Location of maximum moments in a continuous two span beam ...

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Masters Theses Student Theses and Dissertations

1966

Location of maximum moments in a continuous two span beam Location of maximum moments in a continuous two span beam

due to moving loads due to moving loads

Thomas Houng-Yn Yang

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LOCATION OF MAXIMUM MOMENTS IN A CONTINUOUS

TWO SPAN BEAM DUE TO MOVING LOADS

by

THOMAS HOUNG-YN YANG

A

THESIS

submitted to the faculty of the

UNIVERSITY OF MISSOURI AT ROLLA

in partial fulfillment of the requirements for the

Degree of

MASTER OF SCIENCE IN CIVIL ENGINEERING

Rolla, Missouri

1966

ii

ABSTRACT

The objective of this investigation is to determine the

positioning of vehicle wheels so as to cause maximum positive

and negative moments in a two span continuous girdero Several

variables are considered, namely, the ratio of loads to each

other, the distance between M1eels, and ratio of one span

length to the other. The necessary equations are derived and

the results, after being programmed on a computer, are shown

in a series of curveso

These curves serve as a tool for the designer, enabling

him to arrive at design moments by simple substitution in the

appropriate moment expression, within the limits on the

variables established for this studyo

iii

ACKNOWLEDGMENTS

The author wishes to thank Dr. W. A. Andrews for his

assistance and guidance throughout the course of this investi­

gation.

The author also wishes to thank the Unites States Govern­

ment for the privilege of pursuing his graduate studies at

the University of Missouri at Rolla.

LIST OF SYMBOLS

The symbols used are defined where they first occur in

the text and are listed here in alphabetical order for con­

venience:

a (or b) Distance from exterior load to end support.

C Ratio of two loads.

d Distance between two loadso

iv

k (or K) Ratio of smaller span length to larger span length.

1 larger span length.

M Bending moment.

P (or W) Concentrated load.

w2 Equivalent uniform loading"

v

LIST OF FIGURES

Figure Page

1. Moment Diagram for Use In Three Moment Theorem,

Using Single Concentrated Load ...•......•........... 5

2. Support Moment: Case I a+d<l .•....................• 6

3. Support Moment: Case II b+d<kl •....•............••• 6

4. Support Moment: Case III a+d>l, a<l .•............•• 7 5. Positive Moment: Case I a+d<l •......•...........•.. 8

6. Positive Moment: Case II b+d<kl... ... . . . . . . . . . . . . . . . 9

7. Positive Moment: Case III (k+l)l>a+d>l ............. 10

8. Location Of "a(b)" For Maximum Negative Moments at

Support B o ••••••• o ••• o •••••••••••••••••• o •••••••• o • • 12

9. Location of Maximum Positive Moments Under Load P •.. 13

10. Location of Maximum Positive Moments Under Load CP .. 14

11. Location of a(or b) For Maximum Negative Moments

At Support B. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

12. Location of M~ximum Positive Moments Under Load P ... 16

13. Location of Maximum Positive Moments Under Load CP •. 17

14. Location of a(or b) For Maximum Negative Moments At

S u pp or t B •••. o ••••••••••••• o • • • • • • • • • • • • • • • • • • • • • • • • 18

15. Location of Maximum Positive Moments Under Load P ... 19

16. Location of Maximum Positive Moments Under Load CP .. 20

17. Influence Line For Support Moment,~···· · ······· · ·· 51

vi

TABLE OF CONTENTS

Page

Abstractooooooo 0 0 • 0 • 0 • 0 0 0 0 0 • • • 0 0 0 0 0 0 0 • 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 • 0 •

Acknowledgmentoo

List

List

of

of

Symbols.

Figures.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 • 0 0 0 • • 0 0 0 • 0 0 0 • • • 0 • • • • • • • • •

• • • 0 0 • • • 0 • • • • • • • • • • • • • • • • • • • • • 0 • • • 0 • • • • 0 •

• • • • • • • • • • • • • • • • • • 0 • • • • • • • • • • 0 • • 0 • • • • • • • •

ii

iii

iv

v

Introduction ..• o. a. o •••••••••••••• o ••• a •••••••••••••• o... 1

Scope •••••••• a •• a • o •••••• a ••••••••• o ••••••••••••••• o •• o • • 3

Solution of Problem. o • • • • 0 • • • • • • • • 0 • • • • • • • 0 • • • • • • • • • 0 • 0 • • 4

Discussion and Conclusions ••••.•••••••• o••••············· 21

Future Investigations •• • 0 • • • • • • 0 • • 0 • • • • • • • • • • • • • • • • • • • • • •

Appendix I Programming •••••• • • • 0 • • • • • • • • • • 0 • • • • • • • • • 0 • 0 • 0

26

27

Appendix II Numerical Check of Support Moment .•••••••••• o 51

Bibliography a •••••••••• o ••••••••••• o •••• a •• o o. a ••••••• o • o 53

Vi t a • o • • • • • a • • • • • a • o • • • • • • • • • • o • • • o • • a • a • • a • • • • o • • • • o • o • a 54

1

INTRODUCTION

In the design of girders subjected to moving live loads,

such as found in highway bridges and railroad bridges, bending

moments are one of the essential factors in determining the

size of beam or girder. From the standpoint of economy, it is

often rnore de~sirable to use a continuous beam rather than one

simply supported beam.

In a simply supported beam, according to the Ministry of

Transport Loading,* the live loading is regarded as equivalent

to a uniform loading w2 per square foot plus a concentrated

knife-edge load W acting at the mid-point of the span for

maximum moment. That is

WL M =

4 (ft-lb) +

But there is no given method specified for determining

the maximum moments for moving concentrated loading. The

designer can adapt the use of influence lines and, by trial

and error, converge upon a reasonable approximation to the

maximum moments. The beam could be investigated at relatively

close intervals, and with the aid of the influence line at

each specific point chosen, a maximum moment can be found and

a curve of approximate maximum moments plotted for the girder.

-1( See A. W. Legat, "Design and Construction of Reinforced

Concrete Bridges," p. 53.

It is the aim of this investigation to present another

approach to design, enabling the engineer in many cases to

establish the critical moments more rapidly and in more a

direct manner.

2

Equations are derived for moment and for positioning of

wheels so as to cause a maximum moment. It will be necessary,

in solving for the absolute maximum moment within a span,

for a given set of parameters, to look at all cases which

can give algebraic maxima and then select the one giving the

numerical maximumo Once "D" wheel positioning is determined

for a given set of parameters, it is a relatively simple

task to substitute all data into the proper moment expression.

A numerical example will illustrate this. Within the limita­

tions described in the next section, SCOPE, the expressions

were derived and then were programmed for the Computer at the

University of Missouri at Rolla. The program itself is shown

in Appendix I. The results were reduced into convenient form

in a number of graphs, shown in Figures 1 through 9, pages 12

through 20"

SCOPE

For practical purposes, the limits of the problem were

set as follows:

1) Two span continuous beam with variations in span

length ratios of 1:1 to 0.5:1.

2) Constant moment of inertia.

3) Each outer end of the continuous beam is simply

supported.

4) Two Wheels are chosen, varying the distance between

them as a function of the length of the longer side from a

minimum of 0.10 (1) to a maximum of 0.30 (1). The ratio of

one wheel load to the other varies from 4:1 to 1:1 to 1/4:1.

For other values in between, a linear interpolation of the

graphs in Figures 1 through 9 is sufficiently accurate for

practical design purposes.

3

SOLUTION OF PROBLEM

In the following seven pages, the basic derivations

are presented. By using the three-moment equations for

elastic beam continuity, expressions are derived for the

unknown moment, MB' over the intermediate support, shown

on page 5.

Then,utilizing 0~ = 0, or0~ = O, one can determine

the positioning of the wheels that gives the maximum moment

or moments. The parameter "a" is used in span 1, and "b 11

in span kl. These derivations are shown on pages 6 through

11. Within the numerical limitation set on the distance

between the wheels, case 3, page 7, for negative moment at

B itself, was found never to govern.

The equations were then programmed on a computer and

the results shown in graph form that is convenient to use,

in Figures 8 through 6, pages 12 through 20. For any given

set of data, the necessary values are picked off the graphs

and resulting design moments are solved by substitution and

elementary statics.

4

Derivation of Support Moment

r --- =?!fc ---- ___ __kl__ __ l

1 (l+a )/~ i ~.· ~ 4 ------1 I ------ ~ ~

F!Go (1) Moment diagram for use in three moment theorem,

using single concentrated load

Load P on span AB only, and setting up three moment

equation, one gets

= -6P(l-a)a 1

Since MA=Mc=O

MB=-P(l2-a2 )a/212 (1+k)

yields

Load P on span BC at Distance b from C only, same as above

M.i~+2M_al (l.+k )-fMc (kl )=-P [{kl )2-b2 ]b/kl

Since MA = MC = 0 1 ~--P [(k1) 2-b2)b/2kl2 (~+k)

5

Letting c2 = l/2kl2 (l+k) Yields

MB = -PC2~kl)2-b2]b ------------(B)

Derivation of Wheel Positioning for Maximum Support

Moment (MB)

1. a+d<l

1c ~ l .

A B · 1-----------·-1_ .. -----------+ _____ .k_l ___ -I I

FIG: (2): Support Moment: Case I a+d<l

Applying equations (A), yields

~ = -PC1 (12 -a2 )a-CPC1 [12-(a+d)2)(a+d) ------(C)

Taking the partial derivative of M_s with respect to "a"

set

;,MB -= <:Ja

0, and one gets

2. b+d<kl

FIG. (3): Support moment: Case II b+d<kl

Applying equative (B), yields

6

~ = -CPC2 [{kl )2-b2 ]b-PC2 rkl )2 - (b+d )2] (b+d) ---(D)

Taking the partial derivative of Ms with respect to "b '~ setting

dMs as- = 0, one gets

2 2 2 3(l+C)b +6db+ 3d -(kl) (l+C) = 0 --------- (2)

3. a+d>l, a<l,

FIG: (4): Support moment: Case III a+d>l, a<l

Applying equation (A) and (B) yields

2 2 [ )2 2J MB = -Pc1(1 -a )a- CPC2 (kl -b b--------- (E)

Since b = 1 (l+k)-(a+d)

MB= -PC1 (12-a2 )a

-CPC2 ((kl) 2-(l+kl-a-df (l+kl-a-d)

Setting

oMs da = 0, and simplifying, yields

3(cc2 -c1 )a2 -scc2 [l(l+k).o;d]a+C112 }

+ cc2 [3(1-d)2+2(k1)2+6kl(l-d)]= o -------- (3)

7

Derivation of Wheel Positioning for Maximum Positive Moment. 1. a+d<l

7F 4 B ! kl ! c

·-......,..;.--- - - --'

p cp,

FIG• (5): Positive moment: Case I a+d<l

Applying equations (A) and from free body one gets

Under load P

Setting oM cap = 0, and simplifying, yeilds

8

4c1 (l,+C )a3+gcc1 da2+[6cc1 d2-2c112 (l,+C )-2 (l.+C l] a J-- (4)

+l(l.+C) - cd (J.+C1 (12 -d2 )] = 0

Under load CP

Mp = RA(a+d)-pa

= (a+d) [l: .(1-aJ...£! (1-a-d)

: p~l [(12 -a2 ya+C(12-a+a 2 )(a+d)Jj -pa ~-----(G)

9

S . dMP d 1 f ett1ng oa- = 0, an simp i ying, one gets

4c1 (l.+C)a3+3C1_d(l;+4C)a2+2[c1 (6Cd2 -12i.+C)-(l,+C)j a] ---(5)

+l-d(l.+C112 )+Cl-2Cd[l,+C1 (12-2d2 )] = 0

2o b+d<kl

B.Jf ! kl : c --r-------------------'

P CP

Ms ( t-~Lt!? -4 . *f B I c r--- .kl _______ _ , Rc

FIG~ (6): Positive moment: Case II b+d<kl

Applying equations (B), and from free body, one gets

Ms = - CPC2 {(kl )2-h2 )b-PC2 [ (kl)2- (b+d)2 j(h+d)

RC - CP(kl-b) + P(kl-b-d) + MB - kl kl KI

Under load CP

MP = Rtb ------------------------------------------(H) By substituting, then,

dMP Setting os- = 0, and simplifying, one gets

4c2 (\+C )b3+gc2db2+ [sc2d2-2c2 (k1)2(.1jC )-2 (l.+C )]b

+kl(~+C)-d[~+C2(kl '2 -d2 )]= 0

}--(6)

10

Under load P

Mp = Rc(b+d) - CPob -----------------------(I)

By substitution, then

Setting dM ~ = 0, and simplifying, one gets

4C2 (l;+C )b3+3C2d(C+4 )b2+2 [ 6C2d2 -c2 {kl )2 (l;+C )- (l+C)}b J--+Ckl .. 'Cd(l+C2Kl 2 )+kl~12d-2C2d(KI __g2d2 ) = o (7)

3· (k+l)l>(a+d)>l

k _q -- - .. -f--~1-- _Q i A f--- - -L---~-- U ~ C

I

RA Rc

FIGo (7): Positive moment: Case III (k+l)l>a+d>l

Applying equation (A) and (B)

Also

MB = - PC1a(l2-a,7CPC2b «kl)2-b2)

where b = (k+l)l-a-d

a = (k+l)l-b-d

RA=f (1-a)+ ~ } [left side of B . . ~ from free body Rc= ~ (kl-b)~ right side of B

Under load P

Mp = ~oa ---------------------------------------(J) By substitution, then,

CMp Setting ~ = 0, and simplifying, one gets

4(c1-cc2 )a3+9(cc2 (kl+l-d)) a2 - fcc2 (6(12+d2 )

+ 4kl '2+12(12k-dl 0 l,+k)]+2 [1~112 ]ja + cc2 [13+312 (kd-d)+l(2kl"2+3d2 )

+ d2 (3kl-d)-d(2kl"2+612k)] + 1 = 0

-------------------------(8) Under load CP

MP = Rcob ----------------------------------------(K)

By substitution, then

CMp Setting oo- = 0, and simplifying, one gets

4(cc2 -c1 )b3+g[c1 (l+kl-d) Jb2-fc1 (6(kl . 2+d2 )

+ 412+12(12k-dl•I+k)) +2C(l+C2kl.2 )J b

+ c 1 [Ckl)3+3(k1)2 (1-d)+(kl) (212+3d2 )

+d2 (31-d)-d(212+612k)) +Ckl = 0

------------------------(9)

11

a/1

P CP l

- - an td 1

P CP

k1 ~c

1 - -·--------~- -- --· ··- - . -!

c: 4 a/1: -­

b/1: -----

0.6~--------------------------------------~

0.25

--...__.....__ ____ d/1 = 0.10

------------- 0.15 0.20 _,___ _____ . 0. 25

0.30

0.1

o.o~~~uu~~~ .. ~~.-~~~~~~~~~~ o.s 0.6 o.B

K

FIG. (8): Location of "a (b)" for maximum negative moments at support B

1.0

12

A ..

c = 4 a/1:

b/1 ------r- 1 k1 . c '"""1

0.6r---------------------------------------~

t ::::---·-------- . I

a/1 0.20--

0.15---...

d/1 = 0 .10--....

&.---·-·------====--_L-t---t---r--=

0.6 0.8

FIG. (9): Location of maximum positive moments under load P.

13

p

d P CP

I I I L.d --1- _b_J

I

c: 4

all:

K ·-~r - - c

1 .· kl :

r-----------------~--- ------ ---·--1 b/1: -----

. i

·30

0

·20 L------------~'""""""-1

.IsL-------~---l

.Io ·: '· o.s o.s

FIG. (10):

0.8 K

o.g

Location of maxtmum positive

moments under load CP.

l.o

14

.. I

J I'

P CP i ! ~

A# ;g ,__,_ --~---------~-

r ~b_i k1 ; c

c = L.o a/1: -­

b/1: -----

o .

_,

0.6 0.8 o.g K

FIG. (11): Location of a(or b) for maximum negative moments at support B.

15

d/1 = 0.10 \

g:~6~ \ \ 0.25-- \ 0.30 \

0.1

c: 1.o a/1: -­b/1: -----

o.omw~~~~~~--~~~~~~~~~~~~~

Oo5 0.6 0.8 K

FIG. (12): Location of maximum positive

moments under load P.

l.o

16

P CP P CP ,

, , , 1 r 1

L ____ a_ __ ""f-.. I -d ... : ~~-b-.;

I 1 j I I i

A~f----L-_. · . 1 .If kl f'c --.;..-..---

=~

i

i

I I I

c: l.o a/1:

b/1: -----

d/1 = 0.10 ·-------........

0.15 ~'

0.20~' - --

0.25 --~- - . f .4o

a/1

o.s

o. 30-------

d 1 = 0.10

0.6

FIG. (13):

0

0.7 0.8 K

Location of maximum positive

moments under load CP.

I.o

17

1

~-a

A k

p CP P CP

+d I i

i l d t b I r- --!'~ -- ---,

::n_c

c: o. 25 a/1:

b/1: -----

1 k1 l -----l l

K

FIG. (14): Location of a (or b) for maximum negative moments at support B.

18

p CP ;

' I d 1

.~ _a _t .d1 k

A L____j,__ ;a: r C B 1 1

c: o. 25

a/1:

b/1: -----

-1._ __ kL ... -- : . --------- i - - j

! I 0.6

# 10.3 I

b/1

0.2

o.s 0.6

FIG. (15)

K

maximum positive Location o~er load P. moments un

19

.,

,

0.15

·30

4 /.25F---

-------~~~

I b/1

0.6

FIG. (16 ):

0.8 K

Location f 0 ma:x·

moments under l~~CPositive

P.

C: 0 ·25 all:

b/1· . -----

l.o

20

DISCUSSIONS AND CONCLUSIONS

1) Example:

The following example illustrates the use of the curves

of Figures 1 through 9. The loading is one of the standard

loadings of the AASHO (American Associations of State High­

way Officials), called an "H20-44" loading, ("Standard

Specifications For Highway Bridges," AASHO, 1961, p. 11).

21

The wheel spacings vary from 14 1 to 30'o The percentage of

load that goes to an individual girder (or stringer) will

depend upon the number of lanes and number of girders. For

the purposes of the example it will be assumed that one truck

load is applied to one girder. The example is shown on page

23. As can be seen it is necessary to investigate more than

one potential absolute maximum but the procedure is direct

and easy.

2) For two wheel loadings that do not fall within the

scope of the numerical limitations set on this investiga­

tion, and which can not be interpolated if any of the

parameters fall outside the band of limits used, a designer

is still free to use the derived equations Which are indepen­

dent of numerical parameters other than a two wheel loading

and a two span continuous beam.

3) An independent check of the formula derivations

was made for both the positive and negative moments,

utilizing a specific set of numerical parameters. For

example, consider the negative moment at the intermediate

22

support. On page 51. Appendix II, is the influence line

for MB· As can be seen in the example one can quickly

approximate quite accurately the maximum moment by a trial

and error procedure. The "correct" solution is found using

the equations and they are in agreement with the solution

using the method in Appendix II. An analogous but consider­

ably more laborious technique was used for positive moment

expressions. This check, not shown1 was also in complete

agreement to the same degree of accuracy as that of negative

moment.

4) In the event that more economical considerations

should finally result in the use of cover plates or hanuches,

thus destroying the parameter of constant cross section,

the technique is of value in arriving at the preliminary

design.

23

Numerical Example F:or finding Maximum Moments

Load H20-44 (p=32k cp=8k)

1=70~t d=l4~t

C=0.25 k=0.50

c - 1 =--....;;,;1---= 1 1- 212(~+k) 2(70)2(~+0.5) 14700

c - 1 - 1 - 1 2 - 2kl2 (~+k) - 2(0.5)702 (~+0.5) - 735o

Negative Moment

From Fig (14): X=0.5, d/1 = 14/70 = 0.20

we get

b/1 = 0.117 b=O.ll7x70=8.2ft

b/1 = o. 53 a=O. 53x70=37 it

(1) Both loads on span AB (a=37')

From equation (C)

M =-(32) 1 [c7o2 -372 )·37+0.25r7o2 -C37+14)2]C37+14~ B 14700 l )

= -348k-ft

( 2 ) Both loads on span BC (b=8.2~ From equation (D)

1 f 2 2 (. 2 21 MB =- (32) 7350 o. 25 (3.5 -8.2 )8. 2+ L 35 - (8. 2+14) )

.(8.2+14)J= -81.2k-ft.

Maximum Negative Moment = 348k-ft at a=37ft

Positive Moment

1). Both loads on span AB

Under load P

From Fig. (15)

a/1 = Oo4

Subo into eqn. 1

~ = - 32 14700

under load CP

From Fig. (16)

(F)

ft a= 0.4 x 70 = 28 ·

(702-282 )x28+0. 25 [102 - (28+14 )2)

.{28+14)~= -322k-ft J

+ 0 · 2~~32 (70-28-14) - 3~~ = 17.8k

a/1 = 0.28 a=0~.28 x 70 = 19.6ft

Sub. into eqn. (G)

24

MB =-32· 14ioo j(702 -19.6~·19o6+0.25[7o2-(19.6+14)~(19.6+14J = -260k-ft

RA = ~(70-19.6)1°· 2~~32 (70-19.6-14)

Mp = 23.48(19.6+14)-32(19.6)=15lk-ft

2). Both loads on Span BC

Under load P

From Fig. (15)

b/1 = 0.044 b = 0.044 X 70 = 3.1~t Sub. into eqn. (I)

R = Oo25x32 c 35 (35-3.1)+ §; (35-3.1-14) - 7§54 = 21.38k

Mp = 21.38(14+3.1)-0.25·32(3.1)=341.2k-ft

Under load CP

From Fig. (16)

b/1 = 0.16 b = 0.16 X 70 = 11.2 1

Sub. into eqn. (H)

25

~ = ·32 7~50 [o.25(352-11.22)xll.2+[352-(11.2+14)2j(ll.2+14J = -77.8k-ft

Rc = 0 • 2 ~532 (35-11.2)+ ~ (35-11.2-14) - 7~58 = 13.2k

k-ft Mp = 13.2 X 11.2 = 149.2

Maximum Positive Moment = 498k-ft At a = 28 ft

FUTURE INVESTIGATIONS

This problem may be extended in several directions:

1) The use of three spans.

2) The use of other loadings than two Wheel

loadings.

3) The use of cover plates or haunches over the

support.

26

A girder of more than three spans will lead to quite

complex expressions. The use of other loadings, such as three

wheels, or one wheel and uniform load, is also feasible. The

introduction of cover plates and varying moment of inertia

very appreciably increase the amount of programming required,

particularly if either of the other two suggestions are simul­

taneously investigated.

4) A study into the possibility of using or modifying

the technique of this thesis to establish maximum negative

moments at points other than the support would be of bene-

fit in locating steel bend-up points in reinforced concrete

girders. If this does not prove feasible, the technique

itself is nevertheless of value in establishing the pre­

liminary design and utilizing the traditional influence

line plus maximum-minimum moment curve to converge upon the

final design.

PROGRAM NO. 1 EQUATION(l) Al=a

rJ ·-C***9"0863C EXOZ5· - ' THOii\"AS -

H YM .JG 11/10/65 FORI··lO o/q

' f· c RESEARCH FO R LOCATIO N OF HA X I i•iUH rW HENTS OF i'iOVH!G LOADS Of '

c ;CmH I NUOUS BEAH c . . THOMAS H • YANG c ': C=RATIO OF H!O LOADS, K= RA TI O OF nvo SPAf\1-L EN GT H c L=ONE SPAN LENGTH, D=DI S TAI,!C E BEn•I EE N LOADS ··

PUNCH 50 X=lOO.

' c = 16. ' Al=50. DO 1 1=1,3 C=C /4.0 Y=0.4 DO 1 L=1,6

! 0=5.0 : Y=Y+O .1

DO 1 N= 1, 5 ::'- '0=0+5. 0

3';, A= Al Al=SQRTF((( X **21* C1.+Cl-3. * C*(D** 2)~6.*C*D*Al/(3.*(1. ~C )) ) I F ( A B SF ( A- A 1 } -O .• i 5 E -1 } 1 , 1 , 3 .

:il PUNCH 100,C,X,YtP tA1 : ..... ;:

..... STOP {':

·- 5'"0 FORMAT ( 12 X , 1 H C ~\i:zx , 1 H L , 12 X , 1 H K , 12 X , fH D , 12 X , 2 H A l ) 10;0 FEJR 1·,1AT (5F14.1) ' .,

•:•

,._ ' END -,, .. ,-.. , ,;.; \, _ , . 1:;_-_ -~~- ----- · - ···-- -·· ----.- ... .. .. . - -- --- ----- - ·

--; -·· r -0'.

.:::1-. (;$_

4;Q: ·' ·4.JY

. ~ 4 . '{;" ' 4.0 4.0 4.;0 4.0

,. ~;; 4.0 ,,;f4.0

.4.0 i " ;4.-0

f ..

4.0 :M 4.0 . . 4.0

; ~ .o

.,Jft .O

4.0 4.0 4.0 4~0 4.0 4.0 4.0 4.0

PROGRAM NO. l CONTINUED .,_ ... _'[_ ·- ·---·- ':{'"'' ' - ::: ·-~-- - ---·- b- ---·-··-·--l'oo .o 100.0 . lOO ;.O .

100.0 100.0 roo .o· 100 e!o 100~0

100.0 100 ... 0 lO.Q .o 100.0 10Q.o 10;0.0 1 ()('>'. 0 tt'it .o 10t.o

·101);.0 . 1010, .• 0

·· i$j,.o 10G.O lOt> .o 10·0.0 100.0 106·.0 10'0•0 100.0 100.0 100.0

.5

.5

.5

.5

.5

.6

.6

.6

.6

.6

.7

.7

.7

.7 ·

.7

.a

.a

.a

.s

.8

.9

.9

.9

.9

.9 1.0 1.0 1.0 1. 0

10.0 15.0 20.0 25.0 30.0 10.0 15.0 20.0 25.0 30.0 10.0 15.0 20.0 25.0 30.0 10.0 15.0 20.0 25.0

'30·. ·0 10.0 15.0 2 0 .0 25.0 30.0

7:'', 10.0 :;~

15.0 20.0 25.0

· ··-·-A·l

4-9.6 4 5.4

' Lt-1. 2 36.9 32 .5

-49 .6 45.4 41.2

-:- 36.9 32.5 49. 6 45.4 41. 2

. "" •· 36.9 32.5 49.6 45.4 41. 2 36.9 32.5 49.6 45 • .:::, 41.2 36.9 32.5 49.6 4 5 • .:::.. 41.2

·· - -4_!_Q _________ __ 36., 9

1.0 30.0 32 '. r ___ 100.0~-----------~~--------~~~--------~4JU

1\) (X)

PROGRAM NO. 1 CONTINUED

1 ~o - t90 -.-o- ;s 1.0 100.0 .5 :1.0 . 100.0 .5 1.0 ' 100.0 .•' .5 1 : o 100.0 .5 '1.0 100.0 .6 1.0 100.0 .6 1.0 100 .o .6 1.0 100.0 .6 1.0 100e0 .6 1.0 100~0 .7 1.0 I 100.0 .7

~'

lk 1.0 100.0 .7 1.0 100.0 .7 1.0 100.0 .7 .1.0 100 .o .8 1.0 100.0 ' . 8 1.0 100 .o .8 1.0 100.0 .8 1.0 100.0 .a 1.0 100.0 .9 1.0 100 .• 0 .9 1.0 ' 10Q.o .9 1.0 100.0 .9 1.0 100.0 .9 1.0 100.0 1.0

f.-; 1.0 100.0 1.0 •' 1.0 10Q.O 1.0

1.0 100.0 1.0 1.0 _J_QQ .o . ····"--------------- · - . . - .. -- - --···· · l •O

10-.o 15.0 20.0 25. 0 30. 0 10.0 15.0 20.0 25.0 30.0 10.0 15.0 20.0 25.0 30!!0 10.0 15.0 20.0 25.0 30.0 10.0 15. 0 20. 0 25.0 30.0 10.0 ' 15.0 · 20.0 25.0 , ,

·--------------- 3o .. o.-'-'"-- -- --- --

5 2 .5 4 9 .7 4 6. 9 43. 9 4 0. 8 52. 5 4 9. 7 46. 9 43. 9 40. 8 52.5 49.7 46. 9 4~. 9

~Q. 8 • ~ +. - ·

52. 5 49.7 46. 9 43.9 40. 8 52.5 L~ 9 • 7 46. 9 43.9

1+ o. 8 52. 5 49.7 46. 9 L~ 3. 9 ~0. p .

1\) .\0

-~ .3

~,--.... • 3 .3 · .3 .3 • 3 .

.3

.3

.3

.3

.3

.3

.3

.3

.3 • 3 .3 .3 .3 .3 .3 .3 .3 .3 .3

i .3 .3 .• 3 • 3

PROGRAM NO. 1 CONTINUED

!'Q6; 01 100 .o 100 .o . 100 .o·"" 100.0 100.0 100 .. 0 100.0 100.0 100.0 100.0 100.0 100•0 100.0 100.0 100.0 100.0 100.0 100:.0 1001.0 100'• 0 100~•0

1oo;~ o

10~~0 1oar.o 1 OQIZ. 0

,;· 1oar .• o 1oq.o

0 .•• o 1od .o

•. 5. .5 .5 • 5 .5 • 6 .

.6

.6

.6

.6

.7

.7

.7

.7

.7

.8

.8

.8

.8

.8

.9

.9

.9

.9

.9 1.0 1.0 1.0 1.0 l. 0

10.0 15.0 20.0 25 .o 30.0 10.0

15-0 20.0 25.0 30.0 10.0 15.0 20.0 25.0 30.0 10.0 15.0 20.0 25.0 30.0 10. 0 15.0 20.0 25.0 30.0 10.0 15.0 2 0 .0 25.0 30 .0

..

'

. .

,...

5 5.6

5 '""· 4 53. 2 51.9 50. 5

55.6 54. 4

'53. 2 51. 9 50.5 55. 6 54.'" 53.2 51. 9 50.5 55.6 54. L: .

53. 2 51. 9 50.5 55. 6 5-4-. L; .

53. ;: 51. (} 50. :; 55. 6 54.4 5 3 .2 51.9 50 .5

-

lA) 0

PROGRAM NO. 2 : EQUATION (2) Al = b

C C***98674CEX025 THOMA'S YANG 12/03/65 FORMO

C RESEARCH FOR LOCATION OF MAXIMUM MOMENTS OF MOVING LOADS ON C CONTINUOUS BEAM C THOMAS H. YANG C C=RATIO OF TWO LOADS. K=RATIO OF TWO SPAN-LENGTH C L=ONE SPAN LENGTH. D=DISTANCE BETWEEN LOADS

PUNCH 50 X=100. C=16. Al=3le DO 1 I= 1, 3 C=C/4.0 Y=0•4 DO 1 L=1,6 D=35· Y=Y+Oel DO 1 N=1,5 D=D-5.0 Cl=1e/(2•*<X**2)*tl.+Yl l C2=1./(2e*Y*(X**2l*(l.+Yl l

3 A=Al Al=( (Y**2l*(X**2l*(l.+Cl-3•*<D**2l l/(3•*<l•+Cl*A+6e*Dl IFtABSF<A-All-0.5E-lll,l,3

1 PUNCH 10o,c,x,y,D,A1 50 FORMATtl2X,1HC,l2X,lHL,l2X,lHK,I2X,IHD,12X,2HAil

100 FORMAT<5Fl4.2l END

PROGRAM NO. 2 CONTINUED Al = b

c L K D Al 4.00 100.00 .50 30.00 20.27 4.00 100.00 .50 25.00 22.09 4.00 100.00 .so 20.00 23.72 4.00 100.00 .so 15.00 25.22 4.00 100.00 .so 10.00 26.57 4.00 100.00 .60 30.00 26.51 4.00 100.00 .60 25.06 28.18 4.0 0 100.00 .60 20.00 29.68 4.00 100.00 .60 15.00 31.10 4.00 100.00 .60 10.00 32.43 4.0 0 100.00 .70 30.00 32.61 4.0 0 100.00 .70 25.00 34.18 4.00 100.00 .70 20.0 0 35.64 4.0 0 100.00 .7C 15.00 36.95 4.00 100.00 • 7 0 1o.oo 38.24 4.00 100.00 .so 30.00 38.62 4.00 100.00 .so 25.00 40·11 4.00 100.00 .a 0 zo.oo 41·47 4. 00 100.00 • 8 0 15.00 42.82 4.00 100.00 .so 10.0 0 43.99 4.00 100.00 .90 30.00 44.54 4. 00 100.00 .90 25.00 45.97 4. 00 100.00 .90 20.0 0 47.36 4. 00 100.00 .90 15.00 48.64 4.00 100. 00 .90 10.00 49.83 4. 00 100.00 1.oo 30.00 50.45 4.0 0 100 .0 0 1.oo 25.00 51.84 4.00 100.00 1.oo 20.00 53.20 4.0 0 100.00 1.oo 15.00 54.40 4. 0 0 100.00 l. () 0 lf'l . f'l(\ 55.62

VJ 1'\)

PROGRAM NO. 2 CONTINUED

1.00 100.00 • '5""0 3'6.oo 9e67 1.00 100.00 .50 25.00 13.51 1.00 100.00 .50 20.00 17.07 1. 00 100.00 .50 15.00 20·36 1.oo 100.00 .so 1o.oo 23.45 1.00 100.00 .60 30.00 16.24 l.GO 100.00 .60 25.00 19.82 1.00 100.00 .60 20.00 23.1S 1.00 100.00 .60 15.00 26.30 1.0 0 1oo.oo .60 1o.oo 29.30 1.00 100.00 .70 30.00 22.53 1.00 100.00 .70 25.00 25.92 1. 0 0 100.00 .70 zo.oo 29.14 1.oo 100.00 .70 15.00 32.20 1.oo 100.00 .70 10.00 35. 0 8 1.00 100.00 .so 30.00 2S.67 1. 00 10 0 .00 .so 25.00 31.95 1.00 100.00 .so zo.oo 35.08 1. 00 100.00 .so 15.00 3S.o6 1.00 1oo.oo .so 10.00 40e94 1. 0 0 100.00 .90 30.00 34.76 1. 00 100.00 .90 25.00 37.95 1.o o 100.00 .90 zo.oo 4leOl 1. 00 100.00 .90 15.00 43. 94 1.oo 100.00 .90 1o.oo 46.74 1. 00 10 0 . 0 0 1.00 30.00 40.74 1. 00 lO O.CO 1.oo 25.00 43.85 l. OC 100.00 1.oo 20.00 46e S8 le CO 10 0 .00 1.oo 15.00 49.7 7 leO C 10 0 . 0 0 1.oo 10.00 5 2 . 50

w w

PROGRAM NO. 2 CONTINUED

~ 25 100 . 00 . 50 30 . 00 2 . 26 . 25 . 100. 00 . 50 2 5. 00 7 · 08 . 2 5 100 . 00 . 50 20 . 00 11 . 74 • 2 5 100 . 00 . 50 1 5. 00 16 . 25 . 25 100 . 00 . 50 10 . 00 20 . 57 . 25 1 00 . CO • 6 0 30 . 00 8 . 5o · 25 100 . 00 • 6 0 25 . 00 13 el 7 . 25 100 . 00 • 6 0 2o . oo 1 7 . 69 . 25 100 . 00 .6 0 1 5 . 00 22 . 10 · 25 100 . 00 . 60 10 . 00 26 . 39 . 25 100 . 00 . 70 30 . 00 14 . 58 · 2 5 100 . 00 .7 0 25 . 00 19 . 15 · 25 100 . 00 . 70 20 . 00 23 . 63 • 2 5 100 . 00 . 70 15 . 00 2 7 .98 · 25 100 . 00 • 7 0 1 o . oo 32 . 20 • 2 5 100 . 00 . so 30 . 00 2 0. 6 1 . 25 100 . 00 . so 2 5. 00 25 . 10 . 25 lOO . OJ .so 2o . oo 29 . 48 . 25 100 . 00 . so 15 . 00 33 . 8 1 · 25 100 . 00 • 8 0 10.00 38 . 00 • 2 5 1 00 . 00 . 90 30 . 00 26 . 55 • 2 5 100 . 00 . 90 2 5. 00 31 . 00 . 25 100 . 00 . 90 20 . 00 35 . 36 . 25 100 . 00 . 90 15 . 00 39 . 60 • 2 5 100.00 . 90 1o . oo 43 . 79 . 25 1oo . co 1 . oo 30 . 00 3 2 . 48 . 25 100 . 00 1 . oo 2 5 . 00 36 . 85 . 2 5 100 . 00 1 . oo z o . oo 41 . 16 . 25 l oo . no 1. oo 15 . 00 45 . 44 . 25 10 0 . CO 1. 00 10 . 00 49 . 62

PROGRAM NO. 3 : EQUATION (4) Al = a

C C***96887CEX~~~ THOM~~ YANG 11/29/65 FORMO

C RESEARCH FOR LOCATION OF MAXIMUM MOMENTS OF MOVING LOADS ON C CONTINUOUS BEAM C THOMAS H. YANG C C=RATIO OF TWO LOADS, K=RATIO OF TWO SPAN-LENGTH C L=ONE SPAN LENGTH, D=DISTANCE BETWEEN LOADS

PUNCH 50 X=100• C=l6. Al=30e DO 1 I= 1, 3 C=C/4.0 Y=Oe4 DO 1 L=lt6 D=35. Y=Y+O•l DO 1 N=lt5 D=D-5.0 Cl=le/C2e*CX**2l*Cl.+Y)l C2=le/C2e*Y*CX**2l*Cle+Yll

3 A=Al A2=4e*Cl*Cle+Cl*(A**3l+9e*C*Cl*D*IA**2l+X*<l•+Cl A3=c-1.l*C*D*C1.+Cl*CCX**2l-CD**2lll Bl=2e*Cle+Cl+2.*Cl*Cle+Cl*CX**2l-6e*C*Cl*(D**2l A1=CA2+A3>/Bl IFcABSFCA-All-Oe5E-llltlt3

1 PUNCH lQQ,c,x,y,D,Al 50 FORMAT Cl2XtlHC,l2XtlHL,l2X,lHK,l2X,lHD,l2X,2HAll

100 FORMAT C5Fl4e2l END

w IJl

PROGRAM NO. 3 CONTINUED Al = a

c l K D A1 4e00 100.00 .50 30.00 31.82 4e00 100.00 .so 2S.OO 33.20 4.00 100.00 .so 20.00 34e64 4e00 100.00 .50 15.00 36.14 4.00 100.00 .so 10.00 37.68 4e00 100.00 e60 30.00 32·31 4e00 100.00 e60 25.00 33·69 4e00 100.00 .60 2o.oo 35.1S 4e00 1oo.oo .60 15.00 36e67 4.00 100.00 .60 10.00 3ee23 · 4e00 100.00 .70 30.00 32·72 4e00 100.00 .70 25.00 34.12 4.00 100.00 .70 zo.oo 35.60 4.00 100.00 .70 15.00 37.14 4e00 100.00 .70 10.00 3e.71 4e00 100.00 .eo 30.00 33.07 4.00 100.00 .ao zs.oo 34.50 4.00 100.00 .a o 20.00 36.oo 4.00 100.00 .eo 15.00 37eS5 4.00 100.00 .eo 10.00 39.15 4.00 100.00 .90 30.00 33.38 4.00 100.00 .90 25.00 34.e3 4e00 100.00 .90 20.00 36.36 4.00 100.00 .90 1S.OO 37.92 4e00 100.00 .90 1o.oo 39·53 4.00 100.00 1.oo 30.00 33.65 4.00 100.00 1.oo 2-s. 00 35e13 4e00 100.00 1.oo 20.00 36.67 4e00 100.00 1.oo 15.00 38.25 4e00 100.00 1.oo 1o.oo 39e88

VJ (J')

PROGRAM NO. 3 CONTINUED

1.00 100.00 .50 30.00 35.87 1.00 100.00 .50 25.00 36e49 1e00 100.00 .50 zo.oo 37.24 1.00 100.00 .so 15.00 38.06 leOO 100.00 .so 10.00 38.95 1.00 100.00 .60 " 30.00 36.34 1.00 100.00 .60 25.00 37•00 1.oo 100.00 .60 20.00 37.76 1e00 100.00 .60 15.00 38e60 1e00 100.00 .60 10e00 39e50 1.00 100.00 .70 30.00 36.75 1.00 100.00 .70 25.00 37.44 1.00 100.00 e70 20.00 38e22 1.00 100.00 .70 15.00 39e08 1.00 100.00 • 70 10.00 39.99 1.oo 100.00 .so 30.00 37.12 1.00 100.00 .so 25e00 37.83 1.00 100.00 .so zo.oo 38e63 1.00 100.00 .eo 15.00 39e50 1e00 1oo.oo .ao 1o.oo 40e43 1.00 1oo.oo .90 30.00 37.44 1.oo 1oo.oo .90 25.00 38.18 1.00 100.00 .90 20.00 39eOO

! 1.00 100.00 e90 15.00 39.88 1.oo 100.00 e90 10e00 40e82 1e00 100.00 1.oo 30.00 37.73 1e00 1oo.oo 1.oo 25.00 38.49 1.00 100.00 1.oo 20.00 39.33 1e00 100.00 1.oo 15.00 40e22 1.00 100.00 1.00 10.00 41e17

PROGRAM NO. 3 CONTINUED

e25 100.00 .so 3o.oo 39.14 ·25 100.00 .so 25.00 39e30 ·25 100.00 .so zo.oo 39.55 ·25 100.00 .so 15.00 39.84 .2S 100.00 .so 1o.oo 40.16 ·25 100.00 .60 30.00 39e66 .25 100.00 .60 25.00 39.84 .25 100.00 .60 20.00 40e10 ·25 100.00 .60 1S.OO 40e39 ·25 100.00 .6 0 10.00 40e72 ·25 100.00 • 70 3o.oo 40e12 ·25 100.00 .70 25.00 40.31 ·25 100.00 .70 20.00 40.58 ·25 100.00 .70 15.00 .40. 89 .25 100.00 .70 1o.oo 41.22 ·25 100.00 .8o 30.00 40.S3 ·25 100.00 .ao 25.00 40.74 ·25 100.00 .8o 20.00 41.02 ·2 5 100.00 .so 15.00 41.33 .25 100.00 .eo 10.00 ' 41.67 e25 100.00 .90 30.00 40e90 .25 100.00 .90 25.00 41.11 e25 100.00 .90 20.00 41.40 e2S 100.00 .90 1S.OO 41.72 .25 100.00 .90 1o.oo 42.07 .2 5 tmr.co 1.oo 30.00 - 7+1.23 e25 lOOeeo 1.oo 2S.OO 41.45 e2S 1 OOaOO 1.oo 20.00 41·75 .25 lO"OeOO 1.oo 15.00 42.07 ·25 lOOtOO 1· 00 1o.oo 42.43

PROGRAM NO. 4 : EQUATION (5) Al = a

[:::*::~9824 "J:EX025 THOMAS H. YANG 1 2 I 0 2 I 6 5 Ei1 RHO

C RESEAR: H FOR LOCATIJI\J OF MAXItJIUM ,~·10f.1HJTS OF ~10Vlf\JG LIJADS lf'.i C CONTINliOlJS BEAM C THOMAS H. YANG C C=RATIO OF TWO LOADS. K=RATIO OF TWO SPAN-LENGTH

f-- C L-ONE SPAN LENGTH. D-DISIAf\!CE BEH/EEN LOADS PRINT 50 X=lOO. t.l:ln

C=l6. DO l 1=1,3 C=CI4.0 Y=0.4 DO l L=l,6 n -:?. r:;

Y =Y +0. l DO l N=l 7 5 D=D-5.0 Cl=l./(2.,!<(X*'~2)>!:( l.+Y)) C2=1./(2.:;:y:~(Xt;:!~2)::!'( 1.+Yl)

?, !1-:!11 A3=X-D* ( 1. +C 1* ( x:;:::~2)) +C>:: ( X-2. :;: D>!: ( 1. +C 1':' ( (X*':: 2 )-2. >:: ( D:!<>!<2)))) B 1 = ( - l • l ::: ( 4. * C 1 ,;: ( l • + C l * ( A:;::;: 2 l + 3 • :;: C l ::: D::: ( l • + 4 • >::( l :;: A l

I h 2 = ( -1 • ) ::: 2 • ::: ( C 1 ;: ( 6. *C >:: ( [)::: :::2 l - ( x::: * 2 l ::: ( 1. +C l l - ( 1 • +C l l r---1 Al=A3/(Bl+82)

IFCABSF(A-All-0.5E-l) 1,1,3 l PRINT lOQ,C,X,y,p,AJ

5 0 F 0 R r•l Ar ( l 2 X , l H C , l 2 X , l H L , l ? X , l Hi( , l 2 >: , 1 H D , 1 ? X , 2 H A 1 ) 100 FORMAT( 5F 14.2l

w \0

PROGRAM NO. 4 CONTINUED Al = a

c L K D .L\ 1 4.00 100.00 .50 30.00 1 3. 9 7

4.00 100.00 .50 25.00 18.39 4.00 100.00 .50 20.00 22.8 4 4.00 100.00 .50 15.00 27.32 4.00 100.00 .50 10.00 31. ~12

4.00 100.00 .60 30.00 14.54 4 0 4.00 100.00 .60 20.00 23.41 4.00 100.00 .60 15.00 2 7. 8 9 4.00 100.00 60 o.oo 32 39 4.00 100.00 • 70 30.00 15.0 4 4.00 100.00 • 70 25.00 19.46 4.00 100.00 • 70 20.00 23 92 4.00 100.00 .70 15.00 28.40 4.00 100.00 • 70 10.00 32.90

4.00 100.00 .80 30.00 15.48

4.00 100.00 .80 20.00 2+. 3 7 4.00 100.00 .80 15.00 28.85

4.00 100.00 .90 30.00 15. 8 9 4.00 100.00 • 90 25.00 20.32 4.00 100.00 .90 20.00 24.78 4.00 100.00 • 90 15.00 29.26 4.00 100.00 .90 10.00 33.77 4.00 100.00 l. QQ 3Q.QQ 16.25 4.00 100.00 1. 00 25.00 20 .6 e 4.00 100.00 1. 00 20.00 2 '::i .!Lc 4.00 100.00 1.00 15.00 29.A3 4.00 100.00 1.00 10.00 y~. l 1!·

+=" 0

PROGRAM NO. 4 CONTINUED

1.00 100.00 .50 30.00 18.12 1.00 100.00 .50 25.00 21.79 1.00 100.00 .50 20.00 25.51 1.00 100.00 .50 15.00 t-:9.29 1.00 100.00 .50 10.00 53. 12 1.00 100.00 .60 30.00 18.S9 1.00 100,00 .60 25.00 22.36

100.00 .60 15.00 29.87 1.00 100.00 .60 10.00 33,69

00 0 1. 00 100.00 • 70 25.00 22.0 7 1.00 100.00 .70 20.00 26.60 1.00 ]00.00 • 70 15.00 30.38 1.00 100.00 .70 10.00 34.21 1,00 100.00 .so 30.00 19.65 1.00 100.00 .80 25.00 23.33 1.00 100.00 .so 20.00 2 7. 0 6 1.00 100,00 .so 15.00 30.84 1.00 100.00 .80 10.00 31t.:27 1.00 100.00 ,90 30.00 20.05 1.00 100.00 .90 25,00 2 3. 7 4 1.00 100.00 .90 20.00 27. lt8

1.00 100.00 .90 15.00 31.26 1.00 100.00 .90 10.00 35 .08

! 1.00 100.00 ]. 00 30.00 20.42 1.00 100.00 1.00 25.00 24.12 1.00 100.00 1. 00 20.00 27.--i'J 1.00 100.00 1.00 15.00 31 • 64 1.00 100.00 1.00 10.00 35,!1·6

PROGRAM NO. 4 CONTINUED

.25 100.00 .50 30.00 21.72

.25 100.00 .50 25.00 24.80 • 25 100.00 .so 20.00 2 7. 94 .25 100.00 .50 15.00 31 .12 .25 100.00 .50 10.00 3+. 35

.25 100.00 .60 25.00 25.40 .25 100.00 .60 20.00 28.53 .25 100.00 .60 15.00 31.70 .25 100.00 .60 10.00 34.93 .25 100.00 • 70 30.00 22. B 4 .25 100.00 .70 25.00 25 .93 • 25 100.00 • 70 20.00 29 . o 6 .25 100.00 .70 15.00 32.23 .25 100.00 . 70 10.00 35.45 .25 100.00 .80 30.00 23.32 .25 100.00 .80 25.00 26 .40 .25 100.00 .80 20.00 29.53 .25 100.00 .80 15.00 32.70 .25 100.00 .80 10.00 35.92 • 25 100.00 • 90 30.00 23.75 .25 100.00 .90 25.00 26.84 .25 100.00 • 90 20.00 29.9 7 .25 100.00 .90 15.00 33. 13 . 25 100.00 .90 10.00 36.34 .25 100.00 1.00 30.00 24.14 .25 Joo.oo LOO 25 .00 2 7. 2 3 .25 100.00 1.00 20.00 3() . 36 .25 100.00 1. 00 15 .00 33 . 52 • 2 5 - --- 100.00 1.00 10.00 36 .72

PROGRAM NO. 5 : EQUATION (6) Al = b

C C***98255:EX025 THOMAS YANG 12102/65 EORi+J , "f·

C RESEARCH FOR LOCATION OF >1AXli•1Uf·~ ~,10f'/1ENTS OF f·10VING LOADS (1!\

C CONTINI!O\!S BEAt1 i C THOt-1AS H, YANG

C C=RATIO OF TWO LOADS. K=RtHIO OF T!~O SPAN-LENGTII A

Al=15, DO 1 1=1,3

Y=0.4 DO 1 L=1,6

Y=Y+O .1 DO 1 N=1,5

C 1 = 1 , I ( 2 , ,;, ( X':":' 2 l ':' ( 1 • + Y l ) C 2= 1 , I ( 2 , ,;: Y ':' ( X ;;, ':' 2 ) ,;, ( 1 • + Y l )

A 2= 4. ':'C 2 ,;, ( 1 • +C ) ,;, ( A'::* 3 ) +9 • ':'C 2 ':' (}:< ( A':":' 2 l +X':' y,;, ( 1 • + C ) A3=<-l. )'::D*< l.+C2':'< (Y':":'2)':q x':":'2l-< o,:-:,:<2) J l Bl=2.*(1.+C)+2.*C2*(1,+Cl*(X**2l*<Y** Zl-6,*C2*(D**2l Al= ( .6,2+A3) /Bl I F ( A B SF ( A- A l ) -0 • 5 E - 1 ) 1 , 1 , 3

l pI I NC H 1 a a , C , X 7 Y , Q 7 A l 50 F 0 R ~~AT ( 1 2 X , 1 H C , l 2 X , 1 H L , 1 7 X , 1 HI<, 12 X, 1 H D , 1 2 X , 2 H A 1 l

l 0 0 F 0 R t~ Ar ( 5 F 1 4 , 2 ) Ef\J D

PROGRAM NO. 5 CONTINUED Al = b

L K D t-\ 1 4.00 100.00 .50 30.00 20.95 4.0 0 100.00 .50 25.00 21. 11 4.00 100.00 .50 20.00 2 1 • 3Lt-

4.00 100.00 .50 15.00 2 1. 'J2 4.00 100.00 .50 10.00 21. S7

4.00 100.00 .60 30.00 25 .1 0 4.00 100.00 .60 25.00 25. 29 4.00 100.00 .60 20.00 2 5.? 3 4.00 100.00 .60 15.00 25 . P3 4.00 Joo.oo .60 10.00 2 6. ] 8 4.00 100.00 .70 30.00 29 . 20 4.00 100.00 • 70 25.00 2 9. 4(1 4.00 100.00 • 70 20.00 29.66 Lt- • 00 100.00 • 70 15.00 29 . 9 7 4.00 100.00 • 70 10.00 30 . 32 4.00 100.00 .80 30.00 33.24 4.00 100.00 .so 25.00 3 3. <t6 4.00 100.00 .so 20.00 33 . 73 4.00 100.00 . 8 0 15.00 3 't-o 'J5 4 00 100 00 80 10 00 34 . LtO

4.00 100.00 .90 30.00 3 7. 2 4 4.00 100.00 .90 2 5 .00 3 7. Lc8

4.00 100.00 .90 2o.oo 3 7.77 4.00 100.00 .90 15.00 38 . 08 4.00 100.00 .90 10 .00 38 . 44 4.00 100.00 1.00 30.00 Ltl e 2 1 4.00 100.00 1. 00 25.00 It l. 45 4.00 100 .00 1. 00 20.00 4 1.75 4.00 100.00 L 00 15.00 't 2. '] 7 4.00 100.00 1.00 10 .00 42 • L3

-I=" -I="

PROGRAM NO. 5 CONTINUED

1.00 too .do · • 50 30.00 l 7. 45 1.00 100.00 .50 25.00 ] 8 . QG 1.00 100.00 .50 20.00 18 . 8 7 1.00 100.00 .50 15.00 19 .73 1.00 ]00.00 .50 Jo.oo 20 . '27 1.00 100.00 .60 30.00 21. 59 1.00 100.00 .60 25.00 22. 29

1.00 100.00 .60 15.00 2 3. 9 5 1.00 100.00 .60 10.00 24. 90

1.00 100.00 .70 25.00 26. 4-2 1.00 100.00 • 70 20.00 2 7. 2 2 1.00 100.00 .70 15.00 28 .10 1.00 100.00 • 70 10.00 29. 0 5 1.00 100.00 . 8 0 30.00 29.74

1.00 . 80 20.00 3 1.30 1.00 100.00 . s o 15.00 3 2.1 9 1 00 00 0 8 4 1.00 100.00 • 9 0 30.00 33. 7 4 1.00 100.00 .90 25.00 34. 5 1

1.oo 100.00 .90 15.00 36 . 23 1.00 100.00 • 90 10.00 3 7. 1 8 1.00 100.00 1.oo 30.00 3 7. 7 1 1.00 100.00 1. 00 25.00 38 . ~ 0

1.00 100.00 1.00 20.00 39 . 33 1.00 100.00 1.00 15.00 'tl1 e 2/ 1. 0 0 100.00 1. 00 10.00 ,, 1 • l 7

PROGRAM NO. 5 CONTINUED

.25 lQO .66 .5'D 30.00 1 2 . 9 2

.25 100.00 .50 25.00 l+. 44 .25 100.00 .50 2 0 .00 16 . o l .25 100.00 .50 15.00 l 7. s 3 .25 100,00 .50 10,00 19 . 30

.25 100,00 .60 25.00 18 .72 , 25 100.00 ,60 20.00 2 0 . 29

.25 100.00 ,6 0 10 .00 23. ':55 .25 100.00 • 70 30.00 21.44

.25 100.00 • 70 20.00 24 . 4.-7 .25 100,00 • 70 15.00 26 . 0 7

.25 100.00 . so 30.00 25 . ':57 .25 100,00 . so 2 5 .00 27. 05

. 25 100 .00 , 80 15.00 30 . 18 . 2 5 100 .00 . so 10.00 3 1. 'i 2

0 f--4

.25 100.00 • 90 25.00 3 1. 11

.25 100,00 . 90 20.00 32 . 66 .25 100,00 , 90 15.00 3 't. ? 4 .25 100,00 , 90 10.00 35. 8 7 . 2 5 100,00 1. 00 30.00 33, S5 .25 100.00 1,00 2 5 ,00 35 . l 3 .25 100,00 1. 0 0 2 0 . 00 36 , /)7 . 25 100.00 1. 00 15.00 38 . 25 •.. 2 5 lOQ.QQ l. QQ l Q.QQ ·:) 0 ?F

c

c c c c c

PROGRAM NO. 6 EQUATION (7) Al = b

C ~:o:~*98606C EX025 THO!VlAS Y Af0G 12/03/65 FO Rf'10 9fj, J

RESEARCH FOR LOCATION OF I~ A X I 1•1 U i~ tVJO t•1 E NT S OF 1"10 VI NG LOADS C ONT Ii\jU OU S BEAM THOMAS H. YANG C=RATIO OF H.JO LOA OS. K=RATID OF nvo SPAN-LENGTH L=ONE SPAN LENGTH. D= DISTANCE BETI-JEEN LOADS PRINT 50 X-100. A1=3. C= 16. DO 1 l-1,3 C =C /4 .o Y=0.4 LJU l L=l,6 0=35. Y=Y+O .1 LJU l 1\J-l,~

D=D-5.0 C 1 = 1. I ( 2. ~~ ( X:::~~~ 2 ) ~~ ( 1. + Y ) ) C~=l./(~.;~y,,qx~P ... ~p.q l.+Y))

3 A=A1 A2=4.*C2*(1.+C)*(A**3l+3.*C2*0*(C+4.}*(A**2) A 5 -'C ;~ ( Y ~X -I)~., ( l • +C ~ ;~ ( Y ,p,~ ~ ) ~.~ ( X ~Pr l ) ) ) A4=Y*X-2.*D*( 1.+C2*((Y*Xl**2-2.*D*D)) B2=2.*(1.+Cl-2.*C2*(6.*D*D-CY*Xl**2*{1.+C)) A 1- ( /4 L +A 5 +A 4 ) I t3 L IF(AbSF(A-A1)-0.5E-1ll,l,3

l PRINT lOO,C,X,Y,D,Al ·-A '

100 FORMAT(5Fl4.2) Ei'ID

0 f\j

PROGRAM NO. 6 CONTINUED Al = b

c L K D A1 4.00 100.00 .50 30.00 4.41 4.00 rou.oo .50 25.00 7. 33 4.00 100.00 .50 20.00 10.31 4.00 100.00 .50 15.00 13.35

• 4.00 100.00 .60 30.00 8.41 4.00 100.00 .60 25.00 11.38

• 4.00 100.00 .60 15.00 17.47 4.00 100.00 .60 10.00 20.60

• 4.00 100 .oo • 70 25.00 15.39 4.00 100.00 • 70 20.00 18.45

• • 70 .so .so •

4.00 100.00 .so 15.00 25.58 4.00 100.00 .so 10.00 28.75

• I 4.00 100.00 .90 25.00 23.31

4.00 100.00 .90 20.00 26.42 4.00 100.0 • 90 15.00 29.57 4.00 100.00 .90 10.00 32.75 4.00 100.00 1.00 30.00 24.1 4 4.00 100.00 I. 00 25.00 27.~3 4.00 100.00 1.00 20.00 30.36 4. 0 0 100.00 1.00 15.00 33. 5 2 4.uu roo.oo 1. 00 Io.oo 36.72

..f::' co

PROGRAM NO. 6 CONTINUED

1.oo 10Q.OO • 5_0 30.00 -- -1-- --·

.56 ; 1.00 100.00 .50 25.00 4. 11 : I.oo 100 .oo .50 20.00 7.72 1.00 100.00 .50 15.00 11.39 1.00 100.00 .50 10.00 15.13 1.00 100.00 .60 30.00 4.61 1.00 100 .oo .60 25.00 8.20 1.00 100.00 .60 20.00 11.84

• 1.00 100.00 .60 10.00 19.30 1.00 100.00 • 70 30.00 8.61

• 1.00 100.00 • 70 20.00 15.91 1.00 100.00 • 70 15.00 19.64

• 1.00 100.00 .so 30.00 12.5S 1.00 100.00 .so 25.00 16.23

• 1.00 100.00 .so 15.00 23.67 1.00 100.00 .ao 10.00 27.47

• 1.00 100.00 • 90 25.00 20. 19 1.00 100 .oo .90 20.00 23.91 1.oo 100.00 • 90 15.00 2 I. 6 I 1.00 100.00 .90 10.00 31. 4S 1.00 100.00 1.00 30.00 20.43 1.00 100.00 r.oo 25.00 24.12 1.00 100.00 1.00 20.00 27.85

100 .oo 1.00 15.00 31.63 oo.oo 1.00 10.00 3 ".i. t-5

PROGRAM NO. 6 CONTINUED ' - . -25- ---- -- --Too~ oCJ - .50 30.00 - 3 .-95 i

! .25 1-00 .oo .50 25.00 • 43 ' .25 100.00 .50 20.00 4. 8 4 .25 ioo .oo .so 15.00 9.27 • 25- 100.00 .50 10.00 13.74 .25 100.00 .60 30.00 .20 • • . ~9 .25 100.00 .60 20.00 9.01 .25 100.00 .60 15.00 13. 46 • • .25 100.00 • 70 30.00 4.2 8 .25 100.00 .70 2 5. oo· 8 .6 8 • • .25 100.00 • 70 15.00 17.5 7 .25 100.00 • 70 10.00 22.06 • • .25 100.00 .80 25.00 12.72 .25 100.00 .80 20.00 17.17 • • .25 100.00 .80 10.00 26 .13 .25 100.00 • 90 30.00 12. 30 • • .25 100.00 .90 20.00 21. 17 .25 100.00 .90 15.00 25.65 .25 100.00 • 90 1o.oo 30.15 .25 100.00 1.00 30.00 16.25 .25 100.00 1.00 25.00 20 . 68 .25 100.00 r.oo 20.00 25.14 . 2 5 100 .00 1. 00 15.00 2 g • .''-.3

! . 25 100.00 1.00 10. 00 :-3'r • )_ 3 , __ - -·- - - ---------

\11 0

_.... ~ 4-1

I ,.!iG --~

6

5

4

3

2 ·

I 1

tO N

ol N

~ i

FIG• (17) Influence line for support moment, MB

NUMERICAL CHECK OF SUPPORT MOMENT

32k r -)12;! t

A- ¥ A' 70' B I . 35~c ~

I I I

From Influence Line Using "Trial and error method"

We get: Ordinate 9 under 32k

7·75 under ak

Therefore,

Maximum negative moment = 9 x 32 + 7.75 x 8

= 350k-ft ~ 348k-ft

(previous answer p. 23)

53

BIBLIOGRAPHY

{1) A. Wo Legat and W. A. Fairhurst, "Design and Construction of Reinforced Concrete Bridges". Revised 1957.

(2) I. P. Church, "Mechanics of Engineering". John Wiley and Sons, Publisher, 1890.

(3) s. F. Borg and Joseph J. Gennaro, "Advanced Structural Analysis". D. Van Nostrand Company, Publisher, 1964.

{4) Paul Anderson, "Structural Mechanics". 1962

(5) Wen-chi Chen, "R. C. Bridge Design". Paper, 1960.

(6) Hale Sutherland, "Structural Theory". John Wiley and Sons.

(7) Standard Specifications for Highway Bridges, American Association of State Highway Officials, 1961. {AASHO)

VITA

Mr. Thomas Houng-Yn Yang was born in Chiangsu, China,

on March 2, 1939.

He enrolled at the National Taiwan University in 1956

where he obtained his Bachelor degree in Civil Engineering

in 1960.

54

He served as a Second Lieutenant in the Chinese Army for

one and one-half years.

From January, 1962, to December, 1963, he was employed

by Shihmen Dam Project Committee of Taiwan as an inspector

of construction.

In January, 1965, he entered the University of Missouri

at Rolla as a graduate student in Civil Engineering Department

to pursue the Master of Science degree.