## 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 . 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).

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).

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.

#### Springs in Parallel

Apply unit deflection, and calculate load.

#### Axially Loaded Bar

Therefore:

Therefore:

#### Inclined Axial Spring

Apply a in the direction of motion.

Therefore, force in spring when extended is , and thus:

#### Example 1

Stiffness

Apply unit deflection to and calculate the total load.

Therefore:

If , then:

#### Example 2

Flexibility

Apply unit load to and calculate the total deflection.

, therefore:

Therefore :

Thus:

If , then:

#### Equivalent Spring Constant Equations

1. axial springs in parallel

2. axial springs in series

3.Springs in parallel and series

4.Inclined axial spring

5.Rotating bar with spring support

6.Rigid bar supported on two springs

7.Rigid bar supported on three springs

8.Axially loaded bar

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

9.Axially loaded tapered bar

10.Axially helical spring

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

11.Torsion of a uniform shaft

Where is the torsional constant of cross section ().

12.Torsion of a tapered circular shaft

13.Spiral torsional spring

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

14.Cantilever bean, end load

15.Simply supported bean, load at midspan

16.Simply supported bean, load anywhere between supports

17.Fixed-fixed beam, load at midspan

18.Fixed-fixed beam, off-center load

19.Propped cantilever, load at midspan

20.Propped cantilever, load at free end

###### Fixed-fixed beam*

###### Fixed-pinned beam with overhang*

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

###### Pinned-pinned beam with overhang*

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

###### Fixed-fixed beam with lateral displacement

* Axial extensions due to axial end constraints considered negligible

#### Torsional Oscillations

Let . From the strength of the material:

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

For soild circular cross sections: