.

‘ WTMTTW{ES‘HESEARCH

TECHNICAL MEMORANDUMS

NATIONAL ADVISORY COMMITTEE! FOR AERONAUTICS

---------

No. 985-.—-----

TABLES FOR COMFUTING VARIOUS CASES OF BEAM COLUMNS

By J. Cassens

Luftfahrtforschungvol. la, Nos. 2 and 39 March 29, 1941

Verlag von R. Oldenbourg, Mtinc’henand Berlin

--———..

,...- ,..,~. ...,.,, .,

WashingtonAugust 1941

https://ntrs.nasa.gov/search.jsp?R=19930094431 2019-08-20T14:40:15+00:00Z

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NATIONAL ’ADVISORY COMMITTEE:.FOR AERONAUTICS---..-..--+

.,, .,. .. .

TECHl$I.CiL’-’tiEMOiANQtiNO;NO; 985” ,-....,,.. ,.,

-- L----- ‘,4... . ..’TABLES FOR 00MEWl!ING VARIOUS. CASES OF BEAM COLUMNS*

. ..’.,.; ,.’.

BY-J:. Cassens ~~... .

,,: .,,.,

For a better understanding of these tables, the meth-ods by which the cases are computed are discussed first.The importance of the buckling-modulus is pointed out.

I. EXPLANATION OF THE METHODS

Aircraft design methods differ from other structuralengineering methods in.the selection of slender beams.Since the former came into being at a time when structuralengineering had already reached certain standards, it isnot surprising that the stress analysis in aircraft designwas largely influenced by the other. But it is surprisingthat Miiller-Breslau, for instance, annexed the columntests in his work l!GraDhical Statics of Building Construc-

tion’ on the subject o~ aircraft design. The slenderspars of braced wings, the slender members of steel-tubebodies and of the ribs no longer permit the calculationof meiobers in rough approximation first for bending andthen for buckling, but the analysis had to correspond toactual loading conditions, that is, combined bending andcolumn stress. The importance attached to this subjectin past years is readily seen from a glance at the oldZ.F.M. (Zeitschrift ftir Flugtechnik und Motorluftschiffahrt).For example, in 1918 and 1919 these problems were treatedby Giimbel, Prall, Trefftz, Mi.iller-Breslau, Reissner,Ratzersdorfer, Arnstein,. Koenig, etc. But the authorsreally dealt merely with the problem shown in the presentreport below no. 4 with corresponding applications.

It is true that in its principal beams modern air-craft design again tends toward short columns, but evenso, column-effect problems remain to be solved,--———-_—--—-_.____+________________________ ___________________

*tlTafel einiger Knickbiegef~lle. lJ Luftfahrtforschuhg,vol. 18, nos. 2 and 3, March 29, 1941, pp. 86-94.

. .. . . ..—-. —

2 “NACA T,echnic& l’Mernorahdum ITo. 985

Column effect an,d.its counterpart; bend3ngdeals with the case of concurrent longitudinal

tension ,and trans -

verse forces acting on a member. Hiit~e and various othermanuals, contain tables wnich give the transverse forces,moments, deflections, and various other data for bendingforces on the straight member with,constant bending stiff-ness. These forces are supplemented by the effect of themoment FY, where P denotes the longitudinal force andY the deflection at’ any one point. The relation betweencurvature radius p and bending forces reads:

1 day Mx+ - = . -— = - --

P. d~a, EJ

where Mx is the moment of all bending forces dependent

upon beam ordinate x, EJ the bending stiffness, and ythe deflection of the neutral axis. (See fig. 1.) Divid-ing moment Mx into a portion (~x) due solely to thetransverse forces and a portion affected only by P, a,f-fords Mx = lfx 4 Py, which, posted in the foregoing rela-

tion, gives ‘

k2yll+y=-%L. .

P(1)

with k2 = ‘+ . (The investigations ,apply to constant

ben,dii~g stiffness EJ and constant longitudinal forceF,) The simplest way of solving the differential equa-tioil is as follows: ,The homogeneous differential eq-~.a-tion is expressed as y = C eax; the values C are coef-ficients to be defined later and must satisfy the boundaryconditions of the problem, while a establishes the con-nection between the formula: and the original equation” (l).The c.o,mple.tedifferential equation is solved either bypower formulas, the coefficients of which are determinedby comparison, or else. by others, for inst,ance, trigono-metrical ones, depending upon the character of the rightside of equation (l). The majority of problems shown inthe tables were solved. by this method. The right side ofequation (1) shows that the law of superposition hol.d.sfor all loads applied transverse to the neutral axis.Each loading condition is to be computed for app,lied lon-gitudinal force 1?.

... . . ,-

NACA Technical ..Memo? andum”Na. ’985 3

-, . .Th6-’1’aw-”oY-superpo’i%it5.0’n’’”canbe-made clever use of

in the calculation of a ~6am”withcons$ant ‘bending stiff-ness and longitudinal force, the beam””being transversely ,loaded by several, arbitrarily many, concentrated loads.This merely requires expressions for deflections and mo-ments of a beam transversely loaded by one concentratedload, as found under no. 13.. ..

Under a concentrated load 40 the moment and thedeflection to the left of”the concentrated load is

Idx = ‘in‘ELsin C(IQ. k ---–~-a

Q() “

(Y p k ::––:

d= -— sin Cp- - xt )

and to the right

Mu = sin’ ~Q. k --:=-a sin ~

Therefore, if a beam. is under transverse load Q3, Q4,and Q5 in addition to the longitudinal force P (fig..2), the moment at point .xo

(%#’qo = -k-

)is:

and the deflection

k sin CpoYo = ---~—-- (Q3 sin ~3 + Q4 sin Iq4 + Q5 sin T5)

P sln a,.,.,,.,.. .. .“ -“ ~~ (QSd3+ Q.d4+’Q5 d.)

As a result of transverse loads QI and Cja with P

(fig 2), the moment at point U. = t ~ X. (*0 = ~- is\ )

m—mmmnnmlmn 1I II .nnmmmm-- m..m., ,,, , ....

4 NACA Technical ‘Win’b’iandurnNo. 98”5

,.. . ,.

and the deflection .

.

Mxo , U. = Mxo + Muo

and the deflection s at this point X. = Z - U. is:

8 =.yo + V.

This problem is naturally much more difficult if longitu-dinal forces are applied at the same pbints as the trans-verse forces. In this case the use of an expanded.Clapeyron equation such as. cited “Dy lifi21er-3re’s18.uin.his11Graphical Statics> ‘tvol. 2, 2, p. 643, is r.e,commended.Examples for such problems may be found in the Z.F.M’.,1920, p. 283. :

A second method for column stress calculation isgiven because the final results appear ,i.nsubst?.ntially””different form. Instead of trigonometric and’ hyperbolicfunctions, a quotient appears which gives the effect of,F in the denominator. Bleich, for example,’ computes’ the”central moment of a beam loaded, as in case no. 3,accord-ing to formula

g~ ( 1.032)

gt2 PX/P + 0s032Mmax= ~ 1+ ------ -— ——..-—--

FE/p - 1 = ‘~- ‘-~~~p - ~

But no advantage accrues from this nethod except for syrn-metr:ic load cases.

r .,. ,.. .—,.. .— .. . . ——... —-.. ——— -—--- . . . . . -..—-J

—

NACA Technical Memorandum”. No. 985 5

A third- method is bas-ed upon. “t-heapplicatio”ri-’”of‘k-he.,-,“passive energy 11or the pri,nc-iple of virtual speed (I%ppl,llDrang and ZWang* 11vol. I, pa 61) . At first sight theidea seems attractive, since the energy of the longitu-dinal forces X the deflection can be discounted in firstapproximate ion. But the, application. of ,the.method in-volves a great number of difficulties.

According to this principle the deflection of a beamfollows from the relation

J MiMk d=

1 &ik = —.ES

The method is best explained by the example illus-trated in. figure 3.

The work performed in the beam is computed after theactual load Q and P is applied on the “Deam deflectedunder load 1. It affords

Z

o 0

The first integral is the same as that obtained from com-puting the beam deflection under transverse load Q; letus call it 8.. The second integral is a function of the

looked-for deflection ~.

Herewith the relation can be transformed into

.

( ,:P”;X

6 1-

)

~ -~-i– d.. = so

\ oThe expression

(2)

6 NACA Technical Memorandum No.’”985

2 “ “.,.‘~fxdx, .J.

l?J‘o

can be found by trial; ‘y is ex~ressed. i“n a’ formula

which satisfies among others the boundary conditions fordeflection at x = O and X=ltangent at x = t.

as well as for the

,For constant P and XJ it affords

o

The result is a very, close approximation; hence

rE

Vew small values 60 are accompa.nieil by an appreci-

able deflection ~ if the denominator expression is li-ke-mi.se very small. If the denominator becomes eq,ual to zero,it affords an expression for the stability condition ofthis problem, which is generally known.

It gives

Tension bending can also be very readily computed by thismetnod. with P indicating a tension, equation (2) reads:

NACA Techni ca:lMemo randum No ~:985 7

,,. .,“..

: [’ijiE(~:a]”=jo,”:-’”- ~~p)

,,. .

whether a column effect case .accor,dirig,to,’this ‘methodz.is.easy or difficult to compute, depefids upon the success with,.

the formula for The Fouri&r a“nalys’i’kcan “be used by;“ ,

analyzing the expression for the deflection without theeffect of P according to I?ourier.

If the central moment of a beam loaded, as in case no.3,is computed by this method, it affordsfleet ion

5 g14 px/p6m = -— -— — ---—

384 E J PE7P- 1

and for the moment

Mg La F’E/P + 0.0281

max = ‘~- ‘-—--——-—1-—FE/F -

which is in good agreement with Bleichl s

first for the de-

result.

11. MODULUS OF BUCKLING UXIIEI?COLUMN EI?I?ECT!

Occasionally it happens that comparatively short col-Fumns must be analyzed for column effect where o = - isF

higher than the break between !!etmayer~s straight line andEuler?s hyperbola in the ok = f(A) diagram. In such

cases it has been found practical to replace E modulusby the buckling modulus. (See the writer~s articles Column Testings 11Luftfahrtforschung, vol. 1?, no. 10,1940, pp. 306-313.) To retain the E-modulus in such acase would afford much too great a safety relative to thecolumn load according to Euler!s hyperbola, while frompure buckling tests it is known that the member can takeup only one load according to the Tetmayer straight line.

., ..I

8 NACA Technical Memorandum ‘No. 985”.

Under comparatively low longitudinal stresses ~g the

E-modulus can be used even if certain parts of the sectiondue to combined bending and longitudinal stress assumevalues within Tetmayer?s range. In such cases the effectis the opposite to that of pure crippling of short col-umns, that is, the bending factor becomes effective. Thebending factor takes into consideration the experiencewith bending tests where it was found that the measured~~Br was greater than computed according to the expres-

sion ~B ~, where is the material strength from‘Btensile tests.

Translation by J. Vanier,lTational Advisory Committeefor Aeronautics.

I

Problemgraph

iden+if iustion

P_m-1P—. . . . . . . -

1

1

%r--

I

be+ Iec+ion u (P time VC31UC’)Tanya.+ -;$- (P. k ,, .1 )

2. Derivo~iOn $$ (Pjv 1. ,s )

Cc.mpr.zs>irm bendinq

~p.v=B ,,nv

Tension bend img

.“ _ MOT p.y=~.~in~+filo~

Pky’=Bc&sq— MO; P.ky’=Bfhf~+MO;

Pks. y’’=— B.sinp Pk:y’’=+B. Ginp

Y“=*(*-’) Y“=+*t-*)

Y“=-$%-fi%) “,=-%%%’)

p.y=B. sinp —Q. xP.ky’=B. cos~—Q. kPk:y’’=— Bsinp

g lx —xy-g Pp.v=A.cosv+B”sin~-T(

gk 1—2x)Pky’=— Asinq+Bcosq ——2(

Pk2y’’= —Acosq— Bsinp+gks

P.y=A. cosqj-Bsinp

g /.z —z:)— Ml—(M,—Ml) f—gk:—~-(

Pky’=— Asin~+Bcosp

gk 1—24=( ,—— Z( M —M,) ;

Pk*y’’= —Acosy —B. sin~+gkz

P.y=A. cosq+B. sin~

—~ l.z—a?)+itfzZ( ()1—; – gkz

kP.k. y’=-Asin~+B .cosq-~k(i-2z)-M,. ~

Pkzy’’= —Acosp —B, sinq+gkg

{

M.

Moments M.= —1?...l~~=~

SpeckIl Moments

Compression bendksq Rns;on bend;nq

M== MO.; +P. y P= Tcri~io.

Mm= B.sinp=-!&. sin P Mz=”MO:—p. y.

Bfz=Q. x+P. y

M== B.sinq=~.sinv

M==@. z-z’)+P. y

M== Acos~+B-sinp —gk21

Mm= = g kz

()

— 1 at cenfcr

Cos;

M==@wz2)+M1+(M3-MJ-; +P. y

Mz=A. cos~+B. sin~—gkz

Mm,= bei tang ~- = -~ = tang PO at po /nt xo=k. qo

Mm. =&—g‘Z=gk’(+-’)+%i”

. ,. . .

Cons+an+s

A and B

Member buckles under

COmpwx- TasLwS;*

Jfo B.—a:>B=-sm a

ia=lw

@.gly

A=+gks

B=+gkzdg;

lx = 18@

A=+gkz+M1B = gk’+M, gkz+M,

--~;sm aor

B = ‘S —si:acos a

+gk’t g;-

a 181Y

A=+gkz —M,,

B=-+gkztg~+$;

wA

Mm,x = ——1

Cosq.‘ — +-*gk2=gh (C05T0 LK= 257030”

Problemgraph

id.zn+ifical-!on

%*

Find.: MB “’

‘q”%’w’w:it”z’”t!=> - v,: II’*=L- v’,

[n fttld t,: J,,; E. J,==lri[nfield fs:J,;’@. J,=B~[n central field:

‘-’V%‘% 1

P-l? a=%

jll~ht curved member

P

fz21

- -<:>=F=- ‘L,,

Cuwa+qre cqua+!w

for F=41:u-f. slnz+-,

PLI- Eule?lod==E.J $;

1a.—k

S I/qht curved member

Cut-va4ure equationfilr P-o:

4fu-?. (t. it-@) @bl- ~

——Tflec+vw (P +imc value)

-rangentdug

z (P. k I ,, }

2. Derku+ion ‘vm (Pk’ !, I. J

P~y’*— Asinp+Bcosp-g T&(l-2x)

P?hsy’’=—Acosq-B. sip~+&k:

U“k”(1-’2 $) + (M,— T.

—Ml). +

~P’y’’=. -AcosB-shp+gk:k:

‘P.jP,y, =~ .sinh~+.4cosp.+B ,sinp

,——P 1

p.f. ~Pk.y’=~,.; .casx~— Asin~+B.cos’~

+–3

ppyll=_ p’. j—.sinn~—Acosq-B. sinp1 ;’——

E

P.y=”Acos,~+B .ainq— P.~(lz-aP)-2PtksPky’=— A.sin~+B. cosp~P. k$(l —2x)Pksy’’=.—c os@sl Tlsini~+2Pk PE’. E

{

M.

Moments Ma=-E.J~

SpaCia{ Moments

l==$(l. x-x’) -afE+P. y A+ cen+er

4z=Acos~+ B.sinp —gks

()

11MM=g Pb. ~.——

‘E=gk’f-f+)=gk’l-:’:n”;:;

(48 *—5}(

l-coca 1m u )

~ga =gla. -—_a-ma 2

M.=:(2. Z—&) —af*--(a.f*-afJ ;+P. y

iUm=A. cos~+B. sinp-g~3from P.kyl’ and P.ky;

the eauations fnr Ml and M, read

~’P-&)+M’(&-’)=g’(:”@i-:)-y’”~”’

M== P.u+P. y

M== ‘“f—.sinn~+A..cos ~+B~sinq1—;

MP;+

max=— d cenfcr

1—;E

M.= P.ll+P. yM== A.cosp+ B.sinp—2PkS. Eor

(M==2P~. kz cosp+tg~sinp—1)

Con”s+an+s

A and B

-sIember buck(cs under

d=+@ F—ME ~

1$C%, =w g

4=+g’ks —M, i!!

B=+g’&tg;F

_M*—dflcosa ~sin a o

.

rvD=oO*n0miqa40r

ND= d.t..m,nu. t

A= O;B=O

a=lW

-0

a = 1800

!

I

I

Pmab IcmNr. graph

ide nti fic&on

as 9

10~(+

Curvature equation

‘or ‘=O; g.=o 4/u=#(l..z—#)fnlt~=F

?,.-l - -

P“m--~J~--------

11” ix

ICurvature “equati;nfor P.90: c-O. .

u- E.(l. x—ti); t-:

Find : .W.

I S;cjh+ curved merdret

+

m~“s.’ ,,:

-1r“ —--- -rM ‘J

x~1~

Curva+urc equai ionfdr P- O; g-O

U= WAX--*);+Find : MS

—4?- ~=—~~p“y=A”cosv-1-B.sinv ~ ( M.l=:(l. z–&)+P. E(2z–&)+P.y–P. $(z; z-#) -P(g+2P. q

Pky’=— A.sinp+B. cosp M== A.cosq-f-B .sinp-~s(g+2P. @

gk 2—22)

()

Mrm. =F(g+2P. g). ~–—T( —PE. k(l —2z) 1 in ccn *CF

Cos;Pk?y’’=-A. cosq-B. sinp+k’(g+2P, ~)

Deflcc+ion “’ 6P time value)d“v

{

4+!.Tangent (P. k ,< ,, )

Cams+an+*

~ Moments ?&= -EJ~ A and B

2. ckrivu+k ~ (phi - “ ) Specie I Moment M*mber buckles under P

I eI [A=+k’(g+2P. q *

E

I a=lw B

: (la!-&)P.y=A. cosp+B. sinp ——

()—P. E(t. x—&)+Mz 1—; —H(g+2Pf)

P.ky’= —A. sinq+B. cos~

gk 1—22)—T. ( —PRH1-24-ME+

PP. y’’=— A.cwp-B.8hq+ks~ +2P. ~)

: (t. z—a!qP*y=A. cos P+ B.sinp---

—P. g(t. ?–&)+M.—w(g+2P. F)Pky’=-A. sinq+B. cosq

‘k l—2z)— Pk&(l —2z)—F(

PPy’’=—A. cosp-B.sinp+P (g+2Pl)

A=+k’(g+2PF)-ME?

B=k’(g+2Pt)tg+

()MS=:(l. Z—&)+PS(ZZY&)—MX 1-; +p.. y M. P

‘tga ~iKe=A. cosq+B*sin p-P(g+2PF)

““!-T%)= (. )

●

(gt’+2Pk. q. L.. tg;-+ @

$%

1 A=+ P(g+2P.E)-ikfE

Ms=#(lx-#) +PF(Iz-&)-Ms+P.y B=[ka(g+2PF)—Ms]Lx

M*=~. cosq+B. sin~-lP(g+2PF) a+ center : xtg~

Mx=F(g+2PE)

“=F(g+2p’)”(%)or‘(=F’)(s; a ~a )M. -—— ( )=(gl’+2PtP) +.tg;–+

a=38tF’

13

()‘P.y=A1.cosp+B1 .sinp-Q~. x-M1 l-~ -M,?

pkY8=-A,. sinq+B1. cosp-Q~. k+ Mi~-Mt~

Pk*y’’=-A1. cosq-B,. sinp

()P.v =A,,cosy+B,.siny-Q ~.u-M1~-M8 l-~.,

Pku’=-A,.siny+B,. cwy-Q~. k-M, ~+Ms~

pk%’’=-A,. cosy— B%.siny

()M== Q+L+M, 1 —+ +M++P”Y;

(--+)+P.VM.= Q;u+W;+M 1

M== A1. cos~+B,. sin~; iUw-=Ar. c&~{ B2. sin Qwith Q the morncm+ is :

I f%= 18@

()P.Y =A1. cosq+B1. sinp-Q$. a+M, l—~

Pky’= —A1. sin~+B1. cosy —Q~”k—Ml~

Pk:y’’= —A; .cosq —B1.%inq

32=’+,0..,-i-+’=f:;”a~~, {~ci’’lMi;s~~12eflec+i0k V Constants

Tangent Nfoments W. — E. J — A and B

Ma ber buckIcs under

()W= Q;X-JA 1—; +P” Y; JfU=Q;-U-~I; +P’V

B,= Q.k E~,—~

MZ=A1. cosq+B1. sin~ M“=A*. cosp+Bs.8iny

M&fj=Qz(& -+) “ ,

8-%~ +L@?-.+..+.~

P-z u

b d1

y- nnd v- some as under 13.15 V=+: Y=+; but Ml and M, of opws;te siqns.

c=+; q=+

I Find: M, nd M, ]

()M=.==Q;x-M1 l—f —Mi~+P. y

Mu= Q:. u-MIf()

—jif*.l”_; +p.1)

M== A1. cosq?+B1. sinq Ms.=Aa.@Jv+&.sinV

M’o-&)+”’(*-l)=Q’’z(%-+) ‘1)M1(#a-l)+M,(l-&) =QZ(#-~) (i)

ii4~=Q.ksin ~. sin v ‘sii t

—M#&—M*-sin a sm a

Ax= —Ml; A,=— Ms

Bl=Q. k~

M,+lj%-- sm a

sin ~B;= Q.k~

-2++

a=3tMP

16

H, Mz

Find : Ifl .?d M.;p only for field N{

y’1: Y’(m-o); V’s”=u’(w)In field Z1:J1In L, 2*:JSIn Cen+ral fieid

y- und v- same as undc-r13.but Ml ~n”dMl of opposife s;g ~s.

A,= —M,; A,=—M,

B1=Q. k~

+_?&__ M,sin a

B,= Q.k%&

–2+%

ND= :=namin=+orND = d.+..rnima rwr

—

M.

17

18

—

—.Problcrnqraph

~dant~ficd+ien

Moments for P=O

Da?}rc+ion P (P timti value)

{

M.

Moments M,- –E. J$:Sp&l Moments

kna10gou5 with cohmm #cc+,sin ~. sin (aa+ as + W)

‘1 ‘Q’” k sin (%+~+~+at)

+Q,,k $inq. sin (q+-+Q; ,k sinm”sina~Sina 8in a

~,= Q1, ksinq”sin(~+a4)th a

+Q,. ksin(~+~). sin(q+a J+ Q~.ksim(al+ a4)”sina4sin a 6in a

M,=Q,. ksin al, sina4

sin’s

+Q,.kSin ‘aij$ ‘in ~’+Q,.k ‘in ‘“’+$ ”%)‘in@f

M,= XRs-f-P. y M“,=Lul.+P. v[n 81$

gg

il“r

‘#

ii

NI

2

—

2

Def[ec+’ion & (P “time valu~).

.$u&ents ($.,k. ,, ,. )

P~.y=A.aos~+B, sinq-Q.n+P. x~y<.P.k. y’.y-AWp+Bc osq”-Q”k+P. k,y/P. P:#=-Acosq-Bsinq

Intermetiiata values in ronqe @—@te be treated as under Nr. i.

Left sidc~ rarqe * Riqnt side: ranqa D&

P. y=ix. sinq-,$. x P.v=B,. siny—~”u

Pky’-- Blcosg–$”. k P.kv’= B,cos~–~”k

P~y’’~.-Blainp” pk*. V”= B, Sin ~

IM.@u

‘oments ‘= -– E “J =I Sp=i.1 Momen&

M@ =i14,.+Q. (i+P.d

df@=~-I-Q”k”tgt+ P”ayi’

M.= B1-sinp M4=B,. sin~

The momerit a-t S is:

M,=M-cm q. sin c

sin. aM,=

cm ~.. sin q

sin a

a= 18(P

NACA Technical Memorandum No. 985

I?i<,ure1,- Bean on two supports under longitudinal loadP and traiisverseforces g.

Figure 2.- Beam on two supports uznderload P andconcentrated,loads Q1 - Q5,

Qi

p—r

-—~ Problem

#’__~..,_______[“ Actual load case (i)

~~y.l..y

!.-~ .. r

EComparative load case (k.)

Jliip-x-+ ~w-’-------l—----+

Figure 3.- Built-in beam under load P and Q.

I

.

~‘illllll!lll!llnmll[fllllllllllllllll“’”’’:”””’””’=”’“’””-”’”:’”7“;;”31176 Ol&10 7283 ‘“ , ,,

—. ...—. — I

,.