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  • 1.
    • DESIGN OF WELDED PLATE
    • Girder
    • ☺ Submitted by ☺
    • HARSHAL M. WARADE
    • (4C=23)
  • 2. DESIGN A WELDED PLATE GIRDER FOR A SIMPLY SUPPORTED GIRDER OF SPAN 30 M. SUBJECTED TO THE SUPER IMPOSED LOAD EXCLUDING SELF WEIGHT OF GIRDER 120 KN/M. CARRYING TWO POINT LOAD 1000 KN AT 10 M FROM EACH SUPPORT. DATA= A Welded Plate Girder Span of the Girder- 30 M. UDL- 120 KN/M. Points loads- 1000 KN.
  • 3. SOLUTION = Self Weight Of The Girder (w)- W / 300 Total Load On The Beam = 1000+1000+120*30 (w) = 5600 KN w = 5600 / 3000 Self Weight of the Girder = 18.66 KN/M = 19 KN/M CALCULATE REACTION RA + RB = 1000+1000+139*30 RA + RB = 6170 …………………………….(1) Take moment about A - RB*30+1000*20+1000*10+139*30*15 =0 - RB*30 = -92550 RB = 3085 KN Put in a equation (1) RA = 3085 KN
  • 4.
    • CALCULATE MAX. BENDING MOMENT
    • MAX. BM= 3085*15-1000*5-139*15*15/2
    • = 25637.5 KN-M.
    • DESIGN OF WEB PLATE
    • Permissible Avg. shear stress = 108 mpa
    • Permissible Bending Stress = 6cbc = 6bt
    • = 0.6 fy
    • = 165 N/ mm 2
    • Assume Thickness of Web Plate = 12 mm
    • Economical effective depth of girder
    • = 1.1 SQRT BM / 6bt *tw
    • = 1.1 SQRT 25637.5 * 10 6 / 165*12
    • = 3958 mm
  • 5. Average shear stress should be less than permissible Bending Stress hence provide thickness of web 12 mm and and Depth of Web 3500 mm
    • dw = 3958 – 10/100 *3958
    • = 3562 mm
    • Provide dw = 3500 mm
    • Avg shear stress = V / tw * dw
    • = 3085 * 10 3 / 12 * 3500
    • = 75.45 ≤ 108 N/mm 2
  • 6.
    • DESIGN OF FLANGE PLATE
    • Assuming Depth of girder to be 5 % greater than depth of web
    • Depth of girder = 3500 + 5/100 *3500
    • = 3675 mm.
    • Economical depth = k * (m/6bt)⅓
    • When cover plate is provided net area of flange required
    • = m/6bt – Aw/8
    • = (25637.5* 10 6 / 165*3675)*(12*3500/8)
    • =37029.34 mm 2
    • Provide more than this
    • Economical Depth (d) = 5.5 ( m/6bt)⅓
    • = 5.5 (25637.5*10 6 /165) ⅓
    • = 3000mm.
  • 7.
    • Area of Web required = Max.SF/ ζ av
    • = 3085* 10 3 /108
    • = 28565 mm 2
    • Thickness of Web = 28565/3500
    • = 8.16 = 10 mm < 12 mm …………..OK
    • When cover plate is Absent
    • Af = m/6bt*d
    • = 25637.5*10 6 / 165*3500
    • = 44394 mm 2
    • Let us provide 60 mm thickness of flange plate
    • Width of flange plate = Af / t
    • = 44394 / 60
    • = 739.9 mm = 750 mm.
    • Provided area of the flange = 750*60
    • = 45000 mm 2
  • 8.
    • Flange outstand = 375-6
    • =369 mm
    • Permissible flange outstand = 12*t
    • = 12*60
    • =720 > 369 mm ……………OK
    • CHECK FOR BENDING STRESS
    • 6bt = 0.66 fy
    • = 165 mpa
    • F = M/I*Y
    • I = Ixx = MI of plate girder about NA
    • Ixx = bd 3 /12 + 2(bd 3 /12 + Ah 2 )
    • Ixx = 12*3500 3 /12 + 2*(750*60 3 /12 +(750*60)*1780 2
    • Ixx = 3.270*10 11
    • F = 25637.5 *10 6 / 3.270 *10 11 *1810
    • = 141.56 N/mm2 < 165 N/mm 2
  • 9.
    • CURTAILMENT OF FLANGE PLATE
    • Using 50 mm plate section
    • Max. flange outstand = 369 mm
    • Permissible outstand = 12*50= 600 mm
    • Ixx = bd 3 /12 + 2(bd 3 /12 + Ah 2 )
    • Ixx = 12*3500 3 /12 + 2*(750*50 3 /12 +(750*50)*1775 2
    • Ixx =2.791*10 11 mm 4
    • MR = 6bd * I / y
    • = 165 * 2.791*10 11 /1800
    • = 2.558*10 10 N-mm
    • = 2.558*10 4 KN-m
    • BM at 0 is parabolic . The equation of parabolic is with a support as a origin.
    • Y = kx (l-x)
    • Y = Max BM
  • 10.
    • k = constant
    • 25637.5*10 6 = k*15(30-15)
    • k = 113.94
    • y = kx(l-x)
    • y = 118.94 x (30-x)
    • 2.516*10 4 = 3418.33x-113.94x 2
    • x = 14.40 m
    • Plate thickness can be reduced 50 mm from 60 mm at length of 14.40 m from support.
    • Provide 750*50 mm plate up to 14.40 m from support.
    • Using 40 mm plate section
    • Max. flange outstand = 369 mm
    • Permissible outstand = 12*40= 480 mm > 369 mm
    • Ixx = bd 3 /12 + 2(bd 3 /12 + Ah 2 )
  • 11.
    • Ixx = 12*3500 3 /12 + 2*(750*40 3 /12 +(750*40)*1770 2
    • Ixx =2.30*10 11 mm 4
    • MR = 6bd * I / y
    • = 165 * 2.30*10 11 /1800
    • = 2.12*10 10 N-mm
    • = 2.12*10 4 KN-m
    • BM at 0 is parabolic . The equation of parabolic is with a support as a origin.
    • Y = kx (l-x)
    • Y = Max BM
    • k = constant
    • 25637.5*10 6 = k*15(30-15)
    • k = 113.94
    • y = kx(l-x)
    • y = 118.94 x (30-x)
  • 12.
    • 2.12*10 4 = 3418.33x-113.94x 2
    • x = 8.476 m
    • Plate thickness can be reduced 40 mm from 50 mm at length of 8.476 m from support.
    • Provide 750*40 mm plate up to 8.476 m from support.
    • CONNECTION
    • Connection between Flange Plate for different thickness.
    • Horizontal Shear = γ ay / Ixx
    • Max. Shear Force = 3085 KN
    • AY = 750*40*1770
    • = 53.72*10 6
    • Ixx = 230*10 11 mm 4
    • Horizontal Shear /mm = 3085*53.62*10 6 *10 3 / 2.30*10 11
    • Horizontal Shear /mm = 712.23 N/mm
  • 13.
    • SIZE OF WELD
    • Welding is done on both side of web permissible average shear stress
    • S = Horizontal shear / 2*0.7*1*108
    • S = 712.23 / 2*0.7*1*108
    • S = 4.71 mm
    • Min size of plate for 40 mm
    • Thickness of plate = 12mm
    • Let us provide an intermittent Fillet weld
    • The effective weld length is 45 or 40 mm
    • Effective weld length = 4*4.78
    • = 18.84 mm
    • Provide 40 mm long intermittent fillet weld SS
    • Therefore
    • Strength of Weld = 2(0.7*12*40*108) / 712.23
  • 14.
    • = 101.89 =100 mm
    • Permissible pitch = 100 mm
    • So provide pitch of 100 mm
    • DESIGN OF WEB STIFFNERS
    • d/tw = 3500 / 12 =291.66
    • Hence provide one vertical stiffeners and two horizontal stiffeners are provided
    • DESIGN OF VERTICAL STIFFENERS
    • Actual average shear stress of web plate =85.69 N/mm 2
    • d / tw = 291.66
    • as per IS 800 Page no – 73 Table no – 6.6A
    • Spacing of vertical stiffeners = 0.7d
    • = 0.7*3500
    • = 2.45 m
  • 15.
    • = 2.4 m
    • Vertical stiffeners are to be provided within 10 m span. Suitable spacing of vertical stiffeners
    • 10 / 2.4 = 4.08
    • Hence provide 4 no. of vertical stiffeners
    • I ≥ 1.5d 3 *t 3 / c 2
    • I ≥ 1.5*3500 3 *12 3 / 2400 2
    • I = 19.29*10 6 mm 4
    • Max outstand in flange section =12t = 144 mm
    • Using 150 mm wide plate and 12 mm thick
    • Ixx = MI of vertical stiffeners about the face of the web
    • Ixx = (bd 3 /12 + Ah 2 )
    • Ixx = ( 12*150 3 /12 + (12*150) * (150/2) 2 )
    • Ixx = 23.62*10 6 > 19.29*10 6 mm 4
    • Provide size of vertical stiffeners 150*12 mm
  • 16.
    • DESIGN OF WELDED CONNECTION BETWEEN WEB PLATE AND STIFFENERS
    • Shear Force = 125 h 2 / h
    • = 125 *12 2 / 150
    • = 120 KN
    • Provide 5 mm size of intermittent plate
    • Strength of Weld per mm = 0.7*5*1* ζ av
    • = 0.7*5*1*108
    • =378 N /mm
    • Length of intermittent plate weld =10*t = 10*12 = 120 mm
    • c/c spacing of weld = 2*378*120 = 756 mm
    • Max spacing of weld = 16 t or 300
    • = 16*12 or 300
    • = 192 or 300 mm
    • = 190 mm
  • 17.
    • Provide 5 mm size 120 mm long plate weld at c/c spacing of 190 mm on both sides of web .
    • DESIGN OF HORIZONTAL STIFFENERS
    • Provide first Horizontal Stiffeners will be provided at 2/5 d from compression flange
    • 2 / 5 * 3500 = 1400 mm
    • Moment of Inertia = 4ct 3
    • = 4*2400*12 3
    • = 16.58 * 10 6 mm 4
    • Using flat section max outstand 12t = 12*12 = 144 mm
    • Using 140 mm outstand width of plate.
    • Ixx = (bd 3 /12 + Ah 2 )
    • Ixx = ( 12*140 3 /12 + (12*140) * (140/2) 2 )
    • = 10.976*10 6 mm 4
  • 18.
    • Using 14 mm thickness of plate and 140 mm width of plate
    • Ixx = ( 14*140 3 /12 + (14*140) * (140/2) 2 )
    • Ixx = 12.805*10 6 mm 4
    • Hence provide size of horizontal stiffeners 140*14 mm
    • CONNECTIONS
    • Let us provide 5 mm size of 140 mm provide second horizontal stiffeners of the NA at the distance d/2.
    • Comp. Flange =3500/2
    • = 1750 mm
    • I reqd. = dt 3 = 3500*12 3 = 6.04*10 6 mm 4
    • Using 12 mm thickness of plate and 140 mm width of plate
  • 19.
    • DESIGN OF BEARING STIFFENERS
    • End Bearing Stiffeners
    • Design load = 3085 KN
    • Permissible bending stress = 0.75 Fy
    • = 0.75 * 250 = 187.5 N / mm 2
    • Bending area required = load / 6p
    • = 3085*10 3 / 187.5 = 16453 N /mm 2
    • Provide 4 plate of 200 mm wide
    • Thickness of plate = Area / 4*width
    • = 16453 / 4*220 = 18 .69 mm
    • Provide 4 – 220 – 20 mm size plate
    • Actual Outstand = 220 mm
    • Max. outstand = 12*20 > 200 mm
    • = 240 > 200 mm ……………….. OK
    • Bearing Area provided = 4*220*20
  • 20.
    • = 17600 > 16453 mm2 …………………..OK
    • Area of Stiffeners = 4*220 *20 + 4*(20*12) * 12
    • = 29120 mm 2
    • Ixx = 4*(bd 3 /12 + Ah 2 )
    • Ixx = 4*( 20*220 3 /12 + (20*220) * (110+12/2) 2 )
    • Ixx = 307.812*10 6 mm 4
    • Rmin = SQRT Ixx / A
    • = SQRT 307.812*10 6 / 29120
    • Rmin = 102.81 mm
    • λ = leff. / Rmin.
    • = 0.65*3500 / 102.81
    • = 22.13
    • 6cbc = 147.36 N/mm 2
    • as per IS 800 Page no – 39 Table no – 5.1
    • Load carrying capacity = stress * area
  • 21.
    • = 147.36 * 29120
    • = 4291.12 > 3085 ……………………..OK
    • Provide 4 plates of 20 mm thick as a end bearing stiffeners
    • CONNECTION BETWEEN WEB PLATE AND STIFFENERS
    • Shear Force = 125 t 2 / h
    • = 125 * 12 2 /220
    • =81.81 KN/m
    • Provide 10 mm size intermittent plate welds.
    • Strength of Weld / mm = 0.7*5*1*108
    • = 378 N /mm
    • Length of intermittent plate weld =10*t = 10*20 = 200 mm
    • c/c spacing of weld = 378*200 / 81.81 = 924.09 mm
    • Max spacing of weld = 16 t or 300
    • = 16*20 or 300
  • 22.
    • = 320 or 300 mm
    • = 300 mm (whichever is lesser)
    • Max. Spacing = 300 mm
    • Let us provide 5 mm size 200 mm long plate weld at c/c spacing of 300 mm on both sides of web
    • Load Bearing Stiffeners
    • Design load = 2000 KN
    • Permissible bending stress = 0.75 Fy
    • = 0.75 * 250 = 187.5 N / mm 2
    • Bending area required = load / 6p
    • = 2000*10 3 / 187.5 = 10666 N /mm 2
    • Provide 2 plate of 200 mm wide
    • Thickness of plate = Area / 4*width
    • = 10666 / 2*200 = 26.67 mm = 30 mm
  • 23.
    • Actual Outstand = 220 mm
    • Max. outstand = 12*20 > 200 mm
    • = 240 > 200 mm
    • Hence Bearing Area provided = 2*200*30 mm size
    • =12000 > 10666 mm 2 …..OK
    • Area of Stiffeners = 2*200 *20 + 2*(20*12) * 12
    • = 13760 mm 2
    • Ixx = 2*(bd 3 /12 + Ah 2 )
    • Ixx = 2*( 20*200 3 /12 + (20*200) * (100+12/2) 2 )
    • Ixx = 116.55*10 6 mm 4
    • Rmin = SQRT Ixx / A
    • = SQRT 116.55*10 6 / 13760
    • Rmin = 92.03 mm
    • λ = leff. / Rmin.
    • = 0.65*3500 / 92.03
  • 24.
    • = 24.72
    • 6cbc = 146.64 N/mm 2
    • as per IS 800 Page no – 39 Table no – 5.1
    • Load carrying capacity = stress * area
    • = 146.66* 13760
    • = 2011.526 > 2000 ……………………..OK
    • Provide 2 plates of 200 mm wide and 20mm thick as a load bearing stiffeners
    • CONNECTION BETWEEN WEB PLATE AND STIFFENERS
    • Shear Force = 125 t 2 / h
    • = 125 * 12 2 /220
    • =81.81 KN/m
    • Provide 10 mm size intermittent fillet welds.
  • 25.
    • Strength of Weld / mm = 0.7*5*1*108
    • = 378 N /mm
    • Length of intermittent plate weld =10*t = 10*20 = 200 mm
    • c/c spacing of weld = 378*200 / 81.81 = 924.09 mm
    • Max spacing of weld = 16 t or 300
    • = 16*20 or 300
    • = 320 or 300 mm
    • = 300 mm (whichever is lesser)
    • Max. Spacing = 300 mm
    • Let us provide 5 mm size 200 mm long plate weld at c/c spacing of 300 mm on both sides of web

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