**Open Access**-
**Authors :**Bibin Babu , Alan Varghese , Raveena R Nair , Gokulkrishnan R, Bismi M Buhari -
**Paper ID :**IJERTV9IS030553 -
**Volume & Issue :**Volume 09, Issue 03 (March 2020) -
**Published (First Online):**03-04-2020 -
**ISSN (Online) :**2278-0181 -
**Publisher Name :**IJERT -
**License:**This work is licensed under a Creative Commons Attribution 4.0 International License

#### 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.

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.

OBJECTIVES

Generate structural frame work

Creating model using STAAD PRO V8i.

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

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

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 Ã— 500 = 1000mm Dimension of column,B Ã— D = 0.45 Ã— 0.45m Factored load =1704.6kN

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

Pile Cap Dimension

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

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

Check for pile load capacity

Self weight of pile cap=(1.8 Ã— 1.8 Ã— 1 Ã— 25) Ã— 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

Bending moment

(1 0.45)

MANUAL DESIGN AND STAAD ANALYSIS

Mu = 953.38

= 262.18kNm

2

Design of pile cap

Check for effective depth

Mu = 0.138dfckd2 = 262.2kNm

262.2 Ã— 106

d=410mm

d2 =

0.13 Ã— 25 Ã— 450

drequired < dprovided Hence checked.

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 Ã— 103

Fig.7 Pile cap arrangement.

Punching shear =

= 0.399N/mm2 5444 Ã— 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.

Main reinforcement

Mu=262.2 KNm

Mu

k = = 0.17135

bd2

0.12

i.e UCS =58kN/mm2

Ast min =

100

Ã— 1800 Ã— 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.

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) Ã— 1800 Ã— 922 Ã— 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

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

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 %

Moment in X-X Direction

d 40

=

D 450

= 0.09 0.1

Pu

fckbD

2208 Ã— 1000

=

25 Ã— 450 Ã— 450

= 0.436 0.44

p

=

fck

Mu

1.5

25

= 0.06

fckbD2 = 0.07

Mux = 0.07 Ã— 25 Ã— 450 Ã— 4502

= 159.46 Ã— 106Nmm

Moment inY-Y Direction

d 40

Fig. 8 Pile cap detailing

= = 0.09 0.1

D 450

Pu 2208 Ã— 1000

Design of column

= = 0.436 0.44

fckbD 25 Ã— 450 Ã— 450

p

fck

1.5

=

25

= 0.06

Mu

fckbD2 = 0.07

Muy = 0.07 Ã— 25 Ã— 450 Ã— 4502

= 159.46 Ã— 106Nmm

Calculation of Puz

p = 1.5%

100Asc

= 1.5

bD

Asc

1.5 Ã— 450 Ã— 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 Ã— 106

1.82

47.95 Ã— 106

1.82

( )

159.46 Ã— 106

+ ( )

159.46 Ã— 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.

Calculation of reinforcement

p = 1.5%

100Ast

= 1.5

bD

Ast

1.5 Ã— 450 Ã— 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

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

Load calculation

Self weight = b Ã— D Ã— con = 1.75N/mm

Dead load =1.75N/mm

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

Design bending moment

Wl2

BM =

8

= 52.22kNm

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.

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

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

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

Design of slab

Fig.15 Dimensional details of Member 4 in the structure

Slab no.707 Size 4.2 Ã— 2.4m

Fig.19 Slab 707 in the structure.

M25 and Fe415 are used.

Type of slab

ly 4.2

= = 1.75

lx 2.4

Therefore it is a two way slab.

Depth of slab

Ast = 234.422mm2 1000A

l s = A

= 20

d

st

=334.994mm

4200

d =

20 Ã— 1.5

= 136.66 150mm

330mm

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

Effective span = clear span + d

= 4200 + 150

Check for deflection

= 4350mm

Depth ,D = 150 + 20 = 170mm

( l )

d min

= ( l )

d basic

Ã— kt

Ã— kf

Ã— ks

Design load Design load= 1.5(DL + LL)

= 1.5(4.25 + 3)

= 10.875 kNm2

l

( )

d max

= 20 Ã— 1.7

=34

4350

=

150

= 29 < 34

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 Ã— 103

Mx =x wlx2

= 0.1 Ã— 10.875 Ã— 4.3502

= 20.577kNm

My =y wly2

= 0.056 Ã— 10.875 Ã— 4.352

= 11.523kNm

Shear Force= wl

2

= 10.875Ã—4.35

2

= 23.65kN

Check for depth

Mx

d = 0.138fckb

20.577 Ã— 106

u = bd = 1000 Ã— 150

=0.157

Percentage of steel = 100Ast

bd

Design of stairs

= 0.264.

=

0.138 Ã— 25 Ã— 1000

= 77.229mm

Hence , drequired < dprovided.

Calculation of Ast (short san)

(Mu)lim

= 0.87fyAst

Astfy

d (1 )

bdÃ—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.

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

bdÃ—f

bdÃ—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

Loads on landing slab

Nominal shear stress, v

Vu

= = 0.417N/mm2

bd

Self weight of Slab = 0.15 Ã— 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 Ã— 9 = 13.50 kN/m2

Loads on waist slab

Dead load of waist slab = thickness of waist slabÃ—25Ã—R2+T2

T

= 0.15Ã—25Ã—0.1452+0.32

0.3

= 4.165 kN/m2

Self weight of step = 0.5 Ã— R Ã— 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 Ã— 1

= 11.228kN/m

Factored load, Wu = 1.5 Ã— 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

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

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 Ã˜ 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 Ã˜ bars at 80 mm c/c.

Maximum Spacing = 3d = 3 Ã— 119

=357mm or 300mm (less)

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

Distribution steel= 0.12% cross sectional area=142.8mm2 Provide 8mm Ã˜ 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