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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

    \[\text{Stress} = E \cdot \text{Strain}\]


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

    \[F = \frac{EA\Delta}{L}\]


    \[k_e = \frac{EA}{L}\]

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

Apply unit deflection to m and calculate the total load.

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

    \[F = \frac{k_1}{4} + k_2\]


    \[k_e = \frac{k_1}{4} + k_2\]

If k_1 = k_2 = k, then:

    \[k_e = \frac{5}{4}k\]

Example 2

Apply unit load to m and calculate the total deflection.

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

    \[\Delta_C = \frac{4}{k_1}\]

Therefore :

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


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

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)}

Fixed-fixed beam with lateral displacement

\Delta = \frac{P\ell^3}{12EI}
k = \frac{12EI}{\ell^3}
y = \frac{P}{12EI}(3\ell x^2-2x^3)

* Axial extensions due to axial end constraints considered negligible

Torsional Oscillations

    \[\begin{split}+\circlearrowright \sum{M_o} &= J_o\ddot{\theta} \\&= -\hat{k}\theta\end{split}\]

    \[J\ddot{\theta} + \hat{k}\theta = 0\]

Let P = \sqrt{\frac{\hat{k}}{J}}. From the strength of the material:

    \[\tau = \mu \alpha I\]

Where \mu is the modulus of rigidity, \alpha is the angle of twist/unit length and I is the polar second moment of inertia. Therefore:

    \[\tau = \mu \frac{\theta}{\ell}I = \hat{k}\theta\]

    \[\hat{k} = \frac{\mu I}{\ell}\]

For soild circular cross sections:

    \[I = \frac{\pi d^4}{32}\]

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