Lateral Vibration of Beams

• Forced Vibration We obtain

This can be written as:

Multiply this by Wm(x), integrate from 0 to l and using orthogonality condition:

-84-8.5 Lateral Vibration of Beams

• Forced Vibration

Qn(t) is the generalized force corresponding to qn(t)

Eq 8.114 is the equation of motion of undamped single DOF system.

Using Duhamel integral, solution of Eq 8.114 can be expressed as

 

^l

   

_n _^l _n

 

台灣師範大學機電科技學系

-85-• Forced Vibration

First 2 terms on the right hand side represent the transient vibration

3^rdterm represent steady state vibration

Once solution is found, total solution can be determined from

   

^

   

,

1

n

W

n

x q

n

t

t x w

台灣師範大學機電科技學系

-86-Example 8.8

Forced Vibration of a Simply Supported Beam

Find the steady-state response of a pinned-pinned beam subject to a harmonic force f(x,t)=f0sinωt applied at x=a as shown below.

台灣師範大學機電科技學系 C. R. Yang, NTNU MT

-87-8.5 Lateral Vibration of Beams

Example 8.8

Forced Vibration of a Simply Supported Beam Solution

We use Mode superposition method approach.

Normal mode function: where βnl=nπ Generalized force:

Steady state response

 

l

x x n x

W_n sin_n sin 

   

t

l a f n xdx t

x f t

Q_n ^l , sin_{n 0}sin  sin

0 





     

 

sin 2

where

1 sin

0 2 0

2 0

xdx l dx

x W b

d t Ab Q

t q

l n l

n t

n n n n

























 



台灣師範大學機電科技學系 C. R. Yang, NTNU MT

-88-8.5 Lateral Vibration of Beams

Example 8.8

Forced Vibration of a Simply Supported Beam Solution

Solution can be expressed as

Thus response of the beam is given by

 

l t

a n Al t f q

n

n 







 ^sin

2 sin

2 2 0

 

 

t

l x n l

a n Al

t f x

n n

 







 2 1 sin sin sin

,

1 2 2 0



^

 



台灣師範大學機電科技學系

-89-• Effect of Axial Force

Consider an element of a beam shown below

台灣師範大學機電科技學系

-90-• Effect of Axial Force Vertical motion:

Rotational motion about 0:

For small deflections,

    

sin



sin ₂



8.118



-91-8.5 Lateral Vibration of Beams

• Effect of Axial Force

Combine all equations with to give:

For free vibration of uniform beam:

Solution can be obtained using separation of variables

     

xt

-92-8.5 Lateral Vibration of Beams

• Effect of Axial Force

Substitute Eq 8.122 into Eq. 8.121:

Assuming W(x)=Ce^sx, auxiliary equation can be obtained:

Roots of equation:

So, W(x)=C1coshs1x+C2sinhs1x+C3coss2x+C4sins2x

台灣師範大學機電科技學系

-93-Example 8.9

Beam Subjected to an Axial Compressive Force

Find the natural frequencies of a simply supported beam subjected to an axial compressive force.

台灣師範大學機電科技學系

-94-Example 8.9

Beam Subjected to an Axial Compressive Force Solution

Boundary conditions: W(0)=0, W(l)=0, C1= C3= 0

Hence W(x)=C2sinh s1x + C4sin s2x

Applying boundary conditions, we have sinh s1l•sin s2l=0

 

0 0, ₂

 

0

2

2 2



 l

dx W d dx

W d

台灣師範大學機電科技學系 C. R. Yang, NTNU MT

-95-8.5 Lateral Vibration of Beams

Example 8.9

Beam Subjected to an Axial Compressive Force Solution

Since sinh s1l>0 for all s1l≠0, the only roots are s2l=nπ, n=0,1,2,…

Thus where P is negative since it is compressive.

Smallest Euler buckling load for simply supported beam:

Thus



 



 

 EI

Pl n n A EI

n l 2

2 2 4 2

2





 

2 2

cri l

P _ EI



 



 



cri 2 4 2

2

P n P A n EI

n l 

 

台灣師範大學機電科技學系 C. R. Yang, NTNU MT

-96-8.5 Lateral Vibration of Beams

Example 8.9

Beam Subjected to an Axial Compressive Force Solution

Some observations:

• If P=0, ωnis same as that of simply supported beam

• If EI=0, ωnis that of a taut string.

• If P>0, ωnincreases as tensile force stiffens the beam.

• If PPcri, ωn0 for n=1

台灣師範大學機電科技學系

-97-• Effects of Rotary Inertia and Shear Deformation

If cross-sectional dimensions are not small compared to length of the beam, need to consider effects of rotary inertia and shear deformation.

Consider an element of the beam:

台灣師範大學機電科技學系

-98-• Effects of Rotary Inertia and Shear Deformation

where Φ is the slope of the deflection curve due to bending deformation alone.

M = bending moment V = shear force G = modulus of rigidity

k = Timoshenko’s shear coefficient, which depends on the shape of the cross-section

x

-99-8.5 Lateral Vibration of Beams

• Effects of Rotary Inertia and Shear Deformation

Equations of motion Translation in z-direction:

Rotation about a line passing through pt D and parallel to y-axis:

   

-100-8.5 Lateral Vibration of Beams

• Effects of Rotary Inertia and Shear Deformation

Using the relation and ignoring 2^ndpowers

in dx, the equations of motion can be expressed as:

Solve Eq 8.133 for and substitute into Eq. 8.134 to obtain equation of motion:

xdx

台灣師範大學機電科技學系

-101-• Effects of Rotary Inertia and Shear Deformation For free vibration, f=0:

Boundary conditions Fixed end: φ=w=0 Simply supported end:

Free end:

0

-102-Example 8.10

Natural Frequencies of a Simply Supported Beam

Determine the effects of rotary inertia and shear deformation on the natural frequencies of a simply supported uniform beam.

台灣師範大學機電科技學系 C. R. Yang, NTNU MT

-103-8.5 Lateral Vibration of Beams

Example 8.10

Natural Frequencies of a Simply Supported Beam Solution

Let

Equation of motion:

Express the solution as Substitute (E.1) into (E.2),

A

-104-8.5 Lateral Vibration of Beams

Example 8.10

Natural Frequencies of a Simply Supported Beam Solution

台灣師範大學機電科技學系

-105-Example 8.10

Natural Frequencies of a Simply Supported Beam Solution

Note:

If only rotary inertia is considered, Frequency equation reduces to:

If only shear deformation is considered, Frequency equation becomes:

2 0

2 4 2 2 4

4 





 



 





t x I w t A w x

EI w  



 



 



2 2 2 2 4

4 4 2 2

1 l

r l n

n

n 



 

2 0

2 4 2 2 4 4

 



 



 





t x

w kG EI t A w x

EI w  



 



 



kG E l

r l n

n

n

2 2 2 2 4

4 4 2 2

1 



 

台灣師範大學機電科技學系

-106-Example 8.10

Natural Frequencies of a Simply Supported Beam Solution

Note:

If both rotary inertia and shear deformation are disregarded, equation of motion becomes:

Frequency equation:

2 0

2 4

4 



 



 t A w x EI w 

4 4 4 2 2

l n

n









C. R. Yang, NTNU MT

在文檔中 8台灣師範大學機電科技學系 (頁 21-27)