GSPL203-73-1517 R-0

MAYTAS NAFTOGASBUD JV GSPL203-73-1517 Rev 0

BOUNDARY WALL DESIGN CALCULATIONS - SV2 STATION AT CHELA Page 2 of 29

CONTENTS PAGE NO1.0 INTRODUCTION........................................................................................................................................................... 3

1.1 GENERAL ....................................................................................................................................................................... 31.2 DESIGN PHILOSOPHY ..................................................................................................................................................... 31.3 REFERENCES.................................................................................................................................................................. 3

2.0 DESIGN CALCULATION ............................................................................................................................................ 5

2.1 DESIGN OF PLINTH BEAM .............................................................................................................................................. 72.2 DESIGN OF GRADE BEAM .................................................................................................................................................... 72.3 DESIGN OF COLUMN ............................................................................................................................................................ 82.4 DESIGN OF RC WALL....................................................................................................................................................... 112.5 DESIGN OF FOOTING .................................................................................................................................................... 14

3..0 CONCLUSION .................................................................................................................................................................. 29



1.0 INTRODUCTION

1.1 GeneralThe objective of this document is to define the minimum requirements for the design and

engineering of the BHARUCH JAMNAGAR PIPELINE PROJECT. The general scope of work is for

engineering procurement and construction of the Bharuch Anand– GSPL pipeline system and

related facilities including Despatch Station, SV Stations, Tap-off station and receiving station.

This pipeline broadly consists of approximately 103.50 KMs of 30” line pipe from existing station

(Receiving Station) at rajkot on GSPL 24” Anand Rajkot Pipe-line to village Pipli at chainage

89.67 Km (Receiving Station at Jamnagar) and subsequently to Reliance Receiving Station (at

Reliance premises)

The purpose of this document is for providing the Design and Details of Boundary wall for

SV4 Station in Chela at Chainage 79.12 Km which consists of Brick Masonry above and below the

Ground Level . Beams are provided at Top of Footing Level, Plinth Level, and at Top of wall over

which wire fencing is supported

1.2 Design Philosophy

The RCC Wall below the GL is supported on Grade Beam and the Brick Masonry above the GL

is supported by Plinth Beam respectively. The tie beam has been designed for wind load and

fencing. The brick masonry has been checked for Compressive Stress and Shear Stress as per IS

1905-1987. The external loads are transferred through these beams to Columns and Foundations.

The height of wall is 2.5m from FGL. The Column Foundations has been designed as Isolated

Footing at a depth of 1.15m below NGL for SBC of 500 KN/ m2 as per Soil Report. Wind

Pressure Calculation shall be based on basic wind speed of 50m/s. Seismic Shear has been

calculated for Zone IV . Wind load is found to be critical and also Earth Pressure has been

considered including Surcharge of 10 KN/ m2. The earth pressure due to 2.0m height of soil has

been considered .

1.3 References

IS : 456-2000 Code of practice of plain and reinforce concrete.

IS 1905-1987 Code of Practice for Structural Use of Un-reinforced Masonry

SP -16 Design Aids for Reinforced Concrete to IS:456-1978



.

IS : 875-1987 (Parts I to V) Code of practice for Design loads (other than earthquake)

for Buildings and Structures

IS : 1893 Pt.I-2002 Criteria for Earthquake Resistant Design of Structure.

IS :13920 Code of practice for ductile detailing of RCC Structures subjected to

Seismic forces

IS: 1080 Code of practice for design and construction of shallow

foundations in soils (other than raft, ring and shell).

IS:1904 Code of practice for design and construction of foundations in soils - General requirements.

Job 20/07-07 Geotechnical / Soil Investigation Report of M. K.. Soil Test

Laboratory, Ahmedabad-7



2.0 Design Calculation

Calculation of Wind Load at FGL of Compound Wall Tie Beam345x230

Brick Wall(230 Tk)

5 KN/m2

RCC Wall (150Tk)

27.44 KN/m2

Vb 50 m/sK1= 1.05K2= 1.05K3= 1Design wind speed Vb*K1*K2*K3 = 55.125 m/sDesign wind pressure 0.6*Vz2 = 1.82 KN/M2

Force coeffiicient (Table 24)B/h=1.16<12 = 1.2Total wind pressure = 2.19 KN/M2

Length of wall above FGL = 3 MWind shear 2.19*2.27 = 4.97 Kn/MShear stress 4.97/0.23 = 21.59 Kn/M2

= 0.022 N/mm2

Allowable shear stress 0.1+fd/6 = 0.1+0.685/6= 0.214 N/mm2 Hence, SAFE

Height of wall above FGL 2.27 M

Self weight of Brick wall 0.23*20*2.27 = 10.442 KN/MCompressive stress on the wall 10.442/0.23 +

5.64x6/1x0.2302 = 684.76 KN/M2

= 0.685 N/mm2

Slenderness ratio 2.27/0.23 = 9.87Redn. Factor (as per Tab 9) = 0.89Height / Width ratio 2.27/2.655 = 0.85Shape Modification Factor = 1.20Allowable Compressive Stress ( as per Tab8 of IS 1905) =8.35x0.89x1.2 8.92 N/mm2 > 0.685 N/mm2

for Mortar Type M1

CALCULATION OF SEISMIC FORCE ON COMPOUND WALL (As per IS 1893-2002)Zone factor for Zone IV Z = 0.24Importance factor I = 1.5Reduction Factor R (as per IS 1893-2002, Tab 7) = 1.5Sa/g For Rock 1.00/T = 1.11Time Period=T 0.09*h/d^0.5 = 0.90Horizontal Seismic Coefficient

Earth Pressure Diagram

Moment at top of Plinth Beam due to wind 4.97*(2.27/2) = 5.64 KN.M/M

2.0 m

2.5 m

GRADE BEAM(345 x 300)

PLINTH BEAM(345x300)

WIND

FGL +99.55

h=4.8m d=0.23mα = Z*I*Sa/2*R*g = 0.13Weight of wall

= 10.44 KN/m0.23*2.27*20



Seismic shear 0.13*10.44 = 1.39 KN/mShear stress 1.39/0.23 = 6.05 KN/M2

= 0.006 N/mm2

Allowable shear stress 0.1+fd/6 = 0.1+0.224/6= 0.14 N/mm2 SAFE

Height of wall above FGL = 2.27 M1.39*(2.27/2) = 1.58 KNm/M

Compressive stress on the wall 10.44/0.23 + 1.58x6/1x0.2302 = 224.48 KN/M2

0.224 N/mm2

Design of Brick Masonry:

Boundary wall is supported on all four sides

Panel size Length L = 2.7 m Height H = 2.27 m

Height/Length = 0.84Refer table 14 of IS 1905Bending moment PL2/24 = 4.97*2.72/24

= 1.51 KNmBending Stress = 1.51*10^6*6/(2700*(230+15+15)^2)

= 0.05 N/mm2

< 0.05*1.25= 0.06 N/mm2

SAFE Refer Cl 5.4.2 of IS 1905-1987 (Note 2)Design of Tie BeamFor Vertical LoadAssume Size of Tie beam = 345x230 Length of ISA 50X50X6 450+300x1.414 = 0.87 mWt of ISA 50x50x6 @ 3.8 Kg/m = 3.32 KgWt of Barbed Wire @ 0.1 Kg/m for 15m = 1.5 KgTotal Wt. of Fence post (ISA 50x50x5) and Barbed wire = 4.82 Kg

0.05 KNSelf Weight of Tie Beam 0.23*0.345*25 = 1.98 KN/mBending Moment 1.98*3*3/12+0.05*3/4 = 1.52 KN .mFactored Moment 1.52*1.5 = 2.28 KN.mFor Wind Load on FencingArea of ExposureFor ISA 50x50x5 = 0.05x0.75

= 0.0375 m2

For Barbed wire = 0.002x5= 0.01 m2

Total area = 0.0475 m2

Wind Pressure = 2.19 KN/m2

Total Load 2.19x0.0475 = 0.10 KNThis will act at half of the projected height of fenceProjected Ht. of fence = 0.75 mTotal Moment 0.1*0.375 = 0.039 KN.m < BM due to Vertical Load

Size of Tie beam = 345x230 mmEffective Depth (230-40-10) = 180 mm

K= M/bd2 = 0.20Provide Minimum reinforcementPercentage of reinforcement = 0.205 %Area of reinforcment = 162.6675 mm2

Provide 12 dia 2 nosArea provided 2*3.14*12*12/4 = 226.08 mm2

Max. Shear 1.98x1.5+0.05/2 = 3.00 KNFrom IS 456, Tab 62Vus/d 1.5x3/18 = 0.25

Moment at top of Plinth Beam due to Seismic

Wind Shear Governs

Provide stirrups #8 @ 150 c/c

Calculation of Earth PressureDensity of Soil w = 27.6 KN/m3

Co-eff. of Earth Pressure at rest Ko for SandKo = 1-sin ∅ ∅ = 30 (Assumed) = 0.5Surcharge q = 10 KN/m2

Lateral Earth Pressure due to Surcharge = 0.5x10 `= 5 KN/m2

Lateral Earth Pressure due to Earth for a height of 2m = 0.5x27.6x2= 27.6 KN/m2

Average Earth Pressure = (5+32.6)/2= 18.8 KN/m2



2.1 Design of Plinth BeamFor lateral earth pressure:Assume Size of Plinth beam = 345x300 Self Weight of Plinth Beam 0.3*0.345*25 = 2.59 KN/mAverage Earth Pressure = 18.80 KN/m2

Load due to Earth Pressure 18.8 x 2/2 = 18.80 KN/mBending Moment 18.8*3*3/10 = 16.92 KN.mFactored Moment 16.92*1.5 = 25.38 KN.mFor vertical Load :Total weight of Brick panel = 10.442 KN/mSelf Weight of Plinth Beam 0.3*0.345*25 = 2.59 KN/mTotal Udl = 13.03 KN/mBending Moment 13.03*3*3/10 = 11.73 KN/mSize of Plinth beam = 345x300 mmEffective Depth ( 345-40-8) = 297 mm

K= M/bd2 = 0.96Percentage of reinforcement = 0.216 %Area of reinforcement = 192.456 mm2


Max. Shear 18.8 x3/2 = 28.20 KNFrom IS 456, Tab 62Vus/d 1.5x28.2/29.7 1.42Provide stirrups #8 @ 150 c/c

< BM due to lateral earth pressure

2.2 Design of Grade Beam

For lateral earth pressure:Assume Size of Grade beam = 345x300 Average Earth Pressure = 18.80 KN/m2

Load due to Earth Pressure 18.8 x 2/2 = 18.80 KN/mTotal Load = 18.80 KN/mBending Moment 18.8*3*3/10 = 16.92 KN.mFactored Moment 16.92*1.5 = 25.38 KN.m

For vertical Load :Total weight of RCC panel 0.15x1.4x25 = 5.25 KN/mSelf Weight of Grade Beam 0.3*0.345*25 = 2.59 KN/mTotal Load = 7.84 KN/mBending Moment 7.84*3*3/10 = 7.05 KN/mSize of Grade beam = 345x300 mmEffective Depth ( 345-40-8) = 297 mm

K= M/bd2 = 0.96From SP 16, Tab 3Percentage of reinforcement = 0.216 %Area of reinforcement = 192.456 mm2


Max. Shear 18.8 x3/2 = 28.20 KNFrom SP 16, Tab 62Vus/d 1.5x28.2/29.7 = 1.42Provide stirrups #8 @ 150 c/c

< BM due to lateral earth pressure



2.3 Design of ColumnSelf Wt. of Centre column 0.475x0.3x4.5x25 = 16.03 KNSelf Wt. of Centre column @ Expan 0.345x0.3x4.5x25 = 11.64 KNSelf Wt. of Corner column 0.410x0.410x4.5x25 = 18.91 KNMoment due to wind at the top of Pile cap 4.97x (2.5/2+2) x 3+0.1x4.875 55.56 KN.m +2.19x3x0.23x(4.27+0.115)Moment due to wind at the top of Plinth Beam 4.97x (2.5/2) x 3+0.1x(2.5+0.375)+2.19x0.23x3x2.39 = 22.53 KN.mCentre Column 475mm x 300 mm

Fx Fy Fz Mx My MzSLSDL+WL+EP 0.00 78.37 129.31 168.36 0 0.00

ULS1.5DL+1.5WL+1.5EP 0 117.56 193.97 252.54 0 0.00

For Centre Column (Uni-axial bending)Total Vertical Load on Column16.03+7.84x2.7+13.03*2.7+1.98x3+0.05 = 78.37 KN

168.36 KNmLength /depth ratio 2500/300 8.33 Short columnSize of column 475x300 mm (300=parrellel to wall, 475=Perpendicular to wall)Concrete grade Reinforcement grade 415 N.mm2

Factored Pu = 117.56 KN

Reinforcement will be equally distributed on all four sides with eff. cover of 52.5 mmAssume reinforcement percentage = 3.5 %

Area of Reinforcement = 4987.5 mm2

p/Fck = 0.14UNIAXIAL MOMENT CAPACITYXX direction d'/D 52.5/475 = 0.11Use chart 45 Pu/fck bd 117.56*1000/25*300*475 = 0.03

Muy1/fckbd2 = 0.165Muy1= 0.165*25*300*475*475 = 279.21 KNm > Muy

Hence,SafeProvide 4 # 32 + 4 # 25Area of Reinforcement provided = 5180 mm2

At FGL:Total weight of brick wall and beams at Plinth Level = 47.99 KN2.59x3+10.44x3+0.3x0.475x2.5x25Total moment due to wind at bottom of Plinth beam =

22.53 KNmLength /depth ratio 2500/475 5.26 Short columnConcrete grade M25Reinforcement grade 415 N.mm2

Factored Pu = 71.99 KNMuy 33.80 KN.m

Reinfocement will be equally distributed on all four sides with eff. cover of 52.5 mmAssume reinforcement percentage = 1 %

Area of Reinforcement = 1425 mm2

p/Fck 1.0/25 = 0.04UNIAXIAL MOMENT CAPACITYYY direction d'/D 52.5/475 = 0.11Use chart 45 Pu/fck bD 71.99*1000/25*300*475 = 0.02

Mux1/fckbD2 = 0.06

Muy1= 0.06*25*300*475*475/10^6 = 101.53 KNm > Muy

Total moment due to Seismic at top of footing at perpendicular direction to wall 55.56+18.8x3x2x1

Factored moment perpendicular to wall Muy = 252.54 KNm

Factored moment perpendicular to wall

4.97x (2.5/2) x 3+0.1x(2.5+0.375)+2.19x0.23x3x2.39

Centre Column at Expansion Jt. 475mm x 300 mmFx Fy Fz Mx My Mz

SLSDL+WL+EP 0.00 45.94 64.66 84.18 0 0.00

ULS1.5DL+1.5WL+1.5EP 0 68.92 96.99 126.27 0 0.00

Total Vertical Load on Column11.64+7.84x1.5+13.03*1.5+1.98x1.5+0.05 = 45.94 KN

84.18 KNmLength /depth ratio 2500/300 8.33 Short columnSize of column 475x300 mm (300=parrellel to wall, 475=Perpendicular to wall)Concrete grade Reinforcement grade 415 N.mm2

Total moment due to Seismic at top of footing at perpendicular direction to wall (55.56+18.8x3x2x1)/2 =



Factored Pu = 68.92 KN

Reinforcement will be equally distributed on all four sides with eff. cover of 52.5 mmAssume reinforcement percentage = 1.5 %


p/Fck = 0.06UNIAXIAL MOMENT CAPACITYXX direction d'/D 52.5/475 = 0.11Use chart 45 Pu/fck bd 68.92*1000/25*300*475 = 0.02

Muy1/fckbd2 = 0.08Muy1= 0.145*25*300*475*475 = 135.38 KNm > Muy

Hence,SafeProvide 8 # 20Area of Reinforcement provided = 2512 mm2

At FGL:Total weight of brick wall and beams at Plinth Level = 26.01 KN2.59x1.5+10.44x1.5+0.3x0.345x2.5x25Total moment due to wind at bottom of Plinth beam =

11.27 KNmLength /depth ratio 2500/475 5.26 Short columnSize of column 300x475 mm (300=parrellel to wall, 345=Perpendicular to wall)Concrete grade M25Reinforcement grade 415 N.mm2

Factored Pu = 39.02 KNFactored moment parrellel to wall Mux = 16.90 KN.m

Muy = 16.90 KN.mReinfocement will be equally distributed on all four sides with eff. cover of 52.5 mm

Assume reinforcement percentage = 1 %

Area of Reinforcement = 1425 mm2

p/Fck 1.5/25 = 0.04UNIAXIAL MOMENT CAPACITYXX direction d'/D 52.5/475 = 0.13Use chart 45 Pu/fck bD 39.02*1000/25*300*475 = 0.01

Mux1/fckbD2 = 0.055Muy1= 0.055*25*300*475*475/10^6 = 93.07 KNm > Muy

Corner Column 410mm x 410 mm

SLS Fx Fy Fz Mx My MzDL+WL+EP 64.66 87.51 64.66 84.18 0 84.18

ULS1.5DL+1.5WL+1.5EP 96.99 131.27 96.99 126.27 0 126.27

(4.97x (2.5/2) x 3+0.1x(2.5+0.375)+2.19x0.23x3x2.39)/2


126.27 KNmFactored moment perpendicular to wall Muy =

Total weight of brick wall and beams at top of Pile Cap = 87.51 KN18.91+7.84x3+13.03x3+1.98x3+0.05Total moment due to wind at top of footing at perpendicular direction to wall

84.18 KNm84.18 KNm

Length /depth ratio 2500/345 7.25 Short columnSize of column 410x410 mm (410=parrellel to wall, 410=Perpendicular to wall)Concrete grade M25Reinforcement grade 415 N.mm2


Muy = 126.27 KN.mReinfocement will be equally distributed on all four sides with 20 mm bars and eff. cover of 52.5 mmAssume reinforcement percentage = 3.2 %



Mux1/fckbD2 = 0.155Mux1= 0.155*25*410*410*410/10^6 = 267.07 KNmYY Direction

d'/D 52.5/410 = 0.13Use chart 45 Pu/fck bd 131.27*1000/25*345*345 = 0.03

Muy1/fckbd2 = 0.155Muy1= 0.155*25*410*410*410/10^6 = 267.07 KNm

(55.56+18.8x3x2x1)/2


Total moment due to wind at top of footing at perpendicular direction to wall =



Refer chart 63 for calculation of Pzp=3.2% fy=415 fck=25

Puz/Ag = 20.8 N/mm2

Puz = 3496.48 KNPu/Puz 131.27/3496.48 = 0.04Mux/Mux1 126.27/267.07 = 0.47Muy/Muy1 126.27/267.07 = 0.47Referring to chart 64 For Muy/Muy1 and Pu/Puz Mux/Mux1 = 0.52 > 0.45

SafeProvide 12 # 25 Area of Reinforcement provided = 5892 mm2

At FGL:Total weight of brick wall and beams at top of Pile Cap = 49.59 KN2.59x3+10.44x3+0.3x0.475x2.5x25Total moment due to wind at top of footing at perpendicular direction to wall =

11.27 KNm

Length /depth ratio 2500/345 7.25 Short columnSize of column 345x345 mm (345=parrellel to wall, 345=Perpendicular to wall)Concrete grade M25Reinforcement grade 415 N.mm2


Muy = 16.90 KN.mReinforcement will be equally distributed on all four sides with eff. cover of 52.5 mmAssume reinforcement percentage = 1.25 %



Mux1/fckbD2 = 0.075Mux1= 0.075*25*410^3/10^6 = 129.23 KNmYY Direction

d'/D 52.5/345 = 0.15Use chart 45 Pu/fck bd 74.39*1000/25*410*410 = 0.03

Muy1/fckbd2 = 0.075Muy1= 0.075*25*410*410*410/10^6 = 129.23 KNmRefer chart 63 for calculation of Pz

p=1.25 fy=415 fck=25Puz/Ag = 14.8 N/mm2

Puz = 1761.57 KNPu/Puz 74.39/1761.57 = 0.04Mux/Mux1 16.9/129.23 = 0.13Muy/Muy1 16.9/129.23 = 0.13Referring to chart 64 For Muy/Muy1 and Pu/Puz Mux/Mux1 = 0.85 > 0.13

SafeProvide 12 # 16 Dia.Area of Reinforcement provided = 1600 mm2

KNm


(4.97x (2.5/2) x 3+0.1x(2.5+0.375)+2.19x0.23x3x2.39)/2 =Total moment on top of footing at parellel direction to wall = 11.27



2.4 Design of RC WALLRC wall has been designed as slab discontinuous in all four sides

Thickness of Wall, Thk = 150 mmClear cover = 30 mmTop Reinforcement:Assumed Dia of Main bar = 8 mmAssumed Dia of secondary bar = 8 mmEffective length in shorter-direction, Eff Lx = 1.40 mEffective length in Longer direction, Eff Ly = 2.70 mDead load 3.75 kN/m2

Live load, LL due to Earth Pressure = 18.80 kN/m2

Unit weight of concrete = 25.00 kN/m3

Grade of concrete,fck = 25.00 N/mm2

Grade of steel,fy = 415.00 N/mm2

Factor of Saftey = 1.50

Total Factor Load = 33.83 kN/m2

Analysis :

Consider One meter breadth of the wall.

coefficient calculations for Moments :

= 1.93

Condition = = All Edges Discontinous

= 0.1070

= 0.0560

Select sides of the slab :

DL+LL+FL

discontinuous discontinuous

1.40 m in lx-direction

DL + LL+FL

discontinuous discontinuous

2.70 m in ly-direction

Input:

Ratio of eff.Ly/ eff.Lx,r

shorterside of edge2,α x +

longer side of edge2,α y+

(IS 456-2000 -Annexure-D-

Table-26)



cl2

cs1 cs2

1.40

2.70

cl1

Note: Refer to Text Book "R.C Design by unnikrishna pillai & devdas menon"

Dead load '+ve' bending moment, Mx+ = αx+ x w x Lx2

= 7.09 kN mDead load '+ve' bending moment, MY+ = αy+ x w x Lx2

= 3.71 kN mDead load Shear force in X-dir Vu = w x lx /2

= 23.68 kNDead load Shear force in Y-dir Vu = w x ly /2

= 45.66 kN

Check for wall thickness Assume dia of bar = 8 mmEffective Thickness of the Slab = 150 - 30 - 8-8 / 2for shorter direction = 116 mm

Effective depth required for Slab = v(Mux/(0.138 x fck x 1000)) d-reqd = 34.24 mm

Calculation of main steel

Effective Depth of slab in X dir dx = 116.000 mm(150-30-8/2)Effective Depth of slab in Y dir dy = 108.000 mm(150-30-8-8/2)

(a) Area of Steel Mx+ = ( 0.5 fck/fy ) x [ 1-v( 1 - (4.6Mux/(fck x 1000 x d2)) ) ] x 1000d AstR = 173.78 mm2 / m

Required spacing of steel S 289.10 mm

(b) Area of Steel My+ AstR = ( 0.5 fck/fy ) x [ 1-v( 1 - (4.6Mux/(fck x 1000 x d2)) ) ] x 1000d

= 96.70 mm2 / mRequired spacing of steel S = 519.56 mm

discontinuous

Provided depth is ok

discontinuous

discontinuous

discontinuous

x

y



Minimum percentage of steel is = 0.12Minimun area of the steel = 0.12 % x b x D

= 0.12x 1000 x 150/ 100 AstM = 180 mm2 / m

Higher value of Ast is provided, AstP = 180.00 mm2 / m

Provide 8 mm dia bar at 200 mm c/cMaximum allowable spacing = 3d or 300 mm whichever is lesser Cls 26.3

Provide8 mm Dia bar @ 200 mm c/c as main steelProvide 8 mm Dia bar @ 250 c/c as DistributorShear Check

Check for shear in X-direction:Designed Shear Force = k τc bd , k and τc are calculated below (IS 456-2000-Table -19)

Ast provided = 300 mm2

Percentage of Steel Provided Pt = 100xAst/bxd= 0.2586

k = 1.300 Refer Cls 40.2 τc = 0.360 N/mm2

(Refer IS 456-2000-Table 19)

Vult = 50.54 KN Depth k300 1.00275 1.05 `175 1.25

Vu = 23.68 KN 150 1.30

Vult > Vu Therefore No shear reinforcement is required

Check for Deflection(Refer RCC Design by Unnikrishnapillai and devodos menon)

Basic span depth ratio for S.S slab = 20 (From IS 456-2000 Clause :23.2.1.a)fs = 0.58 x fy x Ast required / Ast provided

= 240.7 (Ast provided = Ast required)

Effective Span length = 1.4

Percentage of Steel Provided Pt = 0.259

Modification Factor kt = 2 (From IS 456-2000-Fig-4 )

(l/d)max = 40

(l/d)provided = 12.07

(l/d)provided < (l/d)max

Hence it is Safe.

on positive and negative reinforcement

Spacing is less than maximum spacing, OK



2.5 Design of Footing For Corner Footing

Foundation Dimension:

Base Lx (m) 2.50 0.00

Lz (m) 2.50

Pedestal Px (m) 0.410 2.65

Pz (m) 0.410

0.65

Design Summary:

Max Bearing Press.: 335.25 KN/m2

Min Bearing Press.: 68.59 KN/m2

Min Contact Length: 91%

Soil PropertiesSafe Bearing Capacity of Soil q safe = 500.0 KN/m2

At depth h2 q all = q safe + γ x h2 569.4 KN/m2

At depth h2 q all = 1.25 x q safe + γ x h2 694.4 KN/m2

Material Properties:Minimum 28 Days Compressive of Concrete : FC 25000 KN/m2

Yield Strength of Reinforcing Steel : FY 415000 KN/m2

Density of Structural fill 26 KN/m3

Density of Concrete 25 KN/m3

Load Combination:Unfactored Loads (in KN m)

LC No. Fx Fy Fz Mx Mz

Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 64.66 87.51 64.66 84.18 84.18

Case III 0 0 0 0 0 0

Factored Loads (in KN m)


Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 96.99 131.27 96.99 126.27 126.27

Case III 0 0.00 0.00 0.00 0.00 0.00

FyFz

FxFGLx h1 (m)=

h2 (m)=

D (m)=

X-Axis

Z-Axis

Mz1

Mx1



Check For Bearing Pressure:

Total Substructure Load:

Weight of Pedestal = 8.4 KN

Weight of Footing = 101.6 KN

Weight of Imposed Earth above footing = 318.7 KN

TOTAL = 428.7 KN

Case I:Axial load P = Fy + SubStructure Load = 428.7 KN

Moment at base Mx1= Mx + Fz x (h1+h2) = 0.00 KNm

Moment at base Mz1= Mz + Fx x (h1+h2) = 0.00 KNm

Eccentricity,

ex = Mz1 / P = 0.000 m < Lx / 6 Contact Length= 3 x (Lx /2 - ex) / Lx = 100%

ez = Mx1 / P = 0.000 m < Lz / 6 Contact Length= 3 x (Lz /2 - ez) / Lz = 100%

ex / Lx = 0.000 From Table, K= 1.00ez / Lz = 0.000 Max. Bearing Press.=qmax =K x P / Lx * Lz = 68.59 KN/m2 < q all = q safe + γ x h2

=> Hence Satisfactory

Case II:Axial load P = Fy + SubStructure Load = 516.2 KN



Eccentricity,

ex = Mz1 / P = 0.495 m > Lx / 6 Contact Length= 3 x (Lx /2 - ex) / Lx = 91%

ez = Mx1 / P = 0.495 m > Lz / 6 Contact Length= 3 x (Lz /2 - ez) / Lz = 91%

ex / Lx = 0.198 From Table, K= 4.06ez / Lz = 0.198 Max. Bearing Press.=qmax =K x P / Lx * Lz = 335.25 KN/m2 < q all =1.25*q safe + γ x h2

=> Hence SatisfactoryCase III:Axial load P = Fy + SubStructure Load = 428.7 KN

Moment at base Mx1= Mx + Fz x (h1+h2) = 0 KNm


Eccentricity,



ex / Lx = 0.000 From Table, K= 1.00ez / Lz = 0.000 Max. Bearing Press.=qmax =K x P / Lx * Lz = 68.59 KN/m2 < q all = 1.25*q safe + γ x h2




Design Of Base (Ref. IS-456)Downward pressure on Base: ws=(Weight of Footing + Weight of Imposed Earth above footing) / Lx . Lz= 67.24 KN/m2

Cover to Main Reinf. = C = 65.00 mm

Effective Depth d =(D - C - φ/2)= 0.58 m

Projected Length of Base:

ax = 1.045 m

az = 1.045 m

Case I:Factor for Substructure Load = 1.5

Factored ws = fws = 100.86 KN/m2

Axial load P = Fy + SubStructure Load = 643.0 KN


Moment at base Mz1= Mz + Fx x (h1+h2) = 0.000 KNm Base Pressure Distribution Diagram

Eccentricity,

ex = Mz1 / P (m) = 0.000 ex / Lx = 0.000 From Table, K= 1.00

ez = Mx1 / P (m) = 0.000 ez / Lz = 0.000 qmax = K x P / Lx * Lz (KN/m2)= 102.88

qmin = (KN/m2) = 102.88

Moment at face of Pedestal=

BM in X- Direction = (qm - fws) x ax 2/ 2 + ( qmax - qm ) x ax

2/3 = 1.10 KNm / Unit Width

BM in Z- Direction = (qm - fws) x az 2/ 2 + ( qmax - qm ) x az


Shear at 'd' distance from face of Pedestal=

SF in X- Direction = (qS - fws) x (ax -d) + ( qmax - qS ) x (ax -d)/2 = 0.94 KN / Unit Width

SF in Z- Direction = (qS - fws) x (az -d) + ( qmax - qS ) x (az -d)/2 = 0.94 KN / Unit Width

Case II:Factor for Substructure Load = 1.5





Eccentricity,


ez = Mx1 / P (m) = 0.495 ez / Lz = 0.198 qmax = K x P / Lx * Lz (KN/m2) = 502.88

qmin = (KN/m2) = -170.49









Case III:Factor for Substructure Load = 1.5




Moment at base Mz1= Mz + Fx x (h1+h2) = 0 KNm

Eccentricity,



qmin = (KN/m2) = 102.88

d Flexure Shear

Flexure Section

ax

qmaqm qs











Bottom Reinforcement:X-Direction:Maximum Moment Mx = 177.28 KNm / Unit WidthRequired q n = Mu/bd2 = 0.53 N/Sqmm.

ρ = 0.00152 Ast = ρ x b x d = 880.7 mm2Providing Bar : # 12

ρmin = 0.0012 Ast (min) = 694.8 mm2Spacing Required = 128.40 mm

Hence Adopt #12 @ 128mm c/c both sidesHowever, Provide #12 @ 125 c/c BW

Z-Direction:Maximum Moment Mx = 177.28 KNm / Unit WidthRequired q n = Mu/bd2 = 0.53 N/Sqmm.



Hence Adopt #12 @ 128mm c/c both sidesHowever, Provide #12 @ 125c/c BW

One-way Shear Check : Maximum Shear Force at 'd' distance from Pedestal face = V = 163.23 KN / Unit Width

Nominal Shear Stress 0.28 N/mm2

Design Shear Srength of Concrete 0.29 N/mm2

'=> Hence Satisfactory

Top Reinforcement:Maximum fws = 100.86 KN/m2

BM in X - Direction = (fws x ax2/2) = 55.07 KNm

BM in Z - Direction = (fws x az2/2) = 55.07 KNm

X-Direction:Maximum Moment Mx = 55.07 KNm / Unit WidthRequired q n = Mu/bd2 = 0.16 N/Sqmm.



Hence Adopt #10 @ 113mm c/c both sidesHowever, Provide #10 @ 100c/c







Punching Shear Check :

Shear at 'd/2' distance from the col face = 1235.05 KN

Punching Shear Stress (N/mm2) = 0.539

Allowable Shear Stress (N/mm2) = 1.25

'=> Hence Satisfactory.

STABILITY CHECKOverturning CheckCase 2Overturning moment in' Z' direction= 255.5 KNm

Restoring moment in' Z' direction= 645.2 KNm

Factor of safety= 2.5 >1.5 Hence SatisfactoryOverturning moment in' X' direction= 255.5 KNm

Restoring moment in' X' direction= 645.2 KNm

Factor of safety= 2.5 >1.5 Hence Satisfactory

Sliding checkCoefficient of friction= 0.5

Case 2Total sliding force= 91.4 KN

Resisting frictional force= 258.1 KN




For Centre Column Footing


Base Lx (m) 2.70 0.00

Lz (m) 2.70


Pz (m) 0.30

0.70

Design Summary:













Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 0.00 78.37 129.31 168.36 0.00

Case III 0 0 0 0 0 0



Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 0.00 117.56 193.97 252.54 0.00

Case III 0 0.00 0.00 0.00 0.00 0.00

FyFz

FxFGLx h1 (m)=

h2 (m)=

D (m)=

X-Axis

Z-Axis

Mz1

Mx1








TOTAL = 509.2 KN




Eccentricity,








Eccentricity,







Eccentricity,










Projected Length of Base:ax = 1.1125 m

az = 1.2 m






Eccentricity,



qmin = (KN/m2) = 104.78














Eccentricity,



qmin = (KN/m2) = -115.72














Eccentricity,



qmin = (KN/m2) = 104.78

d Flexure Shear

Flexure Section

ax

qmaqm qs









































STABILITY CHECKOverturning CheckCase 2Overturning moment in' X' direction= 517.5 KNm


Factor of safety= 1.5 >1.5 Hence SatisfactorySliding check

Coefficient of friction= 0.5






For Expansion Jt. Column Footing


Base Lx (m) 2.70 0.00

Lz (m) 2.70


Pz (m) 0.30

0.70

Design Summary:













Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 0.00 91.88 129.31 168.36 0.00

Case III 0 0 0 0 0 0



Case I 0 0.00 0.00 0.00 0.00 0.00

Case II (DL+WL+EP) 1 0.00 137.82 193.97 252.54 0.00

Case III 0 0.00 0.00 0.00 0.00 0.00

FyFz

FxFGLx h1 (m)=

h2 (m)=

D (m)=

X-Axis

Z-Axis

Mz1

Mx1








TOTAL = 509.3 KN




Eccentricity,








Eccentricity,







Eccentricity,










Projected Length of Base:

ax = 1.1775 m

az = 1.2 m






Eccentricity,



qmin = (KN/m2) = 104.80














Eccentricity,



qmin = (KN/m2) = -112.92














Eccentricity,



qmin = (KN/m2) = 104.80

d Flexure Shear

Flexure Section

ax

qmaqm qs









































STABILITY CHECKOverturning CheckCase 2Overturning moment in' X' direction= 517.5 KNm



Sliding checkCoefficient of friction= 0.5






3..0 CONCLUSION

SIZE (IN MM) AT BOTTOM AT TOP

FOOTING

COLUMNS At Centre 2700 X 2700 X 700 # 12 @ 100 C/C BW #10 @ 100 C/CBW At Exp.Jt. 2700 X 2700 X 700 # 12 @ 100 C/C BW #10 @ 100C/CBW At Corner 2500 X 2500 X 650 # 12 @ 125 C/C BW #10 @ 100C/CBW

COLUMNS At Centre 475 X300 4 # 32+ 4 # 25 Stirrups #8 @ 200 c/c At Exp. Jt 475 X300 8 # 20 Stirrups #8 @ 200 c/c At Corner 410 X410 12 # 25 Stirrups #8 @ 200 c/c

BEAMS PLINTH BEAM 345 X300 2 #16 (TOP & Stirrups #8 @ 150 c/c BOTTOM)

GRADE BEAM 345 X300 2#16 (TOP & Stirrups #8 @ 150 c/c BOTTOM)

TIE BEAM 345 X230 2#12 (TOP & Stirrups #8 @ 150 c/c BOTTOM)

RC WALL 150 Tk. #8 @ 200 C/C #8 @ 250 c/c Vertical Horizontal

Documents

GSPL203-73-1517 R-0