Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 8 pps

They were employed as the diagonal ele- ments of a Square matrix in the matrix multi- plications, but are written here as a column to conserve space, Member loads q, and q, were selected

Note that in this case [#10] consisted of only

one column, {nasmucn as there was only a single

external load

In art A7.9 the strain energy was written

= Ls | fad {39}

where [145] is the matrix of member flexibility

coefficients (Art A7.10) If now eq (14) and

its transpose are used to substitute into (15)

the expressicn becomes (note the use of (1,3),

The reader may satisfy himself that th

"cross product" term in the middle of the above

result is correct by observing that, because of

the symmetry of tua}

LJ [2] Ess) Esa] {Fa}

L?a_| [Bes] :3] Ess) ¢%s}

The various matrix triple products occurring above are assigned the following symbols, each

having the interpretation given (compare with

eq (24) of Art A7.9)

(ea) * [xs] 2:2) Era] ~ the mitrix

INDETERMINATE STRUCTURES

of externali-point influence coefficients in the

"cut" structure: deflection at point m per uit load at point n

With the above notation one may Write

=| Pa | ml {Pap * ® Le | [en] {Pap

Now according to the Theorem of Least Work

SU/a_ or = 0 for continuity Then, differentiating

eq (19):

ap 70 (Sn) {Pah rs]{as} ~~ 20

This last result may be verifted by writing

eq (19) out in expanded form, differentiating and then recombining in matrix form Rearrangec,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

If now eq

(14) one gets (w

m for n)

(ecb » [Ea] {?a}> [Ecz) [Ear] Be] 9}

= ( {cal - Ez) sé] Beal ) {=}

-( Hal* Gel Bal {3}

The matrix set off in parentheses above,

gives the internal force distribution per unit

value of the external loads It is given the

Eqs (23), (24) constitute the major re-

sult, inasmuch as they present the means for

computing the internal force distribution in a

redundant structure

Example Problem 13

The doubly redundant beam of Fig A8.27 (a)

is to be analyzed for the bending moment dis—

tribution The beam is loaded by couples over

the supports as shown

The choice of internal gensralized forces

ts shown in Fig A€.27 (b>) The appropriate

member flexibility coefficients were arranged in

matrix formas (ref Art A7.10 for coefficient

both beam halfs

The following matrix products were formed:

Gre} = Ga] Gal & 2]

The inverse of fa ] was found (ref though the matrix of member flexibility coe+

appendix) rs fficients was axnanded to 4 6 x 6, the

coefficients for qd, and q, were zero Thus,

= 2 |¬ 286L | ~.428L

5 |~ 429 | 858 4| .148L |~.286L

(The tabular form of presentation of the matrix Gi above, {s used here only to indicate

Clearly the functioning of the subscript nota-

tional scheme In general, it should be unnec-

essary to call out the subscripts in this

fashion excepting for the larger matrices, for

the handling of which, the tabular form may

prove helpful.)

Example Problem 13a

The redundant beam problem of Fig 48.27 is

to be re-solved using the redundant reactions as

tively, positive up These forces did not enter

into the strain energy expression so that, al~

1 9

L9 Then, multiplying out per eqs (17) and (18)

[1 5

Gala f

> fis 8]

[ors] = a [*

The inverse was found:

[ers-*] = SEI [Sil Finally,

[Ess] - Ga] - Ez] (Fe) Ee]

mple Problem 14

The continucus truss of Fig Ag.28 1s

teice redundant it is desired to analyze it

for stress distributions under 4 vartety of

loading conditions consisting of concentrated

vertical Loads applied at the four external

The internal generalized forces (q;;4;)

employed were the axial loads in the various

members These were numbered from one to

thirty-one as shown on the figure The member

flexibility coeffictents in this case were of

the forma, = Mết: (Ref Fig A7.35a) The

coeffictents ar¢ written as a colum matrix be-

low (They were employed as the diagonal ele-

ments of a Square matrix in the matrix multi-

plications, but are written here as a column to

conserve space,)

Member loads q, and q, were selected as

redundants With q, and q, Set equal to zero

("out"), unit loads were applied successively

at external loading points one through four,

the four stress distributions thus found being

arranged in four columns giving the matrix

Bia] (below) By way of illustration, the

loading figure used to obtain the second colum

of [Ea] is shown in Fig À8,29,

Next, unit forces were applied successively

at the redundant cuts "three“ and “five” as shown

in Figs AS.30a and A8.30b These loads were

arranged in two columns to give the matrix [Sz]

NOTE: VOIDS INDICATE ZEROES Multiplying out gave

Next; the values of the redundant forces, for unit values of the applied loads, were

found per eq (22)

[Pea] - Ea!) Gea]

O59 -111 0065 0035

ˆ 20035 O065 111 «059 The calculation was completed as per eq

(23) to give [Sa] ; the values of the member

forces for unit applied external loads

31

Example Problem 14a Fig A@.31 snows the two bays of 4 steel tubular tail fuselage truss which is loaded by tail air loads to be resolved into three con- centrated loads applied as shown The fuselage Dulkhead at the attach-points station (A ?-K)

is heavy enough so that it may be assumed to be rigid in its own plane, Hence, the truss may

be analyzed as if cantilevered from A-E-F-K as shown All members are steel tubes, their lengths and areas being tabulated below

ANALYSIS AND DESIGN OF

The structure was three times redundant,

In @ space framework or p joints, 3p-6 inde-

perncent equations of statics may ce written

(o 2 0), Here, however, stress ¢stails in

the plane ARKF are to be sacrificed; six equa-

tions are lost thereby since only net forces in

ons in this Slane can >e summed

-6 521 equations; 24 member unknown

Members £2, 23 and 24 ere cut

The next step was to compute ths unit

stress distributions [Ft | and fee] Rather

In the method used, the equations of static equilibrium were written for sach of the seven Joints." Summation of forces in three directions

on each joint gave 21 equations in 24 q’s (the unknowns} and the three applied loads P,, P, and P; Some typical equations obtained were:

Note that in each case the equaticns were arranged with the applied loads (P,) and the redundant q’s (qa2, d2s,; daa) grouped on the right hand side of the equal sign This ar- rangement was observed for all 21 equations, after which the equations were placed in matrix

Gea = daa das = Gas

AB, 24

coefficients of the a5:

The matrix equation wes solved for the q,

having as its main advantage the reduction of

the work to a routine mathematical operation,

In the conduct of this work appropriate stand-

ardized techniques may be employed *

The result in this case was

-ð4\6 | 2:727 | 5.571

The member flexibility coe?ficients Ly,

were arranged as the diagonal elements of the

matrix [5] - Then, multiplying out according

te eqs (17) and (18),

177.9 -1137.0 -6430.3

= 195.0 - 151.7 -2263.0 -154.4 161.6 2273.3

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES against warding of the root cross section due

%o any torsion loadings

The generalized forces employed are shown

on the exploded view, Fig 43.33 No forces

were shown Zor the lower surface, its nembers

and forces being equal to those of the upper

surface because of symmetry In the [14]

matrix this fact was accounted fer by doubling

the member flexibility coefficients for the

members of the upper surface

loading points “one” through "six", the [im |

matrix was obtained when the cover sheets are

"cut" the two webs act independently as plane- web beams The details of the stress calculation for such beams are similar to those of Zxample Problem 21, Art A7.7 and are not shown ners

NOTE: VOIDS INDICATE ZEROES

Calculations for [zr] were made by suc-

cessively assigning unit values to the redund~

ants 42, d+ and dia The calculations are illustrated by the exploded view of the end bay

in Fig A8.34 showing the calculation in that part of the structure for qa = 1 Note that

da, = 1 was applied as a self-equilibrating pair

of shear flows acting one on each side of the

"cụt", The ribs were considered rigid in their own planes

Put qdag=l From equilibrium of end rid: (ZM, 3 F)

Qạ 2 +(.5625 ~ 25) = -,2125

Qe = 23125

qa, = O, by hypothesis

So on, into the next bay In the idealization

used here, the ribs have zero stiffness normal

to their own plane so that the axial flange

loads are transmitted directly to the flange

ends of the adjacent bay (see da, 47, de, dio

STATICALLY INDETERMINATE STRUCTURES

The follicwing matrix products Per eq (18):

~,008216 ,008216 ~.QO5150 ,005150 -.002278 ,002278

true sfresses were (per eq 23)

[a] - Fez] Pra]

near root tion

is, the buildup of axial flange stresses the root of a beam under torsion when the

is restrained against warping The solu- for application of a torque is readily

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES A8, 27

Pa = -l Thus, one finds that under this con- 1Í -¬s.o 48.0 9

dition there 1s a root flange load of ESE SE1E17Ẻ mà =- mn] =! Sai} fors} fem] ==] 77° : 9

for a torque of 15 inch lbs

.472 888 0525 ,028

A8.11 Redundant Problem Deflection Calculations by Deflections (and in particular, the matrix Matrix Methods ~ [se] [srs"] [sm] + “E | 10825 0975 1.665 .888 7 1 888 1.665 oes .0525 „0975 .888 0525 472

of influence coefficients) are readily computed

from the results of Art AG.10

Assume that the redundant forces q >have

ceen determined from eq (22) The total de-

flection of external loading points is then

easily computed as the sum of, ONE, the de-

flection due to external loads acting on the

"cut" structure, [mn] {2} (ref eq 16) and,

TwO, the deflections due to the redundant

forces acting on this same cut structure,

[ens | {4s} (ref eq 17) Thus,

{a} Ges} {"}° Esltss}-

Substituting from eq (22}

(5}* Geol (%}- Gal Es] Balto}

‘(Fal CEI El te}

The matrix expression set off in paren-

theses above, giving as it does the deflections

for unit values of the applied loads, is the

matrix of influence coefficients Let

Determine the matrix of influence coeffic-

ients for the redundant truss of example

Solution:

An alternate procedure to that shown by

eq (25) was followed The influence coeffic- lent matrix was formed es in Chapter A~7, Art

A7.9, by the product

Fea] = Pa] Fc] Bs]

This product was formed readily, {nasmuch as [Ea] was available from example problem 15, The result was

Matters of accuracy have to do with the num- ber of significant figures finally obtained in the answer as influenced by the manner of formu-

lation of the problem ‘The accuracy of the re-

Sult may be affected by a muuber o? factors, two

of the most tmportant of which are discussed

These two factors are concerned respective

ly with the left hand side and the right hand

side of eq 31, viz

(Be]{% p= Era] {Fa}

they are discussed in detail below

® ~ ACCURACY OF INVERSION oF [rs | ;

THE CONDITION OF THE MATRIX

The characteristic of the matrix [rs |

Wnich determines the accuracy with which its in-

verse can be computed is its condition The

condition of the matrix is an indication of the

magnitude of elements off the main diagonal

(upper left to lower rfgnt) relative to those

on The smalier are the relative sizes of

elements off the main diagonal, the better is

the condition of the matrix A weJl conditioned

matrix is more accurately inverted than 4 poorly

conditioned one.” Two extreme cases are Tow

given for {liustration:

a) the diagonal matrix Its off-diagonal

elements are Zéro, so that it is ideally con-+

Gitioned Thus, the inverse of

b) a matrix all of whose elements are equal

in each row, All the elements off the main

diagonal are equal to those on.” The determin-

ant of such a matrix ts Zero and hence its

inverse cannot be found (rat condition is terridls,

to describe the strain energy, but the central support reactions were also given symbols as it

was destred to conSider them in the discussion

Application of unit redundant forces Qs zl and q, = 1 gave

The condition of this matrix is poor

Poysically, a unit load at point "3" causes

almost as much deflection at point "4" as at "3"

itself The cros§~coupling is large

SECOND, suppose the moments q, and q, had

Deen chosen as redundants The "cut structure”

in this case being visualized as in Fig

A8.35(d) Application of unit redundant forces

a = Land q, = 1 gave

The condition of this redundant matrix ts

obviously better than that obtained with the

first choice of redundants There is less

cross-coupling between the redundant forces

Thus the analyst, by choice of redundants, determines

the condition of the matrix The choice may be critical in the

case of a highly redundant structure, for it may prove im-

possible to invert a large, ill-conditioned matrix with the

limited number of significant figures available from the initial

data, The following statements and rules-of-thumb may be

useful in the treatment of highly redundant problems

ninimumi-zero, in tact Theoretically ¢ hen, “Sy

proper choice of redundants, the matrix fa vs |

may be reduced to a diagonal matrix (ideally

conditioned)

(1a) The choice of reduncants which gives

zero cross-coupling (orthogonal functions”) is

not readily round in gene In Some special

structures, such as rings and frames, orthogonal

(2) In choosing a set of redundants which

Will yield a well~conditioned redundant matrix, it_is best to make such “cuts” as will leave a statically determinate structure retaining as many of the characteristics of the original structure as Dossible Thus, one may consider that the structure of Fig AS.35(d) retains more

of the features of the original continuous—deam

structure than does that of Fig a8.35(c)

(3) The degree to which one redundant in~

fluences another (extent of cross-coupling) can

de visualized by observing how much their indi- vidual unit-load diagrams "overlap™

ọ elaborate on this point, refer once again to the above {illustrative example The cross-coupling of qs, with q, may be expected to

be large if their unit-load diagrams are drawn as below in Pig A8.36(a) This deduction follows easily if it is recalled that the dummy-unit load equation for such 4 cross-coupling term is of

Mam, ax*

ET Strictly, the comparison is with the terms

the form | » obviously large for ms,

Tạo

gr? at elements of the matrix [rs]

which form the on-diazonal

inca BX sạn

since an intecral of the form

have a contribution from the center stan only

Hence, the J By St is obviously considerably

smaller than J BS or fs se Thus, a

visual insrection of Fig A.36 reveals thet qi,

Qạ 15 4 detter choice of re cundavee than is Iss Dae

* see eq, (8) Art 48.7

"me