 20 N/mm and 30 N/mm are used in series and in parallel; find the equivalent stiffness in each case, and the extension of the series combination under 300 N.

Compression spring ends

End type	Description	Inactive coils	Solid length
Plain	Ends just cut off	0	$d(n+1)$
Plain and ground	Ends flat-ground	1	$d\,n_t$ with $n_t=n+1$
Squared (closed)	Ends turned to touch the next coil	2	$d(n+3)$
Squared and ground	Closed and ground flat	2	$d(n+2)$

Squared and ground ends seat properly and give better load transfer and less buckling.

Extension spring ends

An extension spring works in tension, so its ends need hooks or loops (full loop, half loop, cross-over loop, extended hook, threaded plug). The stresses concentrate at the bend of the hook (bending and torsion), so a hook of larger radius lowers the stress and the hook is designed separately from the body. The body coils are wound tight (touching), unlike a compression spring where space is left between coils for deflection.

Initial tension

Extension springs are wound with coils pressed together, which stores a pretension $F_i$ . No extension takes place until the applied load exceeds $F_i$ . Then

F = F_i + k\,y

Initial tension keeps the spring tight, gives a definite starting load and reduces the length needed. It depends on spring index and manufacturing, and is usually limited so that the stress due to $F_i$ stays within the recommended band for the spring index.

Springs in combination

Series (same load, deflections add): $\dfrac{1}{k_s} = \dfrac{1}{k_1}+\dfrac{1}{k_2} = \dfrac{1}{20}+\dfrac{1}{30}$ , so $k_s = 12$ N/mm.

Parallel (same deflection, loads add): $k_p = k_1+k_2 = 20 + 30 = 50$ N/mm.

Extension of the series combination under 300 N: $y = 300/12 = 25$ mm.

Answer: $k_{series} = 12$ N/mm, $k_{parallel} = 50$ N/mm, extension $= 25$ mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Extension springs, ends and spring combinations

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