N o. 18 6 4

TORSION OF CRANKSHAFTS

By S. T im osh en ko ,1 P h ila d e lph ia , P a .Non-Member

In order to apply the dynamics of elastic systems to the designs of engines it is necessary to determine the torsional properties of the crankshaft. In the present paper the author considers the case of a crankshaft with a single throw and establishes the mathematical relations for such a case. He investigates three conditions of such, a crankshaft: (a) no constraint, corresponding to ample clearance in the bearings; (b) complete constraint, corresponding to no clearance in the bearings; and (c) partial constraint, corresponding to ample clearance in the halves of the bearings nearest the web and no clearance in the other halves.

r P H E time has passed when engine design comprises merely the application of statics, and more attention is now paid to the

fact that in various types of engines there is not even an approxi­mation of a state of equilibrium. If there are variations in the torque and in the magnitudes of the force, they are necessarily cyclic, and as such they produce cyclic changes in the state of motion and deformation of the parts. The problem is no longer one of statics, but concerns the dynamics of an elastic system. Variations in the state of motion are associated with inertia forces, and the stresses in the various parts of the engine are no longer due to the actually impressed forces alone. Obviously, calculations which consider only the impressed forces must lead to erroneous results, and frequently a break occurs where the designer thought there was an ample margin of strength. Moreover cyclic changes in the state of deformation of a crankshaft, for example, bring about a cyclic change in the orientation of the various cranks, and the balance of the reciprocating parts, which may exist without such changes, is totally destroyed. With reciprocating engines the cyclic changes in the state of motion, referred to above, concern principally changes of the velocity of rotation. In order to apply the dynamics of

1 Cons. Engr., Vibration Specialty Co.

Presented at the Annual Meeting, New York, N . Y., December 4 to 7, 1922, of T h e Am erican Society of M ech a nica l E n g in e e r s .

653

654 TORSION OF CRANKSHAITS

elastic systems to the type of engine just mentioned, it is necessary to determine the torsional properties of the crankshaft. This is the object of the present paper. This paper deals with the simpler cases and will be followed by other papers on the same subject. In addition to the derivation of the various formulas, it is shown how the amount of constraint affects the torsional properties.

2 On account of the very complex structure of reciprocating engines, the calculation of the torsional vibrations in their crank­shafts is impossible without making some simplifying assumptions; i.e., th a t certain parts, such as shafts, are considered to be elastic, and others, such as flywheels, armatures of generators, etc., are considered to be absolutely rigid. With these assumptions, the engine may be reduced to a system of flywheels situated on a shaft of uniform diameter. This shaft is called the “ equivalent shaft” ; its diameter may be chosen arbitrarily, but its length between each two flywheels must be such tha t it is equivalent as to torsion to the actual shaft between corresponding parts of the actual engine. That is to say, a given twisting moment M must produce the same torsional deformation in both.

3 The length of the equivalent shaft is called the reduced length i0. If we denote by the torsional rigidity C of a shaft, the product of the modulus of shear G into the quantity which measures its resistance to twist, then by elementary mechanics we have for the twist 5 of a straight shaft of length I of whatsoever cross-section (provided it is uniform),

4 In the above, as well as in the following, angles will be expressed in radians and all other units in inches and pounds.

5 For a circular cross-section,(7= ( 7 0 .................................. .... . .[la]

7r •wherein 6 = ^ is the polar moment of inertia of the cross-section.

6 Similarly, we have for the equivalent shaft:

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7 As the foregoing shows, the calculation of the reduced lengths of the equivalent shaft has no difficulty for those portions of shaft in the actual engine which are straight. When the actual shaft is a crankshaft, however, the situation becomes very much more involved, since it becomes necessary to determine the angular deformation 8 brought about by a twisting moment M. Owing to the complex geometric shape of a crankshaft, these calculations can be accomplished approximately only. Various assumptions will have to be made, of which some are only roughly true. I t is assumed that the throw is built up of component shapes, whose deformations are totally independent of each other. Further, it is necessary to make certain assumptions about the nature and amount of con­straint in the bearings. In the present paper only the single-throw crankshaft (Fig. 1) will be considered, and therefore the very con­siderable influence on it of the neighboring cranks will be neglected. Calculations will be made with three degrees of constraint.

DEFINITIONS AND SYMBOLS

8 Fig. 1 shows the conventional type of a single-cylinder two-bearing crankshaft. In addition to the dimensions shown thereon, the following is a list of the definitions and principal symbols used in the paper:

9 Torsional rigidity, generally denoted by C, is the product of the modulus of shear G into the quantity corresponding to its resistance to twist. For a circular cross-section, the latter is the polar moment of inertia of the cross-section in respect to the axis of twist. For a rectangular cross-section, the quantity referred to is more complex. The formula usually used is only approximately true.

10 Flexural rigidity is similarly the product of Young’s modulus into the quantity I corresponding to the resistance of the cross-section against bending. The latter is always the equatorial moment of inertia of the cross-section in respect to the axis of bending.

7TC0 — GOo = g2 o*G is the torsional rigidity of the equivalent

shaft, do being its diameter, lo its length.7rCi = G6 X = diAG is the torsional rigidity of the journal.

7TC2 = Gdi= d^G is the torsional rigidity of the crankpin.

656 TORSION OF CRANKSHAFTS

Ci = GB% is the torsional rigidity of the web in respect to twist around O-O. Owing to the junction of journal and pin to the web, the latter’s cross-section is not clearly defined, but in order to allow for local stresses a t these junctions, a t least in some measure, the assumption is made tha t the cross-section is a rectangle with sides r and c, whence—•

F ig . 1 C o nv en tio n al T y pe op S in g l e -C y lin d eb , T wo-B ea rin g C ra nk sha ft

twist around p-p. The cross-section is rectangular with sides h and c, whence —■

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Trdr ^Fi = is the area of the cross-section of the journal

Fi = is the area of the cross-section of the pin

F3' — h x c is the area of the cross-section of the web taken on 0 -0

Fi = r X c is taken to be the cross-section of the web on line p-p. This quantity is used in the calculation for the deformation due to shear in the plane through p-p per­pendicular to the plane of Fig. 1. Similarly, as in the case of torsion, around 0 - 0 the cross-section is not clearly defined and r x c is used to allow for stresses at the junction of pin and journal to the web.

CASE I NO CONSTRAINT, CORRESPONDING TO AMPLE CLEARANCEIN THE BEARINGS

11 Fig. 2a is a diagrammatic representation pf the throw before deformation with a twisting moment M applied a t the middle cross-section of the journals. Fig. 2b is a side view of the deformed crankshaft. The total twist consists of the sum of the deformations of the portions b (see Fig. 1) of the journals, of the crankpin and of the two webs.

12 All these parts are here subject to torsion only around the axes x-x; x'-x' and 0 -0 , except the webs, which also suffer bending. Now if Si, S2, § 3 are the angles of twist of the above parts, we have

. Mb . Ma s Mh r .~, oi — t — ; »2 — - t t ; o 3 — -zx j-................................. |_4 JOi L/2 t /3

13 Further, the bending of the web produces the angular dis­placement 84, equal to the angle between the tangents to the curve of flexure. Since the bending moment is equal to the twisting moment M and is constant, the curve of flexure of the web is a circle. Considering it a beam of length r, we have —■

s Mr r .5 4 = - ^ - ........................................ [4a]

14 The total relative angular displacement S, measured at the middle section of the journals, will be —

8 = 25i + 6 2 253+ 2 8 4

658 TORSION OF CRANKSHAFTS

F i a . 2 a D ia g r a m m a t ic R e p r e s e n t a t i o n o f S i n g l e - T h r o w C r a n k s h a f t b e f o r e D e f o r m a t i o n

F ig . 2b D ia g ra m m a tic R e p r e s e n t a t i o n o f S id e V iew o f D e fo rm e dC ra nk sha ft

CASE II CONSTRAINT AT BEARINGS COMPLETE, CORRESPONDING TO NO CLEARANCE

16 The deformed crank is shown diagrammatically in Figs. 3a and 3b. The constraint gives rise to a force A and a moment repre­sented by the vector Mi at each of the journals. The forces A and moments Mi act in a plane through the axis of the journals per­pendicular to the plane of the throw.

17 On the assumption that the deformations are small, we derive from the reciprocal theorem that there will be no forces in the plane of the throw, for forces in the plane of the throw cannot produce torsional displacements corresponding to the twisting moment M, and reciprocally M cannot cause displacements, nor can it give rise to forces, in the plane just mentioned.

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18 By symmetry we conclude that the middle point 0 of the crankpin becomes a point of inflection, and if so, there are in this cross-section shearing stresses only. Thus, if we remove half of the throw, as in Fig. 36, the effect of the removed half on the remainder is a single force. And since the remaining portion is in equilibrium,

F ig s . 3a and 36 D iagrammatic R ep r e sen ta tio n o p D efo rm ed C r a n k —C a se II

this force must be equal and opposite to A and its point of application will be at a certain distance k from the axis x-x.

19 Because of the equilibrium there also must be no moments. Taking them round x-x and y-y, this condition requires

Ml = 2h) and Ak = M .................... [6]A20 The angular displacement of the throw measured at the

sections I - I and I I - I I will consist of the following components:1 The torsional displacement of the journals —

4 The displacement due to bending of the webs. In the present case the web is a cantilever fixed at the journal and with a force A acting at a distance k. Taking the length of the web to be r, the angle <V between the tangents to the curve of flexure a t its ends will be —

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A = 0 or k = co we must have the case of no constraint and, indeed, by placing k = oo in [8] we obtain [5]. The effect of the constraint is therefore measured by the quantity k. And in the special case where 2b = a and Ci = ( \ we can immediately compare the case of a complete constraint with no restraint and find that the constraint

decreases the reduced length in the ratio of 1 to ^1 — ■

23 The calculation of k proceeds as follows: Imagine the half of the throw (Fig. 3b) before deformation rotated around x-x through one-half the angle 8 of [7d], and the cross-section at 11-11 thereupon clamped. Under this supposition, the point 0 will be displaced perpendicular to the initial plane of the throw the distance t 8—• If we then apply the force A and the moment M, the displace-

ment just mentioned must vanish, since 0, according to the fore­going, has no displacement perpendicular to the initial plane of the crankshaft.

24 We therefore proceed to calculate the displacements of the point 0 brought about by the various forces and equate their sum to r8J '

25 These displacements are:1 The deflection of 0, or —

^ a3 rg i 24B2 ............................................ L J

due to the flexure of the crankpin in a plane perpendicular to the initial plane of the throw. I t will be noted that

we have here tjie case of a cantilever of lengthA

acted upon by a force A at its end2 The deflection of 0, or —

due to the shearing force A in the crankpin, wherein y is a coefficient taken to be 1.2

3 The displacement of the point 0 —

due to twisting of the web around p-p, brought about

by the force A acting on a lever — • In the above

the length of the web, subject to torsion, has been taken equal to r. This is approximately true in such cases where there is a considerable portion of the web free between the junctions of journal and pin, that is to say,

where ^ is small as compared with r. For heavy

short-stroke engines, the assumption is far from true.4 The displacement of 0 —

662 TORSION OF CRANKSHAFTS

due to bending of the web in the plane through p-p and perpendicular to the plane of Fig. 1.

5 The displacement of 0

and ......................................[M]

due to shear in the web, brought about by the shearing force A.

6 The displacement of O —

due to twist of the web around 0-0.7 The displacement of 0 —

s Mbr rn ,-,r&i = -^ r- ................................................ [9f]

due to the twist of the portion b of the journal.26 Equating the sum of the above displacements to —, we

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27 This equation permits the calculation of k whence l0 follows from [8]. I t is at variance with the equation given by Geiger (Ueber Verdrehungsschwingungen von Wellen), for which reason, and as a check on the above results, the derivation of k and I by means of Castigliano’s theorem has been repeated (see Appendix). This method is particularly useful in more complicated cases where the deformation of the throw is not as easily seen as in the preceding case.

CASJE I I I PARTIAL CONSTRAINT

28 I t is here supposed that there is ample clearance in the halves of the bearings nearest the web, and no clearance in the other half. In this manner we will now have bending, as well as torsion, in the halves of the journal next to the webs, a condition which no doubt prevails in practice.

29 In the present case [7d] remains as before but [10] has to include the bending of the half of the journal. We here obtain —

The ninth to eleventh terms in this equation are due to the bend­ing of the journal, considered as a beam of length b, fixed at the

journal with a force A a t a distance of (a + h)_ gee ^

from the end. The twelfth term of [11] is the deformation a t 0 due to the shear of the journal brought about by the force A.

30 From [6] and [11] we derive —

664 TORSION OF CRANKSHAFTS

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We take ^ = 2.6.(J

The diameter of the equivalent shaft we take equal to the diameter of journal, whence in this example C0 = Ci= C?.

32 With the above we have, when there is no constraint (Case I) from [5]

lo = 37 in.

For the case of a complete constraint (Case II), we obtain from [ 1 0 a ] —

k = 31.6 in.

and since in the present case 2 b = a, the complete constraint re­duces l0 in the ratio —

1 : ( 1 - - s ) = 1 : ( l - ° ' 1 7 4 )

that is, the reduced length is 'reduced by 17.4 per cent.33 In the case of a partial constraint (Case III), we obtain

from [ 1 2 ] —k = 44.2 in.

and the reduced length, as compared with the case of no constraint, is reduced in the ratio —

of 1 : ^ 1 — 2 x 1 ! ^ ) 1 *s’ ky 1^.5 per cent.

34 These examples plainly show the effect of the constraint at the bearings. The more complete it is, the stiffer the shaft and the shorter the reduced length. The calculations further show that an increase in the diameters of journal and pin and an increase in the length of the pin cause an increase of k and thereby a reduction of the effect of the constraint and reduced bearing pressures A. On the other hand, an increase in the thickness of the web brings about a smaller length k and a corresponding augmentation of the bearing reactions.

666 TORSION OF CRANKSHAFTS

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In the above we consider M as given and A as the variable to be determined. On the supposition, then, of a complete constraint at the bearings, the force A at its point of application brings about no displacement, whence by Castigliano’s theorem —

e v a(y.+ 7.+ . . . Vs) dA dA

Substituting and executing the differentiation, we obtain —

ar r*_ Ar

A = M _________________g _________________________r (a + h)2 ar2 . a3 2r3 hr2 y a 2 yr yh

2 C>' Ci 12jB2 3jBj 2Cj FtG Ft'Q F>0

And remembering that M = Ah, we see that the above result is identical with [1 0 a].

37 For the determination of d we have the equation —

dV d (F ,+ V2+ . . . F„) s dM dM

But remembering that now A is a function of M we have —

d V = d V d V dA _ . dM dM dA dM

dVin which is the partial derivative of V in respect to M . Since, however,

dv n i,0 we have —dAd V _ dV _ - dM dM

Hence 5 is equal to the partial derivative of V in respect to M.38 Executing the differentiation, we have —

2 ( M - Ar)a 2M r - Ar2 2 ( A f“ 2 ) A 2Mb ,2 C, + B 3 + Ca + C, 6

Mand with A — this equation becomes identical with [7d].

39 In another paper the author proposes to discuss the two- and three- cylinder two-bearing crankshaft, the multi-throw crankshaft, and the effect of clearances in the bearings.