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Professional Degree Theses Student Theses and Dissertations

1958

Structural analysis by minimum strain energy methods Structural analysis by minimum strain energy methods

John B. Heagler

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ST~UCTURAL ANALYSIS BY ~NI~I

STRAIN El'JERGY ME:l'HODS

BY

JQllll B. HEAGLER, JR.

A

THESIS

submitted to the £acu1ty of the

SCHOVL OF }.fiNES AND l~ALLURGY OF THE UNIVERSITY OF MISSOURI

in partial fulfillment of the work required for the

Degree of

CIVIL ENGINEER

Rolla 1 Jl..i s ~ouri

1958

-----------

Approved by _......,_.xf£~..-tf;,.:-~2f:Z~_...~~~~~~~~-----­Pro!essor o£ CiV1l Engineering

I

TABLE OF CONTENTS

Structural Symbols •••••••••••••••••••••••••••••••••••••••••••••••• III

Preface ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• 1

c~neral Introduction •••••••••••••••••••••••••••••••••••••••••••••• 2

Classical Methods of Approach to lndeterminant Structures ••••••••• 5

An~sis or Structures B,y t~ruum Strain Energy ••••••••••••••••••• 10

Energy or Deformation ••••••••••••••••••••••••••••••••••••••••••••• 11

Derivation of Equation for Strain Energy or Bending ••••••••••••••• 12

Derivation of Equation for Strain Energy or Shear ••••••••••••••••• 13

DeriTation of Equation for Strain Energy or Torsion ••••••••••••••• 14

Caetiglianos First Theorem for Computing Displacements •••••••••••• 15

Example Problems of Denection by Minimum Strain Energy ••••••• •• •• 20

Solution of Statically Indeterminate Structures by Castigliano's Theorem of Minimum Strain Energy •••••••••••••••••••••••••••••••••• 21

Caetigliano's Second Theorem •••••••••••••••••••••••••••••••••••••• 22

EXample Problem ot Twice Redundant Truss •••••••••••••••••••••••••• 25

Deflection of a Redundant Structure ••••••••••••••••••••••••••••••• 28

Example Problem or Deflection or a Redundant Structure •••••••••••• 32

Tabular Arrangement for !.fi.nimum Strain Energy Equations • • • • • • • • • • • 33

~ple Problem ot Four Degree Redundant Structure •••••••••••••••• 34

Table for Co~utation of Forces in Redundant Members •••••••••••••• 42

Table for Deflection Computations ••••••••••••••••••••••••••••••••• 43

Summary or Minimum Strain Energy Method of Solving for Deflection and Redundant Members •••••••••••••••••••••••••••••••••• 46

TABLE OF CONTENTS (Continued)

APPENDIX I: Special Cases ot Strain Energy Equations

General Case - Constant Area with Varying P~ and Varying Un •••••••••••••••••••••••••••••••••••••••••••••••••••••••

General Case - Varying Area with Varying P 0 and Varying un

Strain Energy o! Quadrilateral Panels in Shear •••••••••••

Chart !or Determination ot Str&in Energy o! Qu&drilateral Panels with Uniform Thickness ••••••••••••••••••••••••••••

Chart to Aid in Determining Strain Energy of Quadrilateral Panels with Varying Thickness ••••••••••••••••••••••••••••

48 I-1

51 I-4

56 I-9

II

63 I-14

64 I-15

STRUCTURAL ANALYSIS

In general., structural. symbo1s used in this report are the same as

those given in ANC-5 Bu11etin, ''Strength of ~1eta1 Aircraft El.ements,"

and as follows:

A =- Area or cross section, in.2; panel. area in.2

E =- Y-odul.us of elasticity, lbs./in.2

r =- Internal stress, 1bs./in.2, Subscripts s, t, c, b indicate

type or stress: f's =- shear stress, ft = tensi1e stress,

fc = compressive stress, tb =- bending stress.

G =- ~1odu1us of rigidity or shear modu1us, l.bs ./in. 2

I = Moment of inert;ia, in.4

K = Energy factor

KA =- R&tio or end areas of tapered axial. members

L =- Length 1 in.

Mo • Moment in member due to carrying externally applied l.oading

through static~ determinate 1oad path, in. l.bs.

m =- Inf'1uence coefficient or bending moment in member o~ the

statically determinate structure due to a Wlit 1oading applied

to the structure at some point and/or cut 'n'.

P a Total applied a:rlal 1oad in member, lbs.

P 0 -. Axial. 1oad in member or statically determinate structure due

to carrying ext.ernal.l.y applied loading, through statically

determinate load path, lbs.

11n = Innuence coefficient or axial. load in member or staticall.y

determinate structure due to a unit 1oading applied to the

structure at some point and/or cut 'n'.

Q = Static moment of' cross section, in.3

m

q ~ Total. applied shear now in panel member. lbs./in.

~ ~ Shear .now in panel member or statica1ly determinate

structure due to carry externa.l.ly applied loading through the

statica1~- determinate load path • lbs. /in.

~ ~ Inr1uence coe££icient or shear flow in panel member of

statica.l.ly determinate structure due to a unit loading

applied to the structure at some point and/or cut 'n'.

T = Torsiona1 moment. in.lbs.

U = Strain energy • in. lbs.

V ~ Total applied shear load in member. lbs.

V0 • Shear load in member o£ statically determinate structure

due to carrying external.lJ' applied loading through the

statica11y determinate load path• lbs.

n ~ Shear load in roember of statically determinate structure due

to a unit loading applied to the structure at some redundant

cut 'n'• lbs.

Xn • Actual value of redundant loading at cut 'n' due to some

externally applied loading condition; (Note: the redundant

loading can be either axial• bending. or shear depending on

the nature or the redundant member that is cut and the unit

loading that is applied to it.)

d ~ Partial derivative

6.~d = De£~ection. in.

e ~E. = Strain; de£ormation per unit length

t) = Angular deflection

IY

f14_

PREFACE

Analysis of structure has its conception for the engineer early

in his formal education and £ollo~ a set sequency rrom solving for

equilibrium in simple beams and free bodies to more complicated

aolu~iona for stresses and def1ections in multiply redundant structures.

Difficulty arises for the engineer when he loses the sequence of

the subject and finds himaelf memorizing formulas and ffiethods o£ solution

without a clear and concise understanding of the underlying fUndamental

principles involved in arriving at these formulae and methods o£

solution.

The Aircra.ft Structural Analyst .finds himsel.r in a relatively new

field of engineering structural design. Here the problems encountered

are di!'terent_ more complicated_ and more exacting- as !'ar as a complete

knowledge or underlying assumptions and approximations are concerned•

than structural problems encountered by other engineers.

With the assumption that the reader or this report has a background

in elementary mechanics of' elastic bodies and understands the assumptions

that exist in the basic elastic theories- it is the intent of' this report

to take the reader from start to finish through the method of analysis

or indeterminate structures which- through past experience. has been ob­

served to be the most flexible- a11 encompassing• method available for

the solution o.r the most complicated problems encountered in aircraft

structural design.

G~ llJTRODUCTION

In the study of the equi1ibrium of a coplanar force system it has

been proved that not more than 3 unknown values may be found by statics

when the system is non-parallel nor more than 2 unknown values may be

found by statics when the system is composed or para11el forces. In the

case of beams, these two unlmown forces are usual.ly the reactions. Thus

the two reactions to simple, overhanging, or cantilever be&rlS (Fig. 1) can

be determined by the equations of statics, or these three types are

statically determinate.

A

1 Pc

! ,;(;;m

A

l I R R ~

l---.&..1_1 ___ I

STATICALLY DETERMINATE BEAMS

Figure 1

It, however, a beam rests on ;nore than two supports or in addition

one or both end supports are fixed, there are more than two externa1

reactions to be determined. Since statics offers only two conditions of

equilibrium for a coplanar parallel-force system, and thus only two

reactions can thereby be found; any additional reactions are excessive or

redundant. These reactions cannot be determined by the equations of

statics al.one, and beams w1. th such reactions are cal.1ed statical.ly

indeterminate beams. The degree or indeterminacy is given by the number

or extra, or redundant reactions. A truss is statically determinate if'

it has not mare than 2 reactions, in the case or paral1e1 coplanar force

system. and not more than (2j - 3) members. The first requirement for

statical determinacy is obvious. the second requirement ~ need some

explanation. A truss is just internally stable if it consists of a

series or triangles as shown in Fig. 2.

STATICALLY DEI'ERMINATE SYSTEM OF J.1EMBERS

Figure2

The first triangle is made up of 3 members and three joints: each

successive triangle required two additional members but only one joint

additional. Thus. if ''m" is the number or members and "j" the number or

joints m = 2j - 3 is re~red for statical determinacy and any additional

members added to the structure without adding additional joints result in

making the truss indeterminate. Fig. 3 shows a truss which is indeter­

minate to the third deuree.

TRUSS WITH .3 REDUNDANTS

Figure .3

There are many dirrerent methode of approach to solving ror de-

tlections or stresses in redundant members in en~eering structures.

No attempt will be made ,in this report to elaborate on all. or these

methods since the majority or them are based on the same 1.\mdamenta1

assumptions and basic theory. However, certain ones deserve nentioning

since there are times when any one or them may be more readily adaptable.

to a speci£ic prob1em confronting the engineer, than any of the other

methods and £ami1iari.ty with these methods woul.d be ad'Y&Iltageous.

CLASSICAL METHODS OF APPROACH TO

INDETERMINATE STRUCTURES

(1) Consistent De£or.mation:

The most basic and most genera~ applicable method or an~zing

indeterminate structures is the method o£ consistent de£ormation. The

procedure consists in £irst setting up a basic determinate structure

£rom the given indeterminate structure by removing the redundants and

considering these redundants as 1oad on the basic determinate structure.

There will always be as many conditions o£ geometry as there are re­

dundants. A system ot "N" simultaneous equations 1 where "N" is the de­

gree o£ indeterminacy, can be established under these conditions o£

geometry wi. th the redundant a as unknown. Be~ ore this method can be

applied, it is necessary to understand methods of solving ror defor­

mation, deflection, or rotation of statical.ly determinate structures.

( 2) Doub1e Integration:

The double integration method tor obtaining the equation tor the

deflection o£ the elastic curve of a loaded beam is general.ly appli­

cable. The engineer should be thoroughly ramiliar with the entire

derivation or this theory since it contains ever.y basic tundamental

used in most or the other methods and gives a clear understanding or

the concepts invo1ved. Numerous problems have been solved by this

method and are available in C1 vi.l or J..iechanical engineers handbooks.

Defiections or beams subjected to several. loading conditions are usual.l.y

synthesized using the principle of superposition.

(3) J.ioment Area 1-lethod:

The lw1oment Area Method is considered to be another o£ the classical

methods ot approach to solutions for defiecti.on and redundant supports

o£ beams and £rames. This method evolved from the double integration

method and may be consiqered a semi-graphical interpretation of the

mathematical operation or solving the di£ferential equation involved in

the double integration method. The Moment Area Kethod for finding de­

flections of beams is much less difficult to use where the moment of

inertia is not a continuous mathematical function along the entire length

of the beam. Where these discontinuities exist the moment area methods

are adaptable with very little more work than where continuity exists.

This method can be used to determine values for redundant reactions with

relative ease, unless the degree o£ redundancy is too high in which case

the arithmetic becomes long and tedious with a high probability or human

error. The method has two important limitations. (1) It gives the de­

f1ection at only one point on the beam and (2) only the deflection due

to bending stresses are considered and those deJ:lections due to shear

stresses are disregarded.

(4) The Slope Deflection Method:

The slope deflection method uses the relationship of angle change

vs. end moments in its basic form, and is another of the classical

methods of solution for redundant supports in beams and !"rames. In this

method of solution a1l connections between members in a continuous

structure arc considered rigid and non-changing so that the angle between

members in rotation remain consta4t. This method rray be used to advantage

where computers are available to solve the simultaneous equations which

are encountered in a redundant structure analysis. Again this method

does not take into consideration any def1ections due to shear stresses

in the :nenbers.

( 5) Virtual ilork:

The principle of virtual. work is one or the most .t\mdantenta1 and

coMprehensive principles of rational mechanics. For a rigid body it nay

be stated: If a rigid Qody under the action of any set of forces in

equilibrium be given a vecy srrall "virtual" displacement (i.e., possible

but not necessari~ actual), the sun of the work done by the force systew

will equal zero 1 and, converse~: If 1 lmen a rigid body tmder the action

of an~r set of :forces is given a verJ small 11virtua1" 1isplacement, the

total work of the forces vanishes, then the system of :forces is in

equili bri UJTl.

The corresponding theorem tor de~ormable bodiea, ~ch as will be

needed in analysis o£ statically indeter~nate stresses, may be stated:

If a structure in equilibrium under a set or forces be ~iven a ver.y

small, virtual def'orr . .at:ion (i.e., one consistent with continuity and

e1astic behavior but not necessarily actual), the total suromation or

internal ~ external work will vanish.

The deflection of structures w~ be obtained from a consideration

of tr e work done by the f'orces acting on the structures. The external

work done by a :force acting through a deforrration of the structure will

be stored in the structural raterial as potential energy of' deforwation,

or strain energ:r. Tris energy is recovered as the structure returns to

its ori~inal FOsition, as the load or loads are re~oved. This statement

follows thQ law of the conservation of energy. In calculating some dis-

placements a numerical solution can be obtained by equating the actual

external work to the corresponding internal strain energy. This is the

basis of the virtual work method whict .. is further explained as :follows:

rr two loads A and Fi are placed on beam I=JL3 I as shown, load

will cause a deflection .a; at A and .6~at ~ and lead Pz will cause a

defiection ~; at ~ and A~' at location or 1oad ~ , it can be shown

by the Naxwell-lO:ohr reciprocal theorem that the ext.erna1 work is equal to

R l~ ~ ! - B

"*'""' l f "' 2 t t :...-A " --- A' LJ./ - - , / ' -- 1-11 z - / '',,, A"T-----------------f-~ .,"'/

........ 1 ---AZ .L'l:f/ j"" --&., __ ..... t....;:>lloJC.~- -.... -l:: -

.___ -------- - -- -

Figure 4

In general. however. if the external work·due to two forces such as

R and ~ is set equal to the corresponding internal energy caused

by the bending morrents only • the equation .L R4 f.L ~~ :.:~ re Mdw is z z ~ 2£r obtained. The right siJe or this equation can be evaluated but as there

are two unknowns A/ and 4 on the left side, no solution is possible.

If only £!!!. displacement is involved, a nwnerica1 solution can be tr.adeo

This we can accomplish by the simple expedient or placing an auxiliary

(virtual) force system on the structure be~ore the actual forces are

applied. The external work done by the virtual force acting through the

actual displacement is equated to the corresponding internal energy due

to the internal virtual forces times the actual internal displacements.

The virtual force system must necessarily consist of a force that is

applied at the point and in the direction or the displacement that is

desired, together with the reactions that are necessary to for:n an

equilibrated force systemo

The Virtual Work differs from the l.fi.ni.mwn Strain Energy theory

'Which is discussed in the next paragraph. However. in the actual

operation of solving problems they are ver.y much alike. It is therefore

important that the reader thoroughly Wlderstand that the "Virtual Work"

method is based on the p~emise that the external work done by the loading

is equal. to the corresponding interna1 energy caused by the loading.

(6) l.fi.nimum Strain Energy:

This method of sol~ng for deflections is the most all encompassing

method available to structural engineers. Although at times it becomes

somewhat tedious and long, it offers a method of solution which is en­

tirely general and may be used for aqy problem in aircraft structures no

matter how complex. It is different .from the Virtua1 load methods only

in basic theory and lends itself to a tabular type set-up as well as the

virtua1 load method. For these reasons it has been selected by MAC engi­

neers as the method to be used for structural a.nalJsis and is outlined in

detail in the body or this report.

Discussion:

Numerous methods have been developed £or solving indeterminate

structures o£ which the above written methods are a few. However, the

ones mentioned are gener~ considered to be the classical methods £ om

which others have been derived. As instances o£ this, the Conjugate

Beam method sometimes called the Elastic Weight method, is derived !'rom

the Moment Area method~ and the Hardy Cross Moment Distribution method,

which is used extensive]Jr in building trame analysis, is simply a method

or solving f'or the unlmown values in the simultaneous equations arrived

at by the Slope Det1ection Equation~ using successive approximations.

All of' the above mentioned methods are capable of' determining de­

nections due to bending stresses. However, the energy methods, Virtual

Work and Minimum strain Energy~ are capable of handling detormations due

to any physical stress~ shear stress~ axial stress~ bending stress or

torsional stress. Since the problems arising in aircraft design may in­

clude all of these orders of stress then it is again emphasized that the

energy methods, minimum strain energy in particular, is the most superior

method.

ANALYSIS OF STRUCTURES

BY MINIMUM STRAIN ~ERGY

It has been stated that the minimum strain energy 1 sometimes called

Castigliano's Least Work method o£ analysis, is superior to other methods

for the composite type structures met in the aircraft structura1 design

field. Before attacking prob1ems by this method it is imperative that

the reader have a complete understanding of the basic fundamentals of

work and energy.

(1) Definition of Work and Energy:

When a body is capable of overcoming resistances it is said to

possess energy. Energy may be divided into two classifications: (1)

Kinetic Energy: Kinetic Energy is that energy possessed by a body in

motion by virtue ot its ability to overcome resistance by motion, alone;

(2) Potential Energy: Potentia1 Energy is that energy ~ch is attri-

butable solely to the configuration, or re1ative position, of' the

particles of -which the body or system is composed. An elastic body which

has been strained will overcome resistances if it is permitted to return

to its normal condition. Potential Energy stored in elastic bodies is

often calle1 ''Strain Energy". The units o£ energy are the same as those

of work and are foot or inch pounds.

Work has dimensions of force times distance and is defined mathe­

matically as v =fl'cc.$<9 d.s 1 where ·('~ is the force acting on a body at

an angle '"'-e" with the direction or motion or the body and the term

represents an increment of distance the body moves while the force is

acting.

From Newton's law of conservation of energy, we know that energy

can neither be created nor destroyed. Therefore the tota1 work done on a

1.i.

body during any given interva1, by the external forces, is equal to the

difference between the amounts o~ energy possessed by the boqy at the be-

ginning and at the end or the interval., provided that during the interval

no energy is transferred to or f'rom the body by agencies other than the

external forces. This is an important concept .f'rom the science ot

kinetics and simply states that the work done on a body is equal to the

energy expended on it.

(2) Energy ot Deformations:

As previous~ stated, a torce applied to an elastic material will

cause a corresponding def'ormation so that the f'"orce will do work on the

body and the work will be stored in the body as potential energy 1 which

will in the rest or this report be referred to as strain energy.

As an example or this fUndamenta1 or mechanics consider the bar

shotm in Fig. 5.

I

I I I I

.3d' I

ll~ F = .30x/O~/bs/sr //?ch

vv-= z~ ooo I i>o .

.6 :. PL/~E

Figure5

Since the torce is applied s1owly, increasing .f'rom zero to 20,000

lbs., then the work done on the boqy is:

I ··- -l- · •o·A - - . •-

~~ l ': J.~' · i --:-~=-·· · ·r----~ -:--:-·j-

i i . . : --.·-' • --- ___ J...._._._ .i r--: I 1

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_. ·---·· -------------4---'--L.i:l'!:. -4--:---~~-----r-~~--~- -+-

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11-,.. , T -~- --~---- -+-

. ; ~(, l I l I . ...., ~_/;rt;~' .. ~ · .':./J • I l ' 1 6 , __ ......... ~--=t" ~+-;......, . __ ...,.../II . I I -=~~ ev Fr~-:t IV. ~ (7),'.r/ . i :

---------- - -~-1

.1 ,; : 1

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.. ··- :-- - -- -Y'i - -1..,--- t-- :·:- --1--- -·---: -r . - -- --:------! - - -~,._JI-i~"-~-~ -~~~ ---.-- - · ; I · I ~I I . . . I I f• I I I i I I I I. I . I ~ I

- ~ ______ __;__ --~-!J+- ~ L--~---K4' -'Lf~.l-+---?~;k-7;--t.-.:~+~J-d-~-d-·+---J... ! . ; tj ' ' : i : ! li :. ! I l r. .. :~ i-- l'f . .:,·),",·r,.:__.· iJLV.::::-~,~ • ! -: _ +-+ _ -J i:: __ ~ _ , .- + --'----r ! h~h-~ +-v::+-e:fr::;:t/~--~- ;:_:_::t· ++ _ :-

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1 ~ -

-~- --· ---~-~-_J~J-____ j_ __ _;_ __ ~_- ~ I . ~ I ., __ · _j __ ! I . ~~- I - . : - - ;_ i l I ... ~ : : , I ~! ~- i ... ~~~ --: : · t • I ~-· · . - - ! - - ~ - ~--; i :'A·rt-_47· · L .. -.~1.~ · +. LJ ' ,.C/i..-.-..~ 1/lh · VA'::.> 7./,....,r').A:7~ ! .,.... :;.--. - --. ~ ~~ -- r - ·

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! . I . I /' i I • : • ,Tf I I I . J - L-'- I •• • I : ' : . l . f ~...... : j I

--+-c..:. r"' .~ ,.:74- . r;f->: 1'-"ff .. t~-;· . ?~·-'4' ...o...r.:.....::Z>·t l:,.,·r__;~ y,_~-+---~ I ' I I I . ,. ' ' I I . 1 . I I I . I I I I ~ , ·- ; d ; , I . I I I I • l . I ' l -r~ .ri_a:.c"J -r;7~ ~ u: . ,· .. "l I I I I l : . I t • i - ! •

- -i --- -, - - -t--·- 1 . - --r~, , _------~ - - -~-- i--- ~- - -r--·--4 ___ ,_ ~--+'-- :~ -rT I ' ' . l / I • l I I I I I I . I I . : -: i.; 1 1. 1 : ' . . ! .. 1~- - i - I · : __ .1 _ ___,....·--+- ----t--+-~-- ~-~~ I . ..L...__ --+--. -------r- -- -:r- - · · - -r 1 : ; . I . I __&. ;., I ~ • 1~ ~ .· i - , : ' .- . • f. 4' ' 1 I . I ... -- I :_L

I ' 1 1 i . ..T-=- -:. · t. i ,/71;:--r- .· ,r -ri ~ ' :-·¥t..::·l~~,.~~- . · 1'- l ! . _j_.

- - 1- - : -- --1---r --+- -~ I - --r,c-L L, .. ,-z~,..,_ . _:_-'(-· ·- . -T - t· -L;c~t- . --r,--:-T·--,-.. r i 1 t • ; ! 1 1 I : 1 ' -L · I 1 1 1- · 1- r· :

11 ! .

I ' I . ' I I " I • I . I I l:-£ ' ' I i I . ______ ..,___·-+---r----~----l--Z]-~ ---t----- . - . .._. - -~--1 I I . ' f I • 1 • I ; - ~ ....... "- ~ .... ,., 1 ' t : ! I I ; ; ., ,. • ! ~. 2:!. ~- . I I I -.,...· --t A .- I . l I I I I

j_ I ' 1 : . ! #, I _:_ ~ - . ' ,-• .J' ./.i ' : f " . f . • - -~ t . .. - . • l ' . I ~ -- - J._ _______ T ------- --~ - -- -- -- ,:r_,_~t . ..:..-. -· -~~ ...... - -..,.-- --~--- .- ~- -.-----t- --· 1·- -

; ! l I I : I . - .. i .. ~.. ~-~ '~. : j A-~ ~ ~ ::. : 81 I I . i . 't ·.- j I ; -: I I I . I . . I ; A.+- : I . ~ . ! . ; I . l I I I - -r · -·--·-r- -------.----:--r---.----r---t----+-----+----- -- --~-...._--+- I ~--~.----· . : l • ~ • /. j. . ' . I . : : - • I ~ . l ' I ~ j I

I , tC~.:,;· ; : . ,./,•,,-. J/, .. . ·r ·'!:-;::· ~ ~; _.;1 :::' -?•f:,,A.• .... I ~,;,--1...._ .. · to-- I : . i ~ l I . ' - -· - ~--~.-~.~ .. . . . .,f..,. ..:....of~ Ly...: . ..... L.,.._~.t_ .... ~):,_......,:.c;:: . ..-i.fl ~-tL --i.«....L . .A • .:..~-+--- - -- ----+-- - ---~---,-- ---..- -

1 ' . . . ; 1 ·1 ' ; I - ! ; I i . ! I I I ! . I - i

; . ! . : . I : : - ,rl • I .:.; I : i i ; : I ! L I t - · .-- -~----;----,--- - ;---7'·---+--1 -~-- r-;,---+.t:-t.,f--r-t-7";-:- ::ii:-=J--;:;r··-:--·;- --r·--t·--r--: -;-- - - --7-

: . l : t '7Jr =- ~-----~1 · :i- t. · ' r ! '· . l : +. -! 1 ' I i · i ---- .-- - . - - ·--- - : -- :- -- · - - i..:..::--~ -~7~::,-~--~-- :t--±....---s-:;t-::. __ ----:L- -l--+-~- -1---- ,.---1- ..,-_;_-

' . ; ! . : , ..... . .I I t .; i l .....::. .f;B- I. . t I I , . . . . , .. i • I I l ! . : ; ' I ·.·I i I : : - I l . . . - ! I - i ·-·-r------·----. -r--~---r-------1 -:- 1 ~ :

1

1 • ~--1--1 ..__ ' , rr-T 1- --:-- ·r-- - ~ -- -- --~ - - -: _t[ ___ : ---L-- : -: C --1--r---~i~-+-~ --,--~ ---+, ---t-~-~--- 1~~ -t---1--t-, iI I I ! I ! I : I : ; I ! -t±. l

--~-- -: --r ! ---- -t--r I 1 1 . • r---;-- + r-:-t-+--'-1- :---+-- -. - -;--' I I . ! t ' I I ' ' ! I t l t • I . I ! . I ' • : . ' f '.; ! I jj; I ! I - I ! : t -~- : - - ~----: ---!-- --·-~-!--- t_--:-~--- -r-i'---l~r---~-:- ·· · ·P-r·--:-rT·T-~~ . -;---;--.- - ;---:-~-r

_J..._ : __ _ ~ ____ I _ _L..--4__._- --+- . I ' ·-~~ I-- ..,L__! I ~ ; ·

I . . I I l I I I • I. . ' ! I

; • I • I I t ! I . I I ' I : i ~ ! 1 : • , ! r .: ! _. , . L ~ . , : . : ~ ; ~ ~ . :

---~- -- - -- ,- ·--1- - - -:-- · - , -- - ~ -- ---- --. _._ ___ t-- -~:_,_J_-- -t----. -·-----+----+---+---.---l=_.__. ~-- ~-t-- -- r-- - --- .. -,.."!' --- I . ; ! . : t: . : • ; . t: Ll .. , .. I . ' i . j i ! ~ ! - ; . :II _> ··-; !- . L ~- - . ·; ' -r-· ~-:--t-- f- ---r-----r-- - . --~ . i I ' . . : . . . : - -·- ' r--:-. . ~- - tt

. I t • I . ! : I I • ' . I . . . .. ~ . l ' .. . .. . ; ..

---.------.- ------- - ~----~------. - --------r-

..• · ' ~

i

) _ ; .

---·- --- -- -- ~ -----!-_. ___ ....;_1 __

[---- -. ·- ·

--___ ..______ __ - -------·-----~ ---------r------ - -~- - - -

__ j --- - --, ---· -- 1 I , ,

- ·!--··--:-- ---:- - -~---:---· ' I -

This simp1e example nay be considered a derivation or the amount of

energy stored in an elastic member carrying axial load. By the principle

of superposition we can determine the energy stored in an entire struc-A P."' .-.-#- P' ~L

ture composed of members carrying axia1 load only. v=L.~'~ ~-~-~., ,z,. , '2~ LA...& JAJ£ LA~E

In order to solve problems, which ~ have other forms of stress

than axial, the genera1 derivation of energy equations for shear now,

bending, and torsional stresses f'ollow.

A structure ~ consist of members which contain any one or a11 of

the previously derived energies, and may contain them in &QY number. In

order to determine the total strain energy in such a structure we can

use the law of superposition and add directly the energies contained in

the structure.

Since the energy stored in a member is a function of the load and

shape of the section, there are many other energy .torms which ~ be

needed to solve a particular problem. Some of these forms are in Appendix

"I" of this report, and others may have to be derived at the time they

are needed.

(3) Concept of Displacement From Castiglianots First Theorem:

A general method for computi.ng displacements is given by an im­

portant relation between forces and strain energy which is known as

Castigliano' s first Theorem. This theorem ~ be stated as follows:

Castigliano's First Theorem:

'~en a structure is acted upon by an equilibrated force

system which produces a total internal strai.n energy ''lJ", the

derivative of' nun with respect to ~ force gives the displace-

ment in the direction of that force."

To explain the physical meaning o£ this theorem consider the beam

shown in Figure 9. By app~ing the loads P2 and P3 first~ the elastic

curve will be deflected an' amount as shown by the "Y" ordinates. When

the load P1 is applied to the beam the elastic curve will undergo addi­

tional deformation as shown by the "z" ordinates. Increasing the load

Pl by some increment "dP1" the "z" ordinates will all change an amount

•'dz". The change in the strain energy due to the increment ''dP1" is as

follows:

Figure 9

' ~ ~ ..... , ~,:,, R \, ......... ... Y. ~ Y. yl , .,. ,' ,' R~

l ', ', .... ,.......... • 1.. .,,-- ,' # ' ............... ~'-- ... __ --------------~~," ,' ', ........ ·... I !:1 ~ • , , ,

"- ...... ~ z. - ... t_ -1--- -z_: '"- - - - - - - A~:... , , , ', -~

........... - A.Z._ - _,, - ______ ... --.,.

Since stress is proportional to strain~ then

dA P, dz,: dP,E, d~.:: c/P,c~ <:/ ~ :: :J!, P, P,

the change in strain energy is:

dl7.:: (R"' ~P:)dr, .., ~c/~ ..t/:?d~, --_f d.r?dz, ~ _; (,~ ~R~~ ~R~J)dR ,

but from the Maxwell Mohr reciprocal theorem 7? ea.~ 8 ~ =-P, ,":'

(Ref'erring to Figure 4. Page 6 Maxwell-Mohr reciprocal theorem states:

P1 · A 1 " = P2 • .A 2'; or stated in words: A :force P1 acting through a

displacement A 1 " at 1 produced by a load P2 at 2 does the same amount

of work as a force P2 actiilg through a displacement of 4 2n at 2 due

to a load P1 at 1.)

Neglecting the second order derivative ;_ (dR dz,)

then, du ..lf?.;r. '1'7? if dq' - .P, dP, or p'R

~d 4/£ch /S /o SQ/' s~.: o)~ •• , $".: ~/q/ ofs~/qce/??e/// q/ ~ ~~ ~ 0' ~t:'<!'* .y.sk'" /?., ~..J d-'?c/ ~ ~/'~

r.,PdC_,?QnS /f? t;?na/~

Obvious~~ as this relation holds tor any value o:f P1~ it can be

used when P1 is zero; that is• the increment of load may start from zero.

Consequently~ if the displacement is desired at. a point where no load is

applied~ a force P must be assumed in the direction of the required dis-

placement and its ~tude reduced to zero after the algebraic expression

for the strain energy has been dif'terentiated. This operation implies c))U

that~ although P may be set equal to zero~ ~ is not. In practice the

P value is conmonly referred to as the dummy load and is assigned a

value of unity.

c

Figure 10

If a couple M equal to PC {Figure 10) is applied, the work done by A, ,,

an increase, dP, in both forces is: dv = (~ """Ll '')c/p =( ; 4 :;J C c;/p

A , ..,. 40 ,, - / - / .,-, - / where ~ - o(9 <7"<7 Cc;r,.- : O'A-'? c -

_/ _/ dV ~ .. ~ .. / _.. <§!!:!, • • ... c;T~ :a,-,.,· -e Dr tf9.: dh"' 4/,~~, /s ,..o so/ -e-.: 0>/>t

To anyone familiar with the 11Virtua1 Work Method", it is obvious

that there is a great deal of similarity between it and the Minimum

Strain Energy Method. It is interesting to compare the two methods in

some detail.

Referring to the fundamental equations of internal strain energy,

we have:

For axially loaded members

For bending members

For shear panels

Since structure may be acted on by a number of simultaneously

occurring loads, the total force, moment or shear appearing in any one

member can be represented as f"ollows:

where Pn represents any one of the simu1taneously occurring loads, while

~/ ~ ~ are infiuence coefficients for a unit val.ue of" ~· P,h ~II#, ,,,, , c 17

f represent that portion of the tot.alP,~ ~ that is independent

of ~ • Differentiating the equations of strain energy with respect

~ y- ,_J.. ~p ~~ £4 til~~ to ~ we have:~Jq =~ :;j~• M J (~) ~ = h'r ~ I(

~ , .,_,. 0 £.z-

~u ? :.-:£ ol'l( <?'7c/ r 3.J ~P,., = z (';~

Substituting these values into Equations 1 1 2 1 3 the familiar

deflections equations are deve1oped. i.e.

Axial. members

Bending members

Shear pane1s

--;-----.-----.--------.---~-~- . ---~-~-2G--!

.!.__ -- -- ·--- ··"-·· ··· · · - -+- --·- ·- ·-:-· - --·- __ ! ______ - -·------ - -·

£A'A~PLE: · Z/er'._~/?!;n~ ~A~ k<erhcQ'/ a~4c£orJ # .;44~ . ~,-~7~/~7 W -67~ /~<'7~ ,sAo.u---./7 : . . .

-·- -- - -~ _ _ _ , _ __ _ _ ! __ ____ __ __ __ _ i _______ · - - - -t· . ·· ··- ···--- · --·· - !·--- -- - - ~ -- - ·--- --· -: ·--·

. . . . • _ .. _ _ ___ _ _ _ _ __ _ c- ______ _ ~-- _ c _ _ __ 1 . .-£/..GU R £:--/ /_ ___ :_ _____ -- -~ __ _________ ______ _:_ ___ ~ - - ___ ·-_ _-_: __ __ _

. -~ : ;._ . "-C.-- -t-~:

-ci W£ -~ ~ - --~ - -z · .

: i . --· -- -·- '---- L---------·· - - - - - -~- - - -r-- -- - - --- ------- -· - -- - .-- - --- ·- ---- - -- - -- ~--- - --- --+-- --

- -- ___ l __ _____ _ _ ' _ __ _ _ _ _

. ~ ~ --~~~//?0/~/ ~b~a<a/· M.e.Lb~cbo~ t?T t.?hd//2/~e ¥/""&'cJiton _,2..(_ ~.t1~~L~~;;~.:u_: r-£'0:~ __ ae£~~-~Z2 - - ~ - __________ _____ __ __ __ :_ _ __ --·-· _______ _

!

· / - - ~/r/~aX - : __ : __ · / ~ · ~i- -- : · : .: · {/ :;r -_:-:; T .zc -- , - --Ll:!~/ ~~ ~ -':_ -·. -' "h-~ o: ,-1':?·x~,l<'_ ')_ . - -- ·-

-. .. ~ -- .:z -: '/' ~~~.- : . . :. . . ; 2;_;_ /i -L +< ; . • . - •

__ _ , __ _ .! _ ...QJ~l ____ .{: ____ :_ A : - -· ' "'~- //.x . X .. : , )< - ' ::::J:i- . . ,,, ""' - ~- ~~~¥;-- -~-:- ~ r ---= --- ...-t-:?_:=- ----==--L - JB . -=;z- __ ___ , __ - -- --T----------;

EXA/l;YP;(£-.~ : z;~kr~-~ de4c~l?n . ~ -/~;·us.s. :;A~~.A/r7 --- - - ---- - - ~ ------- - -- - --- -- ---·-·z - :?""7 · r- . , -- - .; --- ------ - · ---rr -- ,._:_ __ _ ·- · -:

· ' l ~Y c t:7 ..SY/ y-/n//'7~.., ~ .;77~~noo. ; .

I . . . -- ~ ···- --- ..

i - .,. ___ _ ____ __ _ . __ . -- - -- - --- · -- --

' '

I . {

~~~--~--~_-::_~_~_iJ_"'-_- ·.._: . ------p--------A.=-2_S-a .. J/U:~_es ___ U//_a}emLJ~rs./ __ _ : ___ .. : _____ :.

I I •

.c.· -,~ /0 ,-· . ,t..- :c..>~·,-: ~ S/ . -

·--- -- -·-· - - -p_<::...·· -· ·-T _r ~ ~ ';-;;-;;: u ......__ ~-'7•-

··· - --- -- ---- --- ---- · - - -' ~ 7-'• ' . L1 -- 2. : -.//:..;. n=--

Member p~ ' u~ L ,{,£ :: -J2_, Pu.,~E= /'1

AB + eo x lo3 +8 8 X 12 5.12 6 b W X 106 180

BD 0 +1 6 X 12 0

l:IJ X 106

DC 0 0 8 X 12 0

60 X 106

BC - 100 X 1o3 - 10 10 X 12 10 b l; l:IJ X lOb 180

(4) 1~ Inches

Solution o~ Statical.ly Indeterminate Structures By tigliano's

Theorem of Minimum Strain Energ:y

A familiar ~ethod used to determine forces in members o£ statically

determinate structures is to cut certain or the members and replace the

interna1 forces of the uncut member by externa1 f'orces acting on the cut

member and then solve by statics for the force on the cut member.

In the solution of indeterndnate structures» a common method of

analysis is much the same as in the solution for determinate structures.

Internal .forces in members of the red\Uldant structure are replaced by

external forces by cutting sections through the structural members.

Since there can be no relative movement between the pieces o£ any member

cut by an imaginary line. and the strain energy can be expressed in

terms or these unknown forces• then -:;, =- A,., = 0 or the redundant

f'oroe " X," will take such values as to make the strain enereY a r.dnin:wr.

Actu~T the first derivative of a fUnction set equal to zero gives

either maxim~ or minimum value or the variable. It can be proved

mathematica1l.y that the total work JTlUst be a rdnimwr• but it is un-

necessary. since it is inconceivab1e that, when nature has its choice.

it wou1d make the total worl<" a rn.aximun. Nature always tends to conserve

work or energy.

Gastigliano•s second theorem is based on the above prerrise and rray

be stated as follows:

"The redundant reaction components of a statically indeterminate

structure are such as to make the total. internal work a L.inimurr.

This second theorem is the basis of a method for solving indeter-

minate structures which is called the ''l·:inimurn Strain Energy l4ethod".

Its versatility is ~ted and ~ be u.ed ~or any \~ er indeter-

ll"inate structure solution.

Application of the theorew to the so1ution of prob1e~ may follow

two slightly different lines of attack. (1) Cut the redundants and

place an external unrncwn value at the location of the cut. Tl:e :forces

in the cut structure Jue to the unknown .force and the external loading

are then deterrainej. From this the ~nimum strain energy is deternined

and the redundant solution made. (2) The most often used nettod is to

apply a unit load at the position of the cut and determine the forces in

all nembers of the cut structure due to this unit 1oad. The forces in

each member of the cut structure due to the real loading are found

separately. The total force in any member is that due to the rea1 load

plus that due to the unit 1oad ti.Jlles the value of the real. load which

exists in the redundant (cut) member. ~ e

P= tP.7"z..u~x~J

/??:~+~~ ~)

Y= (~ _,1:2~ ><~J

? =(r# r z ?'~ .. x'7 J ~ere /:17 Vn ~ ,v' a~- are influence coe.ff"icients or the forces in a

.,~ " ., T ~ ~.

~enber due to unit axial lead, moment, shear or shear f1ow applied at

the redundant, X"' is the real value or the redundant load, and ~, V.and

~~/'?,are forces, rnoJTtents, shears or shear flows due to the real applied

loads acting through the statically determinate paths.

A beam with '~" supports serves as an example of this second wethod.

fx~ r

~~ 11?

fxc. l~

!f ~A

t ~~

_L /!/...-· ;~,.,.,!» ~ /~;'//

Figure 12

It is evident .fro't the above presentation that for nnn redundants

there will always be "n" simultaneous equations which can be solved .for

the unknown redundant reactions {or £orces).

The above cited examp1e involved only strain energies or bending;

however~ other types or applied loading and energy terms do occur

depending on the type of structure involved. i.e.: ;~.,"" L f~';..t. ~ ~~~ ? ~cr:~rJq ]+ x,._L~;:~ L. "f-~; .. <i.,. ~ct;-ct.] ~ xa] [ · ) + x,J [- 1

As an example of the two different methods of attacking a problem

by Gastigliano's method of minimum strain energy, consider the beam shown

below which has one redundant reaction.

Exuxp1e 1.

Determine the redundant reaction RB on the beam shown.

~· L z 8

P, 4~X. ,,

Figure 13 /lf~fiJod /.

M-.~ ~(R-~)X

X"= %D 811

L 2

p&

~ ..... !.-.: -t

A=Pz

The same results can be obtained by placing a unit load at the

location of the redundant so that the moment in the beam is the moment

due to the real load plus that moment due to the unit load times the

value of the redundant.

X13=

It shou1d be emphasized that for this particular type of problem.

where only bending energy :is required, other methods of solution are

more adaptable than the strain energy method. However. the above example

is of academic interest and should help !amiliarize the reader with the

ndninn.un strain energy concept for indeterminate structures.

As an example of the use of Castigliano's theorem for determining

the loads in members composU1g an internally redundant et.ructure, con-

sider the twice redundant truss shown.

Figure 14

~E = Con.:J/Q,/

,f;,. Q// .memkr.s

For the truss shown there are ll members and 6 joints and from the

equationm-(2;-.V ~ No. o£ redundants we have 11 - (12 - 3) ~ 2 redWldant

members. Figure 15 shows th'e structure with two members cut for solution

or the forces in the staticalzy determinate structure.

F

Figure 15

lOt<.

f'o Fo~-~tf/J3 8 J,t:JM"n on

~, t:.v'f ~J,.v~l-v;4f?.

Cutting the two internal redundant members A-C and F-D and replacing

the unknown internal forces in the members cut by ub ~ and ua Xa, where

\1a • Ub a 1, gives rise to forces Ua and ~ in all members o£ the CUt

structure due to the unit loads. The tota1 force in any member is the

sum o£; the Ub va1ues times the true value or Xbj the Ua values times

the true va1ue or Xa; the force P0 in each member o£ the cut structure

due to the real loads acting through the statically determinate load

path. The above terms may be written P a (Po+ X..ua + XbUb)•

Since there can be no relative movement between the two pieces

formed by cutting the two members A-C and F-D then band a are both

The va1uee o£ ua and ub due to one pound loade applied at the

location o~ the cut mer.bers are shown on the ~o11owing £igures.

Figure 16

"707'~~

. ?0~

Figure 17

.. .101

/

F

0-,t

The !ollowing table has been arranged to aid the computer in

keeping the val.ues £or the various terms in an orderly manner and &leo

to minimize the possibility or arithmetical. • stakes during the process

o£ determining the true ~orcea that exist in the members or the truss.

Mem-ber

AB

'AC

CD

DE

EF

FB

EC

FD

AC

FA

FC

z

P0 u.aL PoubL Po Ua ub AE AE Kips - - I in. in.

-.20 0 -.707 0 +14.14

+10 0 -.707 0 -7.07

0 -.707 0 0 0

-1.0 -y107 0 +7.07 0

-10 -.707 0 +7.07 0

28.28 0 +1 +28.28

14.14 +1 0 +l4.14 0

0 +1 0 0 0

0 0 +1 0 0

-10 0 -.707 0 +21.21.

-2C -.707 -.707 +14.14 +14.14

- -- - +42.42 +70.70

~A-= ~ 42 .. 42 + 4-.()ox, ~a .s-o .X 8 = o .A1 • ~ 7P .. ?o ~o .. .s-oXA,. 4ooX • = o

Ua2L ub2L uaubL p

-AE AE AE ¥ips in/l.b. in./lb •in/l.b.

0 .50 0 -8.28

0 .50 0 +21.73

.50 0 0 +6.03

.50 0 0 -3.97

.50 0 0 -3.97

0 1 0 +11.68

1 0 0 +5.61

l. 0 0 -8.53

0 1 0 -16.60

0 .50 0 -18.27

.50 .50 .50 -2.23

4 4 0.5 -

f,.D,., .,./~~~~ XA •- 8,5.sNlcdn,p .. )" X6 = -/6,6o~ (co~p) P.: .li ,-._;r~NA~ X. M 8 in any member by superposition o£

loads.

( 5) Defiection or a Redundant Structure

Very often it is necessary t.o determine the deflection o£ a point

or points in a redundant structure. The basic theory or the strain

energy method or determining def'1ections requires that internal loads in

each member as well. as reactions be determined for the applied loads and

tor the unit load applied at the location and in the direction of the

required detlection. There£ore~ it would appear that the problem would

have to be solved twice - once tor the actual loading condition and once

tor the unit load applied to the structure.

In the following derivation it will be shown that it is not

necessary to treat the structure as a redundant tor both of the afore-

mentioned loading conditions~ but it is only necessar.y to so1ve the

redundant structure once, £or either the actual loads or the unit loads.

Any staticallY balanced path m!Y be used for the other loading.

The case or a truss wi. th "n" interna1 redundants will be used as

an ~le. However. it is readily apparent that the derivation is

valid £or al1 types or structures, and ~ include energies £rom a1l

types or loadings. The twice redundant truss solved in the preceding

example will then be examined to determine the deflection at point "E"

by this method.

A= ?a=

~, ~ •.. _,t/,:: Same as def'ined under structural symbo1s on Pages 1. e X If J Xa ... .. X_.,=-

?== I

P = rnnuence coef'ficient or axial load in member o:r statically 0

determinate structure due to carrying a unit load through

statically determinate load path. This unit load is

applied at the location or ami in the direction or a

desired def1ection.0> Actual value or redundant 1oading at

cut " " due to an applied external unit load. This unit

load is applied at the location ot and in the direction o£

<'> , , v' a desired defiect,i.on~ XA X5 .. · /'.,

"-'" ·--

P' =- Total axial load in member due to unit load applied at location

and in direction or desired det1ection.

For the applied loading -r?:r> -?X ..u~ ..,.X .H.~ .. ,_.~ X ~ ~,... ~rT 134 _,

~ I.L.., /. _j i2.E. z;T: £ .Z~~ ,MT(*) ~X....,= .B',..,

For minimum strain energy ~-~PL.. O>P- L .;19e

a unit load is

applied at "A" and the internal load pt is determined for each member or

the structure, then pp£....

A..q= 2 6'"

but this involves solving tor both P and pt ldlich means the multiple

redundants muet be solved tor twice, once tor each or the two loading

conditions. However,

But ror minimum strain energy s: P~L: ~P~L=O

L._ ~€ 0 L. - ~f!E"

there! ore

From the foregoin~ derivation it is seen that in any multiply

redundant structure, the deflection, by the strain eneru method, may

be computed by using the true internal loads due to the applied loadings

but using any statically balanced load path .for computing loads due to

the unit loading. This method obviously saves time and labor over

computing internal loads for both real and Wlit loading systems and

makes the process or determining de.f1ections at one or more points

.fairly simple.

In some problems, it may be required to determine deflections due

to various loading systems applied to the structure. I£ this is the

case it simplifies the prob1em if the loads acting on the members due to

the unit load are determined first 1 then use any st.aticall.y detenninate

path for the actual loading. The equation then becomes -

?!P'L L1~=Z ~~

.,Q . ~ .

which states that the su.1'JIIl.ation o£ the final. axial loads in members iue

to unit load times the axia1 loads in members due to carrying applied

loading through a statically determinate path times L/AE ~11 give the

de£1ection at the point o£ application o£ the unit virtual load. This

is an important concept for simpli.fying de.f1ection analysis o.f redundant

structures.

For the example truss of two degree indeterminacy, previously

solved ~or the correct internal .forces, the def1ection at Point nEtt rray

be round r om the equation 4.: z ~~

p pt 1bs. L in. ppt L inches iE lbs'. 0

Kips 0

AE Member

AB -8.28 0 3.J3 ~ 0 b... 10 AE

BC 21.7) 2 43.46

CD 6.03 +1 6.03

DE -3.97 +1 -3.97

EF -J.cn 0 0

FB 11.68 0 0

EC +5.61 0 0

FD -8.53 -1.414 +12.06

AC -16.f:A:J -1.414 +23.47

FA -18.27 -1 +18.27

FC -2.23 +1 - 2.23

- l>-97.09 (L/AE)

and the resultant deflection at point E is: 97.09 x 1000 #/kip x 3.33

x 106 = .324 l in.

( 6) Tabular Arrangen:ent for Minimum Strain Energy EQuations

~~ str~ energy solutions are generally long and tedious for

more complex problems. Therefore, it is fortunate that the general .form

o.r the equations lends itself' so well to tabular arrangements which aid

in keeping the numbers straight and in minimizing the possibility or

arithmetical mistakes. A fUrther advantage is, that once the engineer

has set up the table and r·lled in the necessary information, the

solution can be completed by a computer with little di££iculty.

It is not possible to show any one example table which will en-

compass all possible types of problems. It is more expedient to make a

separate tabular form for each particular problem since the energy .forms

~ di.ffer from problem to problem and there may be more terns required

depending upon the number or redundants.

In order to set up the tabular for.m for any problem it is

suggested that the general f"orm of the energy equation, which holds

true for the case being dealt with, be written. Then each term and

combination o.r terms in the general equation may be set into the table.

The example problem to follow shows the advantage or the tabular form

and how it was set up.

In the example it will be noted that the simple energy rorms pre­

viouslY derived are not sufficient to get the energy for all types o.r

loading. For instance where the load varies along the length of a

member, or shear now is transferring load to a member, the general

equations are not suf'ticient. it is shown in the appendix o£ this

report that dividing the members into enough increments or length will

give sufficient accuracy for the energy value.

Example Problem

Given: A rectangular syst~~ or plates and stiffeners loaded as

shown on its outside boundaries. Panel dimensions are 10

inc!_les by 15 inches with thickness or 0.05 inches. Area

or caps is 0.25 inches.

E=l0x1o6

Determine: True magnitude or force in each member and the deflection in

"p•

1 D E F t '

• H ~

j_ Figure 19 4

There are 24 cap members and 9 panels so m ~ 33 and the number or

joints is 16 :a j. m - ( 2j - 3) :a 33 - (32 - 3) =- 4 which is the degree

or i.ndeterminac7·

Ability to select the proper members or a redundant structure to

cut. in order to make the structure determinate. comes through practice.

However. if' the members in the structure can be cut in a symmetrica1 or

anti-w.y.mmetrica1 pattern it genera1ly simplifies later computations. The

example problem was cut in a synmetrical pattern as shown.

_EJ"TER/YAL ~O;:tl)s T#R041C'/-t' ST.-'9//C . ·;v-4TH··

2 ~·;I-· . : ~ .

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..

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: :; .. j J "

17. . .

. .._ ' ' . . : . ~ ' ' : . . .

. - . . : ~ ~ ! .

. :.·· · ; · ~-. - .·- __ ... . ,. 4- -_ . . : . . . . : ~ : ~ : . : • : . . . . : ' . . . ~ ! . : . . . .

.: ;r.({;C/RE :20

0 _ __._.. ___ ()

F - , ~ J

. . . . . • ;I •• '• •

• ~ I I

• -/ •• t

I l .t •

. : :ts :: --· · ~,: ... I• . ' .

__ _ -o-· o - r··;:.: i :; : ·:.:··: : :' : · : ·_: . ; i ·: . . ::- .

' : . . /( " '' . . . :: . . . . ... . . . ' . .. ' . . . . . . .

. ·:··: ; ·. ( . ... : · ·_ . . . . . . . . .. . . - ' . . . . ~ . - ' . .. . . _,' .._ . . . .

. -' • , • • • t ••

.. . . . . .......... . . ::.: . ~ : ;. ~~ ~~ 'for<:.-es ~re then trac~ thrau.gh . the etatic jlath as . eh.own . :

l

-'a+b~. · The :int~nal forces :SHdwn are list~ in the table in column headed· . . ...,.. ..

. :f.e: ~t: _ ... o• _ :'i'he.: ~~ -~tep ie ~0 plaoe·, _~n~ at a t~e, •::~t ~ll~ now _l '

.. 4: . . : : : ! I I . : ;- · ... •• ' • ~ ~ • • 4 • • • ~ •

. .. . fl

: ~4 : : .. · respect~valy at the location o! ~d in tha. ·#rection . .

·ot the required · Unknown r6ree--in the redurida.nt member• ·. The effect .o!

these forcee is -round oh the rest of the

structure by carrying the val.ue through the static path. The resu1ts o£

this action is recorded in the co1umns or the tab1e headed ~A , N,~ Vt;,

Ur, 'i~"tj"c 1 I"~"B*.r • It is imperitive that a consistent sign convention

be established for these 1oading terms for each member and that the same

convention be used for member 1oading due to externa1ly applied 1oads.

The diagrams that .follow show how the f'orces are traced through the

atatically determinate 1oad path.

It has previously been shown that the 1oad in any member of' the

structure can be represented by the following type equations:

The tota1 strain energy in structure is:

AI . au _ so. ~x.-

(Axial Members)

(Shear pane1s)

-=o

:: J: .u,.j;E" {P. +.v,.XA + McXc -r -CI'frXc +..V.r JC,~) + L 'Ju.ft{'lo + ~c..,XA + ~" .. .Xc +ttu~X" + f,3X~) =

= LPo q,. :E + LCJ. 'J.,,. :t +X A { £u; :r + Llf;. ftj + Xc (zu,.~ ~ .. + Xtt .... i .. .c ~)+X,.( zu,.v~ ~~ +Zio .. ,.~tJ

+ X.r { £ u, Vr :.- + Z ~Ju11f~ -fr) < J I , ~U C>U Q)U -JJMI/A,.l/y, dXC = ~<! : Q ci>X&: 4.4:.: 0 ci>Xr-=-~x= 0

= Z P. '"'<~-~ + .E1.~"e t"t +.X,. (2;u,.u.~+JE.'i .... s~g~)+XJzc.~~';.,~ + z ~.:; lr) + Xt;( r Uc q ~ T flft; '!~" ~ )+ Xr (ZI{rl{ :.-

+ L Z".rft~c :~)

f1fF. I . .

i; .·

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: : : :.i : 1 : : : : . . : : : . : :: t: : : ..

·-----· o-----~ ·---~ -6

4 F/G-UR..E 21

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1 o

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·./-/ -:J. : _ . :/'S.Q: . : . o~ LY ~:ts.o.a: ~3o_o.o: 2~0:. ~ o: o l..;·s;:J ."' . .., ,.-soo.oi c_ c.... · ,"1 · :0 . :c. ::Joo_o a . .:a :/-<~~~~-· ~ :. ~_30_ ~-~- . o: :o: :-s.:J.JJ. /_~_c.? . /.:333.

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1

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43.

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0 -

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) . ..

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. .z:z.: . .. . .L-J . . .

J74...9p;-¥

I-f

j;_z .2j1N

~ .. . . "1t . .. . . "4 . . .. "' .. .

: \

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257-2 ;/IY

-t- : . .

.. '

t . . . !041;,:;1

From the values arrived at troo the foregoing tabl.e and the general

equations we have the following tour independent equations:

(1) 1515.5 + 9.8'5 XA + 4.629 Xc + 1.117 Xo- .481 Xr ~ 0

(2) Z763.5 + 4.629 XA + 9.883 Xc - .481. Xc + 1.1.24 Xr = 0

(3) -5459.0 + 1.117 XA - .481. Xc + 9.880 Xc + 4.629 XI ~ 0

(4) -4313.0 - .481 XA + 1.12.4 Xc + 4.629 Xc + 9.886 Xr ~ 0

.rror.: which

XA 2 -67.9 lb/in.

Xc 2 -257.2 1b/in.

Xo 2 424.2 lb/in.

x1 ~ 263.6 lb/in.

and setting these values in the general. equations: P =(Po + XA~ A + Xc

ftc + Xa ,u G + Xr Pr) q = <k> + XAqH. + XcQ..u + Xa<k + jt c. G

Xrqu.r )• the real values for loads in each member are determined.

In order to determine the defiection at point P in the x - x

direction the unit load is applied at point P in the direction o£ the

desired jeflection. This load is then followed through the statically

determinate path and recorded in the preceding table as P0 ' & qo'. Thus

.from the values found for the true load in each member and the equation ~

Ll = 5 P~~L--+ 5 ?7! A the actual deflection at point P is L·AE L G"C

determined. Then L1 = .240 +.070 = .310 in.

SUMMARY

1. For defiection o~ determinate structure•:

a. App~ a unit load at. the point. and in t.he di.recUon o~ the

required derlection.

b. n.tend.ne the internal. strain energr in the structure.

c. Dir~erentiate the eneru equation v.l.th respect to the unit

1oad. This ia t.he defiect.ion under the unit, 1oad.

2. For rotation or a structpre or member.

a. Uae the same steps aa abo-.. except, uae a unit moment rat.her

than a unit force.

3. For determ1Mt:1on or redup'•nta in an :1ndet.ermiDat.e struot.ure.

a. RemoTe all redundant. membera maldn« the atrw:ture statical.q

determinate. Find the forces in t.hia .tati~ determinate

structure due to the extemal. 1owt1ng condition.

b. ~. at a time p1&ce a uni.t 1oad, shear .tlov, moment, ahear,

46

etc. (vhatever one ~ be required) in place of one of the

redundant members. F1M the 1oade on each member or the

statically determinate st.ructure due to the unit 1oadiO«

condition. RemoTe the unit 1oedin& condit:ion from th:l• position

arxi place it in the next redundant po8ition. So1Ye ror the

loads in each !MIIber or the .tatical.q determinate structure

qain. Repeat thia proceaa until the unit 1oadin& condi.tion

baa been plAced on each redundant .-.ber or at each redundant

reacUon po•it:lon.

c. I>Rend.ne the t.otU eneru- in the structure for that l.oadin&

cond:ition which ia equal. to t.he .tatica117 detend.nate 1~

from the real loadin« condit.:lon p1ua the l.oad• due t.o the unit

loading conditione times the unknown ~ues or each redundant.

Dirterentiate the energy equati.on with respect to the unlmown

Talue or each redwxlant. Set, the resul.ti.Jl« equati.ona equa1 to

sero. SolTe the re.u1tin« simul.taneoua equati.ons ror the un­

known Yal.ues or the redundant members. Fi "•,,7 put the Y&l.ues

round ror the redundant torces i.nto the «eneral equation ror

total tinal. torce, ahear, mcaent, etc., whiche-.er i.a requi.red,

and aol?e tor the UDk:nown ~ue.

4. For det'l.ecti.on ot an indeterminate •tructure:

a. F1Dd true rorces actinc on each IMIDber b7 ao1riJl« tor ~ un­

known redrmdanta u expl•ined aboYe.

b. App~ unit load at. the posi.tion and in the direction ot the

~ed det1ection.

c. Remo._ the redundant meiDbers to make the atructure de~rminate.

Thia ia done moat conTenientq 1r the aame members are remoTed

tor this problem u were remoYed in determinin! the redundants.

F1D1 the torce :in the meaber• ot the at.atical.q determinate

atructure due to the unit load which has beal applied at the

point and in the direction ot the deaired defiect:ion.

d. The required defiect,ion ua;r then be determined b7 multit>J.7ing

the true load in the members t!Aes the load in the meabere due

to the unit load time• Ute alpha conat.ant. ror the particular

member and then S1uw1 ng up the re.Ut• or thia operation ror

all members.

APPmmiX I

Special caaee or st,rain energr equation•.

tf"!Q I -1

(1) General case- coMtant area wit,h ~ P0 and Tar7ing .4

Exact ~sis ,. L • I .., X

71 ~ ~ .. ""'

(P- -o, Po.} .L.

Approx:l.ma~e AnaJ7ais

Approxima~• ~sis i.a baaed on &Tera«e 1oada in l/3 1~b incre-

men~• aDd ie as t'ollowa:

P. •

As prerloual7 defined•

Siw1l•r~• &Ter&«• ~ ror ba7 m-n

.: ~ [ 2. .,. ,d ( 11 .. ,. + ')<.__ ) ]

· . .t-..r: Po ... ~ (z,. ;!-) ~ ~ ( .t.t<)

A-..r. ~ = 'i·r~~-1-) J z• (zt_4)

~- ~P.tnJL+X ~ tJX- L AE L-.

J '} ( J! ,. s:;-; 1 ~·(zl ~)

50 l-3

c). I{~" -~> L~ .!_ : ----- - ____ _ ,!/! - - -- - . - - , ""7"-o;x p~ - .__

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-1

/~f, . rJ 7-~;:;,._-. ~oX//~//-/ ,7£ ••

--· .,- / / / / / / ' , //..:~. "?//)--/ J/'Cr- ·1 -;_7.-- /or; , 5 / /? :-7/Je .~ /ds-:r /-ern?

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~~?r /e5~- .~//c//) / 5~. / /:7/ 5 5~c_7,.;/ J--"y--7//G.-r;-cJn

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/?:;'~e/0t?CJ'. ,-:-' c7 /l C// 5/_::.._ .

~ ... . --I-4

A •pecia1 t'orm ot' the ~enera1 cue is vortb7 ot' note: enerna:L

on the aam.

e1ement are zero on the same end• i.e. ~ • -1 • (.3 • For thia special

1~ condition ( c)'!T) = 7? .AI.~-+ ~~ ~' (exact so1utian). CJ>< ,., 3Ae 3 ,4e-

Theret'ore this 1oading condition. using 1: and the non-sero end 1oada tor 3

the total element (L). will 1n £act ,-ield an exact ao1ution.

(2) General. cue- ft.r7in~ area with ~ P0 and ft.r7ing Un

Exact ~sis

;zz :: ~ (/ ~ ~ )(" ) J .M . ~~: =~ (/ ~ c x~.J

Simil•r:q: aee Pa&e 1-1

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~ :::•L {: /"}(t.,u( ) +A;~ [ 1+1'-lof(t-.<1')-/ .J

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J'J f~ 'i '- f 1"'/o 1 ft+ 4' J+:l. ~;;'; L .9 4' /,, (t-~o~) ~

[tAl t '·~;J ·- z.. ~·t.r+r) + / •/,,r.,..,. h l ,t!" )

I-S

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+Xu,"l. f'•r/tdJ[i~-z.lf;-.d-.,J., &~;-:;f(, .. r)~tJ't£T·d]J A. E ¥ ~ )- J

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X 11

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Di.8CWIS:10D

~~ T-9

Aa indicated on Page I-3 the use ot the approximate method tor

constant area members with n.r;ring l.o.:ls w.1ll re8Ul.t in an error of leas

than 1%. Shown on Page I-7 are SUJD1D&r7 plots ot % error in usiDg the

approximate method as compared with the exact method tor tapered area

members v.l.th ftr7ing loads. It is believed that KA. • 5 is the ex­

ception rather than the rul.e for aircraft structure. Therefore it

lA • 5 is taken as the maxi """'• the error ldll be l.O% or less ror that

member (KA • 5). It is rnrther believed that thi.a Nrlnnun % error on a

relative}T tew (sa.y 10%) members 1• acceptable when the oTeral.l. picture

is rleved. Therefore in the tab1e or computati.ons the value recorded

UDder should be _L (A • aTerage area tor that particular !! 3AE 3

increment) and average loads (P0 • m• for that particular I! :increment) 3

should be entered in their appropriate colUJma. Exc..,t.ion to th:l• i• aa

noted an Page I-4.

5. Str.pn Enerq ot 99adril.at•ral. Panel• in Shear:

Veey often the shear pan.U in problema are not rectangul.ar. Where

tbia is the cue the tollow.in& JDethod ahouldbe used to set the proper

energy- value for the panel. 'ftle method is taken rrom MAC Report 2659

Vhich is aYailable tor reference.

The chart• included are 8\UIID&tion chart• ror t.he laat terma in the

seneral. enerQ equation deriTed ror quadrilateral. pane1a which .fit the

conti~ti0111 coTered by- the ftrioua an&].ee shown on the chart. ~ting

the ftl.ue t.alce trom the chart be equal to "K" then the «Cera1 equation.

?7= en,'"dl'.. ~ .c::; [...!_ 5//v(~-rti!..f fJ+ 4G Co~~{o<+Q)7]d dn. .32~~ L L ~ .».n4.t s/," l' e ~ a(_ ""

•=/CJ• ":/o•

mq be vritt.en _ ?~l-dl.K {/- 32 &-c

where q. 1• detined aa a comaon ahear tl.ov at o<. • 450 = 1 ~ and Qo•

Qu, q, etc. ia the known shear nov on the side or the panel noted.

(s(!'~ cf,Q~ ;;,.. ~ ~ n,, ha ~ ~~d)

~...,

--10

~R :I -I (

d ------------------~'

EXAMPLE: Strain eneru coet'fic:ient o~ a ~t,eral pand aa ahovn

harln« conatant 1*1•1 t.bickn•••· (Rer. Chart I p. I-14)

NOI'E: Chart I tabul.atee the total strain eneru ractor (k) ror con.tant

thickneas panels bounded b7 rqs paa•in« thrOQ&h the upper apex or the pane1 .nd the base line (d) extrerdtiea and by rqa &om base

line (d) extremities at~ • ~ =- 100

Area (B-C-D-G-B) :a Area (A-B-C-D-E-F-A) - Area (A-B-G-E'-F-A)

-Area (A'-Q-D-E-P-At) + Area (A'-G-E'-F-At)

For quadrilateral. shear panels enter in t.ab1ea (ret'. pp. 41, 41&, 42)

~ a d2tc and rererence q ...,., ror ~ q, be ~~ t~o. <l'ln• or ~-i6Gt

Nov to determine the tota1 ener&Y f'actor ~or the quadrilateral. panel.

(B-c-D-G-B) b7 using the chart Y&l.uee we DUat u•e superposi.tion o~ the

panel.s included within the boundaries o~ the chart.. There1'ore the number

at the apex of' the quadrilateral •hown in caae l. incl.ude• the ~actor ~or

the true panel. plus that part of' the panel. bounded b7 (A.-B-C-0-E-P~)

C&se 2 figure shows the porticm (At-G-D-E-F~t) and the t'act.or t'rom. the

chart vhi.ch JDQ' be subtracted .trom caae l. condition. Case .3 ~igure

ehowa the portion (A-B-G-E'-P-A) which can al.eo be subtracted t'rom

case l. Tal.ue, but in dofns so the area (A'-G-E-F-A') haa been subtracted

o~t' one too JDanY times, there~ ore the Ya1ue tor this quadrilateral. must

be added back in to the ener§' f'actor ot' case l.. 'ftle correct t'actor

there~ore i.e as f'ollowa:

+ Case l. - Case 2 - CaM .3 + C&ae 4 • Energr tactor f'or quadrl.lateral

pan~ B-C-D-G.

+ 644-2415 - 634.196? - 634.1967 + 624.4613

- -.3094

and the strain ener~ ot' the pm1el. ia

U • qa2d2 ( .3094) • (Cio X 4h).h2)2

.32Gt

where d 1• the base l.ine di•t.aDce and h1 and b2 are aa •hmm.

( ~St!? / 4~1or /J-on? CJA~¥,.~ ~/:: 64¢. Z4/F

/o,.. guCTd,_/·/er/6?r~/' ~ov~d'~/ ~;y

(;( -: /o o a-he/ 6 0 o c;vnd ;:l = /CJ o d,h/ 6 o "

/

c ,"

\

61 I-13

CASE 3. ~chr fron; c/J<?rl ./::~34/~~7 ,br f?qa::Jf/-.

/~/ere// .bo///JdeO LJy DL -=,/0" t;?/?0'/ 4S6 (3=;V 0

c;?/?0 ~0~ /

I D'

I

'

c /'\ II \

I \

\

)D /

/

A' £/

~. :s· ~ :-_10

0 F .~ .J -~ --~~~ --------~~--~-------- ~=~~~

CA5£4 ~;c/or h.m c/l~r,/~=-C:£44c;/3 ~r _7CA-:7c7f-/~{;?/e~o/ .&.4-"fi'O~~d' 4· <X:p· -c:?a0 ".:,.-""5 •

-I?: /0 ° c;?ad ~:75°

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