DESIGN OF WELDED PLATE Girder

☺ Submitted by ☺

HARSHAL M. WARADE

(4C=23)

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.

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

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/ mm2

Assume Thickness of Web Plate = 12 mm

Economical effective depth of girder

= 1.1 SQRT BM / 6bt *tw

= 1.1 SQRT 25637.5 *106 / 165*12

= 3958 mm

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 * 103 / 12 * 3500

= 75.45 ≤ 108 N/mm2

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*106 / 165*3675)*(12*3500/8)

=37029.34 mm2

Provide more than this

Economical Depth (d) = 5.5 ( m/6bt)⅓

= 5.5 (25637.5*106 /165) ⅓

= 3000mm.

Area of Web required = Max.SF/ζav

= 3085* 103 /108

= 28565 mm2

Thickness of Web = 28565/3500

= 8.16 = 10 mm < 12 mm …………..OK

When cover plate is Absent

Af = m/6bt*d

= 25637.5*106 / 165*3500

= 44394 mm2

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 mm2

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 = bd3/12 + 2(bd3/12 + Ah2)

Ixx = 12*35003/12 + 2*(750*603/12 +(750*60)*17802

Ixx = 3.270*1011

F = 25637.5 *106 / 3.270 *1011 *1810

= 141.56 N/mm2 < 165 N/mm2

CURTAILMENT OF FLANGE PLATE

Using 50 mm plate section

Max. flange outstand = 369 mm

Permissible outstand = 12*50=600 mm

Ixx = bd3/12 + 2(bd3/12 + Ah2)

Ixx = 12*35003/12 + 2*(750*503/12 +(750*50)*17752

Ixx =2.791*1011 mm4

MR = 6bd * I / y

= 165 * 2.791*1011 /1800

= 2.558*10 10 N-mm

= 2.558*104 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*106= k*15(30-15)

k = 113.94

y = kx(l-x)

y = 118.94 x (30-x)

2.516*104 = 3418.33x-113.94x2

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 = bd3/12 + 2(bd3/12 + Ah2)

Ixx = 12*35003/12 + 2*(750*403/12 +(750*40)*17702

Ixx =2.30*1011 mm4

MR = 6bd * I / y

= 165 * 2.30*1011 /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*106= k*15(30-15)

k = 113.94

y = kx(l-x)

y = 118.94 x (30-x)

2.12*104 = 3418.33x-113.94x2

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 = γay / Ixx

Max. Shear Force = 3085 KN

AY = 750*40*1770

= 53.72*106

Ixx = 230*1011 mm4

Horizontal Shear /mm = 3085*53.62*106*103/ 2.30*1011

Horizontal Shear /mm = 712.23 N/mm

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

= 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/mm2

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

= 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 ≥ 1.5d3*t3 / c2

I ≥ 1.5*35003*123 / 24002

I = 19.29*106 mm4

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 = (bd3/12 + Ah2)

Ixx = ( 12*1503 /12 + (12*150) * (150/2) 2 )

Ixx = 23.62*106 > 19.29*106 mm4

Provide size of vertical stiffeners 150*12 mm

DESIGN OF WELDED CONNECTION BETWEEN WEB PLATE AND STIFFENERS

Shear Force = 125 h2 / h

= 125 *122 / 150

= 120 KN

Provide 5 mm size of intermittent plate

Strength of Weld per mm = 0.7*5*1* ζav

= 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

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 = 4ct3

= 4*2400*123

= 16.58 * 106 mm4

Using flat section max outstand 12t = 12*12 = 144 mm

Using 140 mm outstand width of plate.

Ixx = (bd3/12 + Ah2)

Ixx = ( 12*1403 /12 + (12*140) * (140/2) 2 )

= 10.976*106 mm4

Using 14 mm thickness of plate and 140 mm width of plate

Ixx = ( 14*1403 /12 + (14*140) * (140/2) 2 )

Ixx = 12.805*106 mm4

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. = dt3 = 3500*123 = 6.04*106 mm4

Using 12 mm thickness of plate and 140 mm width of plate

DESIGN OF BEARING STIFFENERS

End Bearing Stiffeners

Design load = 3085 KN

Permissible bending stress = 0.75 Fy

= 0.75 * 250 = 187.5 N / mm2

Bending area required = load / 6p

= 3085*103 / 187.5 = 16453 N /mm2

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

= 17600 > 16453 mm2 …………………..OK

Area of Stiffeners = 4*220 *20 + 4*(20*12) * 12

= 29120 mm2

Ixx = 4*(bd3/12 + Ah2)

Ixx = 4*( 20*2203 /12 + (20*220) * (110+12/2) 2)

Ixx = 307.812*106 mm4

Rmin = SQRT Ixx / A

= SQRT 307.812*106 / 29120

Rmin = 102.81 mm

λ = leff. / Rmin.

= 0.65*3500 / 102.81

= 22.13

6cbc = 147.36 N/mm2

as per IS 800 Page no – 39 Table no – 5.1

Load carrying capacity = stress * area

= 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 t2 / h

= 125 * 122 /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

= 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 / mm2

Bending area required = load / 6p

= 2000*103 / 187.5 = 10666 N /mm2

Provide 2 plate of 200 mm wide

Thickness of plate = Area / 4*width

= 10666 / 2*200 = 26.67 mm = 30 mm

Actual Outstand = 220 mm

Max. outstand = 12*20 > 200 mm

= 240 > 200 mm

Hence Bearing Area provided = 2*200*30 mm size

=12000 > 10666 mm2 …..OK

Area of Stiffeners = 2*200 *20 + 2*(20*12) * 12

= 13760 mm2

Ixx = 2*(bd3 /12 + Ah2)

Ixx = 2*( 20*2003 /12 + (20*200) * (100+12/2) 2 )

Ixx = 116.55*106 mm4

Rmin = SQRT Ixx / A

= SQRT 116.55*106 / 13760

Rmin = 92.03 mm

λ = leff. / Rmin.

= 0.65*3500 / 92.03

= 24.72

6cbc = 146.64 N/mm2

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 t2 / h

= 125 * 122 /220

=81.81 KN/m

Provide 10 mm size intermittent fillet 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

= 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