Design and Analysis of Flyover

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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

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