This function calculate the 1st, 2nd and 3rd natural frequency, the stiffness and the displacement.
When the end of the bar is fixed and the other end is free, the situation is the same as those of the fixed cutting tool.
When the concentrated load is applied to the tip, the displacement at the tip is calculated by using the following equation.
\(\displaystyle d = \cfrac{FL^3}{3 E I} \)
\(\displaystyle I = \cfrac{\pi D^4}{64} \)
\( d \): Displacement at the tip
\( f \): Concentrated load at the tip
\( L \): Length of the cantilever
\( E \): Young's modulus (modulus of longitudinal elasticity)
\( I \): Second moment of area
\( D \): Diameter of the cantilevered round bar
Above mentioned two equations leads to the following equation.
\(\displaystyle d = \cfrac{64 L^3}{3 \pi ED^4} F \)
By modifying this equation, the following equation is obtained.
\(\displaystyle F = \cfrac{3 \pi E D^4}{64 L^3}d \)
\(\displaystyle \cfrac{3 \pi E D^4}{64 L^3} \) shows the stiffness at the tip.
Next, we calculate the maximum tensile stress at the base of the cantilever.
This value is able to be used to check the strength of the cantilever.
\(\displaystyle \sigma_{max} = \cfrac{F L}{Z} \)
\(\displaystyle Z = \cfrac{I}{\cfrac{D}{2}} = \cfrac{2I}{D} = \cfrac{\pi D^3}{32} \)
\( \sigma_{max} \): Maximum tensile stress at the base
\( Z \): Section modulus
Above mentioned two equations leads to the following equation.
\(\displaystyle \sigma_{max} = \cfrac{32 F L}{\pi D^3} \)
Finally，natural frequency is calculated.
The situation is that "the end of the bar is fixed and the other end is free".
\(\displaystyle f_n = \cfrac{1}{2 \pi} \cfrac{\lambda^2}{L^2} \sqrt{ \cfrac{E I}{\rho A}} \)
\(\displaystyle A = \cfrac{\pi D^2}{4} \)
\( f_n \)：Natural frequency
\( \rho \)：Density
\( A \)：Area of cross section of the cantilevered round bar
Thse two equations and Second moment of area of the round bar lead to this equation.
\(\displaystyle f_n = \cfrac{\lambda^2}{8 \pi} \cfrac{D}{L^2} \sqrt{ \cfrac{E}{\rho}} \)
At the above mentioned equation,
\( \lambda \)=1.8751 leads to the 1st natural frequency,
\( \lambda \)=4.6941 leads to the 2nd natural frequency,
\( \lambda \)=7.8547 leads to the 3rd natural frequency.
Next, same problems are solved on the stepped cantilevered round bar.
Diaplacement and static stiffness are solved accroding to material mechanics.
\(\displaystyle d = ( \cfrac{l_2^3}{3E_1 I_1} - \cfrac{L^3}{3E_1 I_1} - \cfrac{l_2^3}{3 E_2 I_2})F \)
\( d \): Displacement at the tip of the stepped cantilevered round bar
\( F \): Force at the tip of the stepped cantilevered round bar
\( E_1 \): Young's modulus (modulus of longitudinal elasticity) of the base-side bar
\( l_1 \): Length of the base-side bar
\( I_1 \): Second moment of area of the base-side bar
\( E_2 \): Young's modulus (modulus of longitudinal elasticity) of the tip-side bar
\( l_2 \): Length of the base-side bar
\( I_2 \): Second moment of area of the base-side bar
\( L = l_1 + l_2 \): Full length of the stepped cantilevered round bar
Natural frequencies are obtained with the following equation according to the reference.
\(\displaystyle (1+ C_1 \cdot c_1) (1+ C_2 \cdot c_2) \)
\(\displaystyle +(\cfrac{K_2}{K_1})^2 (\cfrac{l_1}{l_2})^4 (\cfrac{\alpha_2}{\alpha_1})^4 (1- C_2 \cdot c_2)(1- C_1 \cdot c_1) \)
\(\displaystyle -(\cfrac{K_2}{K_1}) (\cfrac{l_1}{l_2}) (\cfrac{\alpha_2}{\alpha_1})(S_1 \cdot c_1 - C_1 \cdot s_1)(C_2 \cdot s_2- S_2 \cdot c_2) \)
\(\displaystyle -(\cfrac{K_2}{K_1}) (\cfrac{l_1}{l_2})^3 (\cfrac{\alpha_2}{\alpha_1})^3(S_2 \cdot c_2 - C_2 \cdot s_2)(C_1 \cdot s_1- S_1 \cdot c_1) \)
\(\displaystyle -2(\cfrac{K_2}{K_1}) (\cfrac{l_1}{l_2}) (\cfrac{\alpha_2}{\alpha_1})(S_1 \cdot c_1)(S_2 \cdot s_2)=0 \)
\(\displaystyle \alpha^4 = \cfrac{\rho a \omega^2 l^4}{K} \)
\( S=\sinh \alpha \)
\( C=\cosh \alpha \)
\( s=\sin \alpha \)
\( c=\cos \alpha \)
\( \omega \)：Frequency
\( \rho \)：Density
\( a \)：Area
\( K=EI \)：Stiffness for bending
\( E \)：Young's modulus (modulus of longitudinal elasticity)
\( I \)：Second moment of area
\( l \)：Bar length
＊Suffix 1 shows the base-side of the stepped cantilevered round bar. Suffix 2 shows the tip-side one.
In the reference, shown above equarions are rewritten for understanding them easily.
\(\displaystyle ( 1 + \cosh \alpha_1 \cdot \cos \alpha_1)( 1 + \cosh k\alpha_1 \cdot \cos k\alpha_1) \)
\(\displaystyle + \beta \gamma( 1 - \cosh \alpha_1 \cdot \cos \alpha_1)( 1 - \cosh k\alpha_1 \cdot \cos k\alpha_1) \)
\(\displaystyle - \beta^\cfrac{1}{4} \gamma^\cfrac{3}{4}( \sinh \alpha_1 \cdot \cos \alpha_1 + \cosh \alpha_1 \cdot \sin \alpha_1)( \cosh k \alpha_1 \cdot \sin k \alpha_1 - \sinh k\alpha_1 \cdot \cos k\alpha_1) \)
\(\displaystyle - \beta^\cfrac{3}{4} \gamma^\cfrac{1}{4}( \sinh k \alpha_1 \cdot \cos k \alpha_1 + \cosh k \alpha_1 \cdot \sin k \alpha_1)( \cosh \alpha_1 \cdot \sin \alpha_1 - \sinh \alpha_1 \cdot \cos \alpha_1) \)
\(\displaystyle - 2 \beta^\cfrac{1}{2} \gamma^\cfrac{1}{2}( \sinh \alpha_1 \cdot \sin \alpha_1 \cdot \sinh k \alpha_1 \cdot \sin k \alpha_1) = 0 \)
\(\displaystyle \lambda = \cfrac{l_2}{l_1} \)
\(\displaystyle \beta = \cfrac{a_2}{a_1} \)
\(\displaystyle \gamma = \cfrac{K_2}{K_1} = \cfrac{E_2 I_2}{E_1 I_1} \)
\(\displaystyle \alpha_2 = k \alpha_1 \)
\(\displaystyle k = \beta^\cfrac{1}{4} \gamma^\cfrac{-1}{4} \lambda \)
\(\displaystyle \alpha_1 = l_1 (\cfrac{\rho_1 a_1 \omega^2}{K_1})^\cfrac{1}{4} \)
However, there is one problem. Density is not considered after rewriting.
\(\displaystyle \alpha_2 = k \alpha_1 \)
\(\displaystyle = \beta^\cfrac{1}{4} \gamma^\cfrac{-1}{4} \lambda l_1 (\cfrac{\rho_1 a_1 \omega^2}{K_1})^\cfrac{1}{4} \)
\(\displaystyle = ( \cfrac{a_2}{a_1})^\cfrac{1}{4} \cfrac{K_1}{K_2})^\cfrac{1}{4} \cfrac{l_2}{l_1}) l_1 (\cfrac{\rho_1 a_1 \omega^2}{K_1})^\cfrac{1}{4} \)
\(\displaystyle = l_2 (\cfrac{\rho_1 a_2 \omega^2}{K_2})^\cfrac{1}{4} \)
\(\displaystyle \neq \alpha_2 \)
Thus, the following equation has to be added.
\(\displaystyle P = ( \cfrac{\rho_2}{\rho_1} )^\cfrac{1}{4} \)
\(\displaystyle \alpha_2 = k P \alpha_1 \)
Modified equations are shown as follows.
\( \alpha_1 \) is obtained from this equation by using the numerical calculation method.
Frequency \( \omega \) is obtained from \( \alpha_1 \).
The frequency is the natural frequency.
\(\displaystyle ( 1 + \cosh \alpha_1 \cdot \cos \alpha_1)( 1 + \cosh k P \alpha_1 \cdot \cos k P \alpha_1) \)
\(\displaystyle + \beta \gamma P^4( 1 - \cosh \alpha_1 \cdot \cos \alpha_1)( 1 - \cosh k P \alpha_1 \cdot \cos k P \alpha_1) \)
\(\displaystyle - \beta^\cfrac{1}{4} \gamma^\cfrac{3}{4} P( \sinh \alpha_1 \cdot \cos \alpha_1 + \cosh \alpha_1 \cdot \sin \alpha_1)( \cosh k P \alpha_1 \cdot \sin k P \alpha_1 - \sinh k P \alpha_1 \cdot \cos k P \alpha_1) \)
\(\displaystyle - \beta^\cfrac{3}{4} \gamma^\cfrac{1}{4} P^3( \sinh k P \alpha_1 \cdot \cos k P \alpha_1 + \cosh k P \alpha_1 \cdot \sin k P \alpha_1)( \cosh \alpha_1 \cdot \sin \alpha_1 - \sinh \alpha_1 \cdot \cos \alpha_1) \)
\(\displaystyle - 2 \beta^\cfrac{1}{2} \gamma^\cfrac{1}{2} P^2( \sinh \alpha_1 \cdot \sin \alpha_1 \cdot \sinh k P \alpha_1 \cdot \sin k P \alpha_1) = 0 \)
Calculated 1st natural frequency of this function is compared with calculated results of FEM and another theoretical equation.
Details of the comparison is shown in the following pdf file.
Calculation result check for the calculation function of natural frequency.pdf
The code, which is written in Python for calculating natural frequencies on the stepped cantilevered round bar, is shown below.
If you can use Python, you can use this code by COPY & PASTE.
This code is wrriten for the stepped cantilevered "round" bar.
However, you can change it to "square" and so on.
If you want to use this code, you have to use this code at your own risk.
↓The code starts from here.
import numpy as np
import matplotlib.pyplot as plt
##Information of the cantilever near the fixed end
L1 = 150/1000##Length(m)
A1 = (40/1000)**2*np.pi/4##Area(m^2)
I1 = (40/1000)**4*np.pi/64##Moment of inertia of area(m^4)
Rho1 = 7850##Density(kg/m^3)
E1 = 210*10**9##Young's modulus, modulus of longitudinal elasticity(N/m^2)
##Information of the cantilever near the free end
L2 = 100/1000##Length(m)
A2 = (40/1000)**2*np.pi/4##Area(m^2)
I2 = (40/1000)**4*np.pi/64##Moment of inertia of area(m^4)
Rho2 = 7850##Density(kg/m^3)
E2 = 210*10**9##Young's modulus, modulus of longitudinal elasticity(N/m^2)
freqmin = 0##Minimum frequency on the frequency range(Hz)
freqmax = 10000##Maximum frequency on the frequency range(Hz)
freqresolution = 1##Resolution of the frequency range(Hz)
#####The calculation part is shown below.#################################
lamb = L2/L1
beta = A2/A1
gamma = (E2*I2)/(E1*I1)
P = (Rho2/Rho1)**(0.25)
K = beta**(0.25)*gamma**(-0.25)*lamb
omega = np.arange(freqmin,freqmax,freqresolution)*(2*np.pi)
alpha1 = L1*(Rho1*A1*omega**2/(E1*I1))**(0.25)
alpha2 = K*P*alpha1
term1 = ( 1 + np.cosh(alpha1)*np.cos(alpha1) )*( 1 + np.cosh(K*P*alpha1)*np.cos(K*P*alpha1))
term2 = beta*gamma*P**4*( 1 - np.cosh(alpha1)*np.cos(alpha1) )*( 1 - np.cosh(K*P*alpha1)*np.cos(K*P*alpha1))
term3 = -beta**(0.25)*gamma**(0.75)*P*( np.sinh(alpha1)*np.cos(alpha1) + np.cosh(alpha1)*np.sin(alpha1) )*( np.cosh(K*P*alpha1)*np.sin(K*P*alpha1) - np.sinh(K*P*alpha1)*np.cos(K*P*alpha1) )
term4 = -beta**(0.75)*gamma**(0.25)*P**3*( np.sinh(K*P*alpha1)*np.cos(K*P*alpha1) + np.cosh(K*P*alpha1)*np.sin(K*P*alpha1) )*( np.cosh(alpha1)*np.sin(alpha1) - np.sinh(alpha1)*np.cos(alpha1) )
term5 = -2*beta**(0.5)*gamma**(0.5)*P**2*( np.sinh(alpha1)*np.sin(alpha1) * np.sinh(K*P*alpha1)*np.sin(K*P*alpha1) )
terms = term1 + term2 + term3 + term4 + term5
##The condition "terms = 0" shows natural frequencies.
##When the product of next to each other "terms" is less than or equal to zero, the natural frequency is calculated according to the previous "terms".
##Thus, there is the possibility that the calculation result has the error. The size of the error is less than the frequency resolution.
##If you require the precise calculation result, you have to change the calculation method.
zeropoints = np.where(terms[:-1]*terms[1:]<=0)[0]
#####The display part of calculation results is shown below.#################################
print("Natural frequency(Hz):",omega[zeropoints]/(2*np.pi))
plt.figure()
plt.plot(omega/(2*np.pi),terms,"b-",label="Calculation results")
plt.plot(omega[zeropoints]/(2*np.pi),terms[zeropoints],"ro",label="Natural frequency")
plt.xlabel("Frequency (Hz)")
plt.ylabel("Sum of terms")
plt.legend()
plt.grid()
plt.tight_layout()
plt.show()
↑The code ends here.References:
Hisashi NAKAMURA and Masashi KONNNO: Resonant Frequency of a Stepped-Cantilever, Acoustical Society of Japan, Vol.21, No.2(1965)pp.55-64(in Japanese).