Natural frequency of bending vibration for stepped beam of different geometrical characters and materials

Introduction

The stepped beam is a kind of non-uniform component, which is very common in engineering, such as the stepped piston components in hydraulic diaphragm metering pumps, ¹ ladder-type sucker rod for pumping oil, ² and drilling column. ³ The workpiece in the turning process can also be regarded as a stepped rod, and its time-varying characteristics of the longitudinal natural frequency and torsion vibration mode were studied by Li and Zhang ⁴ and Li et al. ⁵ The ladder-shaped member with bending as its main deformation can be called a stepped beam. It is well known that the vibration problem of stepped beam is a basic problem of mechanical vibration, and the first-order frequency is paid particular attention to in engineering. The existing vibration mechanics ⁶ only gives the solution of the natural frequency of the transverse free vibration of the uniform cross-section bar. Due to the difficulty of solving the natural frequency of the transverse vibration of the stepped shaped bar, scholars have studied it from the aspects of theory, experiment, and numerical simulation. For example, Naguleswaran ⁷ gave a method of solving the root of frequency equation using determinant elements for the stepped beam and considered nine common support conditions for three types of step beams; this study is comprehensive and perfect but did not consider the materials difference of two portions of the stepped beam and did not give the change rule of frequency with beam parameters. The Adomian decomposition method (ADM) is employed by Mao to investigate the free vibrations of the Euler–Bernoulli beams with two uniform sections ⁸ and multiple cross-section steps, ⁹ the proposed ADM method can be used to analyze the vibration of beams consisting of an arbitrary number of steps through a recursive way, and the ADM offers an accurate and effective method of free vibration analysis of multiple-stepped beams with arbitrary boundary conditions. A transfer matrix method is presented by Attar ¹⁰ to investigate the natural frequencies and mode shapes of a stepped beam with an arbitrary number of transverse cracks, and the general form of characteristic equation for the beam is obtained, which is a function of natural frequencies, the locations and sizes of the cracks, boundary conditions, and geometrical and physical parameters of the beam. The differential transformation method (DTM) is employed by Suddoung et al. ¹¹ to solve the governing differential equations of stepped beams made from functionally graded materials (FGMs) in order to obtain natural frequencies and mode shapes. Xiang et al. ¹² and Patil and Teodoriu ¹³ studied the distribution law of the frequency and vibration displacement of the rotating shaft system using the experiment method and put forward the method of controlling the torsional vibration. The application of the composite element method to the free vibration problem of a multi-stepped cantilever beam was verified by Xie et al.; ¹⁴ Li and Zhang ¹⁵ derived the time-varying equation of the longitudinal natural frequency of the rod and gave the time-varying curve of the first-order and second-order natural frequency of the stepped bar under different fine cross-sectional area ratios; Zhang et al. ¹⁶ carried out modal analysis and harmonic response analysis of the stepped shaped amplitude bar on the basis of three-dimensional (3D) modeling. Based on Euler–Bernoulli beam theory and Timoshenko beam theory, Lee ¹⁷ used Chebyshev-tau method to analyze the free vibration of stepped beams. Lin and Ng ¹⁸ developed a novel numerical method and applied to the prediction of vibration modes of general stepped beams with arbitrary steps and general elastic supports. Wang and Wang ¹⁹ proposed a differential quadrature element method for obtaining highly accurate natural frequencies of multiple-stepped beams with an aligned neutral axis. The proposed method is simple and efficient to analyze beams with any step changes in cross-section conveniently. Jaworski and Dowell ²⁰ investigated theoretically and experimentally the flexural-free vibration of a cantilevered beam with multiple cross-section steps, and the three lowest natural frequencies of a multiple-stepped beam are predicted using a global Rayleigh–Ritz formulation, component modal analysis, and the commercial finite element code ANSYS and are matched with experimental results from impact testing data, but this article did not give the concrete theoretical solutions by its equations and the main work focuses on the comparison of numeric solutions of different element types by ANSYS software with impact testing data.

The natural frequency of the stepped beam is influenced by the shape of the cross-section, the constraint condition, the material, and the length of the segment. Most of the existing literatures only give the formula for calculating the natural frequency of the straight cross-section bar. Derivation of natural frequency of flexural vibration of stepped beam, owing to the complexity of the formula and the large amount of calculation, and the natural frequency formula which can be directly applied to the stepped beam is not found in the existing literature.

In this article, a general cantilever step beam and simplified supported step beam are given as examples, and a general derivation method for the natural frequency equation of the bending vibration of the stepped rod under various boundary conditions is given. The obtained frequency equation is applicable to the straight bar with different cross-sectional shapes, composed of the different materials for the two portions, different lengths, and different areas for the two portions of the stepped bar; according to the frequency equation one can get the corresponding frequency by modifying the corresponding parameters in Mathcad software.

Geometric model of cantilever step beam and supported step beam

As shown in Figure 1, the total length of the stepped beam is L, the mass per unit volume of the left part beam is ρ₁, the cross-sectional area is A₁, the minimum axis inertia distance is I₁, the elastic modulus of the material is E₁; the mass per unit volume of the right part beam is ρ₂, the cross-sectional area is A₂, the minimum axis inertia distance is I₂, and the elastic modulus of the material is E₂, that is, the bending stiffness of the two sections are, respectively, E₁I₁ and E₂I₂.

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Figure 1.

Step beam model: (a) cantilever step beam and (b) simplified supported stepped beam.

Natural frequency equation of bending vibration of cantilever beam

Assume y(x, t) is the transverse vibration displacement at the cross-section x at time t, according to the transverse forced vibration equation of the Euler–Bernoulli beam in the existing literature

∂ 2 ∂ x 2 ( E I ∂ 2 y ∂ x 2 ) + ρ A ∂ 2 y ∂ t 2 = q ( x , t ) − ∂ ∂ x m ( x , t )

(1)

where ρ is the mass per unit volume, and q(x, t) and m(x, t) are, respectively, the distributed load and couple per unit length.

Suppose y₁(x, t) and y₂(x, t) are, respectively, the deflection of the left beam and the right beam. According to equation (1), the transverse free vibration equation for the two portions of the stepped beam is

a 1 2 ∂ 4 y 1 ∂ x 4 + ∂ 2 y 1 ∂ t 2 = 0

(2)

a 2 2 ∂ 4 y 2 ∂ x 4 + ∂ 2 y 2 ∂ t 2 = 0

(3)

where a 1 2 = ( E 1 I 1 ) / ( ρ 1 A 1 ) and a 2 2 = ( E 2 I 2 ) / ( ρ 2 A 2 ) .

Based on the separation of variables method, set mode solution as

y 1 ( x , t ) = Y 1 ( x ) e i ω t

(4)

y 2 ( x , t ) = Y 2 ( x ) e i ω t

(5)

where ω is natural frequency. Substituting equations (4) and (5) into equations (2) and (3), one can get

Y 1 ( x ) = ( C 1 cos β 1 x + C 2 sin β 1 x + C 3 ch β 1 x + C 4 sh β 1 x )

(6)

Y 2 ( x ) = ( D 1 cos β 2 x + D 2 sin β 2 x + D 3 ch β 2 x + D 4 sh β 2 x )

(7)

where β 1 4 = ω 2 / a 1 2 and β 2 4 = ω 2 / a 2 2 .

Without loss of generality, the following derivation takes the cantilever step beam as example. For Figure 1(a), according to the boundary condition of the fixed end y₁(0, t) = 0, y ′ 1 ( 0 , t ) = 0 , one gets the following equations

C 1 + C 3 = 0

(8)

C 2 + C 4 = 0

(9)

From the boundary condition of the free end y ″ 2 ( L , t ) = 0 , y ‴ 2 ( L , t ) = 0 , the following equations can be obtained

D 1 cos β 2 L + D 2 sin β 2 L − D 3 ch β 2 L − D 4 sh β 2 L = 0

(10)

D 1 sin β 2 L − D 2 cos β 2 L + D 3 sh β 2 L + D 4 ch β 2 L = 0

(11)

According to the displacement continuity condition at the stepped section, y₁(L₁, t) = y₂(L₁, t), y ′ 1 ( L 1 , t ) = y ′ 2 ( L 1 , t ) , one gets the following equations

C 1 cos β 1 L 1 + C 2 sin β 1 L 1 + C 3 ch β 1 L 1 + C 4 sh β 1 L 1 = D 1 cos β 2 L 1 + D 2 sin β 2 L 1 + D 3 ch β 2 L 1 + D 4 sh β 2 L 1

(12)

β 1 ( − C 1 sin β 1 L 1 + C 2 cos β 1 L 1 + C 3 sh β 1 L 1 + C 4 ch β 1 L 1 ) = β 2 ( − D 1 sin β 2 L 1 + D 2 cos β 2 L 1 + D 3 sh β 2 L 1 + D 4 ch β 2 L 1 )

(13)

According to the bending moment continuity condition E 1 I 1 y ″ 1 ( L 1 , t ) = E 2 I 2 y ″ 2 ( L 1 , t ) at the stepped section, one can get

E 1 I 1 β 1 2 ( − C 1 cos β 1 L 1 − C 2 sin β 1 L 1 + C 3 ch β 1 L 1 + C 4 sh β 1 L 1 ) = E 2 I 2 β 2 2 ( − D 1 cos β 2 L 1 − D 2 sin β 2 L 1 + D 3 ch β 2 L 1 + D 4 sh β 2 L 1 )

(14)

According to the shear force continuity condition at the step section E 1 I 1 y ‴ 1 ( L 1 , t ) = E 2 I 2 y ‴ 2 ( L 1 , t ) , one can get

E 1 I 1 β 1 3 ( C 1 sin β 1 L 1 − C 2 cos β 1 L 1 + C 3 sh β 1 L 1 + C 4 ch β 1 L 1 ) = E 2 I 2 β 2 3 ( D 1 sin β 2 L 1 − D 2 cos β 2 L 1 + D 3 sh β 2 L 1 + D 4 ch β 2 L 1 )

(15)

Specially, for the equal section beam, it has E₁I₁ = E₂I₂, a₁ = a₂, β₁ = β₂, and L₁ = L, substituting these conditions into equations (8)–(15), the following equations can be obtained

C 3 ( ch β L + cos β L ) + C 4 ( sh β L + sin β L ) = 0

(16)

C 3 ( sh β L − sin β L ) + C 4 ( ch β L + cos β L ) = 0

(17)

The necessary and sufficient condition for the non-zero solution of the above equations is that the determinant of the coefficient is zero, and then one can get the frequency equation of the equal section beam

cos β L ch β L = − 1

(18)

The above equation is the bending vibration frequency equation of the equal cross-section beam, which is exactly the same as the solution in Yin. ⁶

While for the step beam, by simultaneous solution of equations (8)–(15), and by elimination method, according to the condition that determinant of the coefficient is zero, the frequency equation of the bending vibration for the cantilever step beam can be developed as follows (the detailed derivation process is attached to the end of the text, Appendix 1)

{ E 0 M − E 1 I 1 β 1 2 [ A ( β 1 g d − β 2 c e ) + B ( β 2 a e − β 1 g b ) ] } ⋅ { β 1 N H − E 1 I 1 β 1 2 [ C ( β 1 h d − β 2 f c ) + D ( β 2 a f − β 1 b h ) ] } − { F M − E 1 I 1 β 1 2 [ A ( β 1 h d − β 2 f c ) + B ( β 2 a f − β 1 b h ) ] } ⋅ { β 1 N G − E 1 I 1 β 1 2 [ C ( β 1 g d − β 2 c e ) + D ( β 2 a e − β 1 g b ) ] } = 0

(19)

In equation (19)

A = ch β 2 L ⋅ cos β 2 ( L − L 1 ) + sh β 2 L ⋅ sin β 2 ( L 1 − L ) + ch β 2 L 1 B = sh β 2 L ⋅ cos β 2 ( L − L 1 ) + ch β 2 L ⋅ sin β 2 ( L 1 − L ) + sh β 2 L 1 C = ch β 2 L sin β 2 ( L − L 1 ) + sh β 2 L cos β 2 ( L − L 1 ) + sh β 2 L 1 D = ch β 2 L cos β 2 ( L − L 1 ) + sh β 2 L sin β 2 ( L − L 1 ) + ch β 2 L 1

E 0 = ch β 1 L 1 − cos β 1 L 1 F = sh β 1 L 1 − sin β 1 L 1 G = sh β 1 L 1 + sin β 1 L 1 H = ch β 1 L 1 − cos β 1 L 1

a = ch β 2 L sin β 2 ( L 1 − L ) − sh β 2 L cos β 2 ( L 1 − L ) + sh β 2 L 1 b = − ch β 2 L cos β 2 ( L − L 1 ) + sh β 2 L sin β 2 ( L − L 1 ) + ch β 2 L 1 c = − ch β 2 L cos β 2 ( L 1 − L ) + sh β 2 L sin β 2 ( L 1 − L ) + ch β 2 L 1 d = ch β 2 L sin β 2 ( L − L 1 ) − sh β 2 L cos β 2 ( L − L 1 ) + sh β 2 L 1

e = cos β 1 L 1 + ch β 1 L 1 f = sin β 1 L 1 + sh β 1 L 1 g = sh β 1 L 1 − sin β 1 L 1 h = ch β 1 L 1 + cos β 1 L 1

Equation (19) is the natural frequency equation of the cantilever stepped beam. In which, β 1 4 = ω 2 / a 1 2 , β 2 4 = ω 2 / a 2 2 , they contain the frequency ω to be solved.

By the same method, for the simplified supported step beam shown in Figure 1(b), one can develop the corresponding frequency equation as shown in equation (20), according to the boundary condition y₁(0, t) = 0, y ″ 1 ( 0 , t ) = 0 at the end at x = 0, boundary condition y₁(L, t) = 0, y ″ 1 ( L , t ) = 0 at the end at x = L, the displacement continuity condition y₁(L₁, t) = y₂(L₁, t), y ′ 1 ( L 1 , t ) = y ′ 2 ( L 1 , t ) , the bending moment continuity condition E 1 I 1 y ″ 1 ( L 1 , t ) = E 2 I 2 y ″ 2 ( L 1 , t ) and the shear force continuity condition E 1 I 1 y ‴ 1 ( L 1 , t ) = E 2 I 2 y ‴ 2 ( L 1 , t ) at the step section

{ M sin β 1 L 1 − E 1 I 1 β 1 2 [ a ( h β 2 sin β 1 L 1 − f β 1 cos β 1 L 1 ) + b ( g β 2 sin β 1 L 1 − e β 1 cos β 1 L 1 ) ] } ⋅ { N ch β 1 L 1 − E 1 I 1 β 1 [ c ( f β 1 ch β 1 L − h β 2 sh β 1 L 1 ) + d ( e β 1 ch β 1 L − g β 2 sh β 1 L 1 ) ] } − { M sh β 1 L 1 − E 1 I 1 β 1 2 [ a ( f β 1 ch β 1 L 1 − h β 2 sh β 1 L 1 ) + b ( e β 1 ch β 1 L 1 − g β 2 sh β 1 L 1 ) ] } ⋅ { N cos β 1 L 1 − E 1 I 1 β 1 [ c ( h β 2 sin β 1 L 1 − f β 1 cos β 1 L 1 ) + d ( g β 2 sin β 1 L 1 − e β 1 cos β 1 L 1 ) ] } = 0

(20)

In equation (20)

a = sin β 2 L 1 − tan β 2 L ⋅ cos β 2 L 1 , b = sh β 2 L 1 − th β 2 L ⋅ ch β 2 L 1 c = cos β 2 L 1 + tan β 2 L ⋅ sin β 2 L 1 , d = ch β 2 L 1 − th β 2 L ⋅ sh β 2 L 1 e = sin β 2 L 1 − tan β 2 L ⋅ cos β 2 L 1 , f = sh β 2 L 1 − th β 2 L ⋅ ch β 2 L 1 g = tan β 2 L ⋅ sin β 2 L 1 + cos β 2 L 1 , h = ch β 2 L 1 − th β 2 L ⋅ sh β 2 L 1 M = E 2 I 2 β 2 3 ( h e − g f ) , N = E 2 I 2 β 2 2 ( h e − g f )

β 1 4 = ω 2 / a 1 2 , β 2 4 = ω 2 / a 2 2 , ω is the frequency to be solved.

Equations (19) and 20 are too complex to be solved directly artificially. Therefore, the Mathcad software is used to solve the equations.

The fundamental frequencies of cantilever step beams with different geometrical shapes and materials

The fundamental frequency of the beam with circular cross-section

The diameter of the left-hand beam in Figure 1 is d₁, the cross-sectional area is A 1 = π d 1 2 / 4 , the axis inertia moment is I 1 = π d 1 4 / 64 ; set the diameter of the right-hand beam is d₂, the cross-sectional area is A 2 = π d 2 2 / 4 , and the axis inertia moment is I 2 = π d 2 4 / 64 .

L₁ and L₂ take different ratios

Freely suppose: d₁ = 40 mm, d₂ = 38 mm, E₁ = E₂ = 210 GPa, ρ₁ = ρ₂ = 7800 kg/m3, the total length of cantilever stepped beam is L = 0.15 m. The fundamental frequencies of the beam with different length ratios are solved as shown in Table 1. According to Table 1, the curve of the fundamental frequencies is plotted as shown in Figure 2. It can be seen from Figure 2 that the fundamental frequency increases linearly with the length ratios.

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Table 1.

The fundamental frequency of the stepped beam under different length ratios (circular cross-section).

L1:L2	8	3.5	2	1.25	1
ω (rad/s)	8729	4908	3276	2355	2027

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Figure 2.

Curve of fundamental frequencies under different length ratios (circular cross-section).

A₁ and A₂ take different ratios

Freely suppose: L₁ = 0.117 m, L₂ = 0.033 m, E₁ = E₂ = 210 GPa, ρ₁ = ρ₂ = 7800 kg/m3. Change the area A₁, A₂, and then we can solve the fundamental frequencies under different area ratios, which are shown in Table 2. According to Table 2, the curve of the fundamental frequencies is shown in Figure 3. As seen from Figure 3, the fundamental frequencies change irregularly with the increase in area ratios, approximately as the sine curve.

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Table 2.

The fundamental frequencies under different area ratios (circular cross-section).

A1:A2	1.184	1.148	1.125	1.111	1.108	1.103
ω (rad/s)	4391	5424	4262	4779	4908	5166

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Figure 3.

Curve of fundamental frequencies under different area ratios (circular cross-section).

E₁I₁ and E₂I₂ take different ratios

Freely suppose: L₁ = 0.117 m, L₂ = 0.033 m, d₁ = 40 mm, and d₂ = 38 mm, we change the elastic modulus E₁, E₂, as shown in Table 3. The fundamental frequencies under different materials are shown in Table 4. According to Table 4, the curve of the fundamental frequencies is plotted as shown in Figure 4. It can be seen from Figure 4 that the fundamental frequencies change irregularly with the increase in bending rigidity ratios, and the largest fundamental frequency of the cantilever step beam is 4605 rad/s.

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Table 3.

Elastic modulus and bending rigidity of different materials (circular cross-section).

E1 (GPa)	E2 (GPa)	ρ1 (kg/m3)	ρ2 (kg/m3)	E1I1 × 104 (N m2)	E2I2 × 104 (N m2)
206	120	7800	7000	2.589	1.228
206	173	7800	7300	2.589	1.771
108	68	8900	2700	1.357	0.696
127	70	8900	2700	1.596	0.7165
145	103	7550	8800	1.822	1.054

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Table 4.

The fundamental frequencies under different bending rigidity ratios (circular cross-section).

E1I1:E2I2	2.108	1.462	1.949	2.227	1.729
ω (rad/s)	3916	4605	4747	4816	3236

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Figure 4.

Curve of vibration fundamental frequencies under different bending stiffness ratios (circular cross-section).

The fundamental frequency of the beam with rectangular cross-section

The width of the left section beam is b₁, the height value is h₁, the area is A₁ = b₁·h₁, and the moment of axis inertia is I 1 = b 1 ⋅ h 1 3 / 12 . The width of the right beam is b₂, the height value is h₂, the area is A₂ = b₂·h₂, and the axis moment of inertia is I 2 = b 2 ⋅ h 2 3 / 12 .

L₁ and L₂ take different ratios

Given the cross-sectional area and material of the left and right section of the cantilever stepped beam, b₁ = 40 mm, h₁ = 30 mm, b₂ = 38 mm, h₂ = 28 mm, E₁ = E₂ = 210 GPa, ρ₁ = ρ₂= 7800 kg/m3, and L = 0.15 m, take the different length ratios L₁:L₂. The fundamental frequencies are solved as shown in Table 5. According to Table 5, the curve of the fundamental frequencies with different length ratios is shown in Figure 5. It can be seen from Figure 5 that the vibration fundamental frequencies of the cantilever stepped beam gradually increase with the increased ratio of the length.

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Table 5.

The fundamental frequencies under different length ratios (rectangular cross-section).

L1:L2	8	35	2	1.25	1
ω (rad/s)	7427	4176	2788	2004	1725

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Figure 5.

Curve of fundamental frequencies under different length ratios (rectangular cross-section).

Cross-sectional area A₁ and A₂ take different ratios

Given the length L₁ and L₂ and material stiffness E₁ and E₂, L₁ = 0.117 m, L₂ = 0.033 m, E₁ = E₂ = 210 GPa, ρ₁ = ρ₂ = 7800 kg/m3. Change the area ratios A₁:A₂ as shown in Table 6. The fundamental frequencies with different cross-sectional areas are shown in Table 7. According to Table 7, the curve of the fundamental frequency is plotted as shown in Figure 6. It can be seen from Figure 6 that the fundamental frequencies are approximately sinusoidal variation with the area ratios.

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Table 6.

The ratios of A₁with A₂ for the cantilever stepped beam (rectangular cross-section).

b1 (mm)	h1 (mm)	b2 (mm)	h2 (mm)	A1:A2
45	35	42	32	1.172
40	30	38	28	1.128
39	29	37	27	1.132
37	27	34	24	1.224
35	25	33	23	1.153

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Table 7.

The fundamental frequencies under different cross-sectional area ratios (rectangular cross-section).

A1:A2	1.224	1.172	1.153	1.132	1.128
ω (rad/s)	3579	4772	3430	4027	4176

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Figure 6.

Curve of frequencies under different cross-sectional area ratios (rectangular cross-section).

E₁I₁ and E₂I₂ take different ratios

Given the length L₁ and L₂ of beam and freely set: L₁ = 0.117 m, L₂ = 0.033 m, b₁ = 40 mm, h₁ = 30 mm, b₂ = 38 mm, and h₂ = 28 mm. Change the material stiffness E₁ and E₂, which means changing the elastic modulus E₁I₁, E₂I₂ of the two beams as shown in Table 8. The fundamental frequencies with different materials are solved as shown in Table 9. According to Table 8, the curve of the fundamental frequencies is plotted as shown in Figure 7. As can be seen from Figure 7, the fundamental frequencies change irregularly with the increase in bending rigidity ratios, and the largest vibration fundamental frequency is 4098 rad/s.

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Table 8.

Elastic modulus and bending rigidity of different materials (rectangular cross-section).

E1 (GPa)	E2 (GPa)	ρ1 (kg/m3)	ρ2 (kg/m3)	E1I1 × 104 (N m2)	E2I2 × 104 (N m2)
206	120	7800	7000	1.854	0.834
206	173	7800	7300	1.854	1.203
108	68	8900	2700	0.972	0.473
127	70	8900	2700	1.143	0.487
145	103	7550	8800	1.305	0.716

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Table 9.

The fundamental frequency under different bending rigidity ratios (rectangular cross-section).

E1I1:E2I2	2.222	1.541	2.056	2.349	1.823
ω (rad/s)	3332	3918	4039	4098	2753

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Figure 7.

Curve of vibration fundamental frequencies under different bending rigidity ratios (rectangular cross-section).

The fundamental frequencies of simplified step beams with different geometrical shapes

Without loss of generality, the geometry parameters are consistent with the cantilever step beam and the elastic modulus is assumed as E₁ = E₂ = 21 GPa. The fundamental frequencies of simplified step beams with different length ratios under different geometrical shapes are also solved as shown in Table 10. According to Table 10, the curve of the fundamental frequencies is plotted as shown in Figure 8. It can be seen from Figure 8 that the fundamental frequency decreases linearly with the length ratios.

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Table 10.

The fundamental frequencies under different length ratios with different geometrical shapes.

Circular cross-section	L1:L2	8.5:0.5	8:1	7.5:1.5	7:2	6.5:2.5
	ω (rad/s)	7200	7229	7309	7484	7895
Rectangular cross-section	L1:L2	8.5:0.5	8:1	7.5:1.5	7:2	6.5:2.5
	ω (rad/s)	6236	6265	6346	6522	6935

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Figure 8.

The fundamental frequency curves of simplified step beams under different length ratios with different geometrical shapes.

Conclusion

Based on the theory of transverse bending vibration, the natural frequency equation of the bending vibration for the step beam is given using the separation variable method. The method is suitable for different supporting ways, different cross-sectional shapes, different segmented materials, and different length ratios of the stepped beam. With the help of Mathcad software, the variation law of the first-order natural frequency of the stepped beam with different supporting ways, different section shapes, different ratios of length, different ratios of thickness, and different materials is given. The research shows that