### Vibration testing method without weighing the specimen

Here, bending vibrations under free–free conditions are considered. In the case of a thin beam with constant cross section, the effect of deflections due to shear and rotatory inertia involved in the bending vibrational deflection (hereafter, SR effect) is negligible, and the Euler–Bernoulli elementary theory of bending can be applied to the bending vibration.

The resonance frequency, represented by *f*_{n0} (*n*: resonance mode number, 0: value without the additional mass), is expressed as follows:

$$ f_{n0} = \frac{1}{2\uppi }\left( {\frac{{m_{n0} }}{l}} \right)^{2} \sqrt {\frac{EI}{\rho A}} , $$

(1)

where *l*, *E*, *ρ*, *I*, and *A* are the specimen length, Young’s modulus, density, the second moment of area, and the cross-sectional area, respectively. *m*_{n0} is a constant that depends on the end conditions and is expressed as follows:

$$ m_{10} = 4.730, m_{20} = 7.853, m_{30} = 10.996, m_{n0} = \frac{1}{2}\left( {2n + 1} \right)\uppi (n > 3). $$

(2)

The resonance frequency is decreased experimentally by attaching the additional mass, while the dimensions, density, and Young’s modulus are not altered. Hence, it can be said that *m*_{n0} changes to *m*_{n}. The resonance frequency after attaching the additional mass is expressed as follows:

$$ f_{n} = \frac{1}{2\uppi }\left( {\frac{{m_{n} }}{l}} \right)^{2} \sqrt {\frac{EI}{\rho A}} . $$

(3)

From Eqs. (1) and (3),

$$ m_{n} = \sqrt {\frac{{f_{n} }}{{f_{n0} }}} m_{n0} . $$

(4)

The frequency equation for the free–free vibration with the concentrated mass *M* placed at position *x* = *al* (*x*: distance along the bar, 0 ≤ *a* ≤ 1, *a* + *b* = 1) on a bar (Fig. 1) is expressed as follows:

$$ \begin{aligned} \left( {\cos m_{n} \cosh m_{n} - 1} \right) - \frac{1}{2}\mu m_{n} \left\{ {\left( {\cos am_{n} \cosh am_{n} + 1} \right)\left( {\sin bm_{n} \cosh bm_{n} - \cos bm_{n} \sinh bm_{n} } \right)} \right. \hfill \\ \left. { + \left( {\cos bm_{n} \cosh bm_{n} + 1} \right)\left( {\sin am_{n} \cosh am_{n} - \cos am_{n} \sinh am_{n} } \right)} \right\} = 0, \hfill \\ \end{aligned} $$

(5)

where *μ* is the ratio of the concentrated mass to the mass of the bar and is defined as

$$ \mu = \frac{M}{\rho Al}. $$

(6)

The measured resonance frequencies *f*_{n0} and *f*_{n} are substituted into Eq. (4) to calculate *m*_{n}, and the calculated *m*_{n} is substituted into Eq. (5) to calculate *μ*. The specimen mass and density can be obtained by substituting the calculated *μ*, the concentrated mass, and the dimensions of a bar into Eq. (6). Young’s modulus can be calculated by substituting the estimated density, the resonance frequency without the concentrated mass, and the dimensions of a bar into Eq. (1) [1,2,3,4,5,6].

The above steps represent the calculation procedure for the VAM. Weighing the specimen is not required for the calculations.

### Goens–Hearmon regression method based on the Timoshenko theory of bending (TGH method)

Young’s and shear moduli can be obtained simultaneously using only the bending vibration test without a torsional vibration test by the following the Goens–Hearmon regression method based on the Timoshenko theory of bending (TGH method) [13,14,15].

The apparent deflection in the bending vibration consists of deflections due to shear and rotatory inertia, as well as pure bending deflection. Timoshenko added the terms deflections of shear and rotatory inertia to the Euler–Bernoulli elementary theory of bending and developed the following differential equation of bending [13]:

$$ \frac{EI}{\rho A}\frac{{\partial^{4} y}}{{\partial x^{4} }} + \frac{{\partial^{2} y}}{{\partial t^{2} }} - \frac{I}{A}\left( {1 + \frac{sE}{G}} \right)\frac{{\partial^{4} y}}{{\partial x^{2} \partial t^{2} }} + \frac{\rho sI}{GA}\frac{{\partial^{4} y}}{{\partial t^{4} }} = 0, $$

(7)

where *G* is the shear modulus, *y* is the lateral deflection, and *t* is the time.

When Eq. (7) is solved under the free–free condition, the resonance frequency corresponding to the *n*th mode *f*_{gn0} can be written as follows:

$$ f_{{{\text{g}}n0}} = \frac{1}{2\uppi }\left( {\frac{{k_{n0} }}{l}} \right)^{2} \sqrt {\frac{EI}{\rho A}} = \frac{1}{2\uppi }\left( {\frac{{m_{n0} }}{l}} \right)^{2} \sqrt {\frac{{E_{{{\text{a}}n}} I}}{\rho A}} , $$

(8)

where *E*_{an} is Young’s modulus from the Euler–Bernoulli elementary theory of bending using the resonance frequency of the *n*th mode.

*k*_{n0} in Eq. (8) is obtained by transcendental equations represented as follows:

$$ \frac{{{ \tan }\frac{{k_{n0} }}{2}\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + A_{\text{t}} k_{n0}^{2} } }}{{{ \tanh }\frac{{k_{n0} }}{2}\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - A_{\text{t}} k_{n0}^{2} } }} = - \frac{{\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + A_{\text{t}} k_{n0}^{2} } }}{{\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - A_{\text{t}} k_{n0}^{2} } }}\frac{{\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - B_{\text{t}} k_{n0}^{2} }}{{\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + B_{\text{t}} k_{n0}^{2} }}\left( {\text{for symmetric modes}} \right), $$

(9)

and

$$ \frac{{{ \cot }\frac{{k_{n0} }}{2}\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + A_{\text{t}} k_{n0}^{2} } }}{{{ \coth }\frac{{k_{n0} }}{2}\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - A_{\text{t}} k_{n0}^{2} } }} = + \frac{{\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + A_{\text{t}} k_{n0}^{2} } }}{{\sqrt {\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - A_{\text{t}} k_{n0}^{2} } }}\frac{{\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} - B_{\text{t}} k_{n0}^{2} }}{{\sqrt {B_{\text{t}}^{2} k_{n0}^{4} + 1} + B_{\text{t}} k_{n0}^{2} }}\left( {\text{for asymmetric modes}} \right), $$

(10)

where

$$ A_{\text{t}} = \frac{I}{{2l^{2} A}}\left( {1 + \frac{sE}{G}} \right) , $$

(11)

and

$$ B_{\text{t}} = \frac{I}{{2l^{2} A}}\left( { - 1 + \frac{sE}{G}} \right). $$

(12)

Goens approximated Eqs. (9) and (10) using a Taylor series into the following formula [14]:

$$ \frac{{m_{n0}^{4} }}{{k_{n0}^{4} }} = \frac{E}{{E_{{{\text{a}}n}} }} = T = 1 + \frac{I}{{l^{2} A}}\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) + 6m_{n0} F\left( {m_{n0} } \right)} \right\} + \frac{I}{{l^{2} A}}\frac{sE}{G}\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) - 2m_{n0} F\left( {m_{n0} } \right)} \right\} - \frac{{4\uppi^{2} \rho sIf_{n0}^{2} }}{GA}, $$

(13)

where

$$ F\left( {m_{10} } \right) = 0.9825, F\left( {m_{20} } \right) = 1.0008, F\left( {m_{n0} } \right) = 1 (n > 2). $$

(14)

Approximately,

$$ T = 1 + \frac{I}{{l^{2} A}}\left[ {\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) + 6m_{n0} F\left( {m_{n0} } \right)} \right\} + \frac{sE}{G}\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) - 2m_{n0} F\left( {m_{n0} } \right)} \right\}} \right]. $$

(15)

Hearmon calculated *E* and *G* using the following procedure after separating Eq. (13) as follows [15]:

$$ Y = E_{{{\text{a}}n}} \left[ {1 + \frac{I}{{l^{2} A}}\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) + 6m_{n0} F\left( {m_{n0} } \right)} \right\} - \frac{{4\uppi^{2} \rho sIf_{n0}^{2} }}{GA}} \right], $$

(16)

$$ X = E_{{{\text{a}}n}} \frac{I}{{l^{2} A}}\left\{ {m_{n0}^{2} F^{2} \left( {m_{n0} } \right) - 2m_{n0} F\left( {m_{n0} } \right)} \right\}, $$

(17)

$$ \alpha = \frac{sE}{G}, $$

(18)

and then from Eqs. (16)–(19):

$$ Y = - \alpha X + \beta . $$

(20)

Therefore, the linear regression between *X* and *Y* provides the *E* and *G* values. This is the Goens–Hearmon regression method based on the Timoshenko theory of bending (TGH method). The value of *s* is 6/5 theoretically [16] and 1.18 experimentally [17] for the rectangular cross section and 10/9 [16] for the circular cross section.