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

휀𝑥𝑥 =𝜕𝑢

𝜕𝑥= −𝑧

𝜕2𝑤(𝑥, 𝑡)

𝜕𝑥2, 휀𝑦𝑦 = 휀𝑧𝑧 = 휀𝑥𝑦 = 휀𝑦𝑧 = 휀𝑧𝑥 = 0

𝜍𝑥𝑥 = −𝐸𝑧𝜕2𝑤

𝜕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 = 𝑧2𝑑𝐴𝐴

,

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(

𝜕2𝑤

𝜕𝑥2 )2𝑑𝑥 + ⋯… . .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

𝜕2𝑤

𝜕𝑥2 + 𝜌𝐴

𝜕2𝑤

𝜕𝑡2= 𝑓(𝑥, 𝑡) (1.1)

And the boundary conditions as

𝐸𝐼𝜕2𝑤

𝜕𝑥2 𝜕𝑤

𝜕𝑥 𝐼0

𝑙 −𝜕

𝜕𝑥 𝐸𝐼

𝜕2𝑤

𝜕𝑥2 𝛿𝑤𝐼0

𝑙 + 𝑘1𝑤𝛿𝑤𝐼0 + 𝑘𝑡1𝜕𝑤

𝜕𝑥𝛿

𝜕𝑤

𝜕𝑥 𝐼0 + 𝑚1

𝜕2𝑤

𝜕𝑥2 𝛿𝑤𝐼0 + 𝑘2𝑤𝛿𝑤𝐼𝑙 +

𝑘𝑡2𝜕𝑤

𝜕𝑥𝛿

𝜕𝑤

𝜕𝑥 𝐼𝑙 + 𝑚2

𝜕2𝑤

𝜕𝑥2 𝛿𝑤𝐼𝑙 =0 (1.2)

At x=0,

𝜕𝑤

𝜕𝑥=constant (so that 𝛿

𝜕𝑤

𝜕𝑥 = 0) or −𝐸𝐼

𝜕2𝑤

𝜕𝑥2 + 𝑘𝑡1𝜕𝑤

𝜕𝑥 = 0 (1.3)

w = constant (so that 𝛿𝑤 = 0 ) or (𝜕

𝜕𝑥 𝐸𝐼

𝜕2𝑤

𝜕𝑥2 + 𝑘1𝑤 + 𝑚1𝜕2𝑤

𝜕𝑥2 ) = 0

At x=l,

𝜕𝑤

𝜕𝑥=constant (so that 𝛿

𝜕𝑤

𝜕𝑥 = 0) or 𝐸𝐼

𝜕2𝑤

𝜕𝑥2 + 𝑘𝑡2𝜕𝑤

𝜕𝑥 = 0 (1.4)

w = constant (so that 𝛿𝑤 = 0 ) or (−𝜕

𝜕𝑥 𝐸𝐼

𝜕2𝑤

𝜕𝑥2 + 𝑘2𝑤 + 𝑚2𝜕2𝑤

𝜕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 𝜕2𝑤 𝑥 ,𝑡

𝜕𝑥2 + 𝜌𝐴 𝑥 𝜕2𝑤 𝑥 ,𝑡

𝜕𝑡2 = 0 (1.7)

For a uniform beam Eq.(1.7) can be expressed as

𝑐2 𝜕4𝑤

𝜕𝑥4(𝑥, 𝑡) +

𝜕2𝑤

𝜕𝑡2(𝑥, 𝑡) = 0 (1.8)

𝑐 = 𝐸𝐼

𝜌𝐴 (1.9)

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,

𝑐2

𝑊(𝑥)

𝑑4𝑊(𝑥)

𝑑𝑥4= −

1

𝑇(𝑡)

𝑑2𝑇(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:

𝑑4𝑊(𝑥)

𝑑𝑥4 − 𝛽4𝑊 𝑥 = 0 (1.12)

𝑑2𝑇(𝑡)

𝑑𝑡2 + 𝑤2𝑇 𝑡 = 0 (1.13)

where

𝛽4 = 𝑤2

𝑐2 = 𝜌𝜌𝐴 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) = 𝐶𝑒𝑠𝑥 (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 = 𝐶1𝑒𝛽𝑥 + 𝐶2𝑒

−𝛽𝑥 + 𝐶3𝑒𝑖𝛽𝑥 + 𝐶4𝑒

−𝑖𝛽𝑥 (1.18)

where C1, C2, C3 and C4 are constants. Equation (1.18) can be expressed more conveniently as

𝑊 𝑥 = 𝐶1𝑐𝑜𝑠𝛽𝑥 + 𝐶2𝑠𝑖𝑛𝛽𝑥 + 𝐶3𝑐𝑜𝑠ℎ𝛽𝑥 + 𝐶4𝑠𝑖𝑛ℎ𝛽𝑥 (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

𝜌𝐴𝑙4 (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 𝜕2𝑊

𝜕𝑥2 (0) = 0 (1.23)

(1.24)

or 𝜕2𝑊

𝜕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)

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

=𝑛2𝜋2(𝐸𝐼

𝜌𝐴𝑙4)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

3𝑎3𝑡 + 2𝑎2𝑏𝑡 = 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 𝜷𝟏𝒍 = 𝟑.𝟏𝟒𝟏𝟔 𝜷𝟐𝒍 = 𝟔. 𝟐𝟖𝟑𝟐 𝜷𝟑𝒍 = 𝟗. 𝟒𝟐𝟒𝟖 𝜷𝟒𝒍 = 𝟗. 𝟒𝟐𝟒𝟖

Equations Euler-Bernoulli calculated in MathCAD , results are in the units : rad/sc simply supported at

both ends

15

n=1

193 109

0.000129

80 103

0.0147 34

0.5

3.14162

159.562

n=2

190 109

0.000129

80 103

0.0147 34

0.5

6.28322

633.268

n=3

190 109

0.000129

80 103

0.0147 34

0.5

9.42482

1.425 103

n=4

190 109

0.000129

80 103

0.0147 34

0.5

12.56642

2.533 103

n=5

190 109

0.000129

80 103

0.0147 34

0.5

15.7082

3.958 103

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


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 𝛽1𝑙 = 4.73 𝛽2𝑙 = 7.8532 𝛽3𝑙 = 10.996 𝛽4𝑙 = 14.731

Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported at

both ends

20

n=1

190 109

0.000129

80 103

0.0147 34

0.5

4.732

358.879

n=2

190 109

0.000129

80 103

0.0147 34

0.5

7.85322

989.279

n=3

190 109

0.000129

80 103

0.0147 34

0.5

10.99562

1.939 103

n=4

190 109

0.000129

80 103

0.0147 34

0.5

14.13722

3.206 103

n=5

190 109

0.000129

80 103

0.0147 34

0.5

17.27882

4.789 103

n=6

190 109

0.000129

80 103

0.0147 34

0.5

20.42042

6.689 103


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

8𝐿 𝑀𝑏 =

9

128𝑤𝐿2 = 729 𝑁

23

𝑀𝑚𝑎𝑥 =1

8𝑤𝐿2 = 1297 𝑁. 𝑚


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 𝛽1𝑙 = 3.9266 𝛽2𝑙 = 7.06686 𝛽3𝑙 = 10.212 𝛽4𝑙 = 13.3518

Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc simply supported –

fixed ends.

25

n=1

190 109

0.000129

80 103

0.0147 34

0.5

3.92662

247.32

n=2

190 109

0.000129

80 103

0.0147 34

0.5

7.06862

801.479

n=3

190 109

0.000129

80 103

0.0147 34

0.5

10.21022

1.672 103

n=4

190 109

0.000129

80 103

0.0147 34

0.5

13.35182

2.86 103


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

2𝑎𝑛𝑑 𝑀𝑎𝑥 𝑀𝑜𝑚𝑒𝑛𝑡 𝑀𝑚𝑎𝑥 =

𝑤𝑙2

2=

1153 ∗ 9

2= 5188 𝑁. 𝑚


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 𝛽1𝑙 = 1.8751 𝛽2𝑙 = 4.6941 𝛽3𝑙 = 7.8547 𝛽4𝑙 = 10.9956

Equations Euler-Bernoulli calculated in Mathcad, results are in the units : rad/sc fixed-free ends

30

n=1

190 109

0.000129

80 103

0.0147 34

0.5

1.87512

56.399

n=2

190 109

0.000129

80 103

0.0147 34

0.5

4.69412

353.452

n=3

190 109

0.000129

80 103

0.0147 34

0.5

7.85472

989.657

n=4

190 109

0.000129

80 103

0.0147 34

0.5

10.99562

1.939 103


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 𝛽1𝑙 = 4.7300 𝛽2𝑙 = 7.8532 𝛽3𝑙 = 10.9965 𝛽4𝑙 = 14.372

We don’t have any reaction force ;

I listed below mode shapes and natural frequencies ,

33

n=2

190 109

0.000129

80 103

0.0147 34

0.5

7.85322

989.279

n=3

190 109

0.000129

80 103

0.0147 34

0.5

10.99562

1.939 103

n=4

190 109

0.000129

80 103

0.0147 34

0.5

14.13722

3.206 103

n=5

190 109

0.000129

80 103

0.0147 34

0.5

17.27882

4.789 103


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.

http://www.me.berkeley.edu/~lwlin/me128/formula.pdf

http://www.awc.org/pdf/DA6-BeamFormulas.pdf

https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec=fixed&page

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

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