The Torsion and Stretching of Spiral Rods

(1)

The Torsion and Stretching of Spiral Rods (III) 1

H. oKuBO 2

Synopsis

The equations of equilibrium with those of compatibility are derived in two-dimensional forms for the present problem. They are simplified in two extreme cases where (1) The angle between the helix and the axis of helix is small, and (2) The angle bweteen the helix and the plane perpendicular to the axis

of helix is small. Using them, expressions for stresses are derived in the forms, which contain two arbitrary functions satisfying the Laplace or a similar equation, and the Airy equation, respectively. As an illustration of the procedure,

a detailed solution for a coiled helical spring with a circular section is obtained

by means of successive approximations.

Introduction

The torsion and stretching problems of spiral rods were treated in earlier papers M. The present paper is a continuation of the previous studies. In the previous calculations, the equations of equilibrium expressed in terms of displace-ments were used, but in this paper the equations of equilibrium expressed in terms

of stresses with those of compatibility are used, since they are more convenient for analysis in the case where the angle between the helix and the plane perpen-dicular to the axis of helix is small.

Equations of Equilibrium

If there are no body forces, the equations of equilibrium can be presented in

the forms Unlvaraity of Technology HvzIewn:^Ltonlos Lzboratori LiLrzry Makelweg 2 - 2628 CD Delft The Netherlands

Phone; 31 15 7a73-Fax: 31 15 781836

Rep. No. 82 (in the European language) of the Institute of High Speed Mechanics. The paper was published in the Journal of Applied Mechanics, AS M E, Vol. 20

(1953), p. 273.

Formerly Professor of the Institute of High Speed Mechanics.

Cl The numbers in the brackets refer to the Bibliography at the end of the paper.

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.and the six conditions of compatibility are

kD (0.,) =0

We shall take the axis of helix as the axis of z and rotate x, y axes with the

increase of z. For any assigned value of z, the directions of the new axes x', y' are given by the equation

x iy efkz (x' _{( 3 )}

where k is a constant which specifies the inclination of the helix. Assuming that

all stresses, referred to the new co-ordinates, are independent of z at every point in a plane perpendicular to z-axis, we have the equations of equilibrium and the

conditions of compatibility in the forms

(1+v)

V--(1+ v) 0 0 _{( 2 )} 0, (1+v)V2az + ex2 = 0 Oyez (1+ v)V2a, + ey2 ' + exez 620 (1 v) Wrz, 0 = 620 0 0, (1+ v)V2a, + az'

where v is the Poisson ratio, and

82 6 02

\ 7 2

exey

a, 4- az =

Ox- By= ez2

gr.,/

eas f 0

ex fly Oz

ea, firm Eh-,

0 _{( 1)} Ely ecr, Oz es-zz

-ex OtZ 0 Oz ex ay kD(r.'z) +

-+ ex' By' flay' _kD(rv'z)

_kri,

4 ) 6y, _ex' + + + + + = .. ..

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H. oKUBO : The Torsion and Stretching of Spiral Rods (III)

820

Vi2o2 + leD2 (as') 4 k'D (r.' n') 2 lea.' + 2 /ea; 1

1 axt2

1 8=0

Wan' + + 4 k=D + 2 k=a.' 2 k=a; + = 0

1+ v Oy'2

1 8'0

+ k2E (r.'1) + 2 leD (az') 2 k2D (an') 4 k2

+al

0

1+1, xia

Vi,az + Pp (az) + D= (0) = 0

1+v

80

Vi=rv'z + (rut.) + 2 leD -r D( 00

ex' J = 65/' k r-D( a0 80 0 V12Zz /OR= (r.,c) - 2 le D (rIz) k2rz \v L ax' ay' J where a" 62 8 8 7,2 -

+, D = y'

x'

axi= fV2 ax' ay'

D2 = (y'

x'±-)(y'

ay' Bx' ay'

The stresses which satisfy the differential Equations ( 4 ) and ( 5 ) are the functions of x' and y'. Hence, if the boundary conditions be given in forms in-dependent of z, we can reduce the problem to a two-dimensional one.

When the angle between the helix and the axis of z is small, k becomes small,

and for torsion all stresses except -1-:Z2, z-,/, become small, since these stresses vanish for a straight rod. Hence, if we neglect small quantities of the second order, Equations, ( 4 ) and ( 5 ) reduce to

8r.', arn'. Ox' -r ay' 5 8a.' r.'n' kD (r.'n) krnr. = 0 kD (ry',) + krlz = 0 ( 6 ) 8x' Oa,' ay' az-z' + ax' +

(a')

(r.' + ) + + =

(4)

1 020 j'az' - o Ox '2 1 629 = 0 1+v Oy''

71"

Vi2a, -- 0 = = 0

The third Equation ( 6 ) with the last Equation ( 7) are the ones for the torsion problem of a straight rod. The remaining equations are the ones for a plane

problem. Hence the stresses which satisfy Equations ( 6 ) and ( 7 ) can be represented

in the forms 020 a _{- 2 Gk (Cf} _Cf) Oy'' 620

all

-= vV1-=c6 + 2 Gk (Cf CI) C 20 . = 2Gk (cf - if) Ox Oy' 1 020 --0 1--v ex' ey' r.rz = G Ci (f-f) - (KY)

r,', = - G

f - ax')

where G is the modulus of rigidity, C, a are constants, and V 140 = 0,

f

f (c), 7 =

_(c), c x' + 1, c

The arbitrary functions f, and the unknown constants C, a contained in the expressions of the stresses are determined from the boundary conditions by a usual way, if the shape of the section be specified, but as the further analysis is

quite similar to that already shown in the previous papers, it will be omitted here.

It is seen from the character of the problem that the stresses at every point on the spiral AB, Fig. 1, are the same. Hence the expressions for stresses at the points A and B, referred to the cylindrical polar co-ordinates, are the same.

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(8)

v + . = =

(5)

H. -OKuB00: The Torsion and Stretching of-Spiral Rods (III) Remembering the relation that 0 =r -1kz, let Us :make

the transformation of co-ordinates; then. ;Equations ) and ( 5 ) reduce to

an, ero 1 698

+

rre

= 0

Or Oz kr Oz

ea, Or.. 1 81.-,o

+ fir az r kr az en z ea: Trz 1 870z ar az r kr Oz 1 02ar

2,

4 erre 1 020' n V129, +

k2e

ir a"

± kr2 az 1+z, ar=

=0 1 Ercre 2

+ (ar-cro)

lee az= r2

(9)

4 Or,o 1 ( 1 0=01 1 00 kr2 az -1- 1-ry k2r2 0z2 r Or 02r,.0. 4 2 aar 000 +

k2e2 az2 IzY2

Oz - az

-1 (020 1 00 ) 0 (1 +v)kreraz r ez

-1 026, 1 020 -\7126. + az= + - = 0 Fer2 1+1, 1 82roi 2

er,

1 020 \71.2roz + re, 0 ,k2r2 022 Oz

(l+v)kr

az= 1 02rTz 2 are, 1 1 020 WIrrt+ + k=r2 az= +kr2 r2 (1+v)0r8z= o where 02 1

062

.0 =: + (re .+ q2 V1 = + 64,

r

Or az=

Equations-( 9) and ( 10 ) are two-dimensional ones referred to r and z.. When

k is small, it is convenient for calculations to specify the shape

of the rod by a

section perpendicular to the axis of the spiral,. but when k is large it becomes favorable to specify the shape of a section cut by an axial' plane. Equations ( 4 ),

( 5 ) become inconvenient in the latter case and Equations ( 9 ), (10) are mote

useful. , -25 (10) ... -( 4 1 2 0 1

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When the angle between the helix and the plane perpendicular to the z-axis

is small, k is large and all stresses except rro, ro become small, since these stresses

vanish in the extreme case where k is infinity.

Neglecting small quantities of the higher order, we have erra era, 2

rro = 0

Br Oz

aar Or" Or - 0 1 Or,o

Ur az r kr az

Orr, ea, 1 1 eroz

Or Oz r rr" kr Oz 4

rra = 0

r-1 7i2roz --roz 0 2 1 09O 4 fh-ro 71.2ar - ± 1±i fi ± kr' az 2 , , 1

180

4 6,-5., v12a0 + r Br kr= ez 1 eV + 1+v ez2 1 ,

1 eV2 aro,

0 7,72. ' 1+verez _ez

Let us take for the shearing stresses rro, ro, the expressions

1 eo, Tr, = Y2 az

=0

(12) (13)

Then, the first Equation (11) and the first and second Equations (12) are satisfied, if 0, satisfies the equation

8'01 _{_} 3 001 e2oi _{= c}

Or2 r Or Oz, (14)

in which C is a constant.

Substituting the expressions for the shearing stresses Equations (13), into the

+ +

0

1

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H. OKUBO : The Torsion and Stretching of Spiral Rods (III) 27

remaining Equations (11) and (12), we have six equations for the other stresses.

Consider now a stress system

arm Ore 7ri 1 12, Of;,21) 41'w, ( 2,7 cct-: dz dz) coo) =

101 1

,90( \ 2 El 1 d

dz z)

log r

kr Or

r

8r er r Or ke 0.,(o)

_1

1 8 2(1+v) 80, 64,0i (1-2v)kr Or r Or 822 1 e,o, 1 0 r 2(1+v) 801 1 d2:1 (1-2v)kr 02" kr2 Or 1 21, Oz

r

Or

We may show that the stresses, Equations (15), satisfy the remaning six equations. Remembering the relation, Equation (14), we have

edr(0) ar(0) 0.,(0) ₁ _040, ₁ ₀ ₁ ₀₀₁ ₁ _001. _{1 8201} Or (1-2v)kr ar'ez' 7-- kr= Or \ r Or I

ke arez'

kri az'

6lprz(°' 1 _{62 ( 62oi} 3 asbi.

1 0( 1

001 1 iPoi

ez (1-2v)kr 02.2\ 8z2 r Or 1

kr' Or\ r

Or I kr erez2

Hence it becomes

_coo) 1 8' 8201 3 80, ±a20, \ 1 0201

Or Oz r (1 -2v)kr az,\ er2 r Or Oz- ez'

1 820i

1 ar

/7--e

az' --Te-r Oz

We have thus proved that the stress system satisfies the second Equation (11), and in a similar way we may show that the stress system satisfies the other five equations. Accordingly, we can take for the remaining stresses the expressions

ar( 0 _{v I c,k!}

or

(

O2)

8802

= (76,(0) + Oz(vV1202 r Or

,

(0)

{ (2 v) _Vi.202 0202} uz-az 6.20,1

7,

_{{ (1 V)}_V1'02 er (16) o-1 g r (15) ke 1 1 r r cc +

(8)

provided that

Vi2 0 (17)

The expression of the stresses, Equations (13) and (16), is the counterpart of Equations ( 8 ) for the extreme case of large k.

If the equation

F (r,z) = 0 (18)

represents the bounding curve of the axial section of the rod, the condition that the surface of the rod is free from traction is

Consider a spiral rod submitted to a pair of axial forces P, as shown in Fig. 2. Then, the further condition is that the resultants of ri0 and so taken over the section and also the bending moment due to so

vanish, and

r oz dr dz = P (20)

If may be seen from Equations (11), (12), (19) and (20), the shearing stresses rro, 1-.0 can be

de-termined quite independently from other stresses,

and these shearing stresses are the ones for a ring

sector submitted to a pair of axial forces. If these stresses be obtained, substituting them into the remaining Equations (11), (12) and (19), we can reduce the problem

to the one for an axially symmetrical stress distribution. The procedure for this case is similar to that for the case of small h, already shown.

Application of Procedure

As an illustration of the procedure, we consider a heavy helical spring with a circular section, taking the helical angle into consideration. We denote the

OF OF OF OF ₀ (19)

DOE

Or OF rrz -73T OF -= Oz 1

arr

OF kr 1 Oz OF rrz Or cz Oz

krr'

Oz = 0 + Or

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H. 6KUBO : The Torsion and Stretching of Spiral Rods (III) 29

radius and the pitch of the coils by R and s,

respectively, and the radius of the wire by a, as is shown in Fig. 3. A case like this has been treated

by several previous authors, though they neg-lected the effect of the helical angle. Those

results, however, immediately can be applicable

to our calculations. We shall take the approxi-mate solution, worked out by G-Aner [2), and proceed with our further calculations by means

of successive approximations similar to his method of approach. The third ap-proximation for the stresses in a ring sector, obtained by him, is

2PRr 5 EC C

16R2 (27 + 5 C2. 10 a,)]

yr° = 7ra4 Lc

4 R

with the condition that

Fig. 3.

2PR r

7r

3 13 E2 9 EC 2 1 (r2

re° 7ra4 + a2) ./-7-2 J

where r = R+ e, z = C.

Substituting the stresses, Equations (21), into Equations (11), (12) and (19),

and assuming Eli? as a small quantity compared to unity, and using the expansions (21)

we shall solve the differential equations by means of successive approximations.

As a first approximation, we take the equations of equilibrium and the

con-ditions of compatibility in the forms

acir(i) _2P arrzo., 0z(i)

(23) ac 1

ay.

7rka4 0 0 8E' ac Vi2o-00.) 0 V 21-7zo.) 1 av(i) 0 1-- v 1 9$2 8'0(1) _ 0 1 + v ' 1v BE 8C 1 1 ( r R

11 (

R 2E R E2 R2 3E2 122 (22) . +

(10)

2P 2P

Ear"' + Cr,") C' 0, E r rz") + Ca,u) + EC 0

7rka4 zka4

on the bounding circle + C2 =a2, in which

02 0' _{oc.) = ar(.) + ao (70} _{4_ czoo,}

ec,

As a first approximation, we take the stresses

2P 4P

= coo.) 7...(1)

n-kal ' n.ka4

which satisfy Equations (23) with the boundary condition, Equations (24).

To get the the second approximation, we shall find a stress system which

satisfies the equations

ear") er,z(2' 1 9 Pe ((7(2)

,1))

_{= 0} eC 2 zka4R a,,(2) _{3 PC} ₀

0$ ±

OC R `" 27rka'R 1 620") 1 80r(1) 8P ,20.,,(2) ₀ 1 Be' R 6$ 7rka4R n = 1,2, 1 eaow 1 Mu> 8P V12a0(2) = 0 R OE 1 + v)R 0 nlea4R 1 020(2) 1 61a(" 7,20.r(2) 0 1±L1 8C2 R 8$ 1 620") 1 6rrz'1)

i2,,,(2)

0 1 + tn OC R and also the boundary condition

9 PeC2 E0r(2)

cr2)

2 zka4R. 0 PC $r,(2) + Ca,(2) 47rka41? (15 $2 3 c2 + 3 a2) = 0

Substituting the stresses for the first approximation into Equations (26), and

remembering the condition that the resultant of ao taken over the section vanishes, we have a stress system satisfying the differential Equations (26) with the boundary

= 0, + + + - 1 + + + (25) (26); -(27)

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"

H. oKUBO : The Torsion and Stretching of Spiral Rods (III) 31.

tradition Equations (27), aS,

ci,co 4ka4 R (1+v)C(7 6rv) + (21 +

18v)2-

(7 + 6)

a)'

v 60(2) 2 ka4R0.+v) CO.+53V 41/2) E2 (11 ± 5v - 40) C2 (3 + 2) a2)

z

0.z(2) _ - C(17 47rka4R(l+v + 20v) - (5 + 4tv) C2 + (5+ 4v) al) P(2+3v) C-27rka4 R(1+v)

For the third approximation, we consider A stress system satisfYifig the

dif-ferential equations Ovr(3) +_ + cl ) err,(3) 1 E o ) ti) you)) ae ac R (63E2 +115 C2.= -10 a2) 8,n-ka4 R2 r,(3) Bago) 1 _rrz(2> E

_rrto

15 P ac R R2 4 7rka4 R2 1 6219(2) 1 862.(2) e1d,-03 2 V (a <1)_ 60(11) 1?' arm + 1+zi 6$2 R RE 82 Re R2 -26 P k a4 R1E-10 1 00.80)

V12 a6(3) I a0(2) + 2(a,-(1)' - cso"))

R OE + (1+v) R RE R2 acect) _ $ e0") 26 P R2 BE R-2 OE + 7--z-, k R,

- 9

1 ova) 1 86r.t2) eaz(i) 712 aim + 1+ v 8C2 R 6E

n

1 1920(3)" 1 Orrz(2, E en.,(0 Drza> 712 prz(1) + 1+ v OC OE R OE

n

3P C k a4 R2 (28) + e2 + + (ar(2) + 2 (29)

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32 Rep. Inst. High Sp. Mech., Japan, Vol. 9 (1958), No. 82

and he bounclarY C:ofidition

P t2 car(2) + Crrz(2)' -nka4R2 (63 E2 5 C2 - 10a) = 0 8 - (30) P $ C $.Fzzca) + Cato) + 8 T. k R2' (43 c2 1 10 a2) P

-Substituting the stresses, Euatiolis (25) and (28) 'into (29), and integrating the differential equations we have a third stress system. The stress system, satisfying the boundary condition,. Equations (30), with the condition that the bending moment due to vanishes, becomes

Pe

qz("

_{6 ka4(1-1-)}

n-1e

( (20' + 21v + 4v2) (E2 = a2) + 6 (11 + 10v) V),

Fe coo) ---6 zka4 R2 (l+v) [ (11 + 17v+ 16± 40) f2 6 (13-1-8v 6V2 - 21)3) C2 - :4121. (37 + 33v + 4v2) P 0-z(3) = 12 7tka4' R2 (1 + v ) ((43 + 57v + 8) - 39 (1+ v)i C2 + 24(1 + v) a2)

Pt

z,C2) = 24 zka4 (1+v) (3(1 +3z) $2 + (13 =-,9v 16v2),C 2(14 + - 8V2) a2)

By using the stress systems in 'Equations (25), (28) and (31), we can find the stresses for the third approximation, from the relation that

= aro.) 0,:r(2). arp), ,010 croci) coo.) ± cop)

aen + ai(2) + az(3),, Trz") rrz(" rrzco" )

Among the stresses, Equatioris (32), az. and are predoMinating,4 and 'they are

2P [(

a2 17 + 20v2 5 + 4v a2 - C2 2 + 2rka4 R2 ) 8 (1+v) R 8 (l+v) R 43 + 57p + 8v2 ES ES 13 C..2 24(1+v) R2 8 R2 J (31) II

t

+ = a, R2 + ₌ (32)

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H. OKUBO : The Torsion and Stretching of Spiral Rods (III) 33

2P r/i

(14 + 3v 8v2)a2 2 + 3v C C

r"

7rka4 24 (14-v) R2 4 (1+v) R

1 + 3v 62 C 13 9v 161/2 C3

16(1+i) R2

48(1±v) R2J

For R 5a, the stresses a and r, on the axes of C and C are calculated from Equations (33), and are shown in Figs. 4 and 5, v being assumed as 0.3. The

normal stress az attains its maximum at the inner end of the horizontal diameter (C = a, c = 0), the amount of which is

Ps 11 12v a 67 + 81v + 8v2 a'

(az)rnacr

={2 +

(34)

7r2a3 4 (1±v) R 24(1+v) R2f

The shearing stress r, attains its maximum amount of

(T.rz)max = Ps (

5 a2

1 - (35)

7r2a3 16R2

Fig. 4. az and r, on the s-Axis,

when R = 5a

Fig. 5. a r, on the C-Axis, when R = 5a

(33) =

(14)

at both ends of the vertical diameter = 0, C = a).

The greatest stress occurring in the rod is the shearing stress ro., at the inner end of the horizontal diameter where a: becomes maximum. From Equation (21)

it becomes 2 P

_R(

1 , 5 a 17a2 (oz)ma z _na, _{4 R} (36) 16 R2

The maximum values of az and rrz, calculated from Equations (34) and (35), for several values of alR, are shown in Table 1, s being assumed as 3a. For

convenience of comparison, the corresponding values of the greatest stress are also given in the same table (unit 2P/7r a2).

Table 1. Maximum values of az, r, and roz

Acknowledgment

A grant for science research has been given for this stuhy by the Education

Ministry of Japan.

Bibliography

( 1 ) H. Dkubo, The Torsion and Stretching of Spiral Rods, ( I ), ( II ),

Quarterly of Applied Mathematics, Vol. 9 (1951), p. 263, Vol. 11 (1954), p. 488.

2 0. Gohner, Schubspannungsverteilung im Querschnitt einer

Schrauben-feder, Ingenieur Archiv, Bd. 1 (1930), S. 619.

R/a 4 5 6 8

(cr:),,,,ct. 1.38 1.28 1.22 1.14

errz),,,,. 0.47 0.47 0.47 0.48

(roz),... 5.52 6.46 7.43 9.38