3-Loads on bridge

Report
Chapter 3
LOADS ON BRIDGES
contents

3.1 Loads on bridges
 3.2 Dead loads on highway and railway
bridges Fig(3-1)
 3.3 Live loads on bridges
 3.4 Impact loads
 3.5 Centrifugal force Fig(3-4)
 3.6 Temperature effect ((5-5),(6-5)) Fig(3-4)
 3.7 Wind pressure ((5-9),(6-9)) Fig(3-5) , Fig
(3-6)
•3.8 Braking force Fig (3-7)
•3.9 Lateral shock effect (6-7) Fig (3-8)
•3.10 Frictional resistance of bearings ((5-10),(6-10))
•3.11 Settlement of supports ((5-11),(6-11)) Fig (3-9)
•3.12 Forces due to erection ((5-14),(6-14))
3.1 Loads on bridges
Loads acting on bridges are divided into:1. Primary loads.
2. Secondary loads.
A load is considered primary or secondary according to
the part of the bridge which shall be designed. Wind loads
are secondary loads in designing the main girders and
primary loads in designing the wind bracings.
3.1.1 Primary loads on highway and railway bridges
1-Dead loads.
2-Live loads.
3-Impact loads (dynamic effect).
4-Centrifugal forces.
3.1.2 Secondary loads on highway and railway bridges
1-Wind pressure or earthquake.
2-Braking force.
3-Lateral shock effect.
4-Temperature effect.
5-Frictional resistance at movable bearing.
6-Forces due to settlement of supports.
7-Effect of shrinkage and creep of concrete.
8-Forces due to erection.
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3.2 Dead loads on highway and railway bridges Fig(3-1)
It consists of the weight of steel structure and the bridge
floor. The weight of the floor is found from the dimensions
and the unit weight of the different materials.
Weight of an open timber floor for a single track railway
bridge = (250+350)=600 kg/m’ (9.3.4(p149)).
The weight of the steel structure is first approximately
determined from similar existing bridges or from empirical
formula.
* Approximate weight of the steel structure for single track
Railway Bridge with open timber floor (standard grade
steel)
- For through bridge
W = 0.75 + 0.50 L t \ m’
- For deck bridge
W = 0.50 + 0.50 L t \ m’.
For deck bridge without stringers and cross girders
W = 0.25 + 0.50 L t \ m’
Where, W = weight of total steel in (t) for one meter of
bridge, L = effective span (leff) of bridge in meters, (for
continuous bridge, leff = (0.70 – 0.80) L = the distance
between two sequence points with zero total moment.
We have to increase the above steel weight by 90 % for
double track and 80 % more for triple track
* For a ballasted floor the weight of steel is 20 – 40 %
greater.
* For roadway bridge with heavy traffic the weight of steel
structure is approximately for 1 m2 of roadway and side
walks as follows:* Outside side walks
W1 = 200 + 4 L + 0.03 L2 kg/ m2
for roadway
W2 = 100 + 3 L
for side walks
W = W 1 + W2
bridge
kg/ m2
kg/ m2
of
* Inside side walks
W = 200 + 4 L + 0.03 L2
for roadway
W2 = 100 + 3 L
for side walks
W = W 1 + W2
of bridge
kg/
L = effective span of bridge in meter
m2
Rail
1.5 m
Rail
Sleeper
Sleeper
h = L/ 10
100 cm
B = Bridge Width
Railway Through Bridge
W2
bs
W1
0.5
B
W2
0.5
bs
a
b'
Roadway Open Through Bridge
Side Walk Outside
W1
bs
B
bs
a
b'
Roadway Open Through Bridge
Side Walk Inside
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Fig(3-1)
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3.3Live loads on bridges
Railway Bridge (p6-1) Fig(3-2)
The type of trains is different for different countries
according to the importance of lines. In Egypt we shall
consider one type of three train types (D, H, and L), train
type “D” is the heaviest train is used in Egypt.
Train type “D” consists of a two locomotives and two
tenders followed on one side only by an unlimited number
of wagons
Locomotive +Tender + Locomotive + Tender +
unlimited number of wagons
Where total weight of one Locomotive = 100 ton, and its
length = 10.50 m(p6-3).
While total weight of one Tender = 80 ton, and its length =
8.40 m.
If two tracks are loaded at the same time, only 90 % of
loads are used. In case of three tracks only 80 % of loads
are used while in case of four tracks we use 75 % of loads
are used(p6-2).
Locomative = 100 t
12.5
12.5
Tender = 80 t
12.5
10
10
10
Locomative = 100 t
12.5
10
6.25 6.25
1.5
1.75
1.75
12.5
Tender = 80 t
12.5
10
10
10
Wagon = 80 t
10
10
10
+ Any Number of Wagons
10
10
6.25 6.25
2
2
1.5
1.5
[100 t] 10.5 m
1.8
1.8
1.8
1.5
1.5
[80 t] 8.4 m
1.75
1.75
2
2
1.5
1.5
[100 t] 10.5 m
[180 t] 18.9 m
1.8
1.8
1.8
[80 t] 8.4 m
[180 t] 18.9 m
9.52 t/m'
1.5
1.5
1.75
5.5
[80 t] 12 m
12 m
6.666 t/m'
Loads of Train Type 'D'
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1.75
1.5
Roadway Bridges Fig(3-3)
Within the kerb to kerb width of the roadway, the standard
vehicles are assumed to travel parallel to the length of
bridge, and to occupy any positions, which produce the
maximum stress. For the standard vehicle, all the axles of a
unit of vehicle are considered as acting simultaneously in a
position causing maximum stresses. The vehicle in adjacent
lanes is taken as headed in the direction producing
maximum stresses. The maximum bending moment and
maximum shear force on the plate girders are found by
longitudinal location of loading.
For main roadway bridge, the L.L shall be that type of
vehicular rolling load and/ or distributed load representing
it.
a.The main girder.
As well as the floor system shall be designed for truck
concentrated axle loads the standard type shown together
with distributed load of 500 kg/m2 cover the main lane of
three meter width lane(p5-1), the second lane of three
meter width each shall be covered with one truck moving
in the same direction and parallel to the axes of the bridge.
The remaining parts of the floor are covered with a uniform
load of 300 kg/ m2 (p5-3). Also, side walks shall be covered
by the same distributed load. The impact will be considered
for the loads on the main lane only.
b.The elements of side walks. Fig(3-4)
It shall be designed for 500 kg/ m2, and then we check
also for a vertical concentrated load of (5 t, acting without
uniform loads) in the position giving maximum stresses(52-2).
The handrail, shall be designed for line distributed load not
less than 150 kg/m' at top level of the handrail(5-2-6).
2
300 kg/m (Without Impact)
10 t
10 t
(With Impact)
500 kg/m2
0.6
500 kg/m2
10 t
0.2
10 t
10 t
10 t
5t
5t
5t
2
300
kg/m (Without Impact)
0.6
2
300
kg/m (Without Impact)
0.2
5t
5t
5t
2
300 kg/m (Without Impact)
Loads on Traffic Lanes of Roadway Bridges
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(With Impact)
Fig(3-3)
Foot Bridges
It shall be designed for uniform load of 500 kg/ m2, without
impact(5-2-4).
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3.4 Impact loads
Impact is the dynamic effect on the bridge due to the
moving loads. If we measure the deflection at a certain
point of the bridge for slowly moving train (static L.L), and
for rapidly moving train (static L.L + Impact), the increase
of deflection in the later case is due to impact.
The chief factors causing Impact are:1.Roughness and unevenness of the track of a railway
bridge or of the roadway surface of a roadway bridge.
The smoother of the surface the smaller will be the
impact. In Railway Bridge the joints of rails increase the
impact. It is recommended to use long rails on bridges or
to weld the joints.
2.Irregular and eccentric wheels are defective springs.
·
The proportion
Im pact
I

Live Load L
is called Impact coefficient
This coefficient depends on the loaded length and on the
type of
· This coefficient depends on the loaded length and on
the type of the bridge.
· In case of bigger loaded length we have smaller
Impact coefficient.
·Rigid parts are more affected by impact than elastic
parts.
· For main truss member, the impact decreases as the
loaded length increases, since the time necessary to cover a
greater length is more and the load is applied less
suddenly.
·
Impact formula for Railway Bridges (6-3) is:-
24
I
24  L
where,
0.25  I  0.75
Where I is the factor by which the live load is to be
multiplied to give the addition due to dynamic effect
·
L = loaded length in meters of track or the sum of
loaded lengths of double or multiple tracks producing
maximum stresses in members.
EXAMPLE
Span of bridge = 50 m.
Distances between X.G. = 5.0 m
1. Single Track Bridge
For stringer
I
24
 0.83  0.75
24  5
use I = 0.75
24
I
 0.704  0.75
24  2  5
use I = 0.704
24
 0.325  0.25
For M.G. I 
24  50
use I = 0.325
For X.G.
2. Double Track Bridge
For stringer
For X.G.
24
I
 0.83  0.75
24  5
24
I
 0.546  0.75
24  2  2  5
use I = 0.75
use I = 0.546
For M.G.
24
I
 0.194  0.25
24  2  50
use I = 0.25
Impact formula for Roadway Bridges (5-2-3) is:,
I  0.40  0.008 L
L = loaded length
·
Where I is the impact coefficient (for the main lane
only) due to vertical concentrated loads (60 t) and uniform
distributed live load (500 kg/ m2).
·
L = loaded length in meters of traffic main lane
producing maximum stresses in members.
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3.5Centrifugal force Fig(3-4)
For bridges in curves, the stresses due to the centrifugal
action and the super elevation of the track must be
considered in designing the members. A vertical load w
moving in a curve of radius R and a speed V.
W V 2
For Railway bridges,
C
(in tons) (6-4)
127R
W = axle load in tons
(R = 400 – 600 m)
(V = 70 – 90 km/ hr)
C is a horizontal force acting at the center of gravity of
masses 2.00 above the top of rail (6.4.1) . It produces an
increase of the vertical reaction on the outside rail and a
decrease of the vertical reaction on the inside rail.
For Roadway bridges,
3000
C
R  150
(in tons) (5-4)
R = radius of curve in m
C = centrifugal force in tons every 50 m
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3.6 Temperature effect ((5-5),(6-5)) Fig(3-4)
When steel structure is not free to expand or contrast under
variation of temperature, the stresses due a variation of 
30 C. From local main must be considered. The
coefficient of expansion for steel and concrete is 0.00001.
If we consider unequal variation of temperature, in some
structures which are not affected by equal changes, we
allowed only for  15 C.
In two hinged arches and suspension bridges the equal
change of temperature has an effect on the internal forces.
In continuous bridges the equal change of temperature has
no effect because the girders are free to expand, but the
unequal change has an effect.
The horizontal displacement at point b in the main system
10 = F1tL
110 =  S1tL
F12  L
11 
EA
10 + X111 = 0
S1    t  L
10

X1  

11
2 L
 S1 EA
 get 10
·
Temperature has large effect.
·
The modulus of elasticity
E = 2100 t/cm2
for steel.
E = 1000 t/cm2
for cast iron.
E = 210 t/cm2
for concrete
bs
bs
B
Curb
Side walk
Stringer
X.G.
Bracket
b'
Loads on Side Walk
C
W
Centrifugal Force
X1= 1t
X1= 1t
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L
Temperature Force
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Fig(3-4)
3.7 Wind pressure ((5-9),(6-9)) Fig(3-5) , Fig (3-6)
For bridges we consider either the case of unloaded bridge
with a wind pressure of 200 kg/m2 or the case of loaded
bridge with a wind pressure 100 kg/m2 on exposed
surfaces of bridge and train. The effective height of a train
in railway bridges is 3.50 m from the rail level, and that for
crowds or road vehicles is 3.00 m. The train is considered
as having on single vertical plane only.
2
200 kg/m
Wind Loads
3.50 m
2
100 kg/m
Unloaded
Loaded
3.50 m
2
100 kg/m
B = Bridge Width
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Loaded
Fig(3-5)
B = Bridge Width
2
3
100 kg/m
2
100 kg/m
2
200 kg/m
2
100 kg/m
3
2
100 kg/m
2
200 kg/m
Wind Loads
Unloaded
Loaded
Unloaded
Loaded
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Fig(3-6)
3.8 Braking force Fig (3-7)
In railway bridges (6-6) we have to consider the stresses
resulting from the application of brakes to the live load
while passing on the bridge. The braking force is equal to
1/7 of the maximum Live Load, without impact, supported
by one track only. In case of several tracks, the braking
force on the second track is equal to 1/14 maximum L.L (of
the second track). The braking force has a great effect on
the design of the towers and also on the abutments and piers
supporting the fixed bearing of bridges (hinged bearing).
In roadway bridges (5-6) the braking forces ( 90 t) = 0.25
Loads on main lane (L)
(= [(L - 6) 30.50 +
60]0.25)
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3.9 Lateral shock effect (6-7) Fig (3-8)
In railway bridges a single force 6t (without impact) is taken
normal to the track at top rail level and in position giving
maximum stresses. The stresses due to the lateral shock of
locomotive wheel are considered in the design of:1-Stringer.
2-Stringer bracing.
3-Wind bracing.
4-End X-frame.
5-The bearings.
6-Rail connections.
7-The piers, the foundation.
(If there is My (due to lateral shock) use B.F.I.B, for
stringer, to support My.)
For railway bridges on a curve, only the greater of the
centrifugal force or the lateral shock must be considered.
Braking force
B
B
B/ 2
X.G.
B/ 2
Width of bridge
B
Width of bridge
M
My max = B*a = M
X.G.
X.G.
B/ 2
B
B
B/ 2
B/ 2
B
B
B/ 2
B
X.G.
B
stringer
stringer
S
B
X.G.
stringer
X.G.
M.G.
B/ 2
M.G.
M.G.
B
M.G.
B
S
B
X.G.
X.G.
B
B/ 2
B
neglect braking force
B/ 2
B
B
B/ 2
B
B
neglect braking force
B/ 2
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X.G.
width of bridge
12.5
12.5
12.5
12.5
10
10
10
12.5
12.5
10
10
6.25
10
10
10
6.25
M.G.
2
2
3
1.8
1.8
1.8
3
1.75
1.75
L=Span of bridge
Max. L.L. = Max. Sum of wheel loads on the bridge span
B = {[Max. L.L./ 7] } / No. of X.G.
2
2
3
1.8
1.8
1.8
Figure (3-7)
X.G.
M.G.
6t
X.G.
6t
stringer
M.G.
M.G.
M.G.
S
6t
M.G.
6t
stringer
stringer
stringer
S
X.G.
stringer
M.G.
S
6t
stringer
stringer
S
Lateral Shock
6t
X.G.
6t
X.G.
X.G.
6t
X.G.
6t
X.G.
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Fig(3-8)
3.10Frictional resistance of bearings ((5-10),(6-10))
Forces due to friction at the expansion bearing under dead
load only must be considered and the coefficient is:F = RD.L.
For roller bearing with one or two rollers  = 0.03
For roller bearing with three or more rollers  = 0.05
For sliding of steel on hard copper  = 0.15
For sliding of steel on cast iron or cast steel  = 0.25
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3.11 Settlement of supports ((5-11),(6-11)) Fig (3-9)
Stresses due to unequal settlement of continuous
structures supported on piers or abutments shall be added
for all members. (Fig. )
 y 2  y1 y 2  y 3 

M1  L12  2M 2  L12  L 23   M 3  L 23  6EI

L 23 
 L12
(Maximum allowable settlement is 2.50 cm.)
Settlement may be lead to the continuous structure to be
simple structure; hence the internal forces will be
increased.
Settlement of Supports
y1
y2
L 12
y3
L 23
Figure (3-9)
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3.12 Forces due to erection ((5-14),(6-14))
1-Erection by cantilever method.
Additional stressed will be exists due to erection by
cantilever method, so it must be considered during the
design of bridge, also the allowable stresses are increased
by percentage of 25 %, (0.58Fy  1.25 = 0.73 Fy) (2.5P8).
If the erection of the bridge is done by the cantilever
method, the biggest possible forces in the members during
the erection must be considered in the design of these
members. A higher working stress may be used (or 0.73 y)
than for the complete bridge.
2-Erection by floating method.
It is used in cases of simple beam bridges. Where a
loaded ship carries the structure up to the site of erection,
then the loads are removed slowly till the structure has in
its required erection level.
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