MATRIX ANALYSIS OF CONTINUOUS PARABOLIC ARCHES

CHAO-SHYONO HSU

Diploma, Taipei Institute of Technology, 1963

A MASTER'S REPORT

submitted in partial fulfillment of the

requirements for the degree

MASTER OF SCIENCE

Department of Civil Engineering

KANSAS STATE UNIVERSITYManhattan, Kansas

1967

Approved by:

Ma l&Major Professor

TABLE OF CONTENTS

page

SYNOPSIS 1

INTRODUCTION 2

SINGLE FIXED ARCH THEORY k

MATRIX ANALYSIS OF CONTINUOUS ARCHES BY THE DISPLACEMENT METHOD .... 9

NUMERICAL EXAMPLE 16

CONCLUSIONS 30

ACKNOWLEDGMENT 31

REFERENCES 32

NOTATION 33

APPENDIX A FLOW DIAGRAM FOR THE DISPLACEMENT METHOD 35

APPENDIX B CHECK RESULTS OF EXAMPLE USING ENERGY METHOD k&

APPENDIX C SIMILE ARCH INFLUENCE LINES $$

SYNOPSIS

The purpose of this report is to present a method of analyzing continuous

arches with varying cross section on slender piers by obtaining influence

lines of the redundants. The analysis is based on the displacement method.

First, the single fixed arch theory is presented. Then, the continuous

arch is analyzed by applying the displacement method. Finally, a numerical

example consisting of two unsymmetrical parabolic continuous arches with a

central slender pier is given to illustrate the use of the method.

Influence lines for the redundants are drawn. The influence coefficients

are checked by using the energy method. The comparison shows that the results

agree closely with each other.

A detailed flow diagram is given to demonstrate the solution of the

continuous arch problem by a digital computer. The flow diagram is based on

the following assumptions: (1) it applies to any number of arch spans, with

interior arch Joints on piers; (2) the equation of the centroidal axis of

each arch can be expressed as y=otx + fix + / with origin at either end;

(3) the ratio of the moment of inertia at any section of the arch to the

moment of inertia at the crown is secQ, where 9 is the angle between the tan-

gent to the arch and the horizontal axis; and (I4) the ratio of the moment of

inertia of the pier to that of the crown is constant.

INTRODUCTION

Matrix analysis is a relatively new approach to structural analysis.

The main advantage of analyzing a structural system by the matrix method is

that the analyses can be performed by a computer conveniently. The matrix

method is particularly easy to handle if a structure must be analyzed for

the effects of several loading patterns, such as that used in determining

influence lines.

Fig. 1 Typical continuous arch on slender piers.

"An arch is a girder (beam or truss) usually curved in form, that develops

reactions with inwardly directed horizontal components under the action of

vertical loads alone." A typical continuous arch on slender piers is shown

in Fig. 1 . The effect of slender piers is, in general, to decrease the hori-

zontal thrusts, to increase the crown moments and to throw stress onto adjacent

arch spans and onto other piers. The slender piers are a necessary provision

for a large span.

1. John I. Parcel and Robert B. B. Moorman, Analysis of StaticallyIndeterminate Structures , John Wiley and Sons, Inc., New York, 1?o2, p. U57.

The matrix method presented in this report is the displacement method.

Basically, it makes use of the single fixed arch theory, namely the method

of analyzing a single fixed arch. The analysis requires a knowledge of the

equations of the centroidal axis of each single arch and of the relative

moments of inertia.

This method also involves application of the Muller-Breslau Principle and

numerical integration. Although it is quite tedious to use for manual computa-

tions, the basic idea of each method is simple. It should be introduced when-

ever a computer is available.

SINGLE FIXED ARCH THEORY

An unsymmetrical fixed arch is shown in Fig. 2(a). It is statically-

indeterminate to the third degree. Figure 2(c) indicates the statically

determinate base structure obtained by making a free end at point 0.

(a) (b) (c)

Fig. 2 Fixed arch analysis by superposition.

In the following discussion, stresses are assumed to be below the elastic

limit, that is, Hooke's law applies.

Superposition can be used as long as Hooke's law applies. From Fig. 2,

the following three equations exist;

Vll +X2S12

+ Vl3" A 10«

*1 *21+ X

2S22

+ X3S23 " A

20 •

h SV * *2 S32

+ X3S33 " A30 *

In matrix notation,

&11 & 12 S13 \o)8 21 622 S23 h A 20

&31 S32 633^ W ,

A30J

(1)

where

3^ » horizontal force at origin point 0,

X„ » vertical force at point 0,

X, moment at point 0,

&.. - displacement in base structure at point in X, direction, due to

X." unit acting only,

Ajq " displacement in base structure at point in X. direction, due to

all external loads acting.

"She & terms can be determined by using the dummy unit load method,

'11

'22

Area

n-m- L 2 .

y dxei cose *

xLds . CL

xgdx

EI cos8 '

;

33dx

ei cose *

xy dxEI cose '

y dxEI

JEI cosO '

^^L^i^-Cra

Defining

Ly2

dx -EI cosO x

x •5

' ^y dxEI cos©

Eq. (1 ) becomes,

o ^ cose y ' )

dxEI cosO

A,

- T CL

t te J" f

LX dx .-V ) EI cosQ ** > > EI cosO " ** *

xy

Ay

xy

Iy

Ax

Ay"* *>

A 10

Ax h A 20

A A ,

A30,

(2)

X|, X., and X,, which are obtained by setting A, "1 , A -0, and ^o" '

will cause a unit displacement in the X. direction only when X. , X~, and X,

are applied simultaneously at 0. In other words, this is the way to determine

the reactions at the origin point due to a unit horizontal displacement at

point 0.

The same argument applies to the other two cases, that is, A, "0, A., -1,

A30=0and A

10=0, A

2Q-0, A

30-1.

Substituting for A 1Q, A 2Q, and A

30in Eq.(2),

Zx V . Ay

*l1

Ixy

Iy

Ax h m 1

Ay Ax A,h\ \

1

then,

1*3

where

xy

Ay

xy

1y

Ax

Ay

Ax

A

*l \ *3

V, 72

V3

>*lK2 ^J

.(3)

E| , V1

, and 1-L, will cause a unit displacement at in the X. direction,

H„, 7„, and VL will cause a unit displacement at in the X. direction,

H,, V,, and H. will cause a unit displacement at in the X- direction.

To obtain the influence lines, Muller-Breslau's Principle plays an im-

portant role. "The ordinates of the influence line for any stress element (

such as axial force, shear, moment, or reaction) of any structure are propor-

tional to those of the deflection curve which is obtained by removing the

restraint corresponding to that element from the structure and introducing in

its place a corresponding deformation into the primary structure which remains.

Furthermore, "In the case of an indeterminate structure, this principle is

limited to structures the material of which is elastic and follows Hooke's

2law." In other words, Muller-Breslau's Principle states that "an influence

line may be drawn by producing artificially a unit displacement corresponding

to the 'stress' for which the influence line is desired. The term 'stress'

3includes reaction, thrust, moment or shear, as the case may be."

„2

2. Charles H. Horrls and John B. Wilbur, Elementary Structural Analysis ,

McGraw-Hill Book Co., New York, i960, p.h93.3. P. C. A. Concrete Information ST id -2, "Concrete Building Frames

Analyzed by Moment distribution," p. 2.

Thus, the influence line for the horizontal force X. is the deflection

curve caused by H, , 7^ and fit. j the influence line for the vertical force X

is the deflection curve caused by IL, V , and M„; and the influence line for

the moment X, is the deflection curve caused by H,, V , and M_.

To find the deflection curves, the conjugate beam method is used. The

deflection curve of the real beam is obtained by double integration of the

area of the elastic weight on the conjugate beam. That is

defl'n • \ fjjj

ds ,

whereI

M-Mj+^x + Ejy,

or K » M2

+ Vgx + Hg y ,

or M M, + V. x + H, y .

This can be performed by a digital computer. The Gaussian 5-point

Integration Formula is used for the first integration and Simpson's Rule

is used for the second integration.

A detailed flow diagram for single fixed arch theory is given in the

first part of the flow diagram of the displacement method in Appendix A, to

illustrate the computer approach.

h. S. D. Conte, Elementary Numerical Analysis , HcGraw-Hill Book Comoany,New York, 1965, p. 138.

MATRIX ANALYSIS OF CONTINUOUS ARCHES BY THE DISPLACEMENT METHOD

Fig. 3 The choice of redundants.

Figure 3 shows a series of arches on slender piers. The arches may or

may not be identical and symmetrical. It is assumed that the equation of the

centroidal axis of each arch is given and that the relative moments of inertia

are known. It is also assumed that the material obeys Hooke's law.

The structure shown in Fig. 3 is statically indeterminate to the ninth

degree. In other words, nine redundants must be removed to obtain a statically

determinate structure, or base structure. Among numerous base structures, the

one shown in Fig. 3 is suggested. The reason for this selection is that with

this base structure identical redundant patterns can be assigned to each single

arch.

The redundants for the continuous arches can be analyzed by the displace-

ment method once the stiffness values are known. These values can be con-

veniently determined by applying single fixed arch theory.

10

Fig. U Degrees of freedom.

Figure k shows the four degrees of freedom of the structure shown in

Fig. 3- The degrees of freedom at the joints are indicated by arrows &., d ,

d,, and d, . The positive sense of rotations, displacements, moments, and

forces is as indicated by these arrows. The q terms are the unbalanced joint

moments or unbalanced horizontal forces due to the external load P.

The displacement method of analysis may be divided into the following

operations

:

(a) Analysis of each single arch assuming that they are fixed ended and

determination of the end moments and thrusts due to d • 1, i~1 , 2, 3, U.

This can be done by applying the single fixed arch theory discussed in

the previous section. Furthermore, influence lines for the reactions of each

single arch can also be obtained.

(b) Determination of the moments and horizontal forces on the pier due

to dj- 1, 1-1 , 2, 3, It.

11

"wd-1 V/l

»CB

*»

Fig. 5 Her subjected to a unit rotation.

Figure $ shows a pier subjected to a unit rotation, with no translation

and with the far end fixed.

Then, by the slope-deflection method,

^C " T »KS+V * " if '

Xi

Hd-1Ac

^ac—— f*BFirJ -r

*c?CB

Fig. 6 Pier subjected to a unit translation.

12

Figure 6 shows a pier subjected to a unit translation, with no rotation

and the far end fixed.

Also, by the slope-deflection method,

4 . as i » h - 151-Rn T."'"-' T.' " 9 >"BO

2EIM - sat (_3 1) . - 2SXi

%! " " I(MBC

+ MCB>

* 12EI

I

(c) Determination of•[ S ], the stiffness matrix for a single arch and

pierj and of [ K ], the stiffness matrix of the whole structure.

The values of the S terms are found in step (a). For instance, S is

the value of end moment or thrust for a single arch in X. direction due to

d « 1. Figure 7 shows the determination of S. . due to d " 1, j«1, 2, 3, h.

(a) d, - 1.

Fig. 7 Determination of [ S ].

13

d2-1

Sl42

"xU W°

(b) d2

- 1.

(c) dj- 1.

(d) dj^ - 1.

Pig. 7 Determination of [ S ]. (Continued)

1U

The stiffness coefficients of [ K ] at the joints are then obtained by

combination of the S. . values and the stiffnesses of the piers obtained in

step (b). In other words, K. . is the value of moment or force in the q.

direction due to d= 1 . For example,

K21

S31

+ S61

+ fWr •

(d) Determination of the resultant reactions, ^FT^ .

Since [ K ] is obtained in step (c), the unknown rotations and displace-

ments can be found from ( D ^- [K ]" 5 q I . With the rotations and displace-

ments known, correction moments and forces at the ends of the members may be

obtained, making use of the single arch stiffness [ S ]. The reactions, due

to the correction of { D \ , are given by j FDJ [ S ]{$\ . Then the final

moments and forces are obtained by SftV = {FF}+{ FD^ , Where ^FFJ is the

fixed end moment or force due to the external load P.

Influence coefficients of the redundants are determined by varying the

position of P, where P is a unit load. A numerical example will be given to

demonstrate the method of analysis.

A complete flow diagram for determining the influence coefficient of the

redundants by the displacement method is given in Appendix A. This flow

diagram can be used to analyze any number of arches with interior arch joints

on piers. The flow diagram has been formulated based on the following assump-

tions: (1) the equation of the centroidal axis of every single arch can bo

15

6expressed in the form y dx + #x + J > with origin at each end point}

(2) the ratio of I, the moment of inertia at any section of a single arch,

to I , the moment of inertia at the crown of the arch, is equal to secS,

where 6 is the angle between the horizontal axis and the tangent to the arch

at the corresponding section} (3) the ratio of I , the moment of inertia of

the pier, to I is constant.

16

NUMERICAL EXAMPLE

y,/3

I soo6

70'

I secO

V r30' $1

•J c

Jo

(a) Dimensions.

I.

30-

7frr-

(b) Degrees of freedom.

Fig. 8 Numerical example.

Given: A two-span continuous parabolic arch on a slender pier is given

in Fig. 8.

For arch AB: y1

- -0.008^ + 1.12k.

y2

- -O.OO8X2 + 0.8x2

For arch BC: y - -O.OO8X3 + 0.8x

y^ - -0.008xj; + 0.U8x,

origin at pt. A

origin at pt. B

origin at pt. B

origin at pt. C

Required: Influence lines for M^, M^, H^, Mgg, MCB , and H^.

17

Solution:

Step 1 . Determination of the end moment and thrust for each arch due to

d. - 1 , and d = 1

.

H 1

-0.0021 701 1*2256

0. 0021 701 52 .^ B>——0. 0001 1 30282*3

0.00001 808U63

0.0250000911*

0.075000525

'

0.00076389U6

.002170152

d_-1H0.00085830888

0.007321*1912 ji-

I

0.007321*2701(36

.

0.0001 373301*1

0.1121*9952

0. 00732U1 911 *\L_+ *1

° 1

0. 000231*37587

0.037U999U221*

Fig. 9 End moments and forces of each arch due to d." 1, d- 1.

(in terras of EI )c

18I

The moment and forces of each arch at support B due to d," 1 or dg- 1 are

determined by applying Eq. (3) from the single fixed arch theory. The moment

at the opposite end, such as support A or support C, is then calculated from

the equations of statics. The results are shown in Fig. 9.

Step 2. Determination of the end moment and horizontal force for the

pier due to d.» 1, d 1. The results are illustrated in Fig. 10.

V 1

HB"

0.00222220.033333-

0.666667

0.033333vL«.-i

Fig. 1 End moments and horizontal forces due to &. » 1 , d " 1

.

(in terms of EI )c

Horizontal force due to d. - 1 , is

6 EI m6 E(5I

C )

~30T~ -0.033333 EI .

c

Moment due to <L » 1 , is

), FT h E^Ir )

M IT" " —30^" " °'666667 EIc-

Horizontal force due to d 1 , is

H - izE12 E(gy—T 3"

—

lj 3cr

- 0.0022222 EI .o

19

Moment due to d_= 1 , is

„ 6 EI6 E<*V

30'-0.033333 EI .

c

The values determined in Step 1 and Step 2 are listed in Table I.

Table I Stiffness of single arch and pier.

V 1.

d2-1

ARCH AB

MAB

"bA

^A

-0.025000091 it

+0.075000525

+0.002170152

-0.00217011*2256

+0.002170152

+0.0001130281(3

ARCH BC^BC

^C

-0.0371*9991*221*

+0.1 121+9952

+0.007321*1912

-0.007321*2701(36

+0.007321*1912

+0.00085830888

PIER BD+0.666667

-0.033333

-0.033333

+0.0022222

Step 3. Determination of the stiffness for single arches and pier.

They are determined in Step 1 and Step 2, namely,

[S ] -

-O.O250OOO9II*

+0.075000525

+0.002170152

+0.1 1 21(9952

-0.0371*9991(221*

+0.00732W912

-0.00217011*2256

+0.002170152

+0.0001130261*3

+0.007321*1 91 2

-0.007321*2701*36

+O.OO08583O888

20

Step k. Determination of [ K ], the stiffness matrix of the whole

structure. Hence [ K ]" follows.

Or,

K11

- 0.07500052? + 0.112li9952 + 0.666666667 - 0.85li166712 ETc ,

K21

- 0.002170152 + 0.00732lj1912 - 0.03333333 - -0.0238389901 EIc

,

K^2

= 0.002170152 0.00732U1912 - 0.03333333 - -0.0238389901 EIc ,

K„„ = 0.0001 13028U3 + 0.00085830888 + 0.002222222 - 0.00319355953 SIC.C. C

Then,

[K ] EI

[K 1-1

0.85U166712

-0.0238389901

EI

1.lt788182

11.0389U6

-0.0238389901

0.00319355953

11.0389lj6

395.53273

Step 5. Set up [ q J and [ FF ].

By applying single fixed arch theory, the influence lines for M,-, Kg,,

H_, in single arch AB and influence lines for VL„, M-g, H_. in single arch BC

are obtained in Appendix C.

[ q ], the matrix of unbalanced joint moments and unbalanced horizontal

forces, is set up from the influence coefficients of Mg,, H__, H_,, H-^j

while [ FF ], the matrix of moments or thrust due to fixed ends, is set up

from the influence coefficients of M^, M^, H^, MgC

, Mcg,

H^,. [ q ] and

[ FF ] are given in Table II and Table in respectively.

21

Table H Matrix*U

\ i

3^\ 1 2

1 +0. +0.

2 +0.663706 +0.060li93907

3 +2.1$70U2li +0.2086U22

k +3.8U0006 +0.li0000031

5 +5.21U817 +0.5975311

6 +5.925919 +0.771 605

7 +5.759987 +0.900000

8 +li.6U59 +0.9679008

9 +2.651475 +0.9678999

10 -0.00009 +0.8999978

11 -2.96309 +0.7716017

12 . -5.7361*5 +0.5975266

13 -7.6801

9

+0.399996

.11* -8.01 205 +0.208637

IS -5.8076 +O.O60fj68

16 +0. +0.

17 +li.8599U -0.189866

18 +5.1199U -O.6OOO26

19 +2.93993 -1 .033621

20 -0.00005 -1 .35002U3

21 -2.500032 -1 .U6U86U

22 -3.8lt0O12 -1.3500151

23 -3.780003 -1 .0336038

2U -2.56 -0.6000052

25 -0. 8999988 -0.1 098U51

8

26 +0. +0.

22

Tabla III Matrix FF

\i1 2 3 1* 5 6

1 +0. +0. +0. +0. +0. +0.

2 -5.80789 -0.663706 -0.060U939 +0. +0. +0.

3 -8.01236 -2.1570U2U -0. 20861*22 +0. +0. +0.

h -7.68051 -3.8U0006 -0.1*0000031 +0. +0. +0.

5 -5.73677 -5.21ll817 -0.5975311 +0. +0. +0.

6 -2.963U3 -5.925919 -0.771605 +0. +0. +0.

7 -0.000U -5.759987 -0.9000 +0. +0. +0.

8 +2.65Wj8 -U.6U59 -0.9679008 +0. +0. +0.

9 +U.6U567 -2.65W5 -0.9678999 +0. +0. +0.

10 +5.759802 +0.00009 -0.8999978 +0. +0. +0.

11 +5.925783 +2.96309 -0.7716017 +0. +0. +0.

12 +5.21 U722 +5.736U5 -0.5975266 +0. +0. +0.

13 +3.83995 +7.68019 -0.399996 +0. +0. +0.

Mi +2.1570159 +8.01 205 -0.208637 +0. +0. +0.

15 +O.663698U +5.8076 -0.0601*88 +0. +0. +0.

16 +0. +0. +0. +0. +0. +0.

17 +0. +0. +0. -U.8599U -0.900001 +0.1 89866

18 +0. +0. +0. -5.11991* -2.H9999 +0.600026

19 +0. +0. +0. -2.93993 -3.779982 +1 .033621

20 +0. +0. +0. +0.00005 -3.839956 +1 .350021*3

21 +0. +0. +0. +2.500032 -2.1*99901* +1 .1*61*861*

22 +0. +0. +0. +3.81*0012 +0.00015 +1.3500151

23 +0. +0. +0. +3.780003 +2.91*021 +1 .0336038

2U +0. +0. +0. +2.56 +5.12026 +0.6000052

25 +0. +0. +0. +0.8999988 +1*. 86028 +0.1 8981*51

8

26 +0. +0. +0. +0. +0. +0.

23

Step 6. Determination of [ FT J, the matrix of influence coefficients

for mab> *nk> hk> "bc' "cb'

"ad %o«

[ FT ] is obtained in the following manner.

[ FT ] > [ FF ] + [ FD ]

" [ FF ] + [ S ] [ D ]

- [ FF ] + [ S ] [ K ]

_1[ q ]

The results of [ FT ] are drawn in Fig. 11 through Fig. 16.

These results are checked in Appendix B using the energy method. A

comparison of the influence coefficients obtained frctn the two methods shows

excellent agreement (see Figs 11-16).

Ifaergy

Method

+0.0

+0.27015

+0.8361(7

+1 .1(0257

+1.76510

+1.81372

+1.53117

+0.99320

+0.36862

-0.0807U

+0.0

+0.9U87U

+2.1i0805

+3.85338

It. 88537

+5.22975

+U. 7371(5

+3.381(50

+1.27209

-1 .3731(6

-It. 20065

-6.73289

-8.3681(2 .

-8.38033

-5.91658

+0.0

2U

Displacement //

''

I

Method

+0.0 / cj

+0.2701

8

+0.83658

/+1 .1(0276 h+1.76533

+•

+1.81 397 — +

+1.53139

t c+0.99335 V+O.36869 \ --0.08073 *+0.0 a

1*

+0.91(892

// s

+2.U0850

/_ H

+3.851(15 / c

+1(.8861(3 / 3H+

+5.23106

+k. 73891

+3.38600

H

t-H

.-1

M+1 .27350 — fe.

-1.3722l(

-l(.1997l(

-6.73229

-8.36820

-8.380U5

1 \

-5.91695

+0.0 a)L

25

Energy-

Method

+0.0

-0.10,11,8

-1 .3569U

-2.25258

-2.791,18

-2.80712

-2.27632

-1 .31,632

-0.32121

+0.33531

+0.0

+5.12633

+7. 2835k

+7.3191*2

+5^971 26

+3.86576

+1.51910

-0.66308

-2.38566

-3.1*61109

-3.821,35

-3.50295

-2.61*695

-1.51395

-0.1*7209

+0.0

DisplacementMethod

+0.0

-0.U*152

-1.35705

-2.25276

-2.7910,2

-2.80737

-2.2765U

-1 .31,61,8

-0.32130

+0.33528

+0.0

+5.1263U

+7.28330

+7.31889

+5.97039

+3.86U61

+1.5177U

-0.661,55

-2.38715

-3.1:651,8

-3.82555

-3.50389

-2.61,759

-1.511,29

-0.1,721

8

+0.0

G25

uo

0)

C

cH

to

ft,

Energy-

Method

+0.0

-0.01705

-0.05260

-0.08781

-0.10980

-0.11171

-0.09269

-0.05786

-0.01838

+0.00863

+0.0

-0.08222

-0.23000

-0.1.0673

-0.58201

-0.73176

-0.83812

-0.86952

-0.88065

-0.8121*5

-0.6921 1*

-0.53321

-0.35538

-0.181(68

-0.05338

+0.0

26

DisplacementMethod

+0.0

-0.01705

1

-0.05261

-0.08782 J 4-

1-O.IO98I

-0.11172I *

-O.09270

-0.05787

-0.01838/

"3

/ t31H+0.00862

+0.0 \ m

-0.08222

-0.23002

\ c

—A H

-0.1*0676

-0.58207

-0.731 83

\ 8

\\ «>

\ H\ *H

\ C" \ M

-0.838201

,H

-0.88961

-0.8807U2

j fr«

-0.81253

-0.69221

-0.53326

-0.3551*2

-0.181*70

-0.05338

+0.0 < '

27

Energy-

MethodDisplacemMethod

+0.0 +0.0

-0.10825 -0.10823

-0.55625 -0.55620

-1 .U32U7 -1.U32UO

-2.69689 -2.69680

-U.1 8093 -li. 18085

-5.58755 -5.587U9

-6.U9113 -6.U9111

-6.33757 -6.33761

-h.hhh2h -k.UhklO

+0.0 +0.0

-1.185W -1 .1 8539

-1.11738 -1.11721

-0.21)351 -0.2U316

+1.05lt38 +1 .05U91

+2.U6031 +2.U6100

+3.721*11 +3.72U92

+U.661U0 +I1.66228

+5.15365 +5.15W3

+5.1U810 +5.1U892

+U. 65783 +1..6585U

+3.76173 +3.76228

+2.6oWj8 +2.601*85

+1.39659 +1 .39679

+0.U1Ui0 +o.WWi5

+0.0 +0.0

cq

uo

ID

oc

H<nC

toH

28

Energy-

MethodDisplacementMethod

+0.0 +0.0

+5.61 12U +5.61 152

+7.U5553 +7.1*5579

+6.877US +6.87766

+h.9932h +1*. 9931*0

+2.69081* +2.69096

+0.62983 +0.62991

-0.75855 -0.7581*5

-1.27132 -1.27123

-0.93390 -0.93381

+0.0 +0.0

+0.59136 +0.59135

+0.1(011*3 +0.U0132

-0.27726 -0.27750

-1.19609 -1.1 961(6

-2.15036 -2.1 5081*

-2.97926 -2.97983

-3.56591 -3.56653

-3.83732 -3.8379U

-3.761*1*0 -3.761*99

-3. 361 99 -3.362i*9

-2.68882 -2.68922

-1.81*751* -1 .81*780

-0.981*68 -0.981*83

-0.29072 -0.29076

+0.0 +0.0

Gtq

uo

0)

c

oosa)

c

60

6.

29

EnergyMethod

DisplacementMethod

+0.0 +0.0

+0.09176 +0.09177

+0.29581 +0.29582

+0.52237 +0.52238

+0. 701*56 +0. 701*57

+0.79831* +0.79836

+0.78253 +0.78255

+0.65883 +0.65885

+0.1*5176 +0.1*5178

+0.20873 +0.20871*

+0.0 +0.0

-0.09251 -0.09250

-0.07501 -0.071*99

+0.01215 +0.01218

+0.1 31*63 +0.131*68

+0.261*10 +0.261*17

+0.37823 +0.37830

+O.I16067 +O.U6076

+0.50110 +0.50119

+0.1*9519 +0.1,9527

+0.1*1*1*60 +O.UUJU67

+0.35701 +0.35706

+0.21*608 +0.2U611

+0.1 312*8 +0.13150

+0.03890 +0.03891

+0.0 +0.0

n

I

cq

Wf-l

O<n

aB

coC<*>

3

to

30

CONCLUSIONS

The following conclusions can be made from the work completed in this

report.

(1

)

Since the influence lines for the redundants are obtained, the moment,

shear and thrust at any section due to any loading condition on the structure

can be calculated.

(2) The matrix analysis presented in this report can be preformed con-

veniently by a digit computer rather than by hand computations.

(3) The method, presented in detail in the flow diagram, may be applied

to any continuous arch, symmetrical or unsymmetrical.

(U) The flow diagram, given in appendix A, is based on the assumptions

that the arches are parabolic; that the ratio of I, the moment of inertia at

any section of the arch, to I , the moment of inertia at the crown, is sec9.

However, with minor modifications the method is applicable to any other shape

of arches and applicable to the case where the ratio of I to I is a functionc

of X.

(5) In this study, the influence coefficients of the redundants, obtained

by the displacement method and by the energy method, have been shown to agree

very closely. It is evident that any structure may be analyzed by one method,

and then independently checked by the other.

31

ACKN0WLED3MEHT

The writer wishes to express his sincere gratitude and deep appreciation

to his major professor, Dr. Peter B. Cooper, for his valuable advice, criticism

and suggestions during the preparation of this report.

32

REFERENCES

1

.

James Michalos and Edward H. Wilson, Structural Mechanics and Analysis ,

The Macmillan Company, New York, 1965.

2. Samuel T. Carpenter, Structural Mechanics , John Wiley and Sons, Inc.,

New York, 1960.

3. Alexander Hrennikoff, "Analysis of Multiple Arches," Transactions , ASCE,

Vol. 101, p.388 - p.U21, 1936.

k, John I. Parcel and Robert B. B. Moorman, Analysis of Statically Indeter-

minate Structures, John Wiley and Sons, Inc., New York, 1962.

$. Henry L. Langhaar, Energy Methods in Applied Mechanics , John Wiley and

Sons, Inc., New York, 1962.

6. William G. Godden, Numerical Analysis of Beam and Column Structures ,

Prentice-Hall, Inc., New York, 1965.

7. S. D. Conte, Elementary Numerical Analysis , McGraw-Hill Book Company,

New York, 1965.

8. P. C. A. Concrete Information ST U1-2, "Concrete Building Frames Analyzed

by Moment Distribution," p. 2.

33

NOTATION

B modulus of elasticity of the material.

I moment of inertia at any section of arch.

I moment of inertia at crown.c

I moment of inertia of pier.P

6 the angle between the tangent at any section of the arch

and horizontal axis.

A total area of an arch divided by EI.

x the distance from centroid to y-axis.

y the distance from centroid to x-axis.

I moment of inertia of the area A with respect to the x-axis

.

I moment of inertia of the area A with respect to the y-axis.

I product of inertia of the area A with respect to x- and y-

axes.

X. horizontal force at origin point 0.

Z vertical force at origin point 0.

X, moment at origin point 0.

J, displacement in base structure at in X direction, due touie m

X unit acting only.

A displacement in base structure at in X. direction, due to

all external loads acting,

m. moment anywhere in the structure due to X." unit.

m„ moment anywhere in the structure due to X»- unit.

d ' degree of freedom.

d, , 8, 7" arbitrary constants.

3U

1. unbalanced Joint moment or horizontal force in d. direction,

redundants.^1 f ^2 ! *

Si.

K. . the moment or force required at joint in q^ direction to

S. . the moment or force in X. direction due to d - 1

.

cause d.= 1

.

P external load.

D displacement of joint.

FF moments or forces due to fixed ends.

FD moments or forces due to displacement D.

FT total moments or forces, sum of FF and FD.

matrix notation.

{ \ column matrix.

[ S ] stiffness matrix of single arch.

[ K ] stiffness matrix of whole structure.

3*

APPENDIX A

36

FLOW DIAGRAM FOR THE DISPLACEMENT METHOD

I. Single Arch Analysis

Number of arches &interval length

Constants ofGaussian 5-pt.

IntegrationFomula

Arch span length

START

(READ m7~iT)

te--t., =0.9061 7985 a--^ =0.23692689

t,--t2-0.S38U6931 a,=a

2-O.U7862869

t3»o. a

3=0.56888889

H3-S-1

li'-l

A

37

Initialize [ FT ] <

1 = lUa13 " <P"

Wli'°

<1^>^ 3*3+1

i"i+l

Yes

X- 1

m -tA*i

k » 2i - 1U»-

( READ ST ^, rk )r

A-0 1-0 HA(1)=0 HIx(1)=0

Ax=0 I -0 HAx(1 )=0 HI (1 )-0

Ay-0 I -0 HAy(1)-0 HI (1 )-0

i " 2

HA(j)-0 HAx(j)°0 HAy(j)-0

Klx(i)"0

KLy(i)-0 HI^CjJ-O <D

38

xa

= nx - (3 - 2)h

x. - n, - (J - 1 )h

Gaussianintegrating <process

X - t.jh/2 + (xa

+ xb)/2

GA(i)-1. GI (i)-!2

GX(i)=X GI (i)-X2

GT(i)-I GI (i)-XTxy

i-i+1

HA(J) - mi) - a.

±Gk(i.)h/2

HAx(j) - HAx(j) + ftjCOSDh/a

HAy(;j) - HAy(J) + a^Dh/S-«'-—

»

^x^ " HIx(;)>

+ aiGIx(i)h/2

HIy(j) -HI

y(d) + a^yDh/2

HV J)° "V^* aiGV1)h/2

<£„ 5>i^i-i+1

Yes

c

39

Single arch properties <

A - A +HA(j) V w**wAx= Ax+HAx(j) V VHI

y(3)

Ay=» Ay+HAy(j) VVV13

'

I I Ayx xy

K - I I Axxy y

Ay Ax Ak

\

mv. [Ac]k

w£- \k H2k

V1k

V2k

H3k]

V3k

«1kM2k

M3kj

AH(1 ) = SH(1 )-

AV(1 ) - SV(1 )-

AM(1 ) - SM(1 )-

N2 (K1 + 1 )/2

Uo

Gaussianintergrat.1 ngprocess

AH(J)=0 AV(j)=0 m(i)-o

*»"»«.- H - 2 )h

^ - a^ - (i - 1 )h

i » 1

X - t±h/2 + (x

&+ x^/2

i-i+1

<D

EPAH(j) AH(j) + a.jFH(i) h/2

AV(j) - mi) * a±FV(i) h/2

AM(J) - M(i) + kIK(i) h/2

< ± . ^>lH^i.i+1 |-

M

Simpson's rule <

AH(3) - AH(j) + AH(j-1 )

AV(j) " AV(j) + AV(J-1

)

AH(J) - iffl(j) + AM( j-1

)

SH(i) - SH(i-1) + (AH(j)+lMH(j+1)+AH(j+2)) h/3

SV(i) - SV(i-1) + (AV(j)+UV(j+1)+AV(3+2)) h/3

SM(i) - SM(i-1) + (AM(j)+MM(j+1)+AM(j+2)) h/3

U2

M • 31

N-fgL+2-i -<Z>

K - 3J- 1

,.ft.i -©

Influencecoefficientfor singlearch

M-2,N. K 7k< 'sH(i)l

^M.N.

H3k

V3k

M3k,

SV(i)

SM(i)J

^N"^^^ SH(i)

SV(i)

SM(i)

i-i+1

H. Continuous Arch Analysis

Degrees of freedom

Initialize [ q ] <

£33—(6

<D

f - 2(m - 1 )

i-1

TZ3>-

^J-

<J - K:>*HwiM

s Ho rrtV^i.fip

H

U3

Set up [ q ] <

Height of pier & ratio ( RKAD p. , d. }of I of pier to I

Initialize [ S ] <

Noi-m-p -li-i+1

Yes

i - 1

1-1 -I 3 i t-

1

s±r°

<f"3>-^j-mHYes

Mi

No

Set up [ S ]

<1 = 3m>

YesJ

|J"1|

I

-H3=3+1| ^

i - M " 1/1

2 I

1

k-lfl

<D

Si,3 "

M3,J-1 Vj,j*l

+ Si,3+v -\mVi,M*

Si+1,3"

M3,3+1 Vi^+r^^*!

Si+2,3"

H3,3+1

Si+2,3+l"\3+l

Si+3,3*

M3,3 +2

Sl+3,3+l"\3+2

Si+U,3"

H3,3+2

Si+U,3+1" \j+2

Si+5,3*

lS,***VlT3,#*si+5,3+i" "i,3+2-Vl

V1, j +2

+W3^l^3 +3Vl +^ +3

)H3,3+2

+(^+3^+1+^+3Vl +

^+3)H1,3+2

k$

Pier properties

Initialize [ X ] <

Set up [ K ]

li-1

MPi2

~-6i±/pl

HPi2

- 1 2d±/p|

<i»i^NJf9Ji.i+1 L_

Yes

li»1'

i•

\

v - o L-».*ij

Yes I

3"J +11-

<i^)>^Yes

kh- S21

+sU1

+mpii WVu

K21" S31

+S61+HP

11K22" S

32+S62

+HP12

Yes

Ho

®ll

®

U6

K31

" S51

K32 " S

52

KW " S61 \2

S62

J-f -1

12

K0-2,j

=Si-1,J

Kj-2,j +rs

i-i,j+i

K3-i,3+l"

Si+1 f J+1

2 ' n 2' 2

K3*1 ,

J

-Si+1 ,

J

+Si+U,

J

+HPJil 1 Vi .3*1

°Si+ i .3+1

+Si^, 3+i*-w ,

Yea

J =3

. 31 + 1X

2

- i

(

©

U7

Kj-2,3"

Si-i,3

K3-2,j+1

=Si-1,j+1

KJ-i,d

=si+i,J

Kd-1,5 +l"

Si+1^ +1

\i osi,J

+si+2 , 3

+M,ii

,

2»' 2

»*

2 ' 2 '

Kj+2,j"Si+3,3

K3 +2,j+r

Si+3,3+l

KJ+3,o

=Si^,J

KJ +3,j+1°

Si+U,J+1

Influence coefficientsof redundants ofcontinuous arch

.

1 _ fvl" 1/Hot. [KJfrf L

lj(

[d)'-w*«M' '

(ro)- (s) x_(d)

1'

(tf) - [»] + (fd]

f PUNCH [tf]J

W

APPENDIX B

.

h?

CHECK RESULTS OF EXAMPLE USING ENERGY METHOD

The energy method may be divided into the following operations:

(1 ) Six redundants are treated as external loads acting on a statically

determinate base structure. The energy of arch AB, \L ; of arch BC, U,,j and

of pier BD, U-; is then calculated. Thus,

n, -w \> Jtj),

u2- u

2(xu, x

5, x

6 ),

u3

- u3(x

2, x

3, x

h, x

6).

The total energy is given by U » 1. + U + U_. Thus,

{frj t c i{\\ i-i, 2, ... ,6.

(2) ^%A -[C] ||£-^ i-1,2, ... , 6. Let [B)

denotes

-1

[ C ] ,, then

\h\ - f«i{$y.

(3) The influence line for X. is the deflection curve caused by the

\\l\ ' where i " 1 >2,...,6.

50

Step 1 Set up \VL} - [0]{AJ1 1, 2, ... , 6.

V1=(-X

1+X

2-19.2X

;J

)/120

120'

Fig. B1 Free body of arch AB.

t K2

. (12° M

2& vr ( dx2EI sec9 cos©

c

/I 201 f'w 1 P

h; ) I(x

i

+ V + x3y) to

" IT l I[x

i

+ i^(_xi

+ V 19 ' 2X3

)+ V-0 -008* i.iw

c /0

^ui 1^J. . _^_ (ii0Xl 2ox2

nsaij) ,

dx.

3U1 1

3!l Jg- (1152X, + 1152X

2+ 53065.16X

3) .

51

'v2-(-X

u+X5-12.8X

6)/80

80'

12.8'

Fig. B2 Free body of arch BC.

L «2 r 80 M2

vs;i-s M' cbc

„ 2EI secS oosOc

eTJ 1 <Y V + X

6y )

2<*

C 'U

i -80 - 9 5

c

cbc.

?".

^ _L"

\ ' EI<

l.JL5

EI<

"fT"El" (3U1 -33333X

U+ 3W.33333X

5+ 6990.5o667X

6)

r- (26.666671^ + 13.33333Xj + 3U1 .33333^) ,

r- • «r- (13.33333X. + 26.66667X,, + 3UL33333X,) ,

? x5EI

Cu 5

52

*2 \^X3-^•^730'

.. TT V-^ 6 3

1y/y/y

D

Fig. B3 Free body of pier BD.

rL k2

1(

301

[x2

- Ij, (X. - X,)x]2di .

o 3

7\ t

90Xj - 6X!j

+ 90X6 ) ,

2U3 1 , + 1800X, + 90X, - 1800X6 ) ,

9\ ii- °0X, + 62

u - 9ah) '

?U3 1 ,^ET^2

-- 1 800X, - 90X, + 1800X6 ) .

53

The total energy U is the sum of H, , U2

, and U,. Thus,

on ?U, 2^ 0U, i „ ,

23L, 0X, p, 0X,

?X,^- (20X, + U6X

2+ 1062Z

3- 90X

6) ,

22_.JL^Xj EI

c

2U 1

(1152X, + 1062X2

+ 5U88$.16X3

+ yCBjj - 1800X6 ) ,

— (-6X2

+ QOXj + 32.66667Xlt

+ 13.33333X^ + 25L33333X6

) ,

u c

|x~" IT ( 1 3-33333X

h+ 26.66667X

5+ 3W.33333X

6 ) ,

J c

2|_ - _1_( 9ox

2- 1800XJ + 2S1.33333X

U+ 3M .33333^ + 8790.50667X

6 )

^6 c

In matrix notation,

j

2x2

^x3

1

" EIc

22?x

s

22K

llO 20 1152

20 I16 1062 -6 90

1152 1062 5U885.16 90 -1800

-6 90 32.66667 13.33333 251.33333

13.33333 26.66667 3W .33333

90 -1800 251 .33333 3W .33333 8790.50667

V

*6^ )

Si

Step 2 Find [ B ] - [ C ]

[B ] +0.071011512

+0.017965618

-0.001 9095251

+0.015160U27

+0.01 0591 581

-0.001 U1 9671

3

+0.017965618

+0.061 222623

-0.001 686699

+0.027521586

+0.017U07099

-0.0019095251

-0.0016866988

+0.00009558593U

-0.001 00U2091

-0.0006699698

-0.0021^3^991 5 +0. 000091 568307

+o.oi5i6ol»27 +0.01 0591 581 -0.001U196713

+0.027521586 +0.017U07098 -0.0022i3li99lh

-0.001 00U2091 -0.0006699698)1 +0.00009156831

+0.05U373016 -0.002085218 -0.0019610378

-0.002085218 +0.083138013 -O.0O3U8U017

-0.001 961 0378 -O.OO3li8U017 +0.0003U879189

Step 3 Find influence lines for X., i " 1 , 2, .. . , 6.

The influence line for X is the deflection curve caused by | B. .

V

i - 1, 2, ... , 6. The results are given in Fig. 11 through Fig. 16.

55

APPENDIX C

%

SINGLE ARCH INFLUENCE LINES

o\ NO «H Cs. c\m ft UN cv --tr>- CM o NO CNo W CO c^n NOCO O NO ts UN

u-\ iH OSON O NOon C-- cnCN VN soto w no

!>- c--.

NO u\NO *-<

*-< H CO >A ON ON UN ON UNo\ ON o, o- o o ^3- W O

.* >A c\ UN 4 o CN NO o CMH CM VN Jt UN o VO FN CO r-t

IV ON c^- SO NO o On !>- NO o

o *HON CM rsCN CM oon .d- o^~ NO oo CO oNO o o

ONO

CO CN 00 [>- NOCJ CN CN- tH NO NO c- ono ON ON O CM ON cn COON 00 ON NO UN ON NO .-tIN- Cv. ON tH [>- ON CO oNO NO ON C^ ON ON o NOUN ON CO CN- VN CN CM o

*BA

57

--r J- CN ir\

o o> ON oON on o O>A *-t CN o00 *H o o

o .

o00

COCO

o ONo ONo ONNC ONU"\ CO

*zc

o ON CO >A oo ON ON CN ON

o ON ON ON ONo VTN. C- CN ONON v\ o- CO J-

\s

ccCMO

CN vHNO vo H ^ J* UN.

NO CM CO CM vC iH

CO o NO O CO OON o Ci O -d- oCO o CN, Vf> VO U"\

*H no o CN J- c^

'A UNO J»O COO ONo CONO t-H

*K

MATRIX ANALYSIS OF CONTINUOUS PARABOLIC ARCHES

CHAO-SHYOIC HSU

Diploma, Taipei Institute of Technology, 1963

AN ABSTRACT OF A MASTER'S REPORT

submitted in partial fulfillment of the

requirements for the degree

MASTER OF SCIENCE

Department of Civil Engineering

KANSAS STATE UNIVERSITYManhattan, Kansas

1967

This report presents a matrix analysis of continuous arches on slender

piers by obtaining influence lines for the redundants. The analysis is based

on the displacement method.

The method consists of two main parts. First, each single arch is analyzed

by applying single fixed arch theory. Second, continuous arches are solved

by applying the displacement method.

The procedure is demonstrated by a detailed flow diagram, given in Appendix

A. The flow diagram is determined based on the assumptions that (1 ) the arches

are parabolic, the equations of the centroidal axis of each arch with the origin

at each end are given, (2) the ratio of the moment of inertia at any section

of the arch to the moment of inertia at the crown is secO, where is the angle

between the tangent to the arch and the horizontal axis; the ratio of the moment

of the inertia of the pier to that of the crown is constant, (3) the interior

joints are on piers. The method applies to any number of arch spans.

A numerical example consisting of two parabolic arch spans with a central

slender pier and two fixed ends is then presented. In addition to analyzing

the example by the displacement method, the energy method is used to check the

results. A comparison of the results obtained from the two methods shows ex-

cellent agreement.

The matrix method of analysis used in the report involves the application

of the Muller-Breslau Principle and numerical double integration. In this

study, the Gaussian 5-point Integration Formula is used for the first integra-

tion and Simpson's Rule is used for the second integration.