8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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Scholars' Mine
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1939
Design study of a three span continuous tied-arch bridge
George Perry Steen
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P! % C E%% C2Department:
R%#%$%$ C!S%%, G%% P%, "D%2 2$ ! %% 20! #2 %$-!# $%" (1939). Masters Teses. P!0% 4754.
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8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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Design
Study of a Three Span Continuous
Tied Arch Bridge
George
Perry Steen
A
Thesis
Submitted
to the Faculty of the
School of
Mines
and Metallurgy
of
The University
of Missouri
In p r t l ful f i l lment
of the
work
required
for
the
Degree Of
Master of Science in Civil
Engineering
Rolla Missouri
9 9
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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KNOWLEDGEMENT
o
Professor E W Carlton for
h is valuable cri t ic ism and to
Mr Howard
Mullins
for
his help and suggestions the
writer owes
an expression
of appreci at ion
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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T LE OF CONTENTS
Acknowledgment
Page
Synopsis
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5
Lis t of l lus t ra t ions
• • • • • • • • • • • • • • • • • • • • • •
6
A General
Discussion
of tat ical ly
Indeterminate Structures
7
A Brief H istory o f
Arch i s t o r y
Description
of
Type
• • • • • • • • • • • • • • • • • • • • • • •
The
Principle
of
Least
Work •••••••••••••••
6
Direct
Stresses
Notation
Influence Lines
• •
8
19
tress
Analysis
Diagram For Main Span Equations
Equations For Design Main
Span
Diagram
For
Side
Span
Equations
Equations For
Design
Side
Span
• • • • • • • • • •
•
• • • • • •
•
• • •
25
6
35
36
Evaluation
of
Evaluation
of
±
x
A
f s e ~ ¢ d x
o At
•
•
• • • • • •
•
•
•
•
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T LE OF
CONTENTS
Page
omputations esign
N o
6
on lusions
• • • • • • • • • • • • • • • • • • • • • • • • •
B i b l i o g r a p h y
57
Tables
and l lustr t ions • • • • • • • • • • • • • • •
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Synopsis
In th is study
of
the continuous
t ied arch
r i g e ~
having one main span and two equal side spans the
follow-
ing items
are endeavored:
1
se t
up
the
to ta l
in ternal
work
equations
and
to find
the values
of
the unknown
forces
or
redundants.
2. compute to maximum moments for design and to
complete
the
design of
the
super s truc tu re for
one
bridge.
Detai ls such
as
the gusset plates
r ive t s
connection
angles e tc are
not
design-
ed;
the effect of such b ein g allowed in the
design
for
dead
load as a certain
per
cent for
deta i l s
3.
se t up hypothetical conditions for other
bridges of different span lengths
and
to con-
s t ruct
the
influence
l ines of
the
unknown forces
of
each.
4 compare the
different
bridges upon
the
basis
of
the i r
influence
diagrams.
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List of I l lus t ra t ions
Diagram for
Main Span
Equations
Diagram For
Side
Span Equation
Evaluation
of Equations Design
N o
Evaluation
of Equations Design No 2
Evaluation of Equations Design N o 3
Evaluation
of Equations Design No 4
Influence Diagrams
Design
N o
Influence Diagrams Design No 2
Influence
Diagrams
Des ign No. 3
Influence Diagrams Design No. 4
Influence
Ordinates For Live
Load
Moment Design No.
aximum
Positive
and
Negative
Live Load Moment Design No.
1
Influence
Ordinates for Dead
Load Moment and aximumMoment
for
Design
No.
1
Section
Summary
Design
No. 1
Section
and Weight
Summary Design No.
1
Plate A
Plate
Plate
Plate 2
Plate 3
Plate 4
Plate
5
Plate 6
Plate
7
Plate
8
Plate 9
Plate 10
Plate
11
Plate
12
Plate
12A
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A General
Discussion
of Stat ical ly
Indeterminate
Structures
The
s ta t ica l ly
indeterminate type of struc-
ture
i s
not g enera lly favored by the American Engineer.
Until
recently
this
type
has
been
actually
shunned
be
cause
of
the
lack of information
and the
lack of t ra in
ing in the design of such. The
majority of
st ructural
engineers prefer
to
use
types
with which they are famil
ia r making many
estimates based
on
judgment and experi
ence.
They
contend too
tha t
for
every indeterminate
st ructure tha t is possible
to
design a determinate
type
tha t w il l
carry the
same load with as l i t t l e or
less
material .
A
few
ci ta t ions fol low giv ing the com-
parative
amount
of
material
needed
for
a
type
of bridge
stepping
from
a s ta t ica l ly determinate one to
one
of in
determinateness the f i r s t
degree
or from one toanoth
er
type that i s
s ta t ical ly
indeterminate a higher degree.
trFor
the
Sciotovi l le Bridge
over
the Ohio r iver
a t Sciotovi l le Ohio con sis tin g o f
two
spans a t
77 feet
the saving favor of continuous
spans over
two simple
spans
was
found to
be about
15 per
cent.
ftAccording to comparative studies reported by
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four
spans
is
16
19 and 21
per
cent
respectively
when the
span
length is about 325 feet ;
and
20 24
and
28 per
cent
respectively
when the
length is
about
500
feet .
1
Ierriman and Jacoby investigated the resul t
of placing
a
hinge
a t
the
crown
of
a
550-foot
two-hinged
spandrel-braced arch bridge over the Niagara
River
and
found tha t i f jus t
one hinge
were placed in
the
upper
chord the weight of the arches alone would be decreased
11.8
per
cent. 2
Professor
Dietz
of
Munich s tates
that
a
very careful comparison o f mate ria l required for
a
two
hinged and three -h inged arch for
the
110-foot span of
the Hacker
Bridge in
Munich indicated that 11.3 per
cent
excess
was
required
for
the
l a t te r
type.
2
A chief
objection
to the
use
of a continuous
bridge or of a
h inge le ss a rch has
been that i t
is
sensi
t ive to s ettlemen ts of foundations. Needless to say
1 .
Hool and
Kline
Movable
and Long-Span Steel
Bridges p. 200.
2 .
Parcel and Maney
s truct ion
p. 405
Stat ical ly Indeterminate
Con
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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th is type
of bridge would not
be considered unless
rock
foundation
was
available
and any
s l ight
se t t l e -
ment of
the
foundation would cause
very
l i t t l e change
in
the
s t ress
Another objection has
been
that the
continu-
ous bridge
would be
seriously
affected by temperature
changes.
The effect
of
the expansion and contraction
of intermQdiate piers wil l
cause
l i t t l e
changes
in
s t resses
according to
several
studies
reported.
I t
has also been determined tha t
the
effec t of unequal
temperature
changes
different
parts
of the
span can
safely
be
neglected providing the members
of the
span
have the
same
coeff ic ient
of expansion.
A continuous
bridge of the
indeterminate type
has advan tages r ig id i ty
and
economy and the
fewer
number of expansion jo in ts I t s use permits a decrease
the width of intermediate piers .
In
addition a
continuous
bridge
and especial ly a continuous arch span
has a l l the advantages
of
a cantilever bridge
for loca
t ions
over
navigable
waters
where
clearance
must be
mainta ined during the
construction
period.
The
cont in
uous s t ructure e lim in ate s th e detai l ing
and expense of
hinges. There i s
no question but
that the design of an
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indeterminate
structure involves great deal more work
and
calculat ions
t
is
usually
necessary
to
m ke more
than one t r i l design However i f small per
cent
of
material
and
weight is saved
the
extra
time spent
in
design is
warranted
the f in l analysis
the resul ts
given by
engineers both pro
and
con
on the
st t ic l ly
indeter-
minate
structure
are l ar ge ly t he o re ti ca l and are not
based on
extensive experiments There are not enough
actual
monuments of
i n e t e r m ~ t e
structures that have
approximately
the
s me
s i te
conditions of the
m ny
sim-
ple
structures
Until more
such
bridges are bui l t
affording
us
more re l i ble comparisons the advantages
of one
or of
the other wil l
be
debated largely on the
basis
of
theory t
is probable that the
st t ic l ly
determinate s t ructure wil l
be
favored for
re la t ively
short spans and
the inde te rminate
bridge
will replace
them
for
longer spans
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A Brief
H istory of Arch
Bridges
The
modern period in arch-bridge building be
gan in the year 1716
when
the French ~ e p r t e m e n t des
Ponts
e t
Chaussies
was
formed
although
the
f i r s t
real
emploYment
of the
t rue-arch
type
was
as
early
as
6
B.C.
The f i r s t
arch span
of
the monolithic
type was a
wrought
iron bridge bui l t in 1808
over
the River
Crow a t
St.
Den-
nis However,
was not unt i l several years l a te r in
1879, tha t Weyrauch
developed
the
fundamental equations
upon which
the
elast ic-arch theory is based. Arches of
iron
have
given way to structural s teel spans, which type
the
form of r ibs hinged and
f ixed),
t russ design, or
braced
spandrel
construction i s
being
ut i l ized to a very
large
extent
in
the
f ield
today.
The
concrete
arch
con
s t i tutes the
l a t es t
development in the arch
and
is
a
very
important
type,
especial ly
for
short-span structures, in
which has
nearly
an exclusive f ie ld
Weare more
part icular ly concerned with the
long-span
structure, in
which the s tee l arch supersedes
the
use
of
concrete
for
material .
I t i s per t inent and
interest ing to
follow th e
several
examples
of arch spans tha t
have
been erected.
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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tensively is
the
Eads Bridge
erected
in 1874 across the
Mississippi
River
a t
s t
Louis.
t
consists
of
two
side
spans of 502 feet and a center span
of
520 feet with four
l ines of fixed circular arch trusses
with paral lel
chords.
The
structure
carries an upper deck of a 52 foot highway
and a
lower deck
32 feet wide
for
railway
t raf f ic
In
1889 the
Washington
Arch Bridge was bui l t
over
the
Harlem River a t
w
York. wo spans were erect-
ed
at
509 feet and
had 6
l ines of
circular plate girder
arches 13 feet deep and a 90 foot
r ise
Each girder has
two
hinges.
The
bridge
carries
a highway 80
feet
wide.
The
bridge
Alexander
was constructed
1899 over the River
Seine
a t Paris France. Spans of 352
feet with 15 l ines of circular
plate girder arches
3 feet
deep and a 21
foot r ise
were used. A highway deck was
supported on
top
with a width of 131 fee t .
During 1897 1898 the
Rhine Bridge
a t
Bonn
Germany was erected. t was a through
arch
having a
highway
f loor 46
feet wide. t contained two l ines of
arch
trusses
614
feet
center
to
center
of
hinges with
a
r ise of 97
fee t to
the
bottom chord.
The Wupper Railway Viaduct was completed
1897
a t
Mungsten
Germany. The spans were hingeless-
arch
trusses
588 fe·et long
with
a r ise of 219 feet and
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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carried
a deck of 28 fee t in width. The
t russes
were
16
feet
apart
a t
the
~ o v v
and
84
feet
apart
a t
the
springing l ine
n
of
the f i r s t bridges of the type which is
under
study was a
three span continuous t russ
with a t ied
arch in the
m iddle span
bui l t over the Angora
River
a t
Irkutsk
Russia. The
s t ructure c on sis te d o f
two
side
spans
278 fee t and a
center span of
5 feet with
the
arch
r i s ing 78
fee t
In 1933 the Lachine Bridge was designed and
la te r b uil t
across the
S t
Lawrence
River
near
Montreal.
t consisted of
a
continuous
structure
of several spans
but the main span was a
continuous
ti ed ar ch tr us s 400
feet
long. This
structure was
designed
by the Lake
S t.
Louis Bridge Corporation and constructed by the Dominion
Bridge
Company Ltd.
There is no bridge known that consists o fa
three span
continuous plate
girder with
a t ied arch
r ib
in the
middle
span.
This
scheme was
suggested by
the
l a s t
competition
a t
Mannheim Germany
but
is
not
known whether any has
been
designed and
erected.
The
t ied
arch
i t se l f has
been
used
frequently
in Europe but
no examples can be ci ted
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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D es cr ip tio n o f Type
The type
under study is
a
three span
continu-
ous plate girder bridge with
a t ied arch r ib in
the mid-
dle
span. The s t ructure
i s st t ic l ly indeterminate
in
the th ird degree.
In
an ordinary arch
incl ined react ions
are
exerted
on the
bearings
but
in a
t ied arch the hori-
zontal component of
the reaction is resisted by
the
t ie
connecting
the
two bearings
the t ie
acting as a beam as
far
as
the
react ions
are concerned.
The
stresses
how-
ever are
d ete rm in ed as
in
an
arch.
The
bridge
is
a
through
structure
the roadway being suspended
by
hangers
from the arch
r ib
n
end of
the arch
wil l be
fixed
with
the
remaining bearings
consisting
of
ro l le rs
The arch i t s e l f wil l be parabolic shape
the
r ise
of
the arch being about 1/6 of the span length.
Ample r ig id i ty
must
be secured by making
the
width be-
tween the arch
spans t
le s t 1/20 of the span length.
The side spans are plate girders with
the
roadway fasten-
ed
to
the bottom
of
the
girders . The one
exception is
Design
No The length of the
side spans
do
not allow
the
use of a plate
girder
the weight and s ec ti on r equi r-
ed
being
excessive for
economy
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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side-sway
and
windloads
in
the l a t e r a l direct ion by
ample
bra cing bo th
the
upper
chord
and
in
the sec
t ion of the t i e
I t
i s generally agreed tha t the Warren
t russ
system
i s
the
most
economical for such l a t e r a l
bracing.
Types similar to t h i s a re gener ally designed
on the basis
of
e l a s t i c centers or on
the
basis
of
l e a s t work. The method of l e a s t work
wil l be
follow
ed
in th is
study. This method, commonly known
as
Cas
t i g l ia n o s
second theorem,
was discovered independently
by
A
Castigliano
and
was
developed
i n
his
t r e a t i s e
on
the t1Theorie de
l equil ibrede s
systemes elastiques
A
detai led
explanation of
the
method
wil l follow.
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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The Principle
of
Least Work
The fundamental conception upon which the
method of l eas t work is based
m y
be
presented in an
ideally simple
statement
namely that
a
structure wil l
deform
n
a
manner
consis tent
with
physical
l imitat ions
and
so
tha t the internal
work
of deformation
will
be a
minimum.
nl
n order to apply the
above me.thod the
structure
must
be severed so as
to
obtain a s ta t ica l ly
determinate condition the unknown forces being called
redundants.
Then an
expression for
the
to ta l internal
work
is
se t up in terms of these redundants. The to ta l
in ternal work
i s
equal to one
half
the pro duct
of
the
square of
the
moment and
the
sub division x
divided by
the
product
of
the
moment
of
iner t ia
and
the
modulus
of
e las t ic i ty 2 Part ia l derivatives
of the
internal
work are taken n respect to each redundant and
the
de
rivatives are se t equal
to zero. The unknowns
m y
be
determined y solving the result ing simultaneous
equa-
t ions.
A
m m mum
condition
is obtained
n
set t ing the
1.
2.
Grunter
Theory
of
Modern Steel Structures p.
64.
See Plate
A
for the
mathematical
development of the
to ta l in ternal work.
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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derivative
to
zero.
In
every
case of
s t t ic l inde-
termination
where
an
indefini te nrunber
of
different
values
of the redundant
forces
X
will
s tisfy l l s t t -
ic l
requirements the
t rue
values are
those which
rend-
er the to t l in tern l work of deformation a minimum.
D
The
indeterminate
forces or redundants are
chosen
t
points
where
.any displacements are prevented such as
supports.
Hence
the p r t i l derivative is equal to
zero.
Parcel
and
Maney Elementary
Treatise of t ti-
cal ly
Indeterminate
Stresses p. 123.
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Direct S tr es se s
The
tension
s tress in the
horizontal
t ie
produces a direct stress
in
the arch
ring
equal to
Sec
e
There
i s
also a direct stress produced
in
the
t ie
i t se l f
I t
i s
affected
by
the
camber
the
t i e but
the
angle of
inclination i s so
small
tha t
the
secant of
the
angle i s nearly
unity
There-
fore the d ire ct s tre ss i s equal to The camber
of
the
t ie also causes a direct stress in the
hangers
but
the
value of the stress i s negligible
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Notation
H =
horizontal
thrust t ie
Ml= moment
a t l e f t
end
of arch span over pier
M
moment
a t r ight end
of
arch span over pier
I = moment
of
iner t ia
a t any cross section of arch
1 =
moment of iner t ia of side span cross section
A = arch section in square feet
Ah= hanger sect ion in square feet
At=
horizontal
t ie section in square feet
X =
distance
from
end
of arch
span
to
any
section
Xl= distance
from end of
side
span to any
section
Y
=:
ver t ica l
distance between
arch
and
t ie
axes a t
any
point
= incl inat ion of
arch
rib
¢ = incl inat ion
of
horizontal t ie section
=
Linear
increment
arch
span fee t
£
Sl
=
l inear
increment
side
span
in feet
L
=
length
of
arch span feet
Ll
=
length of
side
span
feet
=:
modull s
of
e las t ic i ty
of
structural s tee l
Eh=
modulus
of
e las t ic i ty
of
hanger i f other
than
s tee l
f
=
camber or
horizontal
t i e feet
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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A length of any ha nger
in
feet
A
spacing
of
h n g r s ~
or
panel
length
in feet
do
depth of section over pier
in feet
d
depth
of arch in
fee t
dl
depth
of side span sect ion
feet
k
posit ion r t io of
any
load
x... .
•
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Influence Lines
influence l ine may be
described
as a
curve any ordinate of which
gives
the value of
the
func-
t ion shear moment horizontal thrust s t ress e tc .
for
which the curve is
drawn
when a load of unity is a t
the
ordinate.
influence l ine
for
shear
or
moment
records
graphically
the value
of
the func tio n a t
a
sin-
gle section fo r
loads a t
a l l sect ions
which
contrasts
greatly
with
a
simple shear or
moment curve
that records
graphically the
value
of the function
a t
a l l
sections of
a
structure under
a f ix ed loading . From
i t s manner of
construction
rol1ows tha t
any
influence l ine gives
the effect o f a concentrated. load or of
a
series of
such
loads by the mult ipl icat ion
of each
load in tensi ty
by
the
value
of
the in flu en ce o rd in ate a t
the
load.
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Stress Analysis
Ordinari ly
the
f i r s t step the analysis
of
a continuous
s t ructure is the construction of an
influence
diagram
for one of the unknown
reactions
known as
the el s t ic curve.
The
influence l ines
for
l l members
of the
s t ructure
are constructed
by
draw-
ing s t r ight
l ines
across the curve. In the
case of
the structure under study
not
only are the ver t ic l
reactions unknown
but the
moments t
the piers and the
horizontal
thrust
in
the
t ie
as
well. In
such
a
case
influence
l ines
must be obtained for the
three
redun
After
these values are found
the
stresses may
be
determined by s t t ics
The sections
are
proportioned
for
a uniform
l ive
load
in combination with
one
concent ra ted load
placed where
i t s
effec t
s
maximum
This
is in ac-
cordance with the
bridge
specificat ions
of
the
American
Assoc ia tion of State
Highway
Offic ia ls .
l
This load is
chosen
of
such magni tude
as
to
give
an
effect
equal
to
tha t produced by the actual
vehicle
on the
s t ructure .
1 .
The
American Association of
State Highway
Officials
Standard Specificat ions for
Highway Bridges
p.
174.
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Live
loads
can
be used as a uniform load across
the
span
and
the value
of
th e fu nc tio n
can
be
obtained
by m ultiplying th e a rea of
th e in flu en ce diagram
by
the
uniform load per foot . This has been varied
sl ight ly
by
breaking up the load into
a series
of con-
centrated loads assumed
to
be
acting a t
the panel
points . The l ive load moment is then computed
by
mul-
t iplying the panel load by the
ordinate
a t tha t
point.
A fter the
s imultaneous equations
are
solved
by determinants and
the values
of
H
Ml
and M are
found
we
tabulate
the values
for
each
a t
th e v ariou s
load
points .
By taking
these
values and subst i tut-
ing in our i n i t i a l moment
equations we calculate
the
influence ordinates
for
the
moment
a t
each
load point.
3
The dead
load
moments are computed direct ly from our
original i nf luence ord inates
for
H Ml and M as the
dead
loads are
stationary and wil l always
be acting.
4
For our maximum negative and positive l ive load moments
we
must load
th e in flu en ce diagrams
whose
o rd in at es a re
given
on
Plate
10.
5
For
the
maximum
positive
moment
2·
See
Plate 5 p. 63
3·S
ee
Plate 9
p.
67
4·S
ee
Plate 11
p.
69
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DIAGRAM
FOR MAIN
PAN EQUATIONS
PLATE A.
o
Q
NOTATION
W TOTAL. INTERNAL woRK
M:: B E ND I NG
MOMEN f
0(
=
ANGULAR. D I S T ORT I ON
S ::
TOTAL.
LINEAR
O :5TORTIO l
C DISTANCE TO
r::-XTR£ME
FIBER
- U N I T STR5 i :SS IN EXTREMe: FIBER
E MOOUl-US OF ELASTIC IT Y
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The to ta l work
will
be since W=.1..JM2.
d
x
E I
Where
the
l a s t three terms represent the work
done by
the
arch t i e
and hangers i n r es is tin g
the
di
r ec t s tr es se s
Different ia t ing
equation
5
with
respect
to
Ml
and which are the unknown forces:
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f
kl
k l ~ · kl
J . li-k {t-J
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®
1
kt
M2.J.. 1
d
- [U-k f Y- fdx k r
l-1.) ldx.
] =0
t
0
I
0 I
I
1
1 l
f X Idx. - M. \ -xhdx
M2.
\2. r ~ x -
-\
xfdx.
1 I
lz.
I
J
I
1
0 I ,
0
These th ree equations
may be expressed the f o l
lowing form by s U bs ti tu ti ng s in g ul ar co efficien ts fo r one
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Instead o f a ctu ally in te gra tin g in order
to
ob
t in
the
various
values
for a structure we
may divide
the stru ctu re in to
increments
or 65 distances obtain
the values for the
various subdivisions
and
make a
summation
of
l l
individual
values
in
order to obtain
an
evaluation for the entire st ructure.
Theoretically
these
subdivisions should be infini tesimally
small
but
fo r p ra ctic al purposes we make
each
division
equal
to a
panel length or case
of
the arch r ing
the
amount
subtended by a
panel
length. The main reason for so
doing
is to
allow computations
to be
made
in
tabular
form. Therefore we subst i tute the subdivi sion ~
for the term
dx and the
sign
of summation
for the
in tegra l
sign
.
For
a
symmetric.al
structure
we
write the following. In computing
these values
we have
made
the subst i tut ion that
the moment of
in r t i
varies
d
o
3
as
the cube
of the depth.
For I we subst i tute
where do
i s
the depth over the pier and d
is t
any sec-
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02.=
b =
t t \ - ~ , , 6 5
- -
@
I 1 1 1 2
b
::
c
~
1
l1s
+ - )
~
X ~ 5 1 ]
- -
-
- -
@
3 1 0 I I
k t
\
A =
\- k ) E J< ltls ~
n-XbJ6S
@
0 I u I
I k t A \ z
A ~ - [ l - k ) ~ \ - . ~ u s + k Z
\ -x
A S ] - - - - - - - ®
2
1 0
I k \ I
k l
\
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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t should
be
noted that the second
term of
is antisymmetrical to the f i r s t term,
i s
antisymmetrical
to the
and A is antisymmetrical to A2.
The word antisymmetrical is applied to the
condi
t ion in
which the
to ta l
sum of the values
for
a certain
term
is
equal to tha t of another term, but as
the
values
for one
are increasing or
decreasing, the
values
for the
other are
increasing
or decreasing
but
the reverse or-
der .
For example,
the values of
x
a re increasing
from
a minimum a t
the
l e f t side of the
main
span to a m ximum
a t
the
r ig ht p ie r
At the
same time,
the
values for
l-x are
increasing
from a minimum a t the r ight end to
a maximum
a t
the l e f t
pier
where x was a minimum. The
summation
for both terms is equal.
Inter-changing some
of the
coe ff ic ient s for
their
equal in order
to
have
as few
different
coefficients as
possible, we have:
Ha.-M Qz. - Mzaz. = t - - - - - - -
- - - - - -
@
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Solving
for the
unknown
values
by determinants
we
have:
A
Q2. Qz.
A2
b
2 - 03
H
=
A 3 b ~ b 2
=
- Q2.
al.
Q1.. PI. b
3
a 2. b ~ - b
A
b ~ -
b:)+A2- a1.
b
Chbl.)
A 3 Q 1 . b ~
a
2
b; .
C l \ b ~ -
b ~ a ~ { 2 b 3 - 2 b : z . .
Simplifying
and
cancelling
the term
we
obtain:
@
Q A -a2.
Qz. A2..-
b
3
Q31-A
3
b l
M
:= -
e l l -
Qz. - ell.
Q1
b
z.
b
3
D
®
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Where
is the
same
as
for H
The
influence ordinates
for
are
antisymmetrical
to those
of MI hat
is the ordinates e ~ e s e or -
crease from the l t side to the r ight in the same man-
ner
and
value as
the
ordinates for
l
decrease
or -
crease from the r ight to
the
l t and the summation
of
the
ordinates
i s the
same
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DIAGRAM
FOR
5 D E - s P A N
E Q U A T I O N S
P L A T E
B.
kt
•
tv
CJl
I
o
I .k t
3
I-
Z
-;
z
f)
1;;}J
D
..Q.,
t
Q
USING UNITY
LOAD
A N D
T A K I NG
MOMENTS fiT
Z.MRa. :
-
I
l-k).Q.
0
:. R : 1-1
W H E N
< . < ~ . t l
Mrn= I-Je.)X
WHf:N
> ~ ~ , M n ~ I - J e ) x - l x - . k . t ) , . k . . Q - ) )
R2
D E V E L O P M E N T OF REACT ION
M O M E N T S
1*. - - I
I -.II:-}J .
,m
n
x
I
x
J-
R
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Equations for Design
Side Span:
Developing
the
p r l le l theory for the side spans
we
have
t he fol lowing :
For a uni t load t a distance k l from
the
end of
the
side
span
we have:
Loaded
Side
Span
w n k l l il l -k x , M, - - - - - - @
I
Unloaded
Side Span
Mx=
M ~ ~ @
\
MJ
M
1-
t
M
l -
Hy
®
The to t l work wil l be:
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W=
Where
th e l a s t
three terms as
before represent
the work
done by th e
arch t i e
an d
hangers in res i s t -
ing
the d ir ec t s tr es se s.
D i f f e r e n t i a t i n g th e eq ua tio n fo r
to ta l
work with
resp ect
to
th e
unknown
values
H
Ml
an d
respective-
ly we have
fo r
part ia l de r i va t i ve s :
®
\
it
fl l
MJl ( l-)( y d x ~ J d ) . I i
El 0
I El.
0 :
E
0
:
+
ll
sec:l e dx
E G A
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aw
={
M
r ~ x
n-x.h.dx
M2 S\ :L
d
X
11 ItX IdX } ®
dM1 ~ J
E1
J I
I E\ I
o
0 0 0
Setting the above ~ s t
derivatives
equal to zero,
multiplying
through by
E, and rearranging, we have:
@
M . r \ t - l < . y d ~ M
1 1\ xydx
0
\ \ J I 2 1
I
o 0
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These three equations
may
be expressed in the fol
lowing
manner:
Ha M b Mz.c =O -
®
Hal M b3 M;tC3=O
®
In the
same
manner
as
before
we
substi tute
~
for
dx
and
the E
sign
for the
sign
of
integration.
For
a
symmetrical structure
we
have:
\ \
\.
l..E \.
:L
Q
~
L: sec l.fA5
sec-:l.0A5
f r l£ _ 1:
A f).s
\ 0 I 0 0 A
\ 4
Eh 0 A ; ;
- @
\.
b
= a = l l
\ x . v ~ s
- - - - -
, 1
\.L.r I
o
l
c
:
a -
1
I6os
, l I
®
t
l..£i
l
J. t. 2 ]
b
=C := l 1 [L
Cl
-
x
)
+/_)
~ ~ @
2
3 l
0
I \
t, 0
II
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Again
we have the term for
b and
aZ antisymmetri-
cal to
that
for
c
and
a3.
Changing
some of
the coefficients for their equals
we
have:
Hal M
b
3
- M2 bl : 0 ®
Solving for
the unknown value
by
the method
of
d e t e r m i n n ~ s
we
have:
Q2.
Q;t
A ~ b 1 b 3
o -
b
3 -
b
H
=
~
a G \ 1 . Q ~
Ct1
b
1
- b
3
- b
a
-
bz.
=
Multiplying through
by
1 and
cancelling
terms of
~ - 0-3 we
have:
A4 t
®
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M=
a O 02
C 2
A
4
b
3
Q2. 0 b:z.
Q I Q2. 02 -
0
1
b
b
3
Q 2 b ~ b:z.
Where D
has
the
same
value
as
for
H.
Q
2
Q:2 .
Q b2 . b
3
Q : 2 . b
3
b 2 .
For the values of the
various
unknowns we wil l
use a tabular form.
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o
t
Evaluation of sec} edx
A
The arch axis
is assumed
to
be in the form
of a
parabola.
f so, the vert ic l
ordinates,
or
y distances,
varies as the square of the horizontal
measurement,
or x
distances, from the vertex of the
curve,
or t the crown
of the arch. In accordance, assume
that
the equation
of
the
parabola
i s
x 1 = l y
•
At
the
springing
l ine,
when x = 110 and
y
=
36.67, which i s the tot l
use
of the
arch, have by
substitution
the equation ~ ~ ~ l a y
l\OZ
Q
. 36 67
a
::
\2. \00
= \65
78 31
Then,
the equation
becomes
The
slope
of
the
curve
t any point
i s
the tangent
or
value
of
the
curve
equation
From the trigOm Qlb et:eie
r e l a t i o n l l i ~ s w e 1lavethe
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which becomes
h n the
value
of x
a t
any point is substi tuted
in
th is equation
the
value of
sec 2 g is
obtained.
For
example
a t
the
mid point
of
the
f i f th
panel
x
= 2 0
sec:J.e = \ { ~ o o 5 Y = \ O.\2.12.Y·
=
\ .0 \47 .
In
tabulating
the val ies w
substi tute s
the
len gth a rch axis subtended by
the
panel.
feet and A
=
108 sq. in . or
0.75 sq.
f t
w
have
sec
1
A s = 1 0141
2 0 5 :.
2.7.75
0 75
Securing
the values for ~ h panel
the
summation
across the entire
span
gives us the value of s e c ~ e A s
o
Since w multiplied a l l our terms by
w
must
\
multiply
our summation of
~
sec.:J.e /15 by
to
ab
o
ta in
the
f ina l resul t
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Evalua
t ion
of \sec
A
dx
o
T
The
horizonta l
t ie
between
the bearings of the
arch i s cambered but
the angle of incl inat ion
i s
so
small
tha t
the
secant of the
angle
approaches unity.
For a l l
pract ica l purposes it i s
safe
to cal l
the
value
of sec
¢
as one.
The cross-sect ion
of
the
t ie is
constant so the
term
may be integrated direct-
ly and evaluated.
I f
At
= 40 sq. n or 0.278
f t
and the l imi t s
are
from to 220 f t have
This value
must
also
be multiplied
by d
to
obtain the
f ina l
correct
resul t
6 ~ f
E
} ~ h
Evaluation of E
h
The ra t ioE /E
h
i s used so tha t
the
value will
apply
to
a l l condit ions. I f the
hangers
were
compos-
ed of a di f ferent material the moduli
of
elas t ic i ty
would
not be the
same and
the ra t io
would
be
used.
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In t hi s p ar tic ula r design
we
have
assumed that
the
hanger
and
arch sect ion are
of ident ical
material
s t ruc tura l
s tee l so the
ra t io
i s equal
to unity.
I f we evaluate the term a t panel point
L o
where
the t ie
is
cambered 0.25 f ee t A
0.21 sq. f t the
panel
spacing i s .20
fee t and the length
of the
hang-
er
i s
36.14
f ee t
we
have
the
following:
0 0000\ \5
The
term
i s negl igible as
fa r as
affecting
the
resul t s
of the
design.
Even af te r multiplying by
the
term
d
,
we
obtain 7.5)3 0.0000115= 0.00485
the value
i s
s t i l l
very
small and can safely
be
neg
l ec ted
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Design No.1 .
Data:
Main span
=
220
-0
center
to
center of
bearings
(11
panels
a t :20
on
Two
side
spans =
100
-0
(5
panels
a t 20
1
on
Rise
ra t io
o ~
parabolic
arch
=
1
6
Theore ti ca l he igh t
o ~
arch
=
36
-8
Thiclmess
of
arch a t piers
=
7
-6
Thiclmess
of
arch a t
crown
=
3
1
-6
Clear w id th of roadway = 24 -0
Impact
= 50
1+125
Floor slab
=
8 slab crowned
to
U
a t
center
With 25 lb .
per
sq. f t . wearing
surface
total ing 150 b p ~ r
sq . f t .
Stringers
assumed to weigh 40 per f t . and spaced
4 -0 c. to c.
40,000
= bending moment due
to
concent ra ted load
0.345
- 50
=
impact
rat io
20+125
13,800
= 40,000
x
0.345
=
bending moment
due
to impact
_.2
32,000 =
150x.4+40
x
10
=
bending
moment due
to
dead
2
load
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85,800
tot l bending moment
12
x 85,SOO
=
18,000
S
S
57.2 (a 16' -
40
wf
beam
is ample)
Use l l stringers
as
of this section.
Design
of Floor
Beams: (Spaced 28' -0
c.
to c.)
Maximum
concentration
on
floor
beam
at
stringer
point
nearest
the
center comes from
15- ton t ruck
when one
12,000 wheel load
s
over the floor beam
and a 3000
load 14 feet
away.
This
makes
a concentration
of
12,000 +
3000 x
12,300
20
I 50 0.327
,28+125
R
= 12,300 (5.75 + 11.75 + 14.75 +
,20.75)
+ 28 =
.23,300
234,900
(23,300
x
13
25 -
12,300
x
6)
=
bending
moment
due
to
l ive load.
71,800 = (234,900
x
0.327)
bending
moment
due to
impact.
3352
=
(150 +
1 )
20 ] 5 2
equivalent
dead
load
per
4
foot .
(152
section
assumed)
3.27,000
=
(3352 ,£ . 2) (13 •
25
x 14.75) bending moment due
•
to dead load.
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633,700
=
to t l
bending moment
12
x
633,700 = 18000 S
S = 425 (a 33 -152 wf beam
is
sufficient)
Equivalent l ive load = 480 x
12
x
0.94)
= 600
lb .
9
per
foot .
Total
panel
load =
600 x 20 =
Impact
= 0.30
x
600
=
180
lb
per
foot.
Total
panel
load
due
to impact =
Total
Concentrated load for moment =
l2,000 lb .
3,600 lb .
15,600 lb.
13,500 x 12 x
0.94
=
16,900
lb .
9
Concentrated load
for
shear
=
19,500
x 16 x 0.94 = 24,400
lb.
9
Dead load
from each
f loor beam =
3352
x
8
=
2
46,900
lb .
Lateral
Forces:
Assume 150
per
l ine r
foot on unloaded
chord
Assume 300
per
l ine r
foot on
loaded
chord
Assume a
l te r l
force
of
200
lb per l inear
foot
acting
6
f t
above
the
f loor due
to moving
l ive load and
wind
pressure upon
the
load.
Load per panel = 500 x
.20
= 10,000
lb
on loaded
chord
Load per panel = 150 x
20
=
3000 lb
on unloaded
chord
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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s t
coeff icient
1 2 3 4 5 6 7 8 9 10
= 2i
11 11
Bottom Chord:
Wind Load Stresses
p /0 , ooo . tan =
0.834-;
sec f = 3 5 ; : Psec e = 1/80:f
/
/
Pian
=
8 3 4
Stress
in
chords =
coeff icient x P t ane
Stress
in
webs coeff icient x P sec e
double
system
of bracing wil l be used in each panel
but
each
diagonal
wil l
be assumed
capable
of
resist ing
tens ion only.
The maximum
shear in
any panel
occurs
when
every panel
point to the r ight of the section is
loaded.
The to ta l
wind load =
110
kips .
Design:
Diagonals
Maximum unsupported
length
may be
taken
as
Ii
feet
less
than the
diagonal distance
c.
to c.
of panel points
be-
cause of the support which is provided
by
flanges of
main girders and gusset
plates
a t connection.
1 1.305
x
4
x
12 - 18 = 357 in .
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This unsupported length f i r ly
large
so wil l be
considered
as each member developing tension only.
The net
a re a r equir ed
is 65,.200/ 16000
=
4.07
Q
n
using
7/8 r iv ts t 3
spacing.
Panel
ab
Net a rea furni shed by
two
5
x
3t X 3/8
angles,
with
5
fl
legs back to back.
Net a rea furnished is 80%
x
6.10 = 4.88 lJ
n
Number
of
r iv ts =
65,200
= 11 r ivets Field driven
bOlO
r iv t
value
single
shear) 12 wil l be used. 6 each
angle.
Net area for
panel bc
53,400/16000
3.34
Area
furnished
by two
4
X
3
X 3/8 angles is
(.2
x
2.48) .
.80
=
3.97
n
53400/6010 9
r iv ts
10
wil l
be used - 5
in
each angle).
Panel
cd
42,700/16000 = .2.67
0
n
Area furnished by two
3
x
,2
x 3/8 angles is
2 x 2.11) . .80 3.38 0 n
No.
of r iv ts
=
42,700/6010
=
7 8
wil l be
used
-
4
each angle).
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Panel de
33,200 ;:
16000
=: .2.08 a
Area
furnished by two 3 x : 2 ~ X 5/l6
n
angles
is
.2
x
1.6.2). .80 =: 2.59 c
Number of r ive ts
33, .200} 6010
=: 6 r ivets (8 wil l be
used
4 in each
angle) .
This
same
s ize
angle
wil l
be
used
panels
remaining
as
it
i s the
minimum
angle
allowed fo r
bracing.
Chords:
Panel
AB
41,700/ 16000 =: .2.60 an required
A 10 [15 .3
wi l l be
used
although
furnishing more
area
than necessary. (4.47 a
nG
. ) (4.47-2xO.24)
=
3.99 °nN.
Panel
e
75,100/ 16000 4.70
required
10
[.20 wi l l
be used
(5.86 °nG.)(5.86-2xO.38)
=
5.08
On
N
fane1
CD
100,100/16000
6.25
n required
10 [
.25
wil l be
used
(7.33
2 x 0.53)
= 6.27 sq . in .
N
Panel
DE
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A 10
[30 will be
used
(8.80 0
G.)
(8.80
.2 x
0.67)
=
7.46 sq.
n
N.
Panel EF
141.8/16000
8.85
A 10 [ 35 wil l be used (10.27 [
G.)
(10.27-2 x
0.82
8.63
•
Panel
FE'
150.1/16000
938 •
A 10
[35
wil l be used (10.27
On
G.) (10.27 .2
x
0.82)
8.63
sq.
in .
N.
See
diagram
No. 12
for
sect ions.
Design of Side Span:
Stringers and f loor beams of
the
same section as for
the main span
wil l
be used.
Lateral Bracing
The
l ter l
bracing in
the plane
of
the
lower flanges
must provide for l l the l te r l forces due to the wind
and to the
effe t
of sway, including the wind force which
i s
normally assigned to the
upper
flanges.
Wind force moving load equal
to
30 per a on one
and
one-half times area
in elevation.
It
x
30
x 5 x I 225
per l inear foot.
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
54/72
Also a 200 load a s a
l a te ra l
force
is
act ing aga in st
the l ive load.
Total
horizontal
force = 425 per l inear foot .
The l a te ra l
system
moving
panel
load
is
425 x2 8500
lb
A double
system
of
bracing
wil l
be
used, but each
diagonal
wil l
be
assumed
capable
of
resis t ing
tension
only.
Firs t coeffic ient
1
2
3
4 =
10 =
2
5 5
P = 8500 ; tan-&= 20
0.834;
sec tt = 1.305;
5
P
sec
24
~
=
2220 ;
P
tan
f r-;707o#
•
Stresses in chords
coeff icient x P s ~
The to ta l
wind
load
i s
42.5
kips.
A minimum
size angle of 3
x
.2t
x 5/16
n
was
used
for
a l l diagonals which furnished
a
net area of 2.59 sq. n
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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Conclusions
In combining
our
influence
diagrams for
dead load moment our
in i t i l
equations are used. The
f i r s t term
l-k
x or
k l-x
in
the
equations for the
main
span is
equivalent
to
the
moment at any section,
t reating
the arch
as
though
were a
simple
beam.
The
sum of the values
for
the
l s t
two terms
Ml l-x
and
1
M2 produces a s tr ai gh t li ne
diagram
for a symmetri-
1
cal
design since Ml
and M2 are
antisymmetrical
and
the
totals are
identical .
In superimposing the diagram for
the simple beam moment upon
that
for
the
end moment
we
obtain
points
of zero
stress t
the
intersections
of the
two
diagrams.
t
is
evident that there can be a considera-
ble
error
the
values
for direct stress
and
the
de-
sign
wil l
not be materia lly af fec ted. The
tension
the t ie causes direc t s tress in the hangers, but the
values are negligible.
can have considerable error
assuming
the areas of the arch ring and horizontal
t ie
and our design wil l not be affected a g rea t deal.
The
arch
ring area may
be
error as much as 5 and
yet the values of
the
redundants
be
affected
but by a
small
percent.
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8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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Increasing the
side
span
past
a certain
point affects the type
of
section used A plate gird-
er section may
be used economical ly
up to lengths
of
125 f t Above tha t box girders
must
be
used
Above
a certain point box
girders
would prove uneconomical
and would be feasible to use a truss for the
side
span In
Design
No 4· the main span is
3 5
feet
and
the side spans are
200
feet
in length The
side
span
section would
of
necessi ty be composed of a box
gird-
er or of a t russ Above
the
point
where a box girder
section proves economical and
a
truss section
must be
used would prove
more satisfactory
to use
a
truss
for the main span
also
The design as
followed
would
be equally applicable
BIBLIOGRAPHY
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
58/72
McCullough C. B. and
Thayer
E. S. Elastic Arch
Bridges.
New York City.
John
Wiley
Sons
Inc.
1931. pp. 8 14; 33 42;
125 141.
Spofford Chas.
M
The Theory of Continuous Struc-
tures and Arches. New York and London. Graw
Hill
Book Company Inc. 1937.
pp. 1
57 58;
184-.203.
Krivoshein G G Simplified
Calculation
of Stat ic-
al ly Indeterminate Bridges. Prague Czechoslov-
akia. Published by author 1930.
pp.
77 80;
86 95;
110 113.
Parcel J and Maney G A An Elementary Treatise
on Stat ical ly Indeterminate Stresses.
New
York
and London. John Wiley Sons Inc. ; Chapman
Hall
Ltd.
1936. pp. 121 122;
320 338; 403 420.
Cross
Hardy and Morgan
Reinforced Concrete.
Wiley Sons Inc. ;
pp. 247 270.
N D
Continuous Frames
Of
New
York and London. John
Chapman
Hall
Ltd. 193.2.
Sutherland Hale
and Bowman
H.
L. An Introduction
to
Structural
Theory and
Design.
ew York nd
London.
John
Wiley
and Sons
Inc.;
hapman
nd
Hall Ltd.
1930.
pp. 184;·217 18; 231 37.
Seely
Fred B.
York City.
.283-.284.
Advanced Mechanics
of
Materials.
John Wiley and
Sons
Ind. 1932.
New
pp
•
The
American
Association
of State
Highway Officials.
Standard Specifications for
Highway
Bridges.
Wash
ington
D.C.
pUblished
y
the Association
1935.
pp. 165... 205.
Kunz F. C.
London.
312 366.
Design of
Steel Bridges. New
York and
McGraw Hil l Book Company Inc. 1915. pp.
8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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BIBLIOGR PHY
Hool G A. and Kinne
W S.
Movable and Long Span
Steel
Bridges. New
York and London. McGraw Hill
Book Company Inc. 1923. pp• .218 258; 393 482.
Hool G A. and Kinne W S. Structural Members and
Connections. New York
and
London. McGraw Hill
Book Company Inc. 1923. pp. 1 9 9 ~ 2 3
£.VALUATION
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EQUATIONS
DES
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8/18/2019 Design Study of a Three Span Continuous Tied-Arch Bridge
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