Series and parallel springs

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

Parallel		Series

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.

Formulas

Equivalent spring

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

Quantity	In Parallel	In Series

Equivalent spring constant	$k_{eq} = k_1 + k_2$	$\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$
Equivalent compliance	$\frac{1}{c_{eq}} = \frac{1}{c_1} + \frac{1}{c_2}$	$c_{eq} = c_1 + c_2$
Deflection (elongation)	$x_{eq} = x_1 = x_2$	$x_{eq} = x_1 + x_2$
Force	$F_{eq} = F_1 + F_2$	$F_{eq} = F_1 = F_2$
Stored energy	$E_{eq} = E_1 + E_2$	$E_{eq} = E_1 + E_2$

Partition formulas

Quantity	In Parallel	In Series

Deflection (elongation)	$x_1 = x_2 \,$	$\frac{x_1}{x_2} = \frac{k_2}{k_1} = \frac{c_1}{c_2}$
Force	$\frac{F_1}{F_2} = \frac{k_1}{k_2} = \frac{c_2}{c_1}$	$F_1 = F_2 \,$
Stored energy	$\frac{E_1}{E_2} = \frac{k_1}{k_2} = \frac{c_2}{c_1}$	$\frac{E_1}{E_2} = \frac{k_2}{k_1} = \frac{c_1}{c_2}$

Derivations

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:

F_b = -k_{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 F_b. This means that

F_1 = -k_1 x_1= F_2=-k_2 x_2 .\,

We can solve for $x_{2}\,$ :

x_2 = \frac{k_1}{k_2} x_1 .\,

Now we just plug this back into F_b:

$F_b \,$	$= -k_{eq}\left( x_1 + \frac{k_1}{k_2} x_1\right) \,$
	$F_b= -k_{eq}\left( \frac{k_2+k_1}{k_2} \right)x_1 \,$

Now, we equate F_b = F₁:

-k_{eq}\left( \frac{k_2+k_1}{k_2} \right)x_1 = - k_1 x_1 .\,

Canceling x₁ we get:

k_{eq} = \frac{k_1 k_2 }{k_2 + k_1} .\,

Which can also be written:

\frac{1}{k_{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:

$F_b \,$	$= F_1 + F_2 \,$
	$= -k_1 x - k_2 x \,$

So the force on the block is

F_b = - (k_1 + k_2) x. \,

Which is why we can define the equivalent spring constant as

k_{eq} = k_1 + k_2 . \,

Compressed Distance

In the case where two springs are in series, the force of the springs on each other are equal:

F_1 = F_2 \,

-k_1 x_1 = -k_2 x_2. \,

From this we get a relationship between the compressed distances for the in series case:

\frac{x_1}{x_2} = \frac{k_2}{k_1}. \,

Energy Stored

For the series case, the ratio of energy stored in springs is:

\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:

\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,

\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

\frac{E_1}{E_2} = \frac{k_1}{k_2}. \,

References

↑ Keith Symon (1971), Mechanics. Addison-Wesley. ISBN 0-201-07392-7

This article is issued from Wikipedia - version of the 11/21/2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.