Design and Analysis of A Bus Terminal Building

Download Full-Text PDF Cite this Publication

Text Only Version

Design and Analysis of A Bus Terminal Building

Bismi M Buhari1, Alan Varghese2 , Bibin Babu3, Gokulkrishnan R4, Raveena R Nair5

1Asst. Professor,

2345 Student Civil Department,

Musaliar College of Engineering & Technology, Pathanamthitta, Kerala, India

Abstract- The design and analysis of bus terminal building is carried out. The manual designing of buildings include the design of foundation, design of column, beam, slab, stair etc using a set of procedures and building codes such as IS 456. Here the structural analysis using STAAD Pro V8i SS6 is carried out.

Keywords: Deflection; Bending moment; Shear force; Assembly building.

  1. INTRODUCTION

    In all spheres of human life buildings are always necessary to satisfy human need. This study mainly focuses on the analysis of an assembly building i.e Municipal bus terminal. Pandalam using STAAD Pro V8i SS6 and manual designing. In this study each structural part is analysed. The bending moment, deflection, shear force etc are analyzed.

  2. OBJECTIVES

    • Generate structural frame work

    • Creating model using STAAD PRO V8i.

    • Carry out the structural analysis and manual design, Thus ensure the structural stability.

  3. BUILDING INFORMATIONS

    The proposed bus terminal is a two storey building. It consist of shops, office, staff room, toilet complex etc. there are two openings provided for the entry and exit of buses.

    Fig.1 Plan of the analyzing structure

    Fig.2 Structural analysis diagram

    Fig.3 3D view of structure

  4. SPECIFYING LOADS

    These are self-weights of the structure to be designed. The dimensions of the cross section are to be assumed initially which enable to estimate the dead load from the known unit weights of the structure. The values of the unit weights of the materials are specified in IS 875:1987(Part-I). Dead load includes self-weight of columns, beams, slabs, brick walls, floor finish etc. The self-weight of the columns and beams were taken automatically by the software. The dead loads on the building are as follows. Here in STAAD the load given to the structure are Dead load ( Self weight and UDL due to brick of 18.6 KN/m), Live load of -3 KN/ m2 and load combinations.

    Assume = 0.9

    58

    Cu = = 29kN/m2 2

    Fig.4 Self weight given into the structure

    Fig. 5 UDL given into the structure

    Therefore, ultimate capacity of a single pile , Qu

    = CuNcAp+ CuAs

    = 965.726kN

    Spacing of pile =2hp =2 × 500 = 1000mm Dimension of column,B × D = 0.45 × 0.45m Factored load =1704.6kN

    Factored moment =101.085kN Safe pile load capacity=965.7kN

    1. Pile Cap Dimension

      Length and breadth of pile cap=1000 + (2 × 250) + (150 × 2)=1.8m

      Depth of pile cap=2hp=1000mm=1m

    2. Check for pile load capacity

      Self weight of pile cap=(1.8 × 1.8 × 1 × 25) × 1.5 = 121.5

      Factored load from column, Pu=1704.6kN Total factored load=121.5+1704.6=1826kN No. of pile=4

      y-ordinate of pile cap=0.5m Mx=101.08kNm

      Pu Compressive load in pile about X axis = +

      n

      = 953.38kN

      Mxy

      y2

      Design working load = 953.38 = 635.58kN < 965.7,

      2

      Hence safe.

      Fig.6 Live Load given into the structure

    3. Bending moment

      (1 0.45)

  5. MANUAL DESIGN AND STAAD ANALYSIS

Mu = 953.38

= 262.18kNm

2

  1. Design of pile cap

    1. Check for effective depth

      Mu = 0.138dfckd2 = 262.2kNm

      262.2 × 106

      d=410mm

      d2 =

      0.13 × 25 × 450

      drequired < dprovided Hence checked.

    2. Check for punching shear

      Punching shear at a distance d/2 from face of column=1704.6kN

      Perimeter of critical section=2(1800+922)=5444mm

      1704.6 × 103

      Fig.7 Pile cap arrangement.

      Punching shear =

      = 0.399N/mm2 5444 × 922

      Datas

      Number of piles = 4

      Pile Diameter = 500 mm Spacing of piles 2 hp = 2 x 400 = 800 m Concrete Mix = M20

      Steel Grade = Fe 415

      Length of pile = 23m. Soil type is weak clayey soil

      Assume unconfined compressive strength as0.058N/mm2

      Allowable shear stress for M25=0.25fck = 1.25N/ mm2 > 0.339/mm2

      Hence safe.

    3. Main reinforcement

      Mu=262.2 KNm

      Mu

      k = = 0.17135

      bd2

      0.12

      i.e UCS =58kN/mm2

      Ast min =

      100

      × 1800 × 922 = 1991.52mm2

      No. of bars =

      1991.52

      = 9.905 10 bars

      201.1

      Column no.380

      Therefore provide 16mm, 10 bars at bottom. Reinforcement at top,

      Minimum Ast =1991.52mm2

      Therefore provide 16mm, 10 bars at top.

    4. Check for one way shear

      Maximum shear force at facing of column=958.38kN Shear stress=0.577N/mm2

      For Pt=0.2%,

      c from IS456 = 0.33N/mm2

      Shear to be carried by stirrups,

      Vu = (0.577 0.33) × 1800 × 922 × 103 = 409.92kN

      Vu

      = 4.446kN/m

      d

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

      spacing.

      Grade of Concrete = M30 Grade of Steel = Fe415

      Characteristic compressive strength of concrete , fck ( N/mm2 ) = 30

      Characteristic yield strength of steel , fy ( N/mm2 ) = 415

      Unit weight of concrete , c( kN/m3 ) = 25 Partial safety factor for concrete = 1.5 Exposure condition = Mild

      Nominal Cover to exposure condition ( mm ) = 40 Assumed effective cover all around, d' ( mm ) =40

      1. Dimensions of the Column Breadth of the column B (mm) = 450 Depth of the Column D (mm) = 450 Least lateral dimension = 450 mm

      2. Design Factors

        Factored load, Pu =2208kN

        Factored moment acting parallel to the larger dimension , Mux = 36kNm

        Factored moment acting parallel to the shorter dimension, Muy =48 kNm

        Assume percentage of steel =1.5 %

      3. Moment in X-X Direction

        d 40

        =

        D 450

        = 0.09 0.1

        Pu

        fckbD

        2208 × 1000

        =

        25 × 450 × 450

        = 0.436 0.44

        p

        =

        fck

        Mu

        1.5

        25

        = 0.06

        fckbD2 = 0.07

        Mux = 0.07 × 25 × 450 × 4502

        = 159.46 × 106Nmm

      4. Moment inY-Y Direction

        d 40

        Fig. 8 Pile cap detailing

        = = 0.09 0.1

        D 450

        Pu 2208 × 1000

  2. Design of column

    = = 0.436 0.44

    fckbD 25 × 450 × 450

    p

    fck

    1.5

    =

    25

    = 0.06

    Mu

    fckbD2 = 0.07

    Muy = 0.07 × 25 × 450 × 4502

    = 159.46 × 106Nmm

      1. Calculation of Puz

        p = 1.5%

        100Asc

        = 1.5

        bD

        Asc

        1.5 × 450 × 450

        = = 3037.5mm2

        100

        Fig. 9 Column no. 380 in the structure

        Ac = Ag Asc = 199462.5mm2

        Puz = 0.45fckAc + 0.75fyAsc = 3189.37kN

        Pu

        Puz

        2208.36

        =

        3189.37

        = 0.69

        Check

        From IS 456, value of = 1.82

        Mux n

        ( )

        Mux

        Muy n

        + ( ) 1

        Muy

        36.03 × 106

        1.82

        47.95 × 106

        1.82

        ( )

        159.46 × 106

        + ( )

        159.46 × 106

        < 1

        = 0.1789

        Fig.12 Shear bending diagram of Member 380 in the structure

        Therefore , the column is safe. 1.5% of steel is sufficient.

      2. Calculation of reinforcement

    p = 1.5%

    100Ast

    = 1.5

    bD

    Ast

    1.5 × 450 × 450

    =

    100

    = 3037.5mm2

    No. of bars = Ast

    A

    Assume 16 mm bars.

    = 15.11 16 bars

    Provide 16 bars of 16 mm diameter.

    The STAAD digrams are the following,

    Fig.10 Dimensional details of Member 380 in the structure

    Fig.13 Deflection diagram of Member 380 in the structure

  3. Design of beam

    Beam no.4

    Fig. 14 Beam no. 4 in the structure

    Clear span=4.2m Live load=12kN/m fck=25N/mm2

    fy =415N/mm2

    1) ( )

    xu 700

    Fig.11 Reinforcement details of Member 380 in the structure

    =

    d 1100 + 0.87fy

    = 0.48

    Assumptions of cross sectional dimensions

    l

    = 15

    d

    d=280mm 300mm

    D=d+50=350mm

    4200

    = d

    15

    D

    = 2

    b

    b=175mm 200mm

    1. Load calculation

      Self weight = b × D × con = 1.75N/mm

      Dead load =1.75N/mm

      Design load =(DL+LL)con=20.63N/mm

    2. Design bending moment

      Wl2

      BM =

      8

      = 52.22kNm

    3. Calculation of limiting value of moment

      (M )

      xu xu

      = 0.36 [1 0.42 ] f

      bd2

      u lim

      BM< (Mu)lim

      d

      =62.08 kNm

      d ck

      Fig.16 Reinforcement details of Member 4 in the structure

      Therefore under reinforced.

    4. Reinforcement calculation

      fyAst

      f

      f

      Mu = 0.87fyAstd [1 ]

      ckbd

      Ast=572.60mm2 Assume 16mm bars.

      total area

      No. of bars = = 2.84 4 bars area of one bar

    5. Design for shear

      Fig.17 Shear bending diagram of Member 4 in the structure

      Wle

      Vu = = 46417.5N/mm2 2

      Vu

      v = = 0.77N/mm2 bd

    6. Design shear strength

      100Ast

      Percentage of steel =

      c = 0.64N/mm2

      c max = 3.1N/mm2

      = 1.0048

      bd

      Fig.18 Deflection diagram of Member 4 in the structure

  4. Design of slab

    Fig.15 Dimensional details of Member 4 in the structure

    Slab no.707 Size 4.2 × 2.4m

    Fig.19 Slab 707 in the structure.

    M25 and Fe415 are used.

    1. Type of slab

      ly 4.2

      = = 1.75

      lx 2.4

      Therefore it is a two way slab.

    2. Depth of slab

      Ast = 234.422mm2 1000A

      l s = A

      = 20

      d

      st

      =334.994mm

      4200

      d =

      20 × 1.5

      = 136.66 150mm

      330mm

      Provide 10mm dia bars at 330mm c/c.

      Effective span = clear span + d

      = 4200 + 150

      1. Check for deflection

        = 4350mm

        Depth ,D = 150 + 20 = 170mm

        ( l )

        d min

        = ( l )

        d basic

        × kt

        × kf

        × ks

        1. Design load Design load= 1.5(DL + LL)

          = 1.5(4.25 + 3)

          = 10.875 kNm2

          l

          ( )

          d max

          = 20 × 1.7

          =34

          4350

          =

          150

          = 29 < 34

        2. M along X and Y

      T he slab is restrained, From IS 456, = 1.75

      Therefore the slab is safe.

      9) Check for shear

      wl

      =0.1

      Vu =

      = 23.65kN

      2

      =0.056

      Vu 23.65 × 103

      Mx =x wlx2

      = 0.1 × 10.875 × 4.3502

      = 20.577kNm

      My =y wly2

      = 0.056 × 10.875 × 4.352

      = 11.523kNm

      Shear Force= wl

      2

      = 10.875×4.35

      2

      = 23.65kN

      1. Check for depth

        Mx

        d = 0.138fckb

        20.577 × 106

        u = bd = 1000 × 150

        =0.157

        Percentage of steel = 100Ast

        bd

  5. Design of stairs

= 0.264.

=

0.138 × 25 × 1000

= 77.229mm

Hence , drequired < dprovided.

  1. Calculation of Ast (short san)

    (Mu)lim

    = 0.87fyAst

    Astfy

    d (1 )

    bd×fck

    Ast = 397.27mm2

    Assume 10mm dia bars.

    1000A

    Fig.20 Stair case details.

    s =

    Ast

    = 197mm

    195mm

    Provide 10mm dia bars at 195mm c/c along shorter span.

  2. Calculation of Ast (long san)

d = effective depth (0.5 dia of mai bar

+ 0.5 dia of main bar)

Material Constants Concrete, f ck = 25 N/mm2 Steel, f y = 415 N/mm2

1) Dimensioning

4.35

Height of each flight =

2

= 2.175m

=140mm

Astfy

Let the tread of steps be 300 mm.

Rise of stair =0.145 m

bd×f

bd×f

(Mu)lim = 0.87fyAstd (1 )

ck

Width of stair =1.9 m

Effective span, le =8.4 m

Let the thickness of waist slab be 150 mm. Use 12 mm bars.

Assume, clear cover= 25 mm Effective depth =119mm

Hence, Provide 8 mm diameter bars at 350 mm c/c.

6) Check for shear

(As per IS 456:2000, Clause 40) Maximum Shear force, V= 49.626 kN

  1. Loads on landing slab

    Nominal shear stress, v

    Vu

    = = 0.417N/mm2

    bd

    Self weight of Slab = 0.15 × 25 = 3.75 kN/m2 Finishes = 1.25 kN/m2

    Live Load on Slab = 4 kN/m2 Total = 9 kN/m2

    Factored load = 1.5 × 9 = 13.50 kN/m2

  2. Loads on waist slab

    Dead load of waist slab = thickness of waist slab×25×R2+T2

    T

    = 0.15×25×0.1452+0.32

    0.3

    = 4.165 kN/m2

    Self weight of step = 0.5 × R × 25 = 1.8125 kN/m2

    Floor finish = 1.25 kN/m2

    As per IS: 875(Part 2),1987, Table-1. Live load = 4kN/m2

    Total service load =11.228 kN/m2 Consider 1 m width of waist slab.

    Total service load / m run = 11.228 × 1

    = 11.228kN/m

    Factored load, Wu = 1.5 × 11.228 = 16.842kN/m As per the UDL of the landing slab and waist slab, Reaction RA =45.414kN

    Reaction RB = 49.626 kN

  3. Bending moment

    To get maximum Bending Moment, take Shear Force at x distance from support B=0. Thus obtained X is 2.95 m.

    Maximum moment at X =2.95m, Mu =73.1187kNm

  4. Reinforcement calculations.

Mu

= 5.16 N/mm2

bd2

Percentage of steel, pt =1.19% (From SP16, Table 3)

ptbd

Therefore, Ast = = 1416.1 mm2

100

Minimum steel=0.12% cross sectional area

= 142.8mm2

Use 12mm Ø bars,

1000A

Max. value of shear stress, max=3.1N/mm2 To get design shear strength of concrete, 100As

= 0.09 < 0.15

bd2

From IS 456: 2000, Table 19

c = 0.64 N/mm2, v < c < c max

So, shear reinforcement is not required.

VI . 3D MODEL OF BUS TERMINAL

The 3D model of the bus terminal building is prepared. The model is prepared as per the plan and other details.

Fig.21 Elevation of model.

Fig. 22 Model of bus terminal

Spacing =

Ast

= 79.82mm

Provide 12mm Ø bars at 80 mm c/c.

Maximum Spacing = 3d = 3 × 119

=357mm or 300mm (less)

Hence, provide reinforcement of 12 mm Ø bars at 80 mm c/c

Distribution steel= 0.12% cross sectional area=142.8mm2 Provide 8mm Ø bars.

1000A

Spacing = = 351.82mm Ast

Maximum Spacing = 4d = 476mm

VII . CONCLUSION

  • The structure is completely analysed using STAAD PRO v8i.

  • The structural components of the building are safe in shear and flexure.

  • Amount of steel provided for the structure is economic.

  • Proposed sizes of the elements can be used in the structure.

REFERENCES

[1]. Mahesh Ram Patel; R.C. Singh; Analysis of a tall structure using staad pro providing different wind intensities as per 875 part-iii, IJERST , International Journal of Engineering Sciences & Research Technology, May 2017.

[2]. Aman; Manjunath Nalwadgi ; Vishal T; Gajendra Analysis and design of multistorey building by using STAAD Pro IS 456-2000,International Research Journal of Engineering and Technology (IRJET),Volume: 03 Issue: 06 , June-2016.

[3]. Anoop A ; Fousiya Hussian; Neeraja R; Rahul Chandran; Shabina S; Varsha S; Anjali A; Planning, analysis and design of multi storied building by staad.pro.v8i, International Journal of Scientific & Engineering Research, Volume 7, Issue 4, April-2016.

[4]. IS: 875 (Part 1) 1987( Dead Load ).

[5]. IS: 875 (Part 2) 1987( Live Load ).

[6]. Design Aids For Reinforced Concrete to IS: 456-1978 []. IS. 456: 2000

Leave a Reply

Your email address will not be published. Required fields are marked *