Principles of Engineering Mechanics (2nd Edition) Episode 6 doc

If a system of particles is in equilibrium then the virtual work done over any arbitrary displacement, consistent with the con- straints, is zero: Application to a system with a single d

7.8 The power equation

If the work-energy relationship is written for a

small time interval At, then we have

AW = A(k.e + V ) + A(1osses)

Dividing by At and going to the limt At-0

leads to

- = - (k.e + V ) +-(losses)

or, power input equals the time rate of change of

the internal energy plus power ‘lost’

Let us consider a simple case of a single particle

acted upon by an external force P and also under

the influence of gravity, then

P - v = -[4rnv-v+rngz]

= mv-a+rngi

d

dt

If the motion is planar,

v = X i + i k

a = x i + z k

and P = Pxi+P,k

so that

PxX + P,i = rn (Xx + iz) + mgi

P x + P z t a n a = rntan2cr+1)z+rngtancr

Hence z may be found without considering the

(7.32)

Figure 7.15

condition for equilibrium is

If z = xtancr (Fig 7.14) then dividing by x gives From the free-body diagram (Fig 7.16) the

C P = P , + P , = w = o

and P1 sin cyl + P2sin cr2 - rng = 0

are not apparent, but the reader is asked to be patient as later examples will show some of its rewards

A virtual displacement is defined as any small

displacement which is possible subject to the constraints The word virtual is used because the displacement can be any displacement and not necessarily an actual displacement which may occur during some specific time interval

The notation used for a virtual displacement of some co-ordinate, say u, is Fu This form of delta

is the same as is used in mathematics to signify a variation of u ; indeed the concepts are closely related

The work performed by the forces in the system over this displacement is the virtual work and is given the notation 621r

Conditions for equilibrium

Let us first consider a single particle which is free

to move in a vertical plane subject to the action of two springs as shown in Fig 7.15

The concept of virtual work is one which saves a

considerable amount of labour when dealing with

complex structures, since there is no need to

dismember the structure and draw free-body

diagrams Basically we shall be using the method

as an alternative way of presenting the conditions

for equilibrium and also to form a basis for the

discussion of stability In the early stages of

understanding the principle the main advantages

If equations 7.31 and 7.32 are multiplied by 6x

i

P,cosa2Fx- P ~ C O S c r , F X = 0

P,sincrlFz+P2sina2Fz-rngFz= 0

or C P - F S = 0 (7.33) These are the equations for the virtual work for the arbitrary displacements Fx and 6z - note,

arbitrary displacements In both cases we may state that the virtual work done by the forces over

an arbitrary small displacement from the equilib-

rium position is zero

If a system comprises many particles then the

total virtual work done on all particles over any

virtual displacement (or combination of displace-

ments) is zero when the system is in equilibrium

Principle of virtual work

We may now state the principle of virtual work as

follows

If a system of particles is in equilibrium then

the virtual work done over any arbitrary

displacement, consistent with the con-

straints, is zero:

Application to a system with a single degree of

freedom

Consider a rigid body freely pinned at A and held

in equilibrium by a spring attached at B

(Fig 7.17) This body has one degree of freedom,

that is the displacement of all points may be

expressed in terms of one displacement such as 8,

the angular rotation If the spring is unstrained

when AB is horizontal, then in a general position

the active forces are the weight and the spring

force; the forces at the pin do no work if friction is

negligible

For a small displacement 68, the displacement

of G is a 68 (Fig 7.18) and the virtual work is

W = Mg (a 68COS 8 ) - kR8 (R de)

For equilibrium,

w = o = (Mgacos8-kR2e)68

Mgacos 8 - kR28 = 0

and, as 68 is arbitrary,

(7.35)

7.9 Virtual work 97

If, as in the previous example, the forces are

conservative then equation 7.34 may be inter-

preted as

gravitational potential energy)

elastic strain energy)

W ' = virtual work done by external forces

where W G = - SVG = -(variation of

W E= - - W E = -variation of

and Therefore

W = W ' - 6 v E - 6 V G = o

or W ' = 6 ( V E + V G ) (7.37)

Reworking the last problem,

0 = 6 [ - M g ~ s i n 8 + 4 k ( R 8 ) ~ ]

0 = [ - M ~ U C O S 8 + kR28] 68

Stability

Consideration of some simple situations shown in Fig 7.19 will show that not all equilibrium configurations are stable However, we cannot always rely on common sense to tell us which cases are stable We have demonstrated that for equilibrium W = 0, but further consideration of

the value of W as the virtual displacement

becomes large will lead to the conclusion that if

W becomes negative then the force will be in the

opposite direction to the displacement , showing

that the forces are tending to return the system to

the equilibrium configuration, which is therefore

one of stable equilibrium In mathematical

notation, for stability

S ( W ) < O

Looking at our previous case once again,

therefore 6 ( W ) = (-Mgasine- kR2)(Se)2

state is stable

configuration is defined by

w = (Mgacose- kR2e)6e

For 0<8<7r it is seen that any equilibrium

For a conservative system, the equilibrium

6 V = 0

and stability is given by

S ( W ) < O

6(6V)>O

But, since W = -SV,

If V can be expressed as a continuous function

of 8, then

av

ae

av=-ae

and

a2v

ae2

a2v=- e se=-(6e)2

ae a iaV ae )

Hence for equilibrium

av

ae

- = o

a2v

ae2

and - > O for stability

(7.39)

Systems having two degrees of freedom

The configuration of a system having two degrees

of freedom can be defined by any two indepen-

dent co-ordinates q1 and q 2 The virtual work for

arbitrary virtual displacements 6ql and 6q2 may

be written in the form

(7.41)

W = QI 6qi + Q2 642

Since the virtual displacements are arbitrary,

we may hold all at zero except for one and, as

W = 0 for equilibrium, we have

Qi = O and Q2 = 0 The stability of a system having two degrees of freedom will be discussed for a conservative system

It will be remembered that constant forces are conservative, therefore the majority of cases may

be considered to be of this type

If the independent co-ordinates - referred to as generalised co-ordinates - are q1 and q 2 , then the

total potential energy (gravitational plus strain) is

V = V(ql , q2); hence

a41 a42 6V = - 6ql +- 6q

and, since SV = 0 for equilibrium, we have

av av _ - - - = o a41 a42

(7.42)

For stable equilibrium we must have @V>O for

all possible values of 6ql and S q 2 The second variation may be written

or, since a2V/aql aq2 = a2V/aq2aq1, then

It is clear that, if 6q2 = 0, then

a2v

->0

%I2

and, if 6ql = 0, then

a2v

->O a4z2 These are necessary conditions for stability, but not sufficient To fully define stability, a2V must

be >O for any linear combination of 6ql and 6q2

a2v

->0

a d

a2v

->0

a422

a2v a2v

>

2

and -. ~

a d a d (aqd:dVq2) > O ,

Consider the conservative system shown in Fig 7.20; the active forces, real and fictitious, are shown in Fig 7.21

W = SV- m,xl ax1 -m2x2Sx2 - ZeSe

= 6 [ t h l 2 + mlgxl - m2gxz + const.]

+ (7.44)

I

r1

= k - x2 +mlg- -m2g Sx2

or m 2 + m l - +> i 2 + k - x2

= m2g - ml (:)g (7.46)

This approach does not involve the internal

forces, such as the tension in the ropes or the

workless constraints, but these may be brought in

by dividing the system so that these forces appear

as external forces

Equation 7.46 could have been derived by the

application of the power equation with a similar

amount of labour, but for systems having more

than one degree of freedom the power equation is

not so useful

Discussion examples

Example 7.1

A block of mass m can slide down an inclined

plane, the coefficient of friction between block

and plane being p The block is released from rest

with the spring of stiffness k initially compressed

an amount x, (see Fig 7.22) Find the speed when

the block has travelled a distance equal to 1 &,

Figure 7.22

Solution If a free-body-diagram approach is

used to solve this problem, the equation of

motion will be in terms of an arbitrary

displacement x (measured from, say, the initial

position) and the acceleration R Integration of

this equation will be necessary to find the speed

If an energy method is used, consideration of

the initial and final energies will give the required

speed The two methods are compared below

a) Integration of equations of motion The

free-body diagram (Fig 7.23) enables us to write

the following equations:

[ C F , = mYG1

W C O S ~ - N = O ‘.N= W C O S ~ (i)

[CFx = m.fG]

W s i n a - p N - T, = mx (ii)

If we measure x from the initial position, the spring tension T, is given by T, = k(x - x , ) and

we shall be integrating between the limits 0 and

1 2 ~ ~ We could, on the other hand, choose to measure x from the position at which there is no force in the spring, giving T, = kx, and the limits

of integration would be from -x, to +O.&,

Using the former, (i) and (ii) combine to give Wsina-pWcosa-k(x-x,) = m.f (iii) Since we are involved with displacements, velocities and accelerations, the appropriate form for R is vdvldx: the direct form f = dv/dt is clearly

of no help here

Hence equation (iii) becomes

1.kc

0

I { W(sin a - pcos a) - k ( x - x , ) } dx

= [‘mvdv 0

I:&=

{w(sin a - pcos a ) + kx, } x - tkx2

= m[+o2]

[

0

{ mg (sin a - p cos a) + kx, } 1 kc - i k (1 h,)’

= imv2 and thus v can be found

The reader should check this result by measuring x from some other position, for instance the position at which the spring is unstrained, as suggested previously

b) Energy method Since energy is lost due to the friction, we use equation 7.29 (see Fig 7.24): [work d~ne],,,,,,~ = [k.e + VG + V E ] ~

- [k.e + V G + VGll + ‘losses’ (iv) where the ‘losses’ will be p N ( 1 2 ~ ~ ) as explained

in section 7.6 For the general case of both p and

N varying, this loss will be Jb2”cpNdx None of the external forces does any work, according to our definitions, and thus the left-hand side of

If, however, the block had been following a known curved path, the spring tension T, could have been a complicated function of position giving rise to difficult integrals, possibly with no analytical solution The energy method requires only the initial and final values of the spring energy and so the above complication would not arise Variation in N could cause complications in both methods In some cases the path between the initial and final positions may not be defined

Figure 7.24 at all; here it would not be possible to define T,

as a general function of position An energy method would give a solution directly for cases equation (iv) is zero

It should be pointed out that the correct result where friction is negligib]e (see, for example,

can be obtained by treating the friction force as problem7.2)

external to the system and saying that this force

does negative work since it opposes the motion Example 7m2

The left-hand side of equation (iv) would then be See Fig 7.25(a) The slider B of maSS rn is

wou1d be Omitted' This is a common way Of

fixed to the slider engages with the slot in link dealing with the friction force but is not OA The moment of inertia of the link about o is

Io and its mass is M , the mass centre being a considered to be a true energy method

Kinetic energy In the initial position ( x = 0) the distance a from 0 The spring of stiffness k is speed and thus the k.e are zero In the final attached to B and is unstrained when 8 = 0 position ( x = 1 2 ~ ~ ) the k.e is frnv2, from

equation 7.26

Gravitational energy, V, The datum for

measuring gravitational energy is arbitrary and

we may take as a convenient level that through

the initial position; thus the initial g.e is zero

Since the block then falls through a vertical

distance of 1.2xCsina, the final gravitational

energy is, from equation 7.27, -rng(l.2xcsina)

Strain energy, VE In the initial position, the

Spring is compressed an amount x, and thus, from

equation 7.28, the strain energy is fkx: In the

final position the spring is extended b an amount

O.&, and so the final s.e is fk(0.2~~)

Note that only the gravitational energy can have

a negative value

Equation (iv) becomes

-pN(1*2rC) and the 'losses' t e m On the right constrained to move in vertical guides A pin P

Figure 7-25

The system is released from rest at 8 = 0 under the action of the torque Q which is applied to link

OA ne variation of Q with e is shown in Fig 7.25(b)

Determine the angular speed of OA when

8 = 45", neglecting friction

9

O = [frnv2 - rng ( 1 &,sin a) + fk (0.2x:)I

- [0 + 0 + dhC2 J + p N ( 1 2 ~ ~ )

We still need a free-body diagram to determine

that N = mgcosa, as in equation (i), and then v

can be found directly

For this particular problem there is little to

choose between the free-body-diagram approach

and the energy method In the energy method we

avoided the integration of the first method, which

however presented no difficulty

Solution This problem has been approached in

example 6.5 by drawing two free-body diagrams and writing two equations of motion involving the

a

- I@, = ;IO I ~ I o A ~ + fmVg2 + Mg-

+mgl+Ikl']-[(I] (ii)

Before we can evaluate wOA we need to express

contact force at the pin P Since we are here

concerned only with the angular velocity at a

given position, and details of internal forces are

not required, an energy method is indicated and

will be seen to be easier than the method of

Chapter 6

Equation 7.29 becomes

V B in terms of 0 0 ~ Since y = ltan 8,

(work done),,,,,,, = [k.e + vG + v E ] 2

- [k.e + VG + VEII (i) since there are no losses The left-hand side of

this equation is the work done by the external

and is thus Jf4 Qd8 This is simply the area under

the curve of Fig 7.25(b), which is found to be

(11/32) r e ,

The normal reaction N between the slider and

the guides is perpendicular to its motion and the

force R in the pin at 0 does not move its point of

application: thus neither of these forces does

work (see Fig 7.26)

and at 8 = d 2 , V B = 21IiIOA

Substitution in (ii) gives forces or couples on the system during the motion 11

- TQ, = &(Io + 4 m l 2 ) wOA2

32

+ g M-+ml ++k12

from which I i I O ~ can be found

Comparison of this method with the free-body- diagram approach and the difficulty associated with integrating the equation of motion shows the superiority of the energy method for this problem

What if the force S on the pin P has been required? This force does not appear in the energy method, but this does not mean that the energy method is of no help Often an energy method can be used to assist in determining an unknown acceleration and then a free-body- diagram approach may be employed to complete the solution

- -

Kinetic energy As the mechanism is initially at Example 7.3

rest, the initial k.e is zero Since the motion of

OA is rotation about a fixed axis, the final k.e of

O A , from equation 7.8, is 41000A2 The slider B

has no rotation and its final kinetic energy, from

equation 7.7, is simply fmvB2

Gravitational energy, V G We will take as

datum levels the separate horizontal lines through

the mass centres of link and slider when 8 = 0 and

thus make the initial value of VG zero When

8 = 45" the mass centre of the link has risen

through a height a / d 2 and that of the slider Figure 7.27

through a height I, and so the final value of VG is C D has a moment of inertia about D of 6 kg m2

M g a l d 2 + mgl and its mass is 4.5 kg BC has a moment of inertia

Strain energy, VE Initially the strain energy is about its mass centre E of 1.5 kg m2 and its mass is zero and in the final position the spring has been 4 kg A t the instant when both AB and C D are compressed an amount I; the final value of V E is vertical, the angular velocity of AB is 10 r a d s and

measured in an anticlockwise sense

A four-bar chain ABCD with frictionless joints is shown in Fig 7.27

Substituting in (i) gives

Neglecting the inertia of AB, determine the From the velocity diagram we see that

vE = - l O i d s ; the component of UE in the same

direction is a E - i = -(50- 2 5 / d 3 ) d s 2 Substitut- ing into the power equation (1) gives

torque T which must be applied at A to produce

the above motion

Solution The velocity of B is o x 3

= 10k x lj = - 1Oi d s The velocity diagram is

shown in Fig 7.28 and it can be seen that link BC

is not rotating (I$ = 0)

T10 = 4(10)(50 - 2 5 / d 3 ) + 6(5)(25)

and

Example 7.4

A slider-crank chain PQR is shown in Fig 7.30 in

its equilibrium position, equilibrium being main- tained by a spring (not shown) at P of torsional

stiffness k Links PQ and QR are of mass m and

2m and their mass centres are at GI and G2

respectively- The slider R has a mass M The

moment of inertia of PQ about P is Zp and that of

QR about G2 is ZG2; also, PGI = G I Q

= QG2 = G2R = a

T = 217.2 N m

b , c, e a, d

Figure 7.28

The kinetic energy of link BG is thus $MvE2 and

that of CD, for fixed-axis rotation about D, is

$ Z D ~ 2 'It would clearly not be correct to write the

power equation (section 7.8) as

d

dt

T - w = T o = - (4MvE2 + $ Z k 2 )

since 4MvZ is not a general expression for the k.e

of link BC (it is the particular value when AB is

vertical) As ZI, and a, do not have the same

direction, the correct power equation is

d

dt

Figure 7.30

Find the frequency of small oscillations of the system about the equilibrium position, 0 = eo,

since + = 0 Solution Equations of motion for the links can

of course be obtained from a free-body-diagram approach, but this would involve the forces in the pins and would be extremely cumbersome Use of the power equation leads directly to

= 5k r a d s and thus the required result In this case we have

power = d(energy)/dt = 0, since the energy is constant for the conservative system and clearly

no power is fed into or taken out of the system Let the link PQ rotate clockwise from the equilibrium position through a small angle /3 as

+ = - = - = 2 cc' 50 5 rads2 shown in Fig.7.31 The new positions of the

T o w = T o = - ( ( ~ M + - Z ) E + ~ I ~ ~ + $ Z D $ ~ )

= M+ ' a E + ID44 (i) neglecting friction

The acceleration of B is

Q B = [-l(5O)i- 1(10)j2] m/s2

From the velocity diagram we find $ = 10W2

ac = [ -242- 2(5)2j] d s 2

The acceleration diagram is shown in Fig 7.29

and 6; is given by

Figure 7.29

Kinetic energy Link PQ has fixed-axis rotation about P and its k.e is thus 4Zp@' By symmetry, the angular speed of QR is also /3 The k.e of QR

is given by 4ZG2fi2 + 4 ( 2 r n ) ~ ~ ~ ~ , where

d

dt

= - [3acos (eo + p)i - asin(Oo + pb]

= -3asin(eo + p)Bi - acos(eo + B ) j

The k.e of slider R is M v R 2 where

d + d

v R = - ( P R ’ ) =-[4acos(Oo+p)i]

= -hsin(e,+p)/%

Gravitational energy A convenient datum level

is the horizontal through PR The mass centre of

each of the links PQ and QR is at a height

asin(eo + p ) above the datum and their gravita-

tional energy is thus, from equation 7.27,

VG = mgasin(80++) + ( 2 m ) g a ~ i n ( 8 ~ + p )

The slider R moves the datum level and thus

has no gravitational energy with respect to this

level

Strain energy The couple applied by the spring

t o the link PQ in the equilibrium position is

clockwise and equal to kyo , where yo is the angle

of twist As link PQ rotates clockwise through an

angle p, the angle of twist is reduced to ( y o - p )

and thus the strain energy, from equation 7.28, is

I k ( y 0 - p ) ’

The total energy E is thus

E = {k.e.} + { V G } + { V E }

= { +zPB + $z,, B + f(2m)

x [9a’sin’(~~ + p ) + a’cos’( eo + p ) ] 8’

+ IM 16u2sin2(8, + p ) B’ >

+ {3mgasin(~o++)} + { b k ( y O - - ~ ) ’ }

= constant (since the system is conservative)

Since the above is a general expression for the

energy, it can be differentiated to give the power

equation The term fi’ arises which is negligible

for small oscillations We note that, since p is

small, sin( 0, + p ) = sin 0, and cos(8, + p ) =cos eo,

but these approximations must not be made

before differentiating After dividing throughout

by B, we find, since b2 is small,

ZB + k p = k yo - 3mga cos 80 ( 9

where Z = Z, + ZG2 + 2m (9a’ sin’ eo

+ a2cos2 0,) + 16Ma2sin2 0,

It can be shown by the method of virtual

work, or otherwise that kyo = 3mgacos eo so that equation (i) reduces to ZB+ k p = 0 Thus, for small p, the motion about the equilibrium position is simple harmonic with a frequency of (1/27r)d( k/Z)

Example 7.5

The mechanism shown in Fig 7.32 is in equilibrium Link AB is light and the heavy link

BC weighs 480 N, its mass centre G being midway between B and C Friction at the pins A and C is

negligible The limiting friction couple Qf in the

hinge at B is 10 N m

Figure 7.32

Pin C can slide horizontally, and the horizontal force P is just sufficient to prevent the collapse of the linkage Find the value of P

Solution This problem has been solved earlier

in Chapter 4 (example 4.3) There a free-body

diagram was drawn for each of links AB and BC and the unknown forces were eliminated from the moment equations It will now be solved by the method of virtual work and the two methods will

be compared

If in the virtual-work method we treat forces due to gravity and springs and friction as being externally applied, the total virtual work done may then be equated to zero In order to obtain the correct sign for the virtual work done by the internal friction couple Qf in the present problem,

we may use the following rule: the virtual displacements must be chosen to be in the same

direction as the actual or impending displacements

and the virtual work done by friction is given a

negative sign

Applying this rule, we let the virtual displace- ment of C be 6x to the right, since this is the direction in which it would move if the mechanism were to collapse

If a mechanism has a very small movement, the displacement vector of any point on the mechanism will be proportional to the velocity vector Thus we can draw a small-displacement diagram which is identical in form with the corresponding velocity diagram This results in

very lengthy means of solution, whereas the virtual-work method disposes of the problem relatively quickly (see, for example, problem)

Example 7.6

A roller of weight W is constrained to roll on a circular path of radius R as shown in Fig 7.34 The centre c of the roller is connected by a Spring

of stiffness k to a pivot at 0 The position of the roller is defined by the angle 8 and the Spring is UnStretched when 8 = 90"-

Fig 7.33, where a b is drawn perpendicular to

AB, bc is drawn perpendicular to BC and oc is of

length Sx

Since the weight W acts vertically downwards

and the vertical component (hg) of the displace-

ment of G is also downwards, the virtual work

done by W is positive and given by + W(hg)

The virtual work done by P is - P ( o c ) , since

the force P is opposite in sense to the assumed

virtual displacement

The virtual work done by the friction couple Qf

is -Qf IS+ I, where IS+ I is the magnitude of the

change in the angle ABC AB rotates clockwise

through an ang1e ab/AB and BC rotates a) Show that the position 8 = 0 is one of stable anticlockwise through an angle bc/BC [If equilibium only if W/(Rk) > 0.293

course the angular speed of AB would have been

given by ab/AB, and so on.] The change in the

angle ABC is thus

Figure 7-34

Fig 7-33 had been a ve1ocity diagram then Of

b) If W/(Rk) = 0.1, determine the positions of stable equilibium

Solution The strain energy V, in the spring is zero when centre C is at B We can also make the gravitational energy VG zero for this position by

ab bc

a+=-+-

Summing the virtual work to zero gives

W(hg) - P ( o c ) - Qf : (: - +- 3 = 0

The virtual displacements ab, bc and hg are

Figure 7.35

From Fig 7.35, the stretch in the spring is

OC - O B = 2R cos (8/2) - R d 2 and C is a vertical distance Rcos8 below AB Thus, using equations 7.27 and 7.28, the total potential energy Vis given

scaled directly from the diagram to give

480(0.1875 S X ) - P ( S X )

0 8 3 8 6 ~ 0.451 Sx

which, on dividing throughout by S X , gives by

P = 40.0 N

Comparing the virtual-work solution of this

problem with that of the normal staticdfree-body-

diagram approach of Chapter 4, it can be Seen

that here we are not concerned with the forces at

A and B and the vertical component of the force

at C However, for this simple problem the more

straightforward approach of Chapter 4 is to be

The virtual-work method comes into its own

when many links are connected together In such

cases, drawing separate free-body diagrams for

each link and writing the relevant equations is a

-10 ~

( 0.2235 + E ) = o

v = v,+v,

= - WR cos 8 + +k [2R cos( 8/2) - R d 2 I 2 The equilibrium positions are given, from

dV/d8 = WRsinO+ k[2Rcos(8/2)- R d 2 ] equation 7-39, by

x [ - R sin( 0/2)]

+ d2sin(8/2)) = o (i) a)

NOW, One solution to equation (i) is clearly 8 = 0