# Correction of Young's Modulus Calculation in the Flexural Mode of Resonant Column Test
## 2 Analysis Method of Flexural Test
### Inertial Effect and Calibration of Electromagnetic Drive Disk  
[PERSON] et al. [6] used the Rayleigh method to analyze the resonant frequency in the flexural mode. The deformation of the calibration rod can be regarded as a cantilever beam in the flexural resonant test. The EDD centroid is achieved when the maximum strain energy is equal to the kinetic energy. Figure 3: Calibration rod resonant test configuration diagram. Figure 2: Resonant column magnetic drive system. As shown in Figure 3, the calibration rod is divided into three parts for the analysis: The bottom plate (regarded as the fixed end at the bottom), the middle section with length L, and the top plate. The horizontal deformation (y) of the rod at different elevations (x) is assumed to be a third-order polynomial:  
\[\mathrm{y(x)}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3} \tag{1}\]  
Given that the bottom of the rod (\(x=0\)) is fixed on the base, its displacement \(\mathrm{y(0)}=0\) and slope \(\mathrm{y^{\prime}(0)}=0\), and coefficients \(a_{0}\) and \(a_{1}\) are zero. In addition, the top of the rod (x = \(L\)) is a free end, and its bending moment is EI\(y^{\prime\prime}(L)=0\). Hence, Equation (1) can be rewritten as:  
\[\mathrm{y(x)}=\alpha x^{2}[3L-x]\quad for\;\;x<L \tag{2}\]  
where \(\alpha=(a_{2}/3L)\). Equation (2) is the deformation equation applied to the middle section of the rod (excluding top plate and bottom plate of the calibration rod). The top plate, EDD, and calibration block can be regarded as the rigid body added to the top of the rod. Hence, their displacement can be expressed in terms of the displacement \(\mathrm{y(}L)\) and slope \(\mathrm{y^{\prime}(}L)\) of the top of the rod:  
\[\mathrm{y(x)}=\alpha L^{2}[2L+3(x-L)]\;for\;x>L \tag{3}\]  
The maximum displacement of the rod occurs at the resonant frequency, and the maximum strain energy of the rod is calculated as  
\[J_{U}=\frac{1}{2}EI_{b}\int_{0}^{L}y^{\prime\prime}(x)^{2}dx=6 EI_{b}\alpha^{2 }L^{3} \tag{4}\]  
where \(I_{b}\) is the moment of inertia of the rod. The rod vibrating back and forth at a fixed frequency (e.g., resonant frequency) can be regarded as a simple harmonic motion with the angular frequency \(\omega_{f}\). As a result, maximum kinetic energy \(J_{T1}\) can be calculated as  
\[J_{T1}=\frac{1}{2}\omega_{f}{}^{2}A\int_{0}^{6}y(x)^{2}dx=\frac{33}{70}\omega_ {f}{}^{2}\alpha^{2}L^{6}m_{T} \tag{5}\]  
where \(A\) is the cross-sectional area of the rod and \(m_{T}\) is the mass of the rod. The kinetic energy of the rigid body with mass (\(m\)) above the top of the rod is calculated according to the distance of its centroid from the top of rod \(h\):  
\[J_{T2}=\frac{1}{2}m\Big{[}\alpha L^{2}(2L+3h)\Big{]}^{2}\cdot\omega_{f}{}^{2}= \mathrm{m}\alpha^{2}L^{6}\bigg{[}2+6\frac{h}{L}+\frac{9}{2}\left(\frac{h}{L} \right)^{2}\bigg{]}\cdot\omega_{f}{}^{2} \tag{6}\]  
The resonant frequency of the system from the flexural resonant test can be obtained when the kinetic energy of the system (\(J_{T1}+J_{T2}\)) is equal to the strain energy of the rod (\(J_{U}\)) according to the conservation of energy:  
\[\omega_{f}=\frac{3 EI_{b}}{\left\{\frac{33}{140}m_{T}+m\Big{[}1+3\frac{h}{L}+ \frac{9}{4}\left(\frac{h}{L}\right)^{2}\Big{]}\right\}\cdot L^{3}} \tag{7}\]  
In Equation (7), the effect of the kinetic energy of the rigid bodies above the top of the rod on the measured frequency can be expressed as the product of their mass and the height of the centroid. Hence, this formula can be rewritten as:  
\[\omega_{f}=\frac{3 EI_{b}}{\left[\frac{33}{140}m_{T}+\sum_{i=0}^{N}m_{i}h(h0_{ i},h1_{i})\right]\cdot L^{3}} \tag{8}\]where \(m_{i}\) is the mass of the individual rigid body above the top of the rod; and \(h0_{i}\) and \(h1_{i}\) are the heights from the bottom and top of the individual rigid body to the top of the rod, respectively.