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Document Title
Fundamentals of Stress and Vibration Chapter Title
[A Practical guide for aspiring Designers / Analysts] 2. Engineering Mechanics

Substituting these values for (M and L ) and a value of 1 for (M and L ) we get:
1
2
2
1
M 1 L 1 −3 1 1 −3
= = 0.06286
M 2 L 2 0.45 0.3048

The density in SI unit is given as follows:
ρ = ρ British = ρ = 498.3 ≈ 7909.52 kg
SI
conversion factor SI 0.06286 m 3

Example 2: A component modeled in millimeters, whose vibration or dynamic

analysis is carried out. Determine the appropriate unit of density for numerical
simulation.

Solution: consider a dynamic force, say, centrifugal force, which is given by mω r .
2

therefore, we have:

2
2
F = mω r , this can be rewritten as F = volume ∗ density ∗ ω r - - - - (2.8)
Let us rewrite equation (2.8) in terms of their respective units.
kg 1
3
N = mm ∗ 3 ∗ 2 ∗ mm - - - - (2.9)
mm s
3
The unit of density is chosen to be kg mm , as the model is in mm.

Simplifying equation (2.9), we get:

kg
N = mm - - - - (2.10)
s 2
Since newton N has the units kg m , we could conclude that the RHS of equation . is not in
s 2
newton.

This shows, that, the density must be entered in tonne mm . Therefore, substituting
3

the new unit of density in equation (2.9), we get:

tonne 1 kg × 10 3 kg m
3
N = mm ∗ ∗ ∗ mm = mm =
mm 3 s 2 s 2 s 2

One could test the fact that, for a dynamic analysis, the computed natural frequencies go
3
completely incorrect if the density is input in terms of kg mm instead of

3
tonne mm .

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