DESIGN OF RCC T - GIRDER USING STAAD RESULTS :(All blue coloured fonts depict inputs) BASIC DESIGN DATA 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 Effective span Leff 24.40 m Angle of skew Ang 0.000 degree Clear carriage way Bcw 7.50 m Spacing of main girder c/c Spmg 2.313 m Spacing of cross girder c/c Spcg 6.10 m Width of crash barier Wkerb 0.550 m Thk of deck slab Df 0.250 m Thk of wearing coat Wc 0.065 m Length of cantilever Lcan 1.987 m Cantilever slab thk at fixed end Dcan1 0.350 m Cantilever slab thk at free end Dcan2 0.200 m No of main girder Nomg 3 m Depth of main girder Dmg 2.500 m Web thk of main girder ( at center ) bwmc 0.325 m Web thk of main girder ( at support ) bwms 0.625 m Length of extra widening ( varrying ) Lwv 0.900 m Length of extra widening ( uniform ) Lwu 0.700 m Top haunch Thw x Thh 0.300 x #### m Bottom haunch Bhw x Bhh 0.150 x #### m Bottom bulb Bbw x Bbh 0.625 x #### m No of cross girder Nocg 3 m Depth of cross girder Dcg 2.100 m Web thk of cross girder bwcg 0.325 m Grade of concrete Cgrade 25 N/mm2 Grade of reinforcement Sgrade 415 N/mm2 Clear cover cov 0.040 m Unit weight of concrete wcon 2.400 t/m3 Weight of wearing course wwc 0.2 t/m2 Weight of crash barrier wrail 1 t/m fc ft m 830 20000 10 t/m2 t/m2

30 Stress in concrete (compression) 31 Stress in steel (tension) 32 Modular ratio

Summary of bending moment for 25m span.From STAAD Analysis Inner girder Type of loading At Deff Span L/4 Span L/2 At Deff Span L/4 Span L/2 1 LANE CLASS 70R (W), wheel over girder 1 LANE CLASS 70R (W), wheel adjacent to girder 1 LANE CLASS 70R (W) 1 LANE CLASS 70R TRACKED 1 LANE CLASS A, wheel 28.32 over girder 1 LANE CLASS A, wheel 29.85 adjacent to girder 1 LANE CLASS A 2nd LANE CLASS A 3 LANE CLASS A 52.44 52.48 127.41 271.66 399.07 82.28 82.96 175.05 351.41 526.46 71.24 73.1 127.19 127.41 173.7 175.05 69.51 95.64 12.82 108.46 139.56 248.02 128.19 201.67 29.39 231.06 268.41 499.47 264.91 356.17 621.08 264.91 167.25 41.16 Outer girder

1L CL A+ 1L 70R (W) 1L CL A+ 1L 70R TRACKED Design live load bending moment73.10 Dead load + SIDL 122.43 Design bending moment (t-m) 195.53

Summary of shear forceFrom STAAD Analysis Inner girder Type of loading At Deff Span L/4 Span L/2 At Deff Span L/4 Span L/2 1 LANE CLASS 70R (W), wheel over girder 1 LANE CLASS 70R (W), wheel adjacent to girder 1 LANE CLASS 70R (W) 31.66 28.97 22.21 20.59 8.15 7.5 22.21 40.09 62.30 12.82 11.45 8.1 7.73 12.82 11.70 24.52 41.27 21.69 4.64 41.27 51.55 92.82 35.03 16.83 4.39 35.03 34.98 70.01 20.15 11.31 2.11 20.15 5.39 25.54 Outer girder

1 LANE CLASS 70R TRACKED 1 LANE CLASS A, wheel 12.41 over girder 1 LANE CLASS A, wheel 11.79 adjacent to girder 1 LANE CLASS A nd 2 LANE CLASS A 3 LANE CLASS A 1L CL A+ 1L 70R (W) 1L CL A+ 1L 70R TRACKED Design live load Shear Force (t) 31.66 Dead load + SIDL 49.35 Design shear force (t) 81.01

Support reactions due to dead load + SIDL

G1 74.58

G2 47.39

G3 74.58

Calculation of section properties LongitudinalsEnd section of Inner girder beff1

y N 2 0.625 Area Distance of cg from top fibre (y) A

beff = = > beff =

lo/5 + bw 5.505 m 2.313 m 2.313 m

[ Cl. 305.15.2 IRC 21 ] lo = 24.4 m [ c/c distance of longitudinal girder]

Moment of inertia of longitudinal girder (IL) d/b for segment 1 K d/b for segment 2 K Torsional moment of inertia = K x b3 x d Span section of Inner girder Member no: 214 to 228 beff 1 y N 3 Area Distance of cg from top fibre (y) Moment of inertia of longitudinal girder (IL) d/b for segment 1 K d/b for segment 2 K d/b for segment 3 K Torsional moment of inertia = K x b3 x d2

= 1.985 m2 = 1.011 m = 1.2365 m4 = 9.252 = 0.309 = 3.600 = 0.274 = 0.162 m4

A

beff = = > beff =

lo/5 + bw 5.205 m 2.313 m 2.313 m

[ Cl. 305.15.2 IRC 21 ] lo = 24.400 m [ c/c distance of longitudinal girder]

= = = = = = = = = =

1.452 m2 0.908 m 4 1.042 m 7.117 0.300 6.154 0.296 2.500 0.249 0.034 m4

Member no: 213,229 Area Moment of inertia of longitudinal girder (IL) Torsional moment of inertia = K x b3 x d (Weighted average) = 2.0049 m2 = 1.297 m4 = 0.1336 m4

End section of outer girder beff 1 N y 2 A beff = = > beff = lo/5 + bw 5.505 m 2.313 m 2.313 m [ Cl. 305.15.2 IRC 21 ] lo = 24.4 m [ c/c distance of longitudinal girder]

0.625 Area Distance of cg from top fibre (y) Moment of inertia of longitudinal girder (IL) d/b for segment 1 K d/b for segment 2 K Torsional moment of inertia = K x b3 x d Span section of outer girder Member no: 180 to 194, 248 to 262 beff 1 y N2 3

= 1.985 m2 = 1.011 m = 1.2365 m4 = 9.252 = 0.309 = 3.600 = 0.274 = 0.162 m4

A

beff = = > beff =

lo/5 + bw [ Cl. 305.15.2 IRC 21 ] 5.205 m lo = 24.400 m 2.313 m 2.313 m

Area Distance of cg from top fibre (y) Moment of inertia of longitudinal girder (Iz) d/b for segment 1 K d/b for segment 2 K d/b for segment 3 K Torsional moment of inertia = K x b3 x d

= = = = = = = = = =

1.452 m 0.908 m

2

1.042 m4 7.117 0.300 6.154 0.296 2.500 0.249 0.034 m4

Member no: 179, 195, 247, 263 Area Moment of inertia of longitudinal girder (IL) Torsional moment of inertia = K x b3 x d (Weighted average) = = 2.005 m 4 1.297 m2

= 0.1336 m4

Member no: 196 to 212, 230 to 246 Area Moment of inertia of longitudinal girder (Iz) Torsional moment of inertia = K x b3 x d = 0.0001 m2 = 0.00001 m4 = 0.00001 m4

Member no: 162 to 178, 264 to 280 0.8305N

0.225 A = 0.187 m2

Area Distance of cg from top fibre (y) Moment of inertia (Iz) Torsional moment of inertia =

= 0.1125 m = 0.0008 m4 = 0.0026 m4

Member no: 145 to 161, 281 to 297 Area Distance of cg from top fibre (y) Moment of inertia (Iz) Torsional moment of inertia = = 0.0001 m2 = 0.00000 m = 0.00001 m4 = 0.00001 m4

TransversalsMember no: 1, 19, 37, 55, 73, 91, 109, 127, 18, 36, 54, 72, 90, 108, 126, 144 0.487 m 0.25 beff = lo/10 + bw [ Cl. 305.15.2 IRC 21 ] 1 = 0.487 m lo = 1.619 m A N1.85 2 0.325 Area Distance of cg from top fibre (y) Moment of inertia of end cross girder (Iz) d/b for segment 1 K d/b for segment 2 K Torsional moment of inertia = K x b3 x d = 0.723 m2 = 0.998 m = 0.2837 m4 = 1.948 = 0.226 = 5.692 = 0.294 = 0.020 m4

Member no: 2, 20, 38, 56, 74, 92, 110, 128, 17, 35, 53, 71, 89, 107, 125, 143 1.8005N

0.25 A = 0.450 m2 = 0.1250 m = 0.0023 m4 = 0.0083 m4

Area Distance of cg from top fibre (y) Moment of inertia (Iz) Torsional moment of inertia =

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 1.525 N 0.25 A = 0.381 m2 = 0.1250 m = 0.002 m4 = 6.100 = 0.296 = 0.007 m4

Area Distance of cg from top fibre (y) Moment of inertia (Iz) d/b for segment K Torsional moment of inertia = K x b3 x d

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 1.963N

0.25 A = = = = = = 0.491 m2 0.125 m 0.003 m4 7.852 0.303 0.009 m4

Area Distance of cg from top fibre (y) Moment of inertia (Iz) d/b for segment K Torsional moment of inertia = K x b3 x d

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 m 0.25 1.85 N 2 0.325 Area Distance of cg from top fibre (y) Moment of inertia of end intermediate girder (Iz) d/b for segment 1 K d/b for segment 2 K Torsional moment of inertia = K x b3 x d = 0.763 m2 = 0.952 m = 0.3132 m4 = 2.595 = 0.252 = 5.692 = 0.294 = 0.021 m4 A1

beff = lo/5 + bw [ Cl. 305.15.2 IRC 21 ] = 0.649 m lo = 1.619 m

Member no: 3, 21, 39, 57, 75, 93, 111, 129 Area Distance of cg from top fibre (y) Moment of inertia (Iz) d/b for segment K Torsional moment of inertia = K x b3 x d = 0.0001 m2 m = = 0.00001 m4 = = 0.000 = 0.00001 m4

Torsional moment of inertia = K x b x d b = Shorter side d = longer side K corresponds to d/b from table. d

3

b

d/b 1.00 1.20 1.50 2.00 2.25 2.50 3.00 4.00 5.00 10.0 > 10

K 0.141 0.166 0.196 0.229 0.240 0.249 0.263 0.281 0.291 0.312 0.333

Torsional moment of inertia = K x b x d b = Shorter side d = longer side K corresponds to d/b from table. d

3

b

d/b 1.00 1.20 1.50 2.00 2.25 2.50 3.00 4.00 5.00 10.0 > 10

K 0.141 0.166 0.196 0.229 0.240 0.249 0.263 0.281 0.291 0.312 0.333

Design of section for flexure Inner girderSECTION DATA M (t.m) h (m) bf (m) df (m) bw (m) Ast (m^2) c (m) Asc (m^2) dc (m) m RESULTS d (m) Asf (m^2) AA (m^2) A (m) B (m^2) C (m^3) n (m) CC (m^2) jd (m) fc (t/m^2) fs (t/m^2) At Deff 195.53 2.500 2.313 0.250 0.325 0.00965 0.096 0.00080 0.064 10.0 2.404 0.00342 0.4970 0.3250 1.2014 -0.5891 0.438 0.1013 2.290 197 -8850 L/4 of span L/2 of span 399.07 2.500 2.313 0.250 0.325 0.00965 0.096 0.00040 0.064 10.0 2.404 0.00339 0.4970 0.3250 1.1942 -0.5886 0.440 0.1019 2.289 405 -18066 526.46 2.500 2.313 0.250 0.325 0.01206 0.136 0.00040 0.064 10.0 2.364 0.00345 0.4970 0.3250 1.2424 -0.6949 0.495 0.1225 2.244 515 -19455 At Deff 248.02 2.500 3.144 0.250 0.325 0.01206 0.104 0.00080 0.064 10.0 2.396 0.00464 0.7046 0.3250 1.6649 -0.7550 0.419 0.1334 2.285 191 -8999

Outer girderL/4 of span L/2 of span 499.47 2.500 3.144 0.250 0.325 0.01206 0.104 0.00040 0.064 10.0 2.396 0.00461 0.7046 0.3250 1.6577 -0.7545 0.421 0.1341 2.285 386 -18126 0.544 621.08 2.500 3.144 0.250 0.325 0.01608 0.168 0.00040 0.064 10.0 2.332 0.00475 0.7046 0.3250 1.7381 -0.9266 0.488 0.1701 2.215 462 -17440 0.657

Cracked moment of inertia Ir (m4)0.434 0.506 d=h-c Asf=(bf*df^2+2*(m-1)*Asc*(df-dc))/(2*m*(d-df)) AA=(bf-bw)*df for As= 1.0

Where K1 = K2 = tco =

1.14 - 0.7 d 0.5 + 0.25 r 0.4

Section consider Girder SL. Nr. ed location Inner 1 At Deff Outer Span Inner 2 (L/4) Outer Span Inner 3 (L/2) Outer

tv Design 2 SF (t) (N/mm ) 81.01 1.04 92.82 1.19 62.30 0.80 70.01 0.899 24.52 0.319 25.54 0.34

r%1.235 1.549 1.235 1.549 1.570 2.122

tcK1 0.5 0.5 0.5 0.5 0.5 0.5 K2 1.00 1.00 1.00 1.00 1.00 1.03 (N/mm2 ) 0.20 0.20 0.20 0.200 0.200 0.206

Reinf provided Reinf required (Asv) (cm2/m) Spacing Spacing Section 2 consider Girder (Asv / Sv) (cm /m) No of Bar dia required Provide legs (mm) SL. Nr. ed location (mm) d (mm) Inner 16.8 2 12 134 125 1 At Deff Outer 19.37 2 12 116.8 100 Span Inner 13.0 2 12 174.6 170 2 (L/4) Outer 14.6 2 12 155 140 Span Inner 11.5 2 12 196 190 3 (L/2) Outer 11.4 2 12 199 190

DESIGN OF CANTILEVERA. At intermediate section :0.4m 5.7 t

0.2

0.35

1.825 m 0.25 0.5 Tyre size

Class A

Summary of moments (DL+ SIDL) Lever arm (m) Moment at the face of support (t-m) 1.5495 1.550 0.6373 0.162 0.82932 0.999 2.71

Sl. Nr. 1 2 3 Total

Items. Loads (t) Crash barrier 1 = 1.000 Wearing coat1.2745*0.2 = 0.255 Deck slab 0.5*(0.35+0.2)*1.8245*2.4 1.204 = = 2.46

Live load moment (Class A) Effective width (beff) = 1.2a + b1 As per clause 305.16.2 Where, a is the distance of load cg from support. b1 is tyre width + twice the thickness of wearing coat beff As beff > 1.2m, hence overlaping occers. Therefore, beff =0.5*(1.4294+1.2) Impact factor As per clause 211.2 IRC-6,1966 IRC-21,1987 = = = = = = = = 0.87 0.38 1.429 1.315 1.5 6.50 5.69 8.40 t/m t-m t-m m m m m

Therefore, live load /m width including impact=(5.7*1.5)/1.315 Moment at the face of support due to live load =6.503*0.8745 Design bending moment (DL + SIDL + live load ) Moment of resistance M= Qbd2 Where , Q = 0.5*scbc * j*k k = m / (m + r) j = 1 - k/3

= 109.81 t/m = 0.29 = 0.90

2

Required effective depth Depth provided Required Ast Provide Provided Ast Distribution reinforcement

=(8.398/109.81)^0.5 =0.35-0.04-0.008

= =

0.28 0.30 15.41 15.71

m m cm /m cm /m2 2

=M/sst*j*d=8.398*10000/20000*0.902*0.302 = 16 f + 12 f @ 200 c/c at span ( Alt placed) = As per clause 305.18 IRC-21,1987

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live load BM] = Required Ast Provide 10 f Provided Ast B. Near expansion joint :0.4m 5.7 t

2.25 4.13 7.85

t-m cm2/m cm /m2

=M/sst*j*d=0.302*10000/20000*0.902*2.248 = @ 200 c/c at top & bottom =

0.35

1.675 m 0.25 0.5 Tyre size

Class A

Summary of moments (DL+ SIDL) Lever arm (m) = 1.000 = 0.225 = 1.407 = 2.631 Moment at the face of support (t-m) 1.3995 1.40 0.5623 0.13 0.83725 1.18 2.70 t-m

Sl. Nr. 1 2 3 Total

Items. Loads (t) Crash barrier 1 Wearing coat1.1245*0.2 Deck slab 0.35*1.675*2.4

Live load moment (Class A) Effective width (beff) = 0.6a + b1 Where, a is the distance of load cg from support. b1 is tyre width + thickness of wearing coat beff Impact factor As per clause 305.16.2 IRC-21,1987 = = = = 0.725 0.315 0.750 1.5 m m m

As per clause 211.2 IRC-6,1966

Therefore, live load /m width including impact=(5.7*1.5)/0.7497 Moment at the face of support due to live load =11.405*0.725 Design bending moment (DL + SIDL + live load ) Moment of resistance M= Qbd2 Where , Q = 0.5*scbc * j*k k = m / (m + r) j = 1 - k/3 Required effective depth Depth provided Required Ast Provide Provided Ast Distribution reinforcement As per clause 305.18 IRC-21,1987

= =

11.40 8.263

t/m t-m

= 10.966 t-m

=(10.966/109.81)^0.5 =0.35-0.04-0.01

2 = 109.81 t/m = 0.293 = 0.902 = 0.316 m = 0.305 m

=M/sst*j*d=10.966*10^4/20000*0.902*0.305 = 20 f + 20 f @ 300 c/c at top ( Alt placed) =

19.9 20.9

cm /m cm2/m

2

Design bending moment [0.2 x (DL + SIDL)BM + 0.3 x live load BM] = Required Ast Provide 12 f Provided Ast =M/sst*j*d=0.316*10000/20000*0.293*3.020 = @ 200 c/c at top & bottom =

3.02 5.49 11.31

t-m cm2/m cm2/m

DESIGN OF CROSS GIRDERIntermediate cross girder

2.313 A B

2.313 C

2.313 D

From STAAD output Hogging moment (DL+SIDL) Max hogging moment Class 70 R Total hogging moment Designed of deep beam [ As per clause 28.2, IS 456-1978 ] For span AB L = 2.286 D = L/D = 1.089 >= Lever arm Z = 0.2*(2.2862+1.5*2.1) 2.1 1 for contineous beam

= 10.67 t-m = 23.34 t-m = 34.01 t-m

= 1.087 m2 = 15.64 cm

Required Ast at top =M/sst*Z =34.01/1.08724*20000 Provide 2 nos 20 f + 2 nos 20 f + 2 nos 20 f Distributed as per clause 28.3.2 (b) IS 456-1978 Provided Ast Minimum Ast at bottom =0.2%bd =0.002*32.5*210 Provide 2 nos 16 f + 2 nos 16 f + 2 nos 16 f at bottom within a depth of (0.25D - 0.05L) from bottom face with a development length of (0.8*35*dia of bar) Provided Ast Hanging reinforcement [ As per clause 28.3.3, IS 456-1978 ] Total shear (DL+Live load) From STAAD output

= 18.85 cm2 = 14 cm2

= 0.409 m = 448 mm = 12.1 cm2

= 18.63 t = 9.3 cm2 = 8.05 cm2/m = 10.5 cm2/m

Required Ast as hanging R/F =18.6*10000/20000 Required Ast per m length =9.3/1.1565 Provide 2 L 10 f @ 150 c/c as vertical reinforcement Provided Ast 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 cm

2

End cross girder 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

74.58 t 2.313

47.39 t 2.313

74.58 t

0.65

1.013 B 0.56 0.00 4.31 24.24 -12.62 0.00 -2.78 13.15

0.65

0.65

1.013

0.65

A DF 1.00 FEM -48.48 0.00 Balance 48.48 CO Balance CO Balance Total M -48.48 48.48

C 0.44 0.44 0.56 -7.70 7.70 0.00 3.39 -3.39 -4.31 -1.69 1.69 -24.24 -9.92 9.92 12.62 4.96 -4.96 1.08 -2.18 1.71 2.17 -13.15 12.67 -12.67

D 1.00 0.00 48.48 -48.48 -2.16 2.16 6.31 -6.31 -48.48 48.48 = 48.5 t-m = ##### t-m

Max support moment (DL+SIDL) Max span moment (DL+SIDL)

Design of deep beam For span AB

[ As per clause 28.2, IS 456-1978 ] 2.1 1 for continuous beam = 0.971 m 2.1 1 for continuous beam = 0.971 m

L = 1.941 D = L/D = 0.924 < Lever arm Z = 0.5*1.941 For span CD L = 1.941 D = L/D = 0.924 < Lever arm Z = 0.5*1.941

Required Ast for max span M =M/sst*Z

=15.639/0.971*20000

= 8.06 cm

2

Minimum Ast at bottom =0.2%bd =0.002*32.5*210 Provide 2 nos 16 f + 2 nos 16 f + 2 nos 16 f at bottom within a depth of (0.25D - 0.05L) from bottom face with a development length of (0.8*35*dia of bar) Provided Ast Required Ast for max support M sst*Z =M/ =48.477/0.971*20000 Provide 2 nos 25 f + 2 nos 25 f + 2 nos 25 f Distributed as per clause 28.3.2 (b) IS 456-1978 Provided Ast Hanging reinforcement [ As per clause 28.3.3, IS 456-1978 ] Total shear =0.000+74.580+47.390+74.580

= 13.7 cm2 = 0.428 m = 448 mm = 12.1 cm2 = 24.97 cm2

= 29.5 cm2

= 196.6 t = 98.3 cm2 = 21.2 cm2/m = 23.7 cm2/m

Required Ast as hanging R/F =196.6*10000/20000 Required Ast per m length =98.3/4.626 Provide 2 L 16 f @ 170 c/c as vertical reinforcement Provided Ast 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 cm2

Design of deck slab excluding cantilever portion Summary of bending moment (t-m)Mem no 5 6 7 8 9 10 11 Node no Dead load 5 6 7 8 9 10 11 -0.712 0.236 0.281 0.276 0.281 0.236 -0.712 SIDL -1.300 0.526 0.870 1.049 0.870 0.526 -1.300 Temp rise Temp fall 0.510 2.040 2.371 2.550 2.371 2.040 0.510 0.321 1.286 1.495 1.608 1.495 1.286 0.321 Live load BM +ve BM -ve BM 2.072 0.598 0.510 0.670 0.623 0.938 2.064 -3.229 -1.524 -2.981 -4.003 -2.982 -1.515 -1.669 dl+sidl+ +ve LL 0.060 1.360 1.661 1.995 1.774 1.700 0.052 dl+sidl+ dl+sidl+ LL+ dl+sidl+LL+t -ve LL BM temp rise emp fall -5.241 -0.762 -1.830 -2.678 -1.831 -0.753 -3.681 -0.466 3.101 3.777 4.210 3.834 3.271 -0.470 -0.655 2.347 2.901 3.268 2.958 2.517 -0.659

Nodes 5 to 11 depict the portion of deck slab excluding the cantilever portions. Design BM Design Provided Reqd Reqd Ast Reqd Ast Min Ast (t-m) BM (t-m) depth (m) depth (m) (cm^2/m) (cm^2/m) (cm^2/m) 0.060 -5.241 0.250 0.234 0.17 -14.45 3.00 3.101 -0.762 0.250 0.191 8.55 -2.10 3.00 3.777 -1.830 0.400 0.206 6.00 -2.91 4.80 4.210 -2.678 0.400 0.215 6.69 -4.25 4.80 3.834 -1.831 0.400 0.207 6.09 -2.91 4.80 2.898 -4.114 0.250 0.213 7.99 -11.34 3.00 3.271 -0.753 0.250 0.195 9.02 -2.08 3.00 0.052 -3.681 0.250 0.204 0.14 -10.15 3.00 Design Reinforcement Refer cantilever design for reinf. Provide 16 @ 200c/c 12 @ 200c/c at bot Alt plcd Ast provided 15.71 cm^2 Provide 12 @ 100c/c at top Ast provided 11.30 cm^2 Refer cantilever design for reinf.

Mem no 5 6 7 8 9 9 10 11

Node no 5 6 7 8 9 9 10 11