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# Design of Main Girder [Compatibility Mode].pdf

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### Design of Main Girder [Compatibility Mode].pdf

1. 1. Design of Bridge g g Main Girders According to ECP According to ECP Assis. Prof. Dr. Ehab B. Matar
2. 2. Design Phases of Main Girders 1. Structural analysis for the main system determining the max. and min. straining actions at critical sections 2. Design of Web Plate g 3. Design of Flange Plate 4. Design of Stiffeners (end bearing and intermediate) 5 Design of Connections between web plate and flange plate 5. Design of Connections between web plate and flange plate 6. Design of Splices 7. Check fatigue for all details
3. 3. Bridges Main systems
4. 4. 1- Structural Analysis  Structural analysis is carried out using either influence lines, grillage analysis or finite elements depending on complexity of grillage analysis or finite elements depending on complexity of structures and its importance.  Max. and Minimum shear, moments, reactions…. are determined at critical sections at critical sections. Stress range Stress range Stress range
5. 5. 2- Design of Web Plate  This includes the following tasks: 1. Determining Web height 2 Determining Web thickness 2. Determining Web thickness 3. Checking shear Buckling
6. 6. 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 3 3 7 . 5 3 . 5 2 3 or b b F M F Mk d h     d= plate girder depth (cm)  M= maximum bending moment (t.cm) k ratio of height to thickness of eb plate 100 b b  k= ratio of height to thickness of web plate  100  Fb= allowable bending stress ≈ 0.58 Fy
7. 7. 2.2 Web Thickness  web thickness, tw:  tw ≥8mm y w F t d In 830 case any *  y w F t d 190 used are stiffeners sverse when tran *  y F t d 320 used stiff. e transvers and d/5) (at al longitudin When *  y w F t
8. 8. 2 3 Shear Buckling 2.3 Shear Buckling
9. 9.  No need to check shear buckling resistance if :  No need to check shear buckling resistance if :  for un-stiffened web  for stiffened web y w w F t d / 105 /  q w w F k t d / 45 /  y w w F 1 0 f ) / 34 5 ( 4 shear for factor buckling K 2 q   where /d d 0 . 1 for ) (4/ 5.34 1.0 for ) / 34 . 5 ( 4 2 2             where /d d1   where d d1
10. 10. Shear Buckling Resistance  If the above mentioned limits are exceeded the shear buckling resistance should be checked as follows: y w F t d Calculate *   q q q K 0.35F q then 8 . 0 For * 57 Calculate y b      y q q q F 9 . 0 35 . 0 ) 0.625 - (1.5 q then 2 . 1 0.8 For * b y      y q q F 35 . 0 * 9 . 0 q then 2 . 1 For * b    
11. 11.  It should be noted that longitudinal stiffener at mid depth would be ff ti f ti bj t d t h th th t more 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 bending support 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 should stresses 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 g stress for flange plates according to the following interaction equation. y act b F q F               36 . 0 8 . 0  The second alternative is to design the girder flanges to y b b q         g g g resist the whole acting bending moment without any participation of web for resisting bending moment without reducing the allowable bending stress.
12. 12. 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. The St. 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 43 2 +171 0 +267 +4.43 3 -305.3 -63.6 -296 -61.65 4 +96 -33.07 +96.69,-98.63 -34.92,+3.36 Solution  Proportioning of plate girder  cm cm F Mk d b 190 186 4 . 1 * 2 100 * 100 * ) 3 . 305 296 ( * 3 2 3 3 3      where k is assumed 100  Design of web plate  t t d 56 1 190 190 /     cm t F t d w y w 56 . 1 122 /      By using vertical stiffeners arranged @2.4m (distance between cross girders), then,  26 . 1 240 1    d  190 d  85 . 7 ) 26 . 1 /( 4 34 . 5 ) / 4 ( 34 . 5 2 2       q k  37 . 81 4 . 2 85 . 7 45 45    y q w F k t d 190  thcikness large a is which 34 . 2 37 . 81 190 cm tw     Assume tw = 16mm and check shear buckling resistance.  2 . 1 8 . 0 15 . 1 85 . 7 4 . 2 57 6 . 1 190 57       q q y w q k F t d   q  2 / 66 . 0 ) 4 . 2 35 . 0 )( 15 . 1 625 . 0 5 . 1 ( ) 35 . 0 )( 625 . 0 5 . 1 ( cm t q x x F q b y q b          safe / 41 . 0 6 . 1 190 25 . 125 2      b w q cm t x A Q q 
13. 13. 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 thickness 3. Checking Bending Stresses
14. 14. 3.1 Flange Area Method MY  Plate Girder    3 2 d d I MY  c tf b           12 2 2 2 3 2 3 2 w f d t d A I c d           6 2 12 2 2 3 2 w f w f A A d d t d A I Welded Plate Girder         6 / w f A A d Y I          w f b A A d M F      6 f
15. 15. A d F M A w b f 6    A d F M A d F M A w w f 8 4 3 * 6 @4 arranged bolts assume BUS riveted or bolted For       d C b t b A where d F d F f f b b % 30 20 2 ; * 8 4 6     Section class Part Stress Profile C tf C tf D b 3.2 Local Buckling limits Section class Part Stress Profile Compact Flange Uniform Comp. Rolled Sec. dw tw dw tw Rolled section Welded section t h t y f F t C / 9 . 16 /  p section g p B.U.S Non-compact section Flange Uniform Comp. Rolled Sec. y f y f F t C / 3 . 15 /  y f F t C / 23 /  section B.U.S y f F t C / 21 / 
16. 16. 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 flange be 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 k precautions in the welding technique are taken.
17. 17. Note that:  When plates with unequal thickness or  Change of flange plate width or thick. q width butt welded together, the thicker or g wider plate should be tapered with a slope 2−4mm 1 4 4 1 not exceeding one to four as shown 2mm R = 6 0 c m R
18. 18. 3.3 Allowable Compressive Stresses The 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 with For 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 connected girders that are rigidly connected to the main girder compression flange, then, Lu = distance between x-girders. For deck railway bridges with For deck railway bridges with ballasted floor or for roadway bridges where the compression flange is supported by continuous reinforced concrete continuous 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 flange lateral force of 2% of the flange force at maximum bending moment, then, Lu = 0.
19. 19. For continuous deck roadway or dec oad ay o railway bridges where the compression flange at interior supports at interior supports is located in the bottom side of the girder away from the g y deck slab, then, Lu = the distance between the centers f i t ti f th of intersection of the lower wind bracing with the compression flange compression flange Lu = if there is no lower wind bracing or transverse bracing g then Lu is taken as the distance from the point of maximum bending moment to bending moment to the point of contra- flexure.
20. 20. •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 a EI Lu 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 2 2 1 3 1 2 3 EI B d EI d     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 I1 shown  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.
21. 21. Continue Example 7.5m 2.5m 1.25 1.0m 1.0m 1.25 2.5m Continue Example Bridge Cross section 24m 24m 1 2 4 3 16.8  Flange area Method  For sec. 3, moment and shear acting together, therefore, checking % 60 62 . 0 66 . 0 41 . 0    b q 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 x b 60 20 20 24m 24m should be carried out as follows          2 / 38 . 1 4 . 2 66 . 0 / 41 . 0 36 . 0 8 . 0 / 36 . 0 8 . 0 cm t F F q q F b y b b       Calculating the flange area at sec.3 as follows:  2 66 . 178 6 6 . 1 190 190 38 1 2 3 . 601 6 cm x E A d M A w f       u y f L cm x F b    6 . 774 4 . 2 60 20 20  u b y f b y f L cm C F d A x x x x x C F d A                   86 . 580 . . 1380 3 . 601 78 . 323 3 . 0 3 . 601 78 . 323 05 . 1 75 . 1 ( 4 . 2 196 60 3 1380 . . 1380 2 6 190 38 . 1 6 x d Fb f  Assuming bf=0.3d=0.3x190=57cm→60cm  Then tf=178.66/60=2.98cm→3cm  y f F t c 21 73 . 9 30 2 / 16 300      Therefore Fbc =1.38t/cm2 & Ft=1.4 t/cm2  / 38 . 1 543 , 43 2 3 . 601 2 cm t E F     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. 2 6 164 E  However, although the flange satisfied the limit of compact section ( y F / 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:  2 114 6 . 1 190 2 438 x E A M A w  Stress range = 2 / 378 . 0 543 , 43 2 6 . 164 cm t E   For ADTT >2500 truck, No. of stress cycles =2E6 and for welded plate girder Class B' is taken therefore the allowable fatigue stress range Fsr=0.85t/cm2 . Therefore, the section is safe.  Checking Actual stresses at section 2      2 3 3 4 2 50 6 1 190 x x    2 114 6 190 4 . 1 6 cm x d F A w b f       Assuming bf=50cm  Then tf=114/50=2.28cm→2.4cm  y f F t c 21 08 . 10 24 2 / 16 250            3 4 2 194 , 32 , 714 , 135 , 3 2 . 1 95 4 . 2 50 12 4 . 2 50 2 12 6 . 1 190 cm Z cm I x x x x x I 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/cm2 y f  Checking Actual stresses at section 3       3 4 2 3 3 543 , 43 , 213 , 267 , 4 5 . 1 95 3 60 12 3 60 2 12 6 . 1 190 cm Z cm I x x x x x I gross gross               Therefore Fbc 1.4 t/cm & Ft 1.4 t/cm  / 36 . 1 194 , 32 2 438 2 cm t E F     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,
22. 22. 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 has bridge 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 f material 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 while Mmax=1060mt, Qmax=160t, while for XG Pl#1000x20/2Pl#300*40mm
23. 23. Solution  For the main girder, tw=20mm, dw=3000mm  Af=M/(dFb)-Aw/6 Af 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+30x4x5 2^2)=815947cm4 2 2)=815947cm  d1= (302.4-100-11.94)=190.46cm  d2= 190.46+100/2+4)=244.46cm  B=930cm
24. 24.  Upon calculating the I1, the section is composed of part of composed of part of the web plate (25 twxtw), vertical stiffener assumed 250x20 mm assumed 250x20 mm (satisfying local buckling) and the b k t i t b bracket 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. 9 9cm  I1=25x2x30.9^2+40^3x 2/12+40x2*9.91^2+50 EI B d EI d   2 3 2 2 2 1 3 1  x2x11^2+2x25^3/12+2 x25x24.6^2=111226c m4 ok m a cm e a EI L cm e y u            25 . 2 379 2 6 . 2 * 225 * 43233 * 2100 * 5 . 2 5 . 2 2 6 . 2 815947 * 2100 * 2 930 * 2 ^ 46 . 244 111226 * 2100 * 3 3 ^ 46 . 190 4 4  
25. 25. 3- Design of Stiffeners  This includes the following tasks: D i f ti l tiff (Wh th d b i 1. Design of vertical stiffeners (Whether end bearing or longitudinal stiffeners) 2. Design of horizontal stiffeners g
26. 26. Importance of stiffeners  Reducing slenderness ratio of web plate  Increase the shear capacity of web plate p 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
27. 27. A- Vertical Stiffeners Compression Flange x Tension Flange The outstanding length of vertical stiffeners 1- stiffeners in pairs x  dw/30 + 5 cm 2- single stiffeners x dw/30 + 10 cm
28. 28.  Vertical intermediate stiffeners composed of single or i f t i l ti hil E d b i tiff pair 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 of considered 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). 6tw 6tw 25tw End Bearing Stiffener Stiffener Intermediate
29. 29. • 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 + 12tw 2  Astiff = (Areq - 12tw 2)  4- Choose stiff. sections to avoid local buckling such that f t x 21   5- Calculate the slenderness ratio of the proposed column as follows  calculate moment of inertia and cross sectional area Ix & A &  Ix h lb 8 0 y stiff f t  6- Calculate the permissible buckling stress Fpb  7 Check that F = d F / A  F A Ix i  i h i l w b 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
30. 30. •Intermediate Vertical Stiffeners  Design Steps:  1- Define the design force d F = act y d Q F Q         1 35 . 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 act b d Q q Q     q  Areq = d.F/1 t/cm2  Astiff + 25tw2  Astiff = (Areq - 25tw2)  4- Choose stiff. sections to avoid local buckling such that f t x 21   5- Calculate the slenderness ratio of the proposed column as follows  calculate moment of inertia and cross sectional area Ix & A &  y stiff f t I h l 8 0   6- Calculate the permissible buckling stress Fpb A Ix i  i h i l w b 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
31. 31. •Horizontal Stiffeners  The Egyptian code of practice requires that the hori ontal stiffeners satisfies the follo ing horizontal 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  y stiff f t 
32. 32. 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 i economic 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 weight y g
33. 33. 1. For Parabolic variation of bending moment for simple or continuous girder 6 . 1 w b f A F d M A   f2 A f1 A 6 . max 2 w b f A F d M A   L x   2 . . 3 2 2 3 1 2 L A x Lx x L L M x A x L A Vol f f           M Mmax 6 . 2 . . 2 max L A L x Lx x L L F d M Vol w b            4 0 .) ( 2 2        L x L Lx x vol 5 A and 3 3 3 0 f1        x Lx x x 9 Af2 
34. 34. 2. For Parabolic variation of bending g moment for cantilever girder 3 2 2 max           wL A x x L L M vol 3 0 ) ( 6 . 2 2 2          b L x L x vol L F d vol 1 3 3 0 ) ( 1       f A x L x x 3 1 2 1  f f A
35. 35. 5- Connection between Flange and web plates  Connection is either by using bolts or welds m 1.5mm
36. 36. Procedure  ECP recommends continuous fillet weld between flange and web plates in case of dynamic loading  The size of welding is given by s f QY q 2   Where  Q = maximum shearing force at support s f I Q q weld 2 .   g pp  Y = first moment of area of flange plate about centroidal axis of the section I 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.
37. 37. Minimum sizes of weld tmax S mm tmax mm S mm 10 4 10-20 5 20 30 20-30 6 30-50 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)
38. 38. 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.  3 370 , 17 ) 95 5 . 1 ( 3 60 cm x x Y    x Y Q 370 17 25 125  cm t x I Y Q q / 51 . 0 213 , 267 , 4 370 , 17 25 . 125 .     Assuming direct contact between the web plate and the flange cm x F Sx q 1 2 . 0 2  therefore, mm cm x x S cm x F Sx q u 6 35 . 0 6 . 3 2 . 0 2 51 . 0 1 2 . 0 2     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 cm t x q / 14 0 370 , 17 29 . 34  The shear flow due to this live load cm t q I ll / 14 . 0 213 , 267 , 4    The allowable stress range based on No. of stress cycles of 2E6 and Detail D Group 3, is 0.71t/cm2 . p  safe O.k 1 . 0 14 . 0 71 . 0 2 2 cm S Sx q SxF I ll sr       
39. 39. 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 y important role in locating spliced joints. Highway road system conditions, maximum limit of legal loads over existed bridges and maximum crane loads over existed bridges and maximum crane capacities limits the maximum length and weight of plate girders to be transmitted in one time.
40. 40. 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 p angular distortion which is increased rapidly as the flange thickness increases.
41. 41. Design forces  Splices in webs of plate girders must be designed to resist the moments and shearing designed 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 f and the corresponding shearing force.  The greatest shearing force at the splice and the corresponding bending moment.
42. 42. q f Fig. (17) Staggering compression and tension flanges butt weld The stresses in butt welds: F th b weld Good for 35 . 0   y F A Q q  For the web weld Excellent for 0.385Fy  y w A  For the flange weld Good for 0.4F 0.7x0.58F i.e. 7 . 0 y y     pt F I My f  For the flange weld Excellent for 0.58Fy  I
43. 43. Field Splicing  Arrangement: Generally each girder of a bridge is assembled of two or three parts at the bridge location cite location 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 the bending resistance of the girder section and the actual shearing force at the splice location.  Execution: Generally welded field splices requires y p q testing either by ultrasonic, X-ray or any other testing technique which increases the cost.
44. 44. Welded field splices the 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 are shop 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 is slightly curved so that no shrinkage stresses will arise when the piece is welded 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 deposited last.
45. 45. Bolted field splice Bolted field splice Q Q Q Q Q Q S2 S2 Q Q S2 S2 Detail of Field Splice in bolted Main Girder Flange Pl. Spl. : Direct Method m m m m m m in welded plate girder Detail of Field Splice g p Flange < Spl. : Two Cuts in welded plate girder
46. 46. Q Q tsp2 S2 S2 m m m m m m Flange Pl Spl : Direct Method Detail of Field Splice in bolted Main Girder m m m m m m Flange Pl. Spl. : Direct Method Flange < Spl. : Two Cuts