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Chapter 14
Roadway Bridge
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contents
Roadway Bridge Floor
Side walks and Railings
Bridge Bracings
Design of lateral support at top chord of through
pony bridge
Cross Sections for wind Bracing End X-frame in deck bridges
Transmission of the braking forces the bearing
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Truss Bridges
A. Types of bridge trusses
B- Determination of forces in truss members
c. Proportioning of truss members
D- Box section for bridge trusses Top chords
Lacing bars, batten plates
Bottom chords
Diagonals
Verticals
Design of compression member
Design of Tension Members
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Design of Bolted Joint
Design of Battens and Diaphragms
Design of End Portals
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Roadway Bridge Floor
The floor of a Roadway Bridges consists of:
1. A wearing surface or Roadway Covering.
2. Sub floor transmitting the loads to the
stringers and X-girders.
The sub floor is similar to the solid floor of a
ballasted Railway Bridges. It may be timber, steel
floor or R.C. floor.
Back
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X.G
P
1-2 cm
5-6 cm
Timber floor (Type 1)
For bridges, generally two layers of flanks are provided.For calculating these flanks we assume that the maximum
wheel load is distributed over two flanks.
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Slab on X.G only
A
A
(1.2-1.5)m Rail way
(2.5-3.5) m
Sec A-A
X.G
(2.5-3.5) m
X.G X.G
Reinforced concrete floors (R.C. floor)
It may be supported by the main girders only, the X.G.
only or by stringers and X. girders. The span of the slab
may be 2.5 to 3.5 m, and thickness of slab to be 20 cm
nearly. The R.C. slab reinforcement, generally 12 bars
are used at least per one meter.
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Slab on M.G only
M.G
L.W.Br
U.W.Br
M.G
(2.5-3.5) m
M.G M.G
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Member to prevent
lateral buckling of web
Truss bracket
Kafre El-ziat Roadway Br.
Span of slab
Side walk stringer
M.G
Stringer
X.G (truss)
X.F
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M.G
Stringer
U.W.Br
X.F
L.W.Br
X.G
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X.G
5 cm Asphalt
X.G every 150 cm
Wearing surface
The wearing surface for roadway covering consists
of timber blocks, hard bricks, asphalt bricks, stone blocks,asphalt or concrete.
The choice of material depends on the traffic, the span of
bridge, the cost and climate
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The side walks are placed either inside or outside of themain girder. If they are arranged outside, they must be
supported on cantilever brackets situated in the plane of
the X.G. so that theve bending moment of the bracket is
transmitted to the X.G. The floor of these side walksshould be a precast R.C. slab (6cm) thick resting on the
side walk stringers. The wearing surface is a 2 cm layer of
asphalt. In through bridges the curb should be at least 50
cm inside the main girder
Back
Side walks and Railings
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Take strip 1 m and statical system as continuous beamsupported on side walk stringer (one way slab), take t = 8
cm and get As. The applied loads are considered 500
kg/m2 or one concentrated load 5 t.
Hand railings and brackets withstand the effect of a
transverse horizontal force of 150 kg/ m in cases of
Railway bridges, Roadway bridges, and foot bridges,
supposed acting at top level of hand rail. This horizontal is
transmitted from the hand rail to the main posts and from
their connections to the cantilever brackets.
Side walks parts
1. Slab
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X.G
Stringer
6
2
110 cm
3
4
5
1
2. Hand rail
Simple beam span distances between two brackets (take angle or
channel X. sec.).
1. Side walk stringer
Simple beam span distance between two brackets (take channel X.
sec.)
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4 Post
Cantilever beam (take 2-angle or 2-channel X. sec.).
yP=M
KgL501=P
OROR
OOR
150 Kg/mM=W.L /82 2
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5 Connections
Double shear bolts
yP=M
KgL501=P
21
1d
dM=F
F1
F2
F2
F1
M
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6 Bracket
Calculate M& N& Q at center of bracket.In case of beam loaded alone we must calculate Fl.t.b and
the check that the actual stress Fc is less than the
S.F
+
+
N.FB.M
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allowable stress Fp.b.
For ST 37
If 100
If 100
2c 000065.040.1F
2
2c/
7500F cmt
= l/ i , where I for compression flange only.
for bracket l/ b 2 l/ b
assume unequal angle 80120
Bolts subjects to shear
21
1d
dM=F 2
221
2 dd2=d
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Bolts subjects to tension
A.Nt
IyM= 0.8 Ft
Nd1
d1
F1
F1
A
y
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Bridge Bracings
The bridge is provided with horizontal and
vertical bracings:-
1. The stringers are connected together bystringer bracing given before.
2. The chords of the main girders are jointed
together by an upper and lower horizontalbracing called wind bracings.
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These transmit to the bearings of the bridge;
a. The lateral forces due to wind.
b. The lateral shock 6t.
c. Centrifugal force.
3 - Special horizontal bracings for the braking forces.
4-Two vertical and transverse bracing called X-frames or
portals (in case of through) transmitting reaction of theupper lateral bracing to the bearings of bridge.
5- Some intermediate vertical transverse bracings called
intermediate X-frames or intermediate portals for therigidity of the structure.
It isnt necessary to find all these bracings in every bridge,
there existence depend upon the type of the bridge, the
span and the floor.
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I-The Deck Bridge
Deck Bridge.
U.W.Br
L.W.Br
End X Fram
The upper wind bracing transmits the wind
pressure WT on the train & WF on the floor & WG
on the wind ward side of the main girder.
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The lower wind bracing transmits the wind pressure
WG on the wind ward side of the main girder.
W = 100 kg/ m2 in case of loaded bridge
W = 200 kg/ m2 in case of unloaded bridge
* The wind pressure WT on the train produces in addition
to horizontal loading of the upper wind bracing.
a vertical loading, to the main girder.b
eW=V T
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M.G.G
W.F
W.T
L.W.Br
End X Fram
Dick
b
X.G
V e
3.5 m
U.W.Br
V=W .ebT
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In case of a truss bridge, only the exposed area of the
members is considered. This area is equivalent to 40 % of
the hole area of the surface of the truss. In all bridges with
an upper and a lower wind bracing, their shall be provided
End X Fram
L.W.Br
M.G
U.W.BrX.G
at each end a X-frame
to transmit to the
bearings,the
horizontal reactions of
the upper wind
bracing. The
horizontal reactions ofthe lower wind
bracing are
transmitted directly to
the bearings
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The end X-frames in deck bridges shall be of rigid
type. In all railways and in roadway deck bridges there
shall be intermediate transverse bracing at least at every
third panel point to increase the stiffness of the bridge.
These intermediate X-frames will release the end X-frame
from a part of the horizontal reaction of the upper wind
bracing. Yet it is recommended not to consider that release
unless the bridge as the space structure.
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ii-The Through Bridge
Through with upper bracing Bridge.
Portal Fram
U.W.Br
L.W.Br
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In through bridge two horizontal wind bracings should
be arranged if possible. In the plate girder through
bridges we cant arrange an upper wind bracing in the
bony truss Roadway Bridge we have only a lower wind
bracing which transmits all the wind loads to bearings.
The force WF on the floor will be considered to act on a
solid surface as the plate girder. The through RailwayBridge shown above is provided with two horizontal
wind bracings.
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W.F
W.G
Dick
3.5 m
U.W.Br
L.W.Br
Pony truss bracing
W.T
L.W.Br
2.5 m
W.G
W.F
W.T
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W.T
L.W.Br
3.5 m
Through
W.G
X.G
The upper wind bracing transmits the wind pressure
WG on the wind ward side of the main girder.
The lower wind bracing transmits the wind pressure (WG), WT on the train & WF on the floor & on the wind
ward side of the main girder.
At the connection ofeach X-G to the main
girder, stiffness
bracket shall be
arranged.
Back
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Design of lateral support at top
chord of through pony bridge C = force in flange = Af Ft
The U-frame formed by the two vertical stiffeners
and the horizontal stiff X-girder is acted upon by ahorizontal transverse force = C/100 at the centroidof compression flange as well as the wind pressurebetween two consequence X-girders. The
maximum stressed section is mn. The compressionstress at point n Fltb. If the stress isnt safe, weeither increase the thickness of the bracket plate oradd a stiffening angle.
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The connection between the X-girder and bracket is
designed on the shearing force A that between X-girder
and the bracket and horizontal gusset of wind bracing on
force B. if the X-G is built up section the bracket
connection is designed as a web splice.
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m X.G
n
m
C.G
n
Cos =1/2002x1/200xC=1/100xC
hord1/100 C
C
C
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B
1/100 C
B
W.G
A
X.G (Rolled section)
A
X.G
Back
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Cross Sections for wind Bracing
The diagonal of wind bracing
The diagonal of wind bracing shall have
stiff section to prevent vibration and to helpin reducing the deflection of main girder
due to eccentric loading (space frame
treatment). The section should have a depthnot less than L/40. The recommended
sections are given in Fig.(5.).
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h < L/40
h
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The choice of the section depends more or less upon the
span of the diagonal, the two channel section is convenient
for too long spans executed in Banha and Samanoodbridges. The two channels are connected together by
latticing or batten plates.
in compression & in tension
Gusset for U.W.Br.
Lacing
End patten plat
h
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In case of one diagonal member only
Case of the warren system which designed on a force S;
forcencompressiofor85.0
forcenfor tensio85.0
=S
call
tall
all
FF
FF
FSin
Q
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Case of the N-system which designed on a force S;
forcenfor tensio85.0
=S
tall
all
FF
FSin
Q
C f th K t Rh bi A d M lti l hi h
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Multiple without vertical system
Rhombic system
K- system
Case of the K-system , Rhombic And Multiple which
designed on a force S;
Q
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allFSin
Q
2=S
If the bracing is made up of crossed diagonal and struts,the calculation is made under the assumption that the
tension diagonal are only acting. The struts here receive
compression force. If the multiple systems of wind bracing
a further reduction of 20 % in the allowable stresses givenbefore, shall be made to account for approximation of
solution that both systems (tension and compression
member) equally share the lateral loads. in case of one
angle in compression the allowable compression stressshall be reduced by 40% of Fc.
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Hence, the approximations in allowable stresses are;
ncompressioinonly)sectionsic(unsemmetractionangleonefor0.6stressesReduce
ancyindeterminforionapproximattodue0.80stressesReduce
effectspacetodue0.85stressesReduce
A&forcencompressiofor6.08.085.0
3
3AA&forcenfor tensio8.085.0
eff.
21
1
1eff
gressbpall
tall
AFF
AAAFF
Back
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End X-frame in deck bridges
The compression diagonal is assumed in acting and we design the
tension diagonal, also we assume that the X-frame is resting at amovable support at one end and the hinge support at other end.
.W.Br
U.W.Br + 6 t X.G
Ru+Rl+6 t
Back
T i i f h b ki f h b i
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Transmission of the braking forces the bearing
In Railway bridges especial bracing should be arranged to
transmit the longitudinal forces from the stringers to thepanel points of the main girder, hence; they are transmitted
through the main girder to the hinged bearings. Some times
a bracing is arranged at every panel point. But generally
two bracings at the quarter points of bridge are sufficient.The braking bracing system shown in sketch is statically
indeterminate but the loads are symmetric about
perpendicular axes to X-X. Therefore the diagonal Bn &
cn are zeros since they correspond to themselves. Also, theloads are antisymmetric about axes Y-Y and thus members
mb & mc & nb & nc are zeros, If special bracing of the
longitudinal forces isomitted, these forces are transmitted
from stingers to main girder by bending of the X girder
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from stingers to main girder by bending of the X-girder.
8
B/4
B/4
B/8
B/8B/8
B/8
x
y
b'
b c
c'
2
2 3
y
B/4
B/8
B/8
B/4
B/4
4
B/8
B/8
xX.G
B/8
B/8
X.G
65
7
7
B/8
B/8
B/4
B/8
B/8
M.G
My/Zy
9
Back
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Truss Bridges
b L/ 20, b h/ 3
Where, b = bridge width = distance between center lines
of two main girders
L = span of bridge
The depth of trusses shall be chosen in such away that the
elastic deflection due to L.L (without dynamic effect)
shouldnt exceed L/800 for Railway bridges and L/600
for Roadway bridges.
Back
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b > L/20 , b
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Either both chords are straight and parallel or only one
of them. In a through bridge the upper chord may
polygonal, in a deck bridge the lower chord may be
polygonal. Curved chord should not be used in bridge
trusses on a account of the additional bending stresses.The loads are transmitted to the panel point of the truss
by a system of stringers and cross girder. No load except
the own weight of the truss members should act between
the panel point.
A. Types of bridge trusses
1 Trusses with horizontal chords
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They suitable for span up to 60 m. the joints are
simpler than in trusses with polygonal chords. The depthis h L/ 8 for Railway Bridge, or h L/ 10 for RoadwayBridge. For continuous and cantilever trusses the depth
may be taken h L/ 10 for Railway bridges, h L/ 12 for
Roadway bridges. Some times a greater depth is used toallow an upper wind bracing. The arrangement of web
members may be N-system or warren system. The warren
system trusses require generally less material than the N-
shaped trusses, since the vertical members have smallerforces, the number of joints and changes of cross section
in warren system are also less. Shop work for warren
trusses will be cheaper.
1.Trusses with horizontal chords
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N shaped
Warrren
h
h
N through
N through
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They are used for spans up to 60 m. the economical depth
at middle is h = L/7. The web system is either N-shapedor warren. The economical inclination of the diagonal to
horizontal = = 40 - 60. A polygonal chord trusseslighter than a truss with horizontal chords since the forces
in the diagonals are smaller. On the other hand the shop
work is more complicated which means a higher cost.
2-Trusses with polygonal chords
(0.5-0.7)h
= 40-60N or Warren
h=1/7 L
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3-Trusses with subdivided panels (e), with
Rhombic diagonal (f) and K-system
These kinds are economical for span over 80 m. The panellength is reduced in all this system and thus the cost of the
floor is less, but the increased number of joint increase the
cost of shop work. A truss with Rhombic diagonal has a
good appearance; a truss with subdivided panels has big
secondary stresses. K-system trusses have the smaller
secondary stress.
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Rhombic system
e- Subdived system
K- system
4 T ith lti l b t
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4- Trusses with multiple web system
These were used in past where the tension diagonals
consist of flat bars. Now they are again used for maingirders, but type h with crossed diagonals is frequently
used for wind bracing. For approximate calculation, the
common assumption is that the truss may be divided up
into two or more component trusses with the same chordsbut with different web system. The loads also are divided
and placed upon this component trusses. Then the stress in
a web member is determined as its stress calculated in the
truss of which it is a part. The chords are a part of allcomponent trusses, hence the stresses in a chord member is
obtained by adding it partial stress from each component
truss.
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= +
5- Trusses with 3 chords
The arch truss with a tie (k) and the truss reinforced by a
hinged arch (Pow string truss) are supported on a hingedbearing at one end and a movable bearing at the other.
They are therefore externally static determinant but
internally they are static indeterminate. These trusses
have good appearance but they more expansive than
trusses with two chords.
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Row String
Arch with atie
Back
b D t i ti f f i t b
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b- Determination of forces in truss members
We determine the forces in the truss members on the
assumption that the member are connected by hinges, so thatloads applied at the panel point produce only axial forces in
the truss member. The secondary stress which are the
bending stress induced by the rigidity in connection, are
generally neglected. In our specification it is required tocalculate the secondary stress in the following cases:
1- For trusses with subdivided panels.
2-For member whose width in the plane of the truss ismore than 1/10 of its length.
3-For loads acting between the panel point.Back
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c. Proportioning of truss members
For the chord member we can use either sections with one
web plate (T section) or sections with two web plates(Box section). T-sections are used only for small bridge.
Box sections have grates moment of inertia about axis y-y
and are better used for the connection of gusset plate. The
sections of all chords and web member should be
symmetrical about axis y-y in the plane of the truss.
Diagonals and verticals are usually symmetrical about axis
x-x also. The required area of the chords change at every
panel point of the truss and in choosing the different cross
section we must try to get simpler connection and splices
at the panel point. Back
y y
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x
y
x
Chord membersxx
y
D- Box section for bridge trusses
Top chords
The minimum section consists of a horizontal plate and 2
channels or a horizontal plate, 2 vertical plates and four
angles.
Depth of top chord h = (1/121/15) of the panels length Width a = (0.751.25) h
10
L
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min
a a
max(3/4 - 5/4)h
h
To avoid local buckling, the minimum thickness of web
and cover plates should be as follows;The unsupported width of a plate measured between
adjacent lines of rivets or welds connection the plate to
other parts of the section should not exceed:
t thi k f i l l t f 2 l t
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t = thickness of a single plate or of 2 or more plate
provided that this plates are adequately tacked together.
YFtb 64
(3/4 - 5/4)hmax
t
max
b
b
t
(3/4 - 5/4)h
b
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(3/4 - 5/4)hmax
t
d1
t
b
yFtd 301
Only excess over this width should not be included in the
effective sectional area in computing direct compressivestresses. The center of gravity axis x-x for the different
section should not change too much. In drawing the truss
we use an average value y = (y1 + y2 + y3 + ....)/ n.
It i d ti t l t th h l
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It is good practice to use a cover plate over the whole
length of the top chord even if the end members have
excessive cross section.
y1
x2
y
x
y
y
y3
y
y
xx x
y
y
x
Back
Lacing bars batten plates
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Lacing bars, batten plates
The two plates of the compression members shall be
connected together by diaphragms and the open side of thebox section shall be provided with batten plate close to the
gusset plate and with intermediate batten plates or lacing
bars to avoid lateral buckling of their component parts.
The slenderness ratio of each component part betweenconsequent connections of lacing bars or batten plates shall
be not more than 50 (and 2/3 l/i of the whole section).
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View A
Lz
batten pl.
y
yz
z
A
Bottom chords
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Bottom chords
The depth of the bottom chord is equal to that of the top
chords h = (1/12 1/15) of the panels length, or slightlymore (2 4 cm). No horizontal plate is provided at the
bottom of the section to avoid water packets. In
continuous and cantilever bridges where some bottom
chord members are in compression, horizontal plate maybe used and it must be provide with drainage holes (4 5
cm) . The two component parts of tension member shallbe connected together by diaphragms and batten plates
similar to these of the compression members, but theirthickness may be taken 25 % lighter (t2 lmember/ 15).
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bottom with cover plate (holes)
hh
Diagonals
For appearance the width of the diagonal should not be
more than that of the top chord and should decrease from
the end to the middle of the bridge. The compression
diagonal at end of the warren truss has a section similar to
that of the top chord.Back
Verticals
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Verticals
In trusses with a N-shape web system, the vertical have
similar sections as diagonals. In warren trusses the verticalmay consist of a web plate + 4 flange plates or an I-beam
(B.F.I.B).
For diagonal or vertical tension member;
t2 lmember/ 15 ( D & V )
t2 (lmember)/ 30 Railway bridges ( D & V & C )
t2 (lmember)/ 35 Roadway bridges ( D & V& C )
t2 (lmember)/ 10 (C )
D Diagonal & V Vertical & C Chord
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< 10 mm
2tt
2t2
2t
Back
Design of compression member
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The slenderness ratio l/ i of compression member of main
girder shall not exceed 90 for Railway bridges and 110 for
Roadway bridges.
E. Effective buckling lengths
Table 4.5
Table (4.5)
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Out of planeIn planeMember
Compression chord
Unbraced
Compressio
n chord
Effectively
braced
0.75 span
(Clause 4.3.2.2
or equation 4.2 if
using U-frames)
0.85 l0.85 lChords
1.20 l0.85 l0.70 lDiagonals- Singletriangulated
web system
l0.75 l0.85 l/2- Multiple
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/Intersected
web
rectangular
systemadequately
connected
- K-system
1.50 l1.20 l0.90 l
1.20 l0.85 l0.70 lVerticalmembers
- Single
triangulated
web system- K-
intersected
web system
(0.9+0.3Ns/
Nl)l
(0.75+0.25N
s/Nl)Ll0.35
Effective buckling lengths
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1.2
0.85
0.7
1.2
0.85
0.85
x
x
x
y
y
y
x
y
0.7
0.85
x x
y
y
Lb out-plan
(No- bracing)
Lb out-plan
(with upper bracing)
Ns/N
Ns/N
2.5 EI a
bracing
4
1.5
1.2
1.0
0.75
x
x
x
x
0.7
0.9
0.85
y
y Lb in plan
Unbraced compression chords
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Unbraced compression chords
a- For simply supported truss, with laterally
unsupported compression chords and with no cross-framesbut with each end of the truss adequately restrained
(Figure 4.1), the effective bucking length (kL), shall be
taken equal to 0.75 of the truss span, (clause 4.3.2.2).
b- For a bridge truss where the compression chord is
laterally restrained by U-frames composed of the cross
girders and verticals of the trusses, the effective buckling
length of the compression chord (Lb) is:
aa..I.E2.50=L 4 yb
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y
y
2dI1
1d
2
B
I
E = the Youngs modulus = 2100 t/cm2
Iy = the moment of inertia of the chord member about the Y-Y axis.
a = the distance between U-frames (cm) = S
23 *dd B
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2
2
1
1
EI*2
*d
EI*3
d=
B
d1 = the distance from the centroid of the compression
chord nearest face of the cross girder of the U-frame = dw
Hx.G.
d2 = the distance from the centroid of the compressionchord to the centroidal axis of the cross girder of the U-
frame = dwHx.G./2
I1 = the second moment of area of the vertical memberforming the arm of the U-frame about the axis of bending.
I th d t f f X G b t th i f
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I2 = the second moment of area of X-G about the axis of
bending = IX
B = the distance between centers of consecutive MainGirders connected by the U-frame
Back
Design of Tension Members
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Tension members shall always be of rigid construction
and their slenderness ratio l/ i shall not exceed 160 for
Railway bridges and 180 for Roadway bridges. Theeffective net section area shall be taken for all tension
members. This area shall be the least that can be
determined from any plane or planes cutting each
component plate or sections to its axis,
g
diagonally or following zig-zag line through adjacent
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diagonally or following zig zag line through adjacent
rivet holes. In each case all holes of line to meet with
shall be deducted from the gross sectional area where any
portion of the sectional area is measured for a diagonalplane adding (S2/ 4g) for each gauge space. The minimum
sectional area should not be less than that obtained by
assuming all the holes to be in one perpendicular plane.
Back
Design of Bolted Joints
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Design of Bolted Joints
Connection of web member to gusset plate and splices of
chord member shall have a strength equal to themaximum strength of the connected members. The bolts
shall be arranged symmetrical about the center line of the
member. The connection to either direct or part of it is
indirectly connected by:-
1- Splice plates or lug angles.
2- If breaking is along section S-S bolts (1 single + 2
double + 3 double) shear, must carry the load.
3- The strength of the splice plate should be enough to
carry a force corresponding to bolts (4 + 5) single shear.
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40 % more
20 % more
Lug angleLug angle
20 % more
10 % more
UnsymmetricalSymmetrical
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gusst pl.
gusst pl.
Splice pl.
S
S
Sec S-S
Connections for members
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Connections for members
The connection shall be designed for a capacity based on
the maximum of:-
1- The average between the actual force and the
maximum strength of the member of not less than 0.75 the
maximum strength of the member.2 The bolts between the chord and the gusset plate must
correspond to the algebric sum of the horizontal
components of the strength of the diagonals S1
, S2
L = S1Cos + S2Cos S2S1
3 The number of bolts should correspond to the
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3. The number of bolts should correspond to the
effective strength of the two diagonal, i.e., the number of
= (number of bolts in member 1 + number of bolts inmember 2)Cos .
Example
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I.L.S1
S1 S2
S3 S4
I.L.S1
/tan
1/tan
196 t
70 t
126 t
94 t
L=750
In a lower chord panel, if the maximum force in chords
and diagonals are as given, design suitable cross section,
connections and splices.
Member S1 = + 94 t
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1
2 [ No 24
Bolts M22, = 24
Anet = 2(42.32.40.95 - 22.41.3) = 67.56 cm2
Fact = 94/ 67.56 = 1.39 t/ cm2
< 1.6 t/ cm2
Maximum force = 67.561.60 = 108.10 t
Rleast = Rs. sh
qb = 0.25 Fub For bolts of grade ( 8.8),
Rs. sh = qbAsn = 0.258.03.03 1 = 6.06 t(n = No. of shear planes)
Bolts connecting diagonal to gusset = (Maximum force/
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g g g (
Rleast) 1.15
= (108.1/ 6.06)1.15 = 20.52 boltsUse 24 bolt M22
Member S2
= - 70 t
2 [ No 22 & Bolts M22, = 24
Agross = 2(37.40) = 74.80 cm2
y = ly/ iy = (0.85750/ 11) 1.20 = 69
(1.20 due to lacing bars)
x
= lx/ i
x= (0.70750/ 8.5) 1.20 = 61
x
y
y
x
max = 69 Fp.b = 1.600.000085 (l/ i)2 = 1.119 t/ cm2
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Fact = 70/ 74.80 = 0.94 t/ cm2 < 1.119 t/ cm2
Maximum force = 74.801.119 = 83.70 tRleast = Rs. sh
qb = 0.25 Fub For bolts of grade ( 8.8),
Rs. sh = qbAsn = 0.258.03.03 1 = 6.06 t(n = No. of shear planes)
Bolts connecting diagonal to gusset = (Maximum force/
Rleast) 1.15
= (83.7/ 6.06)1.15 = 15.88 bolts
Use 16 bolt M22
Member S3 = + 126 t
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y
y
x
2 [ No 30 & Bolts M22, = 24
Anet=2(58.8022.41.00 - 22.41.6) = 92.64 cm2
Fact = 126/ 92.64 = 1.36 t/ cm
2 < 1.6 t/ cm2
Member S4 = + 196 t
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y
y
x
2 [ No 30 + 2 PL 24012 & Bolts M22, = 24
Anet for 2[
= 2(58.8022.41.00 - 22.41.6) = 92.64 cm2
Anet for 2PL = 2(2422.4) 1.2 = 46.08 cm2
Anet for 2[+2PL = 92.64 + 46.08 =
138.72 cm2
Fact = 196/ 138.72 = 1.413 t/ cm2
< 1.6 t/ cm2
Force to be transmitted from gusset plate to bottom chord
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= Fmax
Fmax
= (S1
+ S2)Cos = (67.561.6 + 74.801.119) 1/2
= 135.62 t
Bolts connecting diagonal to gusset = (135.62/ 6.06)1.15= 25.74 bolts
Use 28 bolt M22
The bolts in the framing angle are not counted as they used
to transmit the reaction of the cross girder.
Splice of chord
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Splices of tension or compression chord shall be
designed on the maximum strength of the member. For
straight chord the splice shall be outside the gussetplate. For broken chord the splice will be within the
gusset plate.
Member S3
2 [ No 30 & Bolts M22, = 24
Net area of flange = (1023) 1.60 = 12.30 cm2
Net area of splice plate = (1023) 1.60 = 12.30 cm2
Number of single shear field bolts =
(12.301.6/ 6.06) 1.15 = 3.25 bolts
Net area of web = (3021.60) 1.00 = 22.20 cm2
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Use 2 splice plates = 300.80 + 240.80
Net area of 2 splice plates = (30 22.3) 0.80 + (24 -22.30) 0.8 = 35.80 cm2
Rleast = RD. sh or Rb
Rb = Fb d min t
Fb = Fub
= 0.60 for end distance (S1.5d), (Table 6.2)
Fb = 0.60 8.00 = 4.80 t/ cm2
yRb = 4.80 2.20 1.00 = 10.56
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Splice pl.
x
y
y
x
b
t
qb = 0.25 FubFor bolts of grade ( 8.8),
Rs. sh = qbAsn =
0.258.03.03 2 = 12.12 t
Rb = 10.56 t
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Number of double shear field bolts = (22.201.6/ 10.56)1.15 = 3.87 bolts
Use 4 bolt M22
The splices in the chords are placed in the side of the:-
1. Smaller cross section except in cases where the erectionis done by the cantilever method.
2. Gusset plate; in a polygonal chord in order to avoid the
bending of plates, angles and channels the splice plate isplaced at the brake of the chord on the gusset plates.
3.Gusset shall be proportion to withstand the effective forces
in the web member
The thickness of the gusset is determined from the critical
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section abcd. This section should be at least 15 20 %
stronger than the diagonal it self generally all gusset are
made of the same thickness. The thickness of gusset plateshall be at least 12 mm in Railway bridges and 10 mm in
Roadway bridges.
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2 Pl. 240x12
2 NO.300
12M22(8.8)
14M22(8.8)
Bolts for framingonly.
2 NO.240
4M22(8.8)
4M22(8.8)
a
Gusset pl.for L.W.Br.
d
b c
2 NO.220
8M22(8.8)
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4- At the chord panel point the cover plate should be
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connected to a gusset plate by special connection angle to
make the center of gravity of the rivet group between gusset
plate and top chord nearer to the center of gravity of the topchord.
A
S
S
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n
A
A
A
nn
Sec A-A
S
Sec A-A
S
3
221
4
3 223
1
3
3
S
2
1
4
2
1
S
Back
Design of Battens and Diaphragms
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The two parts of the box section must be connected
together in such away that they act ass one unit. Forcompression member stronger details are necessary than
for ten member.
Diaphragms
Diaphragms are transverse plates or channel connected to
the two webs of the box section by angles. They are
necessary to assume the rectangular shop of the box
section. For the chords wee have at least one diaphragmbetween the two panel points. In the diagonals, we arrange
at least one diaphragm near each end.
Batten plate
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One the open sides of the box section we have batten plates
as close to the gusset plate as possible one intermediate
batten plate or lacing bars to avoid lateral buckling of the
unsupported flange for the calculation of the lattice bars of
a compression member we assume a transverse force = 2 %
of the longitudinal force in the member. If there is acontinuous plate at the upper side of the box section,
latticing will be in the lower side only and transverse force
will be according to cross section of the lower side only. In
tension member a lattice system and batten plates 25 %lighter is used.
A
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Sec A-AA
S S
Sec S-S
B
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B
B D
D
B
Battening of compression member
The number of batten is such that we get at least 3 bays
batten shall be of plates, channels, I-section bolted or
welded to resist the following forces:-
The member as a whole can be considered as avierandeen girder or we can assume hinges at mid
distances and change it to statically determine system.
Shear in batten plate = Qd/ a
Bending moment in batten = Qd/ 2
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Q.d/2
M=Q.d/2
Q.d/4
Q/2
F1
F1
Q.d/a
Q/2
d/2
d/2
Q/2
t
a
Q/2
Q/2
Q/2
Q.d/a
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batten pl.
d
Design of End Portals
End transverse bracing are called portals. The portals areplaced either in the vertical plan of the end post (1), in
the plan of the first vertical (2), or in the inclined plan of
the end diagonal (3). In case (2) the first panel of the
lower wind bracing is affected by the reactiontransmitted by the end portals. The arrangement of end
portal in case (2) is stiffer than in case (3). The portal
must not interfere the clearance line.
Back
The shop of the portal depends on the depth and on the
l li Th l ll i
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1 3
2
W1
clearance line. The portals are generally static
indeterminate closed frames in which the post over subject
to bending stresses.
Approximate calculation of portals
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The point of inflection of the post are situated according
to the relative stiffness of the cross girders, post, and the
upper strut at height between 1/31/2 of the force height
h. We can replace the point of inflection by two hinges
at C and C each of them transmits W1. Then the portal
can be calculated as static determine frame. If the portalis in the plan of inclined end diagonal, the points of
inflection C, C should net be more than h/ 3 from A and
A.
W1 W1
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W1.d/2
W1.d/2+W1.e
W1.h/b
W1.d/2
W1 f/2
h
X.G
h'
b
h'
W1.d/2
W1.h/b
W1 f/2
W1
W1/2
W1
h
d
e
W1.h/b
X.G
W1/2
b
W1/2
W1/2
W1.h/b
f
W1
Back