Two springs of stiffnesses KA and KB are placed one inside the other such that they are compressed by the same amount under axial load. The equivalent stiffness of the two springs will be
1
\(\frac{1}{{{K_A}}} + \frac{1}{{{K_B}}}\)
2
KA + KB
3
\(\frac{{{K_A}{K_B}}}{{{K_A} + {K_B}}}\)
4
\(\frac{{{K_A} + {K_B}}}{2}\)