Transverse vibration of beams

Why it matters

Wings, tailplanes, fuselages, propeller and helicopter blades, antennas and pipes all vibrate mainly in bending. Their bending frequencies decide whether engine, rotor or gust excitation will cause resonance, and the first wing bending mode is one of the two modes that combine in classical flutter. The uniform Euler–Bernoulli beam gives closed-form frequencies that every engineer uses to sanity-check finite-element results.

Key ideas

Equation of motion. For a slender beam with flexural rigidity EI and mass per unit length m, transverse deflection w(x, t) satisfies

EI·∂⁴w/∂x⁴ + m·∂²w/∂t² = f(x, t)

It combines the static relation EI·w⁗ = load with the inertia load −m·ẅ. Assumptions (Euler–Bernoulli): plane sections stay plane and normal to the neutral axis, so shear deformation and rotary inertia are neglected; small deflections; linear elastic material. These are good when the length is more than about ten times the depth and for the lower modes. For deep beams or high modes, Timoshenko theory (adding shear and rotary inertia) gives lower frequencies.

Free vibration. Put w = W(x)·sin(ω·t): W⁗ − β⁴·W = 0 with β⁴ = m·ω²/(EI). The general solution is W = C₁·sin βx + C₂·cos βx + C₃·sinh βx + C₄·cosh βx. Four boundary conditions (two per end) give the frequency equation:

• Clamped: W = 0, W′ = 0.
• Pinned (simply supported): W = 0, W″ = 0 (zero moment).
• Free: W″ = 0, W‴ = 0 (zero moment and zero shear).

Frequency equations and roots (β_n·L).

• Simply supported: sin βL = 0, βL = nπ: 3.1416, 6.2832, 9.4248 …; modes sin(nπx/L).
• Cantilever (clamped–free): cos βL·cosh βL = −1: 1.8751, 4.6941, 7.8548 …
• Clamped–clamped and free–free: cos βL·cosh βL = 1: 4.7300, 7.8532, 10.9956 … (free–free also has two rigid-body modes at zero frequency).
• Clamped–pinned: tan βL = tanh βL: 3.9266, 7.0686 … For large n, βL ≈ (2n − 1)·π/2 for the cantilever and (2n + 1)·π/2 for clamped–clamped.

Frequencies. ω_n = (β_n·L)²·√(EI/(m·L⁴)). Two scaling laws follow: frequency ∝ 1/L² (halving the length multiplies every frequency by four), and frequencies spread out as n² (for the simply supported beam 1 : 4 : 9), unlike strings and bars (1 : 2 : 3). For a cantilever ω₂/ω₁ = (4.6941/1.8751)² = 6.27.

Mode shapes. Mode n has n − 1 interior nodes. Modes are orthogonal: ∫m·W_i·W_j·dx = 0 for i ≠ j, so modal analysis applies as for discrete systems.

Other effects. An axial tensile force raises bending frequencies and compression lowers them; for a simply supported beam ω² = ω₀²·(1 − P/P_cr), so the frequency falls to zero at the buckling load. Rotation (helicopter and propeller blades) adds centrifugal stiffening. Lumped masses (engines, stores) are handled by Rayleigh or Rayleigh–Ritz (earlier topic).

Aeroelastic link. The wing is modelled as a cantilever beam in bending coupled with a torsion member; the uncoupled bending and torsion frequencies and their ratio are the first inputs to divergence and flutter estimates.

Formulas

EI·∂⁴w/∂x⁴ + m·∂²w/∂t² = 0 — free vibration; EI (N·m²), m mass per unit length (kg/m), w (m).

β⁴ = m·ω²/(EI) — β (1/m).

ω_n = (β_n·L)²·√(EI/(m·L⁴)) — natural frequency (rad/s); f_n = ω_n/(2π) (Hz).

Simply supported: β_n·L = n·π. Cantilever: cos βL·cosh βL = −1, β₁L = 1.8751, β₂L = 4.6941, β₃L = 7.8548. Clamped–clamped: cos βL·cosh βL = 1, β₁L = 4.7300, β₂L = 7.8532.

m = ρ·A — mass per unit length from density (kg/m³) and area (m²).

ω² = ω₀²·(1 − P/P_cr) — simply supported beam with compressive axial load P (N), P_cr = π²·EI/L².

Worked examples

Example 1 (standard). A simply supported steel bar (E = 200 GPa, ρ = 7850 kg/m³) of span 2 m has a 40 mm wide × 20 mm deep rectangular section and vibrates in the vertical plane. Find f₁ and f₂.

1. A = 0.04 × 0.02 = 8 × 10⁻⁴ m²; m = ρ·A = 6.28 kg/m.
2. I = b·h³/12 = 0.04 × 0.02³/12 = 2.667 × 10⁻⁸ m⁴; EI = 5333 N·m².
3. √(EI/(m·L⁴)) = √(5333/(6.28 × 16)) = 7.286 rad/s.
4. ω_n = (nπ)²·√(EI/(m·L⁴)): ω₁ = 9.870 × 7.286 = 71.9 rad/s, f₁ = 11.44 Hz; f₂ = 4 × f₁ = 45.8 Hz.

Answer: f₁ ≈ 11.4 Hz, f₂ ≈ 45.8 Hz.

Example 2 (GATE level). A flat aluminium strip (E = 70 GPa, ρ = 2700 kg/m³), 60 mm wide and 6 mm thick, is clamped at one end as a cantilever of length 1.5 m and bends about its thin direction. (a) Find f₁ and f₂. (b) To what length must it be shortened for f₁ = 10 Hz?

1. A = 3.6 × 10⁻⁴ m², m = 0.972 kg/m; I = 0.06 × 0.006³/12 = 1.08 × 10⁻⁹ m⁴, EI = 75.6 N·m².
2. √(EI/(m·L⁴)) = √(75.6/(0.972 × 5.0625)) = 3.920 rad/s.
3. ω₁ = 1.8751² × 3.920 = 3.516 × 3.920 = 13.78 rad/s, f₁ = 2.19 Hz. ω₂ = 4.6941² × 3.920 = 86.37 rad/s, f₂ = 13.75 Hz (ratio 6.27).
4. (b) f₁ ∝ 1/L², so L_new = 1.5 × √(2.19/10) = 0.70 m.

Answer: (a) f₁ ≈ 2.19 Hz, f₂ ≈ 13.7 Hz; (b) L ≈ 0.70 m.

Common mistakes

• Using (β_n·L) instead of (β_n·L)², or applying the 1.8751 root without squaring it.
• Taking m as density instead of mass per unit length ρ·A.
• Using the wrong root set for the boundary conditions (for example 4.730 is the first clamped–clamped root, not a second mode).
• Choosing I about the wrong axis for a flat section.
• Assuming beam frequencies are harmonic multiples like a string; they grow as n².
• Applying Euler–Bernoulli results to short, deep beams or high modes without shear and rotary-inertia corrections.

For GATE AE

Expect frequencies of simply supported and cantilever beams from E, I, ρ, A and L, frequency ratios between modes or boundary conditions, scaling with length and thickness (f ∝ h/L² for a rectangular strip), boundary-condition statements, and the effect of axial load. Practise reading the correct βL root and keeping units consistent.

Quick check

1. For a simply supported beam, what is f₃/f₁?
2. If a cantilever's length is doubled, what happens to f₁?
3. Which two boundary conditions apply at a free end?
4. For a cantilever, what is ω₂/ω₁?
5. For EI = 2000 N·m², m = 5 kg/m and L = 2 m simply supported, find f₁.

Answers: 1. 9; 2. it falls to one quarter; 3. zero bending moment (W″ = 0) and zero shear (W‴ = 0); 4. about 6.27; 5. ω₁ = π² × 5 = 49.3 rad/s, f₁ = 7.85 Hz.