Barber J.R. Intermediate Mechanics of Materials

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12.6 Beams on elastic foundations 535

u = 0, M = 0 ; z = 0,L ,

we obtain the four simultaneous homogeneous equations

+ B

= 0

−B

− B

+ B

= 0

+ B

−

+ B

cos(

L) + B

sin(

L) = 0

−B

− B

−

+ B

cos(

L) + B

sin(

L) = 0 .

Elementary algebraic operations show that B

=0 for all

Ω

, but B

can

take an arbitrary value if sin(

L)= 0 and hence

L= n

, where n is an integer. Thus,

the critical speeds are deﬁned by

Ω

and hence

Ω

For the solid circular shaft,

m =

; I =

giving

Ω

and hence

Ω

Generally only the ﬁrst critical speed is of interest, for which n= 1 and

Ω

. (12.46)

Notice however that as in §12.5, the ﬁrst critical speed can be suppressed by provid-

ing a sufﬁciently stiff central bearing.

Design considerations

For typical machine components such as gearshafts and motors, the operating speed

will be signiﬁcantly lower than the critical speed and hence whirling instabilities will

not be a deciding factor in the design. For example, for a steel shaft (E = 210 GPa,

=7700 kg/m

) of diameter 20 mm and length 500 mm, we have

536 12 Elastic Stability

Ω

×0.02

4 ×0.5

210 ×10

7700

= 1030 rad/s = 9836 rpm,

which is signiﬁcantly higher than any likely operating speed.

Notice however that L

appears in the denominator of equation (12.46), so

whirling instabilities may be important in long transmission shafts. They are also crit-

ical in the design of very large machines, such as turbogenerators, which are therefore

generally designed to operate between the ﬁrst and second critical speeds. Whirling

instabilities differ from the other elastic instabilities considered in this chapter in

that the system is stable at speeds between the critical speeds as well as below the

ﬁrst critical speed. They share this characteristic with other linear vibration prob-

lems. However, it is important not to operate too close to a critical speed, since in-

evitable manufacturing errors would then lead to unacceptably large displacements

and stresses. Also, operating conditions have to be chosen carefully to ensure that

the process of accelerating through the critical speed range during start-up does not

take long enough for the instability to develop to dangerous proportions.

Gears and pulleys

The whirling speed will be signiﬁcantly reduced if the shaft carries a gear or some

other massive body at some point distant from the bearings. A typical assembly is

shown in Figure 12.20, where a gear wheel of mass M

is attached to the shaft at the

mid-point.

Ω

bearing

mass M

Figure 12.20: Shaft with a massive wheel supported at the mid-point

The ﬁrst whirling mode for this system is shown in Figure 12.21, from which it is

clear that the wheel contributes an additional outward force M

Ω

at the mid-point,

where

= u





is the displacement of the mass.

12.6 Beams on elastic foundations 537

u(z)

Figure 12.21: Deﬂected shape for the shaft of Figure 12.20

In many cases, the distributed inertia force due to the shaft mass can be neglected

relative to that due to the wheel, leading to a very simple estimate of the critical

speed. The shaft then acts merely as a spring relating the force at the wheel to its

displacement. Elementary beam deﬂection calculations show that the central force F

in Figure 12.22 produces a central displacement

48EI

Figure 12.22: Loading of a simply supported beam by a central force

Substituting F = M

Ω

, we then obtain

Ω

48EI

and hence the critical speed is

Ω

48EI

. (12.47)

This simple approximation demonstrates that the critical speed can be increased by

increasing the stiffness of the supporting shaft and this is most easily done by locating

the bearings as close as possible to any massive rotating bodies.

The estimate (12.47) is unfortunately non-conservative, since a more exact cal-

culation taking into account the distributed mass of the shaft will give a lower critical

538 12 Elastic Stability

speed. However, a quick check on the probable accuracy of the approximation can

be made by comparing the critical speed (

Ω

) predicted by (12.47) and for the shaft

alone (

Ω

) from (12.46). If (as is usually the case)

Ω

≫

Ω

, the true critical speed

will not be much less than

Ω

. In fact, a reasonable engineering guess at the com-

bined effect is

Ω

, deﬁned by

Ω

, (12.48)

though this is not based on any scientiﬁc argument!

A rigorous but more lengthy way of accounting for the combined effect of the

mass of the shaft and the wheel in Figure 12.20 is to note that the system is symmetric

and hence

= 0 ; V =

Ω

; z =

. (12.49)

Applying these boundary conditions and u = 0 ; M = 0 at z = 0, u =

at z = L/2

in (12.44), and eliminating B

from the resulting equations, yields an

algebraic equation for the critical speed

Ω

12.7 Energy methods

Our discussion of elastic stability has so far been focussed on the neutral stability

or equilibrium method, in which we determine the condition (usually the critical

compressive force) needed to sustain a non-trivial state of deformation in neutrally

stable equilibrium. In this method, we take it on trust that the system will be stable

below the ﬁrst critical force and unstable above it.

A more direct approach is to determine the total potential energy

as a function

of the deformed conﬁguration. We have already seen in §3.5 that

is stationary

at an equilibrium conﬁguration, but it is also clear that the system will be stable if

and only if

is a local minimum, since we can then argue that motion away from

equilibrium will be possible only if an external source of energy is provided.

A simple system illustrating this procedure is shown in Figure 12.23 (a). The

rigid bar AB is pinned at B and loaded by a vertical force P at A, which is also

restrained by a horizontal spring of stiffness k.

The deformed conﬁguration is shown in Figure 12.23 (b) and we conclude that

the potential energy of the force is

Ω

= −PL(1 −cos

) ,

whereas the spring extension is Lsin

, giving a strain energy

U =

sin

12.7 Energy methods 539

(a) (b)

Figure 12.23: (a) A pinned rigid bar supported by a spring; (b) the deformed conﬁg-

uration

The total potential energy is therefore

= U +

Ω

sin

−PL(1 −cos

) (12.50)

and equilibrium conﬁgurations are deﬁned by the condition

= kL

sin

cos

−PL sin

= 0 . (12.51)

The stability of these states can be examined by evaluating the second derivative

= kL

(cos

−sin

) −PLcos

. (12.52)

An equilibrium state is stable if d

>0 and unstable if d

<0.

It is clear from Figure 12.23 (a) that the bar is in equilibrium when it is vertical

and this is conﬁrmed by the fact that

= 0 satisﬁes equation (12.51). To determine

whether this state is stable, we set

=0 in (12.52), obtaining

= kL

−PL , (12.53)

which is positive, indicating a stable state, for P < kL and negative (unstable) for

P >kL. Thus, P

=kL is the critical force for the system of Figure 12.23 (a).

540 12 Elastic Stability

12.7.1 Energy methods in beam problems

The example of Figure 12.23 is simple because the bar is rigid, so the system has

only one degree of freedom

. The method can be extended to systems with more

than one degree of freedom, provided that number is ﬁnite, but for continuous sys-

tems such as elastic beams, it can only be applied in a Rayleigh-Ritz sense. In other

words, we approximate the deformed shape of the beam by a suitable function with

a ﬁnite number of degrees of freedom and determine the condition for stability by

demanding that the total potential energy for this approximate shape be at a local

minimum.

We ﬁrst have to develop an expression for the change in length of a beam due to

lateral deﬂection, since this will enter into the potential energy of the compressive

force P.

δs

u + δu

δz

Figure 12.24: Geometry of beam deﬂection for a small element

Consider the segment

s of the deformed beam shown in Figure 12.24. From the

geometry of this ﬁgure we have

s =

1 +





≈

1 +





since in the limit

z is the slope of the beam, which is assumed to be small. Thus,

z ≈

s −





and the reduction in the distance between the ends of the beam associated with the

deformation of this beam segment is

s −

z =





z . (12.54)

The total change in length of the beam is obtained by taking the limit of equa-

tion (12.54) as

s →0 and then summing the contribution of all inﬁnitesimal beam

segments by integration. For a beam of length L, we obtain

12.7 Energy methods 541

L =





dz , (12.55)

in terms of which, the potential energy of the compressive force can be written

Ω

= −P

L . (12.56)

12.7.2 The uniform beam in compression

We ﬁrst use the energy approach to re-examine the problem of Figure 12.2 (a), in

which a uniform beam of length L is subjected to a compressive force P. In this case,

we know from the preceding analysis that the beam will deform into a sinusoidal

shape, so we assume the Rayleigh-Ritz form

u(z) = C sin





. (12.57)

The strain energy is then

U =





dz =

EIC

sin





dz . (12.58)

This integral can be evaluated using the change of variable x=

z/L, with the result

U =

EIC

. (12.59)

Substituting the displacement u(z) from (12.57) into (12.55, 12.56), we obtain

Ω

= −

cos





dz = −

(12.60)

and hence the total potential energy is

Ω

+U =

EIC

−



−P



. (12.61)

Equilibrium conﬁgurations are deﬁned by the condition d

/dC = 0, which here

yields



−P



= 0 . (12.62)

This condition is satisﬁed for all C if P = P

EI/L

, but it is only satisﬁed by

C=0 for any other value of P. In other words, the trivial undeformed state is the only

equilibrium conﬁguration except at the critical force P = P

, where any function of

the form (12.57) deﬁnes an equilibrium state.

This much was of course established by the equilibrium analysis of §12.1, but we

can now make a more deﬁnitive statement about stability by evaluating

542 12 Elastic Stability



−P



−P) .

This expression is positive for P < P

, showing that

is a minimum and hence that

the equilibrium state (C =0) is stable. However, if P > P

, we ﬁnd d

/dC

< 0,

is at a maximum and the equilibrium state is unstable.

Other approximating functions

Equation (12.62) predicts a critical force

agreeing with the equilibrium argument of §12.1. In general, we shall ﬁnd that the

Rayleigh-Ritz solution is exact if we are lucky enough to guess exactly the right

function to deﬁne the deformed shape. However, usually we don’t know the exact

shape and must use a function that is more or less physically plausible. In this case,

the method always overestimates the critical force. Unfortunately, this means that the

method is non-conservative (i.e. it is not ‘fail safe’), but nonetheless it is extremely

useful as a way of getting an idea of the magnitude of the critical force,

since the

equilibrium method often leads to intractable equations, except for very simple ge-

ometries.

To examine the effect of using an inexact expression for the deformed shape,

suppose we choose to approximate the deformation of Figure 12.2 (b) in the parabolic

form

u = Cz(L −z) , (12.63)

which satisﬁes the kinematic boundary conditions of the problem, as required in all

Rayleigh-Ritz applications. We then ﬁnd

U =





dz =

(−2C)

dz = 2EIC

L (12.64)

Ω

= −





dz = −

(L −2z)

dz = −

(12.65)

= U +

Ω

= 2EIC

L −



12EI

−P



, (12.66)

and comparison with the previous example shows that the critical force is

12EI

, (12.67)

which overestimates the exact value (12.62) by 22%.

Often an order of magnitude estimate is sufﬁcient, since we recall from §§12.2, 12.3 that it

is important to design compression members to be loaded well below the critical force.

12.7 Energy methods 543

A better approximation

We could get nearer to the correct value as in §3.6.2 by choosing an approximate

shape that gives zero curvature (and hence zero bending moment) at the simple sup-

ports z=0, L. However, we can do almost as well without improving the deformation

guess, by using the equation

U =

2EI

dz (12.68)

for the strain energy in place of (12.64) and writing

M = Pu = PCz(L −z) , (12.69)

which also has the effect of forcing the contribution of the bending energy near the

ends to zero.

Substituting (12.69) into (12.68), we ﬁnd

U =

2EI

(L −z)

dz =

60EI

and hence

= U +

Ω

60EI

−



10EI

−P



corresponding to a critical force

P =

10EI

. (12.70)

This overestimates the exact value by only 1.4% and is a considerable improvement

on that obtained in equation (12.67).

12.7.3 Inhomogeneous problems

The Rayleigh-Ritz method can also be used for inhomogeneous problems, such as

those discussed in §§12.2, 12.3. In this case, the expression for total potential energy

will contain both linear and quadratic terms in the degree of freedom C. The condi-

tion d

/dC = 0 will then yield a non-trival solution for C at all forces, which goes

unbounded as P→P

Example 12.5

The simply supported beam of Figure 12.25 is loaded by a compressive force P and

a lateral force F applied at the mid-point. Use the Rayleigh-Ritz method to estimate

the bending moment at the mid-point.

544 12 Elastic Stability

Figure 12.25

We approximate the displacement in the parabolic form of equation (12.63), in

which case





The potential energy

Ω

of the applied forces contains two terms, one due to the

axial force P and given by (12.65), and the other equal to the work done against the

lateral force F. Thus

Ω

= −

−Fu





= −

−

FCL

The strain energy U is still given by (12.64), so the total potential energy is

= U +

Ω

= 2EIC

L −

−

FCL

The equilibrium condition then yields

= 4EICL −

PCL

−

with solution

C =



4EIL −



16EI



1 −



where P

is given by (12.67).

It follows that





64EI



1 −



and the bending moment at z = L/2 is





+ Pu





FPL

64EI



1 −



which can also be written











1 +



1 −







