Harshal

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Harshal

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

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