Vibration Midterm-THEORETICAL SOLUTION AND STATIC ANALYSES STUDY OF VIBRATIONS OF CHANNEL BEAMS

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
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
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
𝑢 = −𝑧
𝜕𝑤(𝑥, 𝑡)
𝜕𝑥
, 𝑣 = 0, 𝑤 = 𝑤(𝑥, 𝑡)
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
𝜀 𝑥𝑥 =
𝜕𝑢
𝜕𝑥
= −𝑧
𝜕2
𝑤(𝑥, 𝑡)
𝜕𝑥2
, 𝜀 𝑦𝑦 = 𝜀 𝑧𝑧 = 𝜀 𝑥𝑦 = 𝜀 𝑦𝑧 = 𝜀 𝑧𝑥 = 0
𝜍𝑥𝑥 = −𝐸𝑧
𝜕2
𝑤
𝜕𝑥2
, 𝜍𝑦𝑦 = 𝜍𝑧𝑧 = 𝜍𝑥𝑦 = 𝜍𝑦𝑧 = 𝜍𝑧𝑥 = 0
And, if we explain the strain energy of the system (π) and Iy=I denotes the area moment of
inertia of the cross section of the beam about the y axis
𝐼 = 𝐼𝑦 = 𝑧2
𝑑𝐴𝐴
,
The kinetic energy of the system (T) , and these are calculated in,
𝑤 = 𝑓𝑤𝑑𝑥
𝑙
0
The application of the generalized Hamiltons principle gives,
𝛿 𝜋 − 𝑇 − 𝑊 𝑑𝑡 = 0 𝑜𝑟 𝛿
1
2
𝐸𝐼(
𝜕2 𝑤
𝜕𝑥2
)2
𝑑𝑥 + ⋯ … . .
𝑙
0
= 0
𝑡2
𝑡1
𝑡2
𝑡1
By setting the expressions are giving by book which is vibration of continuous systems page 320,321

5
We have differential equation of motion for the transverse vibration of the beam as ;
𝜕2
𝜕𝑥2
𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
+ 𝜌𝐴
𝜕2 𝑤
𝜕𝑡2
= 𝑓(𝑥, 𝑡) (1.1)
And the boundary conditions as
𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
𝜕𝑤
𝜕𝑥
𝐼0
𝑙
−
𝜕
𝜕𝑥
𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
𝛿𝑤𝐼0
𝑙
+ 𝑘1 𝑤𝛿𝑤𝐼0 + 𝑘𝑡1
𝜕𝑤
𝜕𝑥
𝛿
𝜕𝑤
𝜕𝑥
𝐼0 + 𝑚1
𝜕2 𝑤
𝜕𝑥2
𝛿𝑤𝐼0 + 𝑘2 𝑤𝛿𝑤𝐼𝑙
+
𝑘𝑡2
𝜕𝑤
𝜕𝑥
𝛿
𝜕𝑤
𝜕𝑥
𝐼𝑙
+ 𝑚2
𝜕2 𝑤
𝜕𝑥2
𝛿𝑤𝐼𝑙
=0 (1.2)
At x=0,
𝜕𝑤
𝜕𝑥
=constant (so that 𝛿
𝜕𝑤
𝜕𝑥
= 0) or −𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
+ 𝑘𝑡1
𝜕𝑤
𝜕𝑥
= 0 (1.3)
w = constant (so that 𝛿𝑤 = 0 ) or (
𝜕
𝜕𝑥
𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
+ 𝑘1 𝑤 + 𝑚1
𝜕2 𝑤
𝜕𝑥2
) = 0
At x=l,
𝜕𝑤
𝜕𝑥
=constant (so that 𝛿
𝜕𝑤
𝜕𝑥
= 0) or 𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
+ 𝑘𝑡2
𝜕𝑤
𝜕𝑥
= 0 (1.4)
w = constant (so that 𝛿𝑤 = 0 ) or (−
𝜕
𝜕𝑥
𝐸𝐼
𝜕2 𝑤
𝜕𝑥2
+ 𝑘2 𝑤 + 𝑚2
𝜕2 𝑤
𝜕𝑥2
) = 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
𝜕2
𝜕𝑥2
𝐸𝐼 𝑥
𝜕2 𝑤 𝑥,𝑡
𝜕𝑥2
+ 𝜌𝐴 𝑥
𝜕2 𝑤 𝑥,𝑡
𝜕𝑡2
= 0 (1.7)
For a uniform beam Eq.(1.7) can be expressed as
𝑐2 𝜕4 𝑤
𝜕𝑥4
(𝑥, 𝑡) +
𝜕2 𝑤
𝜕𝑡2
(𝑥, 𝑡) = 0 (1.8)
𝑐 =
𝐸𝐼
𝜌𝐴
(1.9)

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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,
𝑐2
𝑊(𝑥)
𝑑4 𝑊(𝑥)
𝑑𝑥4
= −
1
𝑇(𝑡)
𝑑2 𝑇(𝑡)
𝑑𝑡2
= a = 𝑤2
(1.11)
where a = 𝑤2
can be shown to be a positive constant Eq. (1.11) can be rewritten as two
equations:
𝑑4 𝑊(𝑥)
𝑑𝑥4
− 𝛽4
𝑊 𝑥 = 0 (1.12)
𝑑2 𝑇(𝑡)
𝑑𝑡2 + 𝑤2
𝑇 𝑡 = 0 (1.13)
where
𝛽4
=
𝑤2
𝑐2
= 𝜌
𝜌𝐴 𝑤2
𝐸𝐼
(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
𝑊(𝑥) = 𝐶𝑒 𝑠𝑥
(1.15)
Where C and s constants. Substitution of Eq. (1.15) into Eq.(1.12) result in the auxiliary equation
𝑠4
− 𝛽4
= 0 (1.16)
The roots of this equation are given by
𝑠1,2 = ∓𝛽 𝑠3,4 = ∓𝑖𝛽 (1.17)
Thus the solution of Eq. (1.12) can be expressed as
𝑊 𝑥 = 𝐶1 𝑒 𝛽𝑥
+ 𝐶2 𝑒−𝛽𝑥
+ 𝐶3 𝑒 𝑖𝛽𝑥
+ 𝐶4 𝑒−𝑖𝛽𝑥
(1.18)
where C1, C2, C3 and C4 are constants. Equation (1.18) can be expressed more conveniently as
𝑊 𝑥 = 𝐶1 𝑐𝑜𝑠𝛽𝑥 + 𝐶2 𝑠𝑖𝑛𝛽𝑥 + 𝐶3 𝑐𝑜𝑠ℎ𝛽𝑥 + 𝐶4 𝑠𝑖𝑛ℎ𝛽𝑥 (1.19) or,

7
𝑊 𝑥 = 𝐶1(𝑐𝑜𝑠𝛽𝑥 + 𝑐𝑜𝑠ℎ𝛽𝑥) + 𝐶2 𝑐𝑜𝑠𝛽𝑥 − 𝑐𝑜𝑠ℎ𝛽𝑥 + 𝐶3 𝑠𝑖𝑛𝛽𝑥 + 𝑠𝑖𝑛ℎ𝛽𝑥 + 𝐶4(𝑠𝑖𝑛𝛽𝑥 − 𝑠𝑖𝑛ℎ𝛽𝑥) (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
𝑤 = 𝛽2 𝐸𝐼
𝜌𝐴
= (𝛽𝑙)2 𝐸𝐼
𝜌𝐴 𝑙4 (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 β in Eq. (1.21) can be determined from the known boundary
conditions of the beam.

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
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
𝜕2 𝑊
𝜕𝑥2
(0) = 0 (1.23)
(1.24)
or
𝜕2 𝑊
𝜕𝑥2
(𝑙) = 0 (1.25)
When used In the solution of Eq. (1.20), Eq.(1.22) and (1.23) yield
C1=C2=0 (1.26)

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(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
=𝑛2
𝜋2
(
𝐸𝐼
𝜌𝐴 𝑙4)1/2
, n=1,2,…….. (1.32)

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 (ρ) : 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
𝐼𝑧 ≈
2
3
𝑎3
𝑡 + 2𝑎2
𝑏𝑡 = 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
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 –Simply Supported Beam
For simply supported both ends
𝐹𝑦 = 0 − 𝑞 + 𝑉𝑎 + 𝑉𝑏 = 0
𝑀𝑏 = 0 3. 𝑉𝑎 − 1.5𝑞 = 0 𝑞 = 1153 𝑁 𝑊 = 3 ∗ 1153 = 3460.9𝑁
Va = 1730N =Vb
Figure – 3 Reaction Force for Simply-supported at both ends (Solidworks 2008)
And Plotting the shear force and Bending moment diagram below:

13
Figure-4 Shear Force and Moment Diagrams for Simply Supported Both Ends
Moment max : 1730*1.5/2 = 1297.5 N.m
𝜍𝑥𝑥 =
𝑀𝑧.𝑦
𝐼𝑧𝑧
=
1297.5∗0.125
0.000126723
= 0.127 Mpa
Figure-5 Stress (normal stress) for z direction
Equation of deflection for simply supported ends
𝛿 =
−5. 𝑤. 𝐿4
384. 𝐸. 𝐼
=
−5.1153.6 ∗ 34
384.190 ∗ 109
∗ 0.000126723
= −0.000051 = 5.1 ∗ 10−5
𝑚

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 𝜷 𝟏 𝒍 = 𝟑. 𝟏𝟒𝟏𝟔 𝜷 𝟐 𝒍 = 𝟔. 𝟐𝟖𝟑𝟐 𝜷 𝟑 𝒍 = 𝟗. 𝟒𝟐𝟒𝟖 𝜷 𝟒 𝒍 = 𝟗. 𝟒𝟐𝟒𝟖
Equations Euler-Bernoulli calculated in MathCAD , results are in the units : rad/sc simply supported at
both ends

15
n=1
193 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
3.1416
2
 159.562
n=2
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
6.2832
2
 633.268
n=3
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
9.4248
2
 1.425 10
3

n=4
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
12.5664
2
 2.533 10
3

n=5
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
15.708
2
 3.958 10
3

And From Solidworks Result we have,
1:39 Monday April 05 2010
Study name:
Study 2
Mode No. Frequency(Rad/sec) Frequency(Hertz) Period(Seconds)
1 311.23 49.534 0.020188
2 454.9 72.4 0.013812
3 714.83 113.77 0.0087898
4 746.45 118.8 0.0084175
5 1181 187.96 0.0053202
6 1382.9 220.09 0.0045436
7 1546.3 246.11 0.0040633
8 2152.4 342.57 0.0029191
9 2305.5 366.94 0.0027253
10 2702.4 430.1 0.002325
11 3311.2 526.99 0.0018976
12 3875.2 616.75 0.0016214
13 4076.8 648.84 0.0015412
14 4082.5 649.75 0.001539
15 4154.1 661.15 0.0015125
16 4309.1 685.81 0.0014581
17 4569.9 727.32 0.0013749
18 4900.2 779.89 0.0012822
19 4969.3 790.89 0.0012644
20 5185 825.21 0.0012118

16
Solidworks = 454 Rd/sc
Theoretical= 633 Rd/sc
Error = %12
Mode shape = 1
Figure-8 Mode Shape 1 For S.S Both Ends
Error = %7
Mode shape = 2
Error = %9
Mode shape = 3
Error = % 2
Mode shape = 4

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 –Fixed-fixed Beam
For fixed-fixed supported both ends
𝐹𝑦 = 0 − 𝑞 + 𝑉𝑎 + 𝑉𝑏 = 0
𝑀𝑏 = 0 3. 𝑉𝑎 − 1.5𝑞 = 0 𝑞 = 1153 𝑁 𝑊 = 3 ∗ 1153 = 3460.9𝑁
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
Moment max= (1153*32
/12)+(1730*1.5/2)=2162.25 N.m
𝜍𝑥𝑥 =
𝑀𝑧.𝑦
𝐼𝑧𝑧
=
2162.25∗0.125
0.000126723
= 2.13 Mpa
Equation of deflection for fixed-fixed ends
𝛿 =
−𝑤. 𝐿4
384. 𝐸. 𝐼
=
1153.6 ∗ 34
384.190 ∗ 109
∗ 0.000126723
= −0.0000101 = 1.01 ∗ 10−5
𝑚
Figure- 15 Deflection on The Fixed-fixed beam ( Solidworks)

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.
, 𝛽𝑛 𝑙 ≅ 2𝑛 + 1 𝜋/2
For n 𝛽1 𝑙 = 4.73 𝛽2 𝑙 = 7.8532 𝛽3 𝑙 = 10.996 𝛽4 𝑙 = 14.731
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported at
both ends

20
n=1
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
4.73
2
 358.879
n=2
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
7.8532
2
 989.279
n=3
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
10.9956
2
 1.939 10
3

n=4
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
14.1372
2
 3.206 10
3

n=5
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
17.2788
2
 4.789 10
3

n=6
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
20.4204
2
 6.689 10
3

Study name: Fixed-
Fixed
1 530.96 84.504 0.011834
2 543.06 86.431 0.01157
3 1265.1 201.34 0.0049667
4 1366.7 217.52 0.0045973
5 1408.6 224.18 0.0044606
6 2208.6 351.51 0.0028449
7 2624.5 417.7 0.0023941
8 3208.8 510.7 0.0019581
9 3333.9 530.61 0.0018846
10 4036.4 642.41 0.0015566
11 4090.4 651.01 0.0015361
12 4321.9 687.85 0.0014538
13 4562.4 726.12 0.0013772
14 4686.2 745.83 0.0013408
15 5114 813.92 0.0012286
16 5251 835.73 0.0011966
17 5299 843.36 0.0011857
18 5380.8 856.39 0.0011677
19 5797.8 922.75 0.0010837
20 6137.6 976.84 0.0010237

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
Therotical= 989.2 rd/sc
Error = %27
Mode=2
Theoretical= 1938.6 rd/sc
Error = %12
Mode=3
Solidworks= 3206 rd/sc
Therotical= 3208 rd/sc
Error = %01
Mode=4
Figure-20 Mode Shape 4 For F-F Both Ends

22
1.5.3 Beam Fixed-Simply Supported Ends
Thirdly, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
frequencies is calculated and compared on same way. For Fixed-Simply Supported
Figure-21-Free Body Diagram –Fixed-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
𝑀𝑏 = 𝑤. 𝑥
3𝐿
8
−
𝑥
2
𝑎𝑛𝑑 𝑀𝑎𝑥 𝑀𝑜𝑚𝑒𝑛𝑡 𝑎𝑡
3
8
𝐿 𝑀𝑏 =
9
128
𝑤𝐿2
= 729 𝑁

23
𝑀 𝑚𝑎𝑥 =
1
8
𝑤𝐿2
= 1297 𝑁. 𝑚
𝜍𝑥𝑥 =
𝑀𝑧.𝑦
𝐼𝑧𝑧
=
1297∗0.125
0.000126723
= 1.27 Mpa
Equation of deflection for simplysupport-fixed ends At x=0.4215l
𝛿 =
−𝑤. 𝐿4
185. 𝐸. 𝐼
= −
1153.6 ∗ 34
185.190 ∗ 109
∗ 0.000126723
= −0.0000209 = 2.09 ∗ 10−5
𝑚
Figure- 24 Deflection on The Fixed-Simply Supported beam ( Solidworks)

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
, 𝛽𝑛 𝑙 ≅ 4𝑛 + 1 𝜋/4
For n 𝛽1 𝑙 = 3.9266 𝛽2 𝑙 = 7.06686 𝛽3 𝑙 = 10.212 𝛽4 𝑙 = 13.3518
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported –
fixed ends.

25
n=1
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
3.9266
2
 247.32
n=2
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
7.0686
2
 801.479
n=3
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
10.2102
2
 1.672 10
3

n=4
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
13.3518
2
 2.86 10
3

Study name:
Study 2
1 399.57 63.594 0.015725
2 486.22 77.384 0.012923
3 817.66 130.13 0.0076843
4 1101.4 175.29 0.0057047
5 1277.1 203.25 0.0049201
6 1597.5 254.25 0.0039332
7 2157.2 343.33 0.0029126
8 2570.3 409.08 0.0024445
9 2862 455.51 0.0021954
10 3345.3 532.42 0.0018782
11 4010.5 638.29 0.0015667
12 4030.2 641.42 0.001559
13 4103.5 653.1 0.0015312
14 4302.6 684.79 0.0014603
15 4479.7 712.96 0.0014026
16 4632.1 737.22 0.0013564
17 5016.4 798.38 0.0012525
18 5219.9 830.76 0.0012037
19 5331 848.45 0.0011786
20 5665.7 901.73 0.001109

26
Solidworks = 399 rd/sc
Theoretical = 247 rd/sc
Error = %61
Mode = 1
Figure-26 Mode Shape 1 for F-S Ends
Error = %2
Mode =2
Solidworks 1597rd/sc
Theoretical = 1672 Rd/sc
Error = % 4
Mode = 3
Error = %0.01

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
frequencies is calculated and compared on same way. For Cantilever Beams
Figure-30-Free Body Diagram –Cantilever 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
And Moment for any point of beam
𝑀 =
𝑤𝑥2
2
𝑎𝑛𝑑 𝑀𝑎𝑥 𝑀𝑜𝑚𝑒𝑛𝑡 𝑀 𝑚𝑎𝑥 =
𝑤𝑙2
2
=
1153 ∗ 9
2
= 5188 𝑁. 𝑚
𝜍𝑥𝑥 =
𝑀𝑧.𝑦
𝐼𝑧𝑧
=
5188∗0.125
0.000126723
= 5.117 Mpa
Equation of deflection for fixed-free ends At free ends
𝛿 =
−𝑤. 𝐿4
8. 𝐸. 𝐼
= −
1153.6 ∗ 34
8.190 ∗ 109
∗ 0.000126723
= −0.000485111 = 4.85 ∗ 10−4
𝑚
Figure- 33 Deflection on The Cantilever Beam ( Solidworks)

29
Mode shapes and Natural Frequencies for Fixed-Free (Cantilever Beams)
Figure-34 Natural frequencies and mode shapes of a cantilever beam
, 𝛽𝑛 𝑙 ≅ 2𝑛 − 1 𝜋/2
For n 𝛽1 𝑙 = 1.8751 𝛽2 𝑙 = 4.6941 𝛽3 𝑙 = 7.8547 𝛽4 𝑙 = 10.9956
Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc fixed-free ends

30
n=1
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
1.8751
2
 56.399
n=2
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
4.6941
2
 353.452
n=3
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
7.8547
2
 989.657
n=4
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
10.9956
2
 1.939 10
3

Study name:
Study 2
1 85.685 13.637 0.073329
2 154.42 24.576 0.04069
3 345.67 55.015 0.018177
4 526.33 83.767 0.011938
5 696.91 110.92 0.0090158
6 1348.4 214.6 0.0046598
7 1427.8 227.24 0.0044006
8 1538.4 244.84 0.0040843
9 2347.7 373.65 0.0026763
10 2554.8 406.61 0.0024594
11 2671 425.11 0.0023523
12 3383.2 538.45 0.0018572
13 3497.6 556.67 0.0017964
14 4044.3 643.66 0.0015536
15 4130.6 657.41 0.0015211
16 4198.3 668.18 0.0014966
17 4482.7 713.45 0.0014016
18 4732.1 753.13 0.0013278
19 4919.5 782.96 0.0012772
20 5372.7 855.09 0.0011695

31
Solidworks = 85.65 Rd /sc
Theoretical = 56.39 Rd/sc
Error = %51
Mode=1
Figure-35 Mode Shape 1 for Fixed-Free Ends
Solidworks = 345.67 Rd/sc
Theoretical = 353.45
Error = %3
Mode=2
Error= %26
Mode=3
Error = %21
Mode=4

32
1.5.5 Beam free-Free Ends
Finally, We are going to find reaction force, moment, stress, deflection, and strain magnitude to
frequencies is calculated and compared on same way. For Free-Free Beam
Figure-39 Natural frequencies and mode shapes of a Free-Free Beam
, 𝛽𝑛 𝑙 ≅ 2𝑛 − 1 𝜋/2
For n 𝛽1 𝑙 = 4.7300 𝛽2 𝑙 = 7.8532 𝛽3 𝑙 = 10.9965 𝛽4 𝑙 = 14.372
We don’t have any reaction force ;
I listed below mode shapes and natural frequencies ,

33
n=2
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
7.8532
2
 989.279
n=3
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
10.9956
2
 1.939 10
3

n=4
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
14.1372
2
 3.206 10
3

n=5
190 10
9
 0.000129 
80 10
3
 0.0147 3
4
 
















0.5
17.2788
2
 4.789 10
3

Study name:
Study 2
1 0 0 1.00E+32
2 0 0 1.00E+32
3 0 0 1.00E+32
4 0.00067445 0.00010734 9316
5 0.0011889 0.00018922 5284.8
6 0.0015522 0.00024704 4048
7 538.93 85.773 0.011659
8 553.57 88.103 0.01135
9 787.16 125.28 0.0079821
10 1445.4 230.05 0.0043469
11 1542.2 245.45 0.0040741
12 1634.6 260.16 0.0038438
13 2488.3 396.02 0.0025251
14 2717.4 432.48 0.0023122
15 3534 562.45 0.0017779
16 3676.2 585.09 0.0017091
17 4028.3 641.13 0.0015597
18 4091.8 651.23 0.0015356
19 4222.8 672.08 0.0014879
20 4301.4 684.58 0.0014607
21 4647.6 739.69 0.0013519
22 4889.3 778.16 0.0012851
23 5088.4 809.85 0.0012348

34
Error = 0
Mode = 0
Figure : 40 Mode Shape 0 for Free-Free Ends
Solidworks = 787.16
Error = % 25
Mode = 1
Error = %18
Mode2
Error = %28
Mode = 3
Error = %17 Mode = 4

35
REFERENCES
1. 1. S.S. Rao, Vibration of Continuous Systems, Wiley, Hoboken, NJ, 2007.
2. 2. http://www.me.berkeley.edu/~lwlin/me128/formula.pdf
3. 3. http://www.awc.org/pdf/DA6-BeamFormulas.pdf
4. 4. https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec=fixed&page
5. =&appendix=beams
6. 5. BEAM FORMULAS WITH SHEAR AND MOMENTDIAGRAMS Copyright © 2007 American Forest &
7. Paper Association, Inc.
6. VIBRATION PROBLEMS IN ENGINEERING TIMOSHENKO SECOND EDITIONFIFTH PRINTING
8.

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