DESIGN OF RCC T - GIRDER USING STAAD RESULTS

DESIGN OF RCC T - GIRDER USING STAAD RESULTS :(All blue coloured fonts depict inputs)

BASIC DESIGN DATA

1 Effective span Leff 24.40 m

2 Angle of skew Ang 0.000 degree

3 Clear carriage way Bcw 7.50 m

4 Spacing of main girder c/c Spmg 2.313 m

5 Spacing of cross girder c/c Spcg 6.10 m

6 Width of crash barier Wkerb 0.550 m

7 Thk of deck slab Df 0.250 m

8 Thk of wearing coat Wc 0.065 m

9 Length of cantilever Lcan 1.987 m

10 Cantilever slab thk at fixed end Dcan1 0.350 m

11 Cantilever slab thk at free end Dcan2 0.200 m

12 No of main girder Nomg 3 m

13 Depth of main girder Dmg 2.500 m

14 Web thk of main girder ( at center ) bwmc 0.325 m

15 Web thk of main girder ( at support bwms 0.625 m

16 Length of extra widening ( varrying Lwv 0.900 m

17 Length of extra widening ( uniform Lwu 0.700 m

18 Top haunch Thw x Thh 0.300 x ### m

19 Bottom haunch Bhw x Bhh 0.150 x ### m

20 Bottom bulb Bbw x Bbh 0.625 x ### m

21 No of cross girder Nocg 3 m

22 Depth of cross girder Dcg 2.100 m

23 Web thk of cross girder bwcg 0.325 m

24 Grade of concrete Cgrade 25

25 Grade of reinforcement Sgrade 415

26 Clear cover cov 0.040 m

27 Unit weight of concrete wcon 2.400

28 Weight of wearing course wwc 0.2

29 Weight of crash barrier wrail 1 t/m

30 Stress in concrete (compression) fc 830

31 Stress in steel (tension) ft 2000032 Modular ratio m 10

N/mm2

N/mm2

t/m3

t/m2

t/m2

t/m2

Summary of bending moment for 25m span.From STAAD Analysis

Type of loadingInner girder Outer girder

At Deff Span L/4 Span L/2 At Deff Span L/4 Span L/2

71.24 127.19 173.7 - - -

73.1 127.41 175.05 - - -

1 LANE CLASS 70R (W) 69.51 128.19 264.91

1 LANE CLASS 70R TRACK - - - - - -

28.32 52.44 82.28 - - -

29.85 52.48 82.96 - - -

1 LANE CLASS A - - - 95.64 201.67 167.25

- - - 12.82 29.39 41.16

3 LANE CLASS A - - - - -

1L CL A+ 1L 70R (W) - - - - - -

1L CL A+ 1L 70R TRACKED - - - - - -Design live load bending mom 73.10 127.41 175.05 108.46 231.06 264.91Dead load + SIDL 122.43 271.66 351.41 139.56 268.41 356.17Design bending moment (t-m) 195.53 399.07 526.46 248.02 499.47 621.08

1 LANE CLASS 70R (W), wheel over girder

1 LANE CLASS 70R (W), wheel adjacent to girder

1 LANE CLASS A, wheel over girder

1 LANE CLASS A, wheel adjacent to girder

2nd LANE CLASS A

Summary of shear force

From STAAD Analysis

Type of loadingInner girder Outer girder

At Deff Span L/4 Span L/2 At Deff Span L/4 Span L/2

31.66 22.21 12.82 - - -

28.97 20.59 11.45 - - -

1 LANE CLASS 70R (W) - - - 41.27 35.03 20.15

1 LANE CLASS 70R TRACK - - - - - -

12.41 8.15 8.1 - - -

11.79 7.5 7.73 - - -

1 LANE CLASS A - - - 21.69 16.83 11.31

- - - 4.64 4.39 2.113 LANE CLASS A - - - - - -1L CL A+ 1L 70R (W) - - - - - -1L CL A+ 1L 70R TRACKED - - - - - -Design live load Shear Force (t 31.66 22.21 12.82 41.27 35.03 20.15Dead load + SIDL 49.35 40.09 11.70 51.55 34.98 5.39Design shear force (t) 81.01 62.30 24.52 92.82 70.01 25.54

Support reactions due to dead load + SIDL G1 G2 G374.58 47.39 74.58

1 LANE CLASS 70R (W), wheel over girder

1 LANE CLASS 70R (W), wheel adjacent to girder

1 LANE CLASS A, wheel over girder

1 LANE CLASS A, wheel adjacent to girder

2nd LANE CLASS A

Calculation of section properties

LongitudinalsEnd section of Inner girder

beff

y beff = [ Cl. 305.15.2 IRC 21 ]

= 5.505 m lo = 24.4 m

> 2.313 m [ c/c distance of

beff = 2.313 m longitudinal girder]

0.625

Area = 1.985Distance of cg from top fibre (y) = 1.011 m

= 1.2365d/b for segment 1 = 9.252 K = 0.309d/b for segment 2 = 3.600 K = 0.274

Torsional moment of inertia = K x b3 x d = 0.162

Span section of Inner girderMember no: 214 to 228

beffbeff = [ Cl. 305.15.2 IRC 21 ]

y = 5.205 m lo = 24.400 m




= 1.042d/b for segment 1 = 7.117K = 0.300d/b for segment 2 = 6.154K = 0.296d/b for segment 3 = 2.500K = 0.249


lo/5 + bw

m2

Moment of inertia of longitudinal girder (IL) m4

m4

lo/5 + bw

m2


m4

N A

3

2

1

N A

1

2

Member no: 213,229

Area = 2.0049

= 1.297


(Weighted average)

End section of outer girder

beff

beff = [ Cl. 305.15.2 IRC 21 ]y = 5.505 m lo = 24.4 m



0.625


= 1.2365d/b for segment 1 = 9.252 K = 0.309d/b for segment 2 = 3.600 K = 0.274


Span section of outer girderMember no: 180 to 194, 248 to 262

beffbeff = [ Cl. 305.15.2 IRC 21 ]

y = 5.205 m lo = 24.400 m

> 2.313 m

beff = 2.313 m


= 1.042d/b for segment 1 = 7.117K = 0.300d/b for segment 2 = 6.154K = 0.296d/b for segment 3 = 2.500K = 0.249


m2


m4

lo/5 + bw

m2


m4

lo/5 + bw

m2

Moment of inertia of longitudinal girder (Iz) m4

m4

N A

1

2

3

N A

1

2

Member no: 179, 195, 247, 263

Area = 2.005

= 1.297


(Weighted average)

Member no: 196 to 212, 230 to 246

Area = 0.0001

Moment of inertia of longitudinal girder (Iz) = 0.00001


Member no: 162 to 178, 264 to 280

0.8305

0.225

Area = 0.187

Distance of cg from top fibre (y) = 0.1125 m

= 0.0008

Torsional moment of inertia = = 0.0026

Member no: 145 to 161, 281 to 297

Area = 0.0001


= 0.00001


m2


m4

m2

m4

m4

m2

Moment of inertia (Iz) m4

m4

m2


m4

AN

Transversals

Member no: 1, 19, 37, 55, 73, 91, 109, 127, 18, 36, 54, 72, 90, 108, 126, 144

0.487 m0.25 beff = [ Cl. 305.15.2 IRC 21 ]

= 0.487 m lo = 1.619 m1.85

0.325


= 0.2837d/b for segment 1 = 1.948K = 0.226d/b for segment 2 = 5.692K = 0.294


Member no: 2, 20, 38, 56, 74, 92, 110, 128, 17, 35, 53, 71, 89, 107, 125, 143

1.80050.25


= 0.0023


Member no: 4, 22, 40, 58, 76, 94, 112, 130, 16, 34, 52, 70, 88, 106, 124, 142 8,26,44,62,80,98,116,134,12,30,48,66,84,102,120,138

1.5250.25


= 0.002d/b for segment = 6.100K = 0.296


lo/10 + bw

m2

Moment of inertia of end cross girder (Iz) m4

m4

m2


m4

m2


m4

N A

1

2

AN

AN

Member no: 5,23,41,59,77,95,113,131,9,27,45,63,81,99,117,135,13,31,49,67,85,103,121,139

7,25,43,61,79,97,115,133,11,29,47,65,83,101,119,137,15,33,51,69,87,105,123,141

1.963

0.25

Area = 0.491


Moment of inertia (Iz) = 0.003d/b for segment = 7.852K = 0.303


Member no: 6,24,42,60,78,96,114,132,10,28,46,64,82,100,118,136,14,32,50,68,86,104,122,140

0.649 m0.25 beff = [ Cl. 305.15.2 IRC 21 ]

= 0.649 m lo = 1.619 m1.85

0.325


= 0.3132d/b for segment 1 = 2.595K = 0.252d/b for segment 2 = 5.692K = 0.294


m2

m4

m4

lo/5 + bw

m2

Moment of inertia of end intermediate girder (Iz) m4

m4

AN

2

1

AN

Member no: 3, 21, 39, 57, 75, 93, 111, 129

Area = 0.0001Distance of cg from top fibre (y) = - m

Moment of inertia (Iz) = 0.00001d/b for segment = -K = 0.000


d/b Kb = Shorter side b 1.00 0.141d = longer side 1.20 0.166K corresponds to d/b from table. 1.50 0.196

d 2.00 0.2292.25 0.2402.50 0.2493.00 0.2634.00 0.2815.00 0.291

10.0 0.312> 10 0.333

m2

m4

m4

Torsional moment of inertia = K x b3 x d

d/b Kb = Shorter side b 1.00 0.141d = longer side 1.20 0.166K corresponds to d/b from table. 1.50 0.196

d 2.00 0.2292.25 0.2402.50 0.2493.00 0.2634.00 0.2815.00 0.29110.0 0.312> 10 0.333

Torsional moment of inertia = K x b3 x d

Design of section for flexure

Inner girder Outer girder

SECTION At Deff L/4 of span L/2 of span At Deff L/4 of span L/2 of span

DATA M (t.m) 195.53 399.07 526.46 248.02 499.47 621.08 h (m) 2.500 2.500 2.500 2.500 2.500 2.500 bf (m) 2.313 2.313 2.313 3.144 3.144 3.144 df (m) 0.250 0.250 0.250 0.250 0.250 0.250 bw (m) 0.325 0.325 0.325 0.325 0.325 0.325 Ast (m^2) 0.00965 0.00965 0.01206 0.01206 0.01206 0.01608 c (m) 0.096 0.096 0.136 0.104 0.104 0.168 Asc (m^2) 0.00080 0.00040 0.00040 0.00080 0.00040 0.00040 dc (m) 0.064 0.064 0.064 0.064 0.064 0.064 m 10.0 10.0 10.0 10.0 10.0 10.0

RESULTS d (m) 2.404 2.404 2.364 2.396 2.396 2.332

Asf (m^2) 0.00342 0.00339 0.00345 0.00464 0.00461 0.00475 AA (m^2) 0.4970 0.4970 0.4970 0.7046 0.7046 0.7046 A (m) 0.3250 0.3250 0.3250 0.3250 0.3250 0.3250

B (m^2) 1.2014 1.1942 1.2424 1.6649 1.6577 1.7381 C (m^3) -0.5891 -0.5886 -0.6949 -0.7550 -0.7545 -0.9266 n (m) 0.438 0.440 0.495 0.419 0.421 0.488

CC (m^2) 0.1013 0.1019 0.1225 0.1334 0.1341 0.1701 jd (m) 2.290 2.289 2.244 2.285 2.285 2.215

fc (t/m^2) 197 405 515 191 386 462 fs (t/m^2) -8850 -18066 -19455 -8999 -18126 -17440

0.434 0.506 0.544 0.657 d=h-c Asf=(bf*df^2+2*(m-1)*Asc*(df-dc))/(2*m*(d-df)) AA=(bf-bw)*df for As<Asf , else 0 A=bw for As<Asf , else bf B=2*(AA+(m-1)*Asc+m*As) C=-(AA*df+2*(m-1)*Asc*dc+2*m*As*d) n=(-B-sqrt(B^2-4*A*C))/(2*A) CC=(bf-bw)*(min(df,n))^2*(3*n-2*min(df,n)) jd=d-(CC+bw*n^3+6*(m-1)*Asc*(n-dc)*dc)/(6*m*As*(d-n)) fs=-M/(As*jd) fc=-(fs/m)*n/(d-n)

Cracked moment of inertia Ir (m4)

Design for shear force

1.75

[ Cl. 304.7.3 IRC - 21 1987 ]Where

1.14 - 0.7 d >= 0.5

>= 1.0

0.4

SL. Nr.

1 At DeffInner 81.01 1.04 1.235 0.5 1.00 0.20Outer 92.82 1.19 1.549 0.5 1.00 0.20

2Inner 62.30 0.80 1.235 0.5 1.00 0.20Outer 70.01 0.899 1.549 0.5 1.00 0.200

3Inner 24.52 0.319 1.570 0.5 1.00 0.200Outer 25.54 0.34 2.122 0.5 1.03 0.206

SL. Nr.

1 At DeffInner 16.8 2 12 134 125Outer 19.37 2 12 116.8 100

2Inner 13.0 2 12 174.6 170Outer 14.6 2 12 155 140

3Inner 11.5 2 12 196 190Outer 11.4 2 12 199 190

Shear stress tv (N/mm2)< tmax = N/mm2

tc = K1 x K2 x tco

K1 =

K2 = 0.5 + 0.25 r tco =

Section conside

redGirder

locationDesign SF (t)

tv (N/mm2)

r % K1 K2

tc (N/mm2

)

Span (L/4)

Span (L/2)

Section conside

redGirder

location

Reinf required (Asv / Sv) (cm2/m)

Reinf provided (Asv) (cm2/m) Spacing

required (mm)

Spacing Provided (mm)

No of legs

Bar dia (mm)

Span (L/4)

Span (L/2)

DESIGN OF CANTILEVER

A. At intermediate section :

0.2 0.35

1.825 m

0.25

0.5Tyre size Class A

Summary of moments (DL+ SIDL)

Sl. Nr. Items.Loads (t)

1 Crash barrier 1 = 1.000 1.5495 1.550

2 Wearing coat1.2745*0.2 = 0.255 0.6373 0.162

3 Deck slab 0.5*(0.35+0.2)*1.8245* = 1.204 0.829318 0.999Total = 2.46 2.71

Live load moment (Class A)

Effective width (beff) = 1.2a + b1 As per clause 305.16.2 IRC-21,1987Where,a is the distance of load cg from support. = 0.87 mb1 is tyre width + twice the thickness of wearing coat = 0.38 mbeff = 1.429 mAs beff > 1.2m, hence overlaping occers.Therefore, beff =0.5*(1.4294+1.2) = 1.315 m

Impact factor As per clause 211.2 IRC-6,1966 = 1.5

Therefore, live load /m width including impact=(5.7*1.5)/1.315 = 6.50 t/mMoment at the face of support due to live load =6.503*0.8745 = 5.69 t-m

Design bending moment (DL + SIDL + live load ) = 8.40 t-m

= 109.81k = m / (m + r) = 0.29

Lever arm (m)

Moment at the face of

support (t-m)

Moment of resistance M= Qbd2

Where , Q = 0.5*scbc * j*k t/m2

5.7 t0.4m

j = 1 - k/3 = 0.90Required effective depth =(8.398/109.81)^0.5 = 0.28 mDepth provided =0.35-0.04-0.008 = 0.30 m

Required Ast =8.398*10000/20000*0.902*0.30 = 15.41Provide 16 f + 12 f @ 200 c/c at span ( Alt placed)

Provided Ast = 15.71

Distribution reinforcement As per clause 305.18 IRC-21,1987

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live load = 2.25 t-m

Required Ast =0.302*10000/20000*0.902*2.24 = 4.13Provide 10 f @ 200 c/c at top & bottom

Provided Ast = 7.85

B. Near expansion joint :

0.35

1.675 m

0.25

0.5Tyre size Class A

Summary of moments (DL+ SIDL)

Sl. Nr. Items.Loads (t)

1 Crash barrier 1 = 1.000 1.3995 1.40

2 Wearing coat1.1245*0.2 = 0.225 0.5623 0.13

3 Deck slab 0.35*1.675*2.4 = 1.407 0.83725 1.18Total = 2.631 2.70 t-m

Live load moment (Class A)

Effective width (beff) = 0.6a + b1 As per clause 305.16.2 IRC-21,1987Where,a is the distance of load cg from support. = 0.725 mb1 is tyre width + thickness of wearing coat = 0.315 mbeff = 0.750 m

=M/sst*j*d cm2/m

cm2/m

=M/sst*j*d cm2/m

cm2/m

Lever arm (m)

Moment at the face of

support (t-m)

5.7 t0.4m

Impact factor As per clause 211.2 IRC-6,1966 = 1.5

Therefore, live load /m width including impact=(5.7*1.5)/0.7497 = 11.40 t/mMoment at the face of support due to live load =11.405*0.725 = 8.263 t-m

Design bending moment (DL + SIDL + live load ) = 10.966 t-m

= 109.81k = m / (m + r) = 0.293j = 1 - k/3 = 0.902Required effective depth =(10.966/109.81)^0.5 = 0.316 mDepth provided =0.35-0.04-0.01 = 0.305 m

Required Ast =10.966*10^4/20000*0.902*0.30 = 19.9Provide 20 f + 20 f @ 300 c/c at top ( Alt placed)

Provided Ast = 20.9

Distribution reinforcement As per clause 305.18 IRC-21,1987

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live load = 3.02 t-m

Required Ast =0.316*10000/20000*0.293*3.02 = 5.49Provide 12 f @ 200 c/c at top & bottom


Moment of resistance M= Qbd2

Where , Q = 0.5*scbc * j*k t/m2

=M/sst*j*d cm2/m

cm2/m

=M/sst*j*d cm2/m

cm2/m

DESIGN OF CROSS GIRDER

Intermediate cross girder

2.313 2.313 2.313A B C D

From STAAD outputHogging moment (DL+SIDL) = 10.67 t-mMax hogging moment Class 70 R = 23.34 t-mTotal hogging moment = 34.01 t-m

Designed of deep beam [ As per clause 28.2, IS 456-1978 ]

For span AB L = 2.2862 D = 2.1L/D = 1.089 >= 1 for contineous beam

Lever arm Z = 0.2*(2.2862+1.5*2.1) = 1.087 m

Required Ast at top =34.01/1.08724*20000 = 15.64Provide 2 nos 20 f + 2 nos 20 f + 2 nos 20 fDistributed as per clause 28.3.2 (b) IS 456-1978


Minimum Ast at bottom =0.2%bd =0.002*32.5*210 = 14Provide 2 nos 16 f + 2 nos 16 f + 2 nos 16 fat bottom within a depth of (0.25D - 0.05L) = 0.409 mfrom bottom face with a development length of (0.8*35*dia of bar) = 448 mm

Provided Ast = 12.1

Hanging reinforcement [ As per clause 28.3.3, IS 456-1978 ]

Total shear (DL+Live load) From STAAD output = 18.63 t

Required Ast as hanging R/F =18.6*10000/20000 = 9.3

Required Ast per m length =9.3/1.1565 = 8.05 Provide 2 L 10 f @ 150 c/c as vertical reinforcement

Provided Ast = 10.5

Side face reinforcement [As per clause 31.4 IS-456, 1978]

0.1 % of web area on either face with spacing not more then 450 mm.

Required Ast =0.001 *210*32.5 = 6.83

=M/sst*Z cm2

cm2

cm2

cm2

cm2

cm2/m

cm2/m

cm2

End cross girder

### 47.39 t 74.58 t

2.313 2.313

0.65 1.013 0.65 0.65 1.013 0.65

A B C DDF 1.00 0.56 0.44 0.44 0.56 1.00FEM -48.48 0.00 0.00 -7.70 7.70 0.00 0.00 48.48Balance 48.48 4.31 3.39 -3.39 -4.31 -48.48C O 24.24 -1.69 1.69 -24.24 -2.16Balance -12.62 -9.92 9.92 12.62 2.16C O 0.00 4.96 -4.96 1.08 6.31Balance -2.78 -2.18 1.71 2.17 -6.31Total M -48.48 48.48 13.15 -13.15 12.67 -12.67 -48.48 48.48

Max support moment (DL+SIDL) = 48.5 t-mMax span moment (DL+SIDL) = ### t-m

Design of deep beam [ As per clause 28.2, IS 456-1978 ]

For span AB L = 1.9412 D = 2.1L/D = 0.924 < 1 for continuous beam

Lever arm Z = 0.5*1.941 = 0.971 m

For span CD L = 1.9412 D = 2.1L/D = 0.924 < 1 for continuous beam

Lever arm Z = 0.5*1.941 = 0.971 m

The end cross girder is designed as a continuous deep beam for bearing replacement condition, continuous over knife supports at the jack locations. The center line of jacks are taken to be 650 mm from the center line of main girders. The reaction of main

Required Ast for max span M =15.639/0.971*20000 = 8.06

Minimum Ast at bottom =0.2%bd =0.002*32.5*210 = 13.7Provide 2 nos 16 f + 2 nos 16 f + 2 nos 16 fat bottom within a depth of (0.25D - 0.05L) = 0.428 mfrom bottom face with a development length of (0.8*35*dia of bar) = 448 mm

Provided Ast = 12.1

Required Ast for max support =48.477/0.971*20000 = 24.97Provide 2 nos 25 f + 2 nos 25 f + 2 nos 25 fDistributed as per clause 28.3.2 (b) IS 456-1978

Provided Ast = 29.5

Hanging reinforcement [ As per clause 28.3.3, IS 456-1978 ]

Total shear =0.000+74.580+47.390+74.580 = 196.6 t

Required Ast as hanging R/F =196.6*10000/20000 = 98.3

Required Ast per m length =98.3/4.626 = 21.2 Provide 2 L 16 f @ 170 c/c as vertical reinforcement

Provided Ast = 23.7

Side face reinforcement [As per clause 31.4 IS-456, 1978]

0.1 % of web area on either face with spacing not more then 450 mm.

Required Ast =0.001 *210*32.5 = 6.83

=M/sst*Z cm2

cm2

cm2

=M/sst*Z cm2

cm2

cm2

cm2/m

cm2/m

cm2

Design of deck slab excluding cantilever portionSummary of bending moment (t-m)

Mem no Node no Dead load SIDL Temp rise Temp fallLive load BM

+ve BM -ve BM

5 5 -0.712 -1.300 0.510 0.321 2.072 -3.229 0.060 -5.241 -0.466 -0.6556 6 0.236 0.526 2.040 1.286 0.598 -1.524 1.360 -0.762 3.101 2.3477 7 0.281 0.870 2.371 1.495 0.510 -2.981 1.661 -1.830 3.777 2.9018 8 0.276 1.049 2.550 1.608 0.670 -4.003 1.995 -2.678 4.210 3.2689 9 0.281 0.870 2.371 1.495 0.623 -2.982 1.774 -1.831 3.834 2.958

10 10 0.236 0.526 2.040 1.286 0.938 -1.515 1.700 -0.753 3.271 2.51711 11 -0.712 -1.300 0.510 0.321 2.064 -1.669 0.052 -3.681 -0.470 -0.659

Nodes 5 to 11 depict the portion of deck slab excluding the cantilever portions.

Mem no Node no Design Reinforcement

5 5 0.060 -5.241 0.250 0.234 0.17 -14.45 3.00 Refer cantilever design for reinf.6 6 3.101 -0.762 0.250 0.191 8.55 -2.10 3.00 Provide 16 @ 200c/c 7 7 3.777 -1.830 0.400 0.206 6.00 -2.91 4.80 12 @ 200c/c at bot Alt plcd8 8 4.210 -2.678 0.400 0.215 6.69 -4.25 4.80 Ast provided 15.71 cm^2 9 9 3.834 -1.831 0.400 0.207 6.09 -2.91 4.809 9 2.898 -4.114 0.250 0.213 7.99 -11.34 3.00 Provide 12 @ 100c/c at top

10 10 3.271 -0.753 0.250 0.195 9.02 -2.08 3.00 Ast provided 11.30 cm^2 11 11 0.052 -3.681 0.250 0.204 0.14 -10.15 3.00 Refer cantilever design for reinf.

dl+sidl+ +ve LL

BMdl+sidl+ -ve LL BM

dl+sidl+ LL+ temp rise

dl+sidl+LL+temp fall

Design BM (t-m)

Design BM (t-m)

Provided depth (m)

Reqd depth (m)

Reqd Ast (cm^2/m)

Reqd Ast (cm^2/m)

Min Ast (cm^2/m)

DESIGN OF RCC T - GIRDER USING STAAD RESULTS

Documents

Transcript of DESIGN OF RCC T - GIRDER USING STAAD RESULTS