matrix analysis of continuous parabolic arches · 19 momentduetod_=1,is „6ei 6e
TRANSCRIPT
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.