Box Girder Computations

BOX GIRDER SUPER STRUCTURE A3 A2

A

A1

T B

M D G C

P

F E L H N O K I

J

I

Q

INPUT DIMENSION (mm) (Designation as per above figure) A= B=

430 200

G= H=

100 250

M= N=

300 315

S= T=

1800 350

S

C= D= E= F=

1800 400 263.5 315

I= J= K= L=

600 3000 200 1700

O= P= Q= R=

150 150 430 600

A1 = A2 = A3 =

350 330 420

420 330 350

Clear Carriage way = 7500 430

0.065 m Wearing Coat 350

200

300 400 100 1800

150

315 263.5 1700 250 315 150 200 600

3000

600

430

1800

400

300 150 685.16

650

200 GEOMETRY OF END CROSS GIRDER

DATA : 1. C/C of span (mm) 25000 2. Effective Span (mm) = C/C Dist.-2 ( Width of End cross girder) 24200 3. C/C of web for outer box span (mm) 3936.5 4. Clear Carriage way (mm) 15000 5. Overall width of decking (mm) 16460 6. Concrete Grade M 30 7. Grade of Steel 415 8. Thickness of wearing coat ( in m) = 0.065 9. Permissible stresses in steel 10. Permissible stresses in concrete 11. Modular Ratio m 12. Density of parapet (t/m)

st (kg / cm2) cbc (kg / cm2)

2000 101.94 10 0.2

Notes:

This box indicate INPUT parameter. This indicate UDL load on span.

16460 OF FOUR LANE BRIDGE

1931.8

7500

7500

CLEAR ROAD WAY

CLEAR ROAD WAY

3936.5 c/c of web of Box girder

4723.5 1800 1700

2200

Elastomeric Bearing RCC Pedestel RCC Pier Cap

2260

mm c/c of pedestel

25000 mm c/c of Pier

2260 mm c/c of pedestel (All Dimensions are in mm)

25000 mm c/c of Pier

Elastomeric Bearing RCC Pedestal RCC Pier cap RCC Pier

RCC Sub Structure

Foundation GROUND LEVEL

RCC Super Structure IN M

30

R

600

25000 mm c/c of Pier

400

200

100

1700

2200

200

C/C of Pier 400

C/C Of Bearing 600 A

SECTIONAL ELEVATION 1-1 2400

9100

OF SYMMETRY

B 263.5

685 315

1

1

3000 c/c of Sofit Box

1843.0 1630

OF BOX GIRDER

315

8660 c/c of Box

685 263.5 25000 Overall Span c/c of Bearing OF PIER

OF bearing A

OF SYMMETRY B

24200 Effective Span c/c of Bearing

PLAN AT SOFFIT LEVEL

JAYESH

DRG-2

BG/DAX/DRG-Section

(2) DESIGN OF CANTILEVER DECK SLAB 430 350

Wearing Coat (m.) 0.065 X

KERB

200

400

1800 X

2.1 DEAD LOAD BENDING MOMENT Dead Load bending moment @ XX, (1) DL due to parapet

0.2

1.8

0.43 2

(2) Parapet kerb = {A*A1 * 2.40 * (C-A/2)} 0.43 0.35 2.4

1.8

0.43 2

(3) Wearing coat = { (C-A) * Thk. Of wearing coat* (C-A/2)} 1.8 0.43 0.065 2.4 1.8 (4) Self weight of slab (a) {(C*B*C/2)*2.40} 1.8 0.2

1.8 2 (b) {1/2*C*(D-B)*(C/3)*2.40} 1 1.8 0.4 2

2.4

0.2

1.8 3

TOTAL DEAD LOAD BENDING MOMENT

2.2 LIVE LOAD BENDING MOMENT 2.2.1 CLASS A Vehicle 0.43

0.43 2

Minimum Clarance (IRC - 6:2000)

2.4

0.317

t.m.

0.573

t.m.

0.146

t.m.

0.778

t.m.

0.259

t.m.

2.073

t.m.

Ground contect Area 0.15

0.5 0.97 1.8

Effective Dispersion width = 1.2 a + b1 (Cl. 305.16.2, IRC-21:2000) a = (C-A) - 0.15 - 0.50/2

=

b1 = 0.25 + 2 (Thk. Of Wearing coat)

0.97 m.

=

Effective Dispersion width bf =1.2 a + b1 1.2 0.97

0.38 m.

0.38

1.544 m.

LIVE LOAD BENDING MOMENT = (Axle load/2) * a * Impact Factor For Class A Axle load 11.40 t Impact factor 50% for cantilever slab as per Fig. 5 Cl. 211.2, IRC-6:2000 LIVE LOAD BENDING MOMENT = (11.40/2) * a * 1.50 11.4 0.97 2

1.5

8.2935 t.m.

2.2.2 CLASS AA Traked Vehicle 0.43

Minimum Clarance in m.(IRC - 6:2000)

Kerb

1.2

Ground contect Area 0.85

1.63

0.17

1.8 As c.g. of loads lying outside, No calculation of B.M. is reqd. Hence class A governs the design. Effective Dispersion width (Cl. 305.16.3, IRC-21:2000) = 0.50(Wheel contact Area) + 2*(Slab thk. + W.C.) Distance between edge to center of load = So, Slab Thk. @ Load center =

0.2

0.43

0.4

0.4

0.2

0.5 2 0.83

0.83 m. 0.292 m.

1.8 Effective Dispersion width = 0.50 + 2 ( Slab thk. + W.C.) 0.50 2 0.292

0.065

1.214 m.

LIVE LOAD BENDING MOMENT / m. Width

=

8.294 1.544

4.423 t.m/m 1.214

When vehicals travels near expansion gap, Eff. Width available across the span. Effective width available across the span, beff. = ( 1.2 x a)/2 + (0.25+W) 1.2 0.97 0.25 0.065 2 LIVE LOAD BENDING MOMENT near expan. gap =

0.897 m. 8.294 0.897

7.613 t.m/m 1.214

(3) SERVICES Service load = 0.2 t/m

(Assumed)

So, B.M. = 0.20 * (Width of Cantilever - Half width of kerb) B.M. 0.20 1.8 0.43 2

TOTAL BENDING MOMENT (D.L. + L.L. + Services)

0.317 t.m/m

=

(L.L.B.M./m. width taken)

2.073

4.423

6.813

t.m.

For M25 Concrete, m = 10 K=

0.338

j = 1- K/3

=

0.887

Q = 1/2 * cbc * k* j

=

15.272

d reqd. =

21.121 cm.

{d reqd. =

d prov. =

36.2 cm

d Prov. =

d reqd. Hence OK...

<

Ast Reqd. Provide

( Total BM / (Q*100)) } 400 362

d prov.

=

10.60 cm2

12

mm dia @

280

mm c/c

16

mm dia @

280

mm c/c

30(cover) - 16/2(half Dia.) mm

0.317

Ast Provided

11.22 cm2

In Cantilever projection of Box slab.

OK….

For End 1 m. near EXPANSION GAP. TOTAL BENDING MOMENT (D.L. + L.L. + Services) (L.L.B.M. taken at Expansion gap) d reqd. =

25.593 cm.

d prov. =

36.2 cm

d reqd. Hence OK...

<

Ast Reqd.

10.003 t.m.

{d reqd. = Sqrt( Total BM / (Q*100)) } d Prov. = 400 - 30(cover) - 16/2(half Dia.) = 362 mm

d prov.

=

Provide

=

15.57 cm2

12

mm dia @

280

mm c/c

25

mm dia @

280

mm c/c

Ast Provided

21.57 cm2

In Cantilever projection of Box slab.

DISTRIBUTION STEEL B.M. = 0.2 DLBM + 0.3 LLBM

(Cl.305.18.2, IRC : 21-2000)

Dead Load BM = DL + Service = Live Load BM = B.M. B.M.

= =

0.2

2.390

0.3

=

2.809 cm2

Ast Minimum

=

3.6 cm2

Ast Reqd.

=

3.60 cm2

10

Ast Provided Provide Ast Provided

4.423

1.805 t.m.

Ast (Dist.)

Provide

2.390 4.423

mm dia @

150

5.24 cm2 10

mm dia @

140

5.61 cm2

( 12% of gross area)

mm c/c

About top & bottom

OK…. mm c/c

About bottom in span direction.

(in Cantilever portion) OK….

OK….

MAIN STEEL : Throughout 12 mm Tor 280 mm c/c

16 mm Tor 280 mm c/c

Throughout 12

20 mm Tor 280 mm c/c

280

10 mm Tor 140 mm c/c

12 mm Tor

16 mm Tor

12

280 mm c/c

280 mm c/c

280

Throughout

DISTRIBUTION STEEEL :

10 mm Tor 150 mm c/c at bottom

215

2185 JAYESH

10 mm Tor

10 mm Tor

150 mm c/c

150 mm c/c

8 mm Tor

8 mm Tor

150 mm c/c

150 mm c/c

3000

1030 Steel Details

JAYESH

Steel Details

16460 OF FOUR LANE BRIDGE

600

7500 CLEAR ROAD WAY

6 0 A

25 20 mm tor At Top. 6 0

300 16 mm Tor 2 Legged Stirps 180 mm c/c.

16 mm Tor 2 Legged Stirrups 180 mm c/c.

25 20 mm tor At Top.

16 mm Tor 2 Legged Stirrups 180 mm c/c.

16 Tor 1900 180 mm c/c on both faces.

16 Tor 180 mm c/c on both faces. 0 6 1800

430

600

A 6 25 0 20 mm tor At bottom.

3000

600

20 25

1800

400 16 mm Tor 2 Legged Stirrups 180 mm c/c.

mm tor At bottom.

SECTION - AA

CROSS SECTION AT END DIAPHARM

JAYESH

ED-1

C/S Of End Diapharm

LONGITUDINAL GIRDER ( 4 ) LIVE LOAD BENDING MOMENT 4.1 Max moment at mid span. ( i ) Class AA Tracked Vehicle 70 Tonne Total Load 70/3.6 = 19.444 t/m. 3.6

12.1 0.4

24.2

Mid span moment =

391.63

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

Moment With I.F. and R.F.

0.4

= =

1.1 1.2

=

(cl.211.3(b), IRC:6-2000)

516.9516 t.m.

( ii ) Class 70R wheeled Vehicle 17 6.6412

17 . 1.37

A

17 . 3.05

17 . 1.37

E

12 . 2.13

12 . 1.52

8 .

3.96

4.1588

B

12.1 0.4

24.2 RA

0.4 RB

c.g. of load from right of first load = Coincide distance = Moment @ E = 441.49 t.m.

5.1238 m. 5.4588 m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

1.18 1.2

= =

Moment With I.F. and R.F.

=

625.15

(From fig. 5 cl.211.3(b), IRC:6-2000)

t.m.

( iii ) Class A Train (Two Lanes) 2.7 2.7 2.66

. 1.10

11.4 . 3.20

11.4 . 1.20.

6.8 . 4.3

A

6.8 . 3.0

6.8 .3.0

E

'6.8 3

2.74 B

12.1 0.4

24.2 c.g. of load from right of first load Coincide distance Distance from A

Moment @ E =

193.28

= = =

4.2 Max moment at quarter span.

9.09 m. 9.44 m. 2.66 m.

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

Moment With I.F. and R.F.

0.4

= =

=

1.18 1.2

547.37

(From fig. 5 cl.211.3(b), IRC:6-2000)

t.m.

( i ) Class AA Tracked Vehicle Quarter of load distance (i.e. 1/4 X 3.6 m) = 0.9

70 T

3.6

6.05 0.4

24.2 Mid span moment =

293.67

0.4

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

Moment With I.F. and R.F.

= =

1.1 1.2

=

( cl.211.3(b), IRC:6-2000)

387.6444 t.m.

( ii ) Class 70R wheeled Vehicle 17 4.68

17 . 1.37

17 . 3.05

17 . 1.37

A

12 . 2.13

12 . 1.52

E

8 .

3.96

6.1200

B

6.05 0.4

24.2

0.4

c.g. of load from right of first load = Coincide distance = Moment @ E = 336.37 t.m.

5.1238 m. 5.4588 m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

1.18 1.2

= =

(From fig. 5 cl.211.3(b), IRC:6-2000)

Moment With I.F. and R.F.

=

476.30

t.m.

( iii ) Class A Train (Two Lanes) 1.75

2.7

2.7

11.4

1.1

11.4

3.2

6.8

1.20.

4.3

A

6.8 3.0 .

6.8 .3.0

'6.8 3.65

3.0

E

B

6.05 0.4

24.2

0.4 23.1

Moment @ E =

164.65

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

= =

Moment With I.F. and R.F.

=

1.18 1.2

(From fig. 5 cl.211.3(b), IRC:6-2000)

466.2888 t.m.

4.3 Max moment at 3 m from left of span. ( i ) Class AA Tracked Vehicle 70 T 3.6

3 0.4

24.2 Mid span moment =

162.80

Give value of impact factor = I. F.

0.4

t.m. =

1.1

(cl.211.3(b), IRC:6-2000)

Give value of Reaction factor = R. F.

=

Moment With I.F. and R.F.

1.2

=

214.90

t.m.

( ii ) Class 70R wheeled Vehicle 17 3

17 . 1.37

17 . 3.05

17 . 1.37

A

12 . 2.13

12 . 1.52

8 .

3.96

7.8000

E

B

3 0.4

24.2 c.g. of load from right of first load Moment @ E =

190.91

=

5.1238 m.

= =

1.18 1.2

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

Moment With I.F. and R.F.

0.4

=

270.33

(From fig. 5 cl.211.3(b), IRC:6-2000)

t.m.

( iii ) Class A Train (Two Lanes) 11.4

11.4 1.2

A

6.8 4.3

E 3

0.4

6.8 3.0 .

6.8 3.0 .

6.8 .3.0

9.5

B 21.2

24.2

0.4

Moment @ E =

90.17

t.m.

Give value of impact factor = I. F. Give value of Reaction factor = R. F.

Moment With I.F. and R.F.

= =

=

1.18 1.2

(From fig. 5 cl.211.3(b), IRC:6-2000)

255.3614 t.m.

RECAPITULATION OF LIVE LOAD BENDING MOMENTS Load Discription

BENDING MOMENT (in tm) .@ Mid Span

@ Quarter Span

@ Widening.

Beginning of

Live load Class AA

516.95

387.64

214.90

Class 70R

625.15

476.30

270.33

Class A

547.37

466.29

255.36

625.150

476.30

270.33

DESIGN BM

Beginning of Widening of section from support (m) =

3

Dead Load Bending Moment Super Imposed Dead Load (SIDL) of Super Structure Wearing Coat ( t ) =Ht.X Clear carriage way X Density=

2.34

t/m.

Parapet ( t ) = 0.2 t/m

0.4

t/m.

0.3612

t/m.

0.2

t/m.

3.3012

t/m.

Kerb

=

= Area X Density

=

Services = 0.1 t/m

= Total SIDL

=

3.3012 t/m.

A

C

D

E

B

3.00 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B =

41.265 t

Bending Moment at mid span (E) Bending Moment at quarter span (D) Bending Moment at Widening (C)

= = =

241.4003 t.m. 180.9842 t.m. 104.7141 t.m.

Dead Load Bending Moment due to self wt. of Super Structure C/S Area of box at mid span = udl

A

C

=

D

8.069

m2

19.37

t/m.

19.37

t/m.

E

3.00 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B =

242.07

t

B

Bending Moment at mid span (E) Bending Moment at quarter span (D) Bending Moment at Widening (C)

= = =

1416.11 t.m. 1061.695 t.m. 614.2768 t.m.

Dead Load Bending Moment due to widening C/S Area of box at End span C/S Area of box at mid span

= =

10.86 8.069

m2 m2

Difference of C/S Area

=

2.791

m2

Wt./R.m.(A X Density)

=

6.6984

t/m.

1 2.4

m. m.

Total Length of END Beam Length of Taperd Section of Beam

= =

6.6984 t/m.

A

C

6.6984

D 2.4

0.6

E

B

3.0 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B =

14.74

t

Bending Moment at mid span (E) Bending Moment at quarter span (D) Bending Moment at Widening (C)

= = =

11.92315 t.m. 11.92315 t.m. 11.92315 t.m.

Total Reaction @ A & B ( i.e. Total DL due to half Span ) = Total DL of Super Structure =

NOTE :

298.07 t 597 t

Put All Geometry in STAAD Analysis and Varify above data.

Summary of DLBM

Sr .No.

LOAD

1

SIDL

2 3

At At MID Quarter At Span (E) Span (D) Widening 't.m' 't.m' (C) 't.m' 241.4003

180.9842 104.71406

Self Wt.of Box (Running Section)

1416.11

1061.695 614.27683

Widning (Self Weight)

11.92315

11.92315 11.923152

TOTAL DLBM

=

1669.433

1254.602 730.91405

SHEAR FORCE Give Value of No. of GIRDER

4

Due To Dead Load Due To SIDL Super Imposed Dead Load (SIDL) of Super Structure Wearing Coat ( t ) =Ht.X Clear carriage way X Density=

2.34

t/m.

Parapet ( t ) = 0.2 t/m

0.4

t/m.

0.3612

t/m.

0.2

t/m.

3.3012

t/m.

Kerb

=

= Area X Density

=

Services = 0.1 t/m

= Total SIDL

=

3.3012 t/m. Y X A

C

D

E

B

3.00 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B =

41.265 t

Section Y = Distance from support to edge of END BEAM Section X = Distance from support to centre of WIDENING Section C = Distance from support to Starting of WIDENING Section D = Distance from support to Quarter Span Shear Force, .@ A .@ Y .@ X .@ C .@ D

41.265 39.945 39.945 39.945 39.945

3.3012 3.3012 3.3012 3.3012 3.3012

0.4 0.6 1.8 3.0 6.05

39.945 37.964 34.002 30.041 19.972

0.6 1.8 3.00 6.05

t t t t t

Dead Load Shear Force due to self wt. of Super Structure 19.37 t/m. Y X A

C

D

E

B

3.00 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B =

242.07

Shear Force, .@ A 242.07 .@ Y 234.324 .@ X 234.324 .@ C 234.324 .@ D 234.324

t

19.37 19.37 19.37 19.37 19.37

0.4 0.6 1.8 3.00 6.05

234.324 222.704 199.466 176.227 117.162

t t t t t

Dead Load Shear Force due to widening 6.698 t/m. Y A

6.698 t/m.

X C

D 2.4

0.6

E

B

3.0 0.4

6.05 12.1 24.2 12.5 25 Reaction at A & B = Shear Force, .@ A 14.74 .@ Y 12.057 .@ X 8.038 .@ C 8.038 .@ D 0.000

14.74

6.70 6.70 5.02 3.35 0.00

t

0.4 0.6 1.2 2.4 6.05

12.057 8.038 2.010 0.000 0.000

t t t t t

Total Reaction @ A & B ( i.e. Total DL due to half Span ) = Total DL of Super Structure =

NOTE :

298.07 t 597 t

Put All Geometry in STAAD Analysis and Varify above data.

Summary of DLSF

Sr .No.

LOAD

1

SIDL

Section A Section Y Section X (t) (t) (t) 39.94

37.96

34.00

2

Self Wt.of Box (Running Section)

234.32

222.70

199.47

3

Widning (Self Weight)

12.06

8.04

2.01

286.33

268.71

235.48

TOTAL DLSF ( t ) =

Due To Live Load FOR Twin Box Take R.F.

=

1.2

AT Support Section ( A ) Class A Two Lane Vehicle 11.4

11.4 . 1.20.

6.8 . 4.3

6.8 . 3.0

6.8 .3.0

'6.8 9.7

3

B 24.2

RA

=

11.4

24.2

23

6.8

18.7

15.7

12.7

24.2 38.20 t S.F.@support with R.F. & I.F. (For Two Lane) 38.20

1.18

1.2

2

108.168 t

( B ) Class AA Traked Vehicle 70 t

A

B 3.60 24.2

RA

=

70

22.4 24.2 64.79 t

S.F.@support with R.F. & I.F. (For Two Lane) 64.79 85.527 t

1.1

1.2

( C ) Class 70R Wheeled Vehicle 17

17 . 1.37

17 . 3.05

17 . 1.37

12 . 2.13

12 . 1.52

8 .

3.96

10.800

B 24.2 RA

RA

=

RB

17

24.2

22.83

19.78

18.41

12

16.28

8

10.8

24.2 78.83 t S.F.@support with R.F. & I.F. (For Two Lane) 78.83

1.18

1.2

111.619 t RECAPITULATION OF LIVE LOAD SHEAR FORCE At SUPPORT Sr. No. 1 2 3

LOADING

S.F. ( t )

Class A ( 2 lane) Class AA Traked 70R wheeled Vehicle

AT Section - ' Y '

(

0.6

108.168 85.527 111.619

) m. From support

( A ) Class A Two Lane Vehicle 11.4

11.4

0.6 . 1.20.

6.8 . 4.3

6.8 . 3.0

6.8 .3.0

'6.8 9.1

3

B 24.2

RA

=

11.4

23.6

22.4

6.8 24.2

36.96 t S.F.@support with R.F. & I.F. (For Two Lane)

18.1

15.1

12.1

36.96

1.18

1.2

2

104.658 t

( B ) Class AA Traked Vehicle 70 t 0.6 A

B 3.60 24.2

RA

=

70

21.8 24.2 63.06 t

S.F.@support with R.F. & I.F. (For Two Lane) 63.06

1.1

1.2

83.236 t

( C ) Class 70R Wheeled Vehicle 17

17

0.60 . 1.37

17 . 3.05

17 . 1.37

12 . 2.13

12 . 1.52

8 .

3.96

10.200

B 24.2 RA

RA

RB

=

17

23.60

22.23

19.18

17.81

12

15.68

8

10.20

24.2 76.35 t S.F.@support with R.F. & I.F. (For Two Lane) 76.35

1.18

1.2

108.109 t RECAPITULATION OF LLSF At Sr. No.

LOADING

0.6

m.From Support. S.F. ( t )

1 2 3

Class A ( 2 lane) Class AA Traked 70R wheeled Vehicle

AT Section - ' X '

(

1.8

104.658 83.236 108.109

) m. From support

( A ) Class A Two Lane Vehicle 11.4

11.4

1.8 . 1.20.

6.8 . 4.3

6.8 . 3.0

6.8

'6.8

.3.0

7.9

3

B 24.2

RA

=

11.4

22.4

21.2

6.8

16.9

13.9

10.9

24.2 34.48 t S.F.@support with R.F. & I.F. (For Two Lane) 34.48

1.18

1.2

2

97.636 t

( B ) Class AA Traked Vehicle 70 t 1.8 A

B 3.60 24.2

RA

=

70

20.6 24.2 59.59 t

S.F.@support with R.F. & I.F. (For Two Lane) 59.59

1.1

1.2

78.655 t

( C ) Class 70R Wheeled Vehicle 17 1.80 . 1.37

17

17 . 3.05

17 . 1.37

12 . 2.13

12 . 1.52

8 .

3.96

9.000

B 24.2 RA

RA

RB

=

17

22.40

21.03

17.98

16.61

12

14.48

8

9.00

24.2 71.39 t S.F.@support with R.F. & I.F. (For Two Lane) 71.39

1.18

1.2

101.087 t RECAPITULATION OF LLSF At Sr. No. 1 2 3

1.8

m.From Support.

LOADING

S.F. ( t )

Class A ( 2 lane) Class AA Traked 70R wheeled Vehicle

AT Section - ' C '

(

3.00

97.636 78.655 101.087

) m. From support

( A ) Class A Two Lane Vehicle 11.4

11.4

3.00 . 1.20.

6.8 . 4.3

6.8 . 3.0

6.8 .3.0

'6.8 6.7

3

B 24.2

RA

=

11.4

21.2

20

6.8 24.2

32.00 t S.F.@support with R.F. & I.F. (For Two Lane) 32.00

1.18

90.615 t

( B ) Class AA Traked Vehicle

1.2

2

15.7

12.7

9.7

70 t 3.0 A

B 3.60 24.2

RA

=

70

19.4 24.2 56.12 t

S.F.@support with R.F. & I.F. (For Two Lane) 56.12

1.1

1.2

74.073 t

( C ) Class 70R Wheeled Vehicle 17

17

3.00 . 1.37

17 . 3.05

17 . 1.37

12 . 2.13

12 . 1.52

8 .

3.96

7.800

B 24.2 RA

RA

RB

=

17

21.20

19.83

16.78

15.41

12

13.28

8

7.80

24.2 66.43 t S.F.@support with R.F. & I.F. (For Two Lane) 66.43

1.18

1.2

94.066 t RECAPITULATION OF LLSF At Sr. No. 1 2 3

AT Section - ' D '

3

LOADING Class A ( 2 lane) Class AA Traked 70R wheeled Vehicle

(

6.05

) m. From support

m.From Support. S.F. ( t ) 90.615 74.073 94.066

( A ) Class A Two Lane Vehicle 11.4

11.4

6.05 . 1.20.

6.8 . 4.3

6.8 . 3.0

6.8

'6.8

.3.0

3.65

3

B 24.2

RA

=

11.4

18.15

16.95

6.8

12.65

9.65

6.65

24.2 25.70 t S.F.@support with R.F. & I.F. (For Two Lane) 25.70

1.18

1.2

2

72.768 t

( B ) Class AA Traked Vehicle 70 t 6.1 A

B 3.60 24.2

RA

=

70

16.35 24.2 47.29 t

S.F.@support with R.F. & I.F. (For Two Lane) 47.29

1.1

1.2

62.427 t

( C ) Class 70R Wheeled Vehicle 17 6.05 . 1.37

17

17 . 3.05

17 . 1.37

12 . 2.13

12 . 1.52

8 .

3.96

4.750

B 24.2 RA

RB

RA

=

17

18.15

16.78

13.73

12.36

12

10.23

8

4.75

24.2 53.83 t S.F.@support with R.F. & I.F. (For Two Lane) 53.83

1.18

1.2

76.219 t RECAPITULATION OF LLSF At Sr. No. 1 2 3

6.05

m.From Support.

LOADING

S.F. ( t )

Class A ( 2 lane) Class AA Traked 70R wheeled Vehicle

72.768 62.427 76.219

RECAPITULATION OF DL & LL SHEAR @ VARIOUS SECTION Sr. No.

S.F. Due To Support

Shear Force At Section in Tonne Sect - Y Sect - X Sect - C Sect - D

1

DL + SIDL

286.33

268.71

235.48

206.27

137.13

2

LIVE LOAD

111.619

108.109

101.087

94.066

76.219

DESIGN S.F.( t )

397.94

376.81

336.56

300.33

213.35

CHECK FOR SHEAR STRESS & REINFORCEMENT CALCULATION At Support Section : MAX. Shear force =

397.94

SF / Girder

=

397.94 4

Shear Stress

=

99.49 65 6.96 21.582

<

Providing

12

mm dia.

4

t 99.49 t

1000 220 Kg / Cm2 Kg / Cm2

OK….

legged stirrups,

Spacing Provide

12

'S'

=

mm dia.

Shear Force Taken =

196.672 mm 4

legged stirrups, @

108.70

>

99.49

180

mm C/C.

Hence OK….

At Section ' Y ': MAX. Shear force =

376.81

SF / Girder

=

376.81 4

Shear Stress

=

94.20 65 6.59 21.582

<

Providing

12 Spacing

Provide

12

mm dia. 'S'

=

mm dia.

Shear Force Taken =

4

t 94.20 t

1000 220 Kg / Cm2 Kg / Cm2

OK….

legged stirrups,

207.700 mm 4

legged stirrups, @

108.70

>

94.20

180

mm C/C.

Hence OK….

At Section ' X ': MAX. Shear force =

336.56

SF / Girder

=

336.56 4

Shear Stress

=

84.14 45 8.50 21.582

<

Providing

16 Spacing

Provide

16

mm dia. 'S'

=

mm dia.

Shear Force Taken =

2

t 84.14 t

1000 220 Kg / Cm2 Kg / Cm2

OK….

legged stirrups,

206.701 mm 2 96.62

legged stirrups, @ >

84.14

180

mm C/C.

Hence OK….

At Section ' C ' : MAX. Shear force =

300.33

SF / Girder

=

300.33 4

Shear Stress

=

75.08 25 13.65 21.582

<

Providing

16 Spacing

Provide

16

mm dia. 'S'

2

=

t 75.08 t

1000 220 Kg / Cm2 Kg / Cm2

OK….

legged stirrups,

231.637 mm

mm dia.

2

Shear Force Taken =

legged stirrups, @

96.62

>

75.08

180

mm C/C.

Hence OK….

At Section ' D ' : MAX. Shear force =

213.35

SF / Girder

=

213.35 4

Shear Stress

=

53.34 25 9.70 21.582

<

Providing

16 Spacing

Provide

16

mm dia. 'S'

2

=

t 53.34 t

1000 220 Kg / Cm2 Kg / Cm2

OK….

legged stirrups,

326.071 mm

mm dia.

2

Shear Force Taken =

86.96

legged stirrups, @ >

53.34

200

Hence OK….

SHEAR REIFOREMENT DETAILS: {1}

{2}

Y

{3}

X

{4}

C

{5}

D

mm C/C.

0.4

0.6

1.8 2.4 3.00 6.05

Portion (No.) 1 2 3 4 5

Dia. (mm) 12 12 16 16 16

Legged (No.) 4 4 2 2 2

C/C Dist. (mm) 180 180 180 180 200

m. m. m. m.

Section C Section D (t) (t) 30.04

19.97

176.23

117.16

0.00

0.00

206.27

137.13

9.7

14.76

9.1

14.16

7.9

12.96

6.7

11.76

3.65

8.71

5.0 Reiforcement calculations and checking stresses at various section. 5.1 AT MID SPAN Dead Load B.M. Design B.M.

= =

1669.43

t.m. for twin box (D.L. + S.I.D.L.)

1459.87

t.m. per box.

(All Dimensions are in mm.) 250

315

79.057 150

 25.46

237.2 270.9431

200 146.9 90.31

461.86

420 881.86

 71.565051 X = 237.17082 mm Y = 79.056942 mm Z = 270.94306 mm X1 = 90.314353 mm T = 146.85647 mm T1 = 461.85647 mm

 25.463345 a= 420 mm Provide

48

Nos. of

Ast =

32

Tor in

3

rows

385.991 cm2

Provide Clear Cover

=

30

mm

Provide dia. Of Stirups

=

12

mm

Provide no. Of bar in 1st Row Provide no. Of bar in 2nd Row Provide no. Of bar in 3rd Row Total

= = = =

18 16 14 48

no. no. no. OK…….

Rein. Spacing (Betwn. In to In ) = Clear Side Cover = Diameter of stirrups = Diameter of Main Bar =

50 40 12 32

mm mm mm mm

NO. OF BAR in First raw

9

Provision of No. of BAR in First Row

=

Total Distance, 9

32 792 mm

8

50 <

2 881.86 mm OK….

12

2

40

Distance between end of Soffit to centre of 1 st row Distance between centre of 1st row to centre of 2nd row

= =

58 64

mm mm

Distance between centre of 2nd row to centre of 3rd row

=

64

mm

c.g. of steel from bottom of girder, = d eff. 180

11.67

cm

= 208.33 cm

420

223

20 20

30 31.5 [2]

n 10

[1]

[5]

43

15

[4] 25 [6] [7]

(All Dimensions are in cm.)

Portion

Length cm

Depth cm

c.g. from Top (cm)

REMARK

NO.

[1] [2] [3] [4] [5] [6] [7]

420 180 223 180 31.5 43

30 20 30 20 10 15 25

15 10 15 26.67 33.33 35 30

Rectangle Rectangle Rectangle Triangle Triangle Triangle

1 1 1 1 2 1 2

Due to Reinforcement

AREA cm2 X n 12600 3600 6690 1800 315.00 322.5 -1500

189000 36000 100350 48000 10500 11287.5 22500

3859.9066 804147.2

[3]

Sum

n2 + 1107.496 n

=

27687.41

1176785

1107.50

47071.39

- 47071.39 .= 0

N.A. from Top of girder =

4.55E-008 Don't Delete this cell, it is useful for

40.9857 cm

For finding out Value of n.

M.I. Of section @ N.A., 180 20 40.99

40

n

20 -0.99 20.99 (All Dimensions are in cm.) Portion

[1] [2] [3] [4]

Length cm

Depth cm

420 30 180 20 223 30 85.56416 20.985743 180 20

Section Modulus at Compression,

c.g. from Top (cm) 15 10 15 26.67

Zc =

REMARK

M.I. cm4

AREA cm2

M.I. + (A x h2)

h

2

Rectangle 945000 12600 675.2588 Rectangle 120000 3600 960.1163 Rectangle 501750 6690 675.2588 Eq. Rectn. 65899.73 1795.6274 220.2007 Triangle 40000 1800 205.0359

9453261.2 3576418.5 5019231.5 461298.14 409064.69

Due to Reinforcement

=

108097517

M.I. Of section @ N.A.,

=

126555493 cm4

3087793 cm3

Section Modulus at Tension, Stresses in Concrete

Stresses in Steel

Zt =

756243.29 cm3

=

47.278631 kg/cm2 < 101.94 kg/cm2

OK…….

1930.4188 kg/cm2 < 2000 kg/cm2

OK…….

= 47.28

40.99

Stress in Outer layer,

167.35

+ 167.35

5.87

= 1998.093 kg/cm2 < 2000 kg/cm2

214.2 167.35 1930.42 5.87 (All Dimensions are in cm.)

5.2 AT QUARTER SPAN Dead Load B.M. Design B.M.

= =

1254.6021 t.m. for twin box (D.L. + S.I.D.L.) 1103.601 t.m. per box.

(All Dimensions are in mm.) 250

315

237.17 79.057 150

1930.41882

OK…….

 25.46 270.9431 200 146.9 90.31

461.86

420 881.86

 71.565051 X = 237.17082 mm Y = 79.056942 mm Z = 270.94306 mm X1 = 90.314353 mm T = 146.85647 mm T1 = 461.85647 mm  25.463345 a= 420 mm Provide

38 Ast =

Nos. of

32

Tor in

305.576 cm2

Provide Cover

=

30

mm

Provide dia. Of Stirups

=

12

mm

18 16 4 38

no. no. no. OK…….

Provide no. Of bar in 1st Row = Provide no. Of bar in 2nd Row = Provide no. Of bar in 3rd Row = Total =

3

rows

Distance between end of beam to centre of 1 st row Distance between centre of 1st row to centre of 2nd row

= =

58 64

mm mm

Distance between centre of 2nd row to centre of 3rd row

=

64

mm

c.g. of steel from bottom of box,

=

d eff. 180

9.84

cm

= 210.16 cm

420

223

20

n

30

(All Dimensions are in cm.) 420 x n x n/2 +

223 411.5

x n x n/2 +

n2

+

180 3055.7594

N.A. from Top of girder

x n x n/2 +

n

-

=

35.97

+

420

.= 10 x

305.576

x(

210.16 .-n )

642192 .= 0

cm

0 Don't Delete this cell, it is useful for For finding out Value of n.

M.I. Of the section @ N.A., 2 420

x 12

30.0

^3

30

35.966

-15

.= 2

6483482

223

x 12

180

30.0

^3

+

223

30

35.966

-15

x 12

20 ^3

+

180

20

35.966

-10

36.31

x 12

15.97 ^3

+

36.31

15.97

15.97 2

143.69

x 36

15.97 ^3

+ 1/2

143.69

15.97

x

210.16 -35.96574

.=

9155494

.=

1560000

.=

49255

.=

146197

2

2

2 15.97 x 2 3

2 10 x

305.576

.= 92720625.3 M.I. Of the section @ N.A., =

Section Modulus at Compression, Section Modulus at Tension, Stresses in Concrete

Stresses in Steel

Zc = Zt =

3061665 cm3 632147 cm3

=

36.04577 kg/cm2 < 101.94 kg/cm2

OK…….

1745.798 kg/cm2 < 2000 kg/cm2

OK…….

=

110115054 cm4

36.05 35.97 Stress in Outer layer, 174.19

174.19

+ 174.19

4.04

214.2 = 1786.309 kg/cm2

1745.79797

< 2000 kg/cm2 1745.80

4.04

5.3 AT BEGINNING OF WIDENING SECTION Dead Load B.M. Design B.M.

= =

730.91405 t.m. for twin box (D.L. + S.I.D.L.) 635.78558 t.m. per box.

(All Dimensions are in mm.) 250

315

237.17 79.057 150

 25.46 270.9431

200 146.9 90.31

461.86

420 881.86

 71.565051 X = 237.17082 mm Y = 79.056942 mm Z = 270.94306 mm X1 = 90.314353 mm T = 146.85647 mm T1 = 461.85647 mm

OK…….

 25.463345 a= 420 mm Provide

26 Ast =

Nos. of

32

Tor in

Provide Cover

=

30

mm

Provide dia. Of Stirups

=

12

mm

Provide no. Of bar in 1st Row Provide no. Of bar in 2nd Row Provide no. Of bar in 3rd Row Total

= = = =

14 12 0 26

no. no. no. OK…….

c.g. of steel from bottom of box, d eff.

20

rows

209.078 cm2

Distance between end of beam to centre of 1 st row Distance between centre of 1st row to centre of 2nd row Distance between centre of 2nd row to centre of 3rd row

180

3

=

8.75

= = =

58 64 64

mm mm mm

cm

= 211.25 cm

420

223 30

n

(All Dimensions are in cm.) 420 x n x n/2 +

223 411.5

x n x n/2 +

n2

+

180 2090.7827

N.A. from Top of girder

=

x n x n/2 +

.= 10 x

n

441669.8 .= 0

-

209.078

x(

211.25 .-n )

30.3194 cm

0.0006016 Don't Delete this cell, it is useful for For finding out Value of n.

M.I. Of the section @ N.A., 2 420

x 12

30

^3

+

420

30

30.319

-15

223

x 12

30

^3

+

223

30

30.319

-15

180

x 12

20 ^3

+

180

20

30.319

-10

87.13

x 12

10.32 ^3

+

87.13

10.32

10.32 2

92.87

x 12

10.32 ^3

+ 1/2

92.87

10.32

.= 3902026.68 2 .= 2071790.35 2 .= 1606363.92

2 .= 31914.5388

2 10.32 x 2 3

.= 25515.4848

2 10 x

209.078

x

M.I. Of the section @ N.A., =

211.25 -30.31942 75850791 cm4

.= 68440690.8 76078301.7

Section Modulus at Compression, Section Modulus at Tension,

Zc = Zt =

2509227 cm3 420492.3 cm3

=

25.33791 kg/cm2 < 101.94 kg/cm2

OK…….

1512.003 kg/cm2 < 2000 kg/cm2

OK…….

Stresses in Concrete

Stresses in Steel

=

25.34 30.32 Stress in Outer layer,

180.93

+ 180.93

2.95

1512.00284

214.2 180.93

= 1536.688 kg/cm2 < 2000 kg/cm2

1512.00 2.95

CALCULATION OF WEB REIFORCEMENT : (Skin Reinforcement)

As per Cl.305.10, IRC - 21:2000, Min. Shrinkage reiforcement shall be 250 mm 2 of Steel area per metre. 2 Skew Web Dimension : Length = Width = Total Steel Req.

=

1897.37

2

1800 250 250

600

1897.3666 mm

mm 474.34

mm2

OK…….

1000 Provide

7

Numebrs

10

mm

at top and bottom.

0

mm

at top and bottom.

+ Provide

0

Ast provided

Numebrs

=

549.710

mm2

7 10 On Each Faces.

Third raw Second ra First raw

7 8 9

32 32 32

OK….

Don't Delete this cell, it is useful for operation of Goal Seek… For finding out Value of n.

Don't Delete this cell, it is useful for operation of Goal Seek… For finding out Value of n.

Don't Delete this cell, it is useful for operation of Goal Seek… For finding out Value of n.