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 .
Springs in Parallel
Apply unit deflection, and calculate load.
Axially Loaded Bar
Inclined Axial Spring
Apply a in the direction of motion.
Therefore, force in spring when extended is , and thus:
Apply unit deflection, and calculate total load.
If , then:
Apply unit load, and calculate total deflection.
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-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
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: