Base excitation and vibration isolation

Base-excited SDOF systems, displacement and force transmissibility, the √2 crossover, the effect of damping on isolation and the design of isolator mounts.

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Why it matters

Much aircraft equipment is not forced directly: it sits on a vibrating floor, rack or engine pylon, and the vibration arrives through its supports. Avionics, inertial sensors, cameras and engine accessories are mounted on isolators so that as little of that motion as possible reaches them, and machines such as pumps and engines are mounted so that their forces do not shake the airframe. Both problems are solved with the same transmissibility curve, and the key design rule is surprising: isolation only begins above √2 times the natural frequency.

Key ideas

Base excitation. A mass m sits on a spring k and damper c whose lower end (the base) moves as y(t) = Y·sin(ω·t). The spring and damper forces depend on the relative motion x − y, so

m·ẍ + c·(ẋ − ẏ) + k·(x − y) = 0, or m·ẍ + c·ẋ + k·x = k·y + c·ẏ

The base acts on the mass through both the spring and the damper. This is why the damper term appears in the numerator of the transmissibility.

Displacement transmissibility. The ratio of mass amplitude to base amplitude is

T_d = X/Y = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²), with r = ω/ω_n

Force transmissibility. For a machine of mass m excited by a force F₀·sin(ω·t) and mounted on k and c, the force reaching the foundation is the sum of the spring and damper forces, F_T = √((k·X)² + (c·ω·X)²). Its ratio to F₀ has exactly the same expression: F_T/F₀ = T_d. So one curve, the transmissibility TR, governs both protecting equipment from a moving base and protecting a structure from a vibrating machine.

Features of the TR curve.

  • r = 0: TR = 1 (the mass moves with the base).
  • r = 1: TR ≈ 1/(2ζ) for light damping: the mounts amplify.
  • r = √2: TR = 1 for every value of ζ. All curves cross here.
  • r > √2: TR < 1, isolation. Undamped, TR = 1/(r² − 1).
  • In the isolation region, more damping increases TR (worse isolation), because the damper becomes a stiff link at high frequency. Below √2, more damping reduces TR.

Design trade-off. Good isolation needs a low natural frequency (soft mounts, large r), but soft mounts mean large static sag, large relative motion when the machine runs up through resonance, and poor control of the equipment under manoeuvre load factors. Some damping is kept to limit the resonant peak during start-up and shut-down. Isolation efficiency is often quoted as 1 − TR.

Static deflection link. For undamped mounts, k·δ_st = m·g gives ω_n = √(g/δ_st), so choosing the required ω_n fixes the static deflection.

Relative motion. Accelerometers and seismometers use the relative displacement z = x − y: Z/Y = r²/√((1 − r²)² + (2ζr)²). This leads to the vibration-measurement topic later.

Assumptions. Linear mounts, rigid mass and rigid base, single-direction motion, steady state. Real isolators have stiffness and damping that vary with frequency, amplitude and temperature, and a mass on several mounts can also rock, so check all rigid-body modes in practice.

Formulas

m·ẍ + c·(ẋ − ẏ) + k·(x − y) = 0 — base excitation; y base displacement (m).

r = ω/ω_n, ζ = c/(2·√(k·m)).

TR = X/Y = F_T/F₀ = √(1 + (2ζr)²)/√((1 − r²)² + (2ζr)²) — displacement or force transmissibility (dimensionless).

TR = 1/(r² − 1) — undamped, r > √2 only.

r_required = √(1 + 1/TR) — undamped design, for a target TR < 1.

F_T = F₀·TR — force transmitted to the foundation (N).

Z/Y = r²/√((1 − r²)² + (2ζr)²) — relative displacement z = x − y.

ω_n = √(g/δ_st) — mounts' natural frequency from their static deflection δ_st (m).

Worked examples

Example 1 (standard). A 4 kg avionics box sits on mounts that give f_n = 10 Hz and ζ = 0.1. The aircraft floor vibrates with amplitude Y = 2 mm at 25 Hz. Find the box amplitude and compare it with the response if the floor shook at 10 Hz.

  1. r = ω/ω_n = 25/10 = 2.5; 2ζr = 0.5; 1 − r² = −5.25.
  2. TR = √(1 + 0.5²)/√(5.25² + 0.5²) = 1.1180/5.2738 = 0.212.
  3. X = TR·Y = 0.212 × 2 = 0.424 mm.
  4. At 10 Hz (r = 1): TR = √(1 + 0.2²)/0.2 = 1.0198/0.2 = 5.10, so X = 10.2 mm.
  5. Note: with no damping at r = 2.5, TR = 1/(6.25 − 1) = 0.190; damping makes isolation slightly worse.

Answer: X ≈ 0.42 mm at 25 Hz (79 % isolation); about 10.2 mm if forced at 10 Hz.

Example 2 (GATE level). A 20 kg fuel pump runs at 3000 rpm. Its mounts (neglect damping) must transmit only 10 % of the unbalance force to the airframe. Find the required natural frequency, the total mount stiffness and the static deflection.

  1. ω = 3000 × 2π/60 = 314.16 rad/s.
  2. TR = 1/(r² − 1) = 0.1, so r² = 1 + 1/0.1 = 11, r = 3.317.
  3. ω_n = ω/r = 314.16/3.317 = 94.72 rad/s (15.08 Hz).
  4. k = m·ω_n² = 20 × 94.72² = 1.794 × 10⁵ N/m (four mounts of about 44.9 kN/m each).
  5. δ_st = m·g/k = 20 × 9.81/1.794 × 10⁵ = 1.09 × 10⁻³ m.

Answer: ω_n ≈ 94.7 rad/s (15.1 Hz), k ≈ 1.79 × 10⁵ N/m, δ_st ≈ 1.09 mm.

Common mistakes

  • Using the forced-vibration magnification 1/√((1 − r²)² + (2ζr)²) for transmissibility and forgetting the √(1 + (2ζr)²) numerator.
  • Thinking isolation starts at r = 1; below r = √2 the mounts amplify the motion.
  • Adding damping to "improve" isolation at high frequency; it does the opposite above r = √2.
  • Mixing rpm, Hz and rad/s: convert rpm with ω = 2π·N/60.
  • Forgetting that the transmitted force includes the damper force, not just k·X.
  • Choosing mounts so soft that the static sag or manoeuvre deflection is unacceptable.

For GATE AE

Expect transmissibility at a given r and ζ, the required stiffness or static deflection for a target isolation, the frequency at which TR = 1 (r = √2), and statements on how damping affects isolation above and below √2. Practise the undamped shortcut r² = 1 + 1/TR and full damped calculations, and keep rpm-to-rad/s conversions clean.

Quick check

  1. At what frequency ratio is TR = 1 for any damping?
  2. Undamped mounts, r = 3. What is TR?
  3. In the isolation region, does increasing ζ raise or lower TR?
  4. What r is needed for 80 % isolation with undamped mounts?
  5. A mount deflects 2.5 mm under static load. What is its natural frequency in Hz?

Answers: 1. r = √2; 2. 1/8 = 0.125; 3. it raises TR; 4. TR = 0.2, r = √6 = 2.45; 5. ω_n = √(9.81/0.0025) = 62.6 rad/s, f_n = 9.97 Hz.

Interview questions

All Structural Dynamics and Aeroelasticity interview questions

Try answering each one aloud before you open it.

  1. 1.What is base excitation in the context of structural dynamics?Concept

    Base excitation refers to the motion or vibration applied to the base or support of a structure. This can occur due to external forces such as earthquakes, machinery vibrations, or other dynamic loads. The response of the structure to this excitation is crucial for understanding its dynamic behavior and ensuring its stability and integrity.

  2. 2.Explain the concept of vibration isolation and its importance in aerospace engineering.Concept

    Vibration isolation involves reducing the transmission of vibrations from a source to a structure or component. In aerospace engineering, it is crucial to protect sensitive equipment and ensure passenger comfort. Effective vibration isolation can prevent fatigue failure, reduce noise, and enhance the performance and lifespan of aerospace components.

  3. 3.How does a spring-damper system work in vibration isolation?Concept

    A spring-damper system works by absorbing and dissipating energy from vibrations. The spring provides a restoring force that opposes displacement, while the damper dissipates energy through friction or viscous resistance. Together, they reduce the amplitude of vibrations transmitted to the structure, enhancing isolation.

  4. 4.Why is it important to consider the natural frequency of a structure in base excitation scenarios?Application

    The natural frequency is the frequency at which a structure tends to oscillate in the absence of damping or external forces. If the frequency of base excitation matches the natural frequency, resonance can occur, leading to large amplitude oscillations and potential structural failure. Therefore, understanding and avoiding resonance is crucial for structural safety.

  5. 5.What happens if a vibration isolator is too stiff or too soft?Application

    Isolation needs the excitation frequency to be above √2 times the mount natural frequency, ideally r of 3 or more. A mount that is too stiff gives a high natural frequency, so r falls below √2 and the mount transmits or even amplifies the vibration, possibly at resonance. A mount that is too soft isolates well in steady running but gives large static sag, large excursions under manoeuvre loads and a big resonant amplitude when the machine runs up or down through its natural frequency, so travel limits and some damping are needed.

  6. 6.In what scenarios would you use a passive vibration isolation system over an active one?Application

    Passive vibration isolation systems are preferred when simplicity, reliability, and low maintenance are priorities. They are suitable for environments where the vibration characteristics are well-known and consistent. Active systems, which require power and control systems, are used when adaptive or precise control of vibrations is necessary, such as in variable or unpredictable environments.

  7. 7.How does damping affect the response of a structure to base excitation?Application

    Damping limits the response near resonance, where the transmissibility is about 1/(2ζ), so it protects equipment when the excitation sweeps through the natural frequency. All transmissibility curves cross TR = 1 at r = √2, and above that, in the isolation region, more damping increases the transmitted motion because the damper force c·ω grows with frequency and couples the mass to the base. Isolators therefore use only modest damping: enough to control the resonant peak, not so much that it spoils high-frequency isolation.

  8. 8.Calculate the natural frequency of a simple spring-mass system with a mass of 5 kg and a spring constant of 200 N/m.Numerical

    The natural frequency (f_n) of a spring-mass system is given by the formula: f_n = (1/2π) * √(k/m), where k is the spring constant and m is the mass. Substituting the given values: f_n = (1/2π) * √(200/5) = (1/2π) * √40 ≈ 1.006 Hz.

  9. 9.Determine the damping ratio of a system with a damping coefficient of 50 Ns/m, a mass of 10 kg, and a spring constant of 500 N/m.Numerical

    ζ = c/(2·√(k·m)). Here √(500 × 10) = √5000 = 70.71, so the critical damping coefficient is 2 × 70.71 = 141.4 N·s/m. ζ = 50/141.4 ≈ 0.354, a fairly heavily damped (but still underdamped) system.

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