Strain Energy Analysis of Swept Boxes with Ribs Normal to the Spars

VUIGTUIGBOUWKUNDi

2 7 JUNI 1953

DELFT

THE COLLEGE OF AERONAUTICS

CRANFIELD

STRAIN ENERGY ANALYSIS OF SWEPT BOXES WITH

RIBS NORMAL TO THE SPARS

by

D. HOWE, D.C.Ae.

This Report must not be reproduced without the permission of the Principal of the College of Aeronautics.

2 7 ,H!H! tt)53

R e p o r t No» 65

May, 1 9 5 5 .

T H E C O L L E G E OF A E R O N A U T I C S

C R A N F I E L D

The Strain Energy Analysis of Swept Boxes with Rihs Normal to the Spars *

hy

-D. Howe, -D.C.Ae. oOo

SUMMRY

The root constraint problem associated with iiniform rectangular swept boxes, having ribs normal to the spars is considered. A strain energy method using self-equilibrating internal end load systems is used.

Solutions are presented for the cases of a single cell box having either all ribs rigid, or the root rib flexible.

In addition, a consideration is made of second order effects combined with a flexible root rib. The case of a box having two equal cells, with all ribs rigid and a built in root is investigated, and the method of dealing with special root connections in this case is Indicated. The effect of the flexibility of the root rib in the two cell box is also considered.

In all cases, the boxes are analysed for loading by a torsion couple, and a normal force applied on the centre-line at the tip,

The basic theory apertaining to this method of solution, together with the equivalent unswept solutions, are given in appendices.

BHF

Work carried out xmder the auspices of a 1951 Clayton

Research Fellowship, awarded by the Institute of Mechanical Engineers.

- 2

LIST OF CONTENTS

k»0 Single Cell Swept Box - Second Order Effects

Page Notation

1.0 Introduction 5 2.0 Single Cell Swept Box 6

2.1. All Ribs Rigid 2.2. Flexible Root Rib

3.0 Two Cell Swept Box 10 3.1. Fixed Root Conditions

3.2. Flexible Root Rib

3.3» Special Root Conditions

20

Appendix 1 - Basic Theory 27 Appendix 2 - Unswept Box Theories 30

1.0. Single Cell 2.0. Two Cell

3.0. Single Cell - Second Order Effects Appendix 3 - Treatment of Triangular

Skins k5

NOTATION

The following notation is used throughout this

report;-Oxyz System of orthogonal axes, based on rib and spar directions. Aj_ ( 1 = 1 J 2 , 3 J R ) Boom area.

S A = Ap + 2A^ Sum of boom areas on one side of plane of symmetry.

a Typical length of shear panel, in particular rib pitch. a^ Rib pitch in first bay of swept boxes.

a. . Strain Energy Coefficients,, defined in Appendix 1 Eqs.(l), ^^ (2) and (3).

^kk' ^kE* ^k-f Particular Values of s-..s, defined in Appendix 1. Joefficlen

Eq. (19).

B Coefficient used to define shear flow, given by Appendix 2 R13 _ 2c/f3b 20"!

' - t / ^t- - t; b Total depth of spar web.

C^,Cp,C Constants used in solution of strain energy equations, and defined in Appendix 1 Eqs. (6), (8) and (9).

c Width of skin panel in direction of y axis. d Typical length of shear panel.

E Young's Modulus of Elasticity. G Shear Modulus

k Particular rib station, used in derivation of formulae.

L^ Torsion Couple (applied at rib i ) .

{ Length of box along the front spar»

n Number of ribs in box, root defined as 0. P. (i=1-6,R) End Load in booms.

SJ (i=:1-6,w,R) Shear flow.

T- """(l^a^bjc) End Load systems used as unknowns in strain "" energy calculations.

t Skin thickness t' Spar web thickness tp . Rib web thickness U Strain Energy

W Loading in Z direction, applied on x axis at tip X Specifically used to denote section between ribs

k, k + 1, (x = 0 at rib k) Z Normal load on box

a Coefficient used in calculating C. Appendix 1, Eq.(5), a = cosh 0

Q Complement of angle of sv/eepback

0 Function used in definition of a , Appendix 1, Eq. (5)

1. INTRODUCTION

The solution of the root problem of a swept wing having ribs normal to the spars, using the fundamental method involves much algebra, and it would seem that a strain energy method is preferable. It is known that one of the more satisfactory ways of applying a strain energy method to the solution of a structure, is to use the method of self equilibrating end load systems,

suggested by Ebner and Roller^ ^ and developed by Hemp^ '

This method has been applied to a single cell sv^rept box with rigid ribs, by Hemp in some unpublished v7ork.

Amongst the advantages of this method are that it provides a solution which is relatively simple to apply, and

that it can be readily modified to take into consideration the effects cf flexible ribs and second order quantities, such as shear lag. Different root conditions can be solved without the necessity of evolving a completely new theory. It must be mentioned that one disadvantage of the method is that it can only consider discrete panels carrying pure shear, and hence only average shears can be calculated. When attempts are made to overcome this, the calculations become lengthy and most of the advantages are lost.

In addition to the solution given by Hemp, which is Included for completeness, this work gives solutions when

shear lag and a flexible root rib are considered. The case of a two cell box with a fixed root is also dealt with, and the method of solving the problem when special root conditions

apertain is indicated. There is evidence that the flexibility of the first rib joining the mainspar root to the front spar is of importance, and the modifications to the solution for the two cell box v/ith fixed root to cover this case are given.

All the solutions are given for loading by a torsion couple, and a normal force applied on the centreline of the box, at the tip.

Appendices contain the basic theory, procedure for solving the strain energy equations, and the solutions for the equivalent unswept boxes.

2.1 All Ribs Rigid

This solution has been given by W.S. Hemp, but is included here for completeness, and because the results are needed for the other cases to be considered.

The geometry of the box structure is shown in Fig. 7. It is basically similar to the unswept case. Appendix 2, § 1 , except for the root configuration. The first bay is triangular in planform, and the hypotenuse of the triangle, which forms the root, is built in. The first rib pitch is a .

The assiomptions made are the same as those for the unswept case, the triangular skin being treated by the method of Appendix 3

Internal Load Systems

As for the unswept case, the basic internal load system is the doubly antlsymmetrical form.

Consideration of the spar and rib boom equilibrium in Bay 0 gives:-T Q ^2 " ^1 ~ a~ ^R = " ^1 ^2 = ^R o . o S J + Sp = 0 T Q . . S^ = - Sg = - S^ = 2i^ .. (1)

The distribution in bay k (k / o) will be as for the unswept case App.2 Eqs. (1) and (2).

Calculation of Strain Energy Coefficients

a = ^ a I for unswept case as there is only half a , = ^ a . j" the structure contributing to the

; coefficients.

.', Using App. 2, Eqs (7),(8) and

(9):-Ü0

--I-

U 1

00 ~ 3EA "^ UGa \t' "*" t>

' / ^ V .. (2)

lo 1 fb

01 ~ 3EA ~ UGa^ \t^ + tj

There will be half the unswept contribution to a.. from bay 0» and a full contribution from bay 1•

•'• ^11 = 3 E A ( - # - * - ^) + 2G (2ÏÏ;'" a)(^r-+ t) •• ^^^

All the other coefficients vi/ill be as for the unswept case.

External Load Systems

(1) Bending by Z v/ise force applied on x axis at tip Z = - W W

^3 = ;

JL

2b

where P is the end load in the upper spar booms at rib k.

w

Statically Correct Solution:- s. = - S, = p:g- 1

m r •• W

P = JL(( - ak)

Strain Energy Coefficients End Load System T .

There will be a contribution from Bay 0 only (a -x)

End Load at section x in Bay 0 = J^^:'-x)+T^—| + T^ . ^

o o T T Shear in Web = ^ + ° - ^

Wa^-C

. a 2b ^ 2a 2a o o 'oE - 2bEA I; 3<j'^ ÜGt^ •• O ; the contributions being from two booms and one web.

End Load System T.

In this case there will be contributions from both bay 0 and bay 1. The bay 1 contribution will be zero, due to

the internal loads being doubly antisymmetric, and the external loads singly antisymmetric about the Oxy plane

Wa

K I

2a \

From Bay 0;- a^^ = ^ (i - - ^ j - ^^^ .. (6) End Load System Tj^ (k j/ 0, 1)

All contributions will cancel.

.'. a^g = 0 (k 7^ 0, 1) .. (7) / (2)

th .„..

(2) Loading by Torsion Couple L. applied at 1 rib

= S.— S^— S-:^— S 2bc - '^1- ^2 k Statically Correct Solution:

-Strain Energy Coefficients End Load System T

a „ = i a - for the unswept case, as there is only half

(8) the structure . ' . F r o m A p p . 2 E q . ( I O ) : - a oE i4Gb

be

{t " F/

. . (9) End Load S y s t e m T 1

From Bay 0 : - C o n t r i b u t i o n = - _ilGbc ^ 1

L i /c Prom Bay 1 : - C o n t r i b u t i o n = ony^r, [T

^1E ~ i+Gb

1_ /c ^ \

be U " F")

; 2Gbc / c b_

It ~ t

^ / . . ( 1 0 ) End Load S y s t e m T, (k / 0 , 1 ) T h i s w i l l b e s i m i l a r t o t h e unsv/ept c a s e '^kE "= ° (k / 0 , 1 , i ) L. ^ i (o b \ 2Gbc \ t F i

I

. . ( 1 1 )

2.2 Flexible Root Rib

Apart from the fact that the flexibility of Rib 1 is considered, the details of this box are the same as those for §2.1

Internal Load Systems

The doubly antls,ymmetric system will still apply. From App.2, Eqs. (I), (2), and (3) we have:-_

Shear in rib webs due to T, :

'R k+1 'R k-1 k 2a a • • S h e a r i n web o f r i b 1 = S^ = ^1 2 a . 2a 2 a 2 a ( 1 2 ) / M o d i f i c a t i o n . . . .

Modification of Strain Energy Coefficients Contribution to a be

0 0

^%'°*E

Contribution to a

where t„ is rib web thickness be /1 1 o1

_^o^h

_{\a + a}

Contribution to a be 11 ~ i+Gt R \a _2. I i o 2 •*• a a ••• „ 2

J

Contribution to a be I 1 . 1 12 UaGtj^ \a -^ a .. (13) Contribution to a be 22 Contribution to a o2 Ua^Gtj^ be Ua^.a.Gt o R

The values of the coefficients now become :-Using Eq.(2) & ( 3 ) :

-2a

a 1 _ ' £ be

00 3EA ^ UGa^ \^^F "*• t "*• a^tj^

1 /b _{, - c be ii 1_1} 'o1 " 3EA ^ ÜGa \ F + t •*• t^ la "^ a. • _{R '•}

a

Jt_

_\J

_{-^ ^/ + 2G i^S; + aji^F + tj}

a a 11 ~ 3EA 2a _ 1 _ [b c _bc l^_ 1_ 12 ~ 3EA " 2Ga I F "*• t ^ 2tj^ |a "*" a^ 8a 1 /b c be \ + T^GtT i~2^a^a^^2 ^ \ a o a a 22 ~ 3EA ^ Ga \ t be o2 Ua^.a.Gt-o K (1U)

There is no external load in the ribs, hence the

external coefficients are unchanged. All the other coefficients will be identical to those calculated in §2.1

TWO CELL SYlTEPT BOX

3.1 Fixed Root Condition

Fig.8 shows the structure of this box. The general arrangement is similar to the two cell unswept box considered in App.2 §2, but there are two triangular skins at the root, which are assumed to be built in. The same assumptions are made as for the iinswept case, and the triangular skins are dealt with by the method of Appendix 3»

Internal Load Systems

All the three load systems of Pig.2 are possible.

Consideration of the spar and rib boom equilibriiom conditions, shows that the load distributions are the same in the root bays, as for the unswept case, App.2$ Eqs. (13),(I6),(17),(20)

(22) and (25).

Calculation of Strain Energy Coefficients End Load System a

(a) Only the front spar booms and skins will contribute to S Q Q • Prom App.2, Eq

(26):-/ N 2a

^00 - 3EA^ •*• Gta^ •• ^^^^

(a)

In the case of a^.' there will be contributions from the same components as above

• „(a) _ _ % c _ /./-N • • % 1 - 3EA^ Gta^ •• ^^^^

Both bays 0, and 1, will contribute to a|^ Front spar booms give:- l+a /3EA^

Main spar booms give;- 8a /3BAp

Bay 0 Skins:- c/Gta Bay 1 skins:- 3c/Gta

• • ^ 1 - 3E U ^ + Agl ^ Gta^ •• ^^f^

The front and mainspar booms and skins will contribute

Booms contribute:- T=; IT- +

_3E

_[A,

_{+ A^^}

and skins:- - 3c/Gta 0

• o(a) % /1 V\ 3c /.ON

•

^ 2

=

3Ë 17+

^ - Gti;

••

^^^^

Bay 2 will contribute fully to ap2 » ^^* *^® ^^-^ "* ^^^^^'^s will not be complete. Bay 2 gives:- i~ ( ~ + -r-] + T^^

Jüj t A. Ap i G t a

2% -^ 1 Ul Bay 1, spar booms c o n t r i b u t e : - -^rr- 1 7 - + j ^ j

Skins g i v e : - 3c/Gta

• • ^22 ~ 3E

(a^ + 2a) U(a + a^)

I + T

-It f - l - (^5)

1 x^2 j ^" \"o

In all other cases, k.,A > 2, App.2 Eq (26) will apply. End Load System b«

(h)

The front booms, web and skins will contribute to a^^'.

* 00

Using Appo 2 Eq.(27),

Booms give 2a /3Bft.^

2

Skins glv3 ^ (1 - B^)^ ana wet ^ ( | ^ ) = ^

O 0 0

2 ^, 2i

' • ''oo - 3EA

^-Gi;lFr(^')'-f(^ -^')'J- (2°^

The shear contributions to a^.^ will be equal to that lor a^ ; but opposite in sign.

I 2

2"-Shears give:- - ^ ] \ r (B^) + |(1 - B^) \ and booms give a /3EA^

• • ^o1 - 3EA 7 - G ^ ) F - ( B ^ ) ' - f(^ - S " ^ ) ' } - ^21)

( b ) . 2a _{1 4 b}

^ora\^^^; Bay o g i v e s : - _ £ , ^ | ^ ( B - ) -. f d - B " )

Bay 1 g i v e s : -

_3E

_(AT^AJ

_{^ Gaj5tT(B}

. a

M - iifo/i

3E

2

) \ ^(1-B^)^j

11

^(i; *Q^ê; l3^(B^)' ^ f (^ - B^

.. (22)

In the case of a i „ % t h e skin e f f e c t s a r e equal t o those from Bay 1 for a ^ . ' , but opposite in s i g n .

• c(^) ^o / 1 h_\ 1 f p b / ^ b x ^ 3c / , T^bx^ . / ^ , x

• • ^ 2 = 3Ê

\^A7

+ I^j "

GS;

j^F" (^ ) + t ^^ - ^ ^ J •• (23)

The contribution to app' from Bay 2 will be complete:

r ^ ^ 2 2f

Bay 1 gives» ! ^ ( X . ^ ) . 5^^ [ s ^ (B^)' . f (1 - B^)'J

• • 22 - 3E|^ A., * Ag"-^*%''f + O f \a * a j * Gt W%)

(21+)

For coefficients where k, -\ 7> 2 , App.2 Eq. (27) will hold.

End Load System c.

The front b o o m s , web and skins contribute to a A p p . 2 E q . ( 2 8 ) : -(o) 2=»-(o) oo Using 1 00 "" 3EA. "^ UGa 1

t)

.. (25) (c)

The shear contribution to a:^' will be equal and opposite to that for a (c) oo a _{o1 ~ 3EA,} (c) ___!c o1 1 ÏÏGa"

t)

.. (26) / The bay

fc) ( c ) The bay O c o n t r i b u t i o n t o ^^^ i s t h e same a s t h a t f o r a^ ' :

-2_%_ 1 | " b A

3EA^ + rjïïH^ [ÏT + tJ

Bay 1 c o n t r i b u t i o n : - 3^^^ ^ -A^ ^ ^ ^ „(c) ^ o _^ 1 f b ^ 2c'\ /P7N

• • ^ 1 = 3ËA^ + 2Gi; VF- +

—J "

(27)

Cc)

The shear contribution to a)^'^ will be eaual and opposite to (c) ^2

that from Bay ^, for a^^^^

* o(c) ^o 1 I b 3cj /r.o\

• • ^12 = 3 Ê A 7 - ü G i ; v F - + t 7 •• (28)

a 1 / b 2c\

a ^ g ' h a s a f u l l c o n t r i b u t i o n from bay 2 : - 3fr~+2Ga(F'"*'T"J 1 \, /

Cc) ita 2a.

(29) Bay 1 c o n t r i b u t e s 3 ^ ^ + - J _ ( ^ ^ + ^ j

^22 - 3EA^ - i+G ^ F ^ | a ^ + a j + t la^ + aj F o r k , ^ > 2 , a ^ ^ and a^V w i l l be a s App.2 Eq. ( 2 8 ) .

External Load Systems

1 ) Bending by Z-wise f o r c e a p p l i e d on x a x i s a t t i p Z = - W

Statically Correct

Solution:-This is assumed to be the same as for the unsv/ept case.

WA^ WAg

S^ = - S^ = =^^ S^ = :^^

where ;^'A = 2A^+ Ap End l o a d i n f r o n t and r e a r upper booms a t r i b k :

-WA. .

^1 = ^3 = bii^-^ - ^ ^)

End load in mainspar upper boom at rib

k:-^2 = Ss^^ • ^ ^'

(30) / Calculation •...•

Calculation of Strain Energy Coefficients End Load System a

All the contributions v\?ill cancel about the Oxy plane, due to the nature of the internal loading system.

/ . a^^) = 0 .. (31)

End Load System b

The front booms and webs Y/ill give the only contributions to a(^)

°E a

r WA. (%''^)

"^^^o^ ( ^o

^ ^ « - x ) - ^ — dx = - ^ ^ ^ H -

^

2 Booms give:- :^rr-° b a^b WA^ _ gb W A ^ B " Webs give :- ^ p . . ^ . ^ . ^ = - ötT^ o '''

(y.\ Wa^.( / a\ WA.B

% E - b E S A y 3 ? j G t ' S A •• ^^'^^

(h)

There will be contributions to a;™' from bays 0, and 1

^ 1E f° WA. Wa i / 2a s Bay 0, f r o n t booms g i v e : - ^ J - _ ^ ( ^ - x ) . ^ = - S Ë ^ A T" i t ) o WA.B"^ F r o n t Web:- ^rrr _Gt'23A a Bay 1 : - F r o n t booms g i v e rrr— - , _ , . ( < - - a ^ - x ) - ^ — ^ = i^^^ J DSA o a^ o Wa /,,) i+a ^ bEZ7A V 3 /

P r ° 2WAp (^o"^^ 2Wa (p ha \ Mam booms g i v e ^ J ^ ^ ( ^ - a ^ - x ) — ^ — d x = ^ j ^ (<- - y - ^ WA^B o ^ b WAo a _ b „ ^ b w ^ - h = . - 1 ^ 2B 2 o WB f^f . N ^®^^- G F ^ A + i r ~ - b C A : G F = G F 2 A (2^^2 ^M ^ • • /^s 2Wa^2 2WB^Ao 1E ~ 3 b E 2 A "*• Gt'SJA • • ^^^^ / The boom

The boom contributions to aA^,"^ in bay 2 , will cancel one another, and the front and main web shears will cancel

from bay 1 t o bay 2 .

WA.B^ Bay 2 : - Rear Web g i v e s - rr-r

G t ' S A a '•' ° WA. Wa^ /., 5 a \

J

WA. wa /* p a \ o a Main b o o m s : - ^ J — ^ ^ ^ ( ^ - . a ^ - x ) d x = ^ — ^ |^^- - 3 - j

o °

511>" ~ r ; 5 F S A •• ^^'^^

• • _{2E ~ b E S A}

w

For a^ % k > 2 App.2 E q . ( 3 1 ) w i l l a p p l y .

End Load System c

(c) The f r o n t s p a r y i e l d s t h e o n l y c o n t r i b u t i o n t o a^ '

Wa -^ / a A a b . WA. WA. F r o n t booms g i v e : - ^ ^ ^ (^ " 3 l ) ^^^ ^^^ G F " ' 2 i ; - b 2 A = 2 G F S A

( c ) ^^o'^ ( ^o'^ ^^1

^OE = bE2A I j ~ 3TJ "^ 2Gt'2A " ^^^^

(c)

In the case of a;^'^ also, only the front spar will contribute, the webs cancelling from bay to bay

Wa -{ / 2a >. Bay 0, booms give ^ g ^ ^ i1 j-]

Wa -( / ka \

Bay 1, booms give : g ^ ^ \^ - -j^j

(r,\ 2Wa i / 2 a ^ \

^1E - bE2A V

^ 1

•• ^^^''

There will b e no contribution to a ( ° ^ ^^""^ ^^^ 2»

^'^'^o fQ 5a \ - WA. Bay 1, front spar booms give ^g—,^' K ^ J , and web 2Gt*"-"^A

• a(°) - - ^ ^ ('>(-

^M

- ^'^ (37)

• • ^ 2 E " bE^l'A l ^ 3 j 2 G t ' S A " ^^'^ For a ^ ( ° S k > 2 App. 2 Eq. (32) a p p l i e s .

2) Loading by Torsion Couple L., applied at the i rib Statically Correct

Solution:-- ^ = Si = S 2 = 8 3 = S^= S3 = Sg .. (38)

End Load System a

All contributions will cancel about the Oxy plane

.-. a ^ ^ = 0 .. (39) (a)

'kE End Load System b

The web and skins contribute to a^.,'^

OJi ^

a b -L. gb L.B ^ Front Web g i v e s : - OF" ' Ijbïï * V = ïjiGbïï ' F*

°a„ -L ,. _ bv L. (1 - B^) Skins give. ^^ • Ubc a - UGbc * t

o

'^B^ = 5 ^

I n I

I B ^

|r - (1 -

B'')

. f] .. (UO)

• . a

The web contributions to a) ' will cancel. L. b e Bay 0 skins give j j ^ ^ (I - B ) :^

Bav 1 Skins give 2oa 'JX d - B ^ ) _ V f ü " i l z B ^ _ " ^ i ^ ^ ' ^ ' ' ^ £ Bay 1 slclns give ^^ . j j ^ ^ Ub^Gt' a ^ T 5 b ^ t

o o

•. a(5^ = 0

1E - "^ .. (U1)

fb^

All the contributions from bay 2 to a^ "^ will cancel about the X axis.

Bay 1 skins give j ^ (I - B^)f , and webs - j ^ . ^

. •

a(^) - 3 ^ ^n - -R^^ £ - R'b _b

.. (42)

End Load System c

The web and skins contribute to a ( c )

oE

^^^^ " ïïöéïï • 2F^ ^""^ ^^^^^ ïöfe • 2Ï

a _oE ( c )

_{8Gbc U " F"y}

_ _ c b \

. . (U3)

fc)

Only the skins contribute to a>„^ as the webs cancel from bay to bay. L. IE '1 ÜGbt

..

(kU)

(c)

The only contribution to a^ '' will be from the rear web, and a "triangular half" skin from bay 2, as the rest will cancel with bay 1.

a _2E (o) _ _8Gbc

_{be It}

1__ /c _b _t

..

(k5)

3.2 Effect of a Flexible Root Rib

There is some evidence that the degree of flexibility of the half rib [joining the mainspar to frontspar at station 1, is of importance, and the modifications to the strain energy coefficients are given in this section.

End Load System b.

The load distribution is given in App.2 Eq. (21) Contribution to a

(b) _

2 ( 1 - 2 B " ^ ) V

^

bc(B^)^

oo

^^A

O^E^O

= Contribution to a^p' and a p'

Contribution to a (b) o1 -U(1-2B^) c^ _ 2bc(B^)

^Vo

3E;i.„a2 R 0 Contribution to a (t) 11 = Contribution to a._ = 8(1-2B^)^C^ ^ 4bc(B^)^ ^^R^o ^^R^o ..

(U6)

/ End Load

End Load System c

This load distribution is given in App.2 Eq. (25).

Contribution to a (c) _ be 0 0

«'Vo

(c) fc) = Contributions to a^p^ and a^p'

Contribution to a (c) _ - be o1 2^*R^o = Contribution to a (c) 12 Contribution to a (c) _ be 11 ^^R^o

>

..

(kl)

There will be no contribution to the coefficients due to the statically correct solution. The contributions of Eqs. (U6) and (kl) must be added tc their respective coefficients as given in §3.1.

3.3 Special Root Conditions

The effect of special root conditions is to reduce the

niimber of redundancies at the section corresponding to the special connection. For example, consider the case of a two cell box, similar to that discussed in §3.1, except that only the mainspar is built in at the root, the front and rear spars being arranged to transfer shear, but not end loads, to their supports.

At station 0, the only internal end load system will be similar to that considered for T in the sins-le cell swept

o

box §2.1, and it will be statically determinate in that it must be equal and opposite to the front spar boom loads given by the statically correct solution at rib 0.

The load system T. , ?;ill also give only one possible

system, which will act in the front and mainspars, and be similar to the load system T°. At rib 2, the load system must fulfil certain equilibrium conditions. There must be equilibrium of the system on the inboard side of rib 2, although there can be no end load corresponding to the rear spar booms. Also there must be no discontinuity of end load across rib 2. The only

end load system which will meet these requirements is one similar to that of T.. This will give no internal loads in the rear spar booms, and an additional statically determinate system must be added to cancel the statically correct loads In the

rear spar booms at station 2%

The distribution of these loads into the individual

structural components, and the calculation of the strain energy coefficients, are made as given in § 3.1.

20

-SINGLE CELL SWEPT BOX - SECOND ORDER EFFECTS

The structure of this box is shown in Fig.8. Except for the omission of the centre web and the half rib joining it to the front spar at the root, it is similar to the two cell box. The centre booms are retained and the root is built in. The flexibility of rib 1, v/hlch In this case is complete, is

considered. All the other assumptions are identical to those of the previous cases. Appendix 3 indicating the method used for dealing with the triangular skin panels.

Internal Load Distribution

All the three load systems are possible, and the load distributions resulting from them will be similar to those for the unswept case. App. 2 §3,

Calculation of Strain Energy Coefficients

The coefficients due to the internal load systems (a) and (b) will be the same for internal loads only.

End Load Systems a and b.

The contributions to a^^'^ will be from the front booms, with

00 '

part contributions from the centre booms, front and rear skins and rib booms.

2

Centre booms give =;;7-^ 2 J

a

° . 2 a

^^^ '^ = 3ÊA^

a

0/2

The front and rear skins will give „ . •2 ° % / 1 \ 2c !

-p-c

„_, " Gta^

a^/ 0

The booms of rib 1 give =^

JiA.

( f . ^ ) '

^ = - ^

•R i o ^^^R^o

• • % o - ^00 - 3E ^A^ + A^^ + Gta^ 3EA a2 •' (^8)

R O

Ca")

In the case of a^.' there will be similar contributions. Centre booms give 2a /3EAp and skins - 2c/Gta

a / 2 Rib booms g i v e g g - 1 - [ ^ « ^

• a ( ^ ) - a ( ^ ) - to (± ^ J-) - _2£_. _ Uc^ f1 ^ ±.) • * 0I - 0I " 3E ^A^ ^ k^' Gta^ 3EA^a^''a "** a^-*

"'^ ' ' ° ' K ^ y 3EAj^a^ ^^ ^ o

^^ . A r> . o ^ I , « 3

(a) There will be the normal unswept contribution to a ) . from bay 1, in addition to the boom and reduced skin

contribution from bay 0.

prombayi ^ (J- , ^ ) , _ k |

2 f 2x 2 7a

Bay 0, centre booms give rrr— (-—) dx = EAg J ^a^^ ^ - 3EA2

^0/2

Skins give 2c/Gta and front spar booms 2a /3EA.

I. l* f rt r>

\/i

^

Rib booms give:

^ -

E^- J^l^a-a--ci

(|(a+i~)«cj

^

^

=

= 3 M j ^ L 2 + a^a

^

^2^

••• ^il ^-^?^= 3i7(2a.a^).3^(8a.7a^).if (a.^ .

^^

(^ifi^?^

(50)

There will be contributions from rib 1 to a)„ j a^„'^ and

(a)

a^p', v/hich will otherwise be as for the unswept solution.

i+c^/1 . J L .

"•^a a

o J

Rib Boom c o n t r i b u t i o n to a^^K ^ ^ _ ( | . Z | £ ^ ^ j ^ ^ = _ - | £ ^ i - .

• a(^) - a(^) - 2a , J _ _2.N _ J^c _ kc^ / i

J_N

/^.

N

• • ^12 - ^12 - 3E ^A^ + Ag^ Gta 3EAjja ^a + a^^ . ' • ^^^ ^

Rib boom contribution to a^„'^ is

^^^

22 EA-n

(a) ,. _k ° - " ^ '-^

_{(? . Ö dy =}

Uc-o ^ • ° 3KA,a2

• • ^22 - ^22 - 3E ^A^ + Ag^ + Gta + 3 2 '• ^^^^

c

Rib boom contribution to

a\^^

= =;^ j (r • ? • ^ • ?) dy

o^ EAj^ J a c a^ c

... a(5) = a(^) = Uc3 °

°2 02

^ ^

.. (53)

K O

For the other bays, k,-^ > 2, the unswept values, App.2

Eqs. (14-6), (iLt.7) will apply.

End L o a d S y s t e m c ( c )

0 0

rib 1 web, together v/ith part contributions from the skins. be

The contribution to a^^'' will be due to the front spar, and

0 0 Jr *

R i b web c o n t r i b u t i o n

2a

2cb (_±_\

_

Gtj^ ^2a^^ =

a _{0 0} (o) _ _{3EA^ "^ l+Ga^} o 1 / u iLC\

^F

b 2c> _t

2 ö t R a 2 b c 2^*R^0

,\

The r i b web c o n t r i b u t i o n t o a ^ ^ ^ i s TTT^JÖ-^ l-^rr ••• TTT—i

01 Gt^^l^ida^

_R

i^a '^^QJ^ be 2Gtt,a^''a R O ^1 1 ^ (a + S-) ..

(5k)

(c) _ ^Q_ _

• • ^o1 -" 3EA^ UGa^

1 - (-4- , 2C) b c _ /i JL) fa_ ^F^ ^ t ^ 2Gt^a^ '^a + aJ

R o o

.• (55)

In addition to the effect due to the rib web, there will

(c)

be a normal unswept contribution to a.^,'^ from bay 1, and a partial contribution from bay 0. Bay 1 : -^^ + ^ ^ ( T T + rr")

2a

B«^°=- 3lA7*I^(f-¥)

El. We.=- i^ (jL , ^ ) ' = 5£^(Jg - A * ^)

Gt

_R

_{R a'} a a _o

a

« ( c ) 2 / ^ , ^ N 1 f b / 2 1 N 2 c / 2 1 NI b e / 1 , 2 1 ^

• • S i - 3ËA7(2a+a^)+ÜG5F-(a+i^)+—(a^i;-)] ^^ 2Gt;^ i ^2 + ^ ^ + ^2 j

(56)

(c) (c) (c) T h e r e w i l l b e r i b web c o n t r i b u t i o n s t o a ) p ' ^ , ^ p ? ^-"-^ ^ p v/hich w i l l o t h e r w i s e b e s i m i l a r t o t h e unsv/ept c a s e . ( c ) 2cb R i b 1 e f f e c t on a 12 Gt _R M2 2a 3EA,

r j_(_ J i_)l _

"i 2a^ 2a 2a ^j " u o J _ _ L . r - b . 2 c N _ b e (I ±_^ 2Ga ^ F ^ "*• t ^ 2 G t ^ a ^a "*• a ^ - b e /I J_^ 2 G t ^ a ^ a "*" a '' R O . . (57) / Contribution

Contribution from rib 1 to a^p' = -rq— (—) JX be . • ts,, '22 8a 3EA,

i_

(i>

Ga

H

2C' + ^ (+T + — ) + be 2Gtj^a' 2Gtj^a' .. (58)

Contribution from rib 1 to a^^^^ = -p— (rrr . "rr—)

Oc. vjT/T-, c.a ^ a _ ^ R o ,(c) o2 be 2Gt^a a R 0 .. (59)

For the other bays, k,-^ 2, the unswept values given by App.2 Eq. (i|.8) will apply,

External Load Systems

1) Bending by Z wise force applied on x axis at tip Z = - W

Statically Correct Solution

This is assumed to be the same as for the unswept case App. 2

Eq. (U9):- W WA^

^1 - " ^i| - 2b ^1 = ^3 " b S A (^ " ^^^ WA^ ,

^2 = b é (^" "^^

g _ g^ _ »g _ _g _ I /i _ ^ x bg _ bg - b3 - ^5 - b ^2 SA''

Where 2?A = 2A^ + Ag

.v(60)

Calculation of Strain Energy Coefficients

End Load System a

T h e doubly symmetric nature of the internal system, together

with the fact that the statically correct solution is antisymmetric about the Oxy plane only, implies that all contributions will

cancel about the plan of symmetry _a

kE

End Lead System b

There will be contributions to a skins and front and centre booms.

(t)

oE from the front and rear

The skins contribute: a 2a^c o Gt

Iri ^1 ^

b*'2 ^^A-* ~ Gtb 2Wc /I ^1 N ^2 ~ SA'' / Main

Main booms c o n t r i b u t e : _EA

2i

° 2 ( a - x) WAo ? ^ ( ^ - x ) d x a 0 / 2 o •bSA^ F r o n t booms g i v e : - "° (a - x) EA. 1 H a . a ( ^ ) - 2WC , i „ ll_.N ^ % ^ • • oE •" Gtb "^2 SA^ 2 b E S A WA,

• bm(^'^^^ =

b E 2 ] A ^ 2 3-ï^ Wa -C a^ _2 Ci 0) b E S A ' _3>^ . . ( 6 1 )

Both bays 0 and 1 w i l l c o n t r i b u t e t o a . „ ( b ) I n bay 1 1 Jii «

the booms will cancel one another, leaving only a skin contribution.

Bav 1- ^^^ j i - ^ ^^^ ^' Gtb 12 S A

The components contributing from Bay 0, will be the same as for

^oE

'

Skins: - 2Wc _Gtb

Ii

tl

I2 " s

a

A

Front booms: -EA _{1 J}

_ Wa.-?

a^^"^"^^' b ^ A • ^ = bEZJA 2a, TT ("1 - 3:f) Centre booms: EA. I o WA. 2x '"^2 ^^o'^ 3 "^^o

~(^-x) ^ ^ J dx = :g^|J^ (f - ^:^)

a _{0 / 2} • c,(^) - 2Wc f i ^1 ] • • 1E ~ Gtb [2 ^ J Wa -C a . 2 (A 2.) ^ 2bESA ^' < ^ . . (62)

a _kE k ?• 1 w i l l be a s f o r t h e unswept c a s e , App.2 Eq. ( 5 1 ) .

End Load System c

The f r o n t booms, web, and s k i n s w i l l c o n t r i b u t e t o a _oE ( c )

F r o n t booms: _EA^ a

^°(a^- x) WA^ Wa^-C a^ -T- • b l c A ^ ' ^ " ^ ^ ' ^ = bESA^'' - T^ F r o n t Web: a b o W G F • 2b • 2a W ÏÏGF^ A 1 F r o n t S k i n s : | | i - ^ | . | 1 - .

tol 1 _ 3Wc f l

Gt ' 2 - " 5Gtb 2 A ^ Z!_ S A Rear S k i n s : • a ( ^ ) -. • cl oE

I fi h-.

•• b 12 ~ S A Y/a I a o / . . 0\ __ b E S A ^ ' ^ " ^ - 1 o Wc 2 Gtb a c . o 1. G f 2 A. Wc ÏÏGtb A, 1 Z3A

{ 2 - r A j

W

. (63)

Cc)

In the case of a : ^ % the contributions from bay 1, will

be self cancelling due to the nature of the load systems.

The bay 0 contribution will be from the front spar and skins.

Wa -^ 2a

Front booms = ^^ËSA

^^ ~

T ^ ^

Front Spar = - ^ ~ r

Front and Rear Skins = —-°

2Gtb

i ^_

2 " E A

• «(«) ^""O^ (A ^^OS WC ri ^ ] W (r,.

• • S E = b l s A (^ •" ~r^ + 2Gtb 12 "* S A J " k^t^ •• ^^^^

^kE * ^ ^ "* ^^-^-^ ^® as for the unsv/ept case App.2 Eqs. ( 5 2 ) , ( 5 3 ) .

2) Loading by Torsion Couple Li a p p l i e d a t i r i b

S t a t i c a l l y Correct S o l u t i o n :

- I i f e = S^ = ^2 = ^3 = S ^ = S3 = Sg . . (63)

Calculation of Strain Energy Coefficients

End Load System a

All coefficients due to this system will be zero.

• a(^) - 0

• • ^kE - °

End Load System b

The contributions will be due to the skins only.

ca„

.

L, 3L.c

Front Skins:

1

_{2 Gt ' a • I4.bc ~ 8bc.Gt}

_ £

JL

_ IL. - _ i

1 ^i ^0° 1 ^1 °

Rear Skins:

^ ^ . ^ j

^ . ^^ . - ^

= g^^-^

•- -^E^ = - r ^ •• (^6)

• •

In the case of ai„' all the bay 1 contributions will be

^ (b)

self cancelling. Bay 0 effects will be similar to a'' '^ but opposite in sign. _{. • . a} ( t ) _

1B ~ ÏÏGbt ( b )

i ) „ , k>>1, vidll be a s t h e unswept s o l u t i o n App.2 Eq. (55).

.. (67)

End Load System c

Both webs and skins will contribute to a

Front Web; Skins: (c) oE L. a b . 1 o 1 Ubc • G F " • 2a ^i 2a^c _ ^

WÖ

• aF~ • 2a^

8Gbc • t' i+Gbc * t ,(c) oE 8Gbc

(1^-

.. (68) (o)

The bay 0 contribution to a).^' will be equal to that (c) ^^ 'oE

be twice as much but of the same sign

for a^^^ but opposite in sign, whilst the bay 1 effect will

a (c)

1E 8Gb c

/2c b N

( — - ^ ) .. (69)

For a ) ^ , k-7l, the unswept values apply, App.2 Eq. (5^), kE REFEREl^CES No. Author 1 Ebner H. and Koller H. Title

Stress distribution in reinforced cylindrical shells

A.R.C. Report No. 3U70 Dec. 1937

2 Hemp, W.3. On t h e a n a l y s i s of s t a t i c a l l y i n d e t e r m i n a t e s t r u c t u r e s .

R & M. No. 2396 May 1947

APPENDIX 1

Basic Theory

In this method, the unknown quantity is the unit of a self equilibrating internal end load system. The end load system is ass-umed to act only in the bays adjacent to the rib at which it is applied, and to fall linearly to zero over the rib pitch. Pig.1 shows the typical form of an end load system. This latter assumption results in the minim\am of overlap of the

effects due to the iinknowns, and a resulting simplification to the final equations.

One or more of these systems are applied to the structure at each rib station. The number of separate systems at each rib station is determined by the number of spanwise booms, m, and is equal to (m - 3 ) . The three systems used for the solutions considered here are shov/n in Fig. 2.

The externally applied loads are distributed in the structure in a statically correct solution. As the internal loads are in self equilibrium, the overall equilibrium

conditions are fulfilled. The actual stress distribution Is that due to the most general combination of the statically correct solution, and the internal systems.

The nximerical values of the units of the self

equilibrating internal end load systems are determined by using the theorem of minimum strain energy.

In general, strain energy, U = |-SSa. .T.T.+ Sa._T^+ constant

^ J 13 1 3 r

1^ 1

where T., T. are the unknown \inits of the internal load systems,

1 Ü

a. . are coefficients dependent upon on the geometry and elastic

1 u

properties of the structure, and a. are coefficients dependent upon the applied load as well as the structural properties.

Using Castiliano's Theorem of Minimum Strain Energy, and differentiating the strain energy with respect to the end load system at rib k, for the case where the 'overlap' of the end load systems is restricted to one

bay:-^k-1,k "^k-l + ^kk '^k + ^k,k+1 \ + 1 + ^kE = ° •' (2)

Particular Cases for calculating the coefficients, a. ., are:-

ads:-^ T T .

1

till

J EA

(1) Members subjected to end loads:-^ T .T .

(a. .) . T.T. = 1

—r^

dx

1 d

(2) Members subjected to shear (panels):-ad J' 1 3

(a,.) . T,T. = S^S. . ^

(3)

whe re, S^, S. are the shear flows resulting from T., T. respectively.

In practice, the coefficient a, may be zero for parts of the structure not adjacent to discontinuities, and Eq.(2)

becomes:-^k-1,k -k-1 + ^kk "^k + ^k,k+1 ^k+1 - °

..

(k)

Also a, _. !{• = a, , . if the structure is uniform.

Use is made of these occurrences in simplifying the solution of the equations. a, Write:- ^kk a _k,k-1 = - 2a 1

i

a = cosh j2J

The solution of the Homogeneous Equations (k) is:-Tj^ = C. cosh K0 + Co sinh igzJ

.. (5)

.. (6)

A further assvimption is that the internal load system falls to zero at the extremity of the structure.

Hence T = 0 and from Eq. (6) :•

= - tanh n

0

.. (7)

.. (8)

• i • , C. = C sinh n 0 ; Cp = - C cosh n 0 [ and Tj^ = C sinh (n - k)jZ5

I

.. (9)

/ Substitution ....

Substitution From Eq. (9) into the equation for the internal load system at the discontinuity, which takes the form of Eq. (2), enable the value of C to be found, and hence the numerical values of T, .

Procedure for Solution

1) Idealise the structure to conform with that dealt with in the solution given.

2) Distribute the applied loads in the manner of the statically correct solution used in the calculation of a, „•

3) Calculate the strain energy coefficients, a^^, a, j, a,

k) Form the elasticity equations of the type of Eqs. (2),(i4-). 5) Solve the equations as shown above.

6) Calculate the load distributions due to the internal load systems.

7) Superimpose these results upon the statically correct solution in the most general way possible.

This will give the corrected distributiono

APPENDIX 2

Unswept Box Theory

1. SINGLE CELL

The geometry is shown in Fig.3. The box is of uniform rectangular section, referred to the axes Oxyz. The skins

z = ± b/2 and the spar webs y = + c/2, are assumed to carry only shear loads, any end load carrying capacity that they possess being integrated with that of the spar booms. The total

effective area of each spar boom is A. The ribs are rigid and placed at a pitch a. The box is built in at the root which . corresponds to rib 0.

Each of the shear panels is assiimed to be subjected to a constant shear flow, this implying a linear variation in end load between the ribs.

Internal Load Systems

In this case, only the doubly antisymmetric load system T of Fig.2 is possible. The shear notation is shown in Pig.5.

Consideration of the equilibrium of internal loads at the spar and ribs booms, due to T., in the bays adjacent to rib k, yields:-At rib (k+1):- S, - S. = - -^ S^ -f S^ = 0 '^k ^k ^ ^k+1 '^k c S R - S. = 0 ^k+1 ^ k

!k I .. (1)

'^k^ \^ " \^ " 'v rf

At rib (k-1):- S^ - S„ = 0 S. + S^, = 0 ^k-1 ^k-1 ^k-1 ^k-1 Sp - S = 0 '^k-l ^k-1 ^ .*. S R ^ _ ^ = + 2K ^ •• (2) k "•k-l" ^k-l" 2j^_^"' \._-,~ 2a At rib k:- - S^ + S^ - S„ = 0 2k-1 2j^ Rj^ _ (3)

• •

\ - a

E x t e r n a l Load S y s t e m s 1) B e n d i n g b y Z w i s e force o n x a x i s a t tip - W S t a t i c a l l y correct s o l u t i o n : - s. = - S , = ^

P = t « - ak) J

w h e r e P is t h e e n d l o a d i n t h e u p p e r s p a r b o o m s at rib k.

.

(k)

,. \th 2 ) L o a d i n g b y T o r s i o n C o u p l e , L . , a p p l i e d at (i) r i b S t a t i c a l l y correct s o l u t i o n : -= S^ -= So -= s, -= s ^1 2b c '1

k

.. (5) C a l c u l a t i o n o f S t r a i n E n e r g y C o e f f i c i e n t s A t a s e c t i o n i n t h e k b a y d e f i n e d b y x from r i b k, the end l o a d the front spar t o p b o o m i s :

-W / i „,, . „N . ,. X . „ (a - x )

a 2b (^ - ak + x) + T^. 1 + T^_^^

In the (k-1)*^ bay:- ^ (f- a(k-l)+ x) + T ^ ^ ^ ^ -f T^_^. f where in this case, x is from rib (k-l),

Shear Flows in k bay:-S = ^ - ^k+1 '1 2a 2a JL 2b 2b c q _ ^ '^k+l W_ ^ ^3 ~ 2a 2a 2b " 2bc

q _ - Ü ^ ^k+l

^^2 ~ 2a •*• 2 a q - - !k ^ !^±i \ - 2a + 2a 2b c 2bo In (k-1)*^ bay:-„ ... ._ ^k ^ ^k-1 ^ w ^1 "^ 2a "^ 2a "^ 2b "^k ^ ^k-1 W ^3 " 2a + 2a 2b The ribs are rigid.

a

^1 2bc ^1 2bo k 2a T S. = 2a k-1 2a ^k-1 2a 2bc 2bc Then a^^. T 2 = | , 2 T' Tj^ . -^ dx + j--^ . ^ (p- + -^; k /ba ca>

(Contribution from U booms, 2 webs, and 2 skins in each of bays k, k-1)

a _{kk 3E/1 "^ Ga ^F" "^}8a _L / b c\ _t^

> (6)

(7)

o « m m. _ A . f rn rn X (a-x)-, _ ^k"^^ 2 /ba C B N

M * ^k-^it - EA J

\*t

• a ^ r ^ ^ ^2 • G H' + t ^

for^ = k+1, k-1

(Contributions from k booms, 2 skins and 2 webs in bay k)

• a 2 a k( ^ 3ÊA ^k^ = O 1 / b 2Ga ^F" + +) iï = k-1, k+1 -C 7^ k-1, k+1 y .. (8)

At the root there is a contribution from bay O only

• • ^oo = 2 ^kk

.. (9)

The a, due to the Z wise force are zero, as the

contributions from the front and rear spars cancel. This implies that for this case T^^ = 0 i.e. the solution assiamed is correct.

Due to the torsion couple L., in bay k:-2T ai,„«Ti, = k L. 'kE'^k ~ G.2a * 2bc (~ ba . ca _c;a.\ (lc<i) L. kE " 2bc i /£ _ Jb_^ .Ga H t'''

There will be an equal and opposite contribution from bay k-1.

.*• aj^g = 0 (k ;^ 0, i)

I

.. (10)

a - - a - ~ 1 - C£ -

X)

oE ~ IE ~ 2Gbc H ÏT^

This allov/s for warping constraint e f f e c t s .

TWO CELL

The structure of the two cell box is shown in Pig.i|. It is similar to the single cell box, apart from the additional centre spar. The same assumptions are made.

Internal Load Systems

s.

h/

There are three possible internal load systems, as shown in Pig.2. All three are considered, although in practice, for the external loads which give rise to stresses which are antisymmetric about the Oxy plane, the doubly symmetric system, T , is zero.

Using the notation of Fig.5 for the resultant shears across the section to be

zero:-j=i-1 'SZl p. j=i-1

Si = Si - - V ^ +

\

i = 2,6

1 = 3,U,5 j

.. (11)

For zero resultant moment on

section;-S^ = S|^ + ^(Sg + S3 + section;-S^ + Sg) .. (12)

End Load System a

p . = P , = p. = p 1 U = "6 a = T Pg = P5 = - 2T . •

(13)

From E q s . ( I I ) , ( 1 2 ) , ( 1 3 ) : -q -q w ^1 -

\ ~

2 w 2 T_ a a q - !W - T I ^5 ~ 2 a w ,a S , = -TT

_{2 + T}

w ,a Sc = - -yr-_{2 +} -. -. (1^4-.) / The strain •••

The s t r a i n energy of t h e s h e a r flows i s : -2 U = - ^ _{^ 2G} 2 2x S„. S,^'

b42 «w %] cfPw T^Y (\ T^V ppw T^Vi

F-if

w au as • as w w w

(ir^f)

c ^ tC>.,.,A^^^ M-^un^JU. « J - * - ! ^ » ^ _{O —» 7 ^ » o} au

Por minimum energy -^— = 0 <!^. ^ ^TÜ jC^f^t^^é^ in^Stj-i . e . S,„ ( I r + r ) = O ^v^.*.^. in^Stj-i^^^r*^ • • S w = O . . (15) From E q s . ( 1 U ) , ( 1 5 ) : -Sg = S^ = - S3 = - -Sg T_ a a . . (16)

End Load System b

P, = P 3 _{- Pi, = - ^6 = - ^} Pg = - P5 = 2T^ From E q s . ( I I ) , (I 2 ) , (17) : -w ^1 = - ^u = - 2" Sw T^ r ^2 = ^6 = " S3 = - S. = - ( ~ - — ) J S t r a i n Energy: ^ = ^ ^ ff]5Lc2 ^w^ c /S ₂ 2 21

^ f - T - X

o r wri

'w (2F" + I ) ~ ^ (s^ — • I - (~) f^

= 0 . • (17) . . (18) . . (19) Prom E q s . (18) and ( 1 9 ) : - S^ = - S^ = S^/2= " \^^ " '^^ Sg = Sg - 8 3 = S5 = V ( l - B ^ ) T ^ . b^ ^ ^20) a / C o n s i d e r a t i o n

Consideration of the equilibrium of the rib booms gives: S, 'R _k-1 = S R. _k+1 P =5 P ^k-1 -^k+l J- ^ b „ b ï^k = - ^ k B

X = !^ (1 - 2B^)

2 a

End Load System c

^1 = ^ = - ^3 P2 = P5 = 0 - Pg = T^ From E q s . ( I I ) , ( 1 2 ) , ( 2 2 ) : -S „ c g _ _ -W T_ ^1 - 2 "^ 2a S mC q - q - - (ZH . 1.) Sg - Sg - ^ 2 + 2a^

!w

T1

]

"= 2 •*• 2a I, S S , = 85 = w T" 1 2ir

i

S t r a i n E n e r g y : - U = a 2G

r

as w 2 (

c ! o

/~w

_{' ! l}

.

T ° V

. o

/'^w

_

Tf.'\

1

+ r l 2 - f + ^ + 2 2a j \ \ l s ^ . (11) L £ g2 . K / T 1 ) F ^ ] 2 ^w + ''2a'' 1+ t l^w + ^''2a'' = 0

l^*

2ci ' w i t . - . S^ = 0 = 0 From E q s . (23) and (21+):- S^ = S^ = T V 2 a Sg - S3 = S3 - Sg - - 2a C o n s i d e r a t i o n o f t h e r i b boom e q u i l i b r i u m g i v e s : -'R. _{k - 1} = S R _{k + 1} S p mC

_fk = Ik

2 2a

I

/ Calculation

Calculation of Strain Energy Coefficients End Load System a

o (a) _ 8a _l6a _8c ^kk ~ 3EA^ "^ 3ËA2 "^ Gta

where the contributions are from the front and rear spar booms, mainspar booms, and skins respectively.

• a^^) 8a / 1 2 X ^ 8c • • ^vv - 3E ^A^ "^_*kk k^^ + Gta a(a) _ i a ( ^ ) ^00 - 2 ^kk

a

(a) 'k; (a) k . ii ^ k-1, k, k+1) > 2a 3E 'A (J- . -2.) _ ji£

1

A

Gta i= k-1, k+1 where the contributions are as for a _kk (a)

. . (26)

End Load System b

The boom c o n t r i b u t i o n s t o a ^ ' ' a r e a s f o r a^^^'^ Shear c o n t r i b u t i o n _{. a} B. 2ab ,

• G F ^ + \

2ac Gt 2ac Gt . „ ( b ) 8a , 1 2 s kfx^rx'^l^ ^ 2c J . „b) • • a^, ^ = ^ (-r- + - ^ j + ?S- 3]B V TT + T - 11 B V kk ~ 3E 'A^ ^ A2 00 2 kk

4V = 0

G^PI/

t 1

(Jl ^ k - 1 , k, k+1)

>(27)

The boom c o n t r i b u t i o n t o a,\. (-C = ^-1 * k+1) i s a s f o r a^a

. / IN . , ï - . ^ x^ . (b) ^^ The s h e a r i s ( - - ) t h e c o n t r i b u t i o n t o a^^^ 2 2

• • ^ k f = ^E^Ij"-" Ag

(^) _ 2a(:L

2 ) - , f e ( 3 ( B ^ ' . f H - f | l ~ B ^ j '

Ga ^ = k - 1 , k+1 / End Load . . . .

End Load System c

There is no end load in the mainspar booms, and the fc)

contributions to a>,' from the front and rear spar booms will be as for the other end load cases.

•

M

8

-^a ^ 1 .b

2c kk ~ 3 EA. ^ Ga ^yr + t )

the contributions being from the front and rear spar booms, front and rearspar webs, and skins respectively.

a(°) - i a ( ° ) 00 - 2 ^kk

4f -

0

X28)

(J( ^ k-1, k, k+1) i)i: (a)

The end load contribution to a^^^ is the same as that for

the front and rear spars to ^IV * ^'^ " ^'^^' k-1). The skin contribution is (- i) that to _a (c)

kk 3EA,

2CN

2Ga H*^ •*• t ^ Jl=

k-1,

k+1

External Load Systems

1) Bending by Z wise force applied on x axis at tip Z = - W Statically correct solution:- S. = - S- WA^ _{F T A}

WA.

s,.. =

_{w ~ bSA}

where >.'A = Ap + 2A. At rib

k:-WA. ^(29)

End Load in upper front and rear spar booms:- - .-.. (-1- ak) WA2

End Load in upper mainspar booms:- =r^TT ("^ ~ ak)

Strain Energy Coefficients End Load System a

The internal loads are symmetric, and the external

loads antisymmetric about the Oxy plane, hence the contributions from members on either side of the plane will cancel.

(a) .

kE = 0 (30)

End Load System b

Due to the choice of the external load distribution, Eq.(29), and the nature of the internal loads, the end load contribution to a^'^ is zero. The mainspar boom effect is cancelled by that of the front and rear spar booms. There is no shear in skins from the external loads, and hence no contribution from the skins.

WA. B^ I n bay k : - C o n t r i b u t i o n from f r o n t web = - , t ^ .

= C o n t r i b u t i o n from r e a r web -2WApB C o n t r i b u t i o n from c e n t r e web = -,. t ^ ;

• M - - 2WB^ (A .. A )

• • ^kE - G F S A ^^1 ^2^

There will be an equal and opposite contribution from bay k-1. .-. a^^) = 0 ( k / 0, n)

,b

^E^ = - -S^= - ÖPIA ^ - V "(31)

^oE

End Load System c

In this case the contribution from the Internal and external loads will be zero, as the former is doubly antisymmetric, and the latter singly symmetric. Cancellation will take place on either side of the x axis.

^kE = 0 .. (32)

2) Loading by Torsion Couple L., applied at 1 rib Statically correct

solution:-- I ^ = Si = 32 = S3 = S^= S5 = Sg .. (33)

Strain Energy Coefficients End Load System a

The statically correct solution gives a doubly antisymmetric load distribution, and as the internal load system is doubly symmetric, there is no resultant contribution.

(a)

/ End Load

End Load System b

In this case the internal load system is singly antisymmetric about the Oxy plane, and contributions will cancel about

the X axis.

•'• ^kE

= °

••

^

End Load System c

There is no external load in the booms, and therefore no contribution from this source.

In bay k:- the contribution from the webs

Is:-L . 0.

1 b

" I+bc * G F "

^ i 2c

and the c o n t r i b u t i o n from the s k i n s : - + r^jrr • pT

There w i l l be an equal and opposite c o n t r i b u t i o n from

bay (k-1)

. ' . a^^^ = 0 (k 7^ 0, i )

a ( ° ) o ( c ) "^1 / 2 C b X /T,r\

%E = - ^lE = 5Gbïï ^ T " F " ) •• ^^^^

SINGLE CELL - SECOND ORDER EFFECTS

The structure of this box is similar to that of the two cell box, Fig.U, except that there is no centre web. The centre booms are retained, and the assumptions are the same as for the single cell case.

Internal Load Distribution

The three systems of Fig.2 are possible. End Load System a

S„, is zero, as there is no centre web, hence from _{w '} Eq.

(16):-S^ = (16):-S^ = O

Sg = S3 = - S3 = - Sg

=41 •

(38)

As there is no spar web shear, there can be no rib web shear. Consideration of the equilibrium at the rib booms

gives:-^k+1 = -^k-l = - -^k

R, 2T

"a

k

..

(39)

for both upper and lower surfaces, where P- is the end load in the rib boom at y = 0, falling to zero at y = ± c»

End Load System b

^1 -^3 = ^ \ = -^e =

- T Pg = - P5 2T

Using Eqs. (20), (21), and writing S = 0

^1

- \

0

Sg - Sg - - S3 - - S3 =

T^ I

_{k V}

..

(ko)

..

(k^)

Again there is no rib web shear. Rib boom loads on top

surface:-R _k+1 • R k-1 Ri 2T c b k a a .. (^2)

These will be opposite in sign on the lower surface,

End Load System c P, = P^ = - P3 P2 = P5 = 0 = - P^ = T' J. Prom Eq. (25):- S^ = s^ = ^ Sc S2 - S3 _{S6 =} Ik _2a

..

(k3)

1

..

(kk)

In this case there is a rib web shear, but no rib boom load:'

^k-1 ~ ^k+1 T,

^ 1

= 2a [ ^k _k a ..

(U5)

Calculation of Strain Energy Coefficients The ribs are assumed to be rigid.

End Load System a

The load distribution is icSentical to the two cell case. Hence:-(a) _ 8a r-L a. -^^ kk 3E ^A A. 8c Gta a(a) _ i a ( ^ ) ^00 - 2 ^kk

4V = °

(i ^ k-1, k, k+1)

4^^' = § ( 1 7 - ^ ) - ^ (^=.H,..1)

(W)

End Load System b

The distribution is similar to the two cell case with B zero. The coefficients are then identical to those obtained for

load system a. a _kk (•^) _ J a )

45'

= a _kk ^k^ ..

(^7)

/ End Load

End Load System c

This is again identical to the two cell case. • o ( c ) _ _ 8 a 3EA a kk ( c ) 'oo

4 l ' = 0

4f

1 1 / b 2CN +

Gi-

( F - + T ^

lAo)

2 ^ k k 2a 3EA (X ^ k - 1 , k, k+1)

7 - 2 f e ( f - f ) ('^ = k-1,k+l)J

^ . .

(k8)

External Load Systems

1) Loading by Z wise force on x axis at tip Z = - W Statically Correct Solution:- S.

'1 ^k " 2b

The end loads assumed to be distributed to give constant stress across the section,

i.e.:- Load in upper booms at rib

k:-WA. . ^1 = ^3 = b ^ A (^-^^^

WAg ^2 "" b Z A where S A = Ag + 2A^

There will also be skin

shears:-q _ shears:-q _ _ shears:-q _ _ shears:-q _ W / i _ ^ S

^2 - ^6 - S5 - S3 - -^ (.2 2 ^ ;

(•^-ak)

>(k3)

Strain Energy Coefficients End Load System a

In this case the internal loads are doubly symmetric, and the external loads antisymmetric about the Oxy plane.

Therefore all contributions will cancel about the Oxy plane.

.'. a ^ ^ = 0 .. (50)

End Load System b

There are no web shears from the internal loads, and the boom contributions will cancel due to the choice of the

statically correct solution.

In bay k:- Skin

shears:-T^

W

fl -^1

I

Sg - Sg - - S3 = - S3 = — + ^ j 2 - ^j^j

A. . .

(b)

_

iwc ri_ A^

I

kE

~

Gtb [2

yZk

j

There will be an equal and opposite contribution from bay (k-1).

•*• ^kE^ = ° (k / o, n)

a(^) - - a(^) - M£ Ii _ tlJl

oE ~ n-E " Gtb [2

S A J

This allows for shear lag e f f e c t .

.. (51)

End Load System c

As the internal loads are doubly antisymmetric, whilst the external loads are antisymmetric about the Oxy plane only, the contributions will cancel, on either side of the x axis.

.-. a^> = 0 .. (52)

2) Loading by Torsion Couple L., applied at i rib Statically Correct

Solution:-jjl^= S^ = S2 = 83 = S^ = S5 = Sg .. (53)

Strain Energy Coefficients End Load System a

The statically correct solution yields loads which are doub:i.y antisymmetric, the internal loads being doubly symmetric.

End Load System b.

Here the internal loads are singly symmetric about the Oxy

plane, and contributions will cancel about x axis.

(^>) _

^kE

.-. ai^^ = 0 .. (55)

End Load System c

The load distributions are the same as for the two cell

case.

Hence:-'*kE

a^°) = 0 (k / o, 1)

Ao)

_ _

Ao)_ _ ^±

/2c _ b X /^gN

oE ~ ~ IE " I+Gbc ^t

APPENDIX 3

Treatment of Triangular Skins

The forces on the triangular skin are shown in Fig. 6. A uniform shear flow, S, is assumed to act along the two perpendicular sides. This assumption corresponds to that made for the rectangular panels previously considered. Equilibrium is maintained by a shear flow, S', along the hypotenuse, and a system of xiniformly distributed normal forces, p'.

• Resolution of forces normal to, and parallel to the hypotenuse

gives:-- 2Sc cos e = S P ' = P' •• ^57)

Where the root is built in, P' will be reacted along the built in edge. For the case where the liypotenuse is not built in, P' will be reacted at the spars.

REPORT No. 63, 'Kt-l E N D LOAÖ S Y S T E M S . Vt-»2

I'iJ:-System (a)

Sysbgfn ^b) - T «

INTERNAL END SYSTEMS

System_Jc)

^(n.0. V \ V V "Bet^o Say/

44

i \ i

J

LL 3 /!-/ Single Cell A) t'A

A

b 4 — - • i 0 3-N FiQ.3 ^«(n-.Ja $ , ^

1

'

i

tajO 3a// I t \ 1 n-i

'S'

I

.fc' / « .

i i \ \ 1 \ i \ \ \ 1 i

-^ i>..-.

Two Cell and Single Celt feeconA Order Effects)

QEOMETKY O F UNSWEPT BOXES

Smjie Cell

. :

J ^

Ttrfo Cell anrf 5 male Cell (Second

ördw EFfects).

•E

s»

SHEAR NOTATION

-^w.

s

ï - • ' S TRIANGULAR SKIKI

SINGLE CELL

Other Detai/sas Fin.^

-**—JC

E!S-I

^/A /i/otaè/o/1 for Swafe Ceff ISox

rh/su h^

Two

CELL AND £>M6LE CELL

^Second Order EFfecfcs).

Other Be Lai Is as Rg ._4.

GEOMETRY O F SWEPT BOXES - "Ribs Morma) To S ^ f s