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Seismic stress fields in a hollow circular elbow or pipe bend
lnt. J. Pres. Ves. & Piping 17 (1984) 29--49
Seismic Stress Fields in a Hollow Circular Elbow or Pipe Bend H. A. L a n g LANG--Research West, 1201 Idaho Avenue, Santa Monica, CA 90403, USA (Received: 1 November, 1983)
ABSTRACT The methods o f toroidal elasticity theory are used to determine the fields o f stress induced by seismic components acting in two directions, X and Y, on a 90 ° elbow or pipe bend o f hollow circular cross-section.
NOMENCLATURE r, ~b, 0 tr r, a o, a o
z,o, %0, ~oo a b q R
Toroidal coordinates. N o r m a l stress components. Shear stress components. Inner radius. Outer radius. 1 + scos qk Toroidal radius.
s
r/R.
X, Y, Z
Seismic forces.
vl
t~2 ~S 2
+
1 ~3 -S~S
+
1 ~2 S 2 ~(~2"
0
O"r "~- 19"0 "Jr-O"O.
~3q
cos q~ ~3 t3s
v
Poisson's ratio.
sin~b s d~b" 29
Int. J. Pres. Ves. & Piping 0308-0161/84/$03.00 © Elsevier Applied Science Publishers Ltd, England, 1984. Printed in Great Britain
30
H. A. Lang
INTRODUCTION Toroidal elasticity was first introduced at the 1980 Conference on Pressure Vessel Technology in London.1 It has since been applied to numerous b o u n d a r y value problems for elbows, pipe bends and ring sectors. For solid circular ring sectors, the methods of toroidal elasticity led to the determination of stress fields due to seismic accelerations. 2"3 These methods are extended in the present paper to the determination of stress fields for (hollow) circular elbows and curved pipe bends. The general equations of toroidal elasticity are expanded in a series in 1/R (where R is the toroidal radius). Then the method of successive approximations is used to obtain the working equations reported in this paper. The specific objective of this paper is to determine the stress fields arising from two of the three constant components X, Y, Z induced by seismic accelerations. The three components X, Y, Z may be viewed as: 1. 2. 3.
4.
Equivalent b o d y forces - X, - Y, - Z. Extreme or upper bounds obtained from seismic response curves. Derived from displacements u o sin ogt, vo sin ~ot and w o sin ~ t so that X =-UoO92, Y = -VoOJ 2 and Z = -WoO) 2 (where the time term sin ~ot is omitted everywhere). Derived from an equivalent static method where components are multiples of gravity, ng, multiplied further by a factor (typically 1.5). ANALYSIS
As shown in Fig. 1, the components in the toroidal elastic frame of reference are related to the components X, Y, Z by the direction cosine scheme:
F, F~ F~
x
Y
Z
cos ~bcos 0 - sin ~bcos 0 - sin 0
cos ~ sin 0 - sin ~ sin 0 cos 0
sin ~b cos ~b 0
In the present paper, only the field of stress due to components X and Y will be considered. Discussion of the Z c o m p o n e n t is deferred to another paper.
Seismic stress fields in hollow circular elbow
31
\
Z
Fig. 1. Toroidal coordinates and seismic axes X, Y, Z.
The stress field of the X c o m p o n e n t can be derived from the stress field for the Y c o m p o n e n t by a simple substitution. Therefore, it is only necessary to consider the Y c o m p o n e n t in detail. The stresses are given by a converging series such that S(total) = S(0) + S(1) + S(2) + . . . (where S is any stress). The equations of motion are O(ars) l Ozr~ tr~ + = ( Y R sin 0) cos 4) -
-
sos
s Oep
s
0tre 1 O(zros 2) _ sOcl, + s 2 Os
( Y R sin 0) sin 4)
(1)
O('c,4,s) 1 OZ~o sos q s Od? = ( Y R c o s O )
These equations are satisfied by the initial field of stress given by: tr,o = ( Y R sin 0) scos tk trio = ( Y R sin 0) scos ~b aoo = ( Y R sin 0) 2vscos q~
(2)
Zr4,o = 0
Z,o° = ( Y R cos O~s "r~,Oo = 0
The expression for tr0 is based on the initial condition, e 0 = 0. It can be shown that the initial stress field satisfied all six conditions of strain compatibility and, therefore, represents an allowable field of deformation.
H. A. Lang
32
To continue the analysis, it is convenient to omit the multiplier (YRsin 0) from the first four stresses. We also omit the multiplier (YRcos 0) from the last two stresses. Thus O o = ( a t + a 4, + go)
(3)
= 2(1 + v) scos 4)
To determine the next state of stress, we require {l-2v'~
No(1 ) = [(a 0 - a,) cos 4) + z~0]o = vs - ~ ) s c o s
2q~
No(2 ) = [(tro - ao) sin 4)]0 - (1 - 2 2v) s sin 24) No(3 ) = - [2Z,oCOS 4) + ao] o = - ( 1 + 2v) scos 4) For the compatibility equations, we require -1 Vo(1 ) - ~~a~ Vo(2) - c3cr, cgq - 1 1
8or o
OO
Vo(3) = ~ q + 1 + v aq
2(l+v)
Vo(4) = 0 ~Tr 0
Vo(5 ) = ~ Vo(6 ) Stress
field for state
+ 2cos 4) = ~cos 4)
r,0 sin 4)
-
2 sin 4) = - ~ sin 4)
(1)
The equations of motion are:
as
s 04) - s
~-+s 2
es
~ =N°(1)
)~=No(2)
\
((~ ~3
) + _1 ~Te°'~ = No(3) s a4) Jl
(4)
Seismic stress fields in hollow circular elbow
33
All terms on the left-hand side refer to the stress state (l). A solution is readily found to be
[
2~
SaSb ~ =~V sZ+--)--u-(s2+s~)
]
-(1-2v)
xE~'.s. ~..~..]cos.o v aq~ = - - 4
2 +SaSg [s2' S2
(s 2 +s 2) -{-
1 sE4 (1
--
x
~.o-,1-2,,,[./ =
"2
--s-i- + C2
2v)
--
+2A2s 2
~.]
+" ] -y+C
2 c0s24~
sin 2~b
,1+v,s., ,l_.v,(,.s.+s~)COS,~ (~)[ .. ] ,#
.o--,,o T,o=
(5)
- F - s.sb s -T - + (s~ + s .2) c o s
Zeo = I ( 1 1 1 6 - 8 v ' ~ ) s 2 - ( 9 \j+8v'~(s2s2"~-(9+8v'](s21,, -i-~]\~T16 ] + s2)l sin ~b
In the above, the b o u n d a r y conditions determine A 2 , B2, C 2 and D 2. The b o u n d a r y conditions are ~r. = 0 at s = s., s b and %, = 0 at s -- s., s b. We find
Az= B2=-
l(s~
2 2 + s~) + 4s.sb
4
(s2 - s2) z
44 2 2 3 SaSb(S a 71- Sb) /-2-£-- L-~~
4 (sb-s.)
2 4b + s .4 +s.sb) 22 1 (s 2 +s.)(s
C2 = 4
02 =
(s~ - s.2) 2
1 S.Sb(Sb 22 4 22 + S.4 + S.Sb) 2 (s~ - s.~) 2
(6)
H. A. Lang
34
The compatibility
equations for stress state (1) are:
p;;a,_c!%L~(,r_(7~)+i:-;~ = - VOU) s2 ac$ s2 1 >
(
4 k#l +$(6,V&, + -~ s2 a4
(
04) +-
1 a0 --+-sas
1 a28 = - V,(2) ? a@ >> ,
(
1
l+v
I)1=-
(V&J, v:r,+$$(o,-bm,-~+-
42
(
= - V,(3)
1 a20 I a0 _---V,(4) I+~ [ sasa4 ?a4 V~~,,-~_-__2 aT$e = - V,(5) 1
(
s 84 >1
vzr
2 aT _Z9e+_*e
040
s2
s2 W
= -
V,(6)
1
These are satisfied by the solution of eqns (5) provided that 2d, = 4A,v - 1 d, = -2D,v
(7)
It is also useful to form 0, where (1 +v42 8, = (a, + a+ + o& = - ; (sf + sf) + do - 2 -(l-2V)(l
+v)~2r’-~]cos2~
Determination ‘of zrOand ~~~for state (2) is first necessary to determine the right-hand side of the equilibrium equation by calculating a new function N(3) as indicated below: It
N,(3) = - 2rrq cos 4 + 2zge1sin 4 - cO1+ s cos 4(22,@,cos 4 + oOO) = -~o-(~)(s:+s:)+(~+;)s2
+
cos2$
2A,v(l - 2v)s2- 2(1 -2~)~
D2v
1
35
Seismic stressfields in hollow circular elbow and the equilibrium equation is ( ! c3(~'°s)t_ 1 c3z,o'~ as s - ~ J 2 = N1(3) A
partial solution results from integrating _ _ 1 '(Zr°S) --
s
d o - t ~{'9) , +s
8v~
a2 + $2)
Os
-I-(9-I-3-~)S2
leading to dos
/9 + 8v\
T,o- 2 t T )
(~6 3V'~ 3 d3
2
sO°+4)+
+-g)s +V
(8)
The boundary condition z,0 = 0 for s = sa, sb determines do = -~
and
d3_(9 +y)s.sb 3v'~ 2 2
(9)
The equilibrium equation still to be solved is
[!
~Z,oS d_!~zO___ool = c os2qb[s2( 3v - ~) + 2A2v(1-2v)s
s Os
2]
s ~o i 2
+cos2q~
s2
s2
j
(10)
The compatibility equations for Lo and zoo are:
Vo%o Z,o s2
s22 0aToo.~ 4 > j 2 = - Vl(5 ) -t- (scos ~b)Vo(5 ) - Wo(5 ) (11)
(7o2T0°
ZOo 2 ~Zro"~ = - V,(6) + (scos 4b)Vo(6) - Wo(6) s2 + ~-~--J2
36
H. A. Lang
The new quantities required are: Wo(5) = - 2%0 +
0"•) 0 C O S
2 ( 0 r --
3%0 cos 2 ~b
~b - -
(cos 4))0o 1 +v
= -¼s - 2vs - s(cos 2q~)(2v + 3) Wo(6) = 2(a o
_
a4,)o sin q~ + 3%0 sin 4) cos 4)
+ (sin ~b)Oo 1+ v
= s(sin 2~b)(2v + 3) V 1(5)
=
~Z,o, ~ +
Zoo' sin 4' + s
1 l+v
I
(~01 ?~s
i/9 + 8v'~sZs~
+4A2s(1 - 2v) + - s 3 - (1 - 2v)
V1(6)
-
~Z,o I ~q
]
Z~o~sin ~b _~ 1 1 ~Ox s 1 + v s ~dp
= sin 2~b [~ + ( ~9 + 8v'~j - ~
+ (1 - 2v) (4A fi - ~-g-j]4Dz'~]
Collecting terms, the right-hand sides of eqns (11) become
(9+ 3v)s +cosZq~
+ 2v s - ~ - ~ - - , ] + (1 - 2 v ) 4 A z s +4(1 - 2 v ) - ~
and - sin 2q~
+2v
s+\~j~-+(l-Zv)4AES-(1-Zv
)
respectively. The partial solution for % is consistent with the compatibility equation because V2TrO
TrO s2 - (9 + 3v)s
Seismic stress fields in hollow circular elbow
37
T o solve the remaining equations, we take
('Cr0)2= COS2~bI~'s3 '~ t~2 l '{-S°~3s+ 70~4 (12)
[
"1
(%o)2 = sin 2~b ~ss 3 + e6 _ e3 s + s 7 The equilibrium e q u a t i o n reduces to
/'9 + 8v'~ 2 2 C t 6 = ~ - - ) S a S b --(1-- 2v)D2v 40q + 20% = (3v - ~) + 2A2v(1 - 2v) The compatibility e q u a t i o n s reduce to: 40q - 4e 5 = (~t +
2v)s + (1
- 2v)4A 2
4~2 +4~6 =(9 8 8 ~ V ) s ~ - 4 ( 1 - 2 v ) D 2 Hence, we find cz1 =
19)
+9--6 +
(1 - 2 v ) ( 2 v -
1)
/'9 + 8v'X 2 2 Ct2=--~-~---)SaSb--(1--2V)(1 +v)D 2
0%=
(ll) v-~
(13) +½A2(1-2v)(v-2)
/'9 +8v'~ 2 2 ~6 =~--i~)S.Sb -- (1-- 2v)D2v The coefficients 0c3 and ~4 are determined by the b o u n d a r y conditions Zro = 0 at s = s., s b:
~3-
1
4 2 2 (s~+s~) [~2+~l(s4+s.+s.sb)]
• "[
SaSb --
O~2 .~2
]
(14)
H. A. Lang
38
D e t e r m i n a t i o n o f stress a 0 for state (2)
The compatibility e q u a t i o n for stress ao2 is: vZtro2 = - V1(3 ) + Vo(3)cos 4 ) - Wo(3 )
(15)
where we require Wo(3) = - a0o + 2(aro - aOo) - 4Zro° cos 4)
0o 1+ v
= - (2 + 6v) s cos 4)
~ao, V 1 ( 3 ) = - ~0q- -
1
c~01
1 +v
Oq
= - (2 + v) s cos 4) - 2dis (1 - 2v) cos 4) - 4A 2s(1 - 2v) cos 4) + ~2d2 T-(l
1 - 2v)cos34) - 2 v ) c o s 3 4 ) - s ~ -4D2 (
T h e total right-hand side is:
3(2 + 3v) s cos 4) + 2dis (1 - 2v) cos 4) + 4A 2s(1 - 2v) cos 4) 4D 2 2d2s3(1 - 2v) cos 34) + - ~ - ( 1
- 2v) cos 34)
We take
~02 =JO$3 COS 4) "~-Jl COS 34) + h a r m o n i c terms S
(16)
Then' fo = ~ [ 3 ( 2 + 3 v ) + 2 d 1 ( 1 - 2v) + 4 A 2 ( 1 - 2 v ) ] f~ =~[2d2(1 - 2v) - 4D2(1 - 2v)] Using the results o f eqns (7), we find Jo = 3 ( 2 + 3v) +½Azv(1 - 2 v ) J] = -½(1 - 2v)(1 + v)D 2 Let the h a r m o n i c terms be
J2 @J'3ScOS 4) "-~-J4$2 COS 24) + f s s 3 COS 34) +J~6~COS 34) s-
(17)
39
Seismic stress fields in hollow circular elbow
The resultant force on a cross-section must vanish: N-
R2
(aOo + ao, + ao2)sdsd¢ = 0 a
•
This determines f2:
f2 = ( - ~ ) ( S a 2+S2) \
(18)
]
The resultant moment on a cross-section must vanish:
(aOo +ao,
M =R3
+ ao2)(sc°s dp)s ds ddp
a
leading to 4
4 3 2 2 s b2 +- -s. j
f3 = - 2v - ~(fo + 176) [ "
(19)
2flvsas b sbas.Z
where/?6 and/77 are defined on p. 48. We may takeJ~, = 0. The remaining constants J5 andj6 will be determined by the compatibility equations. Determination of stress a,,, a~,, z,0 ' for state (2)
The equilibrium equations are: (~(ars)
~
-t
l~zr,
a~) s
2
= N2(1) (20)
s ~ + ~ s s (z'*s2))2 = N2(2) where N2(1) = (ao, - at,) cos ~b+ zr, ' sin ¢ + zro ' + s cos ~b
[(% s~s~ l'3v
0o) c o s
9 "~
-
rOo]
+ s~)] -J
+ + (~_~)
cos~b s2(¼-2A2v ) - A 2 s 2 - 2 C 2 - ( 1 - 2 v ) D2 cos 3q~[ s 2 (¼ - 2AEV)+ A2 s2
2B2 s4
( 3 _ 2 v ) D~2-]
40
H.A. Lang
N2(2 ) = (a,~ - a0, ) sin q$ - zr,, cos q$ + scos q$ [(aOo• 15 sZ.s2(3v = sm~bE(~ + 4)sZ _ ~ - \ 4 - + 9 )
__(3_~+
a4,,,) sin ~b] 9)(s.
Z+s~)l
+2c +(V)sin +
2B 2
s4
(1
1
+ 2v)D 2 -] s2 + 2A 2vs2
J
We separate the solution into two parts and first consider the terms in cos 3¢k and sin 3qk The stress state is assumed to be ar = COS3(b ( d , s 3 q- ~a2 --t--a3- - As 3
/~
Cs - 5 D )
a s = cos 3q$ /~ls 3 + ~3- + - - + 5As3 + + Cs + s J J
r,4 , =
31))
(21)
(2 (3 B sin 3q$ (~1$3 q- ~ -'t- --S + 3"~S3 -- )-5 + C s - ~ -
cos30t,ss+ Here d I to J6 yield the particular solution while .4, /~, (~" and /5 are determined by the boundary conditions• Equating like terms, the first equilibrium equation leads to 4~il + 3Cl --/91 ---2a2 + 3(2
-- b2 =
3(3
-- b3 --
(1
-
2 2v) [A 2 + 5 _ 2A2v ]
-(1
- 2v)B 2
(1
2v) 2
(2v - 3)D 2
(22)
Seismic stress fields in hollow circular elbow
41
The second equilibrium equation leads to -3/~ 1+5gl
(1-2v)[2A2v_A2_¼] 2 (23)
3b2 + ( 2 = ( 2 v - 1)B 2 _3~3
+t?3 _
(1 - 2 2 v ) (1 + 2 v ) D 2
Next we turn to the three compatibility equations. These are"
[
V2ar
s42 az,4 ac~ ,
7
s22 (tr, - tr,) 4 1 + v 8s2 ]2
= - V~(1) + Vo(1)scos ~b- Wo(l)
= - V 1 ( 2 ) + Vo(2 ) scos q~ -
v%~+~(°r-~)-7-+i-~
aOas s2~
= - Vx(4 ) + Vo(4 ) scos ~b- Wo(4) F o r the terms on the right-hand side, we have Wo(1) = - aro + (tr 0 - trr) o 2 COS 2 '(1) + 4 cos ~b Z,o ° = ( 3 v - ½)scos q~
(1 - 2v) s cos 34~
Wo(2) = - tr#o + 2 sin 2 ~b(tro - tr#) o = (v - ~) s cos ck + ( ~ )
s cos 3(o
Wo(4 ) = - 2%00 sin ~b + (tr, - %)0 sin 2~b {1 -
2v'~
= - ( v + ½)ssin ~b + k ~ ) s
.
sin 34)
(24)
Wo(2)
2
42
H. A. Lang
The complete terms on the right-hand side of the compatibility equations are, successively,
v sosb
| ~ - ~ - / s cos q5 + ~ 3 -
cos ~b
(1 - 2v)
-s-2A2s+
(12v) ~cos3q~-s -
"~'
s cos ~ -
y SaS b
[
cos q~ +
cosq5
8B2 lOD2 ] s5 s3 + 2A zs
s + 6A 2s + _-Tw-. cos q5
]
(l --2 2v)[s + ~8B2 2Dz -+~--2Azs 3v
+~-
v sasb
s sin 4~ + ~ - - s ~ sin q5
cos 3q~
(1 - 2v) F_2A2s + 2s-D~] sin $ 2
/
(122v)
s - 2 A z s - 8B= s5
6$3=_Ijsin3~b
where we have used ~r x
Vl(|)=~-q q ~a,, Vt(2)=-•q V1(4) -
~q
2T~, sin4,
s 2z.,, sin q~ s s
Considering only the terms in s 3 cos 34, or s 3 sin 3~b, the three compatibility equations lead to 96v.~ 6 - 1 2 ~ - 26~ + 2/~, - 1 +--~ + ~ - 6 ~ 1 + 6/~ 1 - 4~ 1 -~
96v.4 l+v
(~i~ + / ~ + i s ) -
6 1 +v~6~' + / 3 ~ + j ~ ) -
1 - 2v 2 (1 - 2Az) 1 - 2v 2 (2A z - 1)
43
Seismic stress fields in hollow circular elbow
Since d 1, /31, ~1 is to generate a particular solution, we take f5 = 4v.4 N o t i n g that the sum of the two e q u a t i o n s a b o v e is
~i1 -/31 + 2cl = 0 we determine the coefficients (using the first o f eqns (22) and (23)) as Fl(4v
61 =
2) + F 2 12v
-
/31 = F l ( 1 0 + 4 v ) - 5 F 2 12v
cl -
(25)
6F 1 - 3F 2 12v
where FI _ (1 ~ - 2v) [ 5 _ ( 2 v -
1)A2]
(1 - - 2 v ) F2- ~ (1 - 2,42)(1 + v)
Next, considering terms in l / s 3 COS 3~b, 1/S 3 sin 3~b, the compatibility e q u a t i o n s are: 12 -- 12g2 -- 262 + 2/32 + ] ~ (/32 + tl2) = (1 - 2v) 4B 2 12 --6a2 + 6/32- 4c2 -t-77~, (t~2 q-/32) ----(1 - 2 v ) 4 B 2 1-t-v
where we take J6 = - 4v/). The difference of the two equations is a2 - / 3 2 - 2c2 = 0
H. A. Lang
44
Using the second of eqns (22) a n d (23), we find 42
- (7 + 4 v ) F 3 - 5 F 4 6(a + 2v)
/32 - F4 - F3(1 + 4v) 6(1 + 2v)
(26)
-F 4 - F 3 (2= 2(1+2v) where F 3 = - (1 - 2v)B 2 F , = (1 - 2v)(1 + v)B 2 In eqns(21), the terms involving /1 and C' identically satisfy the compatibility equations. There remain only the terms 1/scos3~b and 1/s sin 3q5. These yield 2 - 1 0 a 3 - 12c3 +2/33 + ] - ~
(ti3 +/33 + J i ) = (1 - 2 v ) 5 D 2
10
-- 10/33 + 12(3 + 2d3 - 1 + v (¢i3 +/33 +J'l) = - (1 - 2 v ) D 2
-12(3-
6 663 + 6/33 + ] ~ - v v (a3 +/33 + J i ) = (1 - 2 v ) 3 D 2
where Ji = - ~ ( 1 - 2v)(l + v)D 2 f r o m eqn (16). F r o m the last o f eqns (22), (23) and (27), we obtain: V
a3 = - 2 F5 V
/33 = ~ F5 ( v - 1) F5 ~32 where F 5 = (1 - 2 v ) D 2
(27)
Seismic stress fields in hollow circular elbow
45
Eliminating the functions F 1 to F 5, the coefficients for the particular solution involving cos 3~b and sin 3~b become 41
=
(1-2v)(llv-4) 48v
/51 _ 5(1 - 2v)(4 v ) 4 8 v+ ~1
(1 - 2 v ) ( 4 -
t
+ (1 24~-2v) [ 1 5 - l l v -
v)
16v
(1-2v) 2 4 ~ [ - 3 + 7v - 8v 2) 8v 2]
3 (1 - v)(1 - 2 v ) A +8vv
2
(1 - 2v)(2 v)B 2 al 6(1 + 2v) -
t52 _ (1 - 2v)(2 + 5v)B 2 6(1 + 2v)
(28)
v(2v - I)B 2 C2--
2(1 + 2v)
43 = -- ~ (1 -- 2v)D 2
53 = 2 (1 - 2 v ) D 2 C3 --
( v - 1) 2 (1 - 2v)D 2
We complete the solution for terms in cos 3~b, sin 3q~ by using the b o u n d a r y conditions to determine coefficients A t o / 5 . Let 2 2
"~1 = (S~ "~ S4a "~- SaSh) "
+ s.)(sb +
)t 3 .~- (S 2 --~ S 2)
and define R 1 ~--41~. 2 -{--a2 -F 43~. 3 R 2 = / ~ 1 2 1 -4- 43
R 4 = 6121 -+- 63
H. A. Lang
46 Then:
A= B 2 b2 SaS
3R123 + 421(R4 2(822 -
-- R2) -
523R 3
92223)
--3).123R ~ + 52321R 3 + R 2 ( 9 / ] . 2 ~ . 3 - 4212) + R4(92223 - 12212) 2(822 - 92223) (29)
~ = R2(1622
--
9,~,223) q- R4(92223 - 1622) - 6RxJ.x23 223(822 -- 92223)
+ 1021,,~3R 3
/) -- ( 422 -- 3)t223)R 1 - 2)L122R 2 + R3(422 - 72223) + 2z122R4 4(822 - 9 2 2 2 3 ) Determination of stresses depending on cos q~ and sin ~b The final stress field is assumed to be
~rr=(c°sO)31Is3+S]S~s- s(s~+ s2)1 0"4~=COS(~Ifl2S3 ~ fl3S2S2 ~" fl4S(
"~ $2) 1 (30)
zr~ = ( s i n ~b)fl5 Is3
O"0=cOS~
+sZs~-s(S2s +s~) 1
E
f16$3 +f17 S~b
E q u a t i n g like terms, the first equilibrium e q u a t i o n leads to llv 4
4ill q- fl5 - fl2 -~ 3v
- 2 f l ' - fls- f14=4 4 ~5 -- ~3 --
3v 4
13 16
9 16
(1-2V)[A2+2Azv_~] 2
C2(1 - 2v)
(sZb+ SZa)
9 16
(1 -
2v)ZD2 2so2s bZ
(31)
Seismic stress fields in hollow circular elbow
47
The second equilibrium equation leads to v 15 5//5 - / / 2 = 4 + 16
(1 --2v)
-//4 The
-
[-¼ + 2A2v- 3A2] 3v 4
3//5 -
9 (1 - - 2 v ) C 2 + 16 (s g + s 2)
(32)
first compatibility equation leads to
6 3 6//1 -4//5 +2//2 + ] ~ v ( / / l +//2 +//6) - 2
-4//5-
7v ( 1 - 2 v ) [ 1 +2A2] 2 -~ 2
2
v
2(//1- //3) q--1--.(_-v(//1 + / / 3 + / / 7 ) = 2
(1 - 2v)D 2 2 2
SaSb
(33)
The second compatibility equation leads to 2
5
6//2 + 4//5 + 2//1+--~--~_---vv(//1+ //2 + //6) -- 2
v 2v 2 + 1T[1
+6A2]
(34)
The fourth compatibility equation leads to 2 1 3v 4//5- 2//1 + 2//2- 1 +-----~[//1-'~- //2 -'~-//6] =2-'~-T"~-A2( 1 --2V) (35) The total set of nine equations is easily reduced by taking sums and differences (eqns (31) to (35)). The solutions for fll to f17 are: fl, = 1 [3F6 _ F7 + 2Fs] - F 7 + 5F8
f12fls-
6
F7 + F 8 6
where F6-
1iv 4
F v = 41q6 v 15
13 -16
(1-2v)[2A:v+A2-5 ] 2
~(1-2v)[2A2v-3Az-¼ ]
Fs = 7 + 3A 2
(1 - 2 v ) 2
H. A. Lang
48
and
3v /3~ = - 3 / 3 , + 7 - +
9
(1
-- 2v)C 2
(sg + S 2 )
/33 =/35 + ( 3 v + ? - ~ + ( l - - - 2 v ~ + (1-2v) z 02 2 2 \4 10/ \ e / 2SaS b /36 = - (2 + v)/3, +
4/35(1 + v ) 6
/32(4+v) 3
+ (3 -2 7v) ( l + v ) + ~1 - 2v (l+2A2v)(l+v) v /37 =/31v +2(1 + v)/35 -/33(2 + v) + 2
(1 - 2 v )
D 2
2
2 2b SaS
CONCLUSIONS The complete field of six stresses may now be summarized by equation number since expressions for all unknown coefficients have been determined in the body of the paper. Thus, we have: a r = e q n (2) + e q n ( 5 ) + e q n (21) + e q n (30)... % = e q n ( 2 ) + e q n ( 5 ) + e q n (21) + e q n ( 3 0 ) . . . cr0 = eqn (2) + eqn (5) + eqn (16) + eqn (17) + eqn (21) + eqn (30)... zr~ = eqn (5) + eqn (21) + eqn (30) z,e = eqn (2) + eqn (5) + eqn (8) + eqn (12) z~0 = e q n (5) + e q n (12)... The procedure used in this paper could be continued to determine further terms in the series for stresses but the algebraic details would be cumbersome. The solution presented in this paper is expressible as a Fortran computer routine. The programming is straightforward. The stress fields for the X c o m p o n e n t of seismic force can be derived by replacing the multiplier YR sin 0 by XR cos 0 in the expressions for a , %, %, and z~. Similarly, the multiplier YR cos 0 for stresses z~ and z~e may be replaced by - X R sin 0.
Seismic stress .fields in hollow circular elbow
49
REFERENCES 1. Lang, H. A., Stress analysis of pressurized elbows for nuclear components using toroidal elasticity, Proc. 4th Int. ConJl Pressure Vessel Technology, London, May 1980, Vol. 2, pp. 251-60. 2. Lang, H. A., Seismically induced stress fields in a circular ring sector, submitted to Res Mechanica. 3. Lang, H. A., Stress fields produced by a vertical seismic force acting on a circular ring sector, submitted to Res Mechanica.
Seismic Stress Fields in a Hollow Circular Elbow or Pipe Bend H. A. L a n g LANG--Research West, 1201 Idaho Avenue, Santa Monica, CA 90403, USA (Received: 1 November, 1983)
ABSTRACT The methods o f toroidal elasticity theory are used to determine the fields o f stress induced by seismic components acting in two directions, X and Y, on a 90 ° elbow or pipe bend o f hollow circular cross-section.
NOMENCLATURE r, ~b, 0 tr r, a o, a o
z,o, %0, ~oo a b q R
Toroidal coordinates. N o r m a l stress components. Shear stress components. Inner radius. Outer radius. 1 + scos qk Toroidal radius.
s
r/R.
X, Y, Z
Seismic forces.
vl
t~2 ~S 2
+
1 ~3 -S~S
+
1 ~2 S 2 ~(~2"
0
O"r "~- 19"0 "Jr-O"O.
~3q
cos q~ ~3 t3s
v
Poisson's ratio.
sin~b s d~b" 29
Int. J. Pres. Ves. & Piping 0308-0161/84/$03.00 © Elsevier Applied Science Publishers Ltd, England, 1984. Printed in Great Britain
30
H. A. Lang
INTRODUCTION Toroidal elasticity was first introduced at the 1980 Conference on Pressure Vessel Technology in London.1 It has since been applied to numerous b o u n d a r y value problems for elbows, pipe bends and ring sectors. For solid circular ring sectors, the methods of toroidal elasticity led to the determination of stress fields due to seismic accelerations. 2"3 These methods are extended in the present paper to the determination of stress fields for (hollow) circular elbows and curved pipe bends. The general equations of toroidal elasticity are expanded in a series in 1/R (where R is the toroidal radius). Then the method of successive approximations is used to obtain the working equations reported in this paper. The specific objective of this paper is to determine the stress fields arising from two of the three constant components X, Y, Z induced by seismic accelerations. The three components X, Y, Z may be viewed as: 1. 2. 3.
4.
Equivalent b o d y forces - X, - Y, - Z. Extreme or upper bounds obtained from seismic response curves. Derived from displacements u o sin ogt, vo sin ~ot and w o sin ~ t so that X =-UoO92, Y = -VoOJ 2 and Z = -WoO) 2 (where the time term sin ~ot is omitted everywhere). Derived from an equivalent static method where components are multiples of gravity, ng, multiplied further by a factor (typically 1.5). ANALYSIS
As shown in Fig. 1, the components in the toroidal elastic frame of reference are related to the components X, Y, Z by the direction cosine scheme:
F, F~ F~
x
Y
Z
cos ~bcos 0 - sin ~bcos 0 - sin 0
cos ~ sin 0 - sin ~ sin 0 cos 0
sin ~b cos ~b 0
In the present paper, only the field of stress due to components X and Y will be considered. Discussion of the Z c o m p o n e n t is deferred to another paper.
Seismic stress fields in hollow circular elbow
31
\
Z
Fig. 1. Toroidal coordinates and seismic axes X, Y, Z.
The stress field of the X c o m p o n e n t can be derived from the stress field for the Y c o m p o n e n t by a simple substitution. Therefore, it is only necessary to consider the Y c o m p o n e n t in detail. The stresses are given by a converging series such that S(total) = S(0) + S(1) + S(2) + . . . (where S is any stress). The equations of motion are O(ars) l Ozr~ tr~ + = ( Y R sin 0) cos 4) -
-
sos
s Oep
s
0tre 1 O(zros 2) _ sOcl, + s 2 Os
( Y R sin 0) sin 4)
(1)
O('c,4,s) 1 OZ~o sos q s Od? = ( Y R c o s O )
These equations are satisfied by the initial field of stress given by: tr,o = ( Y R sin 0) scos tk trio = ( Y R sin 0) scos ~b aoo = ( Y R sin 0) 2vscos q~
(2)
Zr4,o = 0
Z,o° = ( Y R cos O~s "r~,Oo = 0
The expression for tr0 is based on the initial condition, e 0 = 0. It can be shown that the initial stress field satisfied all six conditions of strain compatibility and, therefore, represents an allowable field of deformation.
H. A. Lang
32
To continue the analysis, it is convenient to omit the multiplier (YRsin 0) from the first four stresses. We also omit the multiplier (YRcos 0) from the last two stresses. Thus O o = ( a t + a 4, + go)
(3)
= 2(1 + v) scos 4)
To determine the next state of stress, we require {l-2v'~
No(1 ) = [(a 0 - a,) cos 4) + z~0]o = vs - ~ ) s c o s
2q~
No(2 ) = [(tro - ao) sin 4)]0 - (1 - 2 2v) s sin 24) No(3 ) = - [2Z,oCOS 4) + ao] o = - ( 1 + 2v) scos 4) For the compatibility equations, we require -1 Vo(1 ) - ~~a~ Vo(2) - c3cr, cgq - 1 1
8or o
OO
Vo(3) = ~ q + 1 + v aq
2(l+v)
Vo(4) = 0 ~Tr 0
Vo(5 ) = ~ Vo(6 ) Stress
field for state
+ 2cos 4) = ~cos 4)
r,0 sin 4)
-
2 sin 4) = - ~ sin 4)
(1)
The equations of motion are:
as
s 04) - s
~-+s 2
es
~ =N°(1)
)~=No(2)
\
((~ ~3
) + _1 ~Te°'~ = No(3) s a4) Jl
(4)
Seismic stress fields in hollow circular elbow
33
All terms on the left-hand side refer to the stress state (l). A solution is readily found to be
[
2~
SaSb ~ =~V sZ+--)--u-(s2+s~)
]
-(1-2v)
xE~'.s. ~..~..]cos.o v aq~ = - - 4
2 +SaSg [s2' S2
(s 2 +s 2) -{-
1 sE4 (1
--
x
~.o-,1-2,,,[./ =
"2
--s-i- + C2
2v)
--
+2A2s 2
~.]
+" ] -y+C
2 c0s24~
sin 2~b
,1+v,s., ,l_.v,(,.s.+s~)COS,~ (~)[ .. ] ,#
.o--,,o T,o=
(5)
- F - s.sb s -T - + (s~ + s .2) c o s
Zeo = I ( 1 1 1 6 - 8 v ' ~ ) s 2 - ( 9 \j+8v'~(s2s2"~-(9+8v'](s21,, -i-~]\~T16 ] + s2)l sin ~b
In the above, the b o u n d a r y conditions determine A 2 , B2, C 2 and D 2. The b o u n d a r y conditions are ~r. = 0 at s = s., s b and %, = 0 at s -- s., s b. We find
Az= B2=-
l(s~
2 2 + s~) + 4s.sb
4
(s2 - s2) z
44 2 2 3 SaSb(S a 71- Sb) /-2-£-- L-~~
4 (sb-s.)
2 4b + s .4 +s.sb) 22 1 (s 2 +s.)(s
C2 = 4
02 =
(s~ - s.2) 2
1 S.Sb(Sb 22 4 22 + S.4 + S.Sb) 2 (s~ - s.~) 2
(6)
H. A. Lang
34
The compatibility
equations for stress state (1) are:
p;;a,_c!%L~(,r_(7~)+i:-;~ = - VOU) s2 ac$ s2 1 >
(
4 k#l +$(6,V&, + -~ s2 a4
(
04) +-
1 a0 --+-sas
1 a28 = - V,(2) ? a@ >> ,
(
1
l+v
I)1=-
(V&J, v:r,+$$(o,-bm,-~+-
42
(
= - V,(3)
1 a20 I a0 _---V,(4) I+~ [ sasa4 ?a4 V~~,,-~_-__2 aT$e = - V,(5) 1
(
s 84 >1
vzr
2 aT _Z9e+_*e
040
s2
s2 W
= -
V,(6)
1
These are satisfied by the solution of eqns (5) provided that 2d, = 4A,v - 1 d, = -2D,v
(7)
It is also useful to form 0, where (1 +v42 8, = (a, + a+ + o& = - ; (sf + sf) + do - 2 -(l-2V)(l
+v)~2r’-~]cos2~
Determination ‘of zrOand ~~~for state (2) is first necessary to determine the right-hand side of the equilibrium equation by calculating a new function N(3) as indicated below: It
N,(3) = - 2rrq cos 4 + 2zge1sin 4 - cO1+ s cos 4(22,@,cos 4 + oOO) = -~o-(~)(s:+s:)+(~+;)s2
+
cos2$
2A,v(l - 2v)s2- 2(1 -2~)~
D2v
1
35
Seismic stressfields in hollow circular elbow and the equilibrium equation is ( ! c3(~'°s)t_ 1 c3z,o'~ as s - ~ J 2 = N1(3) A
partial solution results from integrating _ _ 1 '(Zr°S) --
s
d o - t ~{'9) , +s
8v~
a2 + $2)
Os
-I-(9-I-3-~)S2
leading to dos
/9 + 8v\
T,o- 2 t T )
(~6 3V'~ 3 d3
2
sO°+4)+
+-g)s +V
(8)
The boundary condition z,0 = 0 for s = sa, sb determines do = -~
and
d3_(9 +y)s.sb 3v'~ 2 2
(9)
The equilibrium equation still to be solved is
[!
~Z,oS d_!~zO___ool = c os2qb[s2( 3v - ~) + 2A2v(1-2v)s
s Os
2]
s ~o i 2
+cos2q~
s2
s2
j
(10)
The compatibility equations for Lo and zoo are:
Vo%o Z,o s2
s22 0aToo.~ 4 > j 2 = - Vl(5 ) -t- (scos ~b)Vo(5 ) - Wo(5 ) (11)
(7o2T0°
ZOo 2 ~Zro"~ = - V,(6) + (scos 4b)Vo(6) - Wo(6) s2 + ~-~--J2
36
H. A. Lang
The new quantities required are: Wo(5) = - 2%0 +
0"•) 0 C O S
2 ( 0 r --
3%0 cos 2 ~b
~b - -
(cos 4))0o 1 +v
= -¼s - 2vs - s(cos 2q~)(2v + 3) Wo(6) = 2(a o
_
a4,)o sin q~ + 3%0 sin 4) cos 4)
+ (sin ~b)Oo 1+ v
= s(sin 2~b)(2v + 3) V 1(5)
=
~Z,o, ~ +
Zoo' sin 4' + s
1 l+v
I
(~01 ?~s
i/9 + 8v'~sZs~
+4A2s(1 - 2v) + - s 3 - (1 - 2v)
V1(6)
-
~Z,o I ~q
]
Z~o~sin ~b _~ 1 1 ~Ox s 1 + v s ~dp
= sin 2~b [~ + ( ~9 + 8v'~j - ~
+ (1 - 2v) (4A fi - ~-g-j]4Dz'~]
Collecting terms, the right-hand sides of eqns (11) become
(9+ 3v)s +cosZq~
+ 2v s - ~ - ~ - - , ] + (1 - 2 v ) 4 A z s +4(1 - 2 v ) - ~
and - sin 2q~
+2v
s+\~j~-+(l-Zv)4AES-(1-Zv
)
respectively. The partial solution for % is consistent with the compatibility equation because V2TrO
TrO s2 - (9 + 3v)s
Seismic stress fields in hollow circular elbow
37
T o solve the remaining equations, we take
('Cr0)2= COS2~bI~'s3 '~ t~2 l '{-S°~3s+ 70~4 (12)
[
"1
(%o)2 = sin 2~b ~ss 3 + e6 _ e3 s + s 7 The equilibrium e q u a t i o n reduces to
/'9 + 8v'~ 2 2 C t 6 = ~ - - ) S a S b --(1-- 2v)D2v 40q + 20% = (3v - ~) + 2A2v(1 - 2v) The compatibility e q u a t i o n s reduce to: 40q - 4e 5 = (~t +
2v)s + (1
- 2v)4A 2
4~2 +4~6 =(9 8 8 ~ V ) s ~ - 4 ( 1 - 2 v ) D 2 Hence, we find cz1 =
19)
+9--6 +
(1 - 2 v ) ( 2 v -
1)
/'9 + 8v'X 2 2 Ct2=--~-~---)SaSb--(1--2V)(1 +v)D 2
0%=
(ll) v-~
(13) +½A2(1-2v)(v-2)
/'9 +8v'~ 2 2 ~6 =~--i~)S.Sb -- (1-- 2v)D2v The coefficients 0c3 and ~4 are determined by the b o u n d a r y conditions Zro = 0 at s = s., s b:
~3-
1
4 2 2 (s~+s~) [~2+~l(s4+s.+s.sb)]
• "[
SaSb --
O~2 .~2
]
(14)
H. A. Lang
38
D e t e r m i n a t i o n o f stress a 0 for state (2)
The compatibility e q u a t i o n for stress ao2 is: vZtro2 = - V1(3 ) + Vo(3)cos 4 ) - Wo(3 )
(15)
where we require Wo(3) = - a0o + 2(aro - aOo) - 4Zro° cos 4)
0o 1+ v
= - (2 + 6v) s cos 4)
~ao, V 1 ( 3 ) = - ~0q- -
1
c~01
1 +v
Oq
= - (2 + v) s cos 4) - 2dis (1 - 2v) cos 4) - 4A 2s(1 - 2v) cos 4) + ~2d2 T-(l
1 - 2v)cos34) - 2 v ) c o s 3 4 ) - s ~ -4D2 (
T h e total right-hand side is:
3(2 + 3v) s cos 4) + 2dis (1 - 2v) cos 4) + 4A 2s(1 - 2v) cos 4) 4D 2 2d2s3(1 - 2v) cos 34) + - ~ - ( 1
- 2v) cos 34)
We take
~02 =JO$3 COS 4) "~-Jl COS 34) + h a r m o n i c terms S
(16)
Then' fo = ~ [ 3 ( 2 + 3 v ) + 2 d 1 ( 1 - 2v) + 4 A 2 ( 1 - 2 v ) ] f~ =~[2d2(1 - 2v) - 4D2(1 - 2v)] Using the results o f eqns (7), we find Jo = 3 ( 2 + 3v) +½Azv(1 - 2 v ) J] = -½(1 - 2v)(1 + v)D 2 Let the h a r m o n i c terms be
J2 @J'3ScOS 4) "-~-J4$2 COS 24) + f s s 3 COS 34) +J~6~COS 34) s-
(17)
39
Seismic stress fields in hollow circular elbow
The resultant force on a cross-section must vanish: N-
R2
(aOo + ao, + ao2)sdsd¢ = 0 a
•
This determines f2:
f2 = ( - ~ ) ( S a 2+S2) \
(18)
]
The resultant moment on a cross-section must vanish:
(aOo +ao,
M =R3
+ ao2)(sc°s dp)s ds ddp
a
leading to 4
4 3 2 2 s b2 +- -s. j
f3 = - 2v - ~(fo + 176) [ "
(19)
2flvsas b sbas.Z
where/?6 and/77 are defined on p. 48. We may takeJ~, = 0. The remaining constants J5 andj6 will be determined by the compatibility equations. Determination of stress a,,, a~,, z,0 ' for state (2)
The equilibrium equations are: (~(ars)
~
-t
l~zr,
a~) s
2
= N2(1) (20)
s ~ + ~ s s (z'*s2))2 = N2(2) where N2(1) = (ao, - at,) cos ~b+ zr, ' sin ¢ + zro ' + s cos ~b
[(% s~s~ l'3v
0o) c o s
9 "~
-
rOo]
+ s~)] -J
+ + (~_~)
cos~b s2(¼-2A2v ) - A 2 s 2 - 2 C 2 - ( 1 - 2 v ) D2 cos 3q~[ s 2 (¼ - 2AEV)+ A2 s2
2B2 s4
( 3 _ 2 v ) D~2-]
40
H.A. Lang
N2(2 ) = (a,~ - a0, ) sin q$ - zr,, cos q$ + scos q$ [(aOo• 15 sZ.s2(3v = sm~bE(~ + 4)sZ _ ~ - \ 4 - + 9 )
__(3_~+
a4,,,) sin ~b] 9)(s.
Z+s~)l
+2c +(V)sin +
2B 2
s4
(1
1
+ 2v)D 2 -] s2 + 2A 2vs2
J
We separate the solution into two parts and first consider the terms in cos 3¢k and sin 3qk The stress state is assumed to be ar = COS3(b ( d , s 3 q- ~a2 --t--a3- - As 3
/~
Cs - 5 D )
a s = cos 3q$ /~ls 3 + ~3- + - - + 5As3 + + Cs + s J J
r,4 , =
31))
(21)
(2 (3 B sin 3q$ (~1$3 q- ~ -'t- --S + 3"~S3 -- )-5 + C s - ~ -
cos30t,ss+ Here d I to J6 yield the particular solution while .4, /~, (~" and /5 are determined by the boundary conditions• Equating like terms, the first equilibrium equation leads to 4~il + 3Cl --/91 ---2a2 + 3(2
-- b2 =
3(3
-- b3 --
(1
-
2 2v) [A 2 + 5 _ 2A2v ]
-(1
- 2v)B 2
(1
2v) 2
(2v - 3)D 2
(22)
Seismic stress fields in hollow circular elbow
41
The second equilibrium equation leads to -3/~ 1+5gl
(1-2v)[2A2v_A2_¼] 2 (23)
3b2 + ( 2 = ( 2 v - 1)B 2 _3~3
+t?3 _
(1 - 2 2 v ) (1 + 2 v ) D 2
Next we turn to the three compatibility equations. These are"
[
V2ar
s42 az,4 ac~ ,
7
s22 (tr, - tr,) 4 1 + v 8s2 ]2
= - V~(1) + Vo(1)scos ~b- Wo(l)
= - V 1 ( 2 ) + Vo(2 ) scos q~ -
v%~+~(°r-~)-7-+i-~
aOas s2~
= - Vx(4 ) + Vo(4 ) scos ~b- Wo(4) F o r the terms on the right-hand side, we have Wo(1) = - aro + (tr 0 - trr) o 2 COS 2 '(1) + 4 cos ~b Z,o ° = ( 3 v - ½)scos q~
(1 - 2v) s cos 34~
Wo(2) = - tr#o + 2 sin 2 ~b(tro - tr#) o = (v - ~) s cos ck + ( ~ )
s cos 3(o
Wo(4 ) = - 2%00 sin ~b + (tr, - %)0 sin 2~b {1 -
2v'~
= - ( v + ½)ssin ~b + k ~ ) s
.
sin 34)
(24)
Wo(2)
2
42
H. A. Lang
The complete terms on the right-hand side of the compatibility equations are, successively,
v sosb
| ~ - ~ - / s cos q5 + ~ 3 -
cos ~b
(1 - 2v)
-s-2A2s+
(12v) ~cos3q~-s -
"~'
s cos ~ -
y SaS b
[
cos q~ +
cosq5
8B2 lOD2 ] s5 s3 + 2A zs
s + 6A 2s + _-Tw-. cos q5
]
(l --2 2v)[s + ~8B2 2Dz -+~--2Azs 3v
+~-
v sasb
s sin 4~ + ~ - - s ~ sin q5
cos 3q~
(1 - 2v) F_2A2s + 2s-D~] sin $ 2
/
(122v)
s - 2 A z s - 8B= s5
6$3=_Ijsin3~b
where we have used ~r x
Vl(|)=~-q q ~a,, Vt(2)=-•q V1(4) -
~q
2T~, sin4,
s 2z.,, sin q~ s s
Considering only the terms in s 3 cos 34, or s 3 sin 3~b, the three compatibility equations lead to 96v.~ 6 - 1 2 ~ - 26~ + 2/~, - 1 +--~ + ~ - 6 ~ 1 + 6/~ 1 - 4~ 1 -~
96v.4 l+v
(~i~ + / ~ + i s ) -
6 1 +v~6~' + / 3 ~ + j ~ ) -
1 - 2v 2 (1 - 2Az) 1 - 2v 2 (2A z - 1)
43
Seismic stress fields in hollow circular elbow
Since d 1, /31, ~1 is to generate a particular solution, we take f5 = 4v.4 N o t i n g that the sum of the two e q u a t i o n s a b o v e is
~i1 -/31 + 2cl = 0 we determine the coefficients (using the first o f eqns (22) and (23)) as Fl(4v
61 =
2) + F 2 12v
-
/31 = F l ( 1 0 + 4 v ) - 5 F 2 12v
cl -
(25)
6F 1 - 3F 2 12v
where FI _ (1 ~ - 2v) [ 5 _ ( 2 v -
1)A2]
(1 - - 2 v ) F2- ~ (1 - 2,42)(1 + v)
Next, considering terms in l / s 3 COS 3~b, 1/S 3 sin 3~b, the compatibility e q u a t i o n s are: 12 -- 12g2 -- 262 + 2/32 + ] ~ (/32 + tl2) = (1 - 2v) 4B 2 12 --6a2 + 6/32- 4c2 -t-77~, (t~2 q-/32) ----(1 - 2 v ) 4 B 2 1-t-v
where we take J6 = - 4v/). The difference of the two equations is a2 - / 3 2 - 2c2 = 0
H. A. Lang
44
Using the second of eqns (22) a n d (23), we find 42
- (7 + 4 v ) F 3 - 5 F 4 6(a + 2v)
/32 - F4 - F3(1 + 4v) 6(1 + 2v)
(26)
-F 4 - F 3 (2= 2(1+2v) where F 3 = - (1 - 2v)B 2 F , = (1 - 2v)(1 + v)B 2 In eqns(21), the terms involving /1 and C' identically satisfy the compatibility equations. There remain only the terms 1/scos3~b and 1/s sin 3q5. These yield 2 - 1 0 a 3 - 12c3 +2/33 + ] - ~
(ti3 +/33 + J i ) = (1 - 2 v ) 5 D 2
10
-- 10/33 + 12(3 + 2d3 - 1 + v (¢i3 +/33 +J'l) = - (1 - 2 v ) D 2
-12(3-
6 663 + 6/33 + ] ~ - v v (a3 +/33 + J i ) = (1 - 2 v ) 3 D 2
where Ji = - ~ ( 1 - 2v)(l + v)D 2 f r o m eqn (16). F r o m the last o f eqns (22), (23) and (27), we obtain: V
a3 = - 2 F5 V
/33 = ~ F5 ( v - 1) F5 ~32 where F 5 = (1 - 2 v ) D 2
(27)
Seismic stress fields in hollow circular elbow
45
Eliminating the functions F 1 to F 5, the coefficients for the particular solution involving cos 3~b and sin 3~b become 41
=
(1-2v)(llv-4) 48v
/51 _ 5(1 - 2v)(4 v ) 4 8 v+ ~1
(1 - 2 v ) ( 4 -
t
+ (1 24~-2v) [ 1 5 - l l v -
v)
16v
(1-2v) 2 4 ~ [ - 3 + 7v - 8v 2) 8v 2]
3 (1 - v)(1 - 2 v ) A +8vv
2
(1 - 2v)(2 v)B 2 al 6(1 + 2v) -
t52 _ (1 - 2v)(2 + 5v)B 2 6(1 + 2v)
(28)
v(2v - I)B 2 C2--
2(1 + 2v)
43 = -- ~ (1 -- 2v)D 2
53 = 2 (1 - 2 v ) D 2 C3 --
( v - 1) 2 (1 - 2v)D 2
We complete the solution for terms in cos 3~b, sin 3q~ by using the b o u n d a r y conditions to determine coefficients A t o / 5 . Let 2 2
"~1 = (S~ "~ S4a "~- SaSh) "
+ s.)(sb +
)t 3 .~- (S 2 --~ S 2)
and define R 1 ~--41~. 2 -{--a2 -F 43~. 3 R 2 = / ~ 1 2 1 -4- 43
R 4 = 6121 -+- 63
H. A. Lang
46 Then:
A= B 2 b2 SaS
3R123 + 421(R4 2(822 -
-- R2) -
523R 3
92223)
--3).123R ~ + 52321R 3 + R 2 ( 9 / ] . 2 ~ . 3 - 4212) + R4(92223 - 12212) 2(822 - 92223) (29)
~ = R2(1622
--
9,~,223) q- R4(92223 - 1622) - 6RxJ.x23 223(822 -- 92223)
+ 1021,,~3R 3
/) -- ( 422 -- 3)t223)R 1 - 2)L122R 2 + R3(422 - 72223) + 2z122R4 4(822 - 9 2 2 2 3 ) Determination of stresses depending on cos q~ and sin ~b The final stress field is assumed to be
~rr=(c°sO)31Is3+S]S~s- s(s~+ s2)1 0"4~=COS(~Ifl2S3 ~ fl3S2S2 ~" fl4S(
"~ $2) 1 (30)
zr~ = ( s i n ~b)fl5 Is3
O"0=cOS~
+sZs~-s(S2s +s~) 1
E
f16$3 +f17 S~b
E q u a t i n g like terms, the first equilibrium e q u a t i o n leads to llv 4
4ill q- fl5 - fl2 -~ 3v
- 2 f l ' - fls- f14=4 4 ~5 -- ~3 --
3v 4
13 16
9 16
(1-2V)[A2+2Azv_~] 2
C2(1 - 2v)
(sZb+ SZa)
9 16
(1 -
2v)ZD2 2so2s bZ
(31)
Seismic stress fields in hollow circular elbow
47
The second equilibrium equation leads to v 15 5//5 - / / 2 = 4 + 16
(1 --2v)
-//4 The
-
[-¼ + 2A2v- 3A2] 3v 4
3//5 -
9 (1 - - 2 v ) C 2 + 16 (s g + s 2)
(32)
first compatibility equation leads to
6 3 6//1 -4//5 +2//2 + ] ~ v ( / / l +//2 +//6) - 2
-4//5-
7v ( 1 - 2 v ) [ 1 +2A2] 2 -~ 2
2
v
2(//1- //3) q--1--.(_-v(//1 + / / 3 + / / 7 ) = 2
(1 - 2v)D 2 2 2
SaSb
(33)
The second compatibility equation leads to 2
5
6//2 + 4//5 + 2//1+--~--~_---vv(//1+ //2 + //6) -- 2
v 2v 2 + 1T[1
+6A2]
(34)
The fourth compatibility equation leads to 2 1 3v 4//5- 2//1 + 2//2- 1 +-----~[//1-'~- //2 -'~-//6] =2-'~-T"~-A2( 1 --2V) (35) The total set of nine equations is easily reduced by taking sums and differences (eqns (31) to (35)). The solutions for fll to f17 are: fl, = 1 [3F6 _ F7 + 2Fs] - F 7 + 5F8
f12fls-
6
F7 + F 8 6
where F6-
1iv 4
F v = 41q6 v 15
13 -16
(1-2v)[2A:v+A2-5 ] 2
~(1-2v)[2A2v-3Az-¼ ]
Fs = 7 + 3A 2
(1 - 2 v ) 2
H. A. Lang
48
and
3v /3~ = - 3 / 3 , + 7 - +
9
(1
-- 2v)C 2
(sg + S 2 )
/33 =/35 + ( 3 v + ? - ~ + ( l - - - 2 v ~ + (1-2v) z 02 2 2 \4 10/ \ e / 2SaS b /36 = - (2 + v)/3, +
4/35(1 + v ) 6
/32(4+v) 3
+ (3 -2 7v) ( l + v ) + ~1 - 2v (l+2A2v)(l+v) v /37 =/31v +2(1 + v)/35 -/33(2 + v) + 2
(1 - 2 v )
D 2
2
2 2b SaS
CONCLUSIONS The complete field of six stresses may now be summarized by equation number since expressions for all unknown coefficients have been determined in the body of the paper. Thus, we have: a r = e q n (2) + e q n ( 5 ) + e q n (21) + e q n (30)... % = e q n ( 2 ) + e q n ( 5 ) + e q n (21) + e q n ( 3 0 ) . . . cr0 = eqn (2) + eqn (5) + eqn (16) + eqn (17) + eqn (21) + eqn (30)... zr~ = eqn (5) + eqn (21) + eqn (30) z,e = eqn (2) + eqn (5) + eqn (8) + eqn (12) z~0 = e q n (5) + e q n (12)... The procedure used in this paper could be continued to determine further terms in the series for stresses but the algebraic details would be cumbersome. The solution presented in this paper is expressible as a Fortran computer routine. The programming is straightforward. The stress fields for the X c o m p o n e n t of seismic force can be derived by replacing the multiplier YR sin 0 by XR cos 0 in the expressions for a , %, %, and z~. Similarly, the multiplier YR cos 0 for stresses z~ and z~e may be replaced by - X R sin 0.
Seismic stress .fields in hollow circular elbow
49
REFERENCES 1. Lang, H. A., Stress analysis of pressurized elbows for nuclear components using toroidal elasticity, Proc. 4th Int. ConJl Pressure Vessel Technology, London, May 1980, Vol. 2, pp. 251-60. 2. Lang, H. A., Seismically induced stress fields in a circular ring sector, submitted to Res Mechanica. 3. Lang, H. A., Stress fields produced by a vertical seismic force acting on a circular ring sector, submitted to Res Mechanica.