SME
Mechanical Vibrations · Problem type
Transmissibility: how much force a mount lets through
Use this for “A pump runs at 3 times the natural frequency of its mount, with ζ = 0.05. What share of its shaking force reaches the floor?”
Diagram
TR
TR
=
√(1
+
(2ζr)
2
)
÷
√(
(1 −
r
2
)
2
+
(2ζr)
2
)
TR = 0.13
Every
curve
passes
TR
=
1
at
r
=
√
2
:
only
above
it
does
the
mount
isolate.
This is an example. Type your own values to start a new problem.
TR
=
1
+
(2
×
×
)
2
÷
(1
−
r
2
)
2
+
(2
×
×
)
2
I
Isolation (1 − TR)
calculated · %
Formulas
TR
=
√(1
+
(2 ×
ζ
×
r
)
2
)
÷
√(
(1 −
r
2
)
2
+
(2 ×
ζ
×
r
)
2
)
0.1304
=
√(1
+
(2 × 0.05 × 3)
2
)
÷
√(
(1 −
3
2
)
2
+
(2 × 0.05 × 3)
2
)
I
=
100
×
(1
−
TR)
87 = 100 × (1 − 0.1304)
Assumptions
•
Steady state: the force reaches the floor through the spring and the damper together.
•
TR < 1 only above r = √2. Below it I is negative: the mount passes on more force than it gets.
•
Base motion passed to a mounted body follows the same TR.
Step-by-step
WE KNOW
r = 3, ζ = 0.05
FIND
transmissibility (TR), isolation (1 − TR) (I)
Step 1 · Find transmissibility (TR)
Use TR = √(1 + (2 × ζ × r)²) ÷ √((1 − r²)² + (2 × ζ × r)²)
The floor feels the spring force and the damper force; TR is their total over F₀.
TR
=
√(1
+
(2 × 0.05 × 3)
2
)
÷
√(
(1 −
3
2
)
2
+
(2 × 0.05 × 3)
2
)
TR
=
√(1
+
(0.1 × 3)
2
)
÷
√(
(1 − 9)
2
+
(0.1 × 3)
2
)
TR
=
√
1 +
0.3
2
÷
√(
(−8)
2
+
0.3
2
)
TR
=
√
1 + 0.09
÷
√
64 + 0.09
TR
=
√
1.09
÷
√
64.09
TR
=
1.044
8.006
TR = 0.1304
Step 2 · Find isolation (1 − TR): I
Use I = 100 × (1 − TR)
The share of the force kept off the floor, as a percent.
I
=
100
×
(1
−
0.1304)
I
=
100
×
0.8696
I
=
87%
Check
Put the numbers back in. Both sides should match.
0.1304
=
√(1
+
(2 × 0.05 × 3)
2
)
÷
√(
(1 −
3
2
)
2
+
(2 × 0.05 × 3)
2
)
✓
87 = 100 × (1 − 0.1304) ✓
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Rotating unbalance: the shaking of an off-balance rotor
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