Advanced Engineering Dynamics 2010 Part 9 potx

Waves in a uniform beam 155 seen that most of the energy is accounted for before the abscissa value reaches x.. 6.12 Waves in a uniform beam In this section we shall be examining later

Fig 6.29 Wave velocities for positive curvature dispersion curve

upwards It is conventional to plot the curves with k as the independent variable because o

is univalued for a given k, but not vice versa

Figure 6.30 shows four symmetrical pulse shapes and their respective Fourier transforms

The ordinate is the square of the amplitude as this gives a measure of the energy and it is

Fig 6 3 0 Fourier transform pairs

Waves in a uniform beam 155 seen that most of the energy is accounted for before the abscissa value reaches x This gives credence to the approximations r/T, = 1 and L/h, = 1

The practical difficulty is the determination of the width of the peak and hence the appro- priate value of k,,, Measured values tend to be towards the phase velocity but are far from the group velocity Although the pulse-peak velocity is as precisely defined as the other two

it does emphasize the fact that the group velocity is valid only for a narrow frequency band

In a highly dispersive situation the group velocity gives the arrival time of specific wave- lengths which were generated by an impact

Dispersion indicates that short pulses will spread out but the total energy remains con- stant The term is not to be confused with dissipation in which some of the mechanical energy is converted to thermal energy

6.12 Waves in a uniform beam

In this section we shall be examining lateral waves in a long uniform beam, shown in Fig

6.3 1, with a cross-sectional area A and a second moment of area Z about the z axis through the centroid The xy plane is a plane of symmetry The material has a Young’s modulus E, a

shear modulus G and the density is p We are going to use Hamilton’s principle to obtain

the equations of motion because it is easier to modify the model The exact equations are very involved and therefore approximations are required

A first approximation is to consider only kinetic energy due to lateral motion and strain energy due to bending strains Later we shall include rotary inertia and shear strain energy The simple case can be obtained by free-body diagrams and Newton’s laws but we shall use the vari- ational method and subsequently modify the Lagrangian to take into account the extra terms

Refemng to Fig 6.32 we can write an expression for the kinetic energy (note that in this

section v is the deflection in they direction)

Fig 632

and for the strain energy

L~~ a0 z

If we assume that there are no external forces Hamilton’s principle states that

t2

6 J (TI - Vl)dt = 0

tl

or

2

t2 L

pA aV 2 EI 80

0

tl

Carrying out the variation first

L t 2

/,It [ P A g 0 ( g ) - E I ; 0 (z) dxdt = 0 (6.104)

I

For the first term we reverse the order of integration and integrate by parts to obtain

/ [ It pA z 6 ($) dt ] dr =IL[ pA$ 6v It -Itl$ 6 v dt] dx (6.105)

Because GVvanishes, by definition, at t, and t2 the first term in the square brackets is zero

It, I [ EI E 6 ( z)] dx dt =I [ E I g 60 lo - 1; 2 60 dx] dt (6.106)

For the second term we integrate by parts with respect to x to obtain

12 L

Waves in a unifonn beam 157

We have already assumed for this exercise that there are no external active forces Therefore

the end constraints must be workless and this implies that either 60 or a0lax must be zero at

the ends Hence the first term is zero

Combining the two results, equations (6.105) and (6.106), gives

v and B are not independent but are related by geometry

dV

- - - 0

ax

so that

av

60 = 6 ( dx)

(6.107)

(6.108)

(6.109)

Integrating the first term once more by parts and noting that the end forces are workless leads to

and finally since 0 = avBx

(6.110)

Because 6v is arbitrary (except at t , and t2 where it is zero) the expression in the large paren- theses must be zero Thus

This equation is known as Euler 's equation for beam vibration and is widely used This form

can readily be deduced from free-body diagrams in a similar method to that used for longi- tudinal waves in a bar The reason for using Hamilton's principle here is to expose the details

of the method and to form the basis for development of a more refined model

We now add an extra kinetic energy term to take into account rotary inertia For a thin ele- ment of beam the moment of inertia about a z axis through the centroid is pZ dx, where Z is the second moment of area Therefore the kinetic energy is

(6.112)

Thus

t2

(6.1 13)

The rate at which the lateral deflection changes with x is now augmented by y, the shear strain, giving

av

ax

_ - - 0 + y

and the additional strain energy is

(6.1 14)

(6.115)

The constant K (kappa), which is greater than unity, corrects for the fact that the shear stress

is not uniformly distributed across the cross-sectional area For a rectangular cross-section

K = 6/5; this is based on a parabolic shear stress distribution Thus

This time we shall not make the first term zero as we are now admitting external forces

vention is used)

From Fig 6.3 1 the virtual work done by the external forces is (note that a vector sign con-

(6.1 17)

6W = MI601 + MI602 + SI6vl + S26~2

Therefore Hamilton's principle for the modified model including external forces is

t 2

6 I f i ( q + T2 - VI - V2) dt + GWdt = 0 (6.1 18) Using equations (6.105), (6.106), (6.1 13), (6.1 16) and (6.1 17), equation (6.118) may be written

J: Ji { [ -PAY,, + KGA(v, - 0.~1 dv

+ [ -pI0+,, + EI0, + KGA(v, - 0 ) ] 60 } dxdt

(6.1 19) Here the notation a2vldx' = v, etc is used

double integrals must each equate to zero Thus the two equations of motion are

Because 6v and 60 are arbitrary between I , and t2 the factors of 6v and of 60 under the

Waves in a uniform beam 159

and

2

2

(6.120)

(6.121)

Summing each of the coefficients of 60,, 60,, 6v, and 6v2 to zero gives the boundary conditions

(EZ )2 = M2

au

KGA ( dx - 0 ) = -VI

I

au

KGA ( dx - 0 ) = -V2

2

(6.122)

(6.123)

(6.124)

(6.125)

It is now possible to eliminate 0 between equations (6.120) and (6.12 1) Equation (6.120) can be written as

V.lf - KC, V-cr KC, 0, = 0

Therefore

1

0, - v.u - - 2 VJI

KC,

-

Equation (6.12 1) is written as

and differentiating partially with respect to x gives

(6.126)

(6.127)

or

This is known as the Emoshenko beam equation

For a running wave solution

v = ;d(ar - W

so substituting into equation (6.128) and dividing through by the common factor gives

1

2

which is the dispersion equation for bending waves in a uniform beam This equation is a

quadratic in o2 and therefore yields two values of o for any value of k The lower of the two,

the first mode, approximates to Euler's equation for small values of k (k c 0.2, i.e wave-

lengths longer than about five times the beam depth)

Plots of a, cp and cg are shown in Fig 6.33 The phase velocity tends to a maximum value which is close to the velocity of pure shear waves (see section 7.5) The group velocity also

tends to the same value but passes through a maximum for wavelengths of the order of the depth of the beam It follows that after a short-duration impact these wavelengths are the first to arrive at a distant point but most of the energy will follow at longer wavelengths The model becomes invalid when the wavelengths are very short compared with the depth of the beam in which case the wave speed will tend to that of surface waves which have a speed a little less than the shear wave speed

Figure 6.34 is a similar plot but for the higher, or second, mode It is seen that there

is a minimum frequency; below this no travelling wave is possible in this mode The consequence of this is that the phase velocity tends to infinity at very low wavenum- bers but the group velocity remains finite and less than c, The validity of this mode is not as good as the first mode and is probably only witnessed in I-section beams where the end load is camed mainly by the flanges and the shear is carried mainly by the web

c',k ' 4 - - 1 0 2 - ( 1 +$)02k2+o

(I/A 1

Fig 633 Uniform beam mode 1

Waves in periodic structures 16 1

Fig 6.34 Uniform beam mode 2 6.13 Waves in periodic structures

The type of structure envisaged here is the continuous mass-spring system shown in Fig

6.35 Away from any boundary we can use the same form of expression for displacement as

used in the earlier sections but in place of the continuous location x we have a discrete num-

ber of locations n The mass of each body in the system is m and the stiffness of each spring

is s For the nth body the equation of motion is

S ( U , + ~ - u,) - s ( u , - n , - , ) = mu, (6.130) Let us assume

(6.13 1) Substituting equation (6.13 1) into equation (6.130) and dividing through by the common

factor Ue'" we obtain

u, = (,r&"' - kn) = ue'" e - J h

- e-Jkn) - s (e-Jh - e-Jhn-l)) = -mI1zoze-Jh

s (e-JMn+1)

(e-Jk - 1) - (1 - e'') = - mw'/s

Dividing further by KJh gives

and as

e*Jk = cosk f j sink

Fig 6.35

we get

2

2 COS (k) - 2 = - - mw

S

or

2

2( -2 sin2 (W2)) = - =

S

giving

Figure 6.36 is a plot of the dispersion diagram From this we see that there is a cut-off fie-

quency, a,, = 2 J (slm), above which no continuous wave will propagate At this point k =

X SO

1( = vi@ e-jnn = vi" (-1)n

that is, each body is in phase opposition with its neighbours

The phase velocity is

and the group velocity is

d o

g d k

If we impose a vibration above the cut-off frequency then one solution is to assume that each body has the opposite phase to its neighbour and that the amplitude decays exponen- tially,

(6.135)

U" = U(- 1)" e" e-Kn

Substituting equation (6.135) into equation (6.130) leads to

for o > oca Here the disturbance remains local to the point of initial excitation and does not

propagate; such a mode is said to be evanescent

Fig 6 3 6 Dispersion diagram for mass-spring system

Waves in a helical spring 163

6.14 Waves in a helical spring

The helical spring will be treated as a thin wire, that is plane cross-sections remain plane This assumption has been shown to be acceptable by mechanical testing An element of the wire has six degrees of freedom, three displacements and three rotations, so the possibility exists for six modes of propagation If the helix angle is small these modes separate into two groups each having three degrees of freedom, one set consisting of the in-plane motion and the other the out-of-plane motion The effect of helix angle will be discussed later but here

we shall develop the theory for out-of-plane motion for a spring with zero helix angle This,

as we shall see, is associated with axial motion of the spring, that is with the spring being

in its compression or tensile mode

Figure 6.37 defines the co-ordinate system to be used The unit vector i is tangent to the axis of the wire,j is along the radius of curvature directed towards the centre and k com-

pletes the right-handed triad For zero helix angle k is parallel to the axis of the spring The

radius of curvature is R and s is the distance measured along the wire 8 is the angle through

which the radius turns Thus

First we shall consider the differentiation of an arbitrary vector V with respect to s

(6.138)

where the prime signifies differentiation with respect to non-rotating axes and SZ is the rate

of rotation of the axes with distance s Thus

-

- - - d ’ V + R x V

d V

d s d s

de de

d s d s

R = - = - k

and using equation (6.137)

1

R

Fig 637

From equation (6.138) the components of the derivative are

dV, d'V, 1

-

- - -

or in matrix form

where

p -1lR 0

and

d'

p ' d s

(6.140)

(6.141)

(6.142)

(6.143)

(6.144)

(6.145)

A cross-section has a displacement u and a rotation 0 from its equilibrium position The

spatial rate of change of u is due to stretching and shearing of the element of wire and also

to rigid body rotation, so the strain

(4 = [TI1 (u) - [@IX ( W I d s

Now (ds) = (ds 0 0) and therefore

0 0 - 1

ds

M=l 0 1 0

= [T21

(see appendix 1)

The components of strain are, therefore,

axial strain

shear strain

shear strain

= pui - ujlR

= puj + u,lR - 0 k

&k = puk + 0j

These are related to the elastic constants by

(6.147)

(6.148) (6.149) (6.150)

Waves in a helical spring 165

Pi

E = -

(6.152)

(6.153)

where q is a factor to allow for the shear stress distribution not being uniform A typical

value for circular cross-sections is 0.9

The relationship between the elastic constants, Young's modulus E, shear modulus G and Poisson's ratio u is E = 2G(1 + u) , so let

Combining equations (6.148) to (6.154)

Pi = GA (2mpui - 2mujlR)

Pi = GA (VU, + p i I R - 903

p k = GA (qPuk + q O j )

(6.154)

(6.155) (6.156) (6.157)

For bending we use the usual engineering relationships for bending and torsion of s h a h If

the shape of the cross-section has point symmetry then with J being the polar second moment of area

4 = Ik = Ii/2 = J12

Also E = G2m so that E4 = EIk = mGJ, and therefore

d0;

ds

(6.159)

(6.160)

The equations of motion can be derived with reference to Fig 6.38 Resolving forces act- ing on the element, neglecting any external forces,

a

or, letting D = -

at

Fig 638

or

The component equations are

Now considering moments about the centre of mass of the element

(6.165)

( (M) + _ ) - (M) + (W"P) = 7

ds, = constant

where (L) is the moment of momentum For small rotations

and [ : z k ]

- = pJD2 [ z): ] ds = -

pM, - M,IR = pJD'0,

pM, + M,IR - Pk = pJD20,I2

phfk + P, = @0,/2

a(L)

The three component equations are, after dividing by ds,

(6.168) (6.169) (6.170) Substituting the six equations of state ((6.155) to (6.160)) into the equations of motion ((6.162-6.164) and (6.168-6.170)) will yield six equations in the six co-ordinates and these

Waves in a helical spring 167 will separate into two groups of three So, substituting equations (6.157), (6.158) and (6.159) into equations (6.164), (6.168) and (6.169) leads to

m

R

G J ( p 2 g - p0,IR) - G J - + 0JR) = pJD20, (6.171)

GJ

R GJm b 2 D j + p 0 / R ) + - (PO, - 0,IR) - qGA (pup + oj) = pJD28j/2 (6.172)

It is convenient for discussion purposes to put the above equations into non-dimensional which contain only the three out-of-plane co-ordinates

form To this end we define the following terms

(6.174)

- m

c2

Thus equations (6.17 1) to (6.173) may be written in matrix form as

-$(I + m ) 0

(mF2 - 1 - a q ) 2 -qa2j ][ ik] = E ' [ t] (6.177)

4p" 9 j 2

(Note that this matrix equation can be written in symmetrical form; however, we can discuss the manner of wave propagation just as well in the current form.)

In a manner similar to previous cases we shall assume a wave travelling along the axis of the wire Thus

(6.178)

0, = 0 , e

(6.179)

0 1 - J

J(0f - b )

- 0' e ~ ( w f - k,)

Now

and

- R

D B ~ = - (iw)oi

c2

with similar expressions for the other two co-ordinates

Let us define the non-dimensional wavenumber

K = Rk

and the non-dimensional frequency

(6.182)

(6.183)

(6.184)