Design of Bridge g gMain Girders

According to ECPAccording to ECP

Assis. Prof. Dr. Ehab B. Matar

Design Phases of Main Girders1. Structural analysis for the main system determining the max.

and min. straining actions at critical sections2. Design of Web Plateg3. Design of Flange Plate4. Design of Stiffeners (end bearing and intermediate)5 Design of Connections between web plate and flange plate5. Design of Connections between web plate and flange plate6. Design of Splices7. Check fatigue for all details

Bridges Main systems

1- Structural Analysis

Structural analysis is carried out using either influence lines, grillage analysis or finite elements depending on complexity ofgrillage analysis or finite elements depending on complexity of structures and its importance.

Max. and Minimum shear, moments, reactions…. are determined at critical sectionsat critical sections.

Stress range

Stress range

Stress range

2- Design of Web Plate This includes the following tasks:1. Determining Web height2 Determining Web thickness2. Determining Web thickness3. Checking shear Buckling

2.1 Web height Overall depth, h: Lo/18 ≤ h ≤ Lo/12 (highway two lanes)

L /10 ≤ h ≤ L /7 ( il i l t k) Lo/10 ≤ h ≤ Lo/7 (railway single track) For double tracks railway or four lanes roadway, increase

the above limits by 60-85%the above limits by 60 85% If the maximum moment is known; the web height can be

approximately calculated as

33 7.53.523or

bb FM

FMkdh

d= plate girder depth (cm) M= maximum bending moment (t.cm)

k ratio of height to thickness of eb plate 100

bb

k= ratio of height to thickness of web plate 100 Fb= allowable bending stress ≈ 0.58 Fy

2.2 Web Thickness

web thickness, tw: tw ≥8mm

yw FtdIn 830 caseany *

yw Ftd 190 used are stiffeners sverse when tran*

y

Ftd 320 used stiff. e transversand d/5)(at allongitudin When *

yw Ft

2 3 Shear Buckling2.3 Shear Buckling

No need to check shear buckling resistance if : No need to check shear buckling resistance if : for un-stiffened web for stiffened web

yww Ftd /105/

qww F

ktd /45/

yww F

1 0f)/345(4

shearfor factor buckling K 2

q

where

/dd0.1for )(4/5.34

1.0for )/34.5(4 2

2

where /dd1where

d

d1

Shear Buckling Resistance

If the above mentioned limits are exceeded the shear buckling resistance should be checked as follows:

yw Ftd

Calculate*

q

qq K

0.35Fq then 8.0For *

57Calculate

yb

yqq

q

F

9.0

35.0)0.625-(1.5q then 2.10.8For * b

y

yq

q F35.0*9.0qthen 2.1For * b

It should be noted that longitudinal stiffener at mid depth would be ff ti f ti bj t d t h th th tmore effective for sections subjected to pure shear than that

positioned at d/5 for pure bending. For continuous plate girder, the web panel over an interior

support will be subjected to simultaneous action of bendingsupport will be subjected to simultaneous action of bending moment and shearing force. Therefore: If the actual shear stress qact ≤0.6qb then the allowable bending

stresses in the girder flanges will not be reduced and shouldstresses in the girder flanges will not be reduced and should not exceed 0.58Fy.

If the actual shear stress qact >0.6qb then two alternatives may be followed. The first is to reduce the allowable bending gstress for flange plates according to the following interaction equation.

yact

b FqF

36.08.0

The second alternative is to design the girder flanges to

yb

b q

g g gresist the whole acting bending moment without any participation of web for resisting bending moment without reducing the allowable bending stress.

Example Design a continuous two spans welded plate girder as shown below for a

roadway bridge. The cross girders are arranged each 2.4m. The deck slab is 20cm thick and the asphalt cover is 10cm. The steel used for the design of all elements is St. 37. Moreover, design a field splice located at 16.8m from the end support. TheSt. 37. Moreover, design a field splice located at 16.8m from the end support. The bending moments and shearing forces at the critical locations are tabulated as follows:

Sec. Md.l (t.m) Qd.l (t) Mll+I (t.m) Qll+I (t) 1 0 +38.2 0 +50.8 2 +171 0 +267 +4 432 +171 0 +267 +4.433 -305.3 -63.6 -296 -61.65 4 +96 -33.07 +96.69,-98.63 -34.92,+3.36

Solution Proportioning of plate girder

cmcmFMkd

b

1901864.1*2

100*100*)3.305296(*323

33

where k is

assumed 100 Design of web plate

ttd 561190190/ cmtF

td wy

w 56.1122

/

By using vertical stiffeners arranged @2.4m (distance between cross girders), then,

26.12401 d

190d

85.7)26.1/(434.5)/4(34.5 22 qk

37.814.285.74545

y

q

w Fk

td

190 thciknesslarge a is which 34.237.81

190 cmtw

Assume tw = 16mm and check shear buckling resistance.

2.18.015.185.74.2

576.1190

57 q

q

ywq k

Ftd

q

2/66.0

)4.235.0)(15.1625.05.1()35.0)(625.05.1(

cmtq

xxFq

b

yqb

safe/41.06.1190

25.125 2 bw

qcmtxA

Qq

3- Design of Flange Plate This includes the following tasks:1. Determining Flange cross section (using Flange Area

Method)Method)2. Determining Flange width and thickness3. Checking Bending Stresses

3.1 Flange Area MethodMY

Plate Girder

32 ddI

MY

c

tf

b

1222

232

32w

fdtdAI

c

d

62122

232w

fw

fAAddtdAI

Welded Plate Girder

6/ w

fAAdYI

w

f

b AAd

MF

6f

AdF

MA w

bf 6

AdF

MAdF

MA wwf 84

3*6

@4arrangedbolts assumeBUSrivetedor boltedFor

dCbtbAwheredFdF

ff

bb

%30202 ;* 846

Section class Part Stress Profile C tf C tfD b

3.2 Local Buckling limitsSection class Part Stress Profile

Compact Flange Uniform Comp. Rolled Sec.

dw tw dw tw

Rolled section Welded section t

h

t

yf FtC /9.16/ psection

g p

B.U.SNon-compact section

Flange Uniform Comp. Rolled Sec.

yf

yf FtC /3.15/

yf FtC /23/ section

B.U.Syf FtC /21/

Remember Whenever the width / thickness ratio of compression

flange exceeds the aforementioned limits, the flange should be treated as a slender section and the effective flangebe treated as a slender section and the effective flange area should be calculated to account for local buckling.

The thickness of flange plates as well as web The thickness of flange plates as well as web plates should be reduced by 1.0mm in design calculations due to the effect of undercut in welding unless special

ti i th ldi t h i t kprecautions in the welding technique are taken.

Note that: When plates with

unequal thickness or Change of flange plate

width or thick.qwidth butt welded together, the thicker or gwider plate should be tapered with a slope 2−

4mm

14

41

not exceeding one to four as shown

2mm

R=6

0cm

R

3.3 Allowable Compressive StressesThe allowable compressive stresses in the compression flange depend on whether the flange is braced laterally or un-flange is braced laterally or un-braced. The lateral unsupported length of the compression flange Lu is calculated as follows:For deck railway bridges withFor deck railway bridges with open timber floor where there exist upper wind bracing or not, and with the existence of cross girders that are rigidly connectedgirders that are rigidly connected to the main girder compression flange, then, Lu = distance between x-girders.For deck railway bridges withFor deck railway bridges with ballasted floor or for roadway bridges where the compression flange is supported by continuous reinforced concretecontinuous reinforced concrete or steel deck, where the frictional or connection of the deck to the flange is capable to resist a lateral force of 2% of the flangelateral force of 2% of the flange force at maximum bending moment, then, Lu = 0.

For continuous deck roadway or dec oad ay orailway bridgeswhere the compression flange at interior supportsat interior supports is located in the bottom side of the girder away from the g ydeck slab, then, Lu = the distance between the centers

f i t ti f thof intersection of the lower wind bracing with the compression flangecompression flange Lu = if there is no lower wind bracing or transverse bracing gthen Lu is taken as the distance from the point of maximum bending moment tobending moment to the point of contra-flexure.

•Lateral unsupported length for through bridge (+M a e a u suppo ed e g o oug b dge (region)

The lateral unsupported length of the compression flange is taken as the full

4*5.2 aEILu y compression flange is taken as the full girder length if the compression flange is unrestrained against lateral bending. If the cross girders and the stiffeners forming U-frames, then, Wh

2

22

1

31

23 EIBd

EId

Where E= The Young's modulus of steel (t/cm2) Iy= moment of inertia of the chord

member about the Y-Y axis (cm4) as shown

d2

d1

I1shown a = distance between the U- frames

(distance between x- girders) = the flexibility of the U- frame

d1 di t f th t id f th

I2

I1

d1 = distance from the centroid of the compression flange to the nearest face of the cross girder of the U-frame.

d2 = distance from the centroid of the compression flange to the centroidal axis

B

compression flange to the centroidal axis of the cross girder

I1 = Moment of inertia of the vertical member forming the arm of the U-frame about the axis of bending.

I2 = Moment of inertia of the cross girder B = distance between centers of

consecutive main girders connected by U-frame.

Continue Example7.5m

2.5m1.25

1.0m 1.0m

1.252.5m

Continue ExampleBridge Cross section

24m 24m

1 2 4 3

16.8

Flange area Method For sec. 3, moment and shear acting together, therefore, checking

%6062.066.041.0

bq

q , then, a reduction in the allowable bending stress

should be carried out as follows

Assuming the compression flange is braced laterally by the lower wind bracing which has joints @240cm (distance between cross girders),then Lu=240cm.

Checking whether lateral torsional buckling controls the allowable compression stresses or not as follows:

f xb 602020

24m 24m

should be carried out as follows

2/38.14.266.0/41.036.08.0

/36.08.0

cmtF

FqqF

b

ybb

Calculating the flange area at sec.3 as follows:

266.1786

6.1190190381

23.6016

cmxEAd

MA wf

uy

f LcmxF

b 6.774

4.2602020

ub

y

f

by

f

LcmCFd

A

xxx

xxCFd

A

86.580..

1380

3.60178.3233.0

3.60178.32305.175.1(

4.21966031380.

.1380 2

619038.16 xdFbf

Assuming bf=0.3d=0.3x190=57cm→60cm Then tf=178.66/60=2.98cm→3cm

yf Ftc 2173.9

302/16300

Therefore Fbc =1.38t/cm2& Ft=1.4 t/cm2 /38.1

543,4323.601 2cmtEF O.K safe in both tension and

compression. Checking fatigue stress by considering only 60% of the live loads

are acting i.e. 0.6x274.32=164.6tm. 26164 E

However, although the flange satisfied the limit of compact section ( yF/3.15 ) but the web is non-compact, therefore, the whole section will be considered non-compact.

Calculating the flange area at sec.2 as follows:

21146.11902438 xEAMA w

Stress range = 2/378.0543,43

26.164 cmtE

For ADTT >2500 truck, No. of stress cycles =2E6 and for welded plate girder Class B' is taken therefore the allowable fatigue stress rangeFsr=0.85t/cm2. Therefore, the section is safe.

Checking Actual stresses at section 2 2

33 425061190 xx 211461904.16

cmxdF

A w

bf

Assuming bf=50cm Then tf=114/50=2.28cm→2.4cm

yf Ftc 2108.10

242/16250

34

2

194,32,714,135,3

2.1954.25012

4.250212

6.1190

cmZcmI

xxxxxI

gross

gross

Assuming the compression flange is braced laterally by the existence of R.C. deck directly rested over compression flange therefore,Lu=0. Therefore, there is no need to check Fltb.

Therefore Fb =1 4 t/cm2& Ft=1 4 t/cm2yf

Checking Actual stresses at section 3

34

233

543,43,213,267,4

5.19536012

360212

6.1190

cmZcmI

xxxxxI

gross

gross

Therefore Fbc 1.4 t/cm & Ft 1.4 t/cm /36.1

194,322438 2cmtEF O.K safe in both tension and

compression. Checking fatigue stress by considering only 60% of the live loads

is acting. The live load is ranged at this section between +250.5 tm and -73.971 tm. Then,

Example for through bridge A railway through plate girder y g p g

bridge of 27 m simple span has two main girders 9 m apart. The bridge is for double track and hasbridge is for double track and has an open timber floor. The cross girders are arranged every 2.25 m.The bridge is provided by stringer The bridge is provided by stringer bracing, while braking force bracing is not arranged. The

fmaterial of construction is steel 44; the weight of the timber floor for each track is 0.6 t/m’.

Calculate Lu considering, for main girder tw=20mm, dw=3000mm, M =1060mt Q =160t whileMmax=1060mt, Qmax=160t, while for XG Pl#1000x20/2Pl#300*40mm

Solution For the main girder, tw=20mm,

dw=3000mm Af=M/(dFb)-Aw/6Af M/(dFb) Aw/6 Af=1060*100/(300*1.6)-

2*300/6=120.8cm2 Bf=0.2*300=60cm tf>120.8/60>2.01cm c/t <21/sqrt(F )t =2 4cm

d2

d1

I1

c/tf<21/sqrt(Fy)tf=2.4cm Z=(50x2x27.4+60x2.4x1.2)/(60x2.4+50

x2)=11.94cm

I2

B

Iy=2.4x60^3/12+50X2^3/12=43233cm4

I2=2X100^3/12+2x(30x4^3/12+30x4x52^2)=815947cm42 2)=815947cm

d1= (302.4-100-11.94)=190.46cm d2= 190.46+100/2+4)=244.46cm B=930cm

Upon calculating the I1, the section is composed of part ofcomposed of part of the web plate (25 twxtw), vertical stiffener assumed 250x20 mmassumed 250x20 mm (satisfying local buckling) and the b k t i t bbracket is to be defined according to max. distance allowed by train edges.

Calculate the centriod X=(25x2x1+40x2x22+ X=(25x2x1+40x2x22+

50x2x43+25x2x56.5)/(2x(25+40+50+25)=31.99cm

I1=25x2x30.9^2+40^3x2/12+40x2*9.91^2+50

EIBd

EId

23 2

22

1

31

x2x11^2+2x25^3/12+2x25x24.6^2=111226cm4 okmacmeaEIL

cme

yu

25.237926.2*225*43233*2100*5.25.2

26.2815947*2100*2

930*2^46.244111226*2100*3

3^46.190

44

3- Design of Stiffeners

This includes the following tasks:D i f ti l tiff (Wh th d b i1. Design of vertical stiffeners (Whether end bearing or longitudinal stiffeners)

2. Design of horizontal stiffenersg

Importance of stiffeners Reducing slenderness ratio of web plate Increase the shear capacity of web platep y p

Location of stiffeners Vertical stiffeners are located at location of X-

girders and at supports Horizontal stiffeners at a distance of d/5 from

compression flange and at d/2

A- Vertical StiffenersCompression Flange

xTension Flange

The outstanding length of vertical stiffeners 1- stiffeners in pairs x dw/30 + 5 cm2- single stiffeners x dw/30 + 10 cm

Vertical intermediate stiffeners composed of single or i f t i l ti hil E d b i tiffpair of symmetrical sections while End bearing stiffeners

should be in pairs fastened on each side of the web End bearing stiffeners should be well ground or

machined to fit tightly against the top and bottom flange angles and should never be crimped.

Vertical stiffeners act as a compression members with Vertical stiffeners act as a compression members with buckling length equal to 0.8dw

For the sake of design of the vertical stiffeners it will be considered as a column of cross section consisting ofconsidered as a column of cross section consisting of the area of angles and a certain length of the web ( 25 tw for intermediate vertical stiffeners and 12 tw for end bearing stiffeners)bearing stiffeners).

6tw6tw25tw

End BearingStiffenerStiffener

Intermediate

• End Bearing Stiffeners

Design Steps: 1- Define the design force d.F (Max. Reaction at support) 2- Choosing stiffeners in pairs x dw/30 + 5 cm 3- Required area of End bearing stiffeners

A d F/1 t/ 2 A + 12t 2 Areq = d.F/1 t/cm2 Astiff + 12tw2

Astiff = (Areq - 12tw2)

4- Choose stiff. sections to avoid local buckling such that ft

x 21

5- Calculate the slenderness ratio of the proposed column as follows calculate moment of inertia and cross sectional area Ix & A & Ix hlb 80

ystiff ft

6- Calculate the permissible buckling stress Fpb

7 Check that F = d F / A F

AIxi

ih

il wb 8.0

7- Check that Fact = d.F / A Fpb 8- Design the weld between the stiffeners and the web plate such that the weld

in upper and lower thirds can resist the whole design force

•Intermediate Vertical Stiffeners

Design Steps: 1- Define the design force d F = act

yd Q

FQ

135.0

65.0 1 Define the design force d.F =

2- Choosing single stiffeners x dw/30 + 10 cm 3- Required area of Intermediate vertical stiffeners

actb

d Qq

Q

q Areq = d.F/1 t/cm2 Astiff + 25tw2

Astiff = (Areq - 25tw2) 4- Choose stiff. sections to avoid local buckling such that

ftx 21

5- Calculate the slenderness ratio of the proposed column as follows calculate moment of inertia and cross sectional area Ix & A &

ystiff ft

I hl 80

6- Calculate the permissible buckling stress Fpb

AIxi i

hil wb 8.0

p 7- Check that Fact = d.F / A Fpb 8- Design the weld between the stiffeners and the web plate such that the

weld in upper and lower thirds can resist the whole design force

•Horizontal Stiffeners

The Egyptian code of practice requires that the hori ontal stiffeners satisfies the follo inghorizontal stiffeners satisfies the following stiffness:F hl Stiff At d/5 t f i ti >4d(t )3 For hl. Stiff. At d/5, moment of inertia >4d(tw)3

For hl. Stiff. At d/2, moment of inertia>d(tw)3

Local buckling should be checked to satisfy the relation of x 21

ystiff ft

4-Curtailment of Flange Plates

Why curtailment is needed? Bending moment varies along bridge span, therefore it may be

i t h th l t i d ti t t ieconomic to change the plate girder section to get maximum utilization of steel strength.Ho co ld e specif the c rtailment location? How could we specify the curtailment location?

based on developed formulas by Johnson et al [1] to find the minimum volume of steel that will yield minimum steel weighty g

1. For Parabolic variation of bending moment for simple or continuous girder

6.1w

bf

AFd

MA f2Af1A

6.max

2w

bf

AFd

MA

Lx

2

..3223

12

LAxLxxLLM

xAxLAVol ff

M Mmax

6.2

.. 2

max LAL

xLxxLLFd

MVol w

b

40.)( 22

LxLLxxvol

5A and

3330

f1

xLxxx

9Af2

2. For Parabolic variation of bending gmoment for cantilever girder

322max

wLAxxLLMvol

30)(6.

22

2

b

LxLxvolLFd

vol

13

30)(

1

fA

xLxx

31

2

1 f

f

A

5- Connection between Flange and web plates

Connection is either by using bolts or weldsm

1.5m

m

Procedure ECP recommends continuous fillet weld between flange

and web plates in case of dynamic loading

The size of welding is given by

sfQYq 2 Where Q = maximum shearing force at support

sfI

Qq weld 2.

g pp Y = first moment of area of flange plate about centroidal axis of the

sectionI M t f i ti f l t i d I = Moment of inertia of plate girder

s = size of welding f ld = allowable stresses in fillet weld = 0 2Fu fweld allowable stresses in fillet weld 0.2Fu Fu = ultimate strength of the base metal.

Minimum sizes of weld

tmax S mmtmaxmm

S mm

10 4 10-20 5 20 3020-30 630-50 8 50-100 10

Fatigue strength of the fillet weld under the effect of shear flow due to live load plus impact should be checked, the connection is detailed as class D (Group 3 Fasteners)

Example Given: Qt=125 25t& Ql l=57 15t I=4 267 213cm4 & Given: Qt 125.25t& Ql.l 57.15t , I 4,267,213cm4 &

M.G.2Pl#600X30/1900X16mm, St. 37 Design of the connection between the flange plate and the web plate.

Maximum shear flow is at sec 3 where Q=125 25t Maximum shear flow is at sec. 3 where Q 125.25t. Y=first moment of area of flange plate about N.A. 3370,17)955.1(360 cmxxY

xYQ 3701725125 cmtxIYQq /51.0

213,267,4370,1725.125.

Assuming direct contact between the web plate and the flangecmxFSxq 12.02

therefore, mmcm

xxS

cmxFSxq u

635.06.32.02

51.012.02

based on the maximum thickness of

flange plate = 30mm.flange plate 30mm. Checking the fatigue resistance where 60% of live load is only

considered i.e. 0.6x57.15=34.29t. The shear flow due to this live load cmtxq /140370,1729.34 The shear flow due to this live load cmtq Ill /14.0

213,267,4

The allowable stress range based on No. of stress cycles of 2E6and Detail D Group 3, is 0.71t/cm2.p

safeO.k 1.014.071.02

2cmSSx

qSxF Illsr

6- Splices why a plate girder may be spliced?y p g y p Un-sufficient plates lengths (Web plates ≈ 6m long,

Flange plates ≈12-18m).Th d i d i li d j i id The designer may desire to use spliced joints to aid in cambering.

Change the girder cross section to fit the actual Change the girder cross section to fit the actual bending moment.

Transportation of full length plate girders plays an p g p g p yimportant role in locating spliced joints. Highway road system conditions, maximum limit of legal loads over existed bridges and maximum craneloads over existed bridges and maximum crane capacities limits the maximum length and weight of plate girders to be transmitted in one time.

Shop Splicing Generally, flange and web plates are spliced using single or

double V-joints. J and U joints require the least amount of weld of metal but J and U joints require the least amount of weld of metal but

require the plates to be prepared by planning or milling which is impractical in most structural fabricating shops.This limits the preparation to flame beveling giving a V joint This limits the preparation to flame beveling giving a V-joint. Single V-joint may be acceptable if the plate thickness is up to 25mm.

For thicker plates double V-joints are preferred since they require less weld metal.

It should be remembered that a single V-joint will produce more g j pangular distortion which is increased rapidly as the flange thickness increases.

Design forces Splices in webs of plate girders must be

designed to resist the moments and shearingdesigned to resist the moments and shearing forces at that section. The principle stresses in welds are determined from:in welds are determined from:

The greatest bending moment at the splice d th di h i fand the corresponding shearing force.

The greatest shearing force at the splice and the corresponding bending moment.

qf

Fig. (17) Staggering compression and tension flanges butt weld The stresses in butt welds:

F th b weldGoodfor 35.0 yF

AQq

For the web weldExcellent for 0.385F y

ywA

For the flange weldGoodfor 0.4F0.7x0.58F i.e. 7.0 yy ptFI

Myf For the flange

weldExcellent for 0.58F yI

Field Splicing Arrangement: Generally each girder of a bridge is

assembled of two or three parts at the bridge location citelocation cite.

Methods: The girder assembly is carried out either by welding or by bolting.

Design forces: The ECP [2] cl. 7.5 recommends that, splices should be designed on the maximum bending resistance of the girder section and thebending resistance of the girder section and the actual shearing force at the splice location.

Execution: Generally welded field splices requires y p qtesting either by ultrasonic, X-ray or any other testing technique which increases the cost.

Welded field splicesthe German window splices as shown in Fig. (18) Which is assembled as

follows: (2)

(4)

(2)

(3)

(1)

(3)

(1)

(2)

(3)

(1)

(3)

(1)

German Window Fig. (18) The German window for field splice

A short portion of web is omitted and the adjacent parts of the web areshop welded to the flanges (see ends of weld 1).

The flanges welded first where they are not restrained by the web.Moreover, the roots of these welds are fully accessible.

Th i d i th i i i f th b i th i t d It i The window i.e. the missing piece of the web is then inserted. It isslightly curved so that no shrinkage stresses will arise when the piece iswelded to the remainder of the web.

The remainder welds between the web and the flanges are deposited The remainder welds between the web and the flanges are depositedlast.

Bolted field spliceBolted field splice

Q Q QQQ Q

S2S2

QQ

S2 S2

Detail of Field Splice in bolted Main GirderFlange Pl. Spl. : Direct Method

m mmmmm

in welded plate girderDetail of Field Splice

g pFlange < Spl. : Two Cuts in welded plate girder

QQ tsp2

S2 S2

m m m m mm

Flange Pl Spl : Direct Method

Detail of Field Splice in bolted Main Girderm m m m mm

Flange Pl. Spl. : Direct MethodFlange < Spl. : Two Cuts

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