Series and parallel springs

In mechanics, two or more springs are said to be in series when they are connected end-to-end or point to point, and it is said to be in parallel when they are connected side-by-side; in both cases, so as to act as a single spring:

Series	and	Parallel
	and

More generally, two or more springs are in series when any external stress applied to the ensemble gets applied to each spring without change of magnitude, and the amount strain (deformation) of the ensemble is the sum of the strains of the individual springs. Conversely, they are said to be in parallel if the strain of the ensemble is their common strain, and the stress of the ensemble is the sum of their stresses.

Any combination of Hookean (linear-response) springs in series or parallel behaves like a single Hookean spring. The formulas for combining their physical attributes are analogous to those that apply to capacitors connected in series or parallel in an electrical circuit.

The following table gives formula for the spring that is equivalent to a system of two springs, in series or in parallel, whose spring constants are ${\displaystyle k_{1}}$ and ${\displaystyle k_{2}}$.^[1] (The compliance ${\displaystyle c}$ of a spring is the reciprocal ${\displaystyle 1/k}$ of its spring constant.)

Quantity	In Series		In Parallel
Equivalent spring constant	${\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}}$		${\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}}$
Equivalent compliance	${\displaystyle c_{\mathrm {eq} }=c_{1}+c_{2}}$		${\displaystyle {\frac {1}{c_{\mathrm {eq} }}}={\frac {1}{c_{1}}}+{\frac {1}{c_{2}}}}$
Deflection (elongation)	${\displaystyle x_{\mathrm {eq} }=x_{1}+x_{2}}$		${\displaystyle x_{\mathrm {eq} }=x_{1}=x_{2}}$
Force	${\displaystyle F_{\mathrm {eq} }=F_{1}=F_{2}}$		${\displaystyle F_{\mathrm {eq} }=F_{1}+F_{2}}$
Stored energy	${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$		${\displaystyle E_{\mathrm {eq} }=E_{1}+E_{2}}$

Quantity	In Series	In Parallel
Deflection (elongation)	${\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}={\frac {c_{1}}{c_{2}}}}$	${\displaystyle x_{1}=x_{2}\,}$
Force	${\displaystyle F_{1}=F_{2}\,}$	${\displaystyle {\frac {F_{1}}{F_{2}}}={\frac {k_{1}}{k_{2}}}={\frac {c_{2}}{c_{1}}}}$
Stored energy	${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{2}}{k_{1}}}={\frac {c_{1}}{c_{2}}}}$	${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}={\frac {c_{2}}{c_{1}}}}$

Equivalent Spring Constant (Series)

When putting two springs in their equilibrium positions in series attached at the end to a block and then displacing it from that equilibrium, each of the springs will experience corresponding displacements x1 and x2 for a total displacement of x1 + x2. We will be looking for an equation for the force on the block that looks like: ${\displaystyle F_{b}=-k_{\mathrm {eq} }(x_{1}+x_{2}).\,}$ The force that each spring experiences will have to be same, otherwise the springs would buckle. Moreover, this force will be the same as Fb. This means that ${\displaystyle F_{1}=-k_{1}x_{1}=F_{2}=-k_{2}x_{2}=F_{b}.\,}$ Working in terms of the absolute values, we can solve for ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{2}\,}$: ${\displaystyle x_{1}~=~{\frac {F_{1}}{k_{1}}}\,,\qquad x_{2}~=~{\frac {F_{2}}{k_{2}}}}$, and similarly, ${\displaystyle x_{1}~+~x_{2}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}$. Substituting ${\displaystyle x_{1}\,}$ and ${\displaystyle x_{2}\,}$ into the latter equation, we find ${\displaystyle {\frac {F_{1}}{k_{1}}}~+~{\frac {F_{2}}{k_{2}}}~=~{\frac {F_{b}}{k_{\mathrm {eq} }}}}$. Now remembering that ${\displaystyle F_{1}~=~F_{2}~=~F_{b}}$, we arrive at ${\displaystyle {\frac {1}{k_{\mathrm {eq} }}}={\frac {1}{k_{1}}}+{\frac {1}{k_{2}}}.\,}$

Equivalent Spring Constant (Parallel)

Both springs are touching the block in this case, and whatever distance spring 1 is compressed has to be the same amount spring 2 is compressed. The force on the block is then: ${\displaystyle F_{b}\,}$ ${\displaystyle =F_{1}+F_{2}\,}$ ${\displaystyle =-k_{1}x-k_{2}x\,}$ So the force on the block is ${\displaystyle F_{b}=-(k_{1}+k_{2})x.\,}$ Which is how we can define the equivalent spring constant as ${\displaystyle k_{\mathrm {eq} }=k_{1}+k_{2}.\,}$

Compressed Distance
In the case where two springs are in parallel it is immediate that: ${\displaystyle x_{1}=x_{2}\,}$ ${\displaystyle F_{1}=-k_{1}x_{1}}$ and ${\displaystyle F_{2}=-k_{2}x_{2}.\,}$	In the case where two springs are in series, the force of the springs on each other are equal: ${\displaystyle F_{1}=F_{2}\,}$ ${\displaystyle -k_{1}x_{1}=-k_{2}x_{2}.\,}$ From this we get a relationship between the compressed distances for the in series case: ${\displaystyle {\frac {x_{1}}{x_{2}}}={\frac {k_{2}}{k_{1}}}.\,}$

Energy Stored

For the series case, the ratio of energy stored in springs is: ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x_{1}^{2}}{{\frac {1}{2}}k_{2}x_{2}^{2}}},\,}$ but there is a relationship between x1 and x2 derived earlier, so we can plug that in: ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}\left({\frac {k_{2}}{k_{1}}}\right)^{2}={\frac {k_{2}}{k_{1}}}.\,}$ For the parallel case, ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {{\frac {1}{2}}k_{1}x^{2}}{{\frac {1}{2}}k_{2}x^{2}}}\,}$ because the compressed distance of the springs is the same, this simplifies to ${\displaystyle {\frac {E_{1}}{E_{2}}}={\frac {k_{1}}{k_{2}}}.\,}$ GATO Ignace notes about spring associations

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• Duality (mechanical engineering)