Stress fields for an in-plane end shear force acting on a 90° elbow or pipe bend

Int. J. Pres. Ves. & Piping 16 (1984) 263-284

Stress Fields for an In-plane End Shear Force Acting on a 90 o Elbow or Pipe Bend

H. A.

Lang

LANG--Research West, 1201 Idaho Avenue, Santa Monica, CA 90403, USA

(Received: 21 August, 1983)

ABSTRACT This paper develops the stress fields for a 90 ° elbow or pipe bend acted upon by an end shear force, IV, which produces deformation primarily in the plane of the elbow. There are coupled fields of stress because the elbow is held in equilibrium by an end normal force, N, and by a bending moment, M = WR. The solution makes use of a previous result for pure bending of an elbow. The methods used are those of toroidal elasticity which requires the development of certain equilibrium functions N( i) and the development of certain compatibility functions V(i). These are listed in Appendices. There appear to be a total of eight fields of stress. Three initial fields of stress generate .five first-order .fields of stress when the method of successive approximation is applied to the general theory of toroidal elasticity.

NOMENCLATURE r, q~, 0 Or, tT~, o"0

Trq~, TrS, T~bO Trz, a

b

T~z, Gz

Toroidal coordinates. Normal stresses. Shear stresses. Stresses in a cantilever beam. Internal radius of cross-section. External radius of cross-section.

263 Int. J. Pres. Ves. & Piping 0308-0161/84/$03.00 © Elsevier Applied Science Publishers Ltd, England, 1984. Printed in Great Britain

264

H. A. Lang

I K Ko M N N(i) R s V(i) W • , fl, K, 6~, 6z

Moment of inertia. WRZ/I. K/x. Bending moment. Normal force. Equilibrium functions. Toroidal radius. r/R. Compatibility functions. Shear force. Stress parameters.

INTRODUCTION This paper considers one of the ten unit problems of the 90 o elbow. An end shear force, shown in Fig. 2, acts inwardly in the load plane, 0 = 0. The elbow is in equilibrium under this force, IV, and an equal normal

I

;z

L

t

x

W

x*W

" 'rrz

Cross Section

Fig. 1.

End-loaded shear force acting on a cantilever beam of circular cross-section.

Stress.fields for in-plane end shear force on elbow

265

force acting in the plane, 0 = 90 °. Equilibrium also requires a bending m o m e n t WR at 0 = 90 °. An initial field of stress generates, in part, stress z,0 and -r,0 as shown. To determine the first-order stress field, an initial state of stress is introduced. This state of stress is an extension and adaptation of the stress field in a straight cantilever beam acted upon by an end shear force (see Fig. 1), as given by Love. 1 Additional initial stress states must be assumed to satisfy the boundary conditions and toroidal geometry of the elbow. The methods used depend on the general theory of toroidal elasticity first introduced by Lang. 2 The general theory is difficult to use directly for solving boundary value problems. It is necessary to expand the general theory in a series in 1/R (where R is the toroidal radius) and apply the method of successive approximations. This procedure yields the zero-order equations and the first-order equations used in the present paper. There is a coupling of the shear like stress field in the plane 0 = 0 and the normal like stress field in the plane 0 = 90 °. This coupling is discussed in detail in the paper. The determination of the first-order stress field also requires that we

\

M = - W R

=-W

Fig. 2.

N o t a t i o n for 90 ° elbow acted o n by a n end shear force in the plane 0 = 0.

H. A. Lang

266

add a state of stress due to end bending moments to the other fields of stress which arise. The solution for end bending moments (another unit problem of the elbow) was determined in two earlier papers. 3,4 The results are reproduced in their entirety in the present analysis. In applying the method of successive approximations, terms appear on the right-hand side of the stress equilibrium and stress compatibility equations. We call these equilibrium functions, N(i), and compatibility functions, V(i). The required functions are completely developed in Appendices I to 5.

INITIAL STRESS F I E L D F O R Zro A N D ~,0 To develop the initial stress field for stresses Zro and Z~o, we begin with a cantilever beam of circular cross-section acted upon by an end shear force W(as shown in Fig. 1). From Love 1 the stresses on any cross-section are:

Z,z -

W Fc~7. c 2-v 2(1-~--v)/ L~r + °s 4,(~ vx2 + ~ y

2~ • J +sin 4,(2 +v)xy]

%z =

2(1 + v)I L r &k + cos 4,(2 + v)xy- sin 4,

vx 2 +

y2

where the function X is

7. = - ( ~ ) b 2 r c o s 4 , - ( 3 - - - W - ? - ) a 2 r c o s 4 , + ~ r 2 c o s 3 4 , -(312~v)a2b2c°S4,r

Using x = r cos 4,, y = r sin 4,, the stresses become

%~-

2(1+v)I

z4'z-

2 ( wR~ 1+v)I

[( )cos (s2 sin 4' s£ + ~ + ~ / -

s 2 ].J

( 1 - 2 v ) s 2sin4,

(1)

267

S t r e s s f i e l d s f o r in-plane e n d s h e a r f o r c e on e l b o w

The (downward) force is Fx =

(T,z COS 4) -- T,z sin 4))r dr d R = W .)ado

In addition, there is bending stress

W(L - z)x ¢rz =

]

We generalize this result by replacing x by r cos 4) and (L - z) by p sin 0 (where p = R + r cos 4)) to obtain

WR z sin 0 a0 =

I

[s c o s 4)(I + s c o s 4))1

At 0 = 90 °, we find N = Rz

eros ds d4) = - W a

M = R3

f'f? a

(eoS cos 4))s ds d4) = - WR

**

The assumed expression for o"0 thus possesses the correct resultant force and the correct resultant moment. The transition of stresses from the cantilever of Fig. 1 to the hollow toroidal elbow of Fig. 2 is made by identifying the stresses according to the scheme: Trz "~" TrO "C~bz ---+ T~O ff z ~ ¢70

We next modify the stresses rr0 and z¢0 by adding the (respective) stress fields

WR z 2(1 + v)I

s2 j j c o s

4) (2)

W Rv)I2 I,s2( +

2(1

s.sb + +7-..:j 2 z ~l sin 4)

H.A. Lang

268

The (strain) compatibility conditions for strains l~o and 1~,o can be expressed by the single equation

Oz(oo Os + zee° s

1 3Z,o s &k

- 0

and this reduces to

WR 2 [2vs + (3fl - a)s] sin 4~= 0 2(1 + v)I We satisfy this by imposing the condition (3fl - ~) = - 2v The (inward) force is

Fx =

i I?

(z,0 cos ~b- z~,o sin ~b)r dr d~b

With the added stress fields, this becomes

Fx = R2

WR 2 v,°do L" 2 ( l + v ) I

+RZ fS" f]~ I .

3 + 2v 4

e (sZ--(S2.+S~))COS2 49 sdsd(o

WR2 1~{3+2v

2(i

;v)i_][\-

4

~

)(

s2"s~'~

s"2 + s ~ + s2 j

+ [fl -- ~1 -- 2v)]s:} sin E4~s ds dq~ Performing the integration with respect to ~b yields Fx-

2(1+v)I

.

4

WR'*u fsSb[(3 + 2v +2(1 + v)I . 4

a)(s~ + s ~ ) + [ f l - ~ l - 2 v ) ] s 2 1 s d s

Fx-2(iT;3)I F(3+42v ~)(3(s.~_s4))+ W - ~) 2 (v1+

[2(1 + v) - 3~ +

fl]

fl (s'~ -4 s~)

(1 - 2v) (s~ -

16

s.4'1

,j

S t r e s s fields f o r in-plane end shear f o r c e on e l b o w

269

The last modification of the stresses z,o and z~o consists of replacing the factor 2(1 + v) by x where r = [2(1 + v) - 3ct + ill) Thus, finally TrO - -

WR2

cos~bcos0

61 s 2 - ( s 2 + s 2 ) +

xI (3)

z"R xl

Z 4~0 - -

-6 s + 6,

+ sg) +SaSh]] sin cos0 s 2 JJ

where

The (inward) force F x = 2 W. We choose fl = 3~ so that x = 1 + v, one-half the value for the cantilever beam. F r o m the condition 3 f l - c t = - 2 v we have ~t = - v/4 and # = - 3v/4. Then 61 = ~ 1 + v )

and

52 = ~ 1 + v )

and

6 2 __

F o r future use, we note that 6 1 __

~c

The initial stress,

3 4

x

1 4

a o

The stress tr0 consists of two terms WR 2

tr° =

I

WR 2

s cos ~bsin 0

- -I

S 2 COS 2 t~

sin 0

(4)

Comparing with Appendix 2, the first term corresponds to sinusoidally varying pure bending of a m o u n t WR 2 cER = - - I

H. A. Lang

270

The associated stress, satisfying the boundary conditions, m a y be obtained from the stress field of pure bending by simply multiplying by sin 0 (see Appendix 3). For the second term of go, we list the equilibrium and compatibility functions in Appendix 4.

Additional shear stress due to sinusoidally varying bending Bending, varying as sin 0, produces an additional shear field of stress. The equilibrium equation is:

1~

s ~s (~r0s) +

10Z~o

WR 2 = - - s cos 4~cos 0 s ~b I

The compatibility equations for these stresses are

Z~o

V2T'rO -~ V°2%° -

2 ~Z4,o_

S20d~

WR 2 I 0 -+ v) cos ~bcos 0

s• ~ _+~ _ ~3~o_

I(1WR + 2v------)sin ~bcos 0

A solution can be found in the form

(~o11=@--\ 1 + ~/L -(s~+s~)+

cos~cos0

W R 2 [ ( 1 - 2 V ~ s 2 - / 3 + 2 v \ / I | / 2 + s 2 ) + @g )-] sin O cos 0 (5) L,,,,+,,/ ~ , ' T - ~ / ~ (Sa S2 /]

(~0) 1 = - 81

This stress field has a resultant shear force (inward) of a m o u n t F = - Wcos 0 INITIAL NORMAL FORCE FIELD We assume an initial field of stress expressed by

a r = a o sin OIs2-(S2a + S2) + sz~s~]

[

a~ = sin 0 bo s2 - ao(S 2 + s 2) - a o - 7 - 3 a 0 = sin 0(2o + 2o)(S2 +sg) + sin 0(21 + 20s 2

(6)

271

Stress fields for in-plane end shear force on elbow

The strain compatibility conditions may be satisfied by (3 +v)b o - (1 +3v)a o-2v(21 + 2 a ) = 0 and "~'1 "[- ~L1 ~-- V(ao +

bo)

From these two equations we find the useful identity (3 - 2v)bo = (1 + 2V)ao The equilibrium and compatibility functions derived from this stress field are listed in Appendix 5. Because of the coupled stress fields, it is necessary to introduce two terms in the expression for a o. This will be made evident in the subsequent analysis. Shear stress independent of

There is a state of stress, independent of ~b, and given by

s~?Os(Z~oS)+ 1 c~Zeo- c o s O[Ko(61 + 62)s 2 -2Ko61(s 2 +s2)] s ~4~

-[-WR2s2

+ cos o[_- ~

~xs ~ - ~o(S~ + s~)

]

with compatibility equations

WR2s V2ozro z~o 2~Z¢o $2 S2 COt~ = COS 0 [_K_~ (361 -- 62) -+ I(1 +v) V~z, °

Z,o

2 dZ,o

- - 7 + s 2 &k - 0

We assume = do

t

_2S2 \

0

(7)

Z¢o = 0 and find

WR 2 4do = K o ( 6 1 + 6 2 )4

21

)~1

- 2 d o = - 2Ko61 - 2o

8do

WR 2

Ko (361 - 62) -~ =5I(1 + v)

221 v

H.A. Lang

272

The solution (for f-l, do, 7-o, ao a n d bo) is

WR2v 2

f-1--

Ko61V 4(1+v)

2I(1 -- v z)

5Ko62V 4(1-v)

WR 2 Ko do = 8I(1 - v z) + 16(1 + v) [(4 - 3v)61 + (4 + V)6z] WR 2 Ko f-o = 41(1 - v 2) + -8(1- -- v) [ ( 1 3 v - 12)31 + ( 4 + V)6z] a° =

(3 - 2v) (21 + f-l) 4v

b°-

1 +2v 4v (21 + 2 1 )

(8)

The value of 2 a will be d e t e r m i n e d later. Field of stress in (cos 24)/sin 24))

There is a shear field of stress including terms in cos 24) and sin 24). F r o m the Appendices, we have

s

~s

s 04)

[- WR2s2

=c°s0cos24)l V 2oz ro

Z~o

2 t3z~,o

s2

s 2 04)

+ K0(61--

62)s 2

22 +2Ko61SaSbT-S~--j

V WR2s Ko(61 + 6z ) Sa2 Sb2 1 = c o s 0 c o s 24)11=[_(1+ v ) + 2 s - 2Ko61

7-3

V2 z

_ ~

0 obO

2

O'~rO

S 2 ~ S 2 /~4)

= cos 0 sin 2~b[

L

WR2s

I(1 + v)

Ko(~, + 62) 2

s~s~ 7

s - 2Ko61 ~ - - j

Stress .fields for in-plane end shear force on elbow

273

A solution can be found in the form

Zro = cos O cos 2dp[as 3

-fi + cs + a~ (9)

Z,o=cosOsin2qbl-fs3bg

S-3

]

CS--

where: 2 2 g = - Ko6tSoS ~

1

a= - ~ I~o6,4s~ W R 2 (1 - v)

f-

i

Ko

if(lYe,) ~5 -[36~-6~1

a = Ko62 -~ W R 2 (2 + v) 2 I 12(1 + v) x2KofxS.Sb z 2 _ a ( s ~ dl- Sa" - t - S a2S h z) C~

22

SaSb

b=

S a2S b2 ( - - C - -

a(s 2+s2))

Field of stress in (cos 3~b/sin 3~b) There is a field of stress involving terms in cos 3~b and sin 3~b. F r o m the Appendices, we have WR 2 ~(ag) ~ _1 (~Zr~b __ O'~ __ -s 2 cos 3~b sin 0 s Os s &b s 41

1 t3tr4,

WRZs:

1 0

s o4~ + ~ ~ (~r*S2) -- - 41 -

sin 3q~ sin 0

The compatibility equations are:

V2a'

4 dzr, s ~ 04

2 1 020 =0 sZ (a, - %) + 1 + ~ ~Os

4~z,¢ 2 1 (1~O 1 OZo'~ V2'z* + ~ 0~ + U (a" - a*) + 1 + v \ s 0s + s 2 04, 2 J = 0 V2ao = 0

V2Zr, +

(a,, -- a,) -- ~

+ 1 + ~ Od)

Os

s2 0

=0

H . A . Lang

274

A solution can be found in the form (7 r

= cos 3~b sin Or - 4 A s 3

12B S5

[.

6Cs -

10D

W R 2 s 3]

S3

~J

12B 2D tr~ = cos 3~b sin 0 20As 3 + - 7 + 6Cs -~ s3

WR2s3.]

121 J

(io)

,°--sin. sin @ s 3 _ l~.~ + 6~s- q tro =

(

°

)

Fs 3 cos 34) + ~-~cos 3~b sin 0

where: F = 16Av G = - 8Dr 1WR2 I

A=g B_

c_

i

3~,3(s2 + s2) - 4)~2 1 128,~ - 144~(Sa~ + S~)

1 W R 2 [2s2.s~(23 - 23).2(s'2. + s ~ ) ) ] 3

I

L~---~-+~b)_]

3 I

] "128~--144~.3(~+S~)

_r.

-2 2 2 - -2 ,~(s~ +_£~,~_ 7

1WR2[

D = WR2

i

(ll)

)]-3,~.2

1 1 2 8 ~ - 144,~(s~ + sg)J

4 ~ + s~) ~3 = (4 + Sa)(Sb 2 b2 if'2 = S~ "['-S 4 + S aS

C O U P L E D T O R O I D A L ELASTIC STRESS F I E L D The final field of stress consists of a coupling between the modified initial shear stress (of cantilever type) and the normal force field. Using the Appendices, the equilibrium equations are:

Stress fields f o r in-plane end shear force on elbow

1O

1O'c,~ __ o-,

s Os (or,s) +

--

04)

s

275

[ = cos ~ sin 0 Kofx(S.2 + s~) - Ko61 s2 2 2

3 WR2s 2

S2

4

Ko61 s.sb

-

+ (,~ - ao)~

- -

I

+ (2 0 + ao)(S2. + sg)

ao~.~4-1

lc9 lc9 [ s 04) % +~-~ss (T~*s2) = sin 4~sin 0 Ko62S 2 - Ko6~(s ~ + s~,) 2 2

I," ,~ s~sg

-"o"1

1 WR2s 2 q-(b °

21)s 2

s 2 4 4----7---

_ (a ° + 2o)(S 2 + s2).

ao(s~4)] s2

J

The corresponding compatibility equations are: 4 ~'lTr~b

2 (a 1 d20 ( 22 ' % ) + 1 + Y ds 2 =sin0cosq~ -2aos+2aoS;Sb~ S3 ]

4c~z~

2

V2°a" s 2 d4) - ~ -

1

(logo

1 020"~

v°~°*+s-x 04) +~(~'-~*)+iT-~,sFJ-s +s 2 ~4)2) =sinOcos4)

V2oao=sinOcosdpI 2 ~

~

Vo%, + j b ~ ( ~ , - , , , ) -

22,s(1 + v) + ( 2 + 1

~ +i-g-~v

-2bos-

2aoS2Sg\ s3 )

vV~sll

oo]

~20

1

Os~

s ~ ~-

= s i n 0sin 4)[(% - bo)s + 2ao s2"s~]s3j From the compatibility equation for %, we have

% = cos 4) sin 0 kls 3 + kos + k 2

(12)

276

H. A . L a n g

where - -

+v}

and the terms in k o and k 2 a r e harmonic. We assume the stress field

[

~rr = a I s 3 - s ( s 2 + s ~ ) +

~r4, =

:;]

s b cos ~bsin 0 cos q~ sin 0

a2 s3 + aaS(S2. + s ~ ) + a 5

(13)

=a,Es--s(S+sg)+s a sg]

sin ~bsin 0

F r o m the first equilibrium equation, on equating like terms, there results 3 WR 2

4al-a2+a4=-K°61

4 T

+ ( 2 1 - a°)

- 2 a l - a 4 - a 3 = go61 q-)~o + a o a4 - a5 = -

(14a) (14b) (14c)

Ko6 x - a o

The same procedure applied to the second equilibrium equation leads to 1 WR z 5 a 4 - a z = Ko6 2 + -a3

-

3a4 = -ao

-k- b 0 -

~

- )~o -

21

Ko61

a 4 - a 5 = - Ko61 - a o

(14d) (14e) (14f)

We discard eqn (14f) which is the same as (14c), and form the sum and difference of eqns (14b) and (14e) to obtain al + a 3 + 2 a 4 = 0 - a 1 Wag = Ko6~ + 2 0 + a o We also replace eqns (14a) and (14d) by their sum and difference. The result is: K o ( 6 1 + 32) WR 2 2 1 ( 2 v - 1) al - a4 = 4 47 + 4v 2al-a2+3a4=--2

K o --(62-6a)

1 WR 2 1 4 I +2 (b°-a°)

Stressfields for in-planeend shearforce on elbow

277

The first compatibility equation leads to 6al - 4a4 + 2a 2 + ~

6

[al + a2 + kl] = - 2 a o 2a4+a x +a 3 =0

2 - 4 a 4 - 2al + 2as + ] - ~

[al + a2 + k2] --- 2ao

The second compatibility equation generates only one independent equation: 2 6a 2 + 4 a a + 2 a 1 + ] - - ~ (a I

+a 2

+kl) = -2b o

The third compatibility equation does not yield any independent relations. There are only three equations to consider since the second equation above was already derived from equilibrium considerations. The sum o f eqns (14a) and (14d) is reducible to

1 al+a2=

WR 2

4I(l+v)

The difference of the same two equations is reducible to ao a 2 -k- a 4 = 8

3bo 8

F r o m these, we obtain

1 al + a 4 =

3bo + 8-

WR 2

4I(1 +v~

ao 8

We have two expressions for the difference (al - a4), one derived from the equilibrium equations and one derived from the compatibility equations. Equating those two leads to an identity. We also have

21

Ko(61+fz) =

8

WR2( ~-- - ~

v ) ~

and

4ao v 21 +,Tq - 3 - 2 v

4+v + 2(3 - 2v) a0

H.A.

278

Lang

Solving for a o we find

2(3 - 2v) I-.(Sv~ - v)K ~ - - 7 ~ L 8~1 v~)

ao=

f3v-

(llv-1)

1

~ (11~- 1)

1

l'~ Ko6 x

and then 2(1 + 2v) F(5v 2 - v)K (3v-l~Ko~ 8 b ° - ~S-7-~ L~ ~) + \ l - v /

(i-v~ r8~

and 21=

Ko(61 + 62)

Kv

8

8(1 +v)

+f4+v'~F(5v 2-v)g

( 3 v - l ) go61

:v~)

[4--~v)L~

÷ ~ - ~---T~ -

(llv-1)go6z] ~ ~: ~ ~ - "a

The coefficients can now be determined from the relations K 2°-4(1+v)

5Ko61-1Koc52

a 1 -

1

a2=

as

--

3bo 8

7 8 a°

21

+~

5ao ~

bo

-~

K(2+v) 6(l+v)

al a2

2a4 - al

= ao + a4 +

k 2 =

8

K

4(1+v)

ao a4 - 8 a 3 =

3bo K°61

Kobx

(1 + v)(a o - as) + alv - a 2 + 2a,,

ADJUSTMENT OF THE RESULTANT NORMAL FORCE The normal force is given by

279

Stress fields for in-plane end shear force on elbow

reducing to ( W Q Q=

W), where

(-41v 2 +63v-25) (4 - 7v)(1 - v)

-~

( 2 0 - 4 1 v - 15v 2 + 3 0 v a) 2(1 - v2)(4 - 7v)

In the solution of the problem of pure bending, 3 a term cER2'o = c E R ( 2 + v) (s~ + s 2)

8(1 + v) was adjusted to give zero n o r m a l force. We may now readjust this constant to eliminate the term W Q so that the resultant normal force is simply - W(compression) on the plane 0 = 90 °. Let the increment in 2~ be A2~. We require

cJZR3,~(s~ - 4 ) a,~; +,~R2(4 - :.) So +,~o + T +

= o

Using c E R = - W R 2 / I we find

A~; = Q ( 4 + s 2) In the solution for pure bending, the equation for a0, is to be adjusted by replacing the term , 2 + v (s~ + s 2) 2° - 8(1 + v)

by the new value 2+v

8(i q--~)

l _ Q l ( s 2+sa2 )

ADJUSTMENT OF THE RESULTANT MOMENT The resultant m o m e n t (at 0 = 90 °) is given by M = R3

;f?

(60S COS ¢)s ds d e

- WR

a

= R3

c°s 2 ¢ kxs3 + ko s + k2 s____b_b S2 ds d e - W R a

= R 3 n kx

6

+ k°

4

t-k2s, s b

-~

- -WR

H. A. Lang

280

We select k o so the first term vanishes. Hence

2kzs,2 2sb ko =

+

22 2 kl(s4 + s.4 + s,s b)

3

+

CONCLUSION N o w that all coefficients in the stress equations have been determined, we may summarize the stress fields by equation numbers: ar = eqn (6) + eqn (10) + eqn (13) + Bending Term (Appendix 3) a 0 = eqn (6) + eqn (10) + eqn (13) + Bending Term (Appendix 3)

ao = eqn (4) + eqn (10) + eqn (12) + Bending Term (Appendix 3) zr0 = eqn (10) + eqn(13) + Bending Term (Appendix 3) zr0 = eqn (3) + eqn (5) + eqn (7) + eqn (9)

Zoo = eqn (3) + eqn (5) + eqn (9) The procedure used in this paper can be continued but the next term is algebraically cumbersome. We note that computer routines (in F O R T R A N ) are easily generated for the solution given here.

APPENDIX 1

Equilibrium and compatibility functions The three equilibrium functions for the initial shear stresses Zro and Zoo are: .

N(1) = KoIfl(SZ~ + s~) - 61s z

A cos q~ sin 0

N(2) = KoIf2sZ - 6a(s] + s~)

6~s-~S~1 sin 4~sin 0

N(3) = K ocos 0 [ - 2 6 1 ( s j + s 2) + (61 + 62)s 2]

2SaSh

+ K o COS 0 COS 2q~ (61 - 62)s 2 + ~

61

Stress.fields for in-plane end shear force on elbow

281

The c o r r e s p o n d i n g six c o m p a t i b i l i t y functions are: V(1) = 0 v(2) = o V(3) = 0 V(4) = 0

V(5)=KocosO[2(36,-32)+~(b, V(6)=Koc°sOsin2d~

+ 62)scos 2¢,b

-~(al +aa) s

2S2aS261COS

2s.sb 61 s3

APPENDIX 2

Functions for pure bending modified by sin 0 The initial state for pure bending is a 0 = cERs cos tp. W h e n bending varies as sin 0, we have a o = cERs cos tp sin 0 The three equilibrium functions for this stress field are: N(1) = cERscos 2 dp sin 0 N(2) = - cERs cos q~ sin tp sin 0 N(3) = - cERs cos tp cos 0 The c o r r e s p o n d i n g six compatibility functions are: V(1) = 0 V(2) = 0

V(4) = 0 v(5) =

V(6) =

cER cos ~bcos 0 (1 + v )

cER sin ~bcos 0 (l+v)

H . A . Lang

282

APPENDIX 3 Stress fields for pure bending modified by sin 0

0~( 3 + 2V)S 2 (3 + 2v) S a2S b2 tL1 = --Ksin []-6(-1 + ~ + -(1 -+ v) 16s 2

I 1 (Sb2 + S a2)

- K sin 0 cos 2~b

4 "--3-- ~--~2

(S'~ + 4

-

-~s~ ~ Sa~

K s i n 0{~6~112v)s2

- K sin 0 cos 2

16

Sb

L (s~ s~

(z0r) l = - K sin 0 sin 2~b - ~ - -t

3

222 SaSbS

(1 + v)

+v-~J

3 s., 4s~ (s b2 + s ~ ) ] + 4 s ' ( 4 - 41 ~

+

s°)

2 41_$4 (s 2 +Sa)(Sb + s.2 s2 b )

4(s 2 - s2) 2

3 S a4 S4 b (Sb2 + s~.)

4 s4(s 2 - s2) 2

22 4 4 22} SaSb (S b "F"S a "-FSaSb

2 (sb2 - s2) 2 -~

2-~-(~-- s~)

2+v

(s 2 + s 2 )

a o = - Ksin 0 8(1 + v~----)

+ Q ( s 2 + s 2)

(2+v)s 2 4(1 + v)

s2( 1 +V) cos2~b ] 2 I

-

(3 + 2v)s2~s:~

~0

4(s~-s~) 2

. , F - 3s s.sb

K sin 0 cos 2q~

J

S 21

2 4 + s4. + s.)(sb 2 2 + s.sb)

- Ksin v cos z t p / - 2 - ~

-2-~

+ 4-

, (3 + 2v)(sb2 + s 2)

+ v)

4 4 2 2 3SaSt'(Sb + S a ) q

22

4 s (s b - s . )

22 4 "~ S a4 -Jr SaSb) 22 th~SaSb(Sb

=

}

2

s. + s , , s D - - ~ -

(s b - s . )

- K sin 0 cos eL-

a~,

1 /3 + 2v\

16 ~ - i - - ~ v ) (sb + s2)

222 3vs.sbs

(sb2 __ $2)2

22 4 + S.4 + S.Sb)-] 22 v S.Sb(Sb + S2

~ -

"2-2T-2 %-s°)

"

J

283

Stress fields for in-plane end shear force on elbow

APPENDIX 4 Functions for stress

WR 2

tr o =

- - -I

s 2 cos 2 ~bsin 0

The equilibrium functions are:

N(1)=- - -WR2 (~ $ 2 ) - 7 - - sin 0 $2 COS ~b +-~- COS 3tb N(2) = ~

sm 0

WR Is2 N(3) =---~-- cos 0 7 +

s 2 sin ~b + ~ sin 3~b s2cos2~b.] 2

The compatibility functions are: V(1) = 0

v(2) =o V(3) =

2(2 + v) WR z 1 +~ ~ scos 66 sin 0

V(4) = 0 V(5) =

WR 2 s(1 + cos 2~b) cos 0 I(1 + v)

V(6) =

WR 2 - s sin 24, cos 0 I(1 + v )

-

-

APPENDIX 5 Initial normal force stress field

The equilibrium functions are: N(1 ) = sin 0 Cos ~br(2 o + ao)(S2a + s g) + (21 - ao)s 2 N(2) = sin 0 sin ~bI(b o - 21)s 2 - (ao + 2o)(S 2 + s 2 ) N(3) = cos 0 [ - 21s 2 - 20(s ~ + s2)]

?J

aos s ]

H.A. Lang

284

The compatibility functions are: V(1) = 2a o sin 0cos ~b - s

-sT-J

V(2) = 2 sin 0 cos ~b - bos -

a s2s2q o --~-j

V(3)=2sin0cosq~I

V(4) = sin 0 sin ~

21s(l+V)]v

ao -- bo)s +

2a SaSb7 o --~-J

v(5) = - 2 cos 0 21s v

I/(6) = 0

REFERENCES 1. Love, A. E. H., The mathematical theory of elasticity, Oxford University Press, Oxford, 4th Edition, 1927, pp. 330-5. 2. Lang, H. A., Stress analysis of pressurized elbows for nuclear components using toroidal elasticity, Proc. 4th Int. Conf. on Pressure Vessel Technology, London, 1980, Vol. 2, pp. 251-60. 3. Lang, H. A., In-plane bending of a curved pipe or toroidal tube acted on by end couples, Int. J. Pres. lies. & Piping, 15(1) (1984), pp. 27-35. 4. Lang, H. A., Stress fields for a curved pipe subjected to in-plane end couples, Int. J. Pres. Ves. & Piping, 15(2) (1984), pp. 93-104.