Engineering at Alberta Courses » Equivalent spring constants

Review of single and multi-degree of freedom (mdof) systems: Equivalent spring constants

One of the components we need in these equations of motion is the spring constant $k$ . We can often find this for a system using static techniques. This is easily generalized for MDOF systems.

For SDOF systems, we can imagine the static response of the system using one of the approaches:

1.Apply a unit force (moment) to the mass (inertia) in the positive direction of motion then calculate the displacement that occurs (flexibility approach).

$k = \frac{\text{force (moment) in direction of motion}}{\text{displacement (rotation) in direction of motion}}$

2.Apply a unit displacement (rotation) to the mass (inertia) in the positive direction of motion then calculate the force (moment) required to maintain it (stiffness approach).

$k = \frac{\text{force (moment) in direction of motion}}{\text{displacement (rotation) in direction of motion}}$

It depends on the situation to determine the best one to use. Generally, for a series situation, use flexibility, while for parallel, use stiffness.

Springs in Series

Apply unit load, and calculate total deflection.

$\Delta_1 = \frac{1}{k_1} \ \ \Delta_2 = \frac{1}{k_2} \ \ \Delta_\text{total} = \frac{1}{k_1} + \frac{1}{k_2}$

$k_\text{eff} = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2}} = \frac{k_1k_2}{k_1+k_2}$

Springs in Parallel

Apply unit deflection, and calculate load.

$k_\text{eff} = k_1 + k_2$

Axially Loaded Bar

$\text{Stress} = E \cdot \text{Strain}$

Therefore:

$\sigma = \frac{F}{A} \equiv \Delta = E\epsilon = E \cdot \frac{\Delta}{L}$

$F = \frac{EA\Delta}{L}$

Therefore:

$k_e = \frac{EA}{L}$

Inclined Axial Spring

Apply a $\Delta$ in the direction of motion.

Therefore, force in spring when extended is $k\Delta\cos\theta$ , and thus:

$\frac{\text{Force in direction of motion}}{\Delta \ \text{in direction of motion}} = \frac{(k\Delta\cos\theta)\cos\theta}{\Delta} = k\cos^2\theta$

Example 1

Stiffness
Apply unit deflection to $m$ and calculate the total load.

$+\circlearrowleft\sum{M_A} = k_1\Big(\frac{1}{2}\Big)\Big(\frac{1}{2}\Big) + k_2(\text{L}) - F(\text{L}) = 0$

$F = \frac{k_1}{4} + k_2$

Therefore:

$k_e = \frac{k_1}{4} + k_2$

If $k_1 = k_2 = k$ , then:

$k_e = \frac{5}{4}k$

Example 2

Flexibility
Apply unit load to $m$ and calculate the total deflection.

$\Delta_B = \frac{2}{k_1}$ , therefore:

$\Delta_C = \frac{4}{k_1}$

Therefore :

$\begin{split}\Delta \ \text{at} \ M &= \Delta_C + \frac{1}{k_2} \\&= \frac{4}{k_1} + \frac{1}{k_2}\end{split}$

Thus:

$\begin{split}k_\text{eff} &= \frac{1}{\Delta}\\&= \frac{1}{\frac{4}{k_1}+\frac{1}{k_2}}\end{split}$

If $k_1 = k_2$ , then:

$k_\text{eff} = \frac{k}{5}$

Equivalent Spring Constant Equations

1. $n$ axial springs in parallel

$k_e = k_1 + k_2 + k_2 + k_3 + … + k_n$

2. $n$ axial springs in series

$k_e = \frac{1}{\frac{1}{k_1} + \frac{1}{k_2} + … + \frac{1}{k_n}}$

3.Springs in parallel and series

$k_e = \frac{k_1k_3 + k_2k_3}{k_1 + k_2 + k_3}$

4.Inclined axial spring

$k_e = k\cos^2\theta$

5.Rotating bar with spring support

$k_e = k\Big(\frac{L}{a}\Big)^2$

6.Rigid bar supported on two springs

$k_e = \frac{4k_1k_2}{k_1[1 + \frac{a}{L}]^2 + k_2[1 + \frac{a}{L}]^2}$

7.Rigid bar supported on three springs

$k_e = \frac{3k}{1 + \frac{3}{2}(a / L)^2}$

8.Axially loaded bar

$k_e = \frac{AE}{L}$

Where $A$ is the cross-sectional area, and $E$ is the elastic modulus.

9.Axially loaded tapered bar

$k_e = \frac{\pi E D_a D_b}{4L}$

10.Axially helical spring

$k_e = \frac{Gd^4}{G4nR^3}$

Where $n$ is the active number of turns, and $G$ is the elastic shear modulus.

11.Torsion of a uniform shaft

$k_e = \frac{GJ}{L}$

Where $J$ is the torsional constant of cross section ( $J = \frac{\pi d^4}{32}$ ).

12.Torsion of a tapered circular shaft

$k_e = \frac{3\pi}{32}\frac{D_b^4 G}{L\big[\frac{D_b}{D_a}+(\frac{D_b}{D_a})^2+(\frac{D_b}{D_a})^3\big]}$

13.Spiral torsional spring

$k_e = \frac{EI}{L}$

Where $E$ is Young’s modulus. $I$ is the moment of inertia of cross-sectional area, and $L$ is the total length of the spiral.

14.Cantilever bean, end load

$k_e = \frac{3EI}{L^3}$

15.Simply supported bean, load at midspan

$k_e = \frac{48EI}{L^3}$

16.Simply supported bean, load anywhere between supports

$k_e = \frac{3EIL}{a^2b^2}$

17.Fixed-fixed beam, load at midspan

$k_e = \frac{192EI}{L^3}$

18.Fixed-fixed beam, off-center load

$k_e = \frac{3EI(a+b)^3}{a^3b^3}$

19.Propped cantilever, load at midspan

$k_e = \frac{768EI}{7L^3}$

20.Propped cantilever, load at free end

$k_e = \frac{24EI}{a^2(3L + 8a)}$

Fixed-fixed beam*

$y = \frac{Pb^2}{6EI\ell^3}\big[(2b-3\ell)x^3 + 3\ell(\ell-b)x^2\big]$ $(x \leq a)$
$y = \frac{Pb^2}{6EI\ell^3}\big[(2b-3\ell)x^3 + 3\ell(\ell-b)x^2 + \frac{1^3}{b^2}(x-a)^3\big]$ $(x \geq a)$
$k = \frac{P}{y|_{x=a}}$
$k|_{a=\frac{\ell}{2}} = \frac{192EI}{\ell^3}$

Fixed-pinned beam with overhang*

$y = \frac{P}{12EI}\Big[3b\Big(1-\frac{b^2}{\ell^2}\Big)x^2-\frac{b}{\ell}\Big(3-\frac{b^2}{\ell^2}\Big)x^3\Big]$ $(x \leq a)$
$y = \frac{P}{12EI}\Big[3b\Big(1-\frac{b^2}{\ell^2}\Big)x^2-\frac{b}{\ell}\Big(3-\frac{b^2}{\ell^2}\Big)x^3 + 2(x-a)^3\Big]$ $(a \leq x \leq \ell)$
$y =\frac{-pba^2}{4EI\ell}(x-\ell)$ $(x \geq \ell)$
$k = \frac{P}{y|_{x=a}}$
$k|_{a=\frac{\ell}{2}} = \frac{768EI}{7\ell^3}$

Fixed-pinned beam with overhang (P at x = l + a)*

$y = \frac{Pa}{4EI\ell}(x^3-\ell x^2)$ $(x \leq \ell)$
$y = \frac{Pa}{4EI\ell}\Big[x^3-\ell x^2-\Big(\frac{2\ell}{3a}+1\Big)(x-1)^3\Big]$ $(x \geq \ell)$
$k = \frac{P}{y|_{x=\ell+a}} = \frac{12EI}{a^2(3\ell+4a)}$

Pinned-pinned beam with overhang*

$y = \frac{Pa}{6EI\ell}(a^2-\ell^2)(x-\ell)$ $(x \geq \ell)$

Pinned-pinned beam with overhang (P at x = l + a)*

$y = \frac{Pax}{6EI\ell}(x^2-\ell^2)$ $(x \leq \ell)$
$y = \frac{P}{6EI\ell}\big[ax(x^2-\ell^2)-(\ell+a)(x-\ell)^3\big]$ $(x \geq \ell)$
$k = \frac{P}{y|_{x = a+\ell}} = \frac{3EI}{a^2(a+1)}$