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!TECHNICAL MEMORANDUMS,,..

NATIONAL ADVISORY COMMITTEE FOR’’=ONiUTICS

1?0. 999

STRESS ANALYSIS OF CIRCULAR FRAMES

By H. Fahlbusch and W. Wegner

LuftfahrtforschungVol. 18, No. 4, April 22, 1941

Verlag von R, Oldenbourgi IN.inchenund Berlin

,, . .. ., :.. ,- -.,., ,., ,.. .,,, - . . . -., ..,.., ,,Washington

,. . ...

December 1941

1’ .

https://ntrs.nasa.gov/search.jsp?R=19930094418 2018-09-07T23:22:47+00:00Z

—

‘~lllllllllll~l31176014404181

. NAT ION’A!i”ADVI SORf-”COMMITTEE” F.OR:AER.C)~.A~TICS

TECHNICAL MEMORANDUM

,,. .. . . . . . .. ..

NO. 999. . ..;,

STRESS ANALYSIS Ol? CIRCULAR FRAMES+,, .

By H. Fahlbusch and W+ Wegner,..,..... :.:,...,.,:,... ..

:.,.-”.:. .SUMMARY, ~ ~ ‘-::,,..

The stresses .in’”..~.fr~ultid’,frames of,”c;onsta.ntbendingstiffness, as encountered in thin-wall shells, are in-vestigated from the point of view of finite depth of sec-

-tionalarea of frame. Thesolution...is carri~,d-out. forfour fundamental load conditions. The method is illus-t-tia’ted‘on a,worked out example.’ ;’””’ . . .

...,.

,., . . ,., 1. NOTATION,,.., ..,,

P

s

s

I

EI

7s

r

R

static moment

inertia moment.,

. .

bending stiffness

shear flow .,-,, ,.variable angle at center

. .. .... . ,,.“angle at center

distance of neutral :ak;is.~~from’center,>

distance of shear flow from center.,, ., .,,>’..., ;:...’..,.,:.....”’ ,! ‘.distance of shear flow from neutral fiber

.,

e,.. . . . ,,,, ,.,, ,,,....,>.... ... .:,,., :.. . ,.j...,x distance’ of ‘s”hear.:center ‘from “center

M moment

*llBerechnung der Be.a,’fispr~chung.kreisftlrmiger Ringspante. 11

Luftfahrtforscnung , vol. 1’”8,no= 4“,April 22, 1941, pp.122-127.

II . -. —.

2

B

N

2

x

8

larvid10Rcanthe

XAGA Technical Mernorandnrn No. 99? ;:, ;;- .

bending moment .,,

normal force,. .,’.

transverse force.

.. . . . .statically undetermined quantity”

load and coeffi.cient, respect ively

. ,, ..,.:1-1;. THE FRAME IQU.ILIB.RIUM,. ..;, ; ..,’”- . ,..

......... .,

For the.”ayTli:cAtioh of transverse. forces in a circ~-,.

Rshell wit’h large’ ratio ~, circul’i:r‘f.rarnesare Pr”o.-.

ed. They are in equilibrium with the concentratedds and the shear forces from the shell. Eech loadinghe divided into the transverse force passing throughelastic centroid of the shell and the moment (fig. 1).

ingsheala).

With

Theuncle,r fl

transverse force produces, as a result ‘O”f’bendr transverse force, a sinusoidally distributedow in the shell that reaches ‘~o’:the frame” (fig.

,..

?-s = :s

.:

denoting the mfunction

represents the

The distafor sinusoidal

x

ome nt

she

I =ITR3S

of inertia of the

ar

Ts

fl?ow

J?_fiR

va

.nce of tva.riati

heon,

,.she(fro

~

R//Tsdu

“(-) ..=

g2

riat

ar cm th

& ‘Rn

circul

ion.

enter oe cente

= 1.27

Ela

cmo

,.

shell , the

ircular.vnts to

half

—.— .. .-, .- —.

NilCA.T.e;chni’cal;Memo r,andum No-:.99%. 3

,, :~}The. shear” fl.o.w..hp;pli(ed-at$the f:ram~ as result of a~~moment is” con stant.”and ,~mount.s:t.o(fig:.:1.6):

,, ,.,

The, distance of the shear center of a circular half forconstant variation from the center: is.: :

.- -’, . .. ,.

x= ;R~ 1’:57 R, .!.., ,“

The resulting shear, flow (fig. lc) follows from;

Ts. &sin Q+ MRn 2Re Tr

The distance of the shear center ,,?fa circular halffrom the: center &rnounts to : ,.,- I,

,~f.... ,... . . . . . ..””.... ,.. . ,..:,.

,.. . ,, ,,. .. . ,,, .,. . . .. .4++’R

.,, . ‘,’”x=— on one side ,.

17+2,.. ,

and

. . ‘x = 4,-,. ~R”lT - ~

: .,.’

on the other. ,-.

.111. “STRESS ANALys IS ‘(jr“01”’RCU~ARFRAMEs ‘

: -: ,;- LoadHCa$e A~””” “ ~.

Localized-Radial Force Acting on the Frame,.,. ..: ,. ,, i.”,, ,.,:,,. ,. .

Ordinarily the circular frame is threefold staticallyundetermined. but. in this. In.st:ance and in the subsequentload cases the solution can’be considerably simplified bycleverly chosen sectionalization . At point O of the

4 MAGA Te6hnical .Membr&ndum. No~ 9.99;

load in figure’2, the statically undetermined quantity isX3 = o. ‘The.signs for carrying: out the anazysts are given

in figure 4. The depth of the sectional area of the frame

is introduced by means of the ratio ~. The subsequent re-

sults are valid for @ (fig. 3).=

The elasticity equations re~d.

The displacement quantities generally follow at

EI 6ik =J

Bi Bk du

Determination of the bending moment curve B. in the

statically determined principal system referred to neutralfiber (fig. 4). The tangentially applied shear force ele-ment

sets up s.tpoint T in t~eframe the bending moment

dBo= - TS du e

The distance e follows f~om the geometric relation

e = R -r(sin CPo sin V + cos To COSQ)

Then the bending moment B. at Q is:

,9 Q

‘J

,!.’BoG-S sin Q. d~o + ~ sin ~

Jsin 2 To dCPo +

n l-l09

Ei.z&co’~@” sin CPo cos .90 dqo.Trf

.0 “ !..

1--

NA~A, Technica’1 Mernoiandwm

hence .. ..,..:, ... “. ‘:.””.

, .... ~o.l= P.{ ,, .9 -— i.~ ,:’og-=q;+..”r-~..:,~i~~\

No. :99~ ~~5

; ....:,-.“..

)+-R..

Ba=r (l- cos T) as a result of X2 = 1....’. ..: . .. ,,. :

For reasons of symmetry the “integration can te lim-ited to a half frame. .The,following loads and factorsare obtained. ,’ ..,.

fi :, .. . ,,,,...:’,”

EI tjlo =r (

,..

B1 % du= Pr ~- Rl/o

EI “~=o = Pr2” R(% -:2 “ “ ““: ~

EI il~. = ‘EI 623 > ra+’... ., :,. :. ....,.... .. ...

Solution of the ‘elasticity ‘eg.uations”gives the ‘magnitudeof.the. st,a$,i,cally,,undeter,min,edquantities, .r .~. , .:,

..4 :.. . ..:. .. .. . . . .. . ,,. , ,,

.Xl = - =., .X241-r ‘$(’-:-”iJ”’““

whence the ultimate bending moment ..’!. ..=’

. . . .. .. .. . . .. .

T( ——Pr. .g ~:in..~+=i-2TT )

; C,os.?p:-.’.l.’. (1)

., ,,.

............ . . . . . ,., . ,, .,,,,., ,, . ,

6 IJACA,?e’c,hnical Memorandum. No. .999

The final normal force follows from

.N= No +. XINl + X2N2 .,

The normal force No in the statically determined

principal system at po$nt cp %s obtained by’”splittingthe shear force element in the tangential component fol-lowed by integration from O to Q (fig. ”4“)

,,

dNo = - TS du COS (~ - ~o)

The norme.1 force distribution in the statically undeter-mined system then is ,.

and the transverse force variation

(2)

(3)

Figures 5 to 7 “show bending moments, normal force,and transverse force plotted against the frame circumfer-

ence. The ratio. ~ serves as parameter forR $=

1,

r–=le2, ~=o.8. . .R R .,’

Load Case 3

Localized Moment Acting Along a,

Diameter of th;e Frame (fi~i 8)

For this load the frame is simply statically unde-termined at point O. The elasticity equation reads

NACA Technical Memorandum No. ‘999, 7

630 + X3833 = o.

Bending moments, normal force-, and -transverse -force in. thestatically determined principal system are obtained as forcase A. The loads and factors are:

The statically undetermined quantity follows at

The final bending moment is

the final normal force is

—— . ——

N=-&1

sin V

and the final transverse force is

(,4)

(5)

(6)

as illustrated in figur.!es9 ,,t.o11... ,.

Load’ Case C

Localized Tangential Force Acting ,Along the Neutral Fiber,., ,..,

of the Frame at a Distance r from the Center “’

In this instance also the frame is simply staticallyundetermined at point 0. The procedure i’s the same .as

——

8 NACA Technical ~lernorandum”No”. 999

before,..

The intermediate an”d’’final results are;

Qo=& (‘4 sin CP + ,’I; Cos CP - —:~,,1

furthermore

and

——— .—...————.

IN=% ( 2R

Cp Cos q) + —r sin V - ~ )1sin V. )1

(7’)

(8j

(9)

B and Q are plotted in figures 13 and 14. N h~s thesame asp~ct as Q in load case A (fig. 7).

. . . ... .... ——.,

NACA Technical M“emorandum’No”O 99g””’.: 9

Localized Tangential Force Acting Alon&. $he:,outer,,=P.eri.phery,,

of the Frame at a Distance R from the Csnter (fig. 15)

The stresses, in this frame loading are obtained whenload case G id superposed by & moment of magnitudeM:> P (R- r); Then :,-’. ., ..,

(B=%” 2 sin q)‘m

-~cpcoscp-~sinq -q (lo)2R

.

L----

(11)

Figure 16 illustrates the bending moment distribution; N

‘N in &oad case C for ~ = 1has the same aspect as and

as Q in load case A at .:.= 1 (cf,. fig, 7). Q in load,...

case D has the same aspect as Q.jin load case+C at ~ = 1

(fig. 14).

IV. EXAMPLE.,, .,

* ,,,. ,. .,. .,

Find the stres’sks in ~ Ci’rc’ularframe with ratio

.- = 1.1 under the following loa:ds,;

(Fig. lT. )

By division’ of the ,forae’ P we get

. ,.., ,,,

.,

,., ..,., . . ,..’ .,

-——..-—- ——. -—,.. ,——.,.——..,,,

NACA Technical Memorandum No. 999

Load Case A: 0.9 1? = 1800 kg acting radially

c: 0.4 F = 800, kg acting tangentially

B: .40 mkg

,.

The resulting bending moment curve is fonnd, ..

numerically by superposition of the results from equations(1), (7), and (4), or by graphical superpositi~n of thebending moment curves from the basic losds illustrated infigures 5, 13, and 9. The same method applies to thenormal and the transverse force. Of gre~test interest isthe knowledge of the l~ngitudinal stresses from the bend-ing moments and normal forces.

BENDING MOMENTS (mkg) IN FRAME FROM

THE NUMERICAL SOLUTION

i It

90 “i80 180 270 360

A ““ ““””-:7.2 I 53,9’ ‘-1~1.~ -141.5

*

53.9’”’ -i7 .’2-3.0 0 0 3.0 0B., , ~,,. ,: ,2, + : 20””’0c’ ● , 20.0 -:2.7:’ 0 ‘

,..,Result -“37;2 5’3.6’ -1”~~.:’” ‘“-121.5’ 5&.’2 ‘1’-4”7.2

TABLE II

NORMAL FORCES “’(iikg) IN FRAME

FROM N,Ui$ERICAL SOLUTIONS e,.

,,. —

o. “ 90 l_80:’: ~~180 270 360case

A 378 -450 -378 -378 -450 378B o 41 -400 4(!0 –41 oc o 1 -39 0 0 39 0

Result 378 -448 -’778 I 22 1-452 ~ 378

NACA Technical Memorandum No. 999 11

Tables I and 11 give several intermediate values ofthe m.ethematical solution, while figure 18 shows the finalbending moment and ncirmal force distribution in.the, .c.ir.--cular frame; ‘- ‘“” ““

l?or the stress analysis of the rivets or welds be-tween circular frame and shell the shear flow distributionis employed. It is computed by the method indicated insection II and has for the particular frame loading theaspect shown in figure 19. The maximum shear flow amountsto 26 kg/cm.

v. SCOPE OF VALIDITY

1. The solutions hold for circular frames withsmall sectional depth compared to curvature radius r.In this case the curved member acts similar to a straightmember. Hence the stress distribution was assumed linearand the cross sections presumed to remain plane. The ef-fect of the longitudinal and transverse forces on the dis-placement factors was disregarded.

2. The bending stiffness of the shell plate comparedwith that of the circular frame wa,s presumed to be small.

3. The departure of the frame contour from the cir-cular shape due to elastic strain was discounted.

Translation by J. Vanier,National Advisory Committeefor Aeronautics,

3?3

c

IF’Figure 1.-

P‘ 2W

‘~ L. ——-,.,

a

Equilibriumof frame and division intobasic load cases.

;<1

Figure 2.- Load cane A; Figure 3.- Representationradial loading. shell diameter.

$>.f

of ratio r/R by equal Fiyme 4.- Identificationof Bo, I?o,~

and Si@B. ‘

cocoa

,,, , ,.. .... ..—-. —-—

lEACATechnical Memorandum Ho. 999 Figs. 5,6,7,8

afo f \K /“ i /

405 I\ . , ,.

0 i \ /

I 97° Wo0 1 2’70°\

3# o

h

/ \ ‘1

0

–aosL

*

-423’ 1

Figure 5.- Bending moments underradial load.

o

Figure 6.- Normal forces underradial load.

(.M r..B,= —2n Rslnv

—+

NO= — &!&- sinq,

Qo= +$Jcosp -1).

,..??igure8.- Load case B;

moment loWIlg.

Figure 7.- Transverse forces under radialload, concurrent normal forces

under tangential load (distance r).

.. . . .. . . . .

HAOA Technioal Memorandum No. 999 rigs. 9,10,11,3.2

.

a30

K

w

afo

00 0

-am

-azo

–(230

Figure 9.- Bending momente under Pigure 10.- IJormalforces undermoment loading.

6?30

A’ I I I I I I I I I I I 1075

/// I

i-u

-a

-a

moment loading.

Figure 12.- Load case C;tangential loading

(distance r). ‘

Figure 11.- Transverse forces undermoment loading.

NACA Technical Memorandum No. 999 Figs. 13,14,15,16

f%06 / \

Kf \

ao4 , } q

,. --- --- -, / ‘““’ \

o

0

Figure 14. - Transverse forces undertangential loading

(distance r).

–ao81I

i I I I I I I I

Figure 13.- Bending moments undertangential loading

(distance r).

Figure 15.- Load case D;tangential loading

(distance r).Figure 16.- Bending moments under

tangential loading(distance r).

:3,*!‘...

IUCA Technical Memorandum No. 999 rigs. 17,18,19

.,. , --

Figure 18.-Bendingmomentsandnormalforces.

Figure 17.- Sample fr~loading.

/P. 4970 Kg

600- 60 n

/ (kg - mkg \

300 - 30 “ j I \ / /

\ /N -8 \ I I }

\ 78000 -o I

/ ‘,90° h

/z 700 /\ 360°

f \

–300 -–30 I ‘\ \ “./l\ /

/ ‘/@

! ,// \ !

!/ ~. ,’– 600 --60 \

\

/-900 ––90 1. f

\ ,/’/!

– %20

-750 \

R’igure19.-Magnitude ““””andvariationof shearflow atframe.

I