Springs Questions and Answers

Answer: C

The spring stiffness doubles.

The stiffness of a helical spring is \(k = \frac{Gd^{4}}{8D^{3}n}\), where n is the number of active coils. Halving the length halves n, so

\[ k' = \frac{Gd^{4}}{8D^{3}(n/2)} = 2k \]

Put another way, each half carries the full load but stretches only half as much. The stiffness does not stay the same, halve or triple.

Answer: C

It is called spring stiffness.

Stiffness, or the spring rate, is the load needed per unit deflection, \(k = W/\delta\), in N/mm or N/m.

Flexural rigidity is EI, a beam's resistance to bending; torsional rigidity is GJ, a shaft's resistance to twisting; and Young's modulus is a property of the material, not of the spring.

Answer: D

The arrangement is two \(K_{2}\) springs in parallel, in series with one \(K_{1}\) spring.

Springs in parallel add their stiffnesses, and springs in series add their reciprocals. Two \(K_{2}\) springs in parallel give \(2K_{2}\), which then acts in series with \(K_{1}\).

\[ \begin{aligned} \frac{1}{k} &= \frac{1}{2K_{2}} + \frac{1}{K_{1}} \\ k &= \frac{2K_{1}K_{2}}{K_{1} + 2K_{2}} \end{aligned} \]

The series chain in A gives \(\frac{K_{1}K_{2}}{K_{1} + 2K_{2}}\), half the required value. B gives \(\frac{2K_{1}K_{2}}{2K_{1} + K_{2}}\), and the three parallel springs in C give \(K_{1} + 2K_{2}\).

Answer: C

The equivalent stiffness is \(\frac{K}{2}\).

Springs in series carry the same force, and their deflections add. So the reciprocals of the stiffnesses add:

\[ \begin{aligned} \frac{1}{K_{eq}} &= \frac{1}{K} + \frac{1}{K} = \frac{2}{K} \\ K_{eq} &= \frac{K}{2} \end{aligned} \]

Two equal springs end to end stretch twice as far under a given load, so the combination is half as stiff. 2K is the answer for two springs in parallel, where the stiffnesses add directly. K would mean the second spring adds nothing, and K/4 would be four springs in series.