190x

2025-11-13

VE0101 | Modal Analysis of Cantilever

Description

A steel cantilever with a rectangular cross-section is fully fixed on one side and free on the other side. The aim of this verification example is to determine the natural frequencies of the structure. The problem is described by the following parameters.

Material	Steel	Modulus of Elasticity	E	206000.000	MPa
		Poisson's Ratio	ν	0.300	-
		Density	ρ	7800.000	kg/m³
Geometry	Cantilever	Length	L	90.000	mm
		Width	w	10.000	mm
		Thickness	t	5.000	mm

Modal Analysis of Cantilever

Analytical Solution

Longitudinal Oscillations

The natural longitudinal oscillation of a thin bar is described by the following differential equation

\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}

u	Deflection in the longitudinal direction
c	Longitudinal wave propagation velocity

Differential Equation of Longitudinal Oscillations

c = \sqrt{\frac{E}{ρ}}

Longitudinal Wave Propagation Velocity

The solution is supposed to be in the form u(x,t) = U(x) T(t). Using this form, the differential equation can be rewritten as follows:

\frac{\partial^{2} U}{\partial x^{2}} + \frac{ω^{2}}{c^{2}} U = 0

ω	Angular frequency

Differential Equation

This is the second-order differential equation with the following general solution.

U (x) = C_{1} \sin (\frac{ω^{2}}{c^{2}} x) + C_{2} \cos (\frac{ω^{2}}{c^{2}} x)

C₁, C₂

Real constants

General Solution of Differential Equation

The constants C₁ and C₂ can be obtained from the boundary conditions. The deflection on the fixed side and the stress on the free side has to be zero. The wave equation is determined using these boundary conditions.

C_{1} \frac{ω}{c} \cos (\frac{ω}{c} L) = 0

Wave Equation

The set of angular frequencies can then, obviously, be determined.

ω_{k} = \frac{c}{L} (2 k - 1) \frac{π}{2}, k = 1, 2, 3, \dots

Set of Angular Frequencies

The first natural frequency in the longitudinal direction can be calculated.

f_{x 1} = \frac{ω_{1}}{2 π} = \frac{c}{4 L} = 14275.253 H z

First Natural Frequency in Longitudinal Direction

Transversal Oscillations

The natural transversal oscillation of a thin bar is described by the following differential equation.

\frac{\partial^{4} v}{\partial x^{4}} + \frac{ρ A}{E I} \frac{\partial^{2} v}{\partial t^{2}} = 0

A	Cross-section area

Differential Equation of Transversal Equation

Note that this equation has to be solved for both transversal directions. The solution is supposed to be in the form v(x,t) = V(x) T(t). Using this form, the differential equation can be rewritten as follows:

\frac{d^{4} V}{d x^{4}} - λ^{4} V = 0

Differential Equation

Where λ⁴ is the substitution for the following expression.

λ^{4} = \frac{ρ A ω^{2}}{E I}

Substitution

The fourth-order differential equation has the following general solution>

V (x) = C_{1} \sinh (λ x) + C_{2} \cosh (λ x) + C_{3} \sin (λ x) + C_{4} \cos (λ x)

General Solution of Differential Equation

The constants C₁ to C₄ can be obtained from the boundary conditions. The deflection and rotation on the fixed side, and the bending moment M and transversal force T on the free side, have to be zero. The wave equation for the transversal oscillations is determined using the boundary conditions.

\cosh (λ L) \cos (λ L) + 1 = 0

Wave Equation

This is the transcendent equation, and its solution can be found numerically. The natural frequencies can then be determined.

f_{i} = \frac{1}{2 π} λ_{i}^{2} \sqrt{\frac{E I}{ρ A}}, i = 1, 2, 3,

Transversal Natural Frequencies

RFEM Settings

Modeled in RFEM 6.11 and RFEM 5.08
The global element size is l_FE= 0.001 m
Shear stiffness of the members is neglected
Isotropic linear elastic material model is used
Member entity is used

Results

Natural Frequency	Analytical Solution	RFEM 6	Ratio	RFEM 5 - RF-DYNAM	Ratio
f_x1 [Hz]	14275.253	14275.072	1.000	14275.072	1.000
f_y1 [Hz]	1024.900	1024.778	1.000	1024.820	1.000
f_y2 [Hz]	6422.940	6418.894	0.999	6417.154	0.999
f_y3 [Hz]	17984.417	17960.053	0.999	17960.779	0.999
f_z1 [Hz]	512.450	512.413	1.000	512.495	1.000
f_z2 [Hz]	3211.470	3210.490	1.000	3211.004	1.000
f_z3 [Hz]	8992.208	8986.983	0.999	8988.420	1.000

686x

32x

Verification Example 000101 | 1

;