Advanced Engineering Dynamics 2010 Part 8 docx

From this we learn the very important lesson that rigid body behaviour may be assumed when the variation of force is small compared with the time taken for the wave to traverse the body

Fig 6.10

At x = -L the strain is always zero and hence

g:-, (c,t + ( - L ) - (n - 1)2L) - fn ( c , t - ( - L ) - n2L) = 0

or

At x = 0 there must be continuity of velocity For the short bar the particle velocity is super- imposed on the pre-impact speed of J! Thus

(6.30)

V + cJ*, + c,gL = c2F;

and the contact force is

(note that Z = force/velocity)

From equations (6.29), (6.30) and (6.3 1) we obtain

2 v / c , - (1 - z , / Z , ) g : - ,

g'n =

and

(6.33)

)

clzl

c2z2 ( (1 + Z , / Z , )

v / c , + 2gz-1

Since the first waves are go and Fo it follows thatf, = 0 , g - , = 0 and F-, = 0

Let us first examine the waves immediately after the impact, that is for n = 0

- v / c ,

( 1 + Z,/Z,)

czz2 ( 1 + ZI/Z2)

(6.34)

gb =

and for n = 1

Impact of two bars 135

v z,/z, (ZJZ, - 1)

c2 (1 + Z,/Z,) (Z,/Z, + 1)

-2v/c,

If Z , is less than or equal to Z2 then F', is zero or negative This means that the strain is

zero or positive, that is tensile Because a tensile strain is not possible at the interface the

contact is terminated, the contact time being 2L/c,

f i = g b =

The velocity at the interface is

v = V + c,gb = c,Fb

(6.38)

ZJZZ

(1 + z,/z2>

= v

In the special case when Z , = Z,, v = V/2 Figure 6.1 1 shows the progress of the wave mine the wave functions With a little algebra it can be shown that

z, - z 2

If Z , is greater than Z2 then equations (6.32) and (6.33) can be used repeatedly to deter-

and

T:

Fig 6.1 1

ZIYlC, z, - z 2

(6.40) ( z, + z 2 r

F, =

ZI + z2

so that the force transmitted is

(6.41)

from which we see that force decays exponentially Also the velocities decay to zero so the coefficient of restitution, defined in the usual rigid body way, is zero In the case where the

two bars have the same properties and the second bar is the same length as the first the coef-

ficient of restitution is unity, showing that this quantity can range from 0 to 1 even though the process is elastic

z, vz, z, - z 2 Il

( ZI + z2 1

(EA)2F L =

ZI + z 2

6.7 Constant force applied to a long bar

We shall now consider a long bar under the action of a constant force X applied to the face

at x = 0 as shown in Fig 6.12 If we assume that a wave travels into the bar with a speed c then we may use

force = rate of change of momentum

d

dt

X = - ( ~ A v (et)) = ~ A V C

so

- E = X/(AE) = pVdE

By definition

- E = vt I (et) = vlc

Equating the two expressions for E gives

pvclE = v/c

or c2 = Elp as before

Fig 6.12

Constant force applied to a long bar 137

Now let the bar be of finite length L , as shown in Fig 6.13 At x = L the strain has to be zero Therefore at any time

E = -f, + gk-, = 0

S:-1 = f n

or

At x = 0 the force, X , is constant and therefore

X = -EA ( rn + gl)

and

v = c ( f n + gl)

From equation (6.42)

X

f = -

n EA + A-I

Thus

X

o EA

f = -

f 1 = , , + f , = - EA

Hence

(n + l ) X

Substituting into equation (6.43)

(n + 1)X nX

+ - I EA

CX

EA

- -

- (2n + 1)

P A E

Fig 6.13

now time t = n2LIc so the average acceleration is

t EA 2nl

_ - _

X

PAL

-

- - ( I + 1/2n)

As n tends to infinity

t PAL

-

- - -

So we see that the result is that which would have been given by elementary means From

this we learn the very important lesson that rigid body behaviour may be assumed when the variation of force is small compared with the time taken for the wave to traverse the body and return After a few reflections the body behaves like a body with vibratory modes super- imposed on the rigid body modes

The wave method is most suitable when dealing with the initial stages which, in the case

of impacting solids, may well be when the maximum strains occur As mentioned earlier a vibration approach will require a large number of principal modes to be included

In the previous analysis for which impact occurred between plane surfaces it is seen that the leading edge is sharp leading to instantaneous changes in strain and velocity Although these are not precluded in continuum mechanics, in practice some rounding of the leading edge occurs largely due to the impacting surfaces not being plane We shall assume that in the immediate vicinity of the impact point the material behaves as an elastic spring with linear

or non-linear characteristics

Referring to Fig 6.14 we see that the impacting surfaces are convex and the separation of the two reference planes is denoted by (so - a), a being the compression It is assumed

that the compressive force deflection law is of the form X = ka"

The rate of approach of the two reference planes is

and the contact force

Fig 6.14

The efect of local deformation on pulse shape 139

Eliminating g' andf we get

a = V - - - - = V - ka (l/Zl + l/Z2)

Let h = k(l/Z, + l/&) so that equation (6.46) becomes

If m = 1 then the interface behaves like a linear spring and the solution is, with a = 0

at t = 0,

and

(1 - e")

ZlZ2

4 + z 2

from which we see that the maximum force is as given by equation (6.41) with n = 0 The Hertz theory of contact for two hemispherical bodies in contact states that

where R is the radius and p = (1 - u)/(xE) (u = Poisson's ratio) We may write

x = ka3'2

where

-1

3 x

k = [ 4 (PI + P2)\ ($, + i2)]

Equation (6.47) now becomes

6 + ha3/' = v

or

Using the substitution

leads eventually to

(6.49)

(6.50)

Now

x = h 3 ’ 2 = - kV p 3

h

and because as p + a, t + 1 ,

V

(l/Z, + l/ZJ

Thus

X

- = p3

XI,

Introducing a non-dimensional time

213 113

-

leads to a plot of NX,, versus 7 being made Figure 6.15 shows the plot Equation (6.53)

can be rearranged as

- t = ( 2,2 r( 3 3 ( :) (6.54)

and taking u = 0.3 the constant evaluates to 0.478 Note that from equation (6.52) X is proportional to V

3 ~ ( l - u’)

Also shown on Fig 6.15 is a plot of

-

and this shows a reasonably close resemblance to the plot of (6.54) Equation (6.55) is of the same form as equation (6.48) which was obtained from the linear spring model Thus by

equating the exponents an equivalent linear spring may be obtained

Therefore

(6.56)

3 3 113

h , t = ? = h v t

or

k,(l/Z, + l/ZJ = (k(l/Z, + l/Z2))z3 V1’3 (6.57)

where k, and 1, refer to the linear spring model

Figure 6.16 shows a plot of the rise time to three different fractions of the maximum ver-

sus the product of impact velocity and nose radius It is seen that as the nose radius tends to infinity the rise time tends to zero as was predicted for a plane-ended impact Also as the impact velocity (or the maximum force) increases then the rise time decreases

Fig 6.15

Prediction ofpulse shape during impact of two bars 141

Fig 6.16 Rise time based on Hertz theory of contact

We shall consider the impact of two bars having equal properties One bar is of length L

whilst the other is sufficiently long so that no reflection occurs in that bar during the time of contact If we assume a plane-ended impact then the contact will cease after the wave has

returned from the far end of the short bar, that is the duration of impact is 2Llc

Because the rise time, in practice, is finite several reflections will occur before the con- tact force reduces to zero and remains zero in the long bar The leading edge profile has been predicted in the previous section using the Hertz theory of contact where it was also shown that this could be approximately represented by an exponential expression To simplify the computation we shall adopt the exponential form

Figure 6.17 shows the x, t diagram (which is similar to Fig 6.10)

At x = -L, E = 0 and thus

f n - gb-, = 0

or

At x = 0 the difference in the velocity of the reference faces is a,

( V + CA + Cg;) - c F ~ = a,,

or

VJC + f n + g : - Fi = U , / C (6.59) Also, by continuity of force,

-(EL4g', - EAfn) = -(-EAF;) = ka,

Fig 6.17

or

k

and

k

EA

Adding equations (6.59), (6.60) and (6.61) gives

- + 2f, = - + - an

From equations (6.58) and (6.60)

k

n-' EA

n- I

As fo = 0

k

f ; = - -

EA a'

and

f ; = f; - E a , = - - (a0 + a,)

f n = - E C a,

EA

so

(6.63)

k n-'

0

Substituting equation (6.63) into equation (6.62) gives

.PDrediction ofpulse shape during impact of two bars 143

- - - Ea, = - - + - a,,

c dt EA

c EA

We define the non-dimensional quantities

0

- 2kc

a = a -

EAV

and

EA Thereby equation (6.64) can be written as

n - l _-

(6.65)

(6.66)

(6.67)

0

As C has to be continuous

The parameter p is the value of 7 when ct = 2L, that is the time at which the wave in the

short bar returns to the impact point

From equation (6.66)

4Lk

-

Multiplying equation (6.67) by e' gives

d

n - I

e;(l + E;,)= e; (3 + ti,,) = (e' a,,)

0

and integrating produces

- - n - l

-

a,, = e-' [ J e' (1 - 5 d i , ) d'i + constant]

Carrying out the integration a - = e-' - [ J e' - (1 - 0) dI + constant]

= e-' [e' - 11 = (1 - e-')

(the constant = 1 as - a , = e-' - [ J Go e' - [ = 0 when i = 0) -

- (1 - e-')] d? + constant]

(6.69)

(6.70)

(6.7 1)

-

the constant being determined by the fact that a, - = e-' [ J e' (1 - - e-' [ e ' - - GI (0) = Eo@) Continuing the process

- 1 + 'i + Eo@)]} dI + constant]

t

= e-' [ e ' - e-' + I - - - I - ti@) +constant]

2

- -

-2

= e-‘ [ - t + 7 [1 - ~ ~ ( p ) ] + ~ ~ ( p ) l ] (6.73)

In the same way the next two functions may obtained; they are

-2

t

- [2 - &@)I + 7 [ l - &(p) - zl(p)] + &(p) ] (6.74)

a 3 = e - -

-: 1 ?’

-

and

a4 = e-‘ [- - + - [3 - iio(p)l + T [3 - 2ii0(p) - t i , ( p ) l

+ 711 - Eo(P) - 6 ( P ) 1 - 6 ( P ) + G d P j

24 6

(6.75) Figure 6.18 shows the results of the above analysis This figure should be compared with Fig 6.19(a) and (b) which are copies of actual measurements made on a Hopkinson bar The Hopkinson bar is a similar arrangement to that described above In the above case the

contact force can be deduced either by measuring the strain ( E ) by means of strain gauges

I

p = tiww of arrival offirst reflectionjvmfiee endof the impacting baz

Fig 6.18 Pulse shapes for varying lengths of impacting bar

(a)

Fig 6.19(a) Measured pulse shapes showing variation with bar length

Impact of a rigid mass on an elastic bar 145

&)

Fig 6.19(b) Measured pulse shapes showing variation with V

sited away fiom the reflecting surfaces of the long bar, or by measuring the acceleration at the far end of the long bar and integrating to obtain the velocity (v) The contact force is given either by X = -&A or by X = (v/c)EA

By measurements on the impacting bar it is possible to deduce the compression across the contact region and thereby obtain the dynamic characteristics of that region or of a speci- men of other material cemented there This is particularly useful for tests at high strain rate where the bars can be sufficiently long for the reflected waves to arrive after the period of interest

6.10 Impact of a rigid mass on an elastic bar

In this example the impacting body is assumed to be short compared with the bar but of comparable mass The reflections of the strain waves in the body are assumed to be of such short duration that a rigid body approximation is practicable figure 6.20 shows the relevant details; for this exercise the far end of the bar is taken to be fixed

At the far end the particle velocity is zero and thus when x = L

v = cyn + cg;+, = 0

or

At x = 0 the contact force is

a‘u

at

X = -(-EAYn + EAg;) = - M T = -M(c2fl + c’g:)

Let

EA - PAL - CI

_ - - - -

Therefore

Fig 6.20

E + i f n -E-I - E A - ‘

- E-I - i A-1

- ( eHL f n ) = ep’L

Multiplying by eKIL we get

d

dz

Integrating gives

P

A = e - ~ / L ~ e ~ ~ ( ~ - , - - - ~ - , ) c i z + B n ]

where B, is a constant of integration

Asf-, = 0 the first hnction is

(6.77)

yo = e-@‘ (0 + Bo)

v = G (0) = ceo(Bo> = Y

f = - e-’l;’L

When z = 0, v = V and therefore

ThusB,, = V/c and

V

(6.78)

This finction is valid until the wave returns from the far end, that is, when z = 2L or t = 2L/c

at x = 0

0

C

For t > 2LJc

f, = e-P’L [ J e’lLIL ( -E e-’li/L - - pve-p/L ) d~ + B , ]

- e-@‘ [ - - 2 p v z + B, ]

L C

Now the velocity must be continuous at x = 0 so that

Impact of a rigid mass on an elastic bar 147

and hence

(6.79)

(6.80) The same procedure can be employed to generate the subsequent functions The results for the next two are given below

V [ - !! (4 + 2e-”) z + 2 (f rz2 + B2 ]

L

f 2 = -

C

where

and

(2 - 4p) + 312

f 3 = - e-’LL’L - - 2p [ -4p+

(6.81)

(6.82) where

B, = e-6p + e-4’ (1 - 8p) + e-’’’ (1 - 8p + 8p’) + 1

At the impact point, x = 0, the velocity and the strain are given by

and

EO = - f n + gl = -K + A-1) (6.84)

At the fixed end the velocity is zero but the strain is

Figure 6.2 1 gives plots of these functions versus time It is apparent from the graphs that the highest strains occur at the fixed end at the beginning of the periods, that is for z = 0 and x

= L Figure 6.22 gives a plot of the maxima versus Up

As the ratio of the impacting mass to that of the rod increases it is possible to use an approximate method to determine the maximum strain It is well known from vibration theory that a first approximation in this type of problem is to add one-third of the mass of

the rod to that of the end mass For this case we equate the initial kinetic energy with the

final strain energy

Fig 6.21

Fig 6.22 Maximum strain at x = L for various period numbers, n

- (M + pAL/3) V

= - EALE’ = - C’PALE’

Hence

1 ‘I’

V M + pALl3

E = - (

- e o l + j 1

Y 1 1 ‘ I 2

-

where p = pALIM

by adding the result obtained in the previous analysis, which was that the initial strain is Vlc

Therefore

This is adequate for large values of 1/p but not for small A better approximation can be made

Dispersive waves 149

E = - c V [ l + ( i + j ) 1 1 ‘I2 ] (6.86)

A plot of this aproximation is included on Fig 6.22

6.1 1 Dispersive waves

Let us first discuss the sinusoidal travelling wave The argument for a sinusoidal function is required to be non-dimensional so we shall adopt for a wave along the positive axis

2 = k(ct - x)

= (ckt - kx)

where k is a parameter with dimensions lllength

A typical wave would be

Figure 6.23 shows a plot of u against t for x = 0 If the argument increases by 21r, the time

is the periodic time T Thus

ckT = 21.r

or

ck = 2 d T = 2 7 ~ ~ = o

where u is the frequency and o is the circular frequency

responds to a change in x of one wavelength h Thus

Figure 6.24 is a similar plot but this time versus x An increase of 21.r in the argument cor-

or

k is known as the wavenumber

Fig 6.24

Hence

and

0

This quantity is called thephase velocity as it is the speed at which points of constant phase move through the medium In a dispersive medium cp varies with frequency (or wavelength) and it will be shown that this is not the speed at which energy is propagated

Consider now two waves of equal amplitude moving to the right as shown in Fig 6.25

Note that both arguments are zero for t = 0 and x = 0 The displacement is

k cp

u = UCOS (o,t - k l x ) + U COS (o,t - klx)

and using the formula for the addition of two cosines

kl - 2 k2 x i

x ) cos ( "I ; O 2 t -

kl + k2

t -

2

A6.l

= u c o s (wot - k,x) cos ( - t - - k

( w' : a*

u = u c o s

Fig 6.25(a) and (b)