- - WAPD-TM-182

AEC RESEARCH AND DEVELOPMENT REPORT

DEFLECTIONS AND LOAD DISTRIBUTIONS IN LINEAR ELASTIC STRUCTURES-CTAC AND MODE CODES

JANUARY 1960

CONTRACT AT-11-1-GEN-14

BETTIS ATOMIC POWER LABORATORY, PITTSBURGH, PA., OPERATED FOR THE U. S. ATOMIC ENERGY COMMISSION B Y WESTINGHOUSE E L E C T R I C CORPORATION

DISCLAIMER

This report was prepared as an account of work sponsored by an agency of the United States Government. Neither the United States Government nor any agency Thereof, nor any of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government or any agency thereof. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States Government or any agency thereof.

DISCLAIMER

Portions of this document may be illegible in electronic image products. Images are produced from the best available original document.

WAPD-TM-182

U C - 8 0 : Reactors -Genera l

T I D - 4 5 0 0 (15th Ed.)

y *

A DEFLECTIONS A N D LOAD DISTRIBUTIONS I N LINEAR 1

ELASTIC STRUCTURES-CTAC A N D MODE CQDES

Car l M. Fr iedr ich

January 1960

Contract AT-1 1-1 -GEN-14

Price $1.50

Available from the Office of Technical Services,

Department of Commerce, Washington 25, D. C.

N V I t

This document i s an interim memorandum prepored primarily for internal reference ond does not represent a final expression of the opinion of Westinghouse. When this memorandum is distributed externally, it is with the express understanding thot Westinghouse makes no representation os to completeness, accurocy, or usability of informa- tion contoined therein. I .

BETTIS ATOMIC POWER LABORATORY PITTSBURGH, PENNSY LVANlA

OPERATED FOR THE U. 5. ATOMIC EN'ERGY COMMISSION BY ,. , :, WESTINGHOUSE ELECTRI'C CORPORATION

STANDARD EXTERNAL DISTRIBUTION

UC-80: Reactors-General , TID-4500. .15th Edition

SPECIAL EXTERNAL DISTRIBUTION

Manager . Pi t t sburgh Naval Reac to r s Operations Office, AEC Carnegie Institute of Technology, C. Zorowski Westinghouse Patent Department, E a s t Pit tsburgh, D. J. Smith Westinghouse Resea rch .Laborator ies , N. C. .Small

No. Copies

585.. .,

Total 591

cL[CAL NOTICE

T h i s r e p o r t was prepared as an account o f Government sponsored work. N e i t h e r . t h e Un i t ed S ta tes , nor t h e Commission, n o r any person a c t i n g on beha l f o f t he Commission:

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CONTENTS

I. INTRODUCTION

11. FORCE, DEFLECTION, AND ENERGY

A. Force

B. Working Deflection

Reciprocal Theorem

Castigliano's Theorem

Structure Position

C. Strain Energy

Beam Energy

Cantilever Beam

Simple Support Beam

Transformat ion f r o m Cantilever t o Simple Support

111. SYSTEMATIC CALCULATION

A. Cantilever

Influence Coefficients

Energy

B. Simple Support

CTAC Rule

Flexibility

IV. APPLICATION OF CTAC

A. Structure i n WAPD-115

Given Data

Flexibility of a Subdivided F r e e Body

Total Flexibility

Redundant Load Calculation

Systematic Load Calculation

B. SO207 Code (CTAC and MODE)

Discussion

Arrangement of Decks

CTAC without MODE

CTAC and MODE

MODE Alone

Input Cards

Trans fe r C a r d s

Data Card F o r m a t

CTAC Data Deck (Input)

MODE Data Deck (Input)

Output Cards

Output Card Format

CTAC Output Deck

MODE Ch~tput. neck

Sample Problem Using Structure in WAPD-115

Printout of Cards

Resul ts

Sample Prob lem C a r d s

ACKNOWLEDGMENT

REFERENCES

Page No.

1

2

2

. 2

2 -

3

4

6

6

6

7

8

10

10

10

11

11

11

1 3

13

13

13

1 5

16

18 . .

19

2 1

2 1

2 1

2 1

2 1

22

iii

Since t h e s t r e s s a n a l y s i s o f a comp l i ca ted s t a t i c a l l y i n d e t e r - m i n a t e s t r u c t u r e i n v o l v e s t h e h a n d l i n g o f many s i m u l t a n e o u s e q u a t i o n s , t h e SO207 code f o r t h e IBM-650 was d e v e l o p e d a t B e t t i s t o s o l v e t h i s problem by t h e use o f m a t r i x t r a n s f o r m a - t i o n s . The code .has two p a r t s , each w i t h a t i t l e : f a ) the CTAC code g i v e s s t r e s s e s , d e f l e c t i o n s , and r e a c t i o n s ; and f b l t h e MODE code c a l c u l a t e s n a t u r a l f requencies and modes o f v i b r a t i o n . As an a i d t o users o f the code, a rev iew i s g i v e n o f t h e e l a s - t i c i t y and m a t r i x t r a n s f o r m a t i o n t h e o r y u t i l i z e d i n t h i s code.

DEFLECTIONS AND LOAD DISTRIBUTIONS IN LINEAR ELASTIC STRUCTURES-CTAC AND MODE CODES

C. M. F r i ed r i ch

I. INTRODUCTION

The determination of fo rces and deflections for.complicated statically indeterminate s t r u c t u r e s

involves the handling of many simultaneous equations. In o r d e r to keep t rack of the var iables and

equations in success ive s t eps of calculations, a sys temat ic routine is needed.

One such routine was developed by Gabriel Kron (Ref 1 ) f o r e l ec t r i ca l engineering problems.

This routine of ma t r ix t ransformat ions was adapted by ~ o r ~ e Langefors (Ref 2) for e l a s t i c s t ruc tu re

problems. Kron (Ref 3) later. extended Langeforsl adaptations to take c a r e of nonlinear plastic de -

formation. A thorough and detailed discussion with applications on energy principles of s t ruc tu re

analysis, including the effects of temperature and nonlinear s t r e s s - s t r a i n relations, was made by

J. H. Argyr i s (Ref 4).

Thc purpooco of t h l ~ repor t are:

1) T o give a simplified explanation of the ma t r ix t ransformat ion method f o r l inea r e l a s t i c s t r u c -

t u r e s (preceded by a review of e lementary concepts and theorems in e las t ic i ty which a r e used

i n the method). .

2) T o show the application of this method i n determining influence coefficients of s t r u c t u r e s that

may be redundant.

3) T o show how this method is applied in the CTAC code ( f i r s t par t of the Wolontis-650. code

S0207).

4) TO show'how the MODE code (second part 'of code SO2071 can be used to determine the modes

of'.vihration of a n e las t ic , s t ruc tu re .

11. FORCE, DEFLECTION, AND ENERGY

A. Force

The basis of any engineering analysis of a complicated structure is the free body diagram of

each component. Every force o r moment shown on the f ree body diagram is actually a system of

forces and/or moments which a re in equilibrium with each other.

conventional foundation shading lines on a free body diagram a r e used to indicate the remainder

of the force system acting on the body to retain it in equilibrium. For example, the forces shown

. . acting on the structures in Fig. 1A a re actually the force systems shown in Fig. 1B.

The fundamental process of analysis involves the subdividing of the free body diagram of a com-

plicated structure into two or more free bodies of simpler geometry, a s shown in Fig. 2. Care must

be taken to have the force system on each subdivided body in self-equilibrium and to have the force

systems on the subdivided free bodies actually equivalent to the force system on the original free

body, in regard to force and moment distribution throughout the body.

A. FORCES ON STRUCTURES 8. ACTUAL FORCE SYSTEMS

0 . SUBDIVIDED FREE BODIES

- - - v L

v v

M + V ( L I + L 2 ) v v + +

M

M M+ VL

v M+VL2

v

F l q . I Free trody t q ~ ~ r 1 rhr i~r rn f i g . 2 S u b d i v i s i J i i of F ree B o d i e s

In this treatment a force o r force system may be labeled by one of the forces or tinoments in the

system. Thus, a "force" may actually be a true force o r even a bending moment or a torque.

B. Working 'Deflection

Reciprocal Theorem

Considering three forces F1, F2, arid F 3 acting on a s tructurr , i f the structure is assumcd

linear-elastic, the working deflection of each force is a linear function of all forces. That is ,

DF, = A1 1 F 1 +A12 F2 +A13 Fg *

D ~ 2 = A21 F1 ' A22 F2 ' A23 F3 '

and

D ~ 3 = A31 F1 ' A32 F2 ' *33 F3

where

DFi = working deflection of F i .

A.. = influence deflection coefficient, equal to the working deflection of Fi produced by the 1.l

unit load F . = 1, and J

i , j = subscripts having values .1, 2, or 3.

A very important relationship concerning these influence coefficients can be determined by

changing the order of load applications and comparing final s t rain energies. The change in s train

energy when loads change on the linear elastic structure is , by definition of working deflections,

du l = F; d D k l + F; d D ; F ~ + F; d D k 3 , (2)

where U = strain energy of the structure and a prime i s used to denote the value of any quantity

during a change. Substitution of the deflections of Eqs (1) and (2)' gives

d ~ ' = F,: (Al1 d F; + A12 d.F; + A13 d I?;)

+ F; (AZl'd F; + A22 d F; t AZ3 d F;)

+ F; d F; + d F; + d F;) (3)

F i r s t , allow a force, say F;, to increase from zero to full vaiue F1 while the other forces

remain zero. The strain energy of the structure becomes

Next, allow another force, say F;, to increase from. zero to full value F2 while the other forces

remain unchanged. The s train energy of the structure is increased to become

If the order of load application were reversed, the total s t rain energy would be

Since'the final strain energy of a linear-elastic structure depends only on the final shape of the s t ruc-

ture a s determined by the working deflections and the final working deflections depend only -on the

final forces.according to Eqs ( l ) , then the s train energy should be independent of the order of load

applications. Equating the energies of Eqs (4 ) and (5) leads to.'the resuit that A21 = A12.

This very important result states that the working deflection of one force produced by a unit

value of a second force equals the working deflection of the second force produced by a unit value of

the f i rs t force. This relationship is known a s the Reciprocal Theorem. In general,

A . . = A , . 13' J1

for al l values of i and j.

Cagtiglianols Theorem

A useful generalization of .the energy integral can be obtained by application of the Reciprocal

heo or em., cbrresponding to Eqs ( l ) , the generalized deflection equation is

- x . A . . F . D ~ i - , j 1j j

and the generalized s train enei-gjr equation i s

U = j (Zi Fi d DFi) . Substitution of DFi from Eq (7) gives

Now, by using Eq ( 6 ) ,

- A . i F i d F . t F. dFil - 2 1~ 1~ J J

= ~ E . A . . d (Pi F:) ; 1J 1J J

o r af ter integrating and using E'q (7) .

1 1 . u = - Z . . A . . F ; F . = - C F. D . - 2 IJ 1J 1 J 2 . i ': F i ' (8)

A very important relationship can be obtained by differentiation of the s train enerby U with

respect to a force, say F1. Expansion of Eq (8) in te rms containing f i rs t gives

t h y ( + A I i F ~ + A ~ ~ F ~ F ~ * A ~ ~ F ~ F i + A ~ ~ F i F ~ t A ~ i ~ ~ ~ ~ i ) i *

s o that differentiation with respect to F1 gives

Since A. . = Aji , according to the Reciprocal Theorem; 1J

which, according to Eqs (1) o r Eq. (7). is simply Dp.l;. thus,

This resul t is known a s 'castiglianols Theorem. The pa'rtial. derivative of the s train energy of a

linear elastic system. with respect to a force equals'the woi-kin'g deflection of that force. In general,

Structurc Position

The working deflection of a force acting on a structure is actually the total working deflection

of .all components of the force systemt acting on the structure. F o r example, c ~ n e i d e r the spring

shown in Fig. 3. Suppose that while the left end moves through a distance x l , the spring force in-

c reases from 0 to full value F1 . If the spri'ng constant is K1, the total s t retch of the spring sl is

equal to F l / K l and the total movement of the right end of the spring i s xl + s l .

Tlie wdrk WL done by the left force is - L ;* -/F1 F; dx;. . 0

A. ORIGINAL POSITION

8. FINAL POSITION

F i g . 3 Changed P o s i t i o n o f a S p r i n g

and the work WR done by the r ight force is

The s t r a in ene rgy U in the sp r ing is equal to the total work done on the s t ructure . Thus,

U ,= W.L t WR

which is independent of the final position of the spring. Since sl ,= Fl /K1. then

Then, according to E'q (a), the deflection influence coefficierlt of F1 is

A l l = l / K 1 ,

s o that the working deflection of F1 i s , according t o Eq (7),

DF1 = A l l Fl = ( l / K 1 ) F 1 = s1 I

Thus, the working deflection of the fo rce s y s t e m ' F 1 is s imply the s t r e t c h of the spr ing, a s is

expected.

In general , working deflections of force sys tems acting on a l inear e l a s t i c s t ruc tu re a r e inde-

pen'hent of the final position of the s t ruc tu re and dipend only on the final deformation of the s t ruc tu re .

C. Strain Energy

Beam Energy

The influence coefficients for one sys tem of fo rces can be 'determined f rom the <nfil.uence coef-

f ic ients of a n equivalent s y s t e m of forces by consideration of the s t ra in energy i l l Lhe :s't.ructure.

Calculations involved in th i s p rocess will be ca r r i ed out for a cantilever beam and a s imple support . beam in o r d e r to clarify the development of the necessary formulas.

The s t r a i n energy fo r a uniform beam under the usual engineering approximations ,(Ref 5) is

where

L = length of a beam, in.

x = distance along beam, in.

E = tensile modulus of elasticity, lb-in. -2

Es = shear modulus of elasticity, lb-in. - 2

EI = bending stiffness of beam, Ib-in.

EsAs = shear st iffness of beam, lb

EA = tensile st iffness of beam, lb

Ael M = bending moment a t x , in. -1b

V = shear force at x. l b

F ig . 4 Cant i lever Beam F = tensile fo rce a t x , lb

Bending in one plane, s h e a r in the s a m e plane, and tension a r e the only deformations considered in

Eq (10 ) .

Cantilever Bear11

Consider a cantilever beam loaded with force sys tems F, , V , , and MI a s shown in Fig. 4. The

load distribution along this beam is then

and

Substitution of these re la t ions into the beam energy equation gives

wirlch, after iiitcgratisn, be collies

The cantilever orientation consis ts of having the left end of the beam on and tangent t o the r e f e r -

ence line f r o m which deflections a r e measured. With this orientation, the force cofnponents a t the

left end undergo no deflection and do no work.

The working deflection of the force sys tem M1 in Fig. 4 is s imply the slope 8 a t the right end; 1 the working deflection of the force sys tem V1 is the deflection yl at the right end; and the working

deflection of the force sys tem F1 is the elongation sl of the beam. Therefore , application of

Castigliano's Theorem, Eq (9) , to the canti lever beam energy, Eq (12), gives

and

Simple Support Beam

Next, consider a s imple support beam loaded with force s y s t e m s Fa, Mb, M a s shown in C

Fig. 5. The load distribution along.the beam is

and

The s imple support orientation consists of having the ends of the beam on the reference line

f r o m which deflections a r e measured. With this orientation, the shea r loads a t each end undergo

no deflection and do no work.. Thus, the working deflection of the force sys tem Mc is thes lope Bc

a t the right end , the working deflection of the force sys tem Mb is the slope 0 a t the left end, and b the working deflection of the force sys tem Fa is the elongation sa of the beam.

Instead of evaluating the energy in tegral and then differentiating, the p rocess can be r e v e r s e d

in calculating working deflections by Castigliano's Theorem. Thus, f r o m Eq (9) and Eq ( l o ) ,

F i g . 5 S i m p l e Suppor t Beam

F r o m Eq (14),

s o that

aM - - - 0, and - aF - 1 - , - 0 , - - aFa . a ~ , aFa

L s lo [ o t o +(A) (Fa)( l)] d x = (&)Fa

Similarly,

L

Mb aMb 0

4" {(A) [-. (l - z ) ' M C (?)I (' -?)

+ ( ) [(t) (.h - ..)I ( ) + ( ) (F.)(O)) dx

and €I is found the same way. T.he three working deflections thus evaluated a r e C

and

Transformation from Cantilever to Simple Support

By comparison of Figs. 4 and 5, the cantilever beam and the single support beam a r e seen to

have the same loading distributions if

and

Then, the cantilever deflections can be found in te rms of the simple support loads by putting the

loads of Eqs (16) in the cantilever deflections of Eqs (13). After simplification, the equations

and (a) Fa -

Then, sulistitution of deflections f r o m E q s (17) and fo rces f r o m E q s (16) into the energy equation

(8) gives

1

Differentiation with r e spec t to the force s y s t e m s gives

and

which a r e the s a m e a s in E q s (15). Thus , by using the canti lever deflection fo rmulas of Eqs (13)

and the fo rce relationships of Eqs (16) in an ene rgy equation, the working deflections of the s imple

support beam have been obtained.

This t ransformat ion f r o m the influence coefficients of the canti lever f o r c e ' s y s t e m to the influ-

ence coefficients';of the s imple support force sys tem'shows that the canti lever and s imple .support

beams a r e real'ly the s a m e structurC with the s a m e loading. The only r e a l difference is the change

of the reference line f rom which deflections a r e measured , a s shown i n Fig. 6, to correspond with

the change i n fo rce sys tem notation.

According t o E q s (17c) and (19a), s l = sa. This is t rue in Fig. 6 if deflections a r e s m a l l com-

pared to s t ruc tu re dimensions s o that L >> s l , and if s lopes a r e s m a l l s o that s i n 8 8 and

cos 8 r 1, where 8 = 8 ea, o r Ob. A s shown.in Fig. 6,

A. DEFLECTIONS RELATIVE TO 6. DEFLECTIONS RELATIVE TO CANTILEVER REFERENCE LlNE SIMPLE SUPPORT REFERENCE

LlNE

F i g . 6 E q u i v a l e n c e o f C a n t i l e v e r and S i m p l e S u p p o r t Beam

and

substitution of the v a l ~ e s ~ f r o m Eqs (17) .into these equations gives

and

which a r e the same a s obtained in Eqs (15). Thus, the use of strain energy in transforming influ-

ence coefficients from one force system to another equivalent force system is checked by direct use

of geometry.

111. SYSTEMATIC CALCULATION

A. Cantilever

Influence coefficients

In order to keep calculations involving many variables in simultaneous equations under control,

a systematic procedure involving matrices: will be used. . consider thc cantilever deflections bf

Eqs (13) rewritten with all.influence coefficients present:

In matrix form, thcoc equations bccolrlt:

or simply

where

.(PI = = force ,matrix.of cantilever ,:

+Elementary properties of matr ices a r e discussed in Ref 6 .

= deflection mat r ix of canti lever, and

= flexibility ma t r ix of canti lever.

Let (D ).. = element in row i and column j of matr ix (D ). The equations a r e s o a r ranged that P 1J P

(Dn)il is the working deflection of fo rce (P)il f o r i = 1, 2; and 3. With this ar rangement , the f l ex i -

biiity ma t r ix (A ) is symmetr ical , a s rcquired.by Eq (6). D

Energy

According to Eq (81, the canti lever ene rgy is

1 U = 7 + V1yl + F1sl) ;

o r in ma t r ix fo rm,

T where (U) = = the energy mat r ix of 1 r o w a n d 1 column, and (P ) is the t ranspose of (P).

Substitution of the canti lever deflection mat r ix (D ) f r o m Eq (20) into Eq (21a) gives P

Equations (21a) and (21b) a r e matr ix equivalents of E q (8).

B. Simple Support

CTAC Rule

The canti lever ma t r ix deflection equation (20) and the energy equation (21b) can be used to

determine the ma t r ix deflection equations f o r the ,simple support beam, once the relationship between

the force sys te lns is put in ma t r ix fo rm ' Equation (16) i~ equivalent to:

where

(Q) = = force matr ix f o r s imple support beam,

and

(C) = = connection matrix relating cantilever forces (P) 'to 's'imple

support fo rces (Q).

Substituting Eq (22) into Eq (21a) gives

o r , since the transposed product of two mat r ices i s the product of the transposes of the two matr ices

in r eve r se order .

1 T T (U) = 7 x (Q ) x (C x (Dp) . (23)

. . Let

= simple support dcflcctio~i 111tll1.i~ ~ u ~ . ~ . e s p u r ~ d i n g to the simple support

force nlaL1-ix (Q),

s o that, a s in Eq (21a),

Comparison of this matrix equation with Eq (23) shows that

Next, substitution of (D ) from Eq (20) gives P

(Dq) = (c") x (Ap) x (P) ,

and substitutionof (P ) from Eq (22) gives

, tDq) = (cTl x (Ap) x (C) x (Q) ..

Let (A ) be the simple support flexibility matrix relating (D ) and (Q) in the equation Q '2 (Dq) = (AQ) x (Q) . (244

Then, f r o m E~S' (2%) and ( 24~1 ,

'I'his is the most important equation of all- Lhe "CTAC1' equation. It gives the flexibility of a

sys tem of forces acting on a linear-elastic structure in te rms of (1) the flexibility of an equivalent

second system of forces and (2) the connecti'on matrix relating the second system to the f i rs t .

In more simple notation, let A be the flexibi.lity matrix of a force matrix P, and C be the cor -

rection matrix relating P to an equivalent force matrix P1. Then, the-flexibility matrix A' of P ' i s

simply

where P = CP1.

Flexibilitv

As a check, the CTAC equation (Eq 25) will be evaluated f o r the s imple support beam. Thus ,

Substitution into Eq (24c) gives

- -

This is equivalent to the three simultaneous equations

and

which check Eq (15). Thus, the CTAC procedure does lead to the co r rec t resul t .

IV. APPLICATION O F CTAC

A. Structure in WAPD-115

Given Data

The is to determine the influence coefficients and modes of vibratidn of the s t ruc tu re

used a s an example in WAPD-115 ( ~ e f . 7 ) and shown i n Fig. 7. Geometric and flexibility coefficients

a r e given in the following table.

. TABLE O F GEOMETRIC AND FLEXIBILITY COEFFICIENTS *

F i g . 8 Separated Free B o d i e s

Region No.

Do

Di 2 2 A = n(Do - D i ) / 4

2 I = n(Do - ~ 4 ) / 6 4

As = A / 2 o r 3A/4

L

W = (0 .284 pci )AL

6 EI = (30 x 10 ps i ) I

6 E A = ( 1 2 x 10 psi)As s S

L/EsAs

L I E 1

L~ / ~ E I

F i g . 7 Geometry of Sample Prob lem

1 . 2

12

11

18.06

299.2

9 .03

10

51.29

8976

108 .4

0 .09225

0.001.114

0 .005,570 --

3 , 4

12

8

62.83

816.8

31.42

5

89 .22

24,504

377.0

0.01326

0 .000,204

0.000, 510

L~ / ~ E I

( ~ ~ 1 3 ~ 1 ) + L/EsAs

0.03714

0 .1294

5 , 6

6

0

28.27

63.62

21.20

7 . 5

60. 22

1908.6

, 254.4

0.02948

0.003', 930

0.01474

. 0.'001,700

0.01496

0.03018

0.04903

0.07368

0.10316

in. / l b

in. / l b

7 , 8

7 . 5

0

44 .18

155.3

33. 14

7. 5

94. 10

4659

397.7

0.01885

0. 001,610

0.00604 -

Units

in.

: in.

in. 2

in. 4

in. 2

in.

lb

6 10 lb-in. 2

in6 ib

in. / l b

l ~ - ~ / l b - i n .

1 0 - ~ / l b , ---.

The loads, including the statically indeterminate end reaction F between the two cylinders, 5 may be put into the single. force matrix:

Flexibility of a Subdivided F r e e Body

cons ide r region No. 1 a s a separate f ree body with cantilever loads V1 and M a s shown in 1 Fig. '8. These loads can be put into the force matr ix:

The cantilever equations fo r region No. 1 a r e

and

D~~ = (&) I M1 t (& t A) V1 = (5. 57 x 1 0 - ~ / l b ) ~ ~ + (129. 4 x in. / lb)Vl , EsAs

s o that the flexibility mat r ix for (PI) on body No. 1 is

1. 114 in.

129. 4 in.

By inspection of Figs. 7 and 8, the moment a t the right end of body No. 1 is seen to be

M1 = (15 in. ) F 2 + (20 in. )F5 ,

and the Shear load on body No. 1 is

V1 = F1 + F2 + F 3 .

In expanded fo rm, the equations become

M1 = (0)F1 + (1 5 in. )F2 + (0 )F3 + (0 )F4 + (20 in. )F5

and

Vl = ( l ) F 1 + (1)F2 + (0 )F3 + (0 )F4 + (1 )F5 ,

s o that

where (C1) =

The flexibility matrix of (F) produced by deflections of body No. 1 is then, by 'the CTAC rule,

(AF1) = (cT) x ( A I ) x (CII . . , . Fi 20 in. 1

.1.14 in. .5. 57

129. 4 in.

This is the flexibility matrix giving the deflections of the forces in (F) resulting from the flexibility

of body No. 1.

Total Flexibility ,

In the s a g e way, the deflections due to body No. 2 can be determined. As before, le t

The connection equations for (P2) in t e r m s of (F) a r e

M2 = (5 in. )F2 + (10 in. )F5 and

s o that

Carrying through the multiplications of

(AF2) = (cT) x (A2) X (C2)

gives

x in. / l b .

The total flexibility of (F) due to both body No. 1 and body No. 2 is

(AF1, 2) = (AFs1) + (AF2)

x in. / lb i

The final flexibility matr ix for a l l eight pr imi t ives (bas ic beam s t ruc tu res ) is found to be

x in. / l b .

It is to he not.arl t,hat a.l.1. of the flexibility matrices a r e symmet r i ca l , in agreement with . the Recipro-

ca l Theorem.

Redundant Load Calculation

The working deflection DF5 is the net working deflection of F5 on.both 'canti1ever.b-eams in the

s t ruc tu re . Since the re .is no net deflection between the beams connected a t the end, should be

ze ro . Thus ,

0 = DF5 E (DFI5i ' xi (AF)5i .(F)il . , !. . .: , *,,\::. ,< , ,

= (0. 2408F1 t 0. 94471F2 - 2. 95355F3 - 0. 43481F4 t 5. 654975E5) x in. $lb . This equation is used to obtain the nonworking (redundant) fo rce F5 i n t e r m s of the working (non-

redundant) f o r c e s F1 to F4. Solving f o r F5 gives

By use of th is equation, the net working deflections of the o'ther f o r c e s F1 to F4 can be 'obtained;

f o r example ,

= [ U . 1295F1 + 0. 21295F2 t ( 0 ) F 3 t (0)F4

+ 0.2408(-0.042, 582F1 - 0. 167, 058F2 t 0. 522, 292F3 t 0.076, 890F4)] x 1.0-~ in. / l b

= (0. 119, 246F1 t 0. 172, 7,2ZF2 + 0. 1 2 5 , 7 6 8 ~ ) + 0. 018, 5 1 5 ~ ~ ) x l o m 6 in. / Ib . Similar ly ,

DF2 = (0. 172,722F1 t 0.617, 239F3 t 0 . 4 9 3 ~ 4 1 4 ~ ~ t 0.072, 639F4) x in. / l b ,

and s o forth. The' resul t ing flexibility ma t r ix for the net working deflection is thus determined.

Le t

(w) = be the working force ma t r ix ,

and

be the corresponding deflection mat r ix .

Then, the flexibil i ty ma t r ix in the equation

(DW) = (AW) x (W)

x in. / lb

This checks Eq (5) in WAPD-115 to slide rule accuracy.

Systematic Load Calculation

A systematic way to calculate (AW) i s to use the concept of partitioned matrices. The original

flexibility matrix equation for the statically determinate free bodies is

+ L v '+ - ( D ~ ) ~ (A F)NW ( A ~ ) ~ ~ ( F ) N

It is . to be noted that double lines these matrices into working (W) and nonworking (N) blocks.

and that these blocks can be used in separated matrix equations; that is ,

can be expressed a s

a d

Setting (DFIN = (0 ) gives

To solve for the redundant loads multiply each term by the inverse matrix of (AFINN; thus

I T (u) = iAFINN X (AFINW X ( F I W + (AFINN X (AFINN X (FN) n .

s o that

A s a check, t h i s m a t r i x equation wil l be so lved numerically. F i r s t ,

giving the i n v e r s e

t

(AF)NN = ( (AF)55 I = [ 5.654.975 x in. / l b I ,

Then,

x 1 0.2408 1 0.94471 1 -2.95355 1 -0.43481 1 x in. / l b

which checks Eq (26).

Substi tut ion of E q (29) into E q (28a) g ives

(DFIW = (AF)WW x (FIW + (AFIWN x - (AFINN x (AFINW x (FIW i

I

[ I I = [(AFIWW - (AFIWN x (AFINN x (AFINW I x .

Since

then,

(AW) = (AF)WW - (AFIWN X X .(AFINW~

Numerica l ly ,

(AW) =

x in. / l b

x 1 -0. 042,582 1 -0. 167,058 1 t o . 522,292 1 t o . 076,890

which checks Eq (27).

6 r 10 in. / l b

B. SO207 Code (CTAC and MODE)

Discussion

A WOLONTIS-650 code has been developed to calculate the influence deflection coefficients

(flexibility ma t r ix ) fo r an e las t ic s t ruc tu re with up to 10 applied loads and up to 14 loads when the

s t ruc tu re is made statically determinate. An es t ima te is made of the shock response to a 1 -g load-

ing by using weights a s the applied loads, and the fundamental frequency is approximated by the

formula (Ref 8):

where

f = natura l frequency, cps,

g = 386 in. / s e c 2 , usually,

Wi = weight a t point i.

and

Yi = s t a t i c deflection of Wi.

This formula is good if a l l deflections a r e positive. Local fo rces and moments throughout the s t r u c -

ture can a l s o be determined. All th is is progi-ammed in SO207 deck 1 and is called the CTAC code.

In S0207, deck 2 is programmed the MODE code, which uses the i tera t ion procedure of

WAPD-115 to calculate a s many modes of vibration a s des i red. Local fo rces and moments for each

mode can a l so be determined.

Arrangement of Decks

CTAC without MODE: Successive CTAC problems can be run with the following a r r a n g e -

ment of ca rds . (Each arrangement must be specified on the back of the r eques te r c a r d . )

Specified Arrangement Actual Contents of Arrangement on Requester Card (Not Indicated on Requester Card)

SO207 deck 1 On fi le in 650 room

( CTAC t rans fe r c a r d

Input deck F i r s t CTAC data deck

1. ;st CTAC data deck

CTAC and MODE: The MODE code uses the data f r o m the CTAC code input and output to

calculate natura l frequencies, mode shapes , and mode amplitudes of an e la s t i c s t ruc tu re . If the

MODE problem immediately follows the corresponding CTAC problem, the arrangement of decks is

this:

Specified Arrangement Actual Contents of Arrangement ,

on Requester Card (Not Indicated on Requester Card)

50207 deck 1 On fi le in 650 rooin

Iriput deck 1

CTAC t rans fe r c a r d

F i r s t CTAC data deck - -

1 Las t CTAC data deck

(Continued)

Specified Arrangement on Requester Card

SO207 deck 2

Input deck 2

Actual Contents of Arrangement (Not Indicated on Requester Card) , - ..

On file in 6 50 room'

MODE data deck c o r -

responding to l a s t

CTAC data deck

MODE Alone: If the corresponding CTAC problem had been done on a separst:e. previous

occasion, the arrangement of c a r d s is:

Specified Arrangement on Requester Card

SO207 deck 1

Input deck 1

SO207 deck 2

Input. deck 2

h ~ u t Cards

Actual Contents of Arrangement (Not Indicated on Requester Card)

On fi le

CTAC data deck

CTAC output deck

MOUE t rans fe r ca rd

On fi le

MODE data deck

~ r a n s f e r Cards:

1) CTAC Trans fe r Card

Columns Contents

7-10 6850

77-79 xxx ( ~ r b i t r a r ~ 3 -digit problem identification)

2 ) MODE Trans fe r Card

Columns Contents -

7-10 5480

7 7 -79 xxx (Arb i t ra ry Y -digit problem identification)

Data Card Format : Each number (called a "word" on a ca rd) has a sign, t 0.r - , and 10

digits, taking up a total of 11 columns on a card. A "fixed" number is put a t the right end and is

preceded by ze ros ; fo r example, t23 (fix) is put on a c a r d , a s tUUUUUUUUZ3. A number that is not

fixed is considered to be in the floating f o r m i (a. bcdefgh) x 10 ij-50 and is put on a c a r d a s

iabcdefghi j ; fo r example, -23 is put on a c a r d a s -2300000051, and t o . 056 is put on a c a r d a s

t 5600000048. Standard WOLONTIS format c a r d s a r e used a.s shown on the following page.

Columns Contents

Location of f i r s t word of c a r d in 650

s torage drum.

Number of consecutive words on c a r d ,

1 to 6 . If a row of words is m o r e than

6 , the row is broken in groups of 6 o r

l e s s and the groups put on success ive

ca rds . F o r example, i f a row of 9

words is to be put on c a r d s s t a r t ing a t

location 101, 5 words could be put on a

ca rd with location 101 and 4 words

could be put.on the next ca rd with loca-

tion 101 + 5 = 106.

F i r s t word

Second word

Sixth word

I

CTAC Data Deck (Input) . .

F i r s t Word

TRA NO. 1

P(fix)

P(fix)

---

A1 1

*21

A ~ l

C1 1

C2 1

C ~ l

Second Word

W(fix)

TRA No. 2

Location of F i r s t Word

638

540

543

197

197 + N - -

197 - N + N'

246

246 t R - -

246 - R t NR

Number of Words in Row .

1.

3

3

N

N

N

R

R

R

Third Word

R(fix)

Explanation

TRA NO. 1 = t0203000745, i f t he re

is a load calculation f o r th is CTAC

problem.

TRA No. 1 = t0203000548, if the re

is no load calculation and if MODE

follows th is CTAC problem.

TRA No. 1 = t0203000685, if the re

is no load calculation and if MODE

does not follow this par t icular

CTAC problem. - P = total ni imber of e lementary

!primitive) striirt11re.s; 1 s 999.

W = total number of external loads

(weights) on s t ructure : 1 5 W 5 10.

R = total number of loads (FR) on

s ta t ica l ly determinate s t ructure ;

W 5 R 5 14. The f i r s t W loads of

R a r e the original external loads.

Next come H s e t s of c a r d s fo r P primit ives in o rde r f r o m P = 1 to

P = F.

P = number of primitive; 1 ' 5 P 5

'I'HA No. 2 = t0203000602, if p a r -

t ia l flexibility ma t r ix An due to P

pr imit ives is to be punched.

TRA No. 2 t0203000604 otherwise,

1\J = number of loads on primitive. . . , . --

Flexibility matrix A of primitive P

s t ruc tu re P. The f i r s t word of each

row must s t a r t a new card.

Connection mat r ix Cp giving

(Fp) = (Cp) x (FR). The f i r s t word

nf each row mus t s t a r t a new ca rd .

A12

AZ2 .

A ~ 2

12

C22

cNa .

etc.

e t c .

e t c .

e t c .

e tc .

etc.

CTAC Data Deck (Input), Cont.

Location of F i r s t Word

51 5

500

543

197 - -

197 - N + N '

246 - -

246 - R + NR

F i r s t Word

W1 g

P(fix)

A1 1

1

C ~ l

Number of Words i n Row

W

1

3

N

R

R

Second Word

W2

TRA No. 3

A 12

5 2

cR2

Thi rd Word

etc.

,

N(fix)

e t c .

e tc .

e tc .

~ x ~ l a n a t i o n

These c a r d s come a f t e r the P s e t s

of ca rds .

Weights (pounds fo rce )

Gravity = 386 in. / s e c 2

If there a r e to be load calculations

on some pr imit ives , the following

s e t s of c a r d s must be added to the

deck. Otherwise, they a r e not used.

There is one s e t of c a r d s for each

primitive, to be considered, and

these pr imit ives need not be the

s a m e a s before.

TRA, NO. 3 = t0203000769, if th i s is

not the las t primitive on w-hich load

calculations a r e made.

TRA No. 3 = t0203000548, if th i s is

the las t primitive considered, and

if MODE follows.

TRA No. -3 = t0203000685, if th is is

the las t primitive considered and if

CTAC follows o r if the end of the

las t problem.

Flexibility ma t r ix A . P

Conncction 'matrix C . P

MODE Data Deck (Input)

Explanation

E = relative e r r o r for conver- -6 gence in iterations (10 cor -

responds to an e r r o r in the 7th

digit of a number). T = total

number of iterations carr ied out

if resul ts do not converge, say 15.

Vo = maxlmum starting velocity,

ipo

fo = cutoff natural frequency,

CPS M = number of modes to be

calculated; 1 5 M 5 W.

The starting velocity V for each

mode of frequency f is deter-

mined by the formulas

Second Word

T

f o

F i r s t Word

E

vo

Location of F i r s t Word

337

I

3 27

Third Word

M(flx)

Number of Words in Row

2

3

TRA No. 3

A12

A~~

C1 2

NZ

------

P(fix)

*11

A ~ l

Cl 1

C ~ l

543

1Y7 - -

197 - N + N2

246 - -

246 - H + 1'48

3

N

N

R

H

V = V i f f 5 f o , 0

V = f o Vo/f i f f > f o .

N(fix)

etc.

etc.

etc.

etc.

If there a r e to he lnad c a l c l ~ l a -

tions on some primltlves, the

following s e t s of cards must be

used. These cards have pre-

cisely the same format ac In t h r

CTAC data deck.

Output Cards

Output Card Format: Most of, the o ~ t ' ~ u t is in the form of blocks of data (matrices). Before

each ma t r i i is printed out, three words in locations 997 to 999 give, respectively, the machine s tor -

age location of the first element of the matrix, the number of rows, and the number of columns.

If the structure is redundant, one .of the matr ices will contain the flexibility matrix AW of the

applied loads and the coefficients of the redundant forces in te rms of the applied loads. This matrix

is identified by ones along the lower right diagonal. For example, if a structure has R = 4 loads

when made statically determinate and W = 2 applied loads, this matrix will be in the form:

where the corresponding equations a r e

DF1 = A1lF1 + A12F2 a

D ~ 2 = A21 F1 + A22F2

F j = K.. F + K3LFZ , 31" 1

and

F4 = K41F1 t K42F2 + K43F3 .

. CTAC Output Deck . -1 F i r st Second Word 1 Word

Third Word

Fourth Word Explanation I

If TRA No. 2 for primitive P is

t0203000602, the following cards

a r e punched: . . .

00 1 (fix) .R(fix) Flexibility matrix AR of R loads- I due to P primitive.

If the structure is redundant ( (W < R ) , this matrix containing I 4 anrl coefficients I<. . is

13 I punched.

Flexibility matrix A W of applied

loads.

etc. Static deflectiot~s'of weights, in.

Ys = equivalent static deflection

= (ziwiY;) 1 (ZiWiYi),

2af = natural angular frequency I

f = frequency, cps I -

etc. If W <R, these rcdundant loads I produced by 1 g applied loads a r e I punchcd. I If load calculations were made

for a primitive, the following oct

of cards is punched for this

primitive. . ' 1 1 (fix) Matrix of priilritive loads ~ i i ~ d t i '

I g a pplicrl l.oa.13..

1 (fix) Matrix of primitive deflettiona 1 under 1 g applied load.

1

MODE Output Deck

Location of F i r s t Word

330

336

297

307

317

484

997

- - 997

543

997

997

Number of Words in Row

3

3

W

w . .

W

W

3

- - 3 .

3

3

3

F i r s t Word

2/nf

J

FIJ.

F;I J

F::

Y1 J

344 + WM (fix)

- - [ 344 t RM - M]

P(fix)

010(fix)

080(fix)

Second Word

J

E

FSJ

F; J

F"' 2 J

YZJ

l(f ix)

- - -------- l ( f ix)

TRA NO. .3

N(fix)

N(fix)

Th i rd Word

VJ

T

etc .

etc.

,etc.

e tc .

M(fix)

- - M(fix)

N(fix)

M(fix)

M(fix)

Four th Word

J(fix)

- - - -

Explanation

The following c a r d s a r e

punched f o r each mode calcu-

lated.

27rf = Nat ang f r eq , r p s J fJ = Nat frequency, cps

V = s t a r t ing velocity. i p s

J = mode; J = 1 to M.

T = number of i te ra t ions in J calculation of mode J.

FiJ = fo rce FI in mode J a f t e r

T i tera t ions , lb.

FYJ = f o r c e FI i n mode J a f t e r

T J - 1 i te ra t ions , lb.

F l l f = 2 1, (WI/g)(2rfJ) YIJ; lb-

YIJ = deflection of FIJ, in. .

If W < R, the following-dards

a r e punched.

Matrix of FW+l J, f i r s t redun-

.dant load. - -

Matrix of FRJ, l a s t redundant

load.

If load calculations were made

f o r a pr imi t ive , the' following

s e t of c a r d s a r e punched f o r

e a c h primitive.

Matrix of primitive .loads;

( F ~ ) ~ ~ .

Matr ix of primitive 'deflections:

( D ~ ~ ) ~ ~ .

Sample Problem Using Structure in WAPD- 11 5

Printout of Cards: The printout of the cards used in the determination of influence coeffi-

cients and modes of vibration s t a r t s on page 32. The SO207 input cards were sent to the IBM-650

operator with the following note on'the back of the requester card:

Order of Cards

SO207 deck 1 . ,

Input deck 1

SO207 deck 2

Input deck 2

Results: The CTAC output gives an estimated natural frequency of 223.50 cps, which com-

pa re s well with the f i r s t MODE frequency of 222.45 cps. Fo r a natural frequency of 223. 5 cps, the

shock acceleration factor corresponding to a unit starting velocity is

- 1 - 1 zrrfv = 2 (223.5 sec )(1 in. set ) = 3. 658 . N=- E (386 11,. s e c ~ a )

Thus, f rom the CTAC output,, the moment a t the left end of the outer cylinder is

M = (7688. 1 in. -lb)(3.638) = 27,969 in. -1b , 0

and the absolute sum of the four MODE resul ts i s

Mo = (27,315 t 1,118 -1. 4686 .t 375) in. -1b = 33,494 in. -1b . The MODE code gives a higher moment but, in the usual sh.ock calculations, the higher modes have . ,

smal le r s tar t ing velocities so that the resu l t s will be closer. . .

F r o m CTAC output, ' the mokent .a t the left end of the inner cylinder i s

Mi = (2936. 5 in. -1L)(3. 638) = 10,683 in. -1b ,

and the absolute sum of the four MODE resul ts is

M. = (9131 + 1537 + 4500 + 1978) in. -1b = 17,146 in. -1b . 1

As before, taking only two modes would give a 'closer check.

A direct comparison of mode calculations from WAPD-115 and from the code SO207 is.given in .

the table on the following page.

The odd numbered modes a r e not s o accurate in slide rule calculations because of small differ-

ences between large num:bers in the determination of CJ. Otherwise, the agreement is satisfactory.

Sample problem Cards: The actual c a rds used in the solution of the sample . " problem given

in WAPD-115 a r e found bn pages 32 through 43 of this report. . .

COMPARISON TABLE O F MODE CALCULATIONS

Source

F r o m WAPD- 1 15 (Slide - ru le accu racy )

F r o m SO207 (IBM-650 machine)

3

1776

1776

- 691

3 0

1067

116

. - 26

1

60

1779

- 676

15

1061 -

117

- 26

1

59

2

-134. 8

-135

-215

2 59

164

- 18

- 16

18

18

-148

-233

280

188

- 19

- 17

2 0

2 1

Mode J

C f o r V = 1 in. / s e c J J

F~~ = 'J e~~

(lb)

D~~~ = 'J a~~

( l o 6 in. )

F~ J

(lb)

J

6 . (10 i n . ) - . .-

4

-298. 2

-298

518

-644

823

- 11

11

- 14

2 7

-291

534

-662

808

- 11

11

- 13

25

1

121 .8

122

7 55

699

8 0

234

8 37

7 34

130

122

7 57

698

80

235

8 39

733

131

SO207 I N P U T . "

0 0 1 1 685 0 T R CARD CTAC Input

2 638 1 +C203000745

3 540 3 +COOOOOOC08 +0000000004 +0000300005

- 4 543 3- +COOOOOOC01 - +0203000602 +0000300002 F i r s t Primiti-.re

5 197 2 +1114000G41 +5570000041

6 199 2 +5570000041 +1294000043

7 246 3 +0000000000 +1500000051 + 0 0 0 0 ~ ~ 0 0 0 0 0

8 249 2 +0000000000 +2800000051

9 251 5 +1000000050 +1000000053 +0000300000 +0000000000 +1000000050

1 0 543 3 + ~ 0 0 0 0 0 0 0 2 +0Z 03000601 +0000000002 Second Pr imi t ive

1 1 197 2 +1114000041 +557000004 L

1 2 199 2 +5570000041 t 1294000043

1 3 246 5 +0000000000 +5000000059 +OOOOQ000CO +0000000000 +1000000051

1 4 251 5 +OD00000000 +1C00000050 +0000000000 +0000000000 +1000000050

15 543 3 +0300000003 + 0 2 0 3 0 0 0 6 0 ~ +0000000002 Third - Pr imi t ive

1 6 197 2 +2340000040 +5160000049

1 7 199 2 +5100000040 +1496000042

1 8 246 5 +0300000000 -0000000000 +0000000000 +0000000000 +5000000050

- 1 9 251 5 +0300000000 -1600000050 +0000000000 +0000000000 +1000000050 !

t

.. %

4

0

m m

0

0

.O

.O

0

0

0

0

0

C'

c

0

Cl

0

.-I d

I I

0

0

0

0

C' 0

0

'A

0

0

* 0

0

0

0

o

m

0

0

0

0

0

0

0

0

0

0

0

d

+ +

0

0

.a

0

0

> 0

C

00

c 0

m m

0

0

0

0

0

0

0

0

0

0

In

0

N

0

N

4

I '

I

0

0

0

In

C 0

0

C

0

0

0 e

00

h

B

0

0

'G

0

F &

0

0

0

0

+ +

N

CO

N

o

00

0

C'

00

0

C'

C

OO

0

00

0

0

CO

O

G

CO

C

C

CO

O

o

00

0

0

00

0

+ +

++

C.

O.

0 0

00

00

00

00

00

0.

00

00

~

~0

00

00

00

00

00

C,

00

0

00

00

0.

04

00

00

40

0

04

00

00

m*

00

0m

r0

0

00

d0

00

m~

00

00

'4

00

O

Nm

O0

0m

H0

0Q

.m

d0

0

++

++

++

++

++

++

++

+

mN

Nm

ln

mN

Nl

nm

mN

Nm

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0 0

35 543 3 +0300000007 ~ 0 2 0 3 0 0 0 6 0 4 +0000000002 Seventh Primitive

36 1 9 7 2 +1510000041 ;6040000041

Primi tjve ~6040000041

44 251 5 +0300000000 -0608000000 +0000000000 +000000~3000 -1000~00050

4 5 5 1 5 4 +132'5'800052 - i i ~ ~ 4 4 0 0 0 5 2 +1882000052 +1204403052 Weights

Load Calculation 48 197 3 + I 114000041 +5570000841 +0000C00000 (incl~ding the

foundation momen*) 49 200 3 +5.570000(341 +1294000043 . +0000C~r)0000

1

. . 54 543 3 +0000000005 +0203000548 +0000000003 Last Pr,imitive in Load

Calculation (including 5,5 197 3 +3930000841 +1474000Q42 +00Q000.0000 the foundation moment)

56 200 3 +1474000042 +1131680043 +0000000000

57 203 3 +0000000000 +000~000000 +0000000000

58 245 5 +0000000~~00 +0000000000 +1500000051 +0000000000 -2250000051

59 251 5 +0000000000 +0000000000 +1000000050 +0000000000 -1000000050

60 255 5 +0000000800 +COO0000000 +2250000051 +7500000050 -3000000051

. .

' 61 337 2 +1000000044. +1500000051 MOD3 Input

62 327 3 +1@00000050 +lObB080056 +0000000004

63 543 3 +0fi00000b01 +-0233000769 +0000000003 F i r s t Primit ive i n Load Calculst ion (same a s i n CTAC)

64 19.7 3 +1114000041 +55~70000841: +0800000000

70 543 3 +@~00000005 +0233000548 +0000000003~ Last E i m i t i v e i n Load Celculst ion (same a s i n CTAC)

71 197 3 +3930000041 +1474000042 +0000000000

SO207 DATA

0 0 1 1 540 4 +0000000!308 +0000000004 +0000000.C~05 +0000000001 CTAC Output

0 0 1 2 997 3 +0000000001 +0000000005 +0000030005 % due t o 1 pr imi t i re '

0 0 1 3 001 5 .+1294000043 +2129500043 +0000040000 +0000000003 +2408000043

0 0 1 4. OW 5 ' +2129500843 +5471500043 + 0 0 0 0 0 0 0 ~ ~ 0 +0000000000 +6585500043

0 0 1 5 011 5 +0000000000 +0~300000000 +0000000000 +0000000000 +0000000000

0 0 1 6 016 5 +0000000000 +0000000OCO +0000000000 +0000000000 +0000000000

0 0 1 7 021 5 +24080'00043 +6555500043 . +0000000000 . +0000000000 +7978000043

0 0 1 8 540 4 +0000000008 +0000000004 +OOQOO00005 +0000000002

0 0 1 9 997 3 . +0000000001 +0900000005 +0000000005 % due t o 2 primit ives

0 0 1 10 OC1 5 +129400CH)43 +2129500@43 +0000000000 +OOOOOOOOOO .+2408000043

0 0 1 11 006 5 +2129500043 +7501000043 +00'00000000 +0000000000 . +9272000043

0 0 1 12 011 5 +0000000000 +0300000000 +000000000C +0000000000 +0000000000 . .

0 0 1 13 016 5- +0000000000 +0000000800 +00000000OC +0000000000 +0000000000

0 0 1 14 021 , 5 +2408000043 +9272000043 + 0 0 ~ 0 0 0 0 0 0 C +0000000000 . +1150000044

- - ~ -

. .< . .

001 1 5 540 4 +'3000000(308 +CC80000004 +0000000005 +0000000008

0 0 1 1 6 99' 3 +3000000001 +0000000005 + 0 0 0 0 0 0 0 ~ 0 5 AR due to all primit ives

001 1 7 001 5 +8294000843 +2129500043 +0000000000. +0000030000. +2408000043

001 18 006 5 +2129500043 +7750600043 ' +0000000000 +0000030000 +9447100043

001' '19- 0.11.. 5 . . ' +3000000100 ' - ' +00~00000000 +2023962544 ' " . +3242630043 -2953550044 . ,

001 20 018 5 +3000000800 +OOOQO001CO +3242600043 +10316.30043 -43481 00043

001 21 021 5 +2408000443 +9447100@43 -2953550044 -4348100043 +5654975044 . .

001 22 997 3 + 3 0 0 0 0 0 0 ~ 0 1 +0000000@05 +0000000005 % ani K: j,

001 .23 001 5 +[I91462643 +1.7272.23843 +125,7679943 +185153684? +2.408000043

001 24 006 5 +872'7223843. +6i72384443 +493&147743 +7263858042 +9447100043

0 0 25 011 5 +1257679943 +49341.47843 .+4813460043 +9716209042 . -29.53550.044

001 26 016 5 +C851506842 +726385.8 042 +9716208042 . +697275+042 -4348100043

001 27 021 5 -.1258197448 -1670582149 +5222923149 +7688981848 -9999999949 - ' . . .

. .

. .- .. . . . . . . .

i I . .

-. . -..

001 28 540 3 +00000~10808 +0000000004 +0000000005

001 29 997 3 +0000000001 +OOOQ000004 +0000000004 45 001 30 001 4 +I191462643 +I727223843 +I257679943 +I851506842

001 31 005 4 +172721384? +6172384443 +4934147743 +7263858042

001 32 009 4 +I257679943 +4934 147843 +4813460043 +9716208042

001 33 013 4 +1851506842 +7263858042 +9716208042 +6972754042

001 34 501 4 +6894209545 +2?94671446 +2032377346 +4154479245

001 35 529 3 +I957346746 +1404299653 +2235012252 Frequency

001 36 519 1 +7337809751 Redundant Load

001 37 543 3 +0000000001 +9203000769 +0000000033

001 38 997 3 +0000000351 +0000000003 +0000000001

001 39 351 1 +4144161953 Load Calculation for Primitive No. 1

001 40 352 1 +3543981@52

001 41 353 1 +76881s2957

001 42 907 3 +0000000371 +0000000003 +0000000031

001 43 371 1 +6590593844

001 44 372 1 +6894209645

001 45 373 1 +0000000800

0 0 1 46 543 3, +3000000005 +0203000548 +0000000003

0 0 1 47 997 3 +3000000351 +000@000003 +0000000001

0 0 1 48 351 1 +I171992853 Load Calculation for Primitive No. $

0 0 1 49 352 1 +1148219D52

0 0 1 -50 3.53 ' 1- +19,3645.7153

0 0 1 5 1 997 3 +3000000371 +0C00000003 +0000000001

0 0 1 52 371 1 +5298406544

0 0 1 53 372 1 +>912020145

0 0 1 54 373 1 + 3 0 0 0 0 0 0 ~ 0 0

L

I

001 55 330 4 +I397707353 +2224520352 +1000000050 +0000000001 MODE Output

001 56 336 3 +7000000050 +1660000044 +1500000051

001 57 297 4 +I2176281252 +7572647152 +69776.37252 +7961363451. First Mode

+796'136'2351 001 58 '307. 4 +1217628252 +75726&6952 +69776'37352

001. 59 317 4 +12.17628:252 +7572646852 +6977637452 +796136.3451

00.1 60 ,484 4 +2345345446 +8385139946' +73256'06746 +I306085246

001 61 330 4 +5398087653 +a591323352 +1000000050 +0000000002

001 62 336 3 +1600000051 +lC00000@44 +i500000051

001 63 257 4 -14791053'52 -2328211152 +280.2711752. +1883'356652 Second Mode . .

001 64 3C7 4 -1479105352 -2328287652 +2802786152 +I883356652

001 65 3x7 4 -14790863'52 -2328143Q52 +2802635552 . +I883292252 '

001 66' 484 4 -1910021~545 -1728322945 +I972670045 +to71354345

001 67 330 4 +7572882253 +I265261653 +100000005C~ +0000000003

001 68 336.3 +1600000~51 +1~00000644 +1500000051

001 69 25-7. 4 ' +1778515057 -6769884352 +1524057551 . +lo61353153 . Third Mode

001 70 307 4 +1778515@53 -676917?852 +1524057551 +lo61353153

001 '71 317 4 +I778493653 -6761204852 +1530871451 +1061210053 ,

001 72 464 -4 +I166956146 -2556332445 +5474998445 ' +5930557045

. .

. .

- ~

001 73 33C 4 +1009491@54- +I686654853 +1000000050 +000000000L

001 74 336 3 +2000000@50 +lOOQO00044 +~1.500000051

001 75 297 4 -2908838852 +5340940252 -6623016752 +808154'145.2 Fourth Mode

001 76 307 4 -2908.838752 +534~940252 -6623016752 +8081541452

001 -77 3.1.7 4 - ..-2908840652 . -+5340191952 -6621871752. +8079938252

001 78 48-5 4 -1074087G45 +I133566645 -1 332734245 +2541078445

001 79 997 3 +0000000360 +000000001 +0000000004 'Redundrnf. Load

001 80 360 4 +2388659252 +2'060575052 +I267807252 -36061 38652

001 81 543 3 +Q000000~01 +0203000769 +00.80000003 Load Calculation for Prini.?iv.- No. 1

001 82 997.3 +0000000~10 +0~~0000003 , - +0000000004

001 83 010 4 +I613628954 +62~8349052 -7,605712053 +7991333051 I .

001 84 '016 4 +1117893553 -17b6740652 +1229207353, -1174037252

001 85 016 4 ' +2731522454 -1117905'753 +4686360653 -3749040052

001 86 997 3 +0000000~80 +00@0000@03 +0000000004

001 87 08@ 4 +2420249345 -2724124343 -1626078544 +236295554'¶

001 88 08b 4 +2345345546 -1910021345 +I166956046 . -107408"045 , .

001 89 088 4 +0000000000 ~+0000000000 +0000000000 +0000000000

b

I

-

0 0 1 9 0 ' 5 4 3 3 +0000006005 +0203000548 +00000000@3 . Load ~alclculation f o r Primit ive No. 5

0 0 1 9 1 997 3 . +OO~OOOOQ)10 +0000000003 +0000000004

0 0 1 92 010 4 ' + 5 0 9 1 9 7 3 ~ 5 3 -4322280052 -2623957653 -1820713253

001 93 OL4 4 +4588978052 +7421359051 -1 115401452. -3016578152

0 0 1 9 4 018 4 ' +91308081353 +153689'1453 +4499639653 +I977959053

,001 95 937 3 +0000000080 +0000000303 +0000000004

0 0 1 96 030 4 +2677560845 -6047477043 -1195625545. -1160228145

0 0 1 , 9 7 094 4 +1223955846 +I284633244 -5018361645 -5795942745

0 0 1 98 098 4 ,+000000!3000 +0008000000 +0000000000 +0000000000

. . . .

The actual preparation and "debugging" of the CTAC and MODE codes were done-with the able

assistance of Mr. W. D. Long.

. .

REFERENCES

1. G . Kron, Tensor Analysis of Networks, (New York: John Wiley and Sons, 1939).

2. B. Langefors, "Analysis of Elastic Structures by Matrix ~ r a n s f o r k a t i o n with Special Regard to

Semimonocoque Structures, I' Journal of Aeronautical Sciences, Vol 19 (July 19 52), p 451.

3. G . Kron,. "Solution of Complex Non-linear Plastic Structures by the Method of Tearing, Journal

of Aeronautical sciences',' Vo1,23 (June 1956), p 557.

4. J. H. Argyris, "Eriergy Theorems and Structural Analysis, I' Aircraft Eng. ,' Vol 26 (1954),

pp 347, 383, and 410; Vol 27 (1955), pp 42, 80, 125, and 145.

5. S. ~ imoshenko , Stse'ngth of Materials, Vnl T (Princeton: D. Van Nostrand Co. , Inc. , 1141), . .

6 . L. A. Pipes, Applied Mathematics for Engineers and'physicists, (New York: McGraw-Hill,

1946).

7. C. M. Friedrich, "Shock Calculations for an Elastic System with Distrihnterl Mass, Using the

Starting Velocity Concept, I' WAPD-115 (1 954).

8. L. S. Marks, Mechanical Engineers1 Handbook, Fourth Ed. (New York: McGraw-Hill , 1941);