Engineering Acoustics. Malcolm J. Crocker

w(x,t) of a uniform beam is given by the Euler–Bernoulli beam theory as [10, 13]

(2.51)

      where E is the Young's modulus, ρ is the mass density, I is the cross‐sectional moment of inertia, and S is the cross‐sectional area. Assuming harmonic vibrations in the form

(2.52)

and substituting w(x,t) from Eq. (2.52) into Eq. (2.51) we get

(2.53)

The solution of Eq. (2.53) is

(2.54)

      where λ = (ω² ρS/EI)^1/4 and the C's are arbitrary constants that depend upon the boundary conditions (the deflections, slope, bending moment, and shear force constraints). Classical boundary conditions for a beam are

(2.55)

      (2.56)

      (2.57)

      A very important practical case is a cantilever beam (clamped‐free beam) of length L. In this case, the deflection and slope are zero at the clamped end, while the bending moment and shear force are zero at the free end, i.e.

(2.58)

(2.59)

Substituting the boundary conditions Eq. (2.58) and Eq. (2.59) into Eq. (2.54), we find that C₂ = −C₄, and we obtain the equation

(2.60)

The roots of the transcendental Eq. (2.60) can be obtained numerically. The first four roots are λ₁ L = 1.875, λ₂ L = 4.694, λ₃ L = 7.855, and λ₄ L = 10.996. For large values of n, the roots can be calculated using the equation

      (2.61)

      Noting that , we can solve for ω_n so that the first four natural frequencies of the cantilever beam are

      The mode shapes are given by [10, 13]

      (2.62)

where A_n is an arbitrary constant. Thus, the total solution for the free transverse vibration of the cantilever beam is

      (2.63)

Figure 2.16 shows the first four mode shapes for a cantilever beam.

      Example 2.10

      Determine the natural frequencies of a uniform beam which is simply supported at both ends.

      Solution

      Applying the boundary conditions w = 0 and at x = 0 in Eq. (2.54) leads to C₁ + C₃ = 0 and −λ₂ C₁ + λ₂ C₃ = 0. These equations are satisfied if C₁ = C₃ = 0.

      Applying the boundary conditions w = 0 and at x = L in Eq. (2.54) yields

      C₂ sin(λL) + C₄ sinh(λL) = 0 and −λ₂ C₂ sin(λL) + λ₂ C₄ sinh(λL) = 0. Therefore, nontrivial solutions are obtained when C₄ = 0 and sin(λL) = 0, so λ = nπ/L (for n = 1,2,…). Since λ = (ω² ρS/EI)^1/4, we find that the natural frequencies ω_n are given by

(2.64)

      Example 2.11

      Determine the lowest natural frequency for a cantilever steel beam of thickness a = 6 mm, width b = 10 mm, and length L = 0.5 m. Repeat the calculation when the beam is simply supported at both ends.

      Solution

      For steel we have E = 210 × 10⁹ N/m² and ρ = 7800 kg/m³. The cross‐sectional moment of inertia of a rectangular beam is I = ab³/12 = 0.006 × (0.01)³/12 = 5 × 10⁻¹⁰ m⁴. The cross‐sectional area of the beam is S = 0.006 × 0.01 = 6 × 10⁻⁵ m². Then,

If the beam is now simply supported, we use Eq. (2.64) with n = 1,