Determination of the moment-curvaturerelation for a beam of nonlinear material

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Authors Hsu, Teh-min, 1940-

Publisher The University of Arizona.

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DETERMINATION OF THE MOMENT-CURVATURE RELATION FOR A BEAM OF NONLINEAR MATERIAL

byTeh-min Hsu

A Thesis Submitted to the Faculty of the

DEPARTMENT OF CIVIL ENGINEERING In Partial Fulfillment of the Requirements

for the Degree of

MASTER OF SCIENCE In the Graduate College

THE UNIVERSITY OF ARIZONA

19 6 6

STATEMENT BY AUTHOR

This thesis has been submitted in partial fulfillment 6f requirements for an advanced degree at The University of Arizona and is deposited in the University Library to be made available to borrowers under the rules of the Library.

Brief quotations from this thesis are available without special permission, provided that accurate acknowledgement of the source is made. Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or the Dean of the Graduate College when in his judgment the proposed use of the material is in the interests of scholarship. In all other instances, however, permission must be obtained from the author.,

SIGNED 8 ~

APPROVAL BY THESIS DIRECTOR This thesis has been approved on the date shown belows

, i 9 U

Allan jJ. Malvick, Sc, D. DateAssociate Professor of Civil Engineering

ACKNOWLEDGEMENTS

The writer wishes to express his appreciation to his thesis advisor6 Dr, Allan J„ Malvick and to Dr» Ralph M« Richard for their valuable suggestions and guidance in completing the thesis.

Appreciation is also extended to fellow graduate student Mr, Wilbur D« Birchler for his valuable suggestions and to the staff of the Systems Engineering Department for the use of their IBM 1401-7072 digital computer and related equipment.

iii

TABLE OF CONTENTSpage

LIST OF ILLUSTRATIONS . . . . . . . . . . . . . . . . . vCHAPTER 1 - Introduction . . . . . . . . . . . . . . . 1

1.1. Introduction . . . . . . . . . . . . . . 11.2. The Problem Defined . . . . . . . . . . 11.3. Method of Treatment . . . . . . . . . . 2

CHAPTER 2 - Governing Equation of the Relationbetween Stress and Strain . . . . . . . . 3

CHAPTER 3 - Moment-Curvature Relation for a Beamof Nonlinear Material . . . . . . . . . . 5

3.1. Introduction . . . . . . . . . . . . . . 53.2. For a Rectangular Cross Section . . . . 83.3. For a Circular Cross Section . . . . . . 173.4. For a Wide Flange Cross Section . . . . 26

CHAPTER 4 - Application of Moment-Curvature Relation to Determine the Deflection of Beams of Nonlinear Material . . . . . . . . . . . . 32

4.1. Slope and Deflection Due to Bending . . . 324.2. Statically Determinate Beam with

Rectangular Cross Section . . . . . . . 364.3. Statically Indeterminate Beam with

Wide Flange Cross Section . . . . . . . 40

CHAPTER 5 - Conclusions . . . . . . . . . . . . . . . . 49REFERENCES o . . . . . . . . . . . . . . . . . . . . . e 50APPENDIX A — Notation . . . . . . . . . . . . . . . . 5--

B - Fragment of Computer Program . . . . . . 53

iv

LIST OF ILLUSTRATIONS Figure . Title page1. Stress-Strain Diagram <,„„,»<.»<.*»» . 42. Bending of Member of Rectangular Cross Section, 5

3= Stress-Strain Relation 64. Stress and Strain Diagrams for Different

CaSeS e o o e o o e o o o » e o o o o b ® o e 105, Relation between Bending Moment and Curvature

for a Rectangular Cross Section 136» • Unloading Stress-Strain Diagram , 157, Circular Cross Section • » • » » • • « • • • « 1780 Relation between Bending Moment and Curvature

for a Circular Cross Section 259. Wide Flange Cross Section 2610, Half Wide Flange Cross Section 30

11, Relation between Bending Moment and Curvaturefor a Wide Flange Cross Section , , , , , , , 31

12, Curvature of a Beam Along Axis , , , , , , , , 3313 0 Slope Plotted over Length 3414, Relation between Slope and Deflection , , , , 3515, Cantilever Beam o , , , , , , , , , , , , , , , 3616, Curvature, Slope, and Deflection of

Cantilever Beam , , , , , o , , , , , , , , , 3^17, Statically Indeterminate Beam 4018, Relation between the Assumed Reaction and »

its Corresponding End Deflection , , , , , , 41

v

19o ' Bending Moment Diagrams for Different Valuesof End Moment © © © © © © © © © © © © © © © © 5

20© Relation between Bending Moment and Slopeat Simply Support End » © © © © . « © « © © © 46

21© Moment-Deflection Diagram , © © © © © © © » © 4?22© Normalized Deflection Shapes of Linear and

Nonlinear cases © © © © © © © © © © © © © o © 43

CHAPTER I

INTRODUCTION1.1 ~ Introduction

In the analysis of beams* the first step is to determine the shear and bending moment at any point along the beam* then the maximum value of bending moment can be determined. In calculating bending deflections* it is necessary to derive an equation or equations for the bending moment at any point of the beam or to have a plot of the bending moment.

The theory of bending developed in most text books* such as strength of materials and applied mechanics* is applicable when the relationship between the stress and strain is linear. For materials which are strained beyond the proportional range* however* the stress- strain relation is no longer linear. In this thesis a method is developed to analyze beams whose material'is stressed beyond the proportional limit,

1.2 - The Problem Defined

The object of the research presented in this thesis is to determine the moment-curvature relation for a beam of nonlinear material subjected to increasing* then decreasing bending moment.Three types of cross section, that is* rectangular, circular* and wide flange cross sections * are treated.

2

The moment-ourvature relation is very useful as a basis for determining deflections of a beam of nonlinear material and for the advanced analysis of frame structures,

1.3 " Method of Treatment

The assumption that the cross section of the beams remains plane during bending is used. The formulas for moment“Curvature that are derived here are obtained by a method of numerical integration.

CHAPTER 2

GOVERNING EQUATION OF THE RELATION BETWEEN STRESS AND STRAIN

The beams which will be considered here are made of nonlinearmaterial, having the relation between stress and strain as shown inFigure 1, in which o is yield point stress. The curve consistsy*p.of two straight line segments and a parabolic curve between them.The parabolic segment is tangent to the two straight lines at gss a and a as shown in Figure 1.

The equation of the parabolic curve is

a = AE 2+ B6+ Cwhere A, B, C are real constants

The boundary conditions are

£2 = E =dB e=E1-a

d2| = 0de|£=ej+a

Now £2 = 2A8+ B (1)

at 8= 8-j- a da = = 2A( e n- a) + Bda

(2a)

at 6= a dSL = 0 = 2A( 8 ^ a) + B (2b)

3

4

y.pparabpli

Figure 1 Stress-strain diagram

Subtracting (2b) from (2a)

_ y*EjL = - 4Aa or A s - , 7 ’I ? “

gi 4a£i

oFrom Equation (2b) B = -2A(8^+ a) » 2ae^*^

therefore = . g2 ^ *)E + C4aSn 2aEi

Hence

a s Oy p provided

C s a (Gl+ a)1y.p D-- 4ae ]

o s ay.p 4a61 (3)

This is the relation between stress and strain between 6= 6^- a and 6 = a*

CHAPTER 3

MOMENT-CURVATURE RELATION FOR A BEAM OF NONLINEAR MATERIAL

3.1 - Introduction

The theory of the bending of beams is based upon the assumption that cross sections of the beam remain plane during bending, and hence longitudinal strains are proportional to the distance from the neutral axis. With this assumption, the strain for the fiber at a distance z from the neutral surface is

If dg and d^ are the distances from the neutral axis to the

lower and upper surfaces of the beam, respectively, as shown in Figure 2, then the strains for the outer fibers are

w

(5)

N.A.

Figure 2 Bending of member of rectangular cross section

6

The equations of equilibrium are

r rdiJ odA = b J adz = 0

ozdA fdl = b\/do

azdz = M

(6)

(7)

From Equation 1, z=re and dz=rde, therefore

ods = 0 (8a)

orft >) ode = - \ ode J o

if there is symmetry.

(8b)

To determine the neutral axis, the strain A = 5 -*-60 must be

selected so that the shaded areas of Figure 3 which represent the

integrals in Equation 8b are equal.

4 M

Fleura 3 Stress-straln relation

7

In this way the values and Zc are obtained.

Since d * d^+ dg*

<9)

Equation 9 can now be used to define the neutral axis.

From Equation 7

2rEtbr I OEdE as ML

3 2 fe^(^-)(^)J oede = M ................. (10)

For — ss as g ++ g _ = A, therefore d=rA,r r x, c

Placing this value in Equation 8, for a rectangular cross section.

Aofide = M . . (11)

ec

FTTo compare this with the formula — = M for bending of beamsrfollowing Hooke’s law, set

Elaede = — c—

e=where

12 r tEr = - J H aEd£

Jeo

is called the reduced modulus for a rectangular cross section.

8

If the tension and compression portions of the stress-strain

diagram are the same, the neutral axis passes through the centroid of the cross section and the following simplified expressions are

obtained.

any distance z from the neutral axis can be obtained. From a stress-strain diagram, the value of stress, a, corresponding to each

strain may be found. Then by the equation

] azdA = M ........................(12)Jk

the bending moment M can be obtained.

For any given value of r, and by the diagram of Figure 1,every bending moment M corresponding to every given value of r maybe determined. Thus the curve representing M as a function of ris obtained.

3.2 - Moment-Curvature Relation for a Beam with Rectangular Cross Section

The first example beam treated here is a beam with a

rectangular cross section. For a rectangular cross sectiondA = b'dz

therefore Equation 12 becomes

For any given curvature 6= p, from Equation 4 and the strain at

bozdz

9

orM = b I azdz • • . (13)

Z-i

The relation between stress and strain are shown in Figure 4.

Case 1. When £ < g^- a, Hooke's law is valid and Equation 13 becomes

M = b%2

(E'p)zdz for os* - EZ1

M = ~ ( z 2^- Zl3) ............ . . . . (14)

Since the cross section has symmetry, i.e., the neutral axis

passes through the centroid of the cross section, -z^ = Zg = ^ » hence Equation 14 becomes

orEbh3 El

M = “T2? r

thenCasa_2. When e ^ a ^ e s e ^ a or ,

the Hooke's law is no longer valid. For this case, the stress

and strain diagrams are shown in Figure 4b and its corresponding

10

cross section strain diagram stress diagram

N.A.

z0N.A.

(b) e - a< e <e^+

y.p

N.A

(d) perfect plastic

Figure 4 Stress and strain diagrams for the different cases

11

stresses are

a = 6E for z <Zq=( a)r

0 = CTy.p. Cl- — l'~ ~ ] for z>%o

Hence Equation 13 becomes

Zq Zg /z 12M = 2b f E»—»zdz + 2b f o f 1- —£—■— 3--- *1 zdzJ r J y-P'L 4a e1 J0 z« ■L

After carrying out the integration and replacing z , z » and Eby (s.- a)r, and °y.p. ,respectively, the following formula

2 «]_

for computing the bending moment results.

M = 384ar e a^ “ 24r2h2(e1« a)2+ l6rh ( e + a)

- 3h4 ] (15)

Case 3* When 6 > e + a or r<— , the stress and strain diagrams1 £ i” aare shown in Figure 4c, The stresses are

zo.r _a = gE = uLiiii for z < znre i

z x2r r* eT 1

0= ay.P. L ■JLi ± ] for zo< zi

a = Oy for z>z1=(€1+ a);

12

Equation 13 becomes

M = 2b ~zdz + 2b r0

+ 2b

After carrying • out the integration and simplifying, the equation

Case 4. - Perfect Plastic Case

When the M increases, the corresponding stress diagram

(Figure 4c) becomes steeper as the depth e of penetration of plastic

approaches that shown in Figure 4d. Such a material follows Hooke’s law up to the proportional limit and begins to yield under constant

stress.In this case, the bending moment approaches its maximum value

(or ultimate value) which is

becomes

(16)

deformation approaches the value and the stress distribution

(17)

Using Equations 14, 15, and 16, the relation between bending

13

1moment and curvature ~ can be represented. Using Equation 14 and 17 other curves for a perfect plastic case can be obtained, graphically, as shown in Figure 5s> for a rectangular cross section with b= 1,5 in, and h- 2,0 in.. From Figure 5» it can be seen that up to the value M= (M= Mg in perfect plastic case) the deformation is elasticand the curvature of the beam increases in proportion to the bending moment. When M increases beyond the relation between M and ~becomes nonlinear. The value of defines the position of the horizontal asymptote to the curve. As M approaches a smallincrement in M produces a large increase in curvature, so that at M the extreme fiber strains are infinite.

Unloading and Reloading Effects

If a tensile specimen is loaded until the stress has entered the inelastic range and then is unloaded (before failure occurs), the unloading curve will not coincide with the loading curve. Experiments show that the unloading line is parallel with the straight (elastic) portion of the.loading curve as shown in Figure 6a. When the stress has returned to zero, there will remain a permanent strain. This phenomenon occurs both in tension and compression.

If the specimen is loaded ag&in in the same direction to a higher load than that reached in the former test, the behavior will usually be as shown by the dashed line of Figure 6a, The loading curve will practically coincide with the unloading curve up to the load reached in the former loading, after which it will tend to follow

moment

10 (in

-lb )

14

b » 1.5 in.h = 2.0 in.o = 53000 psi y.p. v re^= 0.0031 in. a = 0.0019 in.

8 maximum bending moment

6

41 nonlinear case2 perfect plastic case

2

0 12103 l1curvature

Figure 5 Relation between bending moment and curvature for a rectangular cross section

15

reloadi

^/reloadini

loading

(first loading)permanent straintotal permanent strain

(a). Residual strain-stress

(b)• Unloading strain-stress diagram

Figure 6

16

the diagram obtained in a single loading. For a nonlinear material the unloading curve will not coincide with the original loading curve, but is parallel with the straight elastic portion of the loading curve as shown in Figure 6b.

To determine the equations of residual strain-stress curve, the curve was broken into three parts. If

42 - g ,3 constant (18)

then a = 6E + C^, where is a real constant

(1). If e &

a = £E (19)

at 6= 0 = ^ = 0

hence

therefore

(20)

(3). if £ = E^+ a, then 0= ay#p#y.p

a = ■■■ ( 6 + a) (21)

17

(4). if e ^E]+ a, then a = ayepe

(22)

where in Equation 20 and in Equation 22 are the values of strain which were found from the loading curve.

3,3 - Moment-Curvature Relation for a Beam with Circular

Cross Section

The next example beam treated here is a beam with circular2 2 icross section. In Figure 7, b= 2(R - z ) and dA = bdz.

therefore

Equation 13 becomes

dz

Figure 7

N.A.

dA ■ 2(R^- z^)^dz

R . p 2x4 az»2(R - z )^dzJ -R

j 2 2 4sc 2 \ az(R - z ) dz -R

18

The cross section is symmetric about the neutral axis, therefore

M = 4 I az(R^- z^)^dz (24)

Case 1. If £< 8^- a, then the Hooke * s law is valid and Equation 24

becomes,’R za o o X

M = 4 | 2LlEl z ( r - z ) dz0 r 1

^ ZtBi. r R Z2(R2- z2)^dz ................(25)2 V 0

The value of the integral f z2(R2-z2)^dz isJ 0z2(R2-z2)&dz = [- ^(R2- z2)^ + ^^(R2- z2)^ ♦ g sin"1 |]Q

0

= 0 + 0 + sinT^(l) + 0 + 0 + 0

m 4 = U “

Hence Equation 25 becomesM = ^ZiEjlhR4 ..................(26)4S1r

When the upper limit of the integral is different from R, for convenience, the Simpson's j law is applied, that is

f f(x)dz = j [ fQ+ 4f^+ 2f2+ 4f3+ .... + 2fn.2+ 4fn_1+ fn] . . (27) J a

where h = — ” and n is an even integer#

19

In this problem, the radius R is divided into six equal parts, i.e., seven stations, therefore n= 6 and h= g'.

F(z) = z2(R2- z2)^

and use the following table to compute the value of this integral.

zi F(z1) = z2(R2- z2) N V N x F(zi)0 0 1 c„= 0

I(35)* r3 216 R 4

0=1= (35):

54 R3

2R J8jL r3 2 c = (32)* R36 27 2 27]R _ 0 ± r3 4 c_= (2)* R36 8 3 2 ,4R Jtli)!. r3 2 C,= 8(5)* R36 27 4 27

iR R3 4 Cc= 2?(11)i R36 216 5 54

R 0 1 c6= 0

HencefR . /RJ z2(R2- z2)?dz = — x ( ( + C2+ C3+ Cj) = C

and Equation 25 can be represented by

4aM = x C ..........................(28)rSn

^aSQ 2. If 6 - a < 6 < a, then the stresses are

20

zoo = £E = y*P* for e,- a ^ e

re 1and — e_ — a)

0 = cy.P^1_ - - - --- ] for £r a< e

Then equation 24 becomes

M = 4 / 0 -p^-zCR2- z2)*zdz + 4 f „°y.p.[i _ (f ~ el~ a)2-i0 £lr J z0 4ae1 J x

(R^- z^)^zdz (29)

Now by setting

f° Z2(R2- z2)2dz = I*0 f(z)dz ..................... (30)J 0 J 0

f F(z)dz = [ ^ 4 + 2Ul± ± ) 22_ ( a)2z](R2. ^ (3l)Jzo Jzo r r

By Simpson's i law, the radius R is divided into six equal parts,

hence n= 6 and h =? -^for the first integration and hg= ZQ for

the second integration.

Setting

i.e,Bi= Nix f(zi) where i is indexed from 0 to 6

Bq= 1 x 0 = 0

%!= 4 x f(-~)

21

B2= 2 x f(^)

B3= 4 x f(^)

B4= 2 x f(£!2)

B5= 4 x f(^2)

B6= 1 x f(»o)

then Equation 30 can be represented by

f °(R2- zZ) h 2dz = ^;(B + B + B + B. + B > B,) = B . . (32)J 0 3 1 2 3 4 5 0

Also setting D^= x F(z^) where i is indexed from 0 to 6

i.e. Dq= 1 x F(Zq )

D1= 4 x F(zq+ h2)

D2= 2 x F(z0+2h2)

D^= 4 x F(zo+ 3h2)

D4= 2 x F(zo+ 4h2)

D^= 4 x F(zo+ 5h2)

D6= 1 x F(z0+ 6h2)= F(R)

then Equation 31 becomes

j F(z)6z = ^ ( D q+ D1+ D2+ D-+ D4+ D,+ D6) = D . . (33)z0 3

22

Hence Equation 29 can be simplified as

M = — x B + — x D ................... (34)rsl ae!

where B and D are given by the Equations 32 and 33*

Case 3. If e ^ e^- a, setting ZQ=r(e^- a) and z^=r(e^+ a), then the corresponding stresses are

zaa = ■ ' for zgz

rel °

a = — ^^•[4a61- (-p - e^- a)2] for zQ< z< z^4a i

and o = o for z >zny.p. 1

Equation 24 becomes

M = 4 T CTy»P. z^(R^- z2)2dz V0

+ 4a6 -[ f - (e1+ a)]2]-(R2- z2)zzdzvzn 4ae1 1 r

+ 4 / o,r „ (R2- z2)2zdz

4o re

4

I*0 z2(R2- z2)2"dz +ei J q

23

+ rZl[- 4 + 2z2(61+ a) - ( e,- a)2z].(R2-z2)*dzae 1 J z0 r2 r 1

+ 5 ay.P.(R2- 4 ^ (35>

Simpson’s law is applied again by dividing ZqZ^ into six equal parts, therefore n= 6 and h^= (z- - Zq)/6, and setting Gi= N^x H(z^) where

H(z) = [- % + — — — z2 - ( e1- a)2z]e(R2- z2)z r r -L

H6nCe G0= 1 x H(z0)

G^= 4 x H(zq+ h^)

G2= 2 x H(zq+ 2h^)

G^= 4 x H(zq+ 3h^)

G^= 2 x H(zQt 4h^)

G^= 4 x H(zQ+ 5h^)

and G^= 1 x H(zq+ 6h^) = H(z^)

therefore the integral in Equation 35 can be represented as follow

A [ _ 4 z2 - (6!- a)2z].(R2_ z2 ) M z

Jz0 r

= [Go+ G l+ G 2’*’G3+ G 4+ G 5+ G 6^ = G ............... ^6)

24

The Equation 35 becomes

M = -— ^ L p x B + ^ x G + zi2)2 ] • • • • (37)

where B and G are represented by Equation 32 and 36,

respectively.

Pago 4. - Perfect plastic case

When the bending moment increases and approaches the maximum

value, the bending moment becomes

W - 4 - x °y.p.x

= J R3(Jy.p. ........................(38)

Using Equations 27, 34, and 37, the relation between bending moment and curvature for a circular cross section with a radius equal to 1 inch can be obtained. For the perfect plastic case, by

dropping the second term of Equation 37 arid setting a= 0 and

replacing Zq= £^r, then Equation 27 and 37 give the relation between bending moment and curvature for the perfect plastic case as shown

in Figure 8.

moment

(10

in-lb)

25

R~ 1.0 in. ay.p.= 53000 psie = 0.0031 in.

a = 0.0019 in.

8/maximum bending moment

6

(T) nonlinear case I- I

(2) perfect plastic case

2

0 1410 121 —3 1curvature ~ (10" in.” )

Figure 8 Relation between bending moment and curvature for a circular ' cross section ' \ "

r

26

3A - Moment-Curvature Relation for a Beam with Wide Flango Cross Section

The last example beam treated here is a beam having a wide flange cross section as shown in Figure 9*

h hl

b

z&dz_r__<•— b-j_

N.A.

Figure 9

From Equation 7

M = I azdAA

In the web, dA= b- dz while in the flange portions, dAs= bdz. Therefore

= I azdA + IJ ’LTtiH * f

M = I azdA + I azdAweb ^ flange

ih- r'ih= 2 azb^dz + 2 azbdz

0(39)

27

Case 1. If e e ^ - a, then the Hooke's law is valid and the bending moment is

|hi -|-hM = 2 / z2b d2 2 f z2bdz

J 0 Elr 1 4^ V

Carrying out the integration and simplifying, the following equation

is obtained:

M = ”l § ^ [ bh3-(b-bl)hl3] (40)

Case 2. If e . a< e 1 < 6 -h a, there are several cases for a 1 fh 1

different value of Zq= ( e - a)r,

(1)# if z0> jh^, then the bending moment becomes"ll

M = 2 f ^ y .-P-r.K z2dz + 2 f ° — bz2dzJ 0 1 J 6lr,ih a

* 2f " 61" a)2]zbdzZq 1

After carrying out the integration and simplifying, this becomes

°*' - - - --•ir-M = U ^ 8b*03-(b- bl )hl3] + 1 2 8 % ' x

[- h* + l6zg^+ e^+ a)r(h3_8Zg^)- 8(s^- a)2r2(h2-4z02)]

(2). if Zq = -|h , this is a special case of case (1), and the

equation may be obtained by setting z^^h^ into Equation 41.

. (41)

28

(3) • if Zq< jh-jL* then the bending moment becomes

M = 2 f — /-*-P-*.b-[ z^dzj 0 6ir 1

+ 2 f (— - e - a)^]b_zdzJ z0 4asl 1 1

+ 2 I ~ *1" a 2] bzdz^ ™ 1

After carrying out the integration and simplifying, this bocomos

M = [fbi^o3 + 12ga— { *bh4+ (b- bl)hl4 + l6blz04

+ e^+ a)r[bh^-(b- b^)h^- Gb^z^]

— 8( 6^— a) r [bh —(b— b^)h^ — ^b^z^^J}1 • • • (42)

Case 3. If ei ^ e + a, there are several cases for aah 1

different value of z^ and z^= ( a)r,

(!)• if z^> yh^ and Zq < Jh^, then the bending moment becomes

ih.M " 2/o 6 rP* blz2dz + 2 I Zo - s1-a)23b1zdz

+ 2 / 1, 4ae’P ' a V (f'ei-a 2 bzdz + 2 f Z ± 1

iha,, _ bzdz

29

After carrying out the integration and -simplifying, the equation

becomes

M = — | b1z03 + 12gar { (b- b;L)h14 - l6(bz1<4'-b1z04)

+ ■"( a)r[-(b- + 8(bz^- b^ZQ^)] - 8( 6^-a) x

r^[-(b- b^)h^^ + 4(bz^- b^z^^)j|] + — (h -4z^) . • (43)

M =

(2 ) • if z1 >jh1 and Z q > fh^, then the equation becomes

= 2 f2 b-,z2dz + 2 f'°— ^ r bz2dzJ o n r 1 J&hi *1^

+ 2f - e1- a)2Jbzdz + 2/ z 0y.p.bzdz

(3). if z^< fh^ and z^< |h^f then the equation becomes

M = 2 f ^ . b i z ^ d z + 2 T y • P *[4ae-i-(~ - 6 - a)Jb :J o s i r J 4ael 1 r 1 1

zdz4ae-1 i 'r 1 '" ]'0

ih-, &h+ 2 / 7 1 ° y . p . V dz + 2 / ih °y.p.bzdzv Z- v 2nq

After carrying out the integration, the equation becomes

a.M = I z03 + - ^ 1-3(Z14-Z04) + 8( «i+a)r(Y-zo3)

- 6( e^- a)2r2(z12_z02)]] + bh2_(b_b^)h^2_4b^z^2] . (i

30

Case 4. - Perfect plastic case

Before determining the maximum bending moment, the distance from the centroidal axis of half wide flange cross section to the edge of the web must be found first, i.e. e, as shown in Figure 10.

" T

Figure 10

b(h2- h^2) + b1hT 6 4[b(h -h1)+ b^h^]

Therefore the maximum bending moment is

“max” 2 [b (h-hl) + blhl3 x °y.p.X 25

= [b(h2- h^2) + b ^ 2]

Using Equations 40, 41, 42, 43, 44, and 45, again the relation between bending moment and curvature for a wide flange cross section

will be developed. In Figure 11, an 8WF1? cross section was used for this example and the relation between the bending moment and curvature

for a case of perfect plastic behavior was obtained.

moment

(10

ft-lb)

31

8WF1?h= 8.0 in. ay.p.= 53000 psi

7.384 in. e^= 0,0031 in.b = 5.250 in. a = 0.0019 in.b^= 0.230 in.

8maximum bending moment

6

41 nonlinear caseI I2 perfect plastic case

2

04 8 12 16 20 24 28

curvature — (10** ft.“^)K

Figure 11 Relation between bending moment and curvature for a wide flange cross section

CHAPTER 4

APPLICATION OF MOMENT-CURVATURE RELATION TO DETERMINE THE DEFLECTION OF A BEAM OF NONLINEAR MATERIAL

4,1 - Slope and Deflection Due to Bonding

The basic relationship between curvature and slope is

For initially straight beams with small deflections, the original length dx is very nearly equal to the curved length ds; consequently no appreciable error is introduced by using dx instead of ds for originally straight beams. Therefore Equation 46 becomes

d l = 1 (47)dx rFor a finite element dx, the corresponding change of slope is found

from Equation 47

change of slope between any two points A and B is found by summation of all the elements.

< & = i (46)ds rwhere 0 is in radians

(48)

If a diagram for curvature, against the length, x, is plotted then Equation 48 is an element of the area under the ~ curve. The total

(49)

32

33

This can be stated as■i

$Ab= area of “ curve between x= A and xs= B.

where is the change of slope between A and B and ismeasured in radians,

1r

Figure 12 Curvature of a beam along axis

To determine the values of the slope along a beam, it is necessary to select a starting point, say the left end point, and then perform the above operation for various values of x. Ety this

procedure, a curve for the slope 0 against x, as shown in Figure 13 may be obtained.

34

rad.

x

Figure 13 Slope plotted over length

Having determined the slope at every point along the axis, the

deflection then can be obtained. Slope is defined as rate of change

of deflection, i.e.

tan t* = - l r

For a flat curve, it is sufficiently accurate to assume that

the tangent of the angle $ equals the angle $ in radians.

tan 0 = 0 = -ffr- (50)

Over an element length dx, the change of deflection dy is

dy = 0 dx ............................(51)

35

This represents the area of an element strip as shown in Figure 13. The total deflection between the points A and B is found by summation of these elemental areas.

yAB= ^ a 61 ^ <52)

This can be stated as

= area under 0 curve between x= A and x= B.

Q)C l,O•H axis of beamm

s..ope

dx

figure 14 Relation between slope and deflection

The base line is tangent to the deflection curve at the point which was used as a base in determining the 0 curve. This must be true because at this point the value of 0 is zero. The deflection is measured normal to this base line.

36

4.2 - Application to a Statically Determinate Problem

Example Is Determine the deflection of a cantilever beam with rectangular cross section (1.5 x 2.0 in.^) having stress and strain relation as shown in Figure 1, subject to the end load of 3000 lbs as shown in

Figure 15.3000 lbs 1.5”

!

72.0"

L= 2 ft.

Figure 15<

2For a 1.5 x 2.0 in. cross section and a stress-strainrelation as shown in Figure 1, the relationship given Figure 5 canbe applied directly. To accomplish this, the beam was divided into24 equal parts with cbc= 1 inch and the moment at each station was

computed. From Figure 5> the curvature j corresponding to eachmoment at each station along the length may be found. Thereforea curve of curvature 1 versus the length x as shown in Figure 16arcan be plotted. Using the numerical integration method of Equation 49, the slope of every station along the length was computed and a diagram of slope against the length as shown in Figure 16b was

constructed. Using Equation 52, the deflection at every point of

37

the beam as shown in Figure 16c was obtained. For convenience 9 all these processes can be tabulated as shown in Table 1.

Table 1

station moment M c u r v a t u r e s l o p e $ deflectionx in..

-410 in-lb £ lO^inT1 10~2 rad. y 10"^in.

24 7.2 6.83 0 023 6.9 5.78 0.63 0.0322 6.6 5.09 1.17 0.1221 6.3 4.59 1.66 0.2620 6.0 4.19 2,10 0.4519 5.7 3.85 2.50 . 0.6818 5.4 3.56 2.87 0.9517 5.1 3.29 3.21 1.2516 4.8 3.04 3.53 1.5915 4.5 2.81 3.82 1.9614 4,2 2,58 4.09 2.3513 3.9 2,36 4.34 2.7812 . 3.6 2.15 4.56 3.2211 3.3 1.95 4.77 ' 3.6910 3.0 1.76 4.95 4.179 2.7 1.57 5.12 4.68

. 8 2.4 1.40 5.27 5.207 2.1 1.23 5,40 5.736 1,8 1,05 5.52 6.285 1.5 0,88 5.61 6.834 1.2 0.70 5.69 7.403 0.9 0.53 5.75 7.972 0.6 0.35 5.80 8.551 0.3 0.18 5.82 9.130 0 0 5.83 9.71

383000 lb

L - 24 ft

curv.510

in.

124 15 5 01020

(a) curvature

slope10“2

3

1

(b) slope

defl.10”1in.

10(c) deflection

Figure 16

39

In Table 1, the stations are arranged from x= 24 to x=0. This is because both slope and curvature should be integrated from the fixed end to the free end.

For convenience and more accuracy, the digital Computer may be used to do this computation. First,however, the proper mathematical expressions for a curve shown in Figure 5 must be found, that is, the curvature must be expressed as a function of bending moment.

The curve obtained in Figure 5 for a rectangular cross section can be expressed by the following equations:

(a). If M <20516.128 in-lb, then

Curvature = ~ = 0.5849057 x 10 ^ M . . . . . . . . (54)

(b). if 20516.128< M <65486.793 in-lb, then

i = 3.0522772 x 10~26x M5 - 5.6615856 x 10 21xr+ 4.1213191 x 10*l6x w 3 - 1.4201007 x 10-U,x M

+ 2.8930082 x 10-7x M - 1.4248202 x 10-3 . . . (55)

(c). if 65486.793 in-lb, then

1 /~0.35033r “ / “max' M

where 79500 in-lb

Using these equations, the curvature, slope, and deflection at

any point of the cantilever beam can be obtained. A Fortran computer

program to compute these values is shown in appendix B.

40

4.3 - Application to a Statically Indeterminate Problem

Figure 17 represents a beam fully restrained at A and simply supported at B. The deflection curve of beam is skitched as shown by the dotted line.

M-RL ' Al%R

Figure 1?

For a linear material, the reaction force R equals but

for a nonlinear material beam the reaction force can not be found directly. The problem is statically indeterminate. Different values of bending moment will be considered. The procedure to handle this kind of problem is shown as below.

First, assuming that the reaction force R is known, for3Mconvenience say R= ^ , then the moment diagram along the length of

the beam can be sketched. Next, by Figure 11, the relation between bending moment and curvature of the wide flange cross section is

known and so the curvatures at every point of the beam can be found. Do the same process as the former example of a statically determinate

cantilever beam. Then the deflected shape of the beam corresponding • to the assumed value of the reaction force R may be obtained. The

41

deflection at B should be zero, but the deflection actually obtainedis different from zero, say y^. Ety ijusting the value of R andrepeating the computational process, another value of deflection atB would be obtained, say y^. By connecting the points and Pg andextending the straight line to intersect the R axis at R^ as shown inFigure 18, the value R^ was picked as an assumed reaction force and

the process was repeated again.The full process is repeated until the deflection at right

end B is essentially equal to zero, say < 0.001. The last assumedy|L

value of reaction force R and its corresponding deflection shape are thus the value of the reaction force and deflected shape of this

statically indeterminate beam.

y

Figure 18

42

Before applying Figure 11, the relation between bending moment and curvature for a wide flange cross section, a suitable mathematical expression to express the curvature as a function of bending moment must be found first.

The curve plotted in Figure 11 can be expressed by the

following equations:(a). If 23755.332 ft-lb, then

1 = 0.15154492 x 10-7 M ........................(57)r

(b). if 23755.332<M <65808.156 ft-lb, then

1 = 0.28780133 x 10'24x M5 - 0.54619712 x lo'^x M r

+ 0.40398149 x 10 ^ x M - 0.14389556 x 10 x M2

+ 0.26194906 x 10-5x M - 0.16363522 x 10-1 . . (58)

(e). if M>65808.156 ft-lb, then

1 = / 0.6462763 r / Mmax-M

where 6878I ft-lb

Some numerical results have been obtained by the foregoing

procedure on an IBM digital computer for a beam clamped at one endand simply supported at the other end in which L= 10 ft. with BWFl?

cross section having a stress-strain relation as shown in Figure 1.The results are shown in Figure 19» 20, 21 and Table 2.

43

Table 2

bending moment (ft-lb)..

reaction of - linear case

(lb) ...reaction of non­linear case

(lb)ratio

slope at simple supported end (10”2 rad.)

23755 3563 3563 1.0000 0.9000. 25000 3750 3751 1.0003 0.945630000 4500 4502 1.0005 1.134135000 5250 5253.3 1.0007 1.326540000 6OOO 6005.5 1.0009 1.531145000 . 6750 6760 1.0015 1.747550000 7500 7519 1.0025 1.971855000 8250 8281 1.0038 2.206860000 9000 9047 1.0052 2.467365000 9750 . 9820 1.0072 2.790966000 9900 9976 1.0077 2.870467000 10050 10135 1.0085 3.023068000 10200 10303 1.0100 3.3203 .

68700 10305 10483 1.0163 4.8233

In Figure 19» it can be seen that the reaction of this nonlinear beam when MXMg^-a (i.e. stress-strain relation follows the Hooke9s law) is just the same as the linear ease. When M gradually increases, the ratio of the reaction of nonlinear beam to the reaction of linear beam (^) of the same type and same dimensions increases slightly as shown in Table 2. When M becomes large and approaches the maximum value, the intersection of bending moment diagram and horizontal axis moves inward. This says that a strain regression has occurred. If the bending moment keeps increasing, there will be a further movement until the maximum bending moment is reached and the beam fails.

44

Several values of slope at the simply supported end corres­ponding to various values of bending moment are computed and shown in Table 2„ Using table 29 the diagram of bending moment against slope can be plotted as shown in Figure 20. From Figure 20 9 it may be seen that when the linear relation between the stress and strain is held, the slope at the simply supported end is proportional to the bending moment as shown by the straight line OC. Beyond the proportional limit, the rate of change of the slope increases gradually as the moment increases| when the bending moment exceeds about three-quarters

of the ultimate value, the rate of change in slope rapidly becomes

greater.It may be imagined, when the linear relation between stress

and strain are held, the deflection of the beam is proportional to the bending moment and the point of maximum deflection remains at the same place when the bending moment increases. When the stress-strain relation is no longer linear, the family of deflection curves for various values of bending moment are shown in Figure 21. From this figure, it is clear that the rate of change of deflection increases, the deflection shape becomes steeper, and the point of maximum deflection moves to the right (moves toward the simply supported end) as the bending moment increases. A normalized deflection shape of linear and nonlinear eases is shown in Figure 22.

reaction l(r

lb45

10'68?00

25000moment 10 ft-lb

12510

36130moment at fixed end

ft-lbmoment at simply support end

12510 2500015020 3000017535 3500020055 4000022600 4500025190 5000027810 5500030470 6000033200 6500036130 58700

Figure 19 Bending moment diagram for different valuesend moment

moment

10 ft-lb

46

8WF17A

10 ft.

8

6

4

2

0_ 2

slope 10 rad.

Figure 20 Relation between bending momentand slope at simply supported end

s

Moment ft-lb1 237552 300003 400004 500005 600006 650007 680008 68700

5.5

~nI

Figure 21 Moment-dofloction diagram

- >■ ■: ..

0.2

0.4

0.60.8

1.0

1 : linear case2 : nonlinear case

Figure 22 Normalized deflection shapes of linear and nonlinear cases

CD

49

CHAPTER 5

CONCLUSION

The procedure and the computer program used in this thesis can be extended to other types of cross section and any other stress- strain diagram or diagrams„

The ratio of the reaction of a nonlinear proped cantilever beam to the reaction of a linear beam increases slightly ( about l-§$) as the bending moment increases beyond the moment which corresponds to the limit of the linear relation between stress and strain (i.e. proportional limit of linear range). When the bending'moment increases and approaches the maximum value, the intersection of bending moment diagram and horizontal axis moves inward. This says that a strain regression has occurred.

When the linear relation between stress and strain are held, the deflection of the beam is proportional to the bending moment and the point of maximum deflection remains at the same place when the bending moment increases; when the stress-strain relation is no longer linear, the rate of change of deflection increases, the deflection shape becomes steeper, and the point of maximum deflection moves to the right (moves toward the simply supported end) as the bending moment increases,

50

REFERENCES

lo Timoshenko, Stephen P= and Gere, James M«

"Theory of Elastic Stability", McGRASJ-HlLL, second edition, 196lo

2» Shanley, F, Re, "Strength of Material", McGRAW-HILL, 1957.

3. Malvick, Allan Je and Lawrence, He N» Lee"Buckling Behavior of an Inelastic Column", Journal of Engineering Mechanics Division, ASCE Vol. 91, No, EM3 proc, paper 4372, June, 1965, pp 113-127.

4. Johnston, Bruce G., "Buckling Behavior above the Tangent Modulus Load", Journal of Engineering Mechanics Division, ASCE Vol..87,No. EM6 proc. paper 3019, December, 196l, pp 79-99.

5. Salvador and Baron"Numerical Methods in Engineering", Prentiee-Hall, second edition, 1961.

51

APPENDIX A

NOTATION

The following syribbls have been adopted for use in this thesis:

a One half of the strain range which represents the-parabolic portion of stress-strain diagram,

b Width of the cross section,b]_ Width of the web!.

Notation used to evaluate the value of integration 1 , by Simpson’s law,E Modulus of elasticity in tension and compression, • .h Depth of cross section,h^ Depth of web of wide flange cross section.I Moment of inertia with respect to x-axis,L Length of the beam.M Bending moment.Mmax Maximum bending moment, 'M^_, a Bending moment corresponding to e = e^- a.R Radius of circular cross section, or reaction of the beam.r Radius of curvature.y Deflection of the beam.y^g Total deflection between points A and B,z Distance from the neutral axis.Zq Distance from the neutral axis to z = ( a ) r .z^ Distance from the neutral axis to z=(e^+ a)r.

s Strain,Strain at the limit of linear stress-strain relation. Strain at the edge of convex side of cross section,

s Strain at the edge of concave side of cross section, a Stress,a Yield point stress,y,p,A= 6,+ 8 t c

d0 Rate of change of slope,ds Arc length of infinitesimal element,0£B Slope change between A and B.

53

APPENDIX B

FRAGMENT OF COMPUTER PROGRAM

HSU 2R# COMPILE FORTRAN9 EXECUTE FORTRANC DETERMINATION OF THE MOMENT-CURVATURE RELATION FOR A BEAM OF 'C NONLINEAR MATERIAL WITH RECTANGULAR CROSS SECTION,, EQUATIONSC 14» 15, AND 16 OF THIS THESIS WERE USED*C SIGMA= YIELD POINT STRESS, B= WIDTH, H= DEPTH*C ALPHA= STRAIN AT THE LIMIT OF LINEAR STRESS-STRAIN RELATIONC D= ONE HALF OF THE STRAIN RANGE WHICH REPRESENTS THE C PARABOLIC PORTION OF STRESS-STRAIN DIAGRAM®.C EPS'IL= STRAIN AT THE EDGE OF CROSS SECTION® '

99 READ 100, 200, R100 FORMAT (FlOeOS■ ' - '

SIGMA= 53000*0ALPHA= 0*0031 B= 1*5H= 2*0 „• •D= 0*0019 CURVA = 1*/R EPSIL = H*0URVA/2®

C START OF COMPUTING MOMENT-CURVATURE RELATION®' IF (EPSIL - (ALPHA - D)) 101, 101, 102

101 XM= SIGMA*B*H**3*CURVA/(12®*ALPHA) - PRINT 201 , CURVA, XM

201 FORMAT (3X, 7H CURVA= 1PE20®8, 4X, 4H XM= E25®8)GO TO 99

102 IF (EPSIL - (ALPHA+D)) 103, 103, 104 1030XM= <B'*SIGMA*CURVA**2*( 16»*R**4*<ALPHA~D)**4

1 -24®*R**2*H**2*(ALPHA-D)**2 t 16®*R*H**32 *(ALPHA + D)-3®*H**4))/(384®*ALPHA*D}PRINT 201, CURVA, XMGO TO 99

1040XM' = -B*R**-2*SIGMA*(ALPKA**2*D-**2)/3»1 +B*H**2*SIGMA/4»PRINT 201, CURVA, XM GO TO 99

200 STOP END

*** HSU 2RCOMPILE FORTRAN, EXECUTE FORTRAN

C FOR A CIRCULAR CROSS SECTION, SIMPSON6S 1/3 LAW WAS USED*C EQUATIONS 27, 34, AND 37 OF THIS THESIS WERE USED®

100 READ 122, 500, BATA

54

122 FORMAT {FlOoO)5IGMA= 5 3000 o 0 ALPHA= 0,0031 D= 0,0019 CURVA= 1 ,/BATA R= 1,0ESP IL= R*CURVA P= ALPHA-D Q= ALPHA^D

C START OF COMPUTING MOMENT-CURVATURE RELATION,IF (ESPIL-P) 200f 200* 201

200 XM= 0*7853982*5IGMA*R**4/(ALPHA*BATA;PRINT 301 * CURVA* XM

301 FORMAT (4X, 7H CURVA- 1PE20,89 4X» 4H XM= £20,8)GO TO 100

• 201 Z0= P*BATA Zl= Q*BATABi= Z0**2/54,*SQRTF(36,*R**2-Z0**2)82= 4,*Z0**2/54,*SQRTF(9,*R**2-Z0**2)B3= Z0**2/2,*SQRTF(4»*R**2-Z0**2)64= 8,*Z0**2/27»*SQRTF(9,*R**2-4,*Z0**2)B5= 25,*Z0**2/54,*SQRTF(36,*R**2-25,*Z0*S2)66= Z0**2*SQRTF(R**2-Z0**2) " -SUMB= ZO/18,*(614-82+83+64+85+86)IF (ESPIL - Q) 400* 400*401

400 X= (R-ZO)/6»0C0= (-Z0**3/(BATA)**2+2,*Q*Z0**2/BATA-P**2*Z0)*SQRTF(R**2 1 -Z0**2)0C1= 4,*(-(ZO+X)**3/(BATA)**2+2,*0*(ZO+X)**2/BATA-P**2*1 (ZO+X))*SQRTF(R**2-(ZO+X)**2) -0C2= 2,*(“ <Z0+2,*X)**3/(8ATA)**2+2,*Q*{Z0+2,*X)**2/BATA ' 1 ~P**2*(Z0+29*X5 )*SQRTF(R**2--(Z0+2,*X )**2 5- 0C3= 4o*( — (Z0+3,*X)**3/(BATA)**2+2o*Q*(ZO+3«*X)**2/BATA 1 ~P**2*(Z0+3,*X} 5*SQRTF(R**2~(Z0+3«*X )**2S.0C4= 2,*(-(Z0+4,*X)**3/(BATA)**2+2,*Q*(Z0+4,*X)**2/BATA 1 -P**2*(Z0+4,*X))*SQRTF(R**2-(Z0+4»*X)**2)0C5= 4,*(-{ZO+5,*X)**3/(BATA)**2+2,*Q*(Z0+5,*X)**2/BATA

1 i -P**2*(Z0+5o*X))*SQRTF(R**2”(Z0+5»*X)**2)SUMC= X/3 ,* (C0+C1+C2+C3+.C4+C5 )XM= 4a*SlGMA*SUMB/(ALPHA*BATA5 + SIGMA*SUMC/(D*ALPHA) PRINT 301 * CURVA* XM

■ GO TO 100 ,401 Y= (Zl-ZO)/6,

0D0=(-20**3/(BATA)**2+2,*Z0**2*Q/BATA-P**2*Z0)*SQRTF(R**2 1 -20**2)0D1= 4,*(-(ZO+Y > **3/(BATA)**2+2 **Q*(ZO+Y)**2/BATA~P**2*1 (ZO+Y))*SQRTF(R**2~(ZO+Y)**2)0D2= 2 o*{-(20+2,*Y)**3/(8,. fA)**2 + 2,*Q*(ZO+2,*Y)**2/BATA 1 —P**2*(Z0+2 ©*Y))*SQRTF(R**2~(ZO+2 ®*Y)**2)0D3= 4,*(-(ZO+3,*Y)**3/(BATA)**2+2,*Q*<Z0+3»*Y)**2/BATA 1 -P**2*(Z0+3e*Y))*SQRTF(R**2"(Z0+3,*Y}**2)

' 0D4= 2 »*(-(Z0+4,*Y 5**3/ (BATA)**2+2,*Q*(Z0+4®*Y)**2/BATA

55

1 -P**2* ( ZO-Ho*Y) )*SQRTF(R**2-(Z0+4a*Y)**2 >0 D 5 = 4.*(-(Z0+5«*Y)**3/(BATA)**2+2.*Q*(Z0 + 5**Y)**2/BATA1 _p**2*(Z0+5.*Y))*SQRTF(R**2-(Z0+5a*Y)**2)0D6= (-Z1**3/(BATA)**2+2,*Q*Z1**2/BATA-P**2*Z1)*SQRTF(R'1 • **2-Zl**2)SUMD= Y/3e*(DQ+Dl+0-2+D3+D4+D5+D6)OXM= SIGMA/ALPHA*( .•*SUMB/BATA+1*/0*SUMD +4.*ALPHA*R**3/3 1 «-SQRTF( ( le-<Zl/R)**2S'K-«-3> >PRINT 301e CURVA9 XM GO TO 100

500 STOP END '

HSU 2R* COMPILE FORTRAN9 EXECUTE FORTRANC FOR A WIDE FLANGE CROSS SECTION.* EQUATIONS 409 41* 42* 43*C 44* AND 45 WERE USED®

100 READ 12 9 500* B i 619 H* HI 9 R 12 FORMAT (KFlOaO)

ALPHA= 0=0031D= 0.-0019 " ‘ ,• •SIGMA= 53000*0 CURVA = 12 «/R EPS IL= H/C2.*R)P= ALPHA-D 0= ALPHA-J-D

C START OF COMPUTING MOMENT-CURVATURE RELATION*IF (EPSIL-P) 14* 14 9 15

14 XM=(SIGMA*(B*H**3-(B-Bl)*H1**3)/(12»*ALPHA*R))/12.PRINT 209 CURVA9 XM

20 FORMAT (4X9 6HCURVA” IPEZOeB* 4X* 3HXM= £20 = 85 GO TO 100 "

15 Z0= P*R21= Q*R 'IF (EPSIL-Q) 24* 24* 25

, 24 IF (ZO - HI/2e» 5 34* 35* 35340XM=(SIGMA/(ALPHA*R)*(2 = *B1*20**3/3.+1./1128.*D*R)* f-B*1 H**4+(B-B1)*H1**4+16.*61*20**4+16./3a*Q*R*(B*H**3-2 CB-Bl5 *H1**3-8»*81*20**35-8.*P**2*R**2*(B*H**2-(B-3 B 1 5 *H1**2—4 e * B l*Z0**2)5))/12 e ■ >

PRINT 209 CURVA9 XM .GO TO 100

350XM=(SIGMA/i12 « *ALPHA*R)*(8« *B*Z0**3-(B-Bl5 *H1**3)+B*SIGMA1 /M28.*0*ALPHA*R**2>*(-H**4+16.*20**4+16./3.*0*R*(2 H**3-8«*Z0**3} —8.*P**2*R**2*(H**2—4.*Z0**25 > 5/12o PRINT 209 CURVA9 'XMGO TO 100

25 IF (Z0-H1/2*) 54» 55g 55 54 IF (Zl-Hl/2.) 649 65e 65640XM=(B1*SIGMA/(ALPHA*R)*(2.*Z0**3/36+l«/(24»*D*R)*(-3.*

V V U V V V

V

56

1 (Z1**4-ZO**4)H-8<>*Q*R#<Z1**3-ZO**3}-6«*P**2*R**2#(2 Zl**2-Z0**2)>)+SlGMA/4.*(B*H**2-(B-Bl)*Hl**2-4»*3 B1*Z1**2))/12 e ■PRINT 20, CURVA» XM GO TO 100

650XMs(SlGMA/(ALPHA*R)*(2**Bl*Z0**3/3*+l*/(128,*D*R) *((B1 -B1)*H1**4-16.*{B*Z1**4-B1*Z'0**4) + 16*/3**Q*R*(-(B-2 81)*Hl*43+a0*<B*Z1**3~B1*Z0#*3))-8.*P**2*R**2*(-(3 B-B1)*H1**2+4.*(B*Z1**2-B1*Z0**2))))+B*SIGMA/4.*(4 H**2-4.*Z1*#2>)/12*PRINT 20, CURVA, XM GO TO 100'

550XM=(SIGMA/(12.*ALPHA*R)*(8.*B*Z0**3-(B-B1)*H1**3)+B*SIGMA1 /(24**D*ALPHA*R**2)*(-3«*(Zl**4-Z0**45 + 8 • ( Z12 **3-Z0**3)-6e*P**2*R**2*(Zl**2-Z0**2))+B*SIGMA/4*^3 ■' (H**2-4.*Z1**2)>/12e . PRINT 20, CURVA, XMGO TO 100•

500 STOPEND • ,

*#*■ HSU ■ " ’ • 2R* COMPILE FORTRAN, EXECUTE FORTRAN

DETERMINATION OF THE CURVATURE, SLOPE, AND DEFLECTION OF A CANTILEVER BEAM WITH RECTANGULAR CROSS SECTION, NUMERICAL INTEGRATION METHOD WAS USED®DX= LENGTH BETWEEN- TWO STATIONS ALONG THE AXIS®E= INVERSE OF MODULUS OF ELASTICITY®CA, CB» CC, CD, CE, AND CF ARE THE COEFFICIENTS OF EQUATION 55®DIMENSION XM(26)» CURV(26), SLOP(26),'DEFL(26)READ 100, (XM(I)» 1= 1, 25)

100 FORMAT (8F10e0)DX- 1*0E= 0,00000005849075 CA= 3 ,05 22 772*10®** <-.26 )CB= ~5,6615856*10,**(”21)CC= 4®1213191*10,**(~16)CD= ~1,4201007*10,**(”11)

• CE= 2,8930082*10®**(-7)CF= -1,4248202*10,**(-3)

C FIND CURVATUREDO 500 1= 2, 4

500 CURV(I)= SQRTF(0,35033/(79500, - XM(I)))DO 400 1= 5, 19

4000CURVCI)= CF +XM(I)*(CE+XM(I)*(CD+XM{I)*(CC+XM{I)*(CB+1 XM(I)*CA))))DO 200 1=209 26 '

200 CURV(I)= E*XM(I)C FIND SLOPE

57

SLOP(2)= 0 o 0 DO 600 J= 3» 26 M = J—1

600 SLOP(J 5 = SLOP(Mi + (CURV(J) + CURV(M))Z2e*DX C FIND DEFLECTION

DEFL(25= 0*0 DO 700 J= 3s. 26 M= J-l

700 DEFL( J>= DEFL(M) + {SLOP (J) -5- SLOP(M) )/2.*DXSTOPEND

HSU 2R* x COMPILE FORTRAN» EXECUTE FORTRANC DETERMINATION OF THE CURVATUREs> SLOPE, AND DEFLECTION OFC A STATICALLY INDETERMINATE BEAM WITH WIDE FLANGE CROSSC • SECTION, NUMERICAL INTEGRATION METHOD WAS USED*C .DX= LENGTH BETWEEN TWO STATIONS ALONG THE AXIS,C E= INVERSE OF MODULUS OF ELASTICITY, " 'C . CA, CB, CC» CD, CE, AND CF ARE THE COEFFICIENTS OF C EQUATION 58*

DIMENSION XM<22), CURV(22), SLOP(22), DEFL(22)99 READ 100, 200, AM, AR

100 FORMAT {2F10,0.) -D X” 0,5XM(2 5= AM - AR*20a*DXXM(3>= AM - AR*19,*DXXM(4>= AM - AR*18.*DXXM(5)= AM - AR*17o*DXXM(6)= AM - AR*16«*DXXM(7 )= AM AR*15«*DXXM(8)= AM - AR*14,*DXXM(9)= AM - AR*13,*DXXM<105= AM - AR*12,*DXXM(11)= AM - AR*11,*DXXM(12)= AM - AR*10«*DX ■XM(13)= AM - AR*9**DX XM(14)= AM - AR*8a*DX XM(15 5= AM - AR*7«*DX XM<16$= AM - AR*6,*DX •XM{173 = AM - AR*5a*DX XM(18 3= AM - AR*4»*DX XM(193= AM - AR*3,*DX

■ XM(203= AM - AR*2,*DX - XM(21)= AM - AR*1«*DX

XM122 3 = AME= 0*00000015154492 ■CA= 0.28780133*10.**(-24)

58

CB= -0.54619712*10.**(-19)CC= 0.40398149*10.**(-14)0D= -0.14389556*10.**(-9)CE= 0.26194906*10.**(-5)CF= -0.16363522*10.**(-!)

C FIND CURVATUREDO 10 1= '2* 15

10 CURVC I-)** E*XM( I )DO 11 1= 16» 22

110CURV( I )= CF +XM ( I )* (CE+XiM ( I )*( CD+XM ( I )*( CCtXM ( I )*( CB+ 1 XM{I)*CA)))) '

C FIND SLOPESLOP(2)= 0.0 DO 12 J= 3» 22 M= J-l

12 SLOP(J)= SLOP(M) + (CURV(J5 + CURV(M))/2»*DXC FIND DEFLECTION

DEFL C 2)= 0.0 DO 13 K= 3 9 22 N= K - 1 ,

13 DEFL(K)= DEFL(N) + (SLOP(K) + SLOP(N))/2a*DX*12«DO 14 1= 2® 22 — • •

14 PRINT 15, XM(I) 9 CURVCI), SLOP(1)» DEFL{I)15 FORMAT (5X» 1P4E20.8)

TESTX= DEFL(22)/DEFL(12)*100.PRINT 16* TESTX

16 FORMAT (5X9 6HTESTX* 1PE20.8)GO TO 99

' 200 STOP •END ' . ,