THE WHIRLING OF SHAFTS

CHAPTER XXXII

THE WHIRLING OF SHAFTS 345. Definition of Whirling Speed.-When a shaft is rotating in bearings, it is unlikely that its axis will be mathematically straight, for, apart from initial crookedness, the deadweight of the shaft must cause some deflection. Consequently, the geometrical axis of the shaft not coinciding with the axis of rGtation, centrifugal forces will tend to make the shaft deflect further, until they are balanced by the restoring forces arising from the stiffness of the shaft. The speed which just gives balance between the t,vo sets of forces, is called the whirling speed of the shaft. 346. Unloaded Shaft.-Let w be the angular velocity of the shaft, w the weight per unit length, and I the moment of inertia of the cross section, which is assumed to be uniform. Measuring x along the axis of rotation, let y be the deflection of the centre line. Then the centriequation for the strained central line is

2

•

Hence, from § 96, the

0, .

(1)

•

(2)

fugal force, per unit length of shaft, is wyw g

d 4y wyw 2 EI-4 = = - dx g

d- 4y - a 4y 4

or

dx

where

a4

==

ww 2

== -

gEl

The general solution of (1) is : y == A sin ax + B sinh ax + C cos ax + D cosh ax (3) If the shaft runs in bearings which are ver.y short compared with their distance apart, the conditions approximate to free ends. As in the case of a beam carrying a steady load, this requires y == 0 and d 2yjdx 2 == 0 at each end, and, if the origin be taken at one of the bearings, we find C == D == o. H l be the length of the shaft, the equation becomes

y

==

A(OSIn ax -

sinalo h ax)

~ln

502

(4)

THE WHIRLING OF SHAFTS

503

We then find that the condition d 2y/dx 2 = 0 when x = l requires A sin al = o. Hence, either A = 0, or sin al = o. In the former case there is no deflection. In the latter case al = 'lC, 2'lC. • • • The constant A is then indeterminate and a state
al wro2

or

= F'

gEl

which gives

=

(a)

= 'lC

jt4

'2

n;2 .

from (2)

/gEi.

(5)

V ~

This is the whirling speed for a long shaft running in very short bearings, the cross section of the shaft being uniform and of dimensions which are small compared with the distance between the bearings. If the shaft runs in long bearings the conditions approximate to those of "fixed "ends. In this case we find that the critical speed is given by cos al . cosh al = 1. The smallest root of this is al

= 4·73

or approximately

~,

which

gives a critical speed (a)

= 9,!-2 . /gEI

4'2

V

(6)

W

347. Single Concentrated Load on a Light Shaft.-If we have a very light shaft carrying a single load, the dimensions of which are small compared with the length of the shaft, we can find the whirling speed by equating the centrifugal force to the elastic restoring force. Treating the shaft as a light beam carrying a concentrated load W, we have (p. 226), Wa 2b 2 YI = 3EIl' where a and b are the lengths of the two parts into which W divides the length l, and 'HI denotes the deflection under the load, the ends being freely supported. In the case of the rotating shaft, W is replaced by co2

the centrifugal force Wlh g

• Hence in the critical state we mus.t have

'HI or

=

_ CI) -

Wy l co 2a 2b2 3gEIl

. /3gEI'

V

Wa 2b 2

(7)

Let us consider the effect of an initial deflection It at the point of

STRENGTH OF MATERIALS

504

attachment of the load.

At speed co the centrifugal force will be Wco 2 -(Ys + h), g where Y2 is the extra defleotion. For equilibrium we must have 2

Wco - ( Y2 g or

. + h) == restorIng Y2(WC0

2

_

g

f orce f or d efl ec t·Ion y",

3EIl)

a 2b2

== _ Wco 2h

== 3EIl --Y2 2 2 ab

g

- Wco 2h Y2 == . Wco 2 _ 3gEIl a 2b2 When co is less than the value given by (7) the shaft is quite stable; if co is equal to this value the shaft is unstable and Y tends to infinity. But, when co is increased beyond the critical value, the shaft becomes stable again and the deflection approaches the value - h as the speed is increased. The interpretation of this is that the axis of rotation and the e.g. of the weight approach one another as the speed is increased. The arrangement is perfectly safe provided the speed is not kept near the critical value, en passant, long enough for excessive deflection to develop. This is the principle of the flexible shaft used in a de Laval turbine. If the shaft is running in long bearings so as to approximate to the condition of fixed ends, we have Yl we get for the whirling speed

_

w -

Wa 3b3

Wa 2b 2

3 Il3

3EIl

== - E instead of - - , so that

. /3~EIP Wa 3b 3

(8)

V

We have postulated above that the dimensions of the load are small compared with the distance between the bearings. If we did not make this stipulation we should have to examine the effect of the continual changing of the plane of rotation. The general effect is to introduce a gyroscopic torque which stiffens the shaft and increases the whirling speed. This is important in propeller shafts for aeroplanes, the airscrew having considerable stiffening effect. 348. Single Concentrated Load on a Heavy Shaft.-If, in the last problem, the centrifugal forces on the shaft itself are not negligible, we obtain a solution thus : Let the load be at the centre of the shaft, which is of length I, and weight w per unit length. Taking the origin at one end, the general equation (3) on p. 502 still holds for either half of the shaft. H the bearings are short we must have y fore C

2

== 0 and d y2 == 0 at x == 0, and theredx

== D == 0, and the equation reduces to y

== A sin ax + B sinh ax.

THE WH IRLING OF SHAFTS

505

dy

l

dx

2

On account of symmetry we must have - = 0 when x =-, al al + B cosh--...:. 2 2 Hence the equation for y becomes

= A cos -

••. 0

11

=

A (Sin

ax - cos ;~z

sinh cosh2

ax) .

The shearing force is given by EI:; at the centre the centrifugal force on the load W is supported by the shearing forces on each side of it, and these will be equal by symmetry. Hence, if Yo denote the deflection at the centre, we must have

- 2EI[~J dx

That is

2Ela3A

(

cos ~

1=

s--2

WYoW g

2 •

az)

COS~

+ ~z

cosh-

cosh2"

2

Ww 2

=gA

cosa al 2 . lal) S1ll2--hazslnh2 ( cos •

2

or

2 . al al - cos al. al) - 4Ela 3 • cos al alJ = o. A [ WW - - (SIn -cosh -8Inh-eoshg 22 22 22 H A = 0 there is no deflection; therefore, the condition for whirling is that the expression inside the brackets vanishes, that is

tan al _ tanh al = 4Ela g 2 2 Ww 2 3

or

a.l

a.l

4w

tan - - tanh - = - . (9) 2 2 a.W This is the condition for whirling. When the numerical values of w, W, and l are known, we must solve this equation for a by trial and error. The whirling speed is then given by (2). 349. Shaft Subjected to End Thrust.-If the rotating shaft be subjected to an end thrust P, the thrust will increase the deflecting action of the centrifugal forces. K2

STRENGTH OF MATERIALS

506

Referring to Fig. 377, the bending moment at any point B due to the centrifugal forces is obtained by integrating the distributed load

FIG.

t\vice (§ 96, p. 117).

377.

If M 1 denote this bending moment we have

wyw 2 g The bending Inoment due to the end thrust is Py. d 2M 1 dx 2

-

(i)

Hence we have d2 y - EI-2 == Py + M t • dx If the section be uniform, differentiating this twice, we get d4y El + ]:Jd2 y2 + d 2M2 1 == 0 4 dx dx dx Substituting froln (i) we get 2 d4 y ]:J d2 y -dx 4 + -EI -dx 2 - -ww y =0 . . (ii) gEl The solution of this equation is y = aleml~ + a 2em2z + aaem.z + at.em~z where m 1 , m 2 , m a, 1n. are the roots of the equation m4

+

pm EI

2 _

2

ww gEl

=

0

·

The roots of this are

±

vi{-2;/ + vi(2;/Y+;;;;} and ±

vI{ -2;/- vI(2;/Y+;;;}

Thus there are two real roots, and tion of (ii) may be written

y where

= A cos a1x + B sin U1X + 0

0.2= J:_+ V(__I!_)2 +~W2 2EI

t-\VO

2EJ

gEl

and

U2

2

=-

imaginary roots, and the solu-

cosh U2X

+ D sinh U 2X

2;/ + vi(~;/ Y+;;;

(iii)

(iv)

If the bearings are short so as to offer no constraint to the direction

THE WHIRLING OF SHAFTS

507

2

0 and d y2 = 0 when x = 0 and when dx x == l. These conditions lead to the equ~tion (a 1 2 + a2 2 )B sin all = 0. From (iv) we see that al 2 + a2 2 =1= 0, so that we must have, unless B == 0, in which case there is no deflection, all == 1l, 21l. • • •• Thus the lowest speed at which instability will occur is given by a1 2 == 1l 2 /l 2 , or

of the axis, we must have y

P

lEI

+V

=

/( P)2 lEI

ww

2

n2

+ ~EI = p: ·

·

(10)

The effect of eccentricity of end load, concentrated loads, etc., might be taken into account as in the case of struts. 350. Shaft Subjected to End Thrust and Torque.-If the shaft in the last problem be subjected to torque the problem becomes far 11lore complicated; the analysis is long and tedious, but, fortunately, the results are comparatively simple. We shall give here only the outline of the analysis, and the results, obtained by R. V. South,vell. * When the shaft is deflected, the applied torque gives component bending moments, as in the problem of the torsion of a beam (p. 375). ~raking the axis of x along the unstrained axis of the shaft, whilst the axes of y and z are t\VO axes perpendicular to this and revolving with the shaft, the component bending moments due to a torque Tare (p. dy 375) - 'lldz about the axis of z and T about the axis of y. The dx dx component bending moments due to the thrust Pare Py and pz respectively. The distributed loads. per unit length, due to centrifugal forces, ww 2 w 2w . are y and - z , respectively. g

g

Considering flexure in the xy plane, let M 1 be the bending moment due to the lateral forces. Then we have d2y _ El = M 1 Py _ T dz dx 2

+

dx

Differentiating this, we have, if the section be uniform, _ El d3y _ pdy + T d2Z = dM 1 = F 1 dx 3 dx dx 2 dx Differentiating again we get d 4y d 2y d 3z d 2M wc.o 2 - E l - - P- + T _ = _2 l =W 1 = - - y dx 4 dx 2 dx8 dx g 2 2 d4y Td3Z = El + pd y2 _ wro Y . (i) or dx 3 dx 4 dx g Similarly, considering flexure in the plane XZ, we get 2 _ T d3y == El d4z + pd z _ 'lOW~z . (ii) 4 3 dx dx dx~ g

* British A 8sociation Report, 1921.

STRENGTH OF MATERIALS

508 Now multiply

* (ii)

by i (==

v'-

u == y

Then we get

1), add to (i) and let

+ iz.

d4u + iTd3U3 + pd 2u2 _ woo 2U == o. 4 dx dx dx g The solution of this is u == A l ei ,\l X + A 2ei '\2 X + Aaei'\lX + A 4 ei'\,x, El

where

A H A 2 , A,3' A 4

are the roots of

+ TA 3 _

EIA,4

P)"2 _

2

WOO

== o.

g If the bearings are long, so that the axis is held straight by them, the terminal conditions are

or when x == 0 or l. give

== z == o·

dy == dz == 0, 'dx dx du u == 0 and - == 0, dx For brevity let ei'\l~ == u, etc. Then these conditions

y

+ A + A a + A == 0 + A A + AaA a + A == 0 ~AI+~A2+U~3+~A4=0 UIAIA I + U A 2A + U3AaA3 + U 4A4A 4 == o. AlAI

Al

4

2

2

2

)..4

2

4

2

Eliminating AI' A 2 , A a, A 4 in the form of a determinant and expanding we get ~(UIU2

Dividing through by

+ U sU )(A 4

v'UIU 2U aU4,

v-+ v .

/ UIU2

£

Ua U 4

j'U aU4 U1U2

A2 )(A. a - A4)

1 -

. (iii)

=== 0 .

and noticing that

== 2 cos (1\,1 + '}

'} 1\,2

'} -I\,a

'} l -1\,4)-, 2

(iii) becomes ~(A,1

- A,2)(A,3 - )..4) cos (A. I

+ )..2 -

)..3 -

1

)..4) '2 == 0

· (11)

This is the criterion for stability. If the bearings are short, so that no constraint is offered to bending, the terminal conditions are y==z==O;

d2y dz El 2 _ T = 0, since M 1 and y vanish at the ends. dx dx El

d2z dx 2

+ pdy == 0 dx

'

z

• This elegant method of obtaining equations (11) and (12) from (i) and (ii) is due to H. A. Webb.

THE WHIRLING OF SHAFTS

509

These conditions lead to the equation

- As9)(Aa 2

~(A19

-

A,2) cos (AI

+ As -

Aa -

A4)~ =

0

• (12)

for the criterion of stability. From equations (11) and (12) Southwell obtains the diagrams given in Fig. 378, where IT l2p WQ) 2l4 A = - B=- 0=--. 2EI' 4EI' I6EIg 7 6 t----+---+-~Ilr-IlIr~Ilr_-t-

The numbers t!Jttilched to the CIJI'YeS denote Values of ~ 'lonl' bearihls

~r~~ '4.... ~

~

3

~

2

~

Q.5

-4-

5

1·5

t.lr ~~

c...

The numDel's attached to

~~

the curves denote va/I/, .0 of 4t '5hort' bearings

~ 1·0

()

~ ~ "" ~

0·5

-1·5

-1·0

For a further treatment of this subject see the following : S. Dtmkerley, in Phil. Trans. Roy. Soc., 1894. C. Chree, in Phil. Mag., 1904, 1905. A. Morley, Engineering, July 30 and Aug. 13, 1909; Nov. 22 and 29, 1918. W. Kerr, Engineering, Feb. and March, 1916, and correspondence arising from these articles. A. Fage, Engineering, July 20, 1917. R. V.. Southwell, Phil. Mag., March, 1921. J. Morris, Advisory Oommittee for Aeronautics, R. and M., 551 (1918), W. L. Cowley and H. Levy, ditto, R. and M., 485 (1918). G. Greenhill, ditto, R. and M. 560. J. Morrow, Phil. Mag., 1905, 1906, 1907. F. B. Pidduck, Proc. London Math. Bor-., 1920.

510

STRENGTH OF MATERIALS

P. F. 'Yard, Phil. Mag., 1913. C. H. Lees, Phil. Mag., 1919, vol. i. H. A. Webb, Engineering, Nov. 2,9,16,1917, on the Whirling of Shafts of non-uniform section, with graphical methods.

EXAMPLES XXXII 1. Show that the whirling speed of a long overhanging shaft, projecting from a long bearing, is given by cos ale cosh al + 1 = 0, and that the smallest root of this equation is 1·875, the notation being the same as on p. 502. Hence show that the critical speed is (J) = 3·52 vgE1/wlf. 2. Show that the whirling speed of a long shaft with a short bearing at one end, and a long bearing at the other, is given by al = 3·927, or (J) = 15°4 vgEllwZ& 3. A long shaft is carried by three short bearings, the lengths of tne t\VO spans being II and ll. Show that the whirling speed is given by sin a(ll + l2) sin all. sin al2 sinh a(ll + l~) = sinh all.sinh al2· 4. Find the whirling speed of a steel shaft, 6 ft. long between the bearings, which are very short. The diameter of the shaft is 1". Take· E = 30 X 10 6 lbs. lin. 2, and the weight of steel = 0·283 lbs. jin. 3 5. A light shaft 12" long carries, at its centre, a small flywheel weighing 5lbs. The shaft is made of steel (as in Ex. 4) and is iN diameter. Calculate the whirling speed. If the e.g. of the wheel be 0·0 I" from the axis of the shaft, calculate the maximum stress when running at 8 speed equal to 0·95 of the critical speed. The bearings are short. 6. A steel shaft, 1" diameter and 5' 6 long betwoon its short bearings, is subjected to an end t,hrust of 250 lbs. Calculate the critical speed of the shaft. 7. A steel shaft, iN diameter, is 15" long between short bearings, and carries at its centre 8 pulley weighing 30 lbs. Find the whirling speed, allowing for the weight of the shaft. 8. A light shaft runs in two short bearings, one at one end, and one at a distance a from the other end. At the extremity of the overhanging end is a load of weight w. Show that the whirling speed is "/3b'lgllwa 2 , where 1 is the total length of the shaft. (Morris.) N