continuous systems - ahmed nagibdrahmednagib.com/onewebmedia/vibration course/continuous syst… ·...

Dr.$Bassuny$El-Souhily$/$Dr.$Ahmed$Nagib$$Mechanical$Vibrations$$

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Continuous Systems INTRODUCTION

We have so far dealt with discrete systems where mass, damping, and elasticity were assumed to be present only at certain discrete points in the system. There are many cases, known as distributed or continuous systems, in which it is not possible to identify discrete masses, dampers, or springs. We must then consider the continuous distribution of the mass, damping, and elasticity and assume that each of the infinite number of points of the system can vibrate. This is why a continuous system is also called a system of infinite degrees of freedom.

If a system is modeled as a discrete one, the governing equations are ordinary differential equations, which are relatively easy to solve. On the other hand, if the system is modeled as a continuous one, the governing equations are partial differential equations, which are more difficult. However, the information obtained from a discrete model of a system may not be as accurate as that obtained from a continuous model. The choice between the two models must be made carefully, with due consideration of factors such as the purpose of the analysis, the influence of the analysis on design, and the computational time available.

In this chapter, we shall consider the vibration of simple continuous systems-strings, bars, shafts, beams, and membranes. In general the frequency equation of a continuous system is a transcendental equation that yields an infinite number of natural frequencies and normal modes. This is in contrast to the behavior of discrete systems, which yield a finite number of such frequencies and modes. We need to apply boundary conditions to find the natural frequencies of a continuous system. The question of boundary conditions does not arise in the case of discrete systems except in an indirect way, because the influence coefficients depend on the manner in which the system is supported.


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1-!Lateral Vibration of Beams

Equation of Motion

The beam shown in Figure-a has a length “L”, area moment of inertia “I”, and a mass per unit length “ρ”. To study the lateral vibration of the beam, consider an element of length “dx” at a distance “x” from the left end, Figure-b.

The beam is subjected to a uniform load “p(x, t)”. The positive directions of the moments and the forces are indicated on the element.

The equilibrium of moments on the element gives

M + d M + p (dx)2 / 2 - [V + d V] d x - M = 0

Neglecting second order infinitesimals. That is

(d x)2 ≈ 0

dV dx ≈ 0

y$$$$$$

x$$$$$$

$(a)$$$$$$

$$$$(b)$

$$$$$$


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Then,

V = xM∂

∂ (1)

The equilibrium of forces gives

V + p(x,t) dx – [V + d V] = 0

Or

p(x,t) = xV∂

∂ (2)

From the fundamental theory of beams

M = E I 2

2

xy

∂

∂

Substitute in Eqs. (1) and (2), then

V(x,t) = !"

#$%

&

∂

∂

∂

∂2

2

xyIE

x (3)

p(x,t) = !"

#$%

&

∂

∂

∂

∂2

2

2

2

xyIE

x (4)

In the case of the vibration of the beam, “p(x,t)” is divided into two parts. The first

is the inertia effect and is equal to “- ρ2

2

ty

∂

∂ ”. The second is the external effect and is

equal to “f(x,t)”. If the beam is homogenous and has a uniform section, the equation of motion is given by

E I 4

4

xy

∂

∂ + ρ2

2

ty

∂

∂ = f(x,t) (5)

Analysis of Free Vibrations

To determine the natural frequencies of a beam, the external effect is considered zero. The equation of motion in this case is given by


4$$

E I 4

4

xy

∂

∂ + ρ2

2

ty

∂

∂ = 0 (6)

Let

y(x,t) = Y(x) (A cos ωt + B sin ωt)

Where, “A” and “B” are arbitrary constants which are determined from the initial conditions. Substituting in Eq. (35)

!"

#$%

&ωρ− )x(Y

xd)x(YdIE 2

4

4

(A cos ωt + B sin ωt) = 0 (7)

Let

IE

2ωρ = β4, then

4

4

xd)x(Yd - β4 Y(x) = 0 (8)

The solution of Eq. (8) is determined by assuming that

Y(x) = K esx

Substitute in Eq. (8), then

s4 - β4 = 0

The roots of this equation are “β, - β, i β, - i β”

The general solution of Eq. (8) is

Y(x) = K1 eβx + K2 e-βx + K3 eiβx + K4 e-iβx

Expanding the exponentials and collecting the constants, hence

Y(x) = C1 cos βx + C2 sin βx + C3 cosh βx + C4 sinh βx (9)

The general form of the solution of the differential equation is

y(x,t) = (C1cos βx+C2sin βx+C3cosh βx+C4sinh βx) (Acos ωt+Bsin ωt) (10)


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The natural frequencies are determined by applying the boundary conditions.

Boundary Conditions

The common boundary conditions are

1.! Free end, the moment and the shearing force are zero.

2

2

xy

∂

∂ = 0 Moment =0

3

3

xy

∂

∂ = 0 Shear =0

2.! Simply supported, the deflection and the moment are zero. Deflection =0 y(x,t) = 0

2

2

xy

∂

∂ = 0 Moment =0

3.! Fixed end, the deflection and the slope of the deflection are zero.

Deflection =0 y(x,t) = 0

xy

∂

∂ = 0 Slope =0

4.! The case when the ends are attached to a spring-mass-damper systems is shown in Figure. The positive directions of the forces applied on the element, as shown in Figure-b, are the positive directions.

m1$ $ $ $ $ $$$m2$$$$$$$$$$

$$c1$ $$$$$$$k1$ $ $ $$$$$$$$$$$$$$$$$$$$$$k2$$$$$$$$$$$$$$$$$c2$$$$$$$$$$

$


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E I 3

3

xy

∂

∂ = m 2

2

ty

∂

∂ + c ty∂

∂ + k y for the right end

E I 3

3

xy

∂

∂ = - m 2

2

ty

∂

∂ - c ty∂

∂ - k y for the left end

Figure. Common boundary conditions for the transverse vibration of a beam.


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Figure Boundary Conditions for Beams


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Example: 1

Find the natural frequencies of a beam simply supported at both ends.

Solution

y(x,t) = (C1cos βx+C2sin βx+C3cosh βx+C4sinh βx) (Acos ωt+Bsin ωt)

For a simply supported beam we apply boundary conditions “2” at both ends.

•! y(0,t) gives 0 = #$%× cos 0 + $+× sin 0 + $.× cosh 0 + $0× sinh 0

0 = #$%×1 + $+×0 + $.×1 + $0×0

0 = #$% + $.

•! yxx(x,t)|x=0 gives 0 = #−$% + $.

Thus

C1 = C3 = 0

•! y(x,t)|x=L gives

C2 sin βL + C4 sinh βL = 0

•! Yxx(x,t)|x=L gives

- C2 sin βL + C4 sinh βL = 0

From the above two equations

C4 sinh βL = 0


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“sinh βL” is only zero when “β = 0”. This means that “ω = 0”, and, consequently, there is no motion. Then, we conclude that “C4 = 0”.

Therefore, the characteristic equation is

sin βL = 0

β L = n π n = 1, 2, 3, ....

The natural frequency of mode number “n” is equal to “ρIE

Ln 2

!"

#$%

& π ”.

Example: 2

Find the natural frequencies of a beam fixed at one end and free at the other end.

Solution

For the left end, we apply conditions “3”,

•! y(x,t)|x=0 gives C1 + C3 = 0


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•! yx(x,t)|x=0 gives C2 + C4 = 0

For the right end, we apply conditions “1”.

•!Yxx(x,t)|x=L gives -C1 cos βL - C2 sin βL+ C3 cosh βL+ C4 sinh βL = 0

•!Yxxx(x,t)|x=L gives C1 sin βL - C2 cos βL+ C3 sinh βL+ C4 cosh βL = 0

Substituting the values of “C3” and “C4” in the last two equations, then

- C1 (cos βL + cosh βL) - C2 (sin βL + sin hβL) = 0

C1 (sin βL - sinh βL) - C2 (cos βL + cosh βL) = 0

For non trivial solution, the determinant of the coefficients is zero. Thus

(cos βL + cosh βL)2 + (sin βL - sinh βL) (sin βL + sinh βL) = 0

This is the characteristic equation and can be reduced to the form

1 + cos βL cosh βL = 0

The first four values for “βL” which satisfy this equation are “1.88, 4.7, 7.9 and 11.0”. The corresponding values of the natural frequencies are

“3.15ρIE , 22.1

ρIE , 61.78

ρIE , and 121

ρIE rad/s”.

Example: 3

Find the natural frequencies of a beam supported as shown in Figure.

m1$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$m2$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$m3$$$$

E,$I1,$L1$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$E,$I2,$L2$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

$$$$$$$$$$

k1$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$k2$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$k3$$$$

$(1)$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$(2)$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$


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Solution

Any interruption in the beam either by changing the cross-section, the property, or the loading makes it necessary to divide the beam to separate regions, each region is represented by an equation of motion. In this example, the beam is divided to regions (1) and (2).

The equations of motion are

E I1 41

4

xy

∂

∂ + ρ 21

2

ty

∂

∂ = 0

E I2 42

4

xy

∂

∂ + ρ 22

2

ty

∂

∂ = 0

The solutions are given by

y1(x,t)= (C1cos β1x+C2sin β1x+C3cosh β1x+C4sinh β1x) (Acos ωt+Bsin ωt)

y2(ξ,t)=(D1cos β2ξ+D2sin β2ξ+D3cosh β2ξ+D4sinh β2ξ) (Acos ωt+Bsin ωt)

“x” represents the axial position for the first region and “ξ” represents the axial position for the second region; “ξ= 0” at the right side of “m2”. The application the boundary conditions is performed as follows:

For the left side of region (1), we apply the moment condition in “1” and shear condition “4”.

•! The moment condition

0xyIE

0x21

2

1 =∂

∂

=

- C1 + C3 = 0 (i)

•! Condition 4

110x

21

2

10x

31

3

1 yktym

xyIE −

∂

∂−=

∂

∂

==

E I1 β13 (- C2 + C4) = (C1 + C3) (m1 ω2 – k1) (ii)


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For the right side of region (1) and the left side of region (2), we apply:

•! The deflection is the same y1(L1,t) = y2(0,t)

C1 cos β1L1 + C2 sin β1L1 + C3 cosh β1L1 + C4 sinh β1L1 = D1 + D3 (iii)

•! The slope is the same

0

22

Lx

11

yIExyIE

1 =ξ=ξ∂

∂=

∂

∂

E I1 β1 (- C1 sin β1L1 + C2 cos β1L1 + C3 sinh β1L1 + C4 cosh β1L1)

= E I2 β2 (D2 + D4) (iv)

•! The moment is the same

022

2

2Lx

21

2

1yIE

xyIE

1 =ξ=ξ∂

∂=

∂

∂

E I1 β12

(- C1 cos β1L1 - C2 sin β1L1 + C3 cosh β1L1 + C4 sinh β1L1)

= E I2 β22

(- D1 + D3) (v)

•! The sum of the forces on “m2” is zero

0y)ωmk(ξyIE

xyIE

0ξ22

220ξ

32

3

2Lx

31

3

1

1

=−+∂

∂+

∂

∂−

===

- E I1 β13

(C1 sin β1L1 - C2 cos β1L1 + C3 sinh β1L1 + C4 cosh β1L1)

+ E I2 β23

(- D2 + D4) + (D1 + D3) (m2 ω2 – k2) = 0 (vi)

For the right side of region (2), we apply the moment condition in “1” and condition “4”.

•! The moment condition

0yIE2L

22

2

1 =ξ∂

∂

=ξ

(- D1 cos β2L2 - D2 sin β2L2 + D3 cosh β2L2 + D4 sinh β2L2) = 0 (vii)


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•! Condition 4

2

22

L33L

22

2

3L

32

3

2 yktymyIE

=ξ=ξ=ξ

−∂

∂−=

ξ∂

∂

E I2 β23

(D1 sin β2L2 - D2 cos β2L2 + D3 sinh β2L2 + D4 cosh β2L2) =

(D1 cos β2L2 + D 2 sin β2L2 + D 3 cosh β2L2 + D 4 sinh β2L2) (k3 – m3 ω2) (viii)

The determinant of the coefficients of the above eight equations is zero and represents the characteristic equation for obtaining the natural frequencies.

continuous systems - ahmed nagibdrahmednagib.com/onewebmedia/vibration course/continuous syst… ·...

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