composite steel girder

7/29/2019 Composite Steel Girder

1/10

200

75

DESIGN CRITERIA :

Length of the Girder = M Ultimate Concrete Stress, f 'C = N/mm2

Girder depth = mm Ultimate Steel Stress, f 'y = N/mm2

EUDL for Moment = kN Allow. stress of Concrete in Compression fC = N/mm2

EUDL for Shear = kN Modulus of Elasticity of Concrete Ec = N/mm2

Coefficent of Dynamic Augument (CDA) = kN Modulus of Elasticity of Steel Es = N/mm2

Design EUDL for Moment / Girder = kN = kN / m fc = N/mm2

Design EUDL for Shear / Girder = kN = kN / m fs = N/mm2

Axle Load of Train ( )t

= kN m =

Distribution width of Load Dispersion = M r =

Design wheel Load per meter width = kN/m k =

Tractive effort = kN j =

Braking force = kN R =

Unit weight of Concrete = Kn/m3

Area of Steel Girder = mm2

Bs = mm Bs = 288 mm

mm

d

s

200

mm

d

s

200

= 25 217

= N A n

ys

COMPOSITE SECTION EQUIVELANT STEEL SECTION

Effective width of the Slab :

i) One - third of the effective Span = mm

ii) Distance between centres of the rib = mm

iii) 12 times the thickness of the slab + breadth of the rib = mm

Area of Steel Girder A S = mm2

Area of composite Section A C = mm2

Moment of inertia of the Steel section IS = mm4

Equivalent moment of inertia of the composite section Ic = mm4

1.0 Self wt. of the Steel Girder/m = = kN/m

2.0 Self wt. of the Slab/m = kN/m

3.0 Rail, Sleeper and Ballast load on top of the Slab = kN/m

4.0 EUDL / Girder for Live load with impact = kN/m

32.5

552.0

24.0

88000

12.0

13.0

14.0

82.812

23.417

75.272

(200L + 1000) N

88000

6666.7

2300

2410

16.67

0.32

0.89

75.272

2300

24000.00

11.25

1.45

200000

9.0

318.5

3.0

25.00

415.0

10.00

166.7

8

6.0

4.0

5.0

2272.4

0.458

DESIGN OF COMPOSITE STEEL GIRDER BRIDGE(WELDED)

16761000

CROSS SECTION OF BRIDGE

1050

200

250

6500

INPUT DATA :

1.0 20.0

2.0

3.0

2500

7.0

5.000

21.840

2065.5

1505.4

1656.2

C.G of Steel Section

145500

11.0 735.0

10.0 53.083

8.0

1.1504E+11

1.8735E+11

971

90

1540

150

21002100 2300


2/10

Fatigue considerration:

Total DL Shear per Girder = KN

Total DL + LL Shear per Girder = KN

Ratio of Minimum Shear & Maximum Shear =

Considering Cycle K =

a) Bending Moment due to Self wt. of the Steel section MS = kN -m

b) Bending Moment due to Dead load of Concrete MC = kN -m

c) Bending Moment due to Rail, Sleeper, Ballast MSU = kN -m

d) Bending Moment due to EUDL (Applied load) MA = kN -m

Over all depth of the Composite Section, D = mm

Section modulus of Steel Section at bottom Z b = mm3

Section modulus of Composite Section at bottom Z bc = mm3

Section modulus of Composite Section at bottom Z tc = mm3

Maxi. Stress in Steel at bottom MS MC + MSU + MA

Z b Z bc

= N / mm2 < N / mm

2

Maxi. Stress in Concrete at top MC + MSU + MA

Z bc

= N / mm2 < N / mm

2

Vartical Shear at support F = kN

Design of Shear Connector :-

The horizontal Shearat the top of the Steel section Fh = F * Bs*ds*(n - ds/2)

m

IC

= N/mm

Or = N

Horizontal force resisted by the compressive area of the concrete slab Fh = o.85 * fc' *Ac

2

= N

Design horizontal force Fh = N

Let us considered 150 mm Long 22 MS Rod as Shear connector

H = 150 mm D = 22

H

D

Allowable Shear taken by one Shear connector q = 1.518 * H * d fCK

= N

Total nos. of studs n = Nos

Let 'S' be the size of the welding all arround the bar with the top flange

We know shear strength P = KLS va

= S

S = 3.0 mmPitch of the Shear connector,

Vsr =

Where,

n is the number of Shear connector in Cross - section = 4

Z r Shear fatigue resistance of an individual Shear connector =

V f Vertical Shear force under Fatigue load combination = kN

Q Area moment about N . A =

I C Moment of inertia of composite section = mm4

d Dia of Shear connector = 22

N Number of Cycle =Vsr = N

Z r = * d 2 = 238 -

= =

Pitch of the Shear connector = 132 mm

Minimum Pitch of Shear connector = = 88 mm

Longitudinal Spacing = L 1

2 n/4

Maximum Pitch of Shear connector = 2.5 d s = mm

Shear(mini)

Shear(maxi)

V f* Q

I

c t

Total

500

29.5 log(N)

25225.8937

764.93

38 * d2

22861.7

600

50078350.5 mm3

1.8735E+11

2000000

52.12 19.00

OK

42.832 165.00

4.1978

OK

11.25

1430.85

268.82

1174.2

502.57

2158.8

0.2328

382.47

382.47 20900

0.0183

= 6.8 > 4

n * Z r

Vsr

1E+08

2E+08

2E+08

=

+=bc

4887500.00

4887500.00

14025

2,000,000 0.93

1259.0

4046.9

2700

OK

349

x = 115 115 mm

* d2

4 * d

109958.79


3/10

DESIGN OF RCC DECK SLAB :kN kN

6.24 kN 312 312 kN

kN/m

100 4.8 kN/m 100

A C B

R1 = kN R2 = kN

1.0 Self wt. of deck slab = kN/m

2.0 Self wt. of Ballast, Sleeper & Rail = kN/m

3.0 Load per wheel = kN/m

4.0 Parapet Load = kN/m

Maxi. (-) Moment at A = kN - m

Maxi. (+) Moment at C = kN - m

DESIGN FOR CANTILIVER PART :

Maxi. (-) Moment at A = Kn-M (Top Tension)

Minimum flaxural strength Mcr = 1.2 x Mcr

Mcr = 19.70 x f 'c x Z b

Section modulus, Z b = m3

Mcr = Kn-M

Minimum flaxural strength = Kn-M

Design Moment, MD = Kn-M

d reqd. = mm

d avil. = mm OK

A s = mm2

/ m

Use Y 16 @ 177 mm c/c

DESIGN FOR MID SECTION :

Maxi. (+) Moment at C = Kn-M (Bottom Tension)

Minimum flaxural strength Mcr = 1.2 x Mcr

Mcr = 19.70 x f 'c x Z b

Section modulus, Z b = m

Mcr = Kn-M

Minimum flaxural strength = Kn-M

Design Moment, MD = Kn-M

d reqd. = mm

d avil. = mm OK

A s = mm2

/ m

Use Y 16 @ 199 mm c/c

DISTRIBUTION REINF :

Distribution reinforcement is 67 % of the main reinforcement

A dist. = mm

Use Y 12 @ mm c/c167

53.083

6.24

46.833

138.93138.93

46.833

4.80

53.08

2300

1676

21002100

6.24

53.08

350 350

mm

21.292

675.8

21.29

121.33

142.00

1008.7

21.29

0.0017

5.1914

6.2297

mm

25.93

133.9

142.00

1228.5

-25.93

25.93

0.0017

5.1914

6.2297


4/10

DESIGN OF PARAPET WALL :

200

Moment at base M = Kn-Md reqd. = mm

d avil. = mm OK

A s = mm2

/ m

Use Y 12 mm @ mm c/c

MINIMUM REINF. REQUIRMENT :

A st =

Use Y 12 mm 214 mm c/c ( Both Way )

12 214 mm c/c

12 167 mm c/c

16 177 mm c/c 12 214 mm c/c

12 214 mm c/c

16 199 mm c/c

45

`

45

375 M 375

STRUCTUREAL DESIGN OF STEEL I GIRDER.

1.0 Length of the Girder = M Yield stress of steel fy = N/mm2 (Mpa)

2.0 Girder depth = mm Allowable shear stress, va = N/mm2

3.0 EUDL for Moment = kN Allowable bending stress, bc = N/mm2

4.0 EUDL for Shear = kN Permissible bearing stress, p = N/mm2

5.0 Coefficent of Dynamic Augument (CDA) = kN Allowable stress for tension, t = N/mm2

6.0 Design EUDL for Moment / Girder = kN Allowable stress for compre. C = N/mm2

7.0 Design EUDL for Shear / Girder = kN Shear stress in the weld vf N/mm2

8.0 Tractive effort = kN

9.0 Braking force = kN Weight of Rail = kg/m

10.0 Unit wt. of Timber = Weight of Rail = kn/m

Length of Sleeper = mm

DESIGN OF STEEL GIRDER :

REINF. DETAILS OF DECK SLAB:

@

@

1.0

50

@

@

@

kN/m

0.458

20.00

LONG SECTION OF GIRDER

20.0

2500

DESIGN CRITERIA :

735.0

552.0

150.00

kN/m3

60.00

70.00

108.00

0.589

2750

12.0

2500

1505.4

1656.2

100.002065.5

2272.4

250.00

165.00

187.50

1.47

@

1568

500 mm2

1.543532.668

144.00

72.11


5/10

mm Width of Sleeper = mm

Thickness of Sleeper = mm

t f= 45mm Spacing of Sleeper = mm c/c

b No.of Sleeper on top Slab = nos.

Weight of Rail, Sleeper over = kNa girderLoad per meter over a girder = kN/m

tw = 10 mm=

d1

t f= 45mm

Self wt. of Plate Girder = kN/m

Self wt. of Plate Girder/m = kN/m

Total uniform load, w ' = kN/m

Maxi. Shear force V = kN

Maxi. Bending moment, M = kN - m

Let the thickness of web, tw = 10 mm ( directly exposed to weather and unpainted )

Economical depth of Plate Girder = d1

d1 = mm mm

Selected web Plate size = x 10 mm

Minimum web thickness of web from Shear considaration t w (mini) = < tw OK

Approximate flange area required A f = mm

The flange outstand should not be greater than 12 x t f

b =12 x t f

Area of flange = ( 2b + t w ) x t f

= ( 2 x 12 x t f+8 ) x t f

24 t f2

+ 8 t f =

24 t f2

+ 8 t f - = 0

t f = mm 40 mm

Width of the flange Plate = mm

Calculated area of flange = mm2

< mm2

NOT OK

Increase flange thickness, Let t f = 45 mm

Width of flange = mm

Area of flange = mm2

> mm2

OK

Hance provide a flange Plate of size = x 45 mm

CHECK BY MOMENT INERTIA METHOD :

I x x = t w d3

12

= mm4

M y

I

CONNECTION BY WELDING :

Welds are designed for horizontal Shear

Horizontal shear per unit length, =

Vertical Shear force, V = kN

Area moment of the flange about N - A, A y = mm

I x x = 2 * A f * (d/2) +

A w d2

6 2Horizontal Shear / mm = N / mm

SIZE OF WELD :

Welding is done on both side of the web

The strength of well is given by P = KLS va

Where,

S = Size of the weld in mm

K = Constant = 0.7

L = Effective length of the weld in mm

vf = Shear stress in the weld in N / mm2

= 108 N / mm2

= 2 x ( 0.7 x S x 1 x 108 )

Maxi. bending stress bc

719.6

( A f ) * = 1.115E+11+

V * A y

OKN / mm2

N / mm2 < 165.0

(d/2 + t f)2

40083750

mm4

31365

31365

700

12

2000.9

t w * d3

I x x

DESIGN OF FLANGE :

31365

31365

31365

1.1504E+11

697 700

31500

+2 *+w f* t f

3

12

(w f* t f)*

784.13 600

3080.3

2500

103.28

35.984

2500

250

3.276AN

220

445

48

67.987

8.0038

719.6

240.65

700

mm

2065.50

2500

2000.94

=

12938.1

24000

= = 145.64


6/10

S = mm

But minimum size of a fillet weld for a 45 mm thick Plate is 10 mm

Let us provide an intermittent fillet weld

The ef fect ive we ld length 4 S = 40 mm

Mini effective weld length, L = 40 mm

Provide intermittent fillet welds of 40 mm long

mm

Clear spacing between the welds should not be more than 12 t w or 200 mm

Maxi. Clear spacing = 120 mm, C/C spacing of welds = mm

Provide 40 mm L ong fillets welds at a Pitch of mm (Provide continuous welding)

BEARING STIFFENER :

Maxi. Shear force V = kN

Permissible bearing stress, p = N/mm2

The bearing area required, A =

Let the thickness of bearing stiffener = t bs

Out stand bbs should not be more than 12 t bs

Use 2 bearing stiffener of bbs mm wide 2 * (t bs x 12 bs ) = A

24 t bs =

t bs = mm 30 mm

Therefore, out stand = 360 mm 200 mm

Bearing area provided = > OK

As per I S specification the bearing stiffener should be designed as column with a

distance of the web 20 times the thickness of web ( t w ) on both side as a part

of the stifferers.

Length of web as stiffener = 2 x 20 x t w = mm

30

10

20 t w 20 t w

= 200 mm = 200 mm

Effecti ve area of st if fener A = mm2

I x x = mm

Radius of gyration , r = mm

Effective length of the web = mmSlenderness ratio of the bearing stiffener =

Permissible axial compression stress ac = N / mm2

Safe load = kN > kN

Provide 200 30 mm P late as stiffeners

CONNECTION BY WELDING :

Let us provide 6 mm size intermittent fi llet weld

Strength of weld / mm = N / mm

Required strength of weld / mm = N / mm < N / mm OK

The length of intermittent fillet weld is taken as 10 t bs = mm

C / C spacing of fillet weld = mm

Maxi. c / c spacing of fil let weld = 16 t bs = 480 mm 300 mm

Provide 300 mm long fillet welds at a spacing of 300 mm

INTERMEDIATE STIFFENERS :

For an unstiffened web the minimum required thickness is as below

1.0 t w (mini) = d va

va = N / mm

t w (mini) = mm

2.0 t w (mini) = d fy

3.0 t w (mini) =

680.08

=

453.6

200.09

816

80.04

27.409

1344

453.6

16.864

300

29.411 mm

mm

Horizontal shear / mm length84.047= mm

OK

400

130

160.00

130.00

2000.94

mm2

10671.7

4.7592

Strength of weld on both face of webPitch of welds =

10672

10671.7

200

187.5

148.63

16000

21.087

12000 mm2

200

172300000

2378.04 2000.9

mm x

103.77

1750

d

85= 29.412

mm2


7/10

The web thickness provided is 8 mm, which is quite less than required if it to

be unstiffened. So, intermidiate stiffenrs will be required

Vertical stiffeners to be provided if the web thickness as calculated below is more than the

web thickness provided.

1.0 t w (mini) = d2 fy

2.0 t w (mini) =

The web thickness provided is 8 mm which is less than mm

Therfore, Vertical stiffeners will be provided

A horizontal stiffener will be required at a distance of 2/5 of the distance from the compression

flange to the N . A, if the web thickness as calculated below is more than that provided.


2.0 t w (mini) =

Therefore, a horizontal stiffener will be provided at 2 = mm from the

5

compression flange. Another horizontal stiffener will be required at the N . A if the web thickness

as calculated below is more than the provided one.


2.0 t w (mini) =

Web thickness provided is more than mm. So, horizontal stiffener at N . A is not required.

VERTICAL STIFFENERS :

The smaller clear panel dimenssion of vertical stiffener > 180 t w = mm

Therefore, Vertical stiffener should be provided at a spacing less than mm

Number of stiffeners = 11 Nos.

Actual smaller clear panel dimenssion = mm mmProvide vertical stiffeners at spacing of mm. A horizontal stiffener will be provided at 500 mm

from the compression flange.

d = mm

Spacing of stiffeners c = 0.9 d > 0.7 d = mm c/c

Provide vertical stiffeners at a spacing of mm c/c

d

t w

Shear stress va = N / mm2

From Table 6.6A of I. S : 800 - 1984, for d and c = 0.7 d

tw

va = N / mm2

Minimum required thickness of vertical stiffener for shear consideration.

t vs = mm 10 mm

Maxi. Outstand of vertical stiffener > 12 t vs = mm mm

Required moment of inertia = 1.5 d3

t3

c2

Moment of inertia of the vertical stiffener about the face of the web I x x = > OK

Use I. S. F mm x 10 mm

Connection :

Connection are designed to rasist a shear of 125 t2

Where, h = Outstand of stiffener = mm

Shear force = kN / m = N / mm

Let us provide a 3 mm size of the intermittent filled weld

Strength of weld / mm = N / mm > N / mm OK

The length of the intemittent fillet weld is taken as 10 t = mm

C / C Spacing of welds = mm

But the c / c spacing of the weld 16 t or 300 mm which ever is less 200 L + 1000 N / m

C / C spacing of weld = 160mmProvide 3 mm size weld 100 mm long welds at c /c spacing of 160 mm on both sides of the web. (Provide continuous welding)

HORIZONTAL STIFFENERS :

A horizontal stiffener be provided at mm from the compression flange.

Let the thickness of horizontal stiffener be t = 10 mm ( minimum thickness of the web )Maxi. Outstand = 12 t = 120 mm Consider a filet section = 100 x 10 mm

Moment of inertia required = 4 C t3

= mm4

C = Actual distance between vertical stiffeners = mm

t = Minimum required thickness = mm

Moment of Inertia provided = mm4

> mm4

OK

Connection :

Provide 3 mm size 8 mm long f ill et weld at a c / c spacing o f 128 mm ( 16 t = 160 mm )

1400

3333333.33 2871235.8

100

399.17

500

1400

OK

120

2871235.8

110

4436666.67

hkN / m

113.64

226.80 113.64

113.64

mm4

= 3139106.2 mm4

3139106.2mm4

= 200

12.50

mm

= 12.353 mm3200

d2 = 12.50 mm200

= 9.8821 mm4000

d2 = 10.00250

= 6.1766400

d2 =

1800

1818.2 1800

8.0038

200

8.0038

80.038

=

100.00

1400

1800

500x

2500

2

mm

6.25

mm

4006.25

1800

2000

110

110


8/10

3 mm size continuous fillet welding

VERTICAL STIFFENERS :

CONNECTION :

ISF 100 x 10 mm ( Hor izonta l Sti ffener )

ISF 110 x 10 mm ( Ve rt ical Sti ff ener )

mm

DESIGN OF LATERAL BRACING :

Lateral braching is provided at the compression flange level. The diagonals in tension will be effective only. The influence line diagram

for the bracing system show the diagonals are in tention. The force in the members L 0U1 and L12U9 will be maximum and also equal.

Therefore, L0U1 or L0U9 is designed and the same section is provided for the other diagonals.

Lateral force is condiderd = N/m

U0 U1 U2 U3 U4 U5 U6 U7 U8 U8 U9

M

2.3

L0 L1 L2 L3 L4 L5 L6 L7 L8 L9 L10

20 m @ 2.3 m e ach panal

TOP VIEW:

U0 U1 U2 U3 U4 U5 U6 U7 U8 U8 U9

Mx

2.3

L0 x L1 L2 L3 L4 L5 L6 L7 L8 L9 L10

20 m @ 2.3 m each panal

2

2

x x

10.9

I L D For L0U1 :

Consider the section x - x and taking the moment about L10 (Unit load place at L0)

Sin 45o L0 U1 x 20.0 = 1 x 20.0

L0 U1 = 1 1 x 2

sin 45o

1

2Force in L0U1 = N

Area A = mm2

Since the angle will not be accessible for clearing and painting thickness < 8 mm provide ISA 100 x 65 x 8 mm

is tried from IS Handbook No. 1. The relevant properties of the section, Shear center Cx x = mm

A 1 .= ( 100 - 4 ) x 8 = mm

A 2 .= ( 65 - 4 ) x 8 = mm

=

1= =

k

9000

763.68

3 A1

3 A1 + A 2

1400

HORIZONTAL STIFFENERS:

32.80

= 0.8252

114551.299

768

488

500


9/10

Net area provided = mm2

Load carrying capacity = N > N OK

CONNECTION : PB

45

A B

N

31.0

Le he force resised by he weld near he ou er edge of flange be PB. Taking momen abou A

PB x 100 = x

PB = N

Force resisted by the weld at A

PA = N

Let us provide a size of weld S = 5 mm weld for making the connection.

Length of welds

= 0.7 x A x S x

B = mm

And= 0.7 x B x S x

A = mm

Force in the end strut = N

Area of Angle required = mm2

Let us try 90 x 90 x 8 mm from IS Hand book No. 1. The releveant properties are

A = mm2

Radius of gyration r = mm

The strut will be welded to the flange plate girder

CONNECTION : PB

45

A B

90

N

25.1

Slenderness ratio of the strut =

For = and fy = N/mm2

Axial comprassive stress ac = N/mm2

Load carrying capacity = N > N OK

Let the force resisted by the weld near the outer edge of flange be PB. Taking moment about A

PB x 90 = x

PB = N

Force resisted by the weld at A

PA = N

Let us provide a size of weld S = 5 mm weld for making the connection.

Length of welds

= 0.7 x A x S x 108

B = 70 mm

And

= 0.7 x B x S x 108

A = mm

DESIGN CROSS - FRAME :

U0 L0

N

Bottom strut

A B

2500

99.399 100

203.65

90000

2300

1285.7

1379

17.5

114263.94 90000

111.71

250.00

82.86

108.00

76978.47

114551.30

204

1121.8168265 114551

90000

L0 U0 90000

32.8

0

37572.8

114551.30 32.80

108.00

37572.8

64900.00

25100.0

64900.0

25100.0

76978.47

Area

25.1

111.71

171.69 175

90000 25.10

66.402


10/10

Two end frames one at each end of the plate girder are provided. In cross - frames double diagonals are provided

of which one in tension is assumed to be effective.

= 45 0

Froce in one diagonal U0A Cos 45o

= N

U0A = N

Provide ISA 100 x 65 x 8 mm, Load carrying capacity of the Angle = N > N OK

as design for the lateral bracing. The connection will also be the same as designed in for the section in lateral bracing system.

Provide a nominal section say 90 x 90 x 8 mm for the bottom strut.

Intermediate cross frames at every 4 m are provided. Since the lateral forae = N which is very less so, provide75 x 75 x 8 mm angles for diagonals as well as for struts.

36000

127280

90000

127280

168265

composite steel girder

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