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# Harshal

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### Transcript

• 1.
• DESIGN OF WELDED PLATE
• Girder
• ☺ Submitted by ☺
• (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
• 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