**2. Frequency Calculations**

The transverse vibrations of a slender clamped-pinned-free beam with a mass at free end of interest is shown in Figure 1. The governing equation for the beam using Euler-Bernoulli's beam theory [13] can be written:

**Figure 1.** Schematic of a clamped-pinned-free beam with mass at free end.

$$
\rho A \frac{\partial^2 w}{\partial t^2} + EI \frac{\partial^4 w}{\partial x^4} = 0 \tag{1}
$$

where *E* is the Young's modulus, *I* is the cross-sectional moment of inertia, *w* is the vertical deflection, *x* is the axial coordinate, *ρ* is the density of the beam, *A* is the cross-sectional area, and *t* is time. Equation (1) can be solved assuming a separation-of-variables solution in the standard form:

$$w(\mathbf{x}, t) = X(\mathbf{x})T(t) \tag{2}$$

where *X* is the spatial solution and *T* is the temporal solution. The spatial solution for a two-span beam then is expressed:

$$X(x) = \begin{cases} X\_1(x), & 0 \le x \le a \\ X\_2(x), & a \le x \le L \end{cases} \tag{3}$$

The sub-functions in Equation (3) can be written as the following general solutions:

$$X\_1(\mathbf{x}) = a\_1 \sin(\beta \mathbf{x}) + a\_2 \cos(\beta \mathbf{x}) + a\_3 \sinh(\beta \mathbf{x}) + a\_4 \cosh(\beta \mathbf{x})\tag{4}$$

$$X\_2(\mathbf{x}) = b\_1 \sin(\beta \mathbf{x}) + b\_2 \cos(\beta \mathbf{x}) + b\_3 \sinh(\beta \mathbf{x}) + b\_4 \cosh(\beta \mathbf{x}) \tag{5}$$

where *β* is the beam vibration eigenvalue. Parameter *β* and seven of the eight coefficients can be solved by applying the boundary conditions of the system. For the clamped-pinned-free, the displacement and slope at the clamped end are zero [7]:

$$X\_1(0) = 0\tag{6}$$

$$\frac{dX\_1(0)}{dx} = 0\tag{7}$$

while at the free end, the bending moment and the shear vanish such that:

$$\frac{d^2X\_2(L)}{dx^2} = 0\tag{8}$$

$$EI\frac{d^3X\_2(L)}{dx^3} = m\_{\text{attacked}}\frac{d^2T\_2(L,t)}{dt^2} \tag{9}$$

where *m*attached is the mass attached to the beam at the free end. In addition to these four boundary conditions, four more boundary conditions (displacement and rotation) are found at the pin location *a*:

$$X\_1(a) = 0\tag{10}$$

$$X\_2(a) = 0\tag{11}$$

$$\frac{dX\_1(a)}{dx} = \frac{dX\_2(a)}{dx} \tag{12}$$

$$\frac{d^2X\_1(a)}{dx^2} = \frac{d^2X\_2(a)}{dx^2} \tag{13}$$

Substituting the first transverse displacement (Equation (4)) into the clamped boundary condition (Equations (6) and (7)) gives:

$$a\_2 + a\_4 = 0\tag{14}$$

$$a\_1 + a\_3 = 0\tag{15}$$

Substituting the second transverse displacement (Equation (5)) into the free end boundary condition (Equation (8)) yields:

$$b\_1 - b\_1 \sin(\beta L) - b\_2 \cos(\beta L) + b\_3 \sinh(\beta L) + b\_4 \cosh(\beta L) = 0\tag{16}$$

Additionally, inserting the second transverse displacement (Equation (5)) into Equation (1) and applying the boundary condition at the free end (Equation (9)) results in:

$$b\_1(-\\\\\cos(\beta L) + \beta L \frac{m\_{\text{attached}}}{m\_{\text{beam}}} \sin(\beta L)) + b\_2(\sin(\beta L) + \beta L \frac{m\_{\text{attached}}}{m\_{\text{beam}}} \cos(\beta L))$$

$$b\_1 + b\_3(\cosh(\beta L) + \beta L \frac{m\_{\text{attached}}}{m\_{\text{beam}}} \sinh(\beta L)) + b\_4(\sinh(\beta L) + \beta L \frac{m\_{\text{attached}}}{m\_{\text{beam}}} \cosh(\beta L)) = 0 \tag{17}$$

where *m*beam is the mass of the beam.

Substituting the first transverse displacement (Equation (4)) into the pinned boundary condition (Equations (10) and (11)) results in:

$$a\_1 \sin(\beta L \frac{a}{L}) + a\_2 \cos(\beta L \frac{a}{L}) + a\_3 \sinh(\beta L \frac{a}{L}) + a\_4 \cosh(\beta L \frac{a}{L}) = 0 \tag{18}$$

and

$$b\_1 \sin(\beta L \frac{a}{L}) + b\_2 \cos(\beta L \frac{a}{L}) + b\_3 \sinh(\beta L \frac{a}{L}) + b\_4 \cosh(\beta L \frac{a}{L}) = 0 \tag{19}$$

After, substituting the second transverse displacement (Equation (5)) into the boundary conditions defined by Equations (12) and (13) provides the following expressions:

$$\begin{aligned} a\_1 \cos(\beta L \frac{a}{L}) - a\_2 \sin(\beta L \frac{a}{L}) + a\_3 \cosh(\beta L \frac{a}{L}) + a\_4 \sinh(\beta L \frac{a}{L}) \\ -b\_1 \cos(\beta L \frac{a}{L}) + b\_2 \sin(\beta L \frac{a}{L}) - b\_3 \cosh(\beta L \frac{a}{L}) - b\_4 \sinh(\beta L \frac{a}{L}) = 0 \end{aligned} \tag{20}$$

$$-a\_1 \sin(\beta L \frac{a}{L}) - a\_2 \cos(\beta L \frac{a}{L}) + a\_3 \sinh(\beta L \frac{a}{L}) + a\_4 \cosh(\beta L \frac{a}{L})$$

$$-b\_1 \sin(\beta L \frac{a}{L}) + b\_2 \cos(\beta L \frac{a}{L}) - b\_3 \sinh(\beta L \frac{a}{L}) - b\_4 \cosh(\beta L \frac{a}{L}) = 0\tag{21}$$

Aggregating Equations (14)–(21) into an 8 × 8 matrix (see Appendix A) and solving for the determinant leads to the transcendental equation expressed:

$$\begin{split} 4\cos(\beta L(\frac{d}{L}-1))\sinh(\beta L(\frac{d}{L}-1)) - 4\cosh(\beta L(\frac{d}{L}-1))\sin(\beta L(\frac{d}{L}-1)) \\ + 2\cos(\beta L(2\frac{d}{L}-1))\sinh(\beta L) - 2\cosh(\beta L(2\frac{d}{L}-1))\sin(\beta L) \\ + 4\cos(\frac{d}{L}\beta L)\sinh(\frac{d}{L}\beta L) - 4\cosh(\frac{d}{L}\beta L)\sin(\frac{d}{L}\beta L) \\ + 2\cos(\beta L)\sinh(\beta L) - 2\cosh(\beta L)\sin(\beta L) + 8\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\sin(\beta L(\frac{d}{L}-1))\sinh(\beta L(\frac{d}{L}-1)) \\ + 2\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\cos(\beta L(2\frac{d}{L}-1))\cosh(\beta L) - 2\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\cosh(\beta L(2\frac{d}{L}-1))\cos(\beta L) \\ + 2\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\sin(\beta L(2\frac{d}{L}-1))\sinh(\beta L) + 2\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\sinh(\beta L(2\frac{d}{L}-1))\sin(\beta L) \\ - 4\beta L \frac{m\_{\text{athedral}}}{m\_{\text{beam}}}\sin(\beta L)\sinh(\beta L) = 0 \end{split}$$

where the natural frequencies (in Hz) are given by:

$$f\_{\rm n} = \frac{(\beta\_{\rm n} L)^2}{2\pi L^2} \sqrt{\frac{EI}{\rho A}}\tag{23}$$

To verify Equation (22), the first five natural frequencies were calculated for three well known cases:


The results are tabulated in Tables 1–3. The frequencies of the first five modes (*β*1–*β*5) are compared between what is found in literature against the results from Equation (22) (proposed model). The small differences are due to rounding errors of the beam vibration eigenvalues *β*, which cause large changes in the calculated frequency (*fn* ∝ *β*<sup>2</sup> *<sup>n</sup>*). The precision for *β* in this paper is ±0.0002.

**Table 1.** Comparison of analytical results: clamped-free (Case 1).



**Table 2.** Comparison of analytical results: Clamped-free with mass at free end ( *<sup>m</sup>*attached *<sup>m</sup>*beam = 0.2) (Case 2).

**Table 3.** Comparison of analytical results: Clamped-pinned-free (pinned at *a* = 200 mm) (Case 3).

