Design and Analysis of Flyover

Design and Analysis of Flyover

Bismi M Buhari¹

Assistant Professor Department of Civil Engineering

Musaliar College of Engineering and Technology Pathanamthitta, Kerala,India

Abhijith S2, Aparna B Lal3, Joseph Samuel4, Sarath S5

Student

Department of Civil Engineering Musaliar College of Engineering and Technology

Pathanamthitta, Kerala,India

Abstract- Our project deals with design and analysis of flyover. The manual design of flyover consists of deck slab, longitudinal girder, cross girder, pier, pier cap, abutment, pile cap and pile based on code such as IS: 456-2000 and IRC: 21- 2000. Here the structural analysis is carried out by using STAAD Pro V8i software.

Keywords: Deck Slab; Girder; Pier; Pier Cap; Abutment; Pile; Pile Cap.

1. INDRODUCTION

   Flyover may be referred as an overpass, a h high-level road bridge that crosses over a highway interchange or intersection. Flyover is a grade separated structure connects road at different levels for the purpose of reducing vehicle congestion. This project topic deals with the analysis of flyover at Kayamkulam Haripad road using staad pro v8i and manual designing.

2. OBJECTIVE

   • Analyse need of flyover at the proposed site.

   • Creating model using STAAD Pro V8i.

   • Designing the flyover manually and using STAAD PRO.

3. STRUCTURAL INFORMATION

   The flyover consists of number of spans with columns (piers), deck slab, girders and abutments etc. Total hight of the structure provided as 6.25m. Grade of concrete and steel provided as M35 and Fe415 respectively. Diameter of the pier taken as 2m. Thickness of the deck slab 0.3m provided. Width of the carriage way is 7.5m

   Fig. 2: 3D diagram of the structure

4. LOAD SPECIFICATIONS

Dead load includes self-weight of the column, slab, girders, abutments, etc. Self-weight of the structure was automatically taken from the software. Different kinds of loads may be estimated by using respective Indian Standard Codes of practice.

Fig. 3: Dead load diagram

Fig. 1: Structural diagram Fig.4: Loading diagram of moving load at mid span

Live load moment = 61.52× 3.2 28.222 × 4.36 × 4.36

2 4

= 130.13 kNm

Shear force due to live load

For maximum shear at support the jec class AA

b = La (L a ) + bl

0 lg

A = 4.36 = 2.18

2

l0 = 6.4m

bl= 1.01m

b = 2.728 × 2.18 × (1 2.18

+1.01 = 5.1m

0

0

)

6.4

Fig. 5: Loading diagram of moving load at end span

V MANUAL DESIGN AND STAAD ANALYSIS

1. Design of Deck Slab

   Centre to center span = 20m

   Width of two-way carriage way = 7.5 m (IRC73:1980, clause 6.4

   Width of crash barrier = 450mm (IRC5:1998, Page 50) Wearing coat = 80mm

   Loading = IRC Class AA Loading

   Depth of slab in assumed to be =300mm Wearing coat = 40mm and 20mm channel bar There for, effective depth = 250

   Effective Span =6.4m

   1) Dead load calculation

   Self-weight of deck slab =24× 0.3 = 7.2KN/m2 Self-weight wearing coat =22× 0.08 = 1.76 Total dead load =7.5+1.76 =8.96KN/m2 Maximum dead load of shear force

   Total width = 2.55 +2.05+2.55 =7.15m

   Average Live Load = 700× 1.197 × 4.36 = 26.89 kNm2

   7.15

   R1 = 77.29 kN

   R2 = 40 kN

   Maximum live load for shear force Vq = 77.24kN

   1. Load Calculation

      Design Bending Moment,Mu = 1.35 Mg+1.5 Mq

      = 1.35 × 45.88 + 1.5 × 130.13

      = 257.133 kNm

      Design Shear Force,Vu= 1.35 Vg +1.5 Vq

      = 1.35 × 28.67 + 1.5 × 77.24

      = 154.56 kN

   2. Design

   Mu limit = 0.138 f1u bd2

   d = Mu limit

   0.138 fck b

   d = 257.133 × 106 = 230 < 300mm

   .138×1.5×1000

   Hence safe

   Ast fy Mu = .87 fy Ast d {1 b d f }

   ck

   = wl 6.4 2

   = 8.96 × = 28.67KN

   2

   2

   2 2

   Maximum dead load for Bending Moment =WL

   8

   6.42

   Ast = 2651.88 mm

   Use 20mm Ø bars,

   Ast = × 202 = 314.15 mm2

   4

   = 8.96 ×

   8

   = 45.88 KNM

   Spacing = 1000×314.15 = 100mm

   2651.88

   2) Live load calculation

   Angle = 45°

   Ast

   provided

   1000 × 314.15

   = = 314.15 mm2

   100

   Total width of d = 4.36m

   Moment = 0.3 Mul + .2Mud = 72.3 kNm

   Impact factor (IRC: 6:2016 clause20.6.3 page 50)

   Impact factor = 25 – 2510

   Astd

   Astprovided

   =

   M

   × Moment = 883.2mm2

   Impact factor =1.197

   9.5

   u

   Provide 12mm Ø

   bar @ 110mm c/c

   Effective width load b = × a(1- a ) + b

   • Check for ultimate flexural strength

     Ast fy

     Ast fy

     M = .87 f Ast d {1 } = 298.1 kNm >

     lo l

     lo = 6.4m

     a = lo = 3.2m

     u y b d fck

     Mu

   • Check for ultimate shear strength

     2 V = (. 12 × K (80

     U).33)bd

     bl=0.85+2× 0.08 = 1.01mm

     Width of slab; b = 8.4m

     b = 8.4 = 1.31

     rdc 1 1

     = 197.13 > Vu

     Hence safe

     lo 6.4

     By interpolation, = 2.728

     b = 2.728× 3.2 (1 3.2) + 1.01 = 5.37m

     6.4

     length = 2.075+2.05+2.685 =6.81m

     average live load = 700×1.19 = 28.224kN/m

     6.81×4.36

     K1 = 61.52kN

     K2 =61.52kN

     Fig. 6: Cross section along shorter span

     Fig.7: Cross section along longer span

     Impact allowance 20m span above 9m =10% The live load is placed centrally on the span. Bending moment = (5+4) × 700 = 3185 KNm

     2

     Area of one side = 0.5×1.8×(5+4.1) =8.19 Area of two side = 2× 8.19 =16.38m Mmax = (700/3.6) × 16.38= 3185KNm

     Bending moment including impact and reaction factor for

     outer girder

     = 3185 × 1.197× 0.5536 =2110.6KNm

     Bending moment including impact and reaction factor of inner girder B = 3185 × 1.197 × 0.3333

     =1270.7KNm

     1) Live load shears in girder

     For estimating the maximum Live load shear in the girder, the IRC class AA loads

     Reaction of w2

     Reaction of w1

     on girder B = (350×0.45) = 63 KN

     2.5

     on girder A = (350×2.05) = 287 KN

     2.5

     Fig.8: Bottom span

     Total load on girder B = 350 + 63 =413KN Total load on girder A = 350 – 63 =287KN Maximum reaction in girder B

     Reaction on girder B =(470×18.2) = 375.83 kN

     20

     Girder A

     Reaction on girder A = (287×18.2)

     20

     = 261.17 kN

     Fig. 9: Top span

2. Design of Longitudinal Girder

   1. Reaction factors

      Using Courbons theory,

      • Reaction factor at outer girder(A)

      Maximum Live load shear with impact factor in Inner girder B = 375.83 × 1.197 =449.87 kN Outer girder A = 261.71×1.197 = 313.27 kN

      1. Dead load moments and shear force in main

         girder

         The depth of girder is assumed as 1400mm Depth of rib = 1.4m

         Width of rib = 0.5m

         Weight of rib/m = 24×1.4 ×0.5 = 16.8 kN/m

         The cross girder is assumed to have the same cross section dimensions of the main girder.

         RA = 2W1[1+ 3

         3I×2.5×1.1

         2I×2.52 ]

         Weight of cross girder = 16.8 kN/m

         Reaction on main girder = 16.8 × 2.5 = 42 kN/m

         = 1.107 w1

         • Reaction factor at inner girder (B)

         Reaction from the deck slab = 22.64 kN/m

         Total dead load / m on girder = 22.64 + 16.8 = 39.44 kN/m

         RA= 2W1[1+ 3

         If w = axle load + 700kN

         3I×2.5×1.1

         2I× ]

         Total downward load = (42×3) + (39.44×20) = 914.8 kN

         Ra=Rb= 914.8/2= 457.4 kN

         Where,w1= 0.5w

         p>RA = 1.107 × .5w = 0.5536w [reaction factor =0.5536]

         RB= 2×.5W= 0.3333w

         3

         1. Dead load from slab for girder

            Dead load of the deck slab Parapet railing = 1kN/m

            Wearing coat = (22×1.197×0.08) = 2.1KN/m Deck slab = (24×1.197×0.3) = 8.61KN/m

            Total load = 11.71 KN/m

            Total dead load of deck = (2×11.71) + (8.96×5.3)

            = 70.91 kN/m Considering the girder to be rigid Dead load = 70.91 = 23.64 KN/m

            girder 3

         2. Live load bending moment in girder

            Span of the girder = 20m

            M max at center = (457.4 ×10) – (39.44×10×(10/2)) –

            (42×5) = 2392 kN-m

            Dead load shear at the support = (39.44×20) + 42 + (42/2)

            2

            =457.4 KN

            1. Design moments and shear force

               Table 1: Design Moments and Shear Force

               B.M	DL B.M	LL B.M	TOTAL B.M
               Outer Girder Inner Girder	2392 kNm 2392 kNm	2110.6KNm 1270.7 kNm	4502.6 kNm 3662.7 kNm
               S.F	DL S.F	LL S.F	TOTAL S.F
               Outer Girder Inner Girder	457.4 kN 457.4 kN	313.27 kN 449.87 kN	770.67 kN 907.27 kN

               Max B.M = 4502.6 kN-m Max S.F = 907.27 kN

               Effective depth = 1450mm

               Approximate Lever arm = 1400-100 = 1300mm

               Ast = (4502.6 × 10^6)/(200 × 0.9 × 1300)

               = 19241.88mm2

               Provide 32mm bars

               Live load shear including impact =

               2×271.25 × 1.25

               No of bars = 19241.88/((3.14×322)/4) = 24

            2. Design of section for maximum B.M and S.F

               3

               (reactio

               n on each longitudinal girder) = 226.04KN

               Nominal shear stress, =

               Dead load shear = 51.33 kN Total shear = 226.04+51.33

               = (907.27× 103)/ (500×1400)

               = 1.3 N/mm

               = 100As/bd = (100×19241.88)/ (500×1400) = 2.75

               = 0.6 N/mm2

               >

               Hence safe provide shear reinforcement

               = 277.37 kN

               1. Design moments and shear force Maximum bending moment = 375.05 kN-m Maximum shear force = 277.37 kN

                  Effective depth = 1400 mm Approximate lever arm = 1300mm

                  Assume 2 bars of 32mm Ø is bent up

                  = (sin )

                  Ast = 375.05×10

                  6

                  6

                  200×0.9 ×1300

                  = 1602.782

                  = 45°

                  2 1

                  Provide 32 mm Ø bars

                  No of bars = 1602.78= 2

                  4

                  4

                  = 200×2× × 32

                  × = 227.36KN

                  2

                  ×32 4

                  Balance shear = 907.27 – 227.36 = 679.91 KN

                  Using 10mm Ø, 4 legged stirrups

               2. Design of shear

      3

      3

      Nominal shear stress,

      =

      =

      = 277.37×10

      (277.37×10^3)/ (500×1400) =0.4N/mm2

      = (200×4×(3.14/4)× (102)×1400)/679.91= 130mm

      Provide 10mm Ø 4 legged stirrups at 120 mm c/c

      500×1400

      100 = 100×1602.78) = 0.23

      500×1400

      = 0.224 N/2

      >

      Hence safe provide shear reinforcement Assume 2 bars of 32mm Ø as bent up Us= (in )

      = 45

      Us = 200×2× × 322 × 1 = 227.36 KN

      4 2

      Fig. 10: Detailing of longitudinal girder

      Balance shear = 277.37 – 227.36 = 50.01 KN Using 10mm Ø, 4 legged stirrups

      × ×

      =

3. Design of Cross Girder

   200×4××102×1450 200×4×(3.14 )×(102)×1400)

   )

   )

   = 4 2 = 317mm

   1. Dead load calculation

      256.92

      277.37×103

      Self-weight of the cross girder (same as longitudinal girder size) = 16.8 KN/m Dead load from slab

      = 2×(1/2)×2.5×1.25×8.96 = 28 KN

      Uniformly distributed load = 28/2.5= 14 KN/m Total load on cross girder = 16.8+14 = 30.8 KN/m Assuming the cross girder to be rigid

      Reaction on each cross girder = (30.8×5)/3 = 51.33 kN Dead load shear = 51.33 kN

   2. Live load calculation

      Load coming on the cross girder =271.25KN

      Assuming the cross girder as rigid, reaction on each longitudinal girder is = 2 ×271.25

      Provide 10mm Ø 4 legged stirrups at 300mm c/c

      Fig.11: Detailing of cross girder

      3

      Live load bending moment in

      = 1.25×180.53×1.475

      cluding impact

4. Pier cap

   • Design Procedure

     Design of hammer head portion over circular pier for the

     {for 5m span for c/s is 25%} =332.85KNm

     Dead load bending moment at 1.475m from support

     = 51.33×1.475-(30.8×1.475) × (1.475/2)

     = 42.2 KNm

     Total bending moment

     = 332.85+42.2 = 375.05

     kNm

     following details

     Live load: IRC Class AA Tracked vehicle Materials: M35 grade concrete and Fe 415 steel

   • Data

     Clear projection of cantilever slab = 3200mm Thickness of wearing coat =80 mm

     Materials: M35grade concrete and Fe 415 steel.

     Live load is IRC class AA tracked vehicle.

   • Permissible stresses (IRC: 21): For M35 grade concrete and Fe415 steel.

     = 11.67 N/mm, m =10, = 200N/mm, j = 0.9, Q = 1.93

     1. Calculations of moments

        Total dead load moment,

        Mg = 70.56+105.84+4.2+15.52+30.24+508.032+432.36

        =1166.752 kNm

   • Live load moment

     The live load is IRC class AA tracked vehicle. This is placed with its edge 1200 mm from the kerb.

     Effective width of dispersion perpendicular to span is given by

     = 1.2x + x = 0.1m

     =[0.85+2×0.075] =1m.

     Therefore be= (1.2×0.1) +1=1.12m.

     Live load per meter width including impact =2110.6 KNm

     1. Design moment Design moment, M=1166.752+2110.6 Factored moment = 4916.028 KNm

     2. Reinforcements

        Effective depth required

        2 = maximum bending moment d = [(4916.028×106) ] = 1595.98mm

        (1.93×1000))0.5

        Effective depth required =2200 – 50

        = 2150 mm > 1595.98 mm

        6

        6

        Hence adopted depth is adequate

        Ast =[ (4916.028×10 ) ] = 127022

        (200×0.9×2150)

5. Pier

   Live load: IRC Class AA tracked vehicle Materials: M35 grade concrete and Fe 415 steel

   • Calculation of loads Data

     Effective span of girder bridge = 20 m Clear width of roadway = 7.5 m

     Live load on bridge class AA Height of pier = 6.250 m Height of flood level = 2 m

     Dead load of super structure per span equal to dead load coming outer girders and inner girders.

     Assume weight of bearing, plate etc. as 10 KN Dead load of pier cap:

     The pier cap is divided into two cantilevers and one rectangular section Weight (moment)of two trapezoidal sections = area × unit weight of concrete =322.56 KNm Dead load moment of circular pier

     3.14 × 2

     3.14 × 2

     2

     = [ ] ×6.250×24=471.24 KNm

     4

   • Stresses Due to Live Load

     Reaction due to live load class AA loading including impact = 1.197×70 = 83.79 kN

     Maximum bending moment = 83.79× 0.5 = 41.895 KNm Maximum and minimum stresses at the base due to this load will be

     = 5.74 kN/2 and -0.916 kN/2

   • Stresses Due to Longitudinal Force

   Maximum longitudinal force will occur due to class AA loading = 14 kN

   Moment at base of pier due to this force = 100 × 14 = 140 kNm

   Use 32 mm Ø bars

   Stress at base =140 ×14 = 12.106 kN/2, Assume

   Ast = [( 3.14×32 )

   Ast = [( 3.14×32 )

   2

   ] = 804.242

   4

   No of bars = 12702 = 16

   coefficient of friction as 0.25 at one bearing and 0.22 at

   other bearing Total resistance at one set of bearing with DL

   804.2

   However, provided

   required.

   more for effective reinforcement than

   and LL

   = 0.25 (166.85 + 1.197×700) = 234.212 kN

   Total resistance at other set of bearings due to DL only

   • Top reinforcement:

     Provide 30 numbers of 32mm bars n 2 layers

   • Side reinforcement:

     Provide 10 numbers of 16mm bars on each face equally spaced

   • Inclined reinforcement:

     Provide 10 numbers of 16mm bars on each face equally spaced.

   • Shear reinforcement:

     Provide reinforcement 12mm 4-legged stirrups

     @ 150 mm/cc.

     = 0.22 × 166.85=36.707 KN

     Unbalanced force at top of the pier = 234.212 36.707 =

     197.505 kN

     Moment at base = 197.505 ×1.4 = 13.932 kN/2

     19.847

     • Stresses Due to Wind Loads Exposed height of the structure

     = depth of girder + thickness of slab + height of railing

     = 1.4 + 0.23 + 1.45 = 3.05 m

     Exposed area contributing wind pressure per pier

     = span × height

     = 20 × 3.141 = 62.831 m

     Assume average height of about 5 m

     The wind pressure from table is 76.96kg/2

     1. Wind force on exposed surface = 62.831 ×76.96

        1000

        = 4.83 kN

     2. The design wind load should not be less than 450 kg/m of the loaded chord.

        Hence minimum design wind force = 20 ×450 =9

        1000

        KN

        Fig. 12: Detailing of Pier cap

     3. The wind force for design purposes should be less than 240 kg/m2

     For the unloaded structure = 62.831 ×240 =15.079KN

     d is the clear cover = 60 mm

     By referring chart number 55 of SP 16

     Where P is the percentage of steel reinforcement

     Condition (iii) gives maximu

     1000

     d force. Hence this is

     P = 0.01 × 20 = 0.2

     considered in design.

     m win

     Area of steel = 0.2 × 2000² = 6283.18 mm²

     4

     Assume this to act at mid height,

     3.141 = 1.57m from the top of pier

     2

     Moment at base about Y axis Mw = 15.079 × (10 + 1.57)

     = 174.964 kNm

     The bending takes place about Y axis.

     Maximum stresses at the base at end of straight portion.

     Use 25 mm Ø bar

     2

     2

     Ast =3.14×25 = 490.87 mm²

     4

     Number of bars = 6283.18 = 14

     490.87

     However, provide 32 numbers of 25 mm bars around the

     circular pier.

     = × 3.6 = 174.464

     × 3.6 = 3.345 /2

     Using 10 mm Ø lateral ties

     1

     184.711

     Spacing is the least of the following

     Maximum stress at the edge of the pier = × 5 = 174.464

     1. Least lateral dimension = 2000 mm

        × 5 = 4.722 /2

        • Total Stresses

          Under dry conditions Total stress = DL + LL + WL At end of straight portion,

          184.711

          2. 16×25 = 400 mm

          3. 300 mm

          Hence provide 10 mm bars of lateral ties @ 300 mm c/c

          Maximum = 9.409 + 5.741 + 12.106 + 142.144 – 3.345 +

          4.722 = 177.467 kg/m2

          Minimum = 9.409 – 5.741 – 12.106 – 142.144 – 3.345 –

          4.722 = -158.649 kg/m2

          At end of pier,

          Maximum = 9.409 + 4.722 = 14.131 kg/m2

          Minimum = 9.409 – 4.722 = 4.687 kg/m2

          Under wet conditions, Total stress = DL + LL + WL At end of straight portion,

          Maximum = 9.409 – 6.609 + 5.741 + 12.106 + 142.144 +

          3.345 +1.373 = 1464.673 kg/m2

          Minimum = 9.409 – 6.699 – 0.916 – 12.106 – 142.144 –

          3.345 – 1.373 = -157.174 kg/m2

          At end of pier,

          Maximum = 9.409 – 6.699 + 4.722 – 1.907 = 9.339 kg/m2

          Minimum = 9.409 – 6.699 – 4.722 – 1.907 = -3.919 kg/m2

          Allowable compressive stress in 1:3:6 concrete is 2000 kg/m2& 250 kg/m2 intension. The stresses in pier are within these permissible limits.

          Fig.13: Detailing of pier

6. Design of Pile Cap

1. Data

   Total Load on the column pu =10517.556 KN Diameter of column = 2m

   Pile diameter = 300mm

   Allowable load on each pile = 1350KN Grade of concrete for pile cap = M50

2. Size of pile cap

   Number of piles required, N=10517.5 = 8

   Weight of IRC Class AA tracked vehicle is 700 KN

   Let us provide 8 nos of pile

   1350

   Total load = dead load + live load = 1929.389+700

   = 20629.389 KN

   Total load with impact = 20629.389 ×2= 5258.778kN Factor of safety = 2

   Factored load = 5258.778×2 = 10517.556 kN Factored load =10517.556 kN.

   e is the eccentricity of the wheel load from center.

   e = 1.1m

   Live load = 700×2 = 1400 KN

   Maximum moment = 1400×1.197 = 1675.8 KN Moment with impact = 700×1.197= 837.9 KN Factored moment = 1675.8 ×2.2 = 3686.76KN-m Therefore, factored moment = Mu = 3686.76 KN-m

   1) Non dimensional parameters

   Spacing of piles: (IS 2911:1/2, 2010) S= 4×300=1200mm

3. Overhang portion of the pile cap

   overhang = 2×300=0.6m

   Thus, the total width and breadth of the pile cap is

   Bcap = (2×1.2) +(2×0.6) =3.6 m

   Wcap= Bcap

   Hence let as assume pile cap size 3.6 m x 3.6 m

4. Thickness of pile cap based on shear

Pile reaction; R= 10517.556 = 1314.69 kN

8

One way shear,

The critical section is at a distance d from the face of the column

6

6

PU = 10517.553 × 103

= 0.1

Clear distance c = 2.4-2-0.3=0.1m

Fck d2

35 × 20002

Considering tolerance of 50mm

MU

Fck 3

=3686.76 × 10

35 × 20003

= 0.02

Clear spacing = 100+50=150mm

150+300

Ratio () = 60 = 0.03 r =

2000

300

Where D is the diameter of the circular pier = 2000 mm

=3×R×r = 3× (150+300) × 1.3 × 106

300

=5916105-13146.9d

Assuming percentage of steel 0.25& for M50 grade

Aassumed = 0.0025×b×D = 6156 mm2 Assuming 16 mm bars

concrete =0.38 N/mm2

No of bars required = 6156

= 31

a) One way shear resistance

162

= 0.38×3600×d = 1368d

×( 4 )

Spacing =3600(31×16)2×75 = 98.466 mm

>

D >407.588mm

S = 90 mm c/c

311

therefore d = 408mm Since, computed value of d

e) Development length.

Tbd = 1.9 N/ mm2

c=150 mm c+d=450 mm

=

(0.87×)×16 = 0.475 m

4× 1.6×

d > 487 mm

Hence assumption is correct

1. Two-way shear

   The critical face is at a distance from the face of the

   2

   column.

   Let us assume that

   =c

   2

   Factored shear force.

   2 =8×1314.69 = 10517.52kN

   Two-way shear resistance

   = 1

   Length available; L =0.8 m Ld < L

   f) Transfer of force at column base

   Column to pile cap intraction

   • For column face.

     1 2

     1 2

     = = ×20002 4

     Fckcolumn= 50 N/mm2

     (1) = 1

     2

     Fbr max column = 0.45×50×1 = 22.5 N/2

   • For pile cap face.

   1 =3600×3600

   = 0.25× Fck=1.77 N/2

   2

   2

   = ×20002 4

   2 = Ks×Tc×(4×(2000+d) d) = 14160d+7.08d

   2 >2

   FckPILECAP= 50 N/2

   1 () = 4.125 (limited to 2)

   d=577mm

   > c

   2

   Fbr max pilecap

   = 0.45×

   50×

   2 = 45 N/2

   2

   assumption that full pile reaction is not valid

   2 = R-(R×(1-1)×(1-2))

   Evidently the column face governs Limiting bearing resistance

   Fbr = 9232.5 kN

   =

   = (150+300)

   P= 10517.556 KN

   1 2 300

   2

   2

   =8×1.3×106×(1-(1- 1)2)

   2 >2

   d = 651 mm

2. Around the pile punching shear

   Factored shear force, = 1000kN

   Two-way shear resistance, = 1×1.77×(pi×(300+d)×d)

   >

   Here, Fbr< pu

   Excess force; P = 1285 kN

   If column bars are extended into the footing, then, force per bar

   3

   3

   Fbar = 1285/32= 40.15 kN

   2

   2

   corresponding stress in bar = 40.15 × 10 = 81.79 N/2

   × 25

   4

   development length required to take the stress

   d = 376 mm

   ld = 81.79 ×25

   4 ×1.6 ×1.25 ×1.9

   = 134.5 mm

3. Shear around the pile along ABCD Vcp = 1×1.77× (+300+)+2×600)

   4

   d = 488 mm

   hence the effective depth is considering critical case of two-way shear around the column d = 651 mm

   Available vertical embedment in the footing d=654 mm more than Minimum

   ld= 135 mm

   Astinterface_min=0.005× × 20002= 15.7 ×103 mm2

   4

   Ast interface provided = 32× × 242 =31.41×103 mm2

   D= 651+16+150 =817 mm

   Let us assume D = 850 mm

   hence area provided is reinforcement required.

   4

   well

   over minimum interface

   d = 850-16-150=684 mm

   1. Design of flexural reinforcement

      Mu=3×R(c+)= 1183.221 kNm

      2

      B= 3600 mm

      D = 684 mm

      Ast required=f

      (12 14.598× mu

      1. Transfer of force at pile-to-pile cap

         Factored reaction from pile; Pupile =1000kn Column to pile cap interaction:

         For pile face

         A1= A2 = × 3002

         4

         Fckpile = 50 N/mm2

         A1

         ck×b×d×

         2×fy

         fck×b×d2

         root(

         F

         )= 1

         A2

         = 0.45 50

         1 = 22.5 N/ 2

         = 4871.48 mm2

         Amin = 0.0012×b×D = 2954.88 mm2

         br max pile

         × × mm

      2. Pile cap face

1

1

A =

4

× 12002

2

2

A = × 3002

• Earth pressure

  4 Lateral earth pressure k = 0.408

  FckPILECAP= 50 N/mm2

  root(A1) = 4 (limited to 2)

  A2

  Fbr max pilecap = 0.45×50×2 = 45 N/mm2

  a

  Horizontal component of earth pressure = total earth pressure × cos ( + )

  Total Earth pressure = 1 × × p × cos × k

  Evidently the column face governs 2 a

  Limiting bearing resistance Fbr = 1590.43 kN

  P= 10517.556 KN

  Here, Fbr>pu Hence Safe

  Fig.14: Detailing Pile and pile cap

  E. Design of Abutment Density of soil = 16.42 kN/m3 Coefficient of friction = 0.6

  Angle of repose = 30°

  Live load on bridge = IRC Class AA load

  Angle of friction between soil and concrete = 18° Longitudinal girder = 3 number of 1.4 m × 0.5 m Deck slab = 0.3 m depth

  =0.5×16.42×4.42 ×cos15.25×0.408 = 62.56 KN

  Horizontal component = 62.56 cos (16.42 + 15.255)

  = 53.24 kN

  Vertical component = 62.56 sin (16.42 + 15.255)

  =32.85KN

  R = V2 + H2 = ((32.852)+(53. 852)) × 0.5

  =63.07KN

  • Overturning

    Earth pressure is acting at 0.42 of height from base

    = 0.42 × 4.4 = 1.848 m

    Moments due to overturning = 1.848×53.24= 98.39 kNm Restoring moments= (0.5×1.2×4.4×24×3.0)

    +(0.6×4.4×24×2.3) + (0.8×3.7×24×1.6)

    +(0.5×1.2×3.7×24×0.8) +(858.68×1.6) =1865.984 kNm

    Factor of safety against overturning =1865.984/98.93

    = 18.86

  • Sliding

    Factor of safety = (0.6×858.68)/53.24 = 9.6

  • Base pressure

Distance of the resultant from the toe.

x = (restoring moment moment due to overturning)

resultant

= 1865.98698.39 = 2.058m

858.68

Eccentricity of the resultant from the centre of the base. e= b x = 0.022

2

Maximum pressure, P =858.68 × (1 + 6×.022 ) =233.82k

Span of the bridge =20 m

= 233.82km/m2

max

3.8

3.8

Self- weight,

W1 = ((0.5×1.24×.4) +(0.5×1.2×3.7) +(0.6×4.4)

+(0.8×3.7)) ×24= 251.04 kN

• Dead load W2 = L1 + L2

L1 = Load from longitudinal girder

= 3 × 1.4 × 0.5 × 20 × 24 = 1008 kN

L2 = Load from slab and wearing course

= (0.3 × 8.4× 20 × 24) + (0.8 × 7.5 × 20 × 22) = 3849.6KN

W2 = 4857.6 kN

Considering dead load (super structure) shared by an abutment and a pillar

= 4857 = 2428.8 kN

Pmin= (858.68/3.8) (1-((6×0.022)/3.8) = 218.12 km/m2

VI CONCLUSION

• The proposed project could help rectify the traffic conjunction problems and improve safe driving.

• Project is designed manually and using STAAD Pro V8i.

• Amount of steel provided for the structure is economic.

• Structure is designed on the basis of IRC class AA loading.

REFERENCE

1. S. Chandra; P.K. Sikdar; Factors Affecting Pcu In Mixed Traffic

   2

   Dead load per meter span of abutment W2=

   2428 =

   84

   Situations on Urban Roads P. 40-50: ILL.; Includes Bibliographical References (P. 49). Road & Transport Research.

   Live load reaction = W × L

   = 289.14 KN

   Vol. 9, No. 3 (Sept. 2000)

2. Rasheed Saleem Abed; Experience on Using Total Station

   For Class AA tracked vehicle, W = 350 Kn

   R 20Wheel load

   Surveying for Mapping and Contouring ISSN 0976 6308 (Print) ISSN 0976 6316(Volume 4, Issue 3, May – June (2013), pp. 155-167 IAEME: www.iaeme.com/ijciet.asp Journal Impact

   A=350 × (

   R

   20 )

   20 3.6

   Factor (2013): 5.3277

3. Narabodee Salatoom; Pichai Taneerananon; A Study of a

   A=350 × ( 2 )

   20

   = 318 kN

   Total load carrying on the abutment W = W1 + W2 + RA

   = 251.04+ 289.14 + 318.5 = 858.68 kN

   Flyover Bridge – Improved Intersection ENGINEERING JOURNAL Volume 19 Issue 1 Received 28 May 2014 Accepted

   8 September 2014 Published 30 January 2015

4. A Nithin Chandra; An Overview Towards Flyover Construction for Lessening Congestion of Traffic (IJITR) INTERNATIONAL JOURNAL OF INNOVATIVE TECHNOLOGY AND

   RESEARCH Volume No.4, Issue No.3, April May 2016, 2934

   2937.

5. IRC 21-2000 Standard specification and code for practise for road bridges section II

6. IRC 5-2000 Standard specification and code for practise for road bridges section I.

7. IRC 6-2000 Standard specification and code for practise for road bridges section II.

8. IS 456- 2000 Plain and reinforcement concrete code of practise