Steel Bridges Ppt

8/12/2019 Steel Bridges Ppt

1/44

A

PRESENTATION ON

STEEL BRIDGES

PRESENTED BY:

PATEL VIVEK D.

SD0608


2/44

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.


3/44

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.


4/44

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.


5/44

Bridges are classified as

Highway or road bridges Railway or rail bridges

Road - cum - rail bridges

Classification based on type of traffic carried


6/44

Classification based on the main structural

system

Girder bridges

Trusses steel

bridges


7/44

Typical rigid frame bridge

Typical arch bridges


8/44


9/44

Normal span ranges of bridge system


10/44

DECK BRIDGE

THROUGH BRIDGE

SEMI-THROUGH BRIDGE


11/44

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


12/44

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


13/44

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.


14/44

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.


15/44

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.


16/44

Influence lines for beams and plate

girders


17/44

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)


18/44

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.


19/44


20/44

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)


21/44


22/44

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


23/44

MAX. B.M.at C = loads* ordinates

=(150*3) + (150*4.8+250*3.6+100*2.4)

= 2310 kN-m


24/44

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


25/44

SYSTEM FOR TRUSS BRIDGE


26/44


27/44

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.


28/44

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.


29/44

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.


30/44

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


31/44

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.


32/44


33/44

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



34/44


35/44

CALCULATION FOR L1L2(BOTTOM CHORD MEMBER)

Consider section at 1-1 Moment @ U1 (As member u1L1 and U1U2 passes though U1)


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


36/44


37/44


38/44


39/44

CALCULATION FOR U1L2(DIAGONAL MEMBER)

Consider section at 1-1


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



40/44

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)


41/44

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


42/44

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


43/44


44/44

THANKS

Documents

Steel Bridges Ppt