Question

You are given four springs : 2.5 N/m, 5 N/m, 7.5N/m, and 10 N/m

How do you solve for the smallest equivalent stiffness that can be made using only three of these springs?

Answer 1

For smallest equivalent stiffness, choose the 3 springs with the lowest stiffness. So, choose: 2.5,5,7.5 N/m.
Now, you can either orient these springs in parallel or you can put them in series.
If in parallel:
\[k _{eqv} = k _{1}+k _{2}+k _{3} = 2.5+5+7.5 = 15 N/m\]
If in series:
\[\frac{ 1 }{ k _{eqv} }=\frac{ 1 }{ k _{1} }+\frac{ 1 }{ k _{2} }+\frac{ 1 }{ k _{3} }=\frac{ 1 }{ 2.5 }+\frac{ 1 }{ 5 }+\frac{ 1 }{ 7.5 }=0.73333\]
\[k _{eqv}=\frac{ 1 }{ 0.73333 } = 1.3636N/m\]
So, the springs with stiffnesses 2.5,5,7.5 N/m in a series configuration, will give the lowest possible stiffness of about 1.36 N/m.