Vibration Midterm-THEORETICAL SOLUTION AND STATIC ANALYSES STUDY OF VIBRATIONS OF CHANNEL BEAMS
1. 1
Spring 2010
Department : Mechanical Engineering, City University of Newyork, Newyork, U.S.A
Course : Adv. Mech. Vibrations: ME I6200 4ST [3223] (CCNY)
Subject : Transverse Vibration of Channel Beams
Instructor : Prof. Benjamin Liaw
Student : Mech. Eng. M. Bariskan
2. 2
THEORETICAL SOLUTION AND STATIC ANALYSES STUDY
OF VIBRATIONS OF CHANNEL BEAMS
Abstract; In this paper, studies of thin-walled channel beams with known cross section area and
length were conducted using certain static analyses and verified with known theoretical
solution. The motion of the system is described by a homogenous set of partial differential
equations. Used; Mathcad to solve roots for natural frequency, Used Euler-Bernoulli Beam
Theoretical equations for equation of motion and Solidworks for certain analyses. The result of
the presented theoretical analyses for Channel beams are compared with result taken from
Solidworks.
Keywords: Transverse Vibration of Beams, Channel Beam , Equation of Motion
3. 3
1.1 Introduction
In this paper the free transverse vibration of beams considered. The equations of motion of a
beam are derived according to the Euler-Bernoulli, Rayleigh, and Timoshenko theories. The
Euler Bernoulli theory neglects the effects of rotary inertia and shear deformation and is
applicable to an analysis of thin beams. The Rayleigh theory considers the effect of rotary
inertia, and the Timoshenko theory considers the effect of both rotary inertia and shear
deformation. The Timoshenko theory can be used for thick beams. The equations of motion for
the transverse vibration of beams are in the form of fourth-order partial differential equations
with two boundary conditions at each end. In this paper; The free vibration solution, including
the determination of natural frequencies and mode shapes, is considered according to Euler-
Bernoulli theory.
1.2 Equation of Motion : Euler- Bernoulli Theory
In the Euler- Bernoulli or thin beam theory, the rotation of cross section of the beam is
neglected compared to the translation. In addition, the angular distortion due to shear is
considered negligible compared to the bending deformation. The thin beam theory is applicable
to beams for which the length is much larger than the depth (at list 10 times), and the
deflection are small compared to the depth. When the transverse displacement of the
centerline of the beam is w, the displacement remain plane and normal to the centerline are
given by figure 1,
4. 4
π’ = βπ§
ππ€(π₯, π‘)
ππ₯
, π£ = 0, π€ = π€(π₯, π‘)
where u, v, w denote the components of displacement parallel to x, y, and z directions,
respectively. The components of strain and stress corresponding to this displacement field are
given by
π π₯π₯ =
ππ’
ππ₯
= βπ§
π2
π€(π₯, π‘)
ππ₯2
, π π¦π¦ = π π§π§ = π π₯π¦ = π π¦π§ = π π§π₯ = 0
ππ₯π₯ = βπΈπ§
π2
π€
ππ₯2
, ππ¦π¦ = ππ§π§ = ππ₯π¦ = ππ¦π§ = ππ§π₯ = 0
And, if we explain the strain energy of the system (Ο) and Iy=I denotes the area moment of
inertia of the cross section of the beam about the y axis
πΌ = πΌπ¦ = π§2
ππ΄π΄
,
The kinetic energy of the system (T) , and these are calculated in,
π€ = ππ€ππ₯
π
0
The application of the generalized Hamiltons principle gives,
πΏ π β π β π ππ‘ = 0 ππ πΏ
1
2
πΈπΌ(
π2 π€
ππ₯2
)2
ππ₯ + β― β¦ . .
π
0
= 0
π‘2
π‘1
π‘2
π‘1
By setting the expressions are giving by book which is vibration of continuous systems page 320,321
5. 5
We have differential equation of motion for the transverse vibration of the beam as ;
π2
ππ₯2
πΈπΌ
π2 π€
ππ₯2
+ ππ΄
π2 π€
ππ‘2
= π(π₯, π‘) (1.1)
And the boundary conditions as
πΈπΌ
π2 π€
ππ₯2
ππ€
ππ₯
πΌ0
π
β
π
ππ₯
πΈπΌ
π2 π€
ππ₯2
πΏπ€πΌ0
π
+ π1 π€πΏπ€πΌ0 + ππ‘1
ππ€
ππ₯
πΏ
ππ€
ππ₯
πΌ0 + π1
π2 π€
ππ₯2
πΏπ€πΌ0 + π2 π€πΏπ€πΌπ
+
ππ‘2
ππ€
ππ₯
πΏ
ππ€
ππ₯
πΌπ
+ π2
π2 π€
ππ₯2
πΏπ€πΌπ
=0 (1.2)
At x=0,
ππ€
ππ₯
=constant (so that πΏ
ππ€
ππ₯
= 0) or βπΈπΌ
π2 π€
ππ₯2
+ ππ‘1
ππ€
ππ₯
= 0 (1.3)
w = constant (so that πΏπ€ = 0 ) or (
π
ππ₯
πΈπΌ
π2 π€
ππ₯2
+ π1 π€ + π1
π2 π€
ππ₯2
) = 0
At x=l,
ππ€
ππ₯
=constant (so that πΏ
ππ€
ππ₯
= 0) or πΈπΌ
π2 π€
ππ₯2
+ ππ‘2
ππ€
ππ₯
= 0 (1.4)
w = constant (so that πΏπ€ = 0 ) or (β
π
ππ₯
πΈπΌ
π2 π€
ππ₯2
+ π2 π€ + π2
π2 π€
ππ₯2
) = 0 (1.5)
1.3 Free Vibration Equations
For free vibration, the external excitation is assumed to be zero :
f(x,t)=0 (1.6)
and hence the equation of motion ,Eq.(1.2), reduces to
π2
ππ₯2
πΈπΌ π₯
π2 π€ π₯,π‘
ππ₯2
+ ππ΄ π₯
π2 π€ π₯,π‘
ππ‘2
= 0 (1.7)
For a uniform beam Eq.(1.7) can be expressed as
π2 π4 π€
ππ₯4
(π₯, π‘) +
π2 π€
ππ‘2
(π₯, π‘) = 0 (1.8)
π =
πΈπΌ
ππ΄
(1.9)
6. 6
1.4 Free Vibration Solution
The free vibration solution can be found using the method of separation of variables as
w(x,t)= W(x)T(t) (1.10)
Using Eq. (1.10) in Eq.(1.8) and rearranging yields,
π2
π(π₯)
π4 π(π₯)
ππ₯4
= β
1
π(π‘)
π2 π(π‘)
ππ‘2
= a = π€2
(1.11)
where a = π€2
can be shown to be a positive constant Eq. (1.11) can be rewritten as two
equations:
π4 π(π₯)
ππ₯4
β π½4
π π₯ = 0 (1.12)
π2 π(π‘)
ππ‘2 + π€2
π π‘ = 0 (1.13)
where
π½4
=
π€2
π2
= π
ππ΄ π€2
πΈπΌ
(1.14)
The solution of Eq.(1.13) is given by
T(t) = A.coswt + B.sinwt
where A and B are constant that can be found from the initial conditions. The solution of equation (1.12)
is assumed to be of exponential form as
π(π₯) = πΆπ π π₯
(1.15)
Where C and s constants. Substitution of Eq. (1.15) into Eq.(1.12) result in the auxiliary equation
π 4
β π½4
= 0 (1.16)
The roots of this equation are given by
π 1,2 = βπ½ π 3,4 = βππ½ (1.17)
Thus the solution of Eq. (1.12) can be expressed as
π π₯ = πΆ1 π π½π₯
+ πΆ2 πβπ½π₯
+ πΆ3 π ππ½π₯
+ πΆ4 πβππ½π₯
(1.18)
where C1, C2, C3 and C4 are constants. Equation (1.18) can be expressed more conveniently as
π π₯ = πΆ1 πππ π½π₯ + πΆ2 π πππ½π₯ + πΆ3 πππ βπ½π₯ + πΆ4 π ππβπ½π₯ (1.19) or,
7. 7
π π₯ = πΆ1(πππ π½π₯ + πππ βπ½π₯) + πΆ2 πππ π½π₯ β πππ βπ½π₯ + πΆ3 π πππ½π₯ + π ππβπ½π₯ + πΆ4(π πππ½π₯ β π ππβπ½π₯) (1.20)
where C1, C2, C3 and C4 are different constants in each case. The natural frequencies of the beam can
be determined of Eq. (1.14) as
π€ = π½2 πΈπΌ
ππ΄
= (π½π)2 πΈπΌ
ππ΄ π4 (1.21)
The function W(x) is known as the normal mode or characteristic function of the beam and w is called
the natural frequency of vibration . For any beam, there will be infinite number of normal modes with
one natural frequency associated with each normal mode. The unknown constant C1, C2, C3, C4 in Eq.
(1.19) or Eq. (1.20) and the value of Ξ² in Eq. (1.21) can be determined from the known boundary
conditions of the beam.
8. 8
1.5 Frequencies and mode shapes of uniform beams
The natural frequencies and mode shapes of beams with a uniform cross section with different
boundary conditions are considered in this section. We can find boundary conditions from given tables
in the textbook. Also Is listed below
9. 9
1.5.1 Beam Simply Supported at Both Ends
The transverse displacement and the bending moment are zero at the both ends. Hence boundary
conditions can be started as:
(1.22)
or
π2 π
ππ₯2
(0) = 0 (1.23)
(1.24)
or
π2 π
ππ₯2
(π) = 0 (1.25)
When used In the solution of Eq. (1.20), Eq.(1.22) and (1.23) yield
C1=C2=0 (1.26)
10. 10
(1.26)
(1.27)
(1.28)
These equations denote a system of two equations in the two unknowns and . For a nontrivial
solution of and , the determinant of the coefficients must be equal to zero. This leads to
or
(1.29)
It can be observed that is not equal to zero unless =0. The value of =0 need not be
considered because it implies, according to equation;
=0, which corresponds to the beam at rest. Thus, the frequency equations becomes
(1.30)
The roots of the equations, , are given by
, (1.31)
And hence the natural frequencies of vibration become
=π2
π2
(
πΈπΌ
ππ΄ π4)1/2
, n=1,2,β¦β¦.. (1.32)
11. 11
We are going to find reaction force, moment maximum, stress, deflection and strain magnitude for this
C-beam and natural frequencies of vibration of a channel beam
(material stainless steel AISI 304) 3 m long,0.03 thick and dimensions 0.15 m, 0.25 m respectively. As
shown figure,
Figure1 : C-beam Main Dimension
And the specifications for this material: AISI 304 Stainless Steel
Elastic Modulus (E) : 190*109
Pa,
Density (Ο) : 8000 kg / m3 =8000*9.81 =78480 N/m3
=78.4 kN/m3
Area (A) :2*( 0.15*0.03) + (0.19*0.03) = 0.0147 m2
πΌπ§ β
2
3
π3
π‘ + 2π2
ππ‘ = 0.000126723 m4
from textbook equation
Volume per meter = 0.0147 m3 Total V= 0.0441m3 mass=rho*V*g(9.81 m/sn2)=3460.9 N
12. 12
Firstly, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
compare static analyses both, theoretical and Solidworks result then mod shapes and Natural
frequencies is calculated and compared on same way.
Figure-2-Free Body Diagram βSimply Supported Beam
For simply supported both ends
πΉπ¦ = 0 β π + ππ + ππ = 0
ππ = 0 3. ππ β 1.5π = 0 π = 1153 π π = 3 β 1153 = 3460.9π
Va = 1730N =Vb
Figure β 3 Reaction Force for Simply-supported at both ends (Solidworks 2008)
And Plotting the shear force and Bending moment diagram below:
13. 13
Figure-4 Shear Force and Moment Diagrams for Simply Supported Both Ends
Moment max : 1730*1.5/2 = 1297.5 N.m
ππ₯π₯ =
ππ§.π¦
πΌπ§π§
=
1297.5β0.125
0.000126723
= 0.127 Mpa
Figure-5 Stress (normal stress) for z direction
Equation of deflection for simply supported ends
πΏ =
β5. π€. πΏ4
384. πΈ. πΌ
=
β5.1153.6 β 34
384.190 β 109
β 0.000126723
= β0.000051 = 5.1 β 10β5
π
14. 14
Figure- 6 Deflection on The Simply Supported beam ( Solidworks)
Mode shapes and Natural Frequencies for Simply Supported Both End
Figure-7 Natural frequencies and mode shapes of a beam simply supported beam.
,
For n π· π π = π. ππππ π· π π = π. ππππ π· π π = π. ππππ π· π π = π. ππππ
Equations Euler-Bernoulli calculated in MathCAD , results are in the units : rad/sc simply supported at
both ends
17. 17
1.5.2 Beam Fixed-fixed Ends
We are going to find reaction force, moment, stress, deflection, and strain magnitude to compare static
analyses both, theoretical and Solidworks result then mod shapes and Natural frequencies is calculated
and compared on same way. For fixed-fixed beams
Figure-12-Free Body Diagram βFixed-fixed Beam
For fixed-fixed supported both ends
πΉπ¦ = 0 β π + ππ + ππ = 0
ππ = 0 3. ππ β 1.5π = 0 π = 1153 π π = 3 β 1153 = 3460.9π
Va = 1730N =Vb
And Plotting the shear force and Bending moment diagram below:
Figure-13 Shear Force and Moment Diagrams for Fixed-fixed Beam
18. 18
Moment max= (1153*32
/12)+(1730*1.5/2)=2162.25 N.m
ππ₯π₯ =
ππ§.π¦
πΌπ§π§
=
2162.25β0.125
0.000126723
= 2.13 Mpa
Figure-14 Stress (normal stress) for z direction
Equation of deflection for fixed-fixed ends
πΏ =
βπ€. πΏ4
384. πΈ. πΌ
=
1153.6 β 34
384.190 β 109
β 0.000126723
= β0.0000101 = 1.01 β 10β5
π
Figure- 15 Deflection on The Fixed-fixed beam ( Solidworks)
19. 19
Mode shapes and Natural Frequencies for Fixed-Fixed Both End
Figure-16 Natural frequencies and mode shapes of a beam fixed-fixed end
Natural frequencies and mode shapes of a beam simply supported at both ends.
, π½π π β 2π + 1 π/2
For n π½1 π = 4.73 π½2 π = 7.8532 π½3 π = 10.996 π½4 π = 14.731
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported at
both ends
21. 21
Solidworks= 530.96 rd/sc
Therotical= 358.59 rd/sc
Error = %48
Mode=1
Figure-17 Mode Shape 1 for F-F Both Ends
Solidworks= 1265.61 rd/sc
Therotical= 989.2 rd/sc
Error = %27
Mode=2
Figure-18 Mode Shape 2 for F-F Both Ends
Solidworks= 2208.6 rd/sc
Theoretical= 1938.6 rd/sc
Error = %12
Mode=3
Figure-19 Mode Shape 3 for F-F Both Ends
Solidworks= 3206 rd/sc
Therotical= 3208 rd/sc
Error = %01
Mode=4
Figure-20 Mode Shape 4 For F-F Both Ends
22. 22
1.5.3 Beam Fixed-Simply Supported Ends
Thirdly, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
compare static analyses both, theoretical and Solidworks result then mod shapes and Natural
frequencies is calculated and compared on same way. For Fixed-Simply Supported
Figure-21-Free Body Diagram βFixed-Simply Supported Beam
V(z)= - 1/8 * w *(5L-8x)
Vmax= Vz=0 Va = - 5Lw/8 Vb=3Lw/8
Va= - 5*3*1153/8 = 2161.8 N Vb= 1297.125 N
Figure-22 Shear Force and Moment Diagrams for Fixed-Simply Supported Beam
And Moment for any point of beam
ππ = π€. π₯
3πΏ
8
β
π₯
2
πππ πππ₯ ππππππ‘ ππ‘
3
8
πΏ ππ =
9
128
π€πΏ2
= 729 π
23. 23
π πππ₯ =
1
8
π€πΏ2
= 1297 π. π
ππ₯π₯ =
ππ§.π¦
πΌπ§π§
=
1297β0.125
0.000126723
= 1.27 Mpa
Figure-23 Stress (normal stress) for z direction
Equation of deflection for simplysupport-fixed ends At x=0.4215l
πΏ =
βπ€. πΏ4
185. πΈ. πΌ
= β
1153.6 β 34
185.190 β 109
β 0.000126723
= β0.0000209 = 2.09 β 10β5
π
Figure- 24 Deflection on The Fixed-Simply Supported beam ( Solidworks)
24. 24
Mode shapes and Natural Frequencies for Simply supported-Fixed End
Figure-25 Natural frequencies and mode shapes of a beam simply supported-fixed end
, π½π π β 4π + 1 π/4
For n π½1 π = 3.9266 π½2 π = 7.06686 π½3 π = 10.212 π½4 π = 13.3518
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported β
fixed ends.
27. 27
1.5.4 Beam Fixed-Free Ends (Cantilever Beam)
Fourtly, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
compare static analyses both, theoretical and Solidworks result then mod shapes and Natural
frequencies is calculated and compared on same way. For Cantilever Beams
Figure-30-Free Body Diagram βCantilever Beam
For simply supported-fixed ends
V(b)= - w.l
Vmax= -1153*3 = 3460.9 N
Figure-31 Shear Force and Moment Diagrams for Cantilever Beam
28. 28
And Moment for any point of beam
π =
π€π₯2
2
πππ πππ₯ ππππππ‘ π πππ₯ =
π€π2
2
=
1153 β 9
2
= 5188 π. π
ππ₯π₯ =
ππ§.π¦
πΌπ§π§
=
5188β0.125
0.000126723
= 5.117 Mpa
Figure-32 Stress (normal stress) for z direction
Equation of deflection for fixed-free ends At free ends
πΏ =
βπ€. πΏ4
8. πΈ. πΌ
= β
1153.6 β 34
8.190 β 109
β 0.000126723
= β0.000485111 = 4.85 β 10β4
π
Figure- 33 Deflection on The Cantilever Beam ( Solidworks)
29. 29
Mode shapes and Natural Frequencies for Fixed-Free (Cantilever Beams)
Figure-34 Natural frequencies and mode shapes of a cantilever beam
, π½π π β 2π β 1 π/2
For n π½1 π = 1.8751 π½2 π = 4.6941 π½3 π = 7.8547 π½4 π = 10.9956
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc fixed-free ends
32. 32
1.5.5 Beam free-Free Ends
Finally, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
compare static analyses both, theoretical and Solidworks result then mod shapes and Natural
frequencies is calculated and compared on same way. For Free-Free Beam
Figure-39 Natural frequencies and mode shapes of a Free-Free Beam
, π½π π β 2π β 1 π/2
For n π½1 π = 4.7300 π½2 π = 7.8532 π½3 π = 10.9965 π½4 π = 14.372
We donβt have any reaction force ;
I listed below mode shapes and natural frequencies ,