steel bridges ppt
TRANSCRIPT
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A
PRESENTATION ON
STEEL BRIDGES
PRESENTED BY:
PATEL VIVEK D.
SD0608
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CONTENTES
INTRODUCTION
TYPES OF BRIGES
LOADS ON BRIGES INTRODUCTION OF INFLUENCE LINES
I.L.D ANALYSIS FOR A GIRDER
I.L.D ANALYSIS FOR A TRUSS REFERENCES.
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Steel bridges
The main advantages of structural steel overother construction materials are its strength andductility.
It has a higher strength to cost ratio in tensionand a slightly lower strength to cost ratio incompression when compared with concrete.
The stiffness to weight ratio of steel is much
higher than that of concrete. Thus, structuralsteel is an efficient and economic material inbridges.
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Classification of steel bridges
Steel bridges are classified according to
The type of traffic carried.
The type of main structural system. The position of the carriage way relative to
the main structural system.
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Bridges are classified as
Highway or road bridges Railway or rail bridges
Road - cum - rail bridges
Classification based on type of traffic carried
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Classification based on the main structural
system
Girder bridges
Trusses steel
bridges
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Typical rigid frame bridge
Typical arch bridges
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Normal span ranges of bridge system
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DECK BRIDGE
THROUGH BRIDGE
SEMI-THROUGH BRIDGE
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LOADS ON BRIGES
Dead load
Live load
Impact load
Longitudinal force
Wind load
Seismic load
Forces due to curvature.
Forces on parapets Frictional resistance of expansion bearings
Erection force
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LOADING
IRC class AA loading
consists of either a tracked
vehicle of 70 tonnes or a
wheeled vehicle of 40
tonnes with dimensions asshown in Fig. Tracked vehicle
Wheeled vehicle
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bridges on national highways and state highways are designed for these
loadings.
Bridges designed for class AA should be checked for IRC class A loading
also, since under certain conditions, larger stresses may be obtained
under class A loading.
Class A loading consists of a wheel load train composed of a driving
vehicle and two trailers of specified axle spacing.
This loading is normally adopted on all roads on which permanent bridges
are constructed.
Class B loading is adopted for temporary structures and for bridges inspecified areas.
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ANALYSIS METHOD
Influence line diagram for any structure
elements for reaction at a support, the S.F. at
section, the B.M. at a section is a graphical
representation of its variation due to changingposition of rolling unit load throughout the
span of the structure.
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Influence Line
Definitions
Response Function support reaction, axial
force, shear force, or bending moment.
Influence Line graph of a response function of a structure
as a function of the position of a downward unit load
moving across the structure structure.
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Influence lines for beams and plate
girders
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Consider a section 1-1 and assume unit-rolling
load is at a distance x from L0.Then, from
equilibrium considerations reactions at L8 and
L0 are determined. The reactions are:
Reaction at L8 =(X/L)
Reaction at L0 =(1-X/L)
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A girder span of 20 m is simply supported t its ends. fours wheel loads
150KN,150KN,250KN and 100KN traverse the girder from right to left with
100 KN load leading. Distance between wheel loads is 3m each.
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First 150KN is at c : S.F.= SC1
second 150KN is at c : S.F.= SC2
w1/a > w/l
w1/a = 150/3 =50 and w/l = 660/20=32.5
but w1/a > w/lSC1 > SC2
put first 150 KN load on c.
MAX. S.F. at C = loads*ordinates
=(150*0.6) + (150*0.45+250*0.3+100*0.15)=247.5 KN(8m from left support)
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Average load on left= Average load on right
(1) only 100 KN load on right side of 8m
Average load on left =loads/spans=550/8=68.75
Average load on right=loads/spans=100/8=8.33
(2) 250 KN crosses point c:Average load on left =loads/spans=300/8=37.5
Average load on right=loads/spans=350/12=29.17
(3) 150KN crosses point c:
Average load on left =loads/spans=150/8=18.75
Average load on right =loads/spans=350/12=41.67
Average load on left
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MAX. B.M.at C = loads* ordinates
=(150*3) + (150*4.8+250*3.6+100*2.4)
= 2310 kN-m
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Definitions of truss bridges
A truss is an assemblage of a number of
straight members arranged in a triangle or
combination of triangles to form a rigid frame
work
A joint is a connection where two or more
members of the truss are fastened together
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SYSTEM FOR TRUSS BRIDGE
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If the cross girders are connected to top
panels points, then load is transferred to top
panel points and type of bridge girders are
knows as deck bridge girders.
If the cross girders are connected to bottom
panels points, then load is transferred through
bottom panel points and type of bridgegirders are knows as through bridge girders.
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Influence Lines for truss
Consider the bridge in Fig. 1.
As the car moves across the bridge, the forces
in the truss members change with the position
of the car and the maximum force in each
member will be at a different car location.
The design of each member must be based on
the maximum probable load each member
will experience.
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Bridge Truss Structure
Subjected to a Variable
Position Load
Therefore, the truss analysis for each member
would involve determining the load position that
causes the greatest force or stress in each member.
If a structure is to be safely designed, members
must be proportioned such that the maximum
force produced by dead and live loads is less than
the available section capacity.
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Structural analysis for variable
loads consists of two steps:
1.Determining the positions of the loads atwhich the which response function is
maximum; and
2.Computing the maximum value of the
response function
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Once an influence line is
constructed:
Determine where to place live load on a
structure to maximize the drawn response
function; and
Evaluate the maximum magnitude of the
response function based on the loading.
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CALCULATION FOR U1U2(TOP CHORD MEMBER)
Consider section at 1-1
Moment @ l2 (As member u1L2 and L1L2 passes though l2)
When unit load is in left section or part, consider equilibrium of right part.
P-u1u2= V2 * (L2L4/H)
P-u1u2= V2 * (8/3) COMPRE.
When unit load is at L0 Pu1u2=0
When unit load is at L1, V2=1/4
Pu1u2 =1/4*8/3= 2/3 compression
When unit load is in right section or part, consider equilibrium of left part only.
When unit load is at L2 Pu1u2=4/3 compression
When unit load is at L3 Pu1u2=2/3 compression
When unit load is at L4 Pu1u2=0
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CALCULATION FOR L1L2(BOTTOM CHORD MEMBER)
Consider section at 1-1 Moment @ U1 (As member u1L1 and U1U2 passes though U1)
When unit load is in left section or part, consider equilibrium of right part.
P-L1L2= V2 * (L1L4/H)
P-u1u2= V2 * (12/3)
When unit load is at L0 PL1L2=0 When unit load is at L1, V2=1/4
PL1L2 =1/4*12/3= 1
When unit load is in right section or part, consider equilibrium of left partonly.
When unit load is at L2 L1L2=2/3
When unit load is at L3 L1L2=1/3
When unit load is at L4 L1L2=0
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CALCULATION FOR U1L2(DIAGONAL MEMBER)
Consider section at 1-1
When unit load is in left section or part, consider equilibrium of right part.
P-U1L2sin= V2 compression
P-U1L2= V2 cosec=1.67V2
When unit load is at L0 U1L2=0 When unit load is at L1, V2=1/4cosec=0.417
Pu1u2 =0.417 compression
When unit load is in right section or part, consider equilibrium of left part
only.
When unit load is at L2 Pu1u2=1.67*V1=0.833 tension
When unit load is at L3 Pu1u2=1.67V1=0.417 tension
When unit load is at L4 Pu1u2=0
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FORCE IN MEMBER
As the whole span is greater then the span of the truss thereforewhole truss is loaded at the bottom chords.
Member is compressive throughout so no tensile force isdeveloped.
force in U1U2 =intensity*area covered
=10*1/2*4/3*16
=106.67kn(c)
Moving load acts on any panel point therefore, force in member istensile only.
force in L1L2=intensity*area covered=10*1/2*1*16
=106.67kn(c)
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FORCE IN MEMBER
Member U1L1
Member is tensile in some part an in compression for some part.
Area above base line gives compression line gives compression area
and below baseline gives tensile area.
Max. compression=intensity*compression areaMax. compression=10*1/2*10.667*2/4
=26.67kn
Max. tension = 10*1/2*1/4*5.33
=6.66kn
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FORCE IN MEMBER
Member U1L2
Max. compression=intensity*compression area
Max. compression=10*1/2*0.417*5.33
=11.11kn
Max. compression=10*1/2*0.834*10.667
=44.48kn
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THANKS