DEVELORENTS OF CIARTS FOR DETERMINING SHEAR LMOENT AND NORMAL STRESSES
IN CIRCULAR,
RECTANGULAR AIN
ELLIPTICAL RINGS, SUCH AS ARE USED IN
TRANSVERSE FRAMES OF FUSELAGES.
by
FERNAND LECAVALIER
Submitted in Partial Fulfillment of the Requirements for
the Bachelor of Science Degree in Aeronautical Engineering
frcm the
Massachusetts Institute of Technology
1941
(Signature of Author)'
-
Acceptance:
Instructor in Charge of Thesis, .
,00,..
(Pdte) 24'If
ACKNOWLEDGMEINT
To Professor T.S.Newell, my thesis advisor, I wish
to express my sincerest thanks for his numerous assistances
and suggestions.
F.L.
TABLE OF CONTENTS.
I.
II.
III.
IV.
V.
VI.
VII.
Summary
Introduction
Development of formulas
Results
Discussion
Bibliography
Appendix
1. Symbols
2. Tables
3. Curves
DEVELOHENET
OF CHARTS FOR DETMTTING SIEARMOTEI'7r
AND NORMAL STRESSES IN CIRCULAR,
ELLIPTICAL RINGS,
RECTANGULAR AND
SUCH AS ARE USED IN TRAW>VERSE
FRAIS OF FUSELAGES.
Charts for shear, mament and normal stresses in rings were
developed, by using the so-called "Symetric and Anti-Synmetric
Loadings".The rings are sub jected to a single concentrated load
and the stress due to shear is
distributed ,according to the
ordinary theories of bending and torsion.
The formulas for the circular rings have already been developed
For rectangular rings, the developnent of formulas for shear, moment
and normal stresses is presented. No formulas were derived for
the elliptical case , due to the impossibility of setting up a
relation between the deflection and the moments.
Curves giving the coefficients for the determination of the
bending, shear and axial loads at any point in a circular ring are
presented. For the rectangular case, the curves give the coefficients
for the bending, shear and normal loads at one point on the ring,
and that for various positions of the load.No curve was drawn for
elliptical rings
-1-
INTRODUCTION.
This method substitutes two loading systems for the original one.
In one system, the forces are symmetrically disposed about the plane
of symmetry of the structure, in the other the forces are arranged
anti-symmetrically. There exist 12 relations connecting forces,
moments, slopes, and deflections.
The basic assumptions made are:
1.- The concentrated loads acting on the rings are held in equilibrium
under the distributed shear reaction in the fuselage skin.
2.- The fuselage skin is equally effective in supporting shear loads
at all points.
3.- The rings have a fairly large diameter as compared to depth.
4.- The rings have constant El.
Development of formulas.
Circular Rings.
When a force of magnitude 2P is
acting perpendicular to the plane
of the section, the formulas were derived for the moment and the loads
at point A. (fig.1)
Via
11
((cosek-1)
(MI
(?T -%6-s in o ')
(2)
N (P) ( A d,( - s inecco sd)
S. ,,lv
Obs
ot(l+cos
(3)
CO-
When the force is acting parallel with the plane of symetry
and anti-symmetry, the expressions for moment, normal force
-2-
and
shear force at point A become: (fig.ll)
M.: (PR)
+ coso(-
(1--o)sinc+
sin4 -
'
(4)
N,= (P) (3/2 - sinoL)
(5)
S. - (E) ( -;, c)4
(6)
s in oL -sin 2e(
In these formulas, the variable is o(., P and R for a given problem
being fixed.
Equations (1),
(2), (3),
(4), (5),
(6),
allow us to determine the
moment, normal and shear forces on one section of a circular ring,
for loads acting at various angles with the plane of a section. The
loads are resolved into components acting parallel and perpendicular
to the plane , and the results are added up
Rectangular Rings:
Load acting perpendicular to the plane of section. (fig.5)
U)
The following formulas were derived for the symmetrical loading
M.ds : 0
EI
M (R
(7)
cosf) ds = 0
(8)
EI
Eq. (7) is similar to the expression for the change in slope
obtained from the moment-area relations on a straight beam.
Eq. (8) is a moment-area relation for the tangential deflection at B
taken with respect to the tangent to the elastic curve at A.
Eq. (7) and (8) enable us to solve for the two unknowms M.and Nb.,
by setting up expressions for the moments at any section , in
of these two unknowns and the known quantities.
terms
From figure(36);
for
<(
M,
M, -. M..
-3-
for
d
4
(-t
..
1..
1r
M4
M, - P.a ( cot eo- cot q)
=M
.+ b.N.-P ( acot,, + b
2
2
Also:
for
o4
f ,
X. b
2
ds, = b sec' .d4
2
ds..
a cosec f.d4
2
,
X.
a. cot
T
2
2
where X is the distance froln the section to the axis AB. and is similar
to (Rcosq
) in the case of the circular ring.
Substituting these quantitie
M.ds
(1Q(b.)(cosecf .df) -
0 = vpb)secqdf.
( EI being constant)
in Eq. (7) and (8)
P(a)( coto(- cot9)(
cosec.df
( 1 sec't d+)
M(R cos
ds)ds
M (b) (b-see
2 42
.d?) +
UQ)(a cot? )(a cosec T.df)
2
-JP(a)( cotoa - cotq )(a cot 7)(_a cosec
22
2
sec.d)
+ (-b.)(b
.dy)
Performing the indicated integration and simplifying the equations:
M = Pa(a4b) cotel +
N
(3a4b) cot'o -a(ab). cotel-ba
8b(3atb) (atb)
= P -a cotL + 3ab(2a*b) cotot
i 2b(3a4b)
)
(9)
(10)
4b(3 a+b)
Eq. (9) and (10)
suffice for the determination of the moment and the
normal force at point A, in the case of the symetrically loaded structure.
In the an.ti-symmetrically loaded structure (fig.3a), the concentrated
load P produces twisting and shearing in the skin, and the ring acts as
a transverse stiffener.
-4-_
The twisting moment due to the load P is
(2P)(a cot~o) and is
F
resisted by the torsional couple provided by the four sides of the
ring. For the side OD , this couple is (taking moments about the
center of the figure):
q.\wIa
7
J
dx
F
=x
2
=q, b
a
4f
lni (x+V x+a')
4
-
in,
'+bL +[a
2 4T
b ) -2n, (-b + b +a'
b2
4
1
2
4
4
and for the four sides of the ring:
2P ) ( a cot
2
8
2 ( a - b ) VaJ* b"+i a ln b4 a+b'
-bi--a'
P. a cot W,=
or
where M= 1(2 (a;b)V2a-b
II
q,= P. a.
,so
+ b ln at- fa'W
-a+4ibI
MT (q.)
alin b4Vab 4- b'in a4Vab
-b47(ib"
-a4raW
(11)
cot el
(12)
MT
The longitudinal shear str ess on a section making an angle
with the axis is,
-_and
t I
But
q,
=t
V Qt =VQ
t
I
I
V -- 2 P
I M a _t ( a + 3b
3 )
about an through the plane of
anti-symmetry at A aiid B
Q is
and point A
the moment about A of the area between the section
,
and is
equal to
X.1
-5-
e
For,
A
Q
X
(2ab + a'- b't an
( b+a-b.tane)t
2
4( b
,
it.., <
i-.
e 4T
(b
4=
q
=P
q
P
(a't
4
+ a -4 b tan
a
tan
btane
-
cote)
-1 (2ab + a - b tan
8
)
q - q.+ q
And the total shear force
.
tan e)
(2ab + aT b'tanl')
-- (4rb.+ab tane)t
P.<
1(2ab
8
a
2
(a cot e )t
2
Seq<<,P
- a -b
-
*
1q2a
a cot( + 3 ( 2ab + a' - btan
2a( a i- 3b )
MT
9)
(13a)
(13b)
a cot.4 4- 3 cot e
a +3b
MT
( 2ab +'
F3
- btane)
2 a'(
a +v 3b )
I
a cot. MT
(13c)
For the anti-symmetrically loaded section, the following formula
has been derived:
JM
( R sine) ds
0
(14)
SEI
Eq. (14) expresses the moment relation for the radial component of
the deflection at B, taken with respect to the tangent to the elastic
curve at A.
For,
M, -- L..b. tan
0 .<Pq,
otq
q<
(15a)
2
M
M4,<Sa+
E
11..i,
9
g ( b - a cot 1) b sec'Gde
. 2
2T
M =.- P a ( cott - cotq)
(15b)
(15c)
q
MA= 4 .b tan (T-1) - P ( a cotd + b ) t
( bsec ede)
2
4
2
2
2
Sq a - _b tan (7T-))]a cosec 9-de(15d)
2f
2
2
For the circular ring is the distance from the section
to the
plane of anti-symmetry. Let Y be this distance for the rectangular ring.
-6-
e)
For
o
< P1'
0(r-q
Y
b tanq
Y
a
Y
=
2
b tan (7(-)
Substituting in Eq. (14) these values of Y and assuming EI constant,
M (R sin q) ds =M Y ds
O
) t JM.,( a ) ( a sec& .d4)
b se4.d
M(b tan ')(
-(Pa ( cota- cotf)( a )( a sec
2
L2
d q)
2
J
b tan (t-')
b secq d4
2
2
(16)
Substituting Eq.(13) for the values of q in (15) and in integrating
with respect to 6, M,and M4; substituting Eq.(15) in Eq. (16) and
integrating with respect to
P
S-
'
and simplifying
a ( a 4 2b)
3a cot 4 + cote
2ab ( 2a - 3b )
-
-
3b
MT
(17)
4a ( a + 3b )
In Eq.(17), Mis a quantity depending on the lengths of the sides
a and b , and can be expressed in terms of a or b if the ratio a/b is
known.
The variable oL defines the position of the loads P and can vary
from P,to (
-'1,) and from ( r+qs) to ( 2 it - If)
When the load is acting parallel with the planes of symnetry
and anti-symmetry, (fig.4) the following expressions are derived
Symmetrical loading:
The tangential shear stress is,
q,= ft=
where
VQ t
tI
VQ
I
V .. 2P
I
.:..
bt
( 3a
6
to the plane of symmetry.
4
b
)
with respect to an axis perpendicular
For
t ( b tan&)
2
S.
Q
X.
A.6
t b'tan&
_b
2
t
2
( a* b - a cot e)
t
(
t ( 2ab + b
4 ( a + b
a cot e)
-
-(b~tan6 )
4(a - b + b tane)
a + b 4 b tanb)
2
-
a cot &)
8
t ( -- btan&)
4
2
So, for
,<
q = 3 tan 9.P
3a + b
(18a)
q = 3P ( 2ab
(18b)
4 b - a cot e)
2b''( 3a + b)
< .
- 3P tanG
(3A t b
q
(18c)
Also, for
MI4M
map
(19a)
MT M..- P b ( tan q - t an 4)(19b)
2
1t
M+
M
N,(
b
a cot1)
-
-
P ( _
2 2
M
3M
if
qb b sece.de -
M.+ b N,-
-
b tand)
q, ( b - a cotf)b sece.do
22
q(
2
2
a - b tanf) a cosec''ede
(19d)
(19e)
- P b [ntan'?-tan
For this symmetrical loading: fram Eq. (7) and (8)
Ir
fM.ds - O
M'ds= 0
= M.(
b sec"'d>)
and
M.( R cos4) ds
=
k
b sec 4 d+)
-4
t{M(
J
O
tano()
j(a cosec4d4)
2w
J
2
- P _ ( tan#- tand,) b sec'4.d4. o
b ( b sec+d# )
0
R cosf ) ds
-FPb ( tan4-
*2
t MA
M
-
P b ( tan4- tanetL)
2
b ( b sec.d{)
2
2
a cot+)( a cosec'4 d 4 ) MA (-bsec.d+
2
2
W,
27d ) =0
sec'
b
(
(-b)
{P.b ( tan4 - tano()
0Cf2
-8-
2
(20)
2
(21)
Substituting Eq.(18a,b)
in Eq. (19c,d) and integrating with respect
to e gives Mand MI.
in (20) and in (21) and integrating
Substituting Eqlg9(a,b,c,de)
with respect to4gives the following relations:
( 3a + 5 ab + b1 ) tane&+ 2a ( 3a + Sab
b ( 3a + b ) tan& - 2b
M,.= P
4
b)
4 ( a + b )( 3a - b
22)
N,= P ( 6ab tan o(- 3a)
4b ( 3a i b )
(23)
For the anti-symietrical loading; (fig.4c)
The twisting moment is:
=
2P b tno(
;F
or
Mrq.
q.
(24)
P b tanot
M,
where MThas the same definition as in Ec.(11)
For
o4
M= S.b.tan+
<
(25a)
2
de<4 4A
<<
4,
-.
M= M,,.
M
(<'<
MIM
P b ( tan4- tano.)
2
q),
b - a cot4
S.a - P (a -b tan
2
2
2 +,
*
2
2
".4,
- q.(
de
(, b see
S.b tan(1r-t) 2E
6
3
(25b)
4.
) b secede
(25c)
2
a
T
2
+ b tan)
2
a cosec&de
2
- P b ( tan4.- tant)
(25d)
(25e)
2
From Eq. (14),
JM(R
sin#) ds =M Y ds = 0
M.Y.dS = M,!( b tan4)(b sec'Jdaf)- P b ( tan+- tnd-)( b tan+)(b sec d)
2
2
(iF
2
2
2
(a)( a cosec 4dc4) + Mj(-b tant ( 6 sec d4)
J 2
2
d4.)
tano')(- .4, .2 tan4
2
(26)
P b ( tan4- tan0)(-b sec4d4 ) .
22
Substituting Eq.(24) in Eq.(25) and integrating with respect to e
then substituting in Eq.(26), and integrating with respect tofgives:
-9-
2ab(
SI
.P
a
3b ) -3b
( a + 4b
]+ 2a
a 4 6b )
anOet M-r
(27)
4a ( a t 3b
Results.
Coefficients for bending moment, normal and shearing forces
have been tabulated in the appendix. From these, curves have been
drawn , for various positions of the loads , and also for various
values of
rings
,
a/b , in the case of the rectangular rings. For these
b/a range from 1/5 to 1, so that it is possible to interpolate
between these curves for any value of b/a.
Discussion.
The curves for the circular rings are symmetrical about
the horizontal diameter. Stresses at all points can be determined,
but it is necessary to resolve the external forces into components
normal and parallel to the horizontal diameter.It is to be noted that
the concentrated load for the original loading was 2P. So that when
using the curves
,
the coefficients should be multiplied by half this
load or P, in order to obtain the stresses at any point.
For the rectangular rings,
the curves are not symmetrical about
the horizontal diameter. When the load is perpendicular to the plane of
symmetry, the results for o( varying from 4,to -r should equal to those
2
for at varying frain ( 2r-4.) to 31r . But that is not the case. The
variableo
is
expressed in
these formulas in
the form of a tangent
or a cotangent, and these two trigonometric functions are positive in
the first and third quadrants,
and negative in the second and in the
fourth quadrants. For this reason , the curves give the same results
-10-
when ae is the first or third quadrant, or in the second or fourth quadrant.
All the formulas for rectangular rings have been checked , by using X
and Y as variables instead of e and
c
. It
is possible that the assumptions
made for the shear distribution in the case of rectangular rings should be
different from those made for a circular ring, due to the discontinuity
in a rectangle.
-11-
BIBLIOGRAPHY.
1. N1VELLP
.S. j
.T.A.S.,Vol.6
2. WISE, J.A.,
The Use of Synmetric and Anti-Symmetric Loadings,
no. 6
p. 235
,
April 1939.
Analysis of Circular Rings for Monocoque Fuselages,
J.A.S., Vol.6 p. 460
3. TIMOSHENKO, S.,
Vol. II.
,
p. 459.
Sept. 1938-9.
Strength of Materials,
APPENDDC.
F
G
7
P
B
SmAf
I
Try
AX AYAME)MY
c
IA
-
Pl-
LtrYF G. 4.-
SYMBOLS
2P .. the applied concentrated load.
R= the radius of the centroidal axis of the ring.
( clockwise moments are positive)
M.: the bending moment at point A
Na = the normal force at point A
S,= the shearing force at point A
ot v the angle between the radius at point of application of the load
and the horizontal diameter.
4 -a variable angle used in determining the moment at any section
6= a variable angle used in determining the shear at any section
q= the torsional force in pounds per inch of circumference or perimeter
q,= the variable tangential force in pounds per inch
q
. the total shear force in pounds per inch
. q,*
q,
C = a moment coefficient = Ma, for the circular ring
=M. or Mj.2 for the rectangular ring
P. a
P.b
C-= the normal force coefficient
Na' for the circular ring
PTIF
Na
for the rectangular ring
P
a shear force coefficient
=
S..
for the circular ring
S.-.
for the rectangular ring
P
=tan a
b
RESULTS
Circular rigs.
Loads perpendicular to diameter.
O'.
_
tC
1.500
1.500
3.138
3.114
1.455
1.320
1.120
1.002
0.604
0.307
2.936
2.759
0.856
0.560
2.527
2.241
0.000
-0.042
20
30
40
50
-0.148
-0.281
-0.420
-0.536
3.071
60
90
100
110
-0.614
-0.642
-0.628
-0.571
-0.481
-0.378
120
130
-0.271
-0.176
140
-0.098
-0.044
150
160
170
1.916
1.570
1.226
0.108
1.500
1.470
1.383
1.250
1.088
3.141
2.622
2.129
1.685
1.308
-0.009
0.912
1.ol
0.250
-0.040
-0.297
-0.500
-0.063
-0.077
-0.074
-0.071
0.750
0.795
0.659
0.590
-0.643
-0.074
-0.112
0.614
0.599
0.548
0.457
0.329
9.172
-0.557
-0.516
-0.442
-0.484
1.088
1.250
1.383
1.470
-0.500
0.003
0.044
0.000
-0.004
-0.029
-0091
-0.500
-0.484
-0.442
-0.378
-0.301
-0.225
1.500
1.470
1'383
1.250
1.088
0.912
-0.172
-0.329
-0.457
-0.548
-0.599
0.750
0.620
0.530
0.500
0.530
0.620
-0.614
-0.604
-0.583
-0.570
-0.'590
-0.659
0.750
0.912
1.088
-0.795
-0.015
-0.003
-0.616
-0.516
-0.557
-0.616
-0.678
-0.730
-0.157
-04225
-0.301
-0.378
0.098
-0.209
230
0.176
-0,380
240
0.271
0.378
0.481
0.571
0.628
0.642
-0.614
-0.901
-1.226
-1.570
-1.916
-2.241
-0.750
-0.725
-0.157
-0.643
-0.074
-0.071
-0.074
-0.077
-2.527
-2.759
-2.936
-3.071
-3.114
-3.138
0.250
0.560
0.856
330
340
350
0.614
0.536
0.420
0.281
0.148
0.042
1.120
360
0.000
-3.141
320
0.750
0.583
0.604
-0-750
-0.730
-0.678
0.004
300
310
0.570
0.614
0.380
0.209
0.091
0.029
0.004
0.000
280
290
0.500
0.530
0.620
-0.725
190
200
250
260
270
0.620
0.530
0.901
180
210
220
C S.
as.,
3.141
0
10
70
80
Loads parallel to diameter.
-0.500
-0.297
-0.040
-0.112
-0,063
0.912
0.000
-1.011
1.250
1.455
-0.009
-0.108
0.307
0.604
1.002
1.383
1.470
-1.308
-1.685
-2.129
-2.622
1.500
1.500
1.500
-3.141
1.320
Rectangular rings.
Loads perpendicular to horizontal axis
X
b/a
0.0'b
1
1
1
1
1
1
1
1
1
1
1
0.1 b
0.2 b
0.3 b
0.4 b
0.5 b
0.6 b
0.7 b
0.8 b
0.9 b
1.0 b
0.0 b
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
b
b
b
b
b
b
b
b
0.9 b
1.0 b
0.0 b
0.5
0.6
0.7
0.8
b
b
b
b
0.9 b
1.0 b
Cs5_
0.0000
-0.0314
1.000
0.291
0.9177
0.165
-0.0520
0.824
0.721
0.612
0.054
1.0.041
-0.122
0.500
0.388
-0.187
-0.0630
-0.0660
-0.0625
-0.0540
-0.0420
-0.0280
-0.0135
0.0000
0.279
0.0000
-0.0274
-0.0457
-0.0554
-0.0582
-0.0554
-0.0482
-0.0378
-0.0254
.8 -0.0123
0.0000
.8
1.000
0,833
0.820
.6 -0.0000
-0.0378
-0.0463
.6 -0.0492
,6 -0.0468
.6 -0.0411
.6 -0.0324
.6 -0.0220
.6 -0.0109
.6 -0.0103
.6 -0.0000
1.000
0.915
.8
.8
.8
.8
.8
.8
.8
.8
.8
0.1 b' .6
0.2 b .6
0.3 b
0.4 b
CM,.
0.176
0.082
0.000
0.719
0.611
0.500
0.389
0.291
0.180
0.167
0.000
0.816
0.713
0.608
0.500
0.392
0.286
0.184
0.090
0.000
-0.238
-0.274
-0.295
-0.300
-0.291
0.235
0,137
0.051
-0.024
-0.088
-0.141
-0.183
-0.215
-0.231
-0.239
-0.235
0.184
0.113
-0.Q49
-0,007
-0.056
-0.096
~0.129
-0.155
-0.172
-0.182
-0.184
Loads parnllel to hor. axis
X
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0.0
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0.1 a
0.2 a
0.3 a
0.4 a
0.5 a
0.6 a
0.7 a
0.8 a
0.9 a
1.0 a
1
18
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1
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1
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0.0 a
0.1 a
0.2 a
0.3 a
0.4 a
0.5 a
0.6 a
0.7 a
0.8 a
0.9 a
1.0-a
.8
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.8
.8
.8
.8
.8
.8
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0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
a
a
a
a
a
a
a
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1.250
1.094
0.946
0.808
0.681
0.562
0.455
0.358
0.270
0.193
0.125
-0.562
-0.487
-0.412
-0.337
-0.262
-0.187
-0.112
-0.037
0.037
0.112
0.187
1.595
1.453
1.305
1.162
1.020
0.875
0.737
0.587
0,442
0.300
0.156
1.250
1.093
-0.742
-0.642
-0.542
-0.445
-0.345
-0.247
-0.148
-0.049
0.049
0.148
0.247
1.565
1.420
1.280
1.135
0.996
0.853
-1.040
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-0.766
-0.625
-0.486
-0.347
1.528
1.384
1.242
1.104
0.961
0.821
0.682
0.540
0.402
0.258
0.119
0.942
0.804
0.677
0.558
0.453
0.358
0.244
0.200
0.138
1.260
1.094
0.939
0.797
0.670
0.552
0.448
0.357
0.277
0.210
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-0.069
0.069
0.208
0.347
0.711
0.568
0.427
0.284
0.142
Rectangular rings.
Loads parallel to horizontal axis
Loads prependicular to horizontal axis
X
0.0
0.1
0.2
0.3
0.4
b
b
b
b
b
0.5 b
0.6 b
0.7 b
0.8 b
0.9 b
1.0 b
0.0 b
0.1 b
0.2
0.3
0.4
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b
b
b
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0.0000
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0.909
0.132
0.091
0.0
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0.810
0.672
0.048
1.260
1*088
0.932
0.788
0.659
0.541
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0.769
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-3.046
-2.578
-2.110
0.641
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0.527
-1.171
0.7
0.8
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0.9
1.0
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0.430
0.348
0.285
0.238
0.208
-0.703
-0.234
0.234 0.261
0.703 0.186
1.171 -0.024
0.0
0.6
0.294
-0.036
-0.051
0.194
-0.065
0.095
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0.000
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0.904
0.806
0.705
0.603
0.500
0.397
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1.260
1.080
0.918
0.0882
0.0638
0.041
0.038
0.018
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0.7
0.8
0.9
0.0000
0.7
0.8
0.9
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0.552
-0.101
-0.128
-0.130
-0.132
-0.0314
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-0.0625
-0.0540
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0.0000
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0.178
0.494
0.356
0.214
0.074
0.290
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0.110
0.6
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0.354
0.281
0.223
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0.000
1.330
0.773
0.632
0.357
0.0085
1.470
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0,912
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