Note on the design of redundant structures

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TECHNISCHE HOGESCHOOL DELFT

VLIEGTUIGBOUWKUNDE

Michiel d,

IlIyt.rwe,

10 • PELfT

1

4

feb. 1961

NOTE ON THE DESIGN OF REDUNDANT STRUCTURES

by

E. D. Poppleton

by

E. D. Poppleton

The author is inde.bted to Dr. G. N. Patterson. Director of the Institute. for providing the opportunity to carry out the investi-gation reported herein. and to the Defence Research Board of Canada for financial assistance.

This note is concerned with the redesign of redundant structures having undesirable stress distributions. A matrix equation is derived relating specified stresses in certain structural members to changes in the stiffness distribution necessary to achieve these stresses. The equation is non-linear, in general, but can usually be solved quickly by iteration. The method of Argyris and Kelsey (Ref. 3) is used in the analysis.

TABLE OF CONTENTS

NOTATION ii

1. INTRODUCTION 1

Il. THE MATRIX EQUATIONS OF ARGYRIS AND KELSEY 2 lIl. THE METHOD OF REDESIGN FOR SPECIFIED STRESSES 4

IV. METHOD OF SOLVING REDESIGN EQUATION 6

4. 1 Case With No "f" or "h" Type Members 6 4. 2 Case With No "e" Type Members 8

4.3 General Case 10

V. CONCL UDING REMARKS 11

REFERENCES 13

APPENDIX A, Simple Example Ca1culations 14

A. 1 Basic Structure 14

A. 2 Change Stresses in Flange 3 and Panels 5 and 6

by Changing Their Stiffness 18

A. 3 Change the Stress in Flange 3 Only by Various

A a b, b o b1 c d D, Do E f g G H I k Ks M m n P (1) p (2) t ' t S NOTATION

cross sectional area of member resisting force P or Q; diagonal mat:fix of such areas.

length of flange and adjacent edge of sheet matrices defined by Equations 3 and 4 matrix defined af ter Equation 4.

width of she et

matrices defined af ter Equation 4. Young's modulus

diagonal flexibility matrix (see after Equation 4 and Equation 5)

-1 matrix such that f = g A Shear modulus

column matrix of initial displacements unit matrix

number of members in structure buckling coefficient of sth panel

rectangular matrix with unit elements defined af ter Equation 16.

number of externaHy .applied.lo.ads number of redundant members. Forces at the ends of the t th flange

force in the t th member having constant axial load shear force on the edge of the s th shea,r panel column matrix of external lOq,ds

column of generalised loads in members. Contains elements of type P (1) _{pl2) , P t , Qs}

.t v

x

Subscripts e, f, g, h, m sheet thickness

column of generalised displacements of members column of generalised loads in redundant members stress; column of generalised stresses in members

defined on pages 5 and 11

1. INTRODUCTION

There exist a number of matrix methods for the analysis of highly redundant structures, and most companies concerned with the manufacture of flying vehic1es now have computer programs suitable for carrying out these analyses rapidly. No systematic attempt appears to have yet been made, however, to tackle the problem of designing a redun-dant structure, as opposed to analysing an existing design. Hence it is not possible to ensure at the outset that the stresses in all structural members are below the design value, or that sorne members are not uneconomically understressed. In this connection, it is possible that the ideas of Ref. 1 could be extended to more general structures. To be of general use, how-ever, it would be necessary to devise a computer program to carry out rapidly the lengthy iterative procedure which would be necessary for a structure with the very large number of redundances common in aircraft practice.

At present, the improvement of a design, which analysis has shown to be deficient in some respect, must proceed on a trial and error basis, by strengthening weak members (and vice versa) and

re-calculating the stress distribution and deflections. Strengthening a member automatically stiffens it, so that it tends to accept more load, and conse- . quently more than one trial may be necessary before a satisfactory stress distribution is obtained

Methods have been proposed for determin~ng the effect of structural modification on the stress distribution in a structure (Refs. 2, 3),

and that of Argyris and Ketsey affords a rapid method of carrying out the redesign, although the procedure is still one of trial and error. It is not necessary to make a complete reanalysis after each trial, however, and each trial involves the inversion of a matrix of order equal only to the number of members to be modified.

The purpose of the present note is to point out a simple

extension of the method of Argyris and Kelsey which enables the redesign to be done directly. Here, the required stress in a number of members not exceeding the number of redundancies may be specified, and the necessary adjustments to the stiffness of the same number of members (not necessarily the same ones) may then be calculated directly. This is most conveniently achieved by iteration of an equation containing matrices which are readily formed from those already used in the initial analysis of the unmodified structure.

The present discussion is limited to structures which can be idealised as an assembly of flanges (having constant cross sectional area between structural nodes) and rectangular sheets carrying constant shear flows along their edges. The axial loads in the flanges may thus vary linearly between nodes. Pinjointed frameworks are inc1uded in the analysis as an important class.

11. THE MATRIX EQUA TIONS OF ARGYRIS AND KELSEY

The notation of Ref. (3) will be used here, as far as possible, and usually symbols denote matrices the order of which is apparent from the context. Whenever scalar multipliers appear in the equations,or--'the elementsof a partitioned matrix are shown, the familicir square brackets

and braces will be used for clarit~.

We consider a structure have k structural elements, n of which are redundant, to which are applied m exte'rnal loads.

The generalised stress resultants or loads in the member&2 are such that for _{a flange ha ving a tensile force P t} (1) at one end and Pt \ )

P (1)

at the other we have: -

[J

t th generalised load = St =

p~(2)

(a two row column) (1) For a member in which the load is constant, it is more con-venient to define the generalised load by a single scalar number so that, in this case:

t th generalised load

=

St = Pt (a scalar) (2a) Similarly, for a sheet carrying a shear force Qs along one edge, we have

s th generalised load = Ss = Qs (a scalar) (2b)

The generalised loads in the whole structure are thus repre -sented by a column S which has k elements, some of which are two row columns like

U>.,

and some of which are numbers like (2).

The external forces and the generalised loads in the members

(or combinations of rnembers) chosen as redundancies are represented by columns Rand X respectively.

It is shown in Ref. 3 th at the generalized loads in the members m ay be wri tten

S = boR

+

b1X - blD- 1b'1 H (3 )

=

boR - b1D- 1D o R - _{b 1 D}-1 b'lH

where

D = _{b'l f b 1 (::. Ha,a of Ref. 4)}

Do

=

b'l fbo (=-Hap. of Ref. 4)

H = column of generalised initial displacements (due to thermal effects or lack of fit)

b = _{b o -b 1 D-1 Do (=:S}. ofRef. 4) c = b1D-1b'1

f

=

diagonal matrix of flexibilities of the unassembled members (1t F of Ref. 4).

The matrices b._{o and b1 can be readily written down from equilibrium}

considerations, if the redundancies X are judiciously chosen as indicated in Ref. 3. The matrix f is such

that:-Strain energy of structure

=

1/2 S' f S

and consequently, in the type of structure considered here, has elements of two type s .

For a flange of length at and cross sectional area At with load varying linearly from pt(l) to pt(2) , we

have:-(1) 2 2 Strain energy = at (Pt

+

Pt(1) Pt (2)

+

P t (2) ) 6Et At

=

[ (1)

PPlJ

[~

1

J

~1::]

at P t 2 6 EtAt i. e. at

G

~

(a 2 x 2 submatrix ) f -t - 6E-tA-t

=

at

~ ~J[ ~ïl

:tJ

lI& gt At 1 (5)

--6E t j

For a member carrying a constant load the flexibility is given by

ft = at =. gt At- 1

EtAt

(a scalar number)

and for a rectangular panel (as x ds) having a shear force Qs along the

side as (cross-sectional area As

=

_{asts ) we}

have:-2

Strain energy

=

1/2 Qs d s

G

_s As

and fs

=

(a scalar number)

For the whole structure therefore we have

(6)

The generalised displacements of the members are obtained from the flexibilities and the initial displacements in the form ·

v = fS

+

H (7)

lIl. THE METHOD OF REDESIGN FOR SPECIFIED STRESSES

In Ref. 3 it is shown that it is possible to allow for a change in the stiffness of a member by imposing fictitious initial displacements, H, on the structure and using equation (7) to determine the necessary values of these displacements . Knowing H it is then possible to find the loads from Equation 4.

Here we solve the inverse problem, i. e., we specify the stress we require and determine the change in stiffness necessary to realise this stress under a given load system. Of course, we are at liberty to

specify arbitrarily only a number of stresses not greater than the number of redundancies since the equilibrium condition must always be satisfied.

We now define four subscripts, each denoting a different

Subscript e denotes rnembers in which a stress is specified and the stiffness of which is to be modified. f denotes members in which a stress is specified

but in which there is no change in stiffness.

g denotes members which remain~ unrrndified

h denotes members whose stiffness is to be modified but in which the stress is not specified.

A further subscript m denotes conditions in the modified structure.

As in Ref. 3 we say that initial displacements may be chosen in such a way that the total displacements, under the given load

system R, in the original structure, are the same as those in the modified

structure without initial displacements.

Thus partitioni.!lg Equation (7) we get

fem Sm

=

_{v m}

=

v

=

fe Sm

+

He ffm _ff 0 fgm fg 0 fhm fh Hh i. e. ,

[~:J

=

Ce'O -fe

o

J~emJ

since ffm

=

ff fhm - fh Shm f_gm

=

_{f g}

We now write Equation (4) in partitioned form

[

sem]

=

bR -Sfm Sgm Spm cehl[He] Cfh 0 Cgh 0 Chh Hh (9) (10)

In the above eq'uations, the subscripts on the submatrices of c denote their order, e. g., ceh has e rows and h columns. The elements in these submatrices may themselves by matrices of order 2 x 2, 2 x 1, 1 x 2 or 1 x 1 (see, _{forexample, the matrix c ee in Section A2 of}

the Appendix).

Equation 10 contains all the information required to effect the redesign of the stru cture, although direct:' solution of the equation is

not gene rally feasible. It should be noted that although the stresses in the "e" type members are specified, the loads Sem 'are not, and are thus

unknowns in Equation 10. Some, or all, of loads Sfm are known, however, since the geometry of these members is not altered and the loads are thus

known from the specified stresses. It should be noted, however, that

frequently only the stress at one end of a flange will be specified so that

stresses corresponding to all the elements of Sem and Sfm may not be known. The other unknowns in the equation are the cross-sectional areas of the members which appear in the matrices fem and fhm .

IIlI. METHOD OF SOLVING REDESIGN EQUATION 4. 1 Case With No "f" or "h" Type Members

In this case Equation 10 simplifies to

of which only the top set of equations is necessary for solution, i. e. , Sem

=

Se - c ee [fem - fe] Sem

in general, term s like and

and g _s r.-_{v s}

(11)

where one of the stresses in the first expression is always specüied, but the other may be unknown, arid the other stresses are specified.

Equation 12 may thus be written Sem

=

Se - cee ge

cr

em

+

c ee fe Sem where n- is a column {_ .. , \

crf:')

cr!.1.) \

-v em \ ... 0:;""' . I , · · I~ ... , ,

...

\

]

(13)

This equation is most conveniently solved by iteration. A first approxi-mation to Sern is assurried (Sem = Se say) and, using the unmodified structural properties, the elements in

cr-

em which are not specified can be calculated, and hence a second approximation to Sem is obtained by evaluating RHS of Equation 13. New 'Yi>lues of the unknowns in (Tem can

then be evaluated (for

ex~)le,

if

CT

_{t 1 is unlillown and}

CT

t(2) is specified, Q""t(1) =

cr-é

2 ) p_t(1)!Pt 2 since the flange has constant area). Using these and the second approximation to Sem' the third approximation is found, and so on.

Having found Sem' the loads Sgm are most conveniently found from Equation 9, which

yields:-He

=

c

~~

{ Se - Sem}

Sgm

=

Sg - c ge c

~~

{ Se - Sem}

It should be noted that in the case of framework, all elements of

cr

em are known, and we can solve Equation 13 explicitly, i. e.

Sem = [I - ceefe] -1 [Se - ceege

cr

em}

It will be noted also that it is possible to incorporate stresses which dep end on the geometry of the structure. For example, suppose

that it were required to prevent a shear panel from buckling. In this case we have

o-s

crit = Ks Es (t/b)2 = Ks Es Ag while Us = Qs/As

a 2 b 2 _{s s} 3 2 2 Hence we have As

=

_{Qs as bs} we wish to specify in the ca1culation is

2

CTs

= (Qs

2 as

/ Ks Es and thus, the stress

Consequently, af ter each iteration the appropriate value of Qs would be selected from Sern, and the element

Us

of (Jern re-evaluated before the next iteration.

4. 2 Case VVith No

"e"

Type Members

Here the matrixequation becom.es

Shm

(14)

and for solution we use the first and last set of equations. Using Equation 6, the first set can be written

S~rn

= Sf - cfh gh LA

-

~m

- Ah1] Shm (15)

We can change the order of the matrices in the last term to give

Sfm

=

Sf - cfh gh [ ShmJ {

Ah~

-

An~

(16) where Shm is now a diagonal matrix of loads, and

\

[A~

- A -

fi]

is now a column whose elements are sub-matrices like

r

A

_tm

-

1

-

A _t

-1]

A- 1 -A- 1

tm t

if the corresponding structural m ernber is a flange, and (A t,~

-

At~~.) if the member is a shear panel or a constant axial load member.

We introduce a rectangular matrix M such that

[

_A~

-

_-

_A~

-~

₌

_M

[-1 -1 }

_{Ahm - Ah'}

where [A-1 -

Ah~1

is a true column in which the area of each flange appears oWfy once together with the area of each shear panel and constant load member. For exarnple, in the simple case of two flanges and one skin we might have

~h~

-

Aii1]

=

- - - - -

}

j

1

flange shear panel flange

so that

G

Ahm - Ah -1 -lJ

=

M

{A~

-

Ah~

1 0 0 -1 -1

=

Alm - Al 1 0 0 0 1 0 A-1 - A- 1 0 0 1 2m 2 0 0 1 A -1 _3m _ A- 1 ₃ Equation (16) may thus be written

Sfm ' Sf - Cfh gh [ ShmJ M

r~~

-

A~l}

(17)

The stresses may be specified at one end only of sorne of the flanges

included in the f type members. The nu~ber of specified loads in Sim must equal the number of elements in [A_hm - Ah1

J

so we form a new equation from (17) such that the L. H. S. contains no unknowns, i. e.

Sfm

=

Sf -

cfh gh [Shm] M

r

A~

-

A~11

(18)

where the bars indicate matrices formed by excluding rows corresponding to the unknown elem ents in Sfm'

The unknown areas Ahm may now be found in terms of the unknown loads Shm

[A~

-

Aïn,

[Cfh

gh

[S~ ~

-lr

Sf - Sfm}

(19)

Similarly we can write the last set of Equation 14 in the form Shm

=

Sh - _{c hh} gh

r

_{Ah:n - Ah-}1

J

Shm

=

Sh - _{c hh gh [ShmJ M {A;m -}

A~)

(20)

in which (19) may be substituted to get

Shm = _{Sh - c hh gh [Shm}

J

M~fh

gh

Equation 21 gives the loads in the h-type members which will ensure that the loads in the f-type members are those specified, and substitution of these into Equation 19 gives the required change in cross sectional area. However, Equation 21 may be rather unwieldy to solve as '

it stands. The solution may be most conveniently found by an iterative process, in which Shm is found from (20) for "an assumed Ahm and then an improved value of Ahm is found from (19).

As in the previous case, the loads Sgm are found from Equation 9 in the form

Sgm

=

Sg - cgh

c~! ~

Sh - Shm

J

In the special case of the pinjointed frame work, Sfm and Shm are of the same order, cfh is square and M ::: I so that Equations 19 and 20 become

{A~~

- Ah}

=

L

Shmj-1

g~1

cfh-1 \ Sf - Sfm

J

Shm

~

Sh - chh cfh -1 { Sf - Sfm

J

from which A

hm is determined directly.

It should be noted that there is a limit to the changes that can be made in the loads in the f-type members by modifications to the h-type members. Obviously, one can do no_better than to make a member infinitely stiff so that a limiting set of values of Sfm is obtained by putting Ah _{m ,} = cD in Equation 15 and 20. This yields

which agrees with the formula given in Ref. 3.

4.3 General Case

If all types of ,members are present then the equations become

more complicated and we see that the three sets of equations giving the 80lution are

It appears that the most convenient way to solve these is to assume an initial column Sern (= S , say) and hence find a first approxi-mation to fAhm -

Ah

1) and Shm byeiteration of the last two equations. These values may then be used in the first equation to find an improved S ,and the iteration process continued until convergence is satisfactory.

V. CONCLUDING REMARKS

Some numerical examples are given in the Appendix and, as might be expected, these show that it is more efficient to change the stress in a member by changing its own stiffness. Consequently, it is to be expected that, in the majority of cases, Equation 13 will be used for re-designing a structure.

Even in a very complex structure, the order of the matrices in the equation will probably be quite low, but, if a large amount of

redesign is necessary, inspection of the matrix c should enable it to be divided into a number of zones of small mutual interaction. For example,

in the simple structure analysed in Section A. 2 of the Appendix, it will be

seen that there is small interaction between flange 1 and the inboard end of

flange 3. A change in the stiffness (i. e., the introduction of a fictitious

initial strain) in either flange causes a change in the load in that flange which is about ten times greater than the change in the load in the remote flange. Such a division of the structure would allow low order matrix. equations to be solved for each zone independently, at least as a first approximation.

The foregoing analysis relates to conditions with a given external load system, R. To find the loads under any other load system, it is necessary to find the modified matrix bm' such that

S _.m

=

b _m R

This is achieved by calculating the modified flexibility matrix from the known new are as

in which

-1

f_m ,= gA m

In the above equations, the subscript g refers to all mem-bers which are not modified (includes types f and g, as previously defined) and h to all members which are modified (includes types e and h).

The redesign equation is non-linear in general, and may

this may effect the rapidity of the convergence of the iterative solution, and

the rather ~isappointingly slow convergence in the first example in the

Appendix may be due to this cause. The very rapidly converging solution of the first examp1e in Section A3 is associated with aquadratic equation which has wel! separated roots,

namely:-A 3rn

=

1. 244 sq. ins. A 3m

=

-0. 117 sq. in. (1) P3 m = 10.109 kips pO) =-18.9 kips. 3m

Finally, the restriction to constant area flanges and rec-tangular sheets is removed fairly readily. The assumption of constant

shear flow wouldbe retained, so for a given flange taper, the matrix g

for the flanges could be evaluated fairly easily as indicated in Ref. 5. The

stiffness of a non-rectangular sheet can be evaluated as in Refs. 4 and 5,

the latter reference allowing the effect of taper in the sheet thickness to be

considered.

1. Francis. A. J. 2. Michelsen. H. F. Dijk. A. 3. Argyris. J. H. Kelsey. S. 4. Lang. A. L. Bisplinghoff. R. L. 5. Klein, B. REFERENCES

Direct Design of Non-linear Redundant Triangulated Frameworks. Australian Journalof Applied Science. Vol. 6. No. 1 March. 1955.

Structural Modifications in Redundant Structures. Readers' Forum J ournal of Aeronautical Sciences. Vol. 20. No. 4. April. 195.3.

The Matrix Force Method of Structural Analysis and Some New Applications. Aeronautical Research Council. Rand M No. 3034. Feb.. 1956.

Some Results of Sweptback Wing Structural Studies, Journalof Aeronautical Sciences. Vol. 18. No. 11, Nov. 1951.

A Sim ple Method of Matrix Structural

Analysis, Journalof Aeronautical Sciences, Vol. 24, No. 11, Nov., 1957.

Page 15 Page 16 Pages 18, 19 Page 21 Pages 21, 22 Page 22

+

sign missing between terms in matrix equation Numerical value of Do should be negative

Equation at foot of page should read b

=

_{b o - b1 D-1 Do}

For Equation 10 read Equation 13

For Equation 19 read Equation 21

For Shh read Shm

For Equation 17 read Equation 19

Value of cfh should be ~ 87 - 40:.1 For Equation 18 read Equation 20

APPENDIX A

Simple Example Calculations A. 1 Basic Structure

As a simple example of the method of redesign let us con-sider the simple structure illustrated in the sketch. This has 12 redun-dancies, but due to the symmetry of the problem we need consider only one quarter of the structure, thus reducing the number of redundancies to

3.

Upper and lower skin thickness Vertical web thickness

Area of corner flanges Area of central flanges Ribs infinitely stiff.

0.03" 0.05" 1 sq. in.

We can then write down the 'equation S = boR

+

blX, choosing as our re-dundant members the panels 4, 5, 6, i. e.

S

=

'P.~) 0 0 0

_VI

0 0 0 [ Q4 I "p~)

,

8 0 0 _V2 -1 0 0 _Q5 - p'{I) 8 0 0 _V3 -1 0 0 _Q6 ~

ft)

_{16 8 0 -1 -1 0}

- --p;C')

_{16 8 0 -1 -1} 0 ~~). _{24 16 8 -1 -1 -1} -~(h- _{,0 0 0} 0 0 0

-el.)

_{0 0 0 1 0 0} -~ä)~ 0 0 0 1 0 0 '5' ~) ; 0 0 0 1 1 0

-

~r.)-

1 ' 0 0 0 .. 1 1 0 "(L) '0 :.0 0 1 1 1

-..,;, ft;,..

Q, 8 0 0 0 0 0 Q2- 8 8 0 0 0 0 Q3 8 8 8 0 0 0 Q4 0 0 0 1 0 0 Q~ 0 0 0 0 1 0

qb

0 0 0 0 0 1

Assuming G

=

O. 4 E, the flexibility matrix is f

=

a 2 1 6 A3E 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 2 1 1 2 0.9375

o.

9375=

o.

9375 6.;25 6.25 6. 25

From these we form the matrices D=b' fb =_a_ 1 1 6 A3 E D

=

b' f b

=

a o 1 0 6 A3 E and hence D- 1 D =-o

~

'

4.25 18.00 6. 0"0 [ 208 160 64 [ 3.88 3.54 :.1. 90 18.00 22.25 6.00 96 88 40 1. 16 2. 55 1. 73

We may now solve for the redundancies using

x

= - D-1 Do R.

and hence get the matrix

b

=

_{b o}

+

bI D-1 Do

=

0 0 4. 12 -1. 16 4. 12 -1. 16 8.56 4.29 8.56 4. 29 14.68 10.56 0 0 3.88 1. 16 3.88 1. 16 7.42 3.71 7.42 3.71 9. 32 5.44 8.00 0 8.00 8.00 8.00 8.00 3.88 1. 16 3.54 2.55 1. 90 1. 73 6.00J 6.00 10.25 24

J

24 16 0'17

U

0.65 1. 08

o

-0.17 -0.17 -0.82 ~0. 82 6. 10

o

0.17 0.17 0.82 0.82 1. 90

o

o

8.00 0.17 0.65 1. 08

We also require certain eIernents of the matrix bI D- 1 b'land we repro-duce the entire matrix ON thé. IollowiIig page.

I~

0 0 0 0 0 0 0 0 0 0 0 0 0 0 3752 a _{192 192 44. 44 18} 0 -192 -192 -44 -44 -18 0 0 0 -192 149 26 0 192 192 44 44 18 0 -192 ::192 -44 -44 -18 0 0 0 -192 149 26 0 44 44 210 210 87 0 -44 -44 -210 -210 -87 0 0 0 -'44 -167 123 0 44 44 210 210 87 0 -44 -44 -210 _ -210 -87 0 0 0 '7"44 -167 123 I -315 0 18 : 18 87 87 402 0 -18 -18 -87 -87 -402 0 0 0 ₊₁₈ -69 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -192 -192 -44 -44 -18 0 192 192 44 44 18 0 0 0 192 -149 26 0 -192 -192 -44 -44 -18 0 192 192 44 44 18 0 0 0 192 -149 26 0 -44 -44 -210 -210 -87 0 44 44 210 210 87 0 0 0 44 167 -123 0 -44 -44 -210 -210 -87 0 44 44 210 210 87 0 0 0 44 167 -123 0 -18 -18 -87 -87 -402 0 18 18 87 87 402 0 0 0 18 69 315 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

o

·

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -192 -192 -44 -44 -18 0 192 192 44 44 18 0 0 0 192 -149 -26 0 149 149 :::167 -167 -69 0 -149 -149 167 167 69 0 0 0 -149 315 -98

-43~

... 0 26 26 123 123 -315 0 -26 -26 -123 -123 315 0 0 0 -26 -98 -.J

-As a concrete example we take VI

=

500 pounds

V 2

=

1500 pounds V 3

=

2000 pounds

S

=

0 -0.02 -0.02 9.09 9.09 35.38 0 4.02 4.02 10. 92 10. 92 16.62 4 16 32 4.02 .6.90 5.71

A. 2 Change Stresses in Flange 3 and Panels 5 and 6 by Changing Their Stiffness

Suppose that the stresses in flange 3 and panels 5 a,nd 6 are

too high, and we wish to find the required modifications to the structure necessary to reduce these stresses to an acceptable value. Here we use Equation 10 and we form the submatrix c ee by selecting th~ the appropriate rows and columns of c, i. e. ,

We also have 6A3E 3752 a f

=

-e 6EA 3 210 87 I -167 I 123 87 _- 402 _- I -_{1- -}69 _{_ _} 1-315 _{_} ~ 167 - 69 I 315 I - 98 123 -315-'-=98

~

438 2 1 1 2 6.25 6.25

and _{a ' .. . , ..}

ge

=

6E 2 1

1 2

7.5

7.5

Hence Equation 10 becomes

Sem

=

p (1) 3m

=

9.09 0.1352 O. 1025 -0.3330 0.2459 CT3

0

p (2) 3m 35.38 O. 1536 0.2376 -0. 1381 -0.6298 ([3(2) Q5 6. 90 -0. 1072 -0.0812 0.6298 -0. 1949

Cf

₅ Q6 5.71 -0.0184 -0. 1352 -0. 1949 0.8757 cr6

+

O. 1352 O. 1025 -0.2775 0.2049 pO 3 O. 1536 0.2376 -0. 1150 -0.5248 P (2) 3 -0.1072 -0.0812 0.5248 -0. 1624 _Q5 -0. 0184 -0. 1352 -0. 1624 0.7297 _Q6 (A. 1)

in which all loads are in kilopounds.

Let us suppose first that we require to make the stresses

0-

3 (2)

=

30 ksi,

er

5

=

cr

6

=

4 ksi then the most convenient form of the

equation is

--Sem = 6.36 O. 1352

V

3 (1)

+

O. 1352 O. 1025 -0.2775 0.2049 -p,(I) !b,) 31. 32

-

O. 1536 _{O. 1536 0.2376 -0.1150 -0.5248 t>~} 7.60 -0. 1072 _{-0.1072 -0.0812 0.5248 -0. 1624} 7. 04 -0.0184 _{-0.0184 -0.1352 -0. 1624 0.7297}

It was found that convergence was not very rapid with th is

equation and about a dozen iterations was necessary to get a satisfactory solution. However, each iteration is performed fairly rapidly on a desk machine. The final result is

Qs Q~

•

Sem = 8~ 74 kilopounds

39.62

8.33

1. 12

which leads to _{A 3m} == 1. 32 sq. in.

A = 2.08 sq. in. (t", 0.052 in. )

A

5m _6m _. 0. 28 sq. in. (t= O. 007 in. )

This solution gives a very thin skin in panel 6. Suppose

that ,we impose the condition that neither of the two panels shall buckle and

take the buckling coefficient as Ks '"" 6, Then we have, from section 4. 1, 2 2 1/3 2/3

0;

=-

(Qs Ks Es/a.s _{b s )} ::; 0.454 Qs

with E == 104 ksi and

0-

₃(2) = 30 ksl as before. The equation is th en

==[ 6. 01 -

~-

O. 1.352 -0.3330

o

.

2

45

~~~1110'

1352 O. 1025 28. 25 O. 1..36 -0. 1381 ..0. 6298 CJ; O. 1536 O. 2376 9.33 -0.1072 0.6298 -{). 1949 ~ -0. 1072 -0. 0812 9.76 -0.0184 -{).1949 0.8757

Ob

-0.0184 -0.1352 -0. 2775 -0. 1150 0. 5248 -0. 1624 O.

2049~~r~l)

1

-0. 5248

P/'

-0. 1624 CSJ) -0.7297 Qb

The convergence is still rather slowand the final result is

Sem _ [ 8.84

['cri) :

8.

5

~

31. 07

"""5

1. 98 9. 16 2.07 9.77 _ ~ which yields A3 = 1. 036 sq. in. A5

=

4.626 sq. in. {t ::r 0. 113 in.) Af) = 4.720 sq. in. ( t == 0.118 in. )

It should be noted th.at in this solution we have assumed

symmetry throughout and ignored the favourable interaction of tensile loads

on the buckling of the lower skin. There is no difficulty in incorporating

interaction effects at each stage of the iteration, however, except that the

problem is then no longer doubly symmetrical, and more elements appear

in the matrices. In practice of cource, buckling of the skin would be

restrained by using stringers since the increase in skin size found above

A. 3 Change the Stress in Flange 3 Only by Various Modifications Suppose that we make

cr

<ik

30 ksi by changing A3 only.

[

::i::J~vr:·.:~a~i~l::::tai:e::f:jr~:::j:o~::.:::n::

::::SJ

~ :j~l

The solution to which is found very rapidly (after four iterations) and gives

[~~::;)J

= [

;~: ~~

A 3m = 1. 248 sq. in.

Let us now change only the stiffness of sheet 6J still specifying

cr

3 (2)

=

30 ksi. We use Equation 19 and since all the terms occurring are scalar numbers we get the simple result

S _{hh =} Q _{6m =} Q _6- chh (P3(2) - P3,m (2» ëfh

From 'matrix_ on p. 17 J we obtain

so we have Q6rn = 13.2 kilopounds.

Frorn Equation 17 we then get

A6~

-

AS1= P (2) _ E(2) 3 3m cfu &1 Q6m which gives A 6rn

=

5.35 sq. in or t 6rn

=

0.134 in.

Finally let us modify the area of flange 6. From maÛ'-ix-on p. 17 J we select the appropriate rows and columns to form the matrix

=

6A3E -210 -87 -87 -402

-3752a 210 87 87 402 •

We also have

Hence, Equation 17 becomes

5.38

=

-Also Equ,ation 18 beco.mes

r

p 6nP)/::

rIO.

92l - [0. 1352

~6~2j ~6.

62J 0.1536 M=

m

C

U)

1]

~-I

402J [2

IJ P

6m

~2)

1

~ ~(2)_

p(2»)

°

P6 1 6E

~3

3m 1 2

m

putting A,3= 1 B2) = 30. 3m

An iterative solution of these equations is obtained very rapidly to yield

[

P6~)] ~

r

13. 93J