INST.

18 NOV1935

A NOTE ON THE PRESENT KNOWLEDGE

ABOUT TORSIONAL STIFFNESS OF AIRPLANE WINGS

By

Hideo Tsukada

M. S. Tokyo Imperial University

1931

Submitted in Partial Fulfillment of the Requirements

for the degree of Master of Science from the

Massachusetts Institute of Technology, 1935

Signature of Author Signature redactedDepartment of Aeronautical Engineering, May 15, 1935

Signature of Professor

in charge of Research

Signature of Chairman of Departmental Committee

on Graduate Students

I

Cambridge, Mass.

May 16, 1935

Prof. G. W. Swett

Secretary of the Faculty

Massachusetts Institute of Technology

Cambridge, Massachusetts

Dear Sir:

In accordance with the requirements for

a degree of Master of Science in Aeronautical

Engineering, I submit herewith a thesis ---

"A Note on the Present Knowledge About Torsional

Stiffness of Airplane Wings."

Respectfully submitted,

This is to acknowledge my indebtedness

to Professor J. S. Newell, under whom

this work was done, for his guidance

and suggestions.

TABLE OF CONTENTS

Introduction - - - - - - - - - - - - - - - - - 1

Analysis of the Loading of a Wing - - - - - - - 2

Angle of Twist - - - - - - - - - - - - - - - - 7

I. John B. Wheatley - - - - - - - - - - - 7

II. H. R. Cox - - - - - - - - - - - - - - 9

III. Paul Kuhn - - - - - - - - - - - - - 16

IV. Hans Ebner - - - - - - - - - - - - - 18

Conclusion - - - - - - - - - - - - - - - - - 33

Appendix I - - - - - - - - - - - - - - - - - 34

Appendix II - ------ - -------- 48

(~( ~4.JL-

~-1 U ~IIiI

1. INTRODUCTION

We find many modern airplanes which have stress-

skin wings. As a result of remarkable progress of airplane

in a few years, it becomes necessary to get such a wing

which has high torsional rigidity to prevent from flutter-

ing and twist during flight, as well as good aerodynamic

characteristics. As nearly half of torque applied to wings

are carried by stress-sking, even in conventional two spar

construction, we cantt divide simply the wing load to both

spars in inverse ratio to the distance to C.P. Farther

more the increased torsional rigidity of stress skinned

wing has the effect upon the design of spars. Therefore

it is essential to calculate the torsional stiffness of

wings, but it is very difficult to calculate accurately

the torsional rigidity of even simple form shell and always

needs some experiments to design wings. I want to explain

in this paper the fundamental principles and a summary of

the essential formulas which may be applied to decide the

torsional stiffness of cantilever wings with two spars.

As a system of normal loading on a wing may be divided two

parts one of which cause only flexure and the other part

causes twist of a wing as described in Section 2. Only the

analysis of the stresses due to torque will be dealt with

in the following investigation.

-a m~ui

2.

2. Analysis of the loading on a wing.

Now we want to divide any system of normal load.

ing on a wing structure into

a system of loads which would alone cause nothing but

flexure and

a system of torques.

How this partition of the external load system

should be made? There is a great difficulty of correctly

allowing for the contributions of the skin to the spar

flexural moments of inertia.

Let us introduce an assumption that the skin,

though capable of resisting torsion, gives no assistance

to the spars in bending. (This assumption is reasonable

when corrugated metal or thin three-ply is used).

Suppose two non-parallel spars of unrelated and

varying cross section as shown in Fig. 1 and that pure

flexure of the wing occurs when the loadings per unit run

of OX on the front and rear spars are respectively

w r and w(l-r) where r = f(x)

YXp

*fK

Fi<. I

3.

0 YR an axis defined by the wing root

YT YT an axis at the wing tip parallel to YR YR

0 X an axis normal to OYR

Xf Xr distances from YT T to the general section

measured along the front and rear spar respec-

tively.

Zf Zr are the deflections of the front and rear spars

respectively at the general section

9 the angle of twist of the wing at any section

c the distance between the spars measured parallel

to OYR at the general section

Af Ar the inclinations of the front and rear spars

respectively to OX

If Ir the moments of inertia of the front and rear

spars respectively

kf kr the stiffness coefficients for front and rear

spars respectively

K the tortional tiffness coefficient for a cross

section of the wing

w the loading per unit run of OX applied normal

to the plane of the wing structure along the

C.. locus.

t the torque per unit run of OX due to non-coin-

cidence of the C.P. locus and the flexural line

of the wing

4.

Mf Mr the bending moments on the front and rear spars

respectively at the general section

5.

Owing to the pu'e flexure, must be

Zf = Zr

for all values of x.

d2Zfd2Zr

(2.1)

with help of the simple theory of bending, we get

f o r r (d x)ifC f.0 0

JO) X;L) (x=

where

If COS30

If C0O30% + Ir CoB30lr

By double differentiation, the front spar loading for pure

flexure is defined by

=R Lkr + p d -R nrdx +zR Ad (dK61 X fo 0 OAf,*O

(2.4)

Then we can get the flexural line, which distant ahead of

the rear spar a variable amount r c. If both spars vary

their moment of inertia along OX, but keep the ratio Ir/If

constant, must be

r = R

or

Cr(- Y) (d X)

(2.2)

(2.3)

6.

The same result may be got when an isolated load

is applied. Now a loading of w per unit run of OX applied

in the direction of lift along the C.P. locus AB into

a loading of w per unit run of OX applied in the

same direction along the flexural line C.D.

and

a distributed torque of w c (n-r) per unit run of OX.

7.

3. Angle of Twist.

(I.) John B. Wheatley T.N. 366

The first idea people may get is that "how to

contribute torque to each member of a wing?" Wheatley

discussed this question for a simple case using the method

of least work and several approximate ones.

t 2

The assumed corss section is not distorted after the load-

ing and also neglected the work done in shear. Suppose

Pu , L' F and PR are applied to members of a wing as a

result of torque M. We have three equations of equilibrium

as

Pu cos n - PL = 0

PU h, cos r + c.PR = M = 1 Moment at B (3.1.1)

P, hi cos t+ c.PF

But as there are four members, we have on redund-

ancy. He used methods of least work to determine this

redundancy.

8.

W = Bending work + shearing work

= f (PU)

= f(Pu) = 0 (3.1.2)b u

From this Pu determined and from the equations of equili.

brium the remaining three unknowns may be found.

And then angle of twist is given

d;y -d: S, Y , d', Cas - I.A (_ cf)o-'s--S (3.1.3)C.

where S' = beam deflection

= angle between sloping drag spar and spar

opposite.

As an approximate method he advised to use the Membrane

Analogy Metvod, in which basic form, the equation is

= T

S' = shearing stress per unit length of

perimeter of the cross section

T = torque on section

A = area enclosed by the centerlines of the

side

The value of S' need then be multiplied only by

the width of the side to obtain the running load P .

9.

(II) H. R. Cox R&M 1617

Considering the equilibrium of internal with

external actions at the general section defined by x, we

obtain (Fig. 1)

d d Mr- + = 0 (3.2.1)d Xf d Xr

and

ftOd X = -C' - MdSmox - M,.SmOI- (3.2.2)

By integrating (3.2.)) and remembering that there

cannot be any met moment about any axis parallel to OY

Mf coscf = -Mr coso(r = (say) M (3.2.3)

Therefore (3.2.2) can be written

Ix =- K -19 + c- -M(tano( tono) (3.2.4)

This may be regarded as the general eq. defininf the equil-

ibrium of the wing. In general If, Ir, K, t and a are all

variables, but it is deferred on account of the great com-

plication. Therefore we will simplify the general equation.

At first let

This assumption is reasonable in the average two-spar

cantilever wing.

10.

Then

Mf = -4 r =M

and (3.2.4) becomes

tdx == - C (3.2.5)

Differentiating and rearranging

d2 M +t (3.2.6)

Even this equation is extremely laborious to solve graph-

ically. Therefore give to If, Ir, K and t simple

functional forms which could be made to represent approxi-

mations. Assume

EI f= N (a + b x)n

K = C (a + b x) n-2 4 (3.2.7)

N +1

2

p Ir/ If const.

and through a treatment (3.2.5) becomes

d'e 6n d2e _ Ce _ tdd x +1+ bx d x2 ~ (a + bx) 2 d X c (a+bx)"

(3.2.8)

P

11.

For the case of an isolated torque at the wing

tip.

tdx = TT say, and we have clearly

dce bn cde I d e T - =-- _0d1 X3 a+ bx d x" (a+ bx),2 d x C C(0+ bxK)"(32)(ai-b) 2 dx(3.2.9)

For the case of uniformly distributed torque tdx = tx

it follows that

J + 3 b d'e I ded X, a+ bx dx (ci+bx)2 dx

+ ta I 0b C (aI+ bx)" ""

put T = total applied torque

then TT T

t T/

The solution of (3.2.9) is given as

Tg =, F(x, n) -a

1: 1bC (c+L~X)'"'

(3.2.10)

(3.2.11)

and that of (3.2.10)

(x,n) F +(x,n) T

(3.2.12)

~-q m in~-

12

where

F3(x, n) = 1 aj. I n-3

b (n-2+:x )(a+bt)n- n-3

(n+an-2) - (n-2) ( a)

(n+a.2) + (a a n+an-3n- (an-1) (a )

+ 1 (n+an-2) - (n-2) (a )n+an-3 j () n+2an-3

(n~an-2 + (Vn

(a+bx ) 3-n-anMRb

a+bxa+be

mij

IFn(x, n)

b3(2n-6+") (a+b e)bL

n-- a+bx

a an+1(n+an-2) - (n-3) )("'a )

a (n+2an-3(n+an-2) + (an-1) (a$b)

ta+b an

an+1 (n+an-2) - (n-3) ( a) l

+1n+ano&3a+ n2n3(n+an-2) + (an-1)( a )+ 2 a-3

a+bx )3-n-an.1t (a+b -)

an=,2(3-n+ ~1n2+4b

1an

.m~j

1+ an

where

mij

13.

When the sections of the two spars vary independently and

continuously, we can get the successive approximation by

assuming K is constant.

In this case (3.2.6) can be written

=2 M + (3.2.13)

where - (3.2.14)E 0 T,

(3.2.13) may be written

4 JrM dxdx + Ax + B

In the process of solving this equation by suc-

cessive approximations any curve of M may be taken as a

starting point. Let such a first approximation be Mj. A

second approximation is then given by

MeV= MI dxdx + Ax + B

X X

a third by Ms = J M.dxdx + Ax + B.

and so on.

If we choose M1 = Ax + B the solution takes the

simple form

14.

M == R x + J10fKx (dxY f + iJ4x(d X)4 + -

And the relative deflection of fore and rear

spar is given

E d- +- -)d Ix a

Then angle of twist e be given

Cr

Those processes are executed graphically.

In above discussion ribs are assumed rigid

against flexure in their own plane.

If, as is in most general cases, ribs are tor-

sionally elastic, rigid against flexure, but flexibly con-

nected to the spars yA4 will take the following form

where E If :1 f J1f f t -e

n = a finite number of ribs

f= the torque which, if applied to the front

spar by the rib, would cause 1 radian twist

of the rib about the spar.

15.

tr = be similarly defined for the rear spar.

T/c = torque necessary to produce 1 radian twist

between the ends of a rib, including its

connections to the spars.

The foregoing analysis is introduced for the two

spar wing under the assumption that the spars are ldng in

comparison with the cross section of wing or the wing has

corrugated skin-covering. Because in a tube, the value

of K at any Piven cross-section

= function (the effective shear modulus of the

material of the tube) (the cross sectional

geometry)

provided that the tube is long in comparison with its cross-

sectional dimensions or is incapable of resisting axial

forces.

Therefore, if this analysis is used for a wing

with uncorrugated skin, the following assumptions must be

implied.

i) the skin is not attached to the fuselage at the

wing root,

ii) relative axial movement between skin and spars

is permitted.

These assumptions may tend to produce overestimates of

the spar bending moments.

16.

(III) Paul Kuhn's method T.N. 500

Paul Kuhn introduced a method to get the torsional

stiffness of thin shells subjected to large torques under

following assumptions.

i) the wall thicknesses are small

ii) the cross section do not change their shape.

When concentrated torques T act on the ends of

a tube of length L the resulting angle of twist is shown

by the expression

TL=-(3.3.1)CT J

where G is the modulus of shear and

J is the torsion constant of the section.

For a thin-wall tube the torsion constant is known to be

4 A2J A 9 (3.3.2)

where A = the area bounded by the median line of the

cross section

ds = a differential element of the perimeter

t = the thickness of the element.

In the case of box spars and stress-skin wings

the sides will buckle at a very small load and will be

transformed into diagonal-tension field beams, because the

wall thicknesses are very much smaller in relation to the

dimensions of the cross section than in the case of tubing.

17.

Hence, if the sheet does not buckle, the actual

thickness of wall may be used in the expression for J,

but if the sheet does buckle this value must be corrected

as

t ETrL

where

E = the modulus of elasticity

= an efficiency factor depending on the

average stress and on the flexibility

of the flanges and can be assumed to

decrease linearly with increase of

stress from unity at zero stress.

*u

p O *

T

0 0

0

*EE

EE

EE

**C

A

%M

0

0',

mhh

iimim

i

00

18.

(IV) Hans Ebner

In Germany this problem is discusses by Eggen-

schnyler, Reissner, Herter, Ebner etc. Among all Ebner

treated the torsion of box spar in the most general case.

i) Elasticity Equation

a = spacing of bulkheads

b, c = sides of bulkheads

db, dc , d = wall thickness of horizontal and upright

longitudinal walls and bulkheads.

1 = length of system

F = cross section of corner stiffener

Mr = individual torque about the longitudinal

axis at the rth bulkhead

mr = distributed torsional load in rth cell

Mr = torsional moment in rth cell = - 1M= MIi=0 r

M at the

Dividing the box spar loaded by individual torque

bulkhead into cells, we get a "principal system",

19.

wherein the cells themselves are in equilibrium when the

twisting moments Mr to be transmitted between each pair

of cells are aprlied as external torque. The stress con-

dition of this system is readily defined. But we must

consider the sectional buckling induced by the distortion

of individual cells.

If two consecutive cells are

unlike as regards load, form

or size, this cross-sectional

' "MOMbuckling at the intersecting

Figure 4.- Corner deformation bulkhead is generally not thewith and without

individual bulding of longi- same. Therefore to reestab-tudinal walls of cell.

lish the connection of the

cells in the box spar, it is

necessary to consider normal stresses applied between the

cells so as to balance the buckling differences.

In the case of square bulkheads, for reason of

equilibrium, the longitudinal stresses transmitted between

the cells must be such that their resultant effect can be

represented in groups of four equal assymmetrical normal

stresses of amount X in the corners of the bulkheads.

Then we can take the rational assumption that at

each bulkhead group of forces X are statically indeter-

minate. Therefore the analysis of boxspar is reduced to

that of statically indeterminate system whose static inde-

termination with n cells and free-to-buckle connection of

20.

end bulkhead is (n-1) times; for one sided and two sided

buckling resistant fixation is n and (n+l) times. The

effect of each group xr = 1 on the principal system is

confined to the two contiguous cells r and (r+l). Thus

the determination of quantities Xr with elastic bulkheads

is effected by means of a elasticity equation of five terms.

dl? x + .2

d X, + d'X2 + d 3 X, + d4 X-4 = r2 -0

C3. X1 + d.2X2 + d3.3 X 3 + d'3 4 X4 +0'.s X,-df>

Y-2 +& Y r X, I+Jy. y Xy I +d tXr r+ Y+,2 X rt2 r.O0

21.

By assuming the bulkheads as being rigid in

their plane, the equations contain three terms. The

absolute terms d'rO and the factors dcr, L denote the

mutual buckling of the right-hand bulkhead of cell r

relative to the left-hand bulkhead of cell (r+l) as a

result of the outside load and condition Xi..

Therefore it becomes necessary to decide the

values of fr,o and 'rL to solve the problem.

ii) Condition of Stress and Deflection of Principal

System.

a) For square cells

Conditions of stress of principal system due to

torque 11, uniformly distributed torsional load m and

longitudinal force X are given in following tables, in

which

T0r is the longitudinal stresses at the edges

due to Xr=l

et () C*()Y C2 are the uniformly distributed shear stresses

due to the edge loading tr

2 r are the additional shear stresses pertinent

to the normal stresses

Stc are the maximum shear stresses which occur

in the center of the horizontal and the

vertical walls due to Xr=l

0- the longitudinal stresses at the edges due

to V

T' are the shear stresses due to M

22.

'4 , tN and i;1 are stresses due to intermedi-

ate loads. Those are additional to the

stresses produced by twisting moments I7

which are formed by the torque Mi at the

bulkhead supports due to distributed tor-

sional load m.

Mi = a+ m+1 1 ai+1

Lt and VI = the linearly increasing tension in flanges

and uprights of square cells with tension

resistant walls, due to the directly ap-

plied components of the tension stresses

in the same direction

L" and V" = the constant tension due to the support

reaction transmitted by the flanges and

uprights of the vertically acting compon-

ents of the tension stresses.

The deflection factors of square cells with shear-resist-

ant walls follow from (with linear stress distribution

for a )

do'* LT K C(dx y + dx ]

c C/ M, + Ap me-

.2b C1/) 2z b . #c d . cd -a,

-Sicar eVistant > 3 b mb

Wa//5 421

dt TI 49& +cS (1 - -W + cdtdeI24c -al ----- -

?~ bolc "(/ 12- r. e/r Oirej

(2-c -~ -) Ar WA, c6~~~ c &rf zye niom

6 c d, A .- da tb te u if r /

mqare Cc// with Shear 00 0 WC4,-7 - * ' CX (a-X);

OC47167ant Walls and (... X) d-3de{/-

Tein forced Corne rs Af -j

Z"Vrc cell wim (2 7) (29Q)

7-erp si or Re ti start

Va //S ~b ec

(28) (30)

4:r/ m -f .L

11 }lhe da d.S 4F t, (/4S)

abcd anddAa

S,.5peclnof or a/keeds d,,. a., g7 wa/ thichne ss oo hori zo.~a Mn ymsy u ogeaotte lawri- R,,frioe omeof jn Vrh cem#

,4wies of 6w/khwadG Ym14 oedda wtsad lieds.-umal axis of the ro% bo/-Aeood a~. ope f IMPRmaNfol 401d JrtM1 A~eS

A , Pth1 of Systeop IF Cross seelton pe corner Swlyfewer -Pdsnue Asoe/a .*.Ce-# ,1\epnt 04de/a

ExWern41 or?(4c Mflis assumed efo-ollo

di vded in the horlZOnta/and V&r fre (C9~ We,(d+ ~, ith /mear stros s distri-

.GCT-r C2/ wa rA 6 d d S (bdA+ c d,)y

Shear Reslstwnf (2 3)

hc (bd + c d]d

If -. 4 _4_ k'A CJO' d- '20r de de S (64&dac&

I f Me bulk/>mads are n7e.d

Mf Alj..$ Mer and 6mlkeads are c/aohe m -*heirrp/one,* en5 o;rmOwjr

/oum IW.n* rMs MM5s he Qaded When 6al/rheads are e/osbc amd ryidow-%Lo disappear. /V -hw y are,

- = -^f~r Me a- / 6- + /-+ /A-btfdt 44p4 Cf MCMbe.S Of/ICOO;Fa Z n r,7da/s#?wr

CT =( + -- L-) .4- or== -2 E , psorg o .- ste aye

-r Mar P-2 t7# pndro2 Ao O bu/lhe"ds repesent

-tiloc WIMdI rstf-'rowc S

547"0 Cc// an b c ke p/fn 20' M-c haiv C4sGOr=r-g 2sa Ep 0 s2/*2 jff>' o edngkTr E.TS oo

4 d, A 304c 6ar s *.*- ppq9hA, 6%w

dc -i a 4, C M'WSt 'be rep/laer

W ilhg. R 3 ) ron wdnn f ( 2 5 ) s u hs ylfa &e / 4 e ?.( .2 S W St s b t o 06C 6 ac I

Cone e.+ 2 3<) rem O'nedo (ipa) b cd)+-_6F 1&sleaW ofi Odl a+ cd )

E d =2[ b - c .J+

54?"Ore Cells with c)f -- + +b c~ d c 6 4 0 1 O ro,)

al) _q.l _ b bc (31 *, . /4i uk/ed

Forekgualc&/s ad ?1fld bulkhCOds 460 40ded 71 A ad o,"v

I- ar~eb1~hra& rt C =n

E~&L. AC A=fr ')

23.

For square cells with tension-resistant walls,

they are

(i) c l Xa n d * ( I )L

6 -, K E F

is the proportion of one longitudinal flange of section F.

When the box has symmetrical size, the results

of tlose integration are shown in Table 2.

b) For Blunt Wedge Cells.

Assumptions

1) Shear stresses distribute uniformly

2) The longitudinal flanges aid in taking up the shear

from the bulkheads.

Abbreviation

x br - br-1 cr -r-l3 g Pbr-.l T='-

bx = br-l (1 + (3 ex) , cx = cr-1 (1 + T3)

F = F + g (bx db + c dc)

Fb = bxdb , Fc = ex dc

K = br cr-j + Or br-l

Then the deformation factors are given

cr)di>

ab,

~~-7 T 4 Xb

K~ bi CT F

in which the transverse forces

Tb = Qb + and T

and the forces of the longitud

L=+ Qb-- br-1 1+(33

a,

an, i

of the wall

= QcI+1

inal flange

0rQ l ; It1+T 3

When F = const.

&,, K

kL)

a c Qe.iQc. -CT C, dc

L- FE Fe L i, (+2

c y . C KC a< -r 3Y-. 7

y -4

.2 &(p+-1)

-2 6 (r+1

Qb,, QCK+QC- Qb,K [2 3 -r)

+ &((r-1)YI ( r-P I3

24.

CT~ F

( /+- Y2

L L , d

a, Qb Qb + 2

CT br- d b (+ P)2

P'

25.

The cross stresses are given

At the bulkheac'(r-) At the bulkhead r.

due to Xr = 1 due to Xr. 1 = 1 due toM

Qb.r br-l br Cr gb.r-1 a -br br- .r -b o = X yAr Kr Ar-1 Kr-l Kr

c or = Cr-1 br Cr Qc.r.- = Cr br-l Cr-1 Q -Ar Kr A- Kr- Kr

iii) Solution of the Elastic Equation.

With the dIx factors as defined in the pre-

ceding table, the elasticity equation can now be written

and solved.

But I want to discuss here an approximate solu-

tion as it is sufficient enough in most cases . In general

it is reasonable to assume bulkheads are rigid in their own

plane. (Under this assumption the unequal application of

torque at the horizontal and vertical longitudinal walls

(Mb Mc) results in discrepancies of the redundacies from

the exact values, which become so much greater as the

successive bulkheads differ in size and loading.) Beside,

the effect of redundancy at the remote bulkheadscan ordi;

narily be neglected. Therefore the system may be divided

into partial systems with two, or at most, three cells each,

and the redundancies determined from one equation with one

unknown or from two equations with two unknowns. Thereby

26.

only the partial system adjacent to the fixation and at

the points of sudden change of load or size of the cells

need to be taken into consideration. And the solutions

for the box spar with equal cells can be applied approxi-

mately to systems having unequal cells by assuming the

the cells to have throughout the size of the cells at the

point of constrained sectional buckling.

In the special case of equal cells and bulkheads

rigid in their plane, the elasticity eq. may take the

following form

Xr1 - 2 o( Xr + Xr+1 =r

When this spar is constrained against buckling on one side,

the solution of this equation may take the following form.

For a torque MO at the free end

+ smh r + r+n V aMX, r - cosh n bC0 (34)

where o / or o<<-/

for uniformly distributed torsional loading m

XV Sinh YIP ( )cosh n f

I_ _ C -bb -) 1 2 mco.5o h Y i 2(N - b c

(35)> or a<<

27.

For intermediate loading the following term must be added

(z)r = 1

cosh(n-r) Pcosh n 51

9(z)

2 (e -1) be

In above solutions the following abbreviations are used.

For the box spar with shear-resistant,

S= cosh

I-~,

16a 1 E

+ () (b+d,+cd)

longitudinal walls

(36)

m

b 0

db d_-1-- (37)

For the box spar with tension-resistant walls and corner

stiffener F

+ P + P2

S- + P2.

+ 6+ c,

-2- Pa +fIy 4+ .

-I a= 3

-11+ )F

== cosh-'Io(I

I F,{ +-)F

as above

I(36a)

(37a)

Wt f2

A,

With great value of nT'

a sinh Cf ^, cosh 1)

such that

and ycosh ni c?

the longitudinal forces produced at the root bulkhead

r = n for *( > 1 or o( < -1 are due to the torque at the

free end

Xn = ambC

(38)

due to uniformly distributed torsional load

aembe (39)X/n -AM n (a( -1) A

When rigid bulkheads spaced so close that

become very small,

(34) become

sInh ' -Cosh

f and T

't ~-10

t "C

and (35)

COSh *(I )C x

c-os h * J*(35b)bc

where

x = ra

1 = na

xP = (n ) =48 1 E

(b/db + c/dc)(bdb

28.

e\. 0

XY (34b)

(40)

X, = -I-

+ cdc)

29.

iv) Angle of Twist - Torsional Stiffness

The angle of twist A of a cell is given by

'6 9, = 's 5- +- X r,- x-, + "' F, x X

where C angle of twist of the cell in the

principal system due to Mr

a ri,. angle of twist of the cell due to

Xr-1 " I

SC~r = angle of twist of the cell due to

Xr 1

In the case of square cells with shear resistant

walls, this may be written

2 61Cb ci y r b C l

Then the angle of twist of the box spar between any two

bulkheads i and K is:

Q&9 RLK

K

y= 3

Therefore the angle of twist of the whole box with con-

stant dimension between its end bulkhead is given

C c _b C

2 6 c b c

(46)9-- 5 )

5-b c~- (X0-X )bcd ~ *

IL Mr Qrr

(X,-X,)i c (47)

30.

in which

F1, , = - n a M.

a M

due to torque at the end

due to uniformly distributed

torsional loading

The torsional rigidity is given

2 6 C2 Cr

t b i

For the box spar with only tension-resistant,

walls and corner stiffener F,

of the end bulkheads

the mutual angle of twist

is

+bdb de

1

+ (b+c) 2

8F

be en

+ be - C8F

n.,Z r ar1

1bc

E 9T' (o-)

db ac 8 F

where

(X - Xn)

6C (xo-Xn)

n

(49)

longitudinal

E n =

ddb

cdo

31.

The torsional rigidity is

b 2 C 2 E

db c1 F

v) Final Stresses and loads

Final stresses of rth cell due to torque M at

the bulkhead are given as below.

A) Box spar with shear-resistant walls

i) Shear stress

T, == , +, + '0,, X (41)

For square cells

+ + (X -- X,..) (41a)

' from (15)

' from (9) and (10) or (11) or (26)

ii) longitudinal stress

c-== _ ,I X,.., + crY X,

for square cell (8) and (13)

If the intermediate loading is taken into account

for uniform torsional load distribution m, stresses t*,,z,

and 0".2 (171 must be added.

B) Box spar with tension-resistant longitudinal walls

diagonal tension stress.

r

32.

== 0 + 0-YY Xr -I + c-., x r

For square cells

C Fn, l

C) Final loads

+ 'Y,yr ( X, - xr,)

fro m ( 2 T) cr, from (29)

in longitudinal flanges

L =L , + LY- XY-. + L,., X,

= (L'+ L),0 + (L'+-L")r, r-X,+ (L+ L)Xr(44)

For square cells at the left and right bulkhead

of the rth cell.

L, =L', . X,

X

+ L y, (X -XrI)

+ L: r.y(X, - Xr-, )

L r

from (28)

from (30)

D) Final stress in shear resistant bulkhead

,-, ,, -, f ,., X, + i-, X,.LOLrf Itr.y-g +d tV* + (45)

For square cells with equal length

Tur.O 4. (x,-, - z X,

't,,0 from

r,.,, from

4- x,.)

(18)

(19)

For tension resitant bulkhead with rigid edge members

tension stress of double the amount occur.

U; (43)

and _L #'I+

dipgonal

33.

CONCLUSION

I explained several methods of analysis, but

it is not easy to get accurate results even for simple

cases. There are many factors to affect upon the tor-

sional stiffness of wings, among all

stiffness of ribs

contribution of stress-skin

irregular cross section of box cells

efficiency factor deoending on the average stress

and on the flexibility of the flanges,

make the problem intricate. But until additional experi-

mentsl and theoretical research introduce more accurate

methods of analysis, above mentioned methods may serve

as a guide.

54.

Simple Examples

'S I lop

i) By method of Kuhn

Calculation of fs.

A = 12 X 5.03

fsT 2 At

T

2 x 12 x 5.03 x .016

1000 2000 3000 4000 5000

fs 518 1036 1554 2072 2590

Calculation of zfs 518 1036 1554 2072 2590

T

T

1.93

.92 .875 .825 .785.95

35.

effective thickness

te = 5 t = X .016 = .01

d = 2(12

1000

+ 5.03) =

2000

34.06

te

3000 4000 5000

te IL .0095 .. 0092 .00875 .00825 .00785

l 03590 3710 3900 4130 4350

3 4.06 3.93 3.74 3.53 3.35

Lo .37 .382 .401 .425 .448-TII

.0037 .0076 .0120 .0170 .0224

ii) By method

db = do

a = 12"

b = 12"

C = 5.03"

of Ebner

*016"

G 216a E 16 x 122 x 2E - LL 3.172b + 0 )(bdb + ad0) (12 + 5.03)2

12 - 5.03 6.97 .40912 + 5.03 17.03

1+ R

.-/1 + 3.172

= -7.121 - 1.586

T

S,4,4

36.

= .409 = -. 7091 - / -. 586

= co5hIO(1 2.65

+ sh 5 x 2.65

ch 5 x 2.65

-. 709 12

F7.122 - 1 12 x 5.03

= t.02 M 0

Mr r = mo a n

= 60 Mo

b + cdb =

2b c2

b - odb 2bc

12 + 5.03

2 x .016 x

6.97

.032 x 12 x 5.03

= .146122 x 5.032

= 3.61

CT 9s = .146 x 60 Mo + 3.61 X5

= 8.76 Mo + 3.61 x .02 Mo

= 8.83 Mo

8.3 Mo 4.415 x 10Cr

Mo 1000 2000 3000

MO

4000 5000

.0132 .0176(r2 .0044 .0088 .0220

37.

With corner stiffener

b*c (6+c) 2

dt 8 F

1 v C2

Z x 17.03 ( 6\ .016

I4 4 x

+ I T. o3+ F

2 5. 2 8

For F = .05

.075

.10

.15

.20

)(V()( + b 4-)bec2bC d, 8 F

=

6.1773_ x 6.97( .0116 +0 F7/

a 2.x . o3

21.95

20.95

19.33

17.72

16 * 90

For F = .05

.075

.10

.15

.20

o( x 6 o o +-

b9.44

51.42

== 47.38

43.35

n

S1

Moo

4-.L3 0

=

.985

.850

.784

.718

.685

p3=

(:x .o2 mo

= .05

.075

.10

.15

.20

4F 1000 2000 3000 4oo 5000

.05 .0059 .0119 .0177 .0236 .0295

.075 .0051 .0102 .0153 .0204 .0255

.10 .oo47 .0094 .0141 .0188 .0235

.15 .0043 .oo86 .0129 .0172 .0215

.20 .oo41 .0082 .0123 .o164 .0205

iii) By mermbranE

S'= T

2 A

analysis

= 8. 1000

Lwith corner stiffener

IL

web

Lee Pf = 5.03 S'

Pn = 12 S'

t=1

= .75

xx= *291

= .70

-. 70) 2]

Are

-- ----- .7 x I

= [.291 + .7 x (5.2

.5033 .01612

= 41.67 looo

= 100 T000

x4 = 11, o48

38.

I~ = .170

39.

I = L + IW = 11.218

Deflection

41.67 x 6033 x 10 x 11.219

12

T1000

T

1000

= .0267

= .00445

-f'0

/5 /5-..-

K,

Blunt-wedge cells with shear-resistant walls

s . -, S.

- 60 -

r ~ r-1- r-1

=b(O, + or -1)

I

Dimensions:

a = 15"

b = 20"

1

Ho = 1000

2\ 3 4o, 6 7.5 9 10.5 12

db=d, .02 .02 .02 .03 .03

1 = 15.02"

r_ 1 1 2 3 1 LI-

390 450

T1000

1000

330 1K, 1270 1

40.

The shears are

'b. r -1 Qb.o

1 .740 -. 222 -. 593 .222 -74 -28

2 .727 -. 273 -. 6o6 .273 -61 -27

3 .718 -. 323 -. 615 .323 -51 -27

4 .711 -. 373 -.622 .373 -44 -27

The deformation factors are given

a QbA. Qb,Kb d

c, Q. Q

(L) C Qb,i Qb6,kCT ,K = E fe 3 b

I Y-I-2QC Q

9 bi QCK -i- Qc,i, Q,ix

b cr-i Y21.. a -1+-

6

and the values of

+-CX db)

are

k 1 2 3 4

1 23.9 20.56

2 26.84 22.25 13.10

3 24.04 13.96 12.48

4 14.40 13.27

T)

(~)2

n2(+ 4

where

QO.O

Then the values ofS(T)

are shown

K j 1 2 3 4

1 20.56 20.204.44 4.63

2 20.20 19.70 13.225.45 5.69 3.90

3 19:54 13.05 12.926.74 4.61 4.70

4 12.92 12.805.34 5.43

1 2 3- 4Fel .09

and the values of'

.094 .149

C C .J

.156

are

iK 1 2 4

1 48.9o 45.39

2 52.49 47.64 30.22

3 50.32 31.62 30.10

4 32.66 31-50

As absolute term are given by

41.

2 3

V> -2056 -1662 -915 -792CTUv0 + 560 + 564 +387 +393

-d, -26 + 32 + 15 + 25

-1522 -1066 -513 -364

Then we get the elasticity equations

49.9o xi + 52.49 X2

45.39 xl + 47.64 x2 + 50.32 x3

30.22 x2 + 31.62 x3 + 32.66 x4

30.10 13 + 31.50 x 4

= +1522

* +1066

- + 515

+ 364

The solutions of those equations are

1

'2

x3

x4

- +27.06

- + 3.1

- + . 9

-+19.09

r

42.

C d 0

Stress conditions are given by

o ----~ d6 K,

o_ Q re n

of which values are given in

___C C,-,y

hcln u

the following tubes

r M Lb.o0.0

1 -1000 -185 -1-I83

2 -1000 -152 -150

3 -1000 - 85.5 - 85

4 -1000 - 74 - 74

b dr br c.r1 3 C dx c

T.=C. x'

661 .740 .4 1.85 -.222 .- 135 -1.463

-7-52 .727 .4 1.82 -. 273 25 .165 -1.503

3 .718 .6 1.194 -.323 97.5 .2925 -1.020

1054 .711 .6 1.186 -. 373 112.5 .3375 -1.031

43.

.trr

44.

r

1

2

3

4

r

Qbr-1

-. 593

-. 606'

-. 615

-. 622

bdb

.4

.4

.6

.6

b.r-1

-1*482

-1.515

-1.025

-1.037

,

r-

xI- I

1 C., I'Cc cr)

1*463

1*503

1.020

1.031

'az, r x r x

Xr

1 -185 -1.482 1.85 27.06 50.1 -134.9

2 -152 -1.515 27.06 -41.0 1.82 3.81 6.9 -186.1

3 -85.5 -1.025 3.81 - 3.9 1.194 -6.83 -8.2 - 97.6

4 - 74 -1.037 -6.83 7.1 1.186 18.09 21.4 - 45.5

r cre 0 C,,t -

1 -183 -1.463 27.06 -39.6 -222.6

2 -150 -1.503 -23.25 +35.0 -115.0

3 - 85 -1.020 -10.64 +10.9 - 74.1

4 74 -101 24 I 92 -25. - 9

-1

.......... &-

x r -X I.-I I -t,,r (xr-x,-,)

24.924 74 -1.031 -25.7 - 99.7

45.

Angle of twist

The final angle of twist of rth cell is given

~ or -o ,r- v- i + i r,r Xr

_ A q r,- -~- bdrI L rI+~

d Kr CX \+

and the angle of twist of the whole box spar between its

end bulkheads is:

AT r

(9)

-(ef3o. I*

(10)

- (4)

(I U

'4.-I

1 7.5 6 6.58 7.41 5.93 6.60 5.27 26.58 -797400 12.59

+843300

-t I I t $ I

-888900

11.07

9*56

27*06

3.81

ti)

Crr'

(Z)

c r1

(4)cV-

ca

(F.)

al V-

C 2I~

2

(4)mI *-rs

C7.)

UIhr

9.0

(a)

V'+ (3)

7.5 8.11

3 10.5

8.93

9.0

7*44

9.63

8.12

( I -)

.-00oX 1Cr40

10.44

6.76 28.11

8*97 9.65 8.29 29.63

4 12.0 10.5 11.15 11.95 10.45 11.15 9.75 31.15 -934500 8.05 -6.83 +23080

-90000

-13100

( 4) ( 5)

(13) (4) 4.0

(56) ( T)

'4 2

(0 s)

(16)

71 )

1 14.07 27.06 114200 -68320 1458 -469

2 12.56 3.81 17210 -916090 2179 -420

3 11.03 -6.83 -31620 -933620 4563 -204

4 9.55 18.09 83000 -828420 6074 -136

P4 = -1229

=-.307 x 10-3 ra]..

r

( 3)

b -- (O

47.

48.

APPENDIX II

REFERENCES

Strength tests of thin-walled duralumin cylinders intorsion E.E. Lundquist NACA. TN. 427

Buckling of a cylindrical shell under torsionSezawa Z.A.M.M. 5 1925

Metal covering of airplanes G. Mather T.M. 592

Problems involving the stiffness of aeroplane wingsJ.A.S. Feb. 1934

Torsional box wings J.B. Wheatley T.N. 366

The torsional stiffness of thin duralumin shellssubjected to large torques Paul Kuhn T.N. 500

Torsional loading on stripped aeroplane wingsH. R. Cox R. & M. 1436

Distorsions of stripped aeroplane wings and torsionalloading D. Williams R. & M. 1507

Summary of the present state of knowledge regardingsheet metal construction H. R. Cox

Some developments in aircraft constructionH. J. Polland J.R.A.S. July, 1934

Stresses in metal-covered planes E. H. AtkinAircraft Engineering July, 1933

Die Beanspruchung dun n wandiger kastentrager aufDrillung bei behinderter QuerschnittswolbungHans Ebner Z.F.M. Nr23 1933

Neuer Probleme der Flugzengstatik H ReissnerZ.F.M. Vol. 17 1926

Aufgaben aus der Flugzengstatik Thalau-Teichman

Zur Berechnung raumlicher Fachwerke im FlagzengbauH. Ebner D.V.L. yearbook 1929

49.

Die Berechnung regelmassiger, vielfach statisch unbestimmterRaumfachwerke mit Hilte von DifferenzengleichungenD.V.L. yearbook 1931

Verdrehsteifigkeit und Verdrehfestigkeit von FlugzengbauteilenH. Hertel D.V.L. yearbook 1931

Torsion of members having sections common in aircraftconstruction G. W. Trayer & H. W. March T.R. 334

Stiffness determination in certain cantilever wingsH. R. Cox R. & M. 1617

The torsion and flexure of cylinders and tubesW. J. Duncan R. & M. 1444

Recent aspects of stressed skin constructionE. E. Blount J.A.S. Vol. 1 No. 4

Some aspects of torsion in multispar cantilever wingsA.C.I.C. 627

Experiment on the distorsion of a stripped two-spar metalwing under torsional loading D. Williams R. & M.1571

On the effect of stiff ribs on the torsional stiffness ofaeroplane wings H. R. Cox R. & M. 1536

Analysis of two-spar cantilever wings with specialreferences to torsion and load transferencePaul Kuhn N.A.C.A. T.R. 508