## Exact nonrotating eigenfrequencies

In Weaver et al. [305] it has been shown that the general solution for the transverse vibration of a uniform, non-rotating beam can be written as:

Y(x) = Ci (cos(kx) + cosh(kx)) + C2 (cos(kx) — cosh(kx)) (4.1) + C3 (sin(kx) + sinh(kx)) + C4 (sin(kx) — sinh(kx))

where Y defines the shape of the natural mode of vibration. The constants C1, C2, C3, and C4 in this expression are determined by satisfying boundary conditions at the ends of the beam. In the derivation of Eq. (4.1) it is assumed that the material is homogeneous, isotropic, and that it follows Hooke's law. Furthermore, it is assumed that the displacements are sufficiently small that the response to dynamic excitations is always linearly elastic.

In our case the left end (x = 0) is built in as illustrated in Fig. 4.1, implying zero deflection and zero slope at the fixed end. That is, the boundary conditions are given by

EI = constant

A p = constant

Figure 4.1: Euler-Bernoulli beam (clamped-free situation).

At the free right end (x = L), because the beam is uniform, both the bending moment and the shear force vanish, resulting in d 2Y \ , (d 3Y \

From the first two boundary conditions it follows that the constants C\ and C3 must be equal to zero, so that Eq. (4.1) reduces to

Y(x) = C2 (cos(kx) — cosh(kx)) + C4 (sin(kx) — sinh(kx)) (4.2)

From the remaining two boundary conditions we obtain

C2 cosh(fcjL) + cos(kjL) sinh(kjL) + sin(fcj L) C4 sinh(kjL) — sin(kjL) cosh(fcjL) + cos(kjL)

for each mode i. Rewriting Eq. (4.3) in matrix form gives sinh(kj L) — sin(kjL) cosh(fcjL) + cos(fcjL) cosh(kj L) + cos(kjL) sinh(kjL) + sin(kj L)

The determinant of the matrix must be zero for a non-trivial solution to exists, that

(sinh(kjL) — sin(kjL)) • (sinh(kjL) + sin(kjL)) + (cosh(kjL) + cos(kjL))2 = 0

The above equation reduces to the frequency equation for transverse vibrations of an Euler-Bernoulli beam

C2

0

C4

cos(kjL)cosh(kjL) = —1 for i = 1,2,3, •••, m (4.5)

using cos2(kjL) +sin2(kjL) = 1 and cosh2(kjL) — sinh2 (fcjL) = 1

Equation (4.5) must be solved numerically (using e.g. a mathematical manipulation package like MAPLE V [45]) and yields an infinity of solutions fcj. The first six non-zero positive roots of this equation are presented below in tabular, rather than graphical form so that the full ten-figure accuracy can be retained:

 k1 * L= 1.8751 k2 * L= 4.69409 k3 * L= 7.85476 * L= 10.9955 k5 * L= 14.1372 ke * L= 17.2788

The eigenfrequencies in radians per second corresponding to these values of k are obtained as where E the modulus of elasticity, I the area moment of inertia, p the mass density of the material, and A the cross-sectional area of the beam. Hence, the frequency of vibration of each mode is inversely proportional to square of the length, and proportional to the radius of gyration of the cross-section (i.e. \JI/A). Thus for geometrically similar beams of the same material, the eigenfrequencies vary in direct proportion to the dimensions.

The form of the mode shapes is thus given by

Yi(x) = C2 (cos(kix) — cosh(kjx)) + C4 (sin(kjx) — sinh(kjx))

C2 1

— • (cos(kix) — cosh(ki x)) + sin(kix) — sinh(kix) • C4 (4.7) C4

These modes can be shown to be orthogonal. Notice that although the ratio C2/C4 is uniquely given by Eq. (4.3), C4 cannot be determined. This remaining coefficient becomes the arbitrary magnitude of the mode shape. The first three mode shapes for this beam are depicted in Fig. 4.2 @ - ©.

Figure 4.2: Illustration of the first three mode shapes for the Euler-Bernoulli beam.

The Euler-Bernoulli beam considered has length L = 50 m, modulus of elasticity E = 21 • 1010 N/m2, area of beam cross-section A = n m2, mass density p = 7850 kg/m3, and area moment of inertia I = 1 n m4. The first four exact frequencies are listed in Table 4.1, and will serve as reference solution.

 Mode Exact eigenfrequencies 1 3.637 2 22.79 3 63.82 4 125.1

Table 4.1: The first four exact eigenfrequencies in radians per second of an Euler-Bernoulli beam with length L = 50 m, a constant flexural rigidity EI of 1.6493 • 1011 Nm2, a uniformly distributed mass density of 7850 kg/m3, and a cross-sectional area of n m2.

Table 4.1: The first four exact eigenfrequencies in radians per second of an Euler-Bernoulli beam with length L = 50 m, a constant flexural rigidity EI of 1.6493 • 1011 Nm2, a uniformly distributed mass density of 7850 kg/m3, and a cross-sectional area of n m2.

## Solar Stirling Engine Basics Explained

The solar Stirling engine is progressively becoming a viable alternative to solar panels for its higher efficiency. Stirling engines might be the best way to harvest the power provided by the sun. This is an easy-to-understand explanation of how Stirling engines work, the different types, and why they are more efficient than steam engines.

Get My Free Ebook