Combined Isolated Footing

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  • 8/11/2019 Combined Isolated Footing

    1/65

    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=1 Col B=2

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 307.5 kN 2

    Ultimate load carrrid by column B = 303 kN 1

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 205 kNWorking load carrrid by column B = 202 kN

    Self wt of footing (10% of column load)= 41 kN

    Total working load = 448 kN

    Length of footing = 3.50 m

    Requried area of footing = 2.24 m2

    Width of footing = 0.64 m

    Provide width of footing = 0.75 m

    Provide Footing size of 3.50 m x 0.75 m = 2.63 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 308 x 0.115 ) + ( 303 x 3.385 )

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    ( 308 + 303 )

    = 1.74 m

    C.G. of footing = 1.75 m

    Eccentricity of load w.r.t c.g. of footing = 0.01 m

    Moment due to eccentricity = ( 407 x 0.01 ) = 5 kN m

    Total Moment = 6 kN m

    Pressure calculation :

    Intensity of pressure due to Axial load = P / A =

    = ( 205 + 202 ) / 2.63 = 155 kN/m2

    Intensity of pressure due to Moment = M / Z =

    = 6 / ( 0.75 x 3.50 x 3.50 ) / 6

    = = 4 kN/m2

    Pmax= ( 155 + 4 ) = 159 kN/m2

    OK

    Pmin= ( 155 - 4 ) = 151 kN/m2

    OK

    1

    0.115 0.115

    151

    159 157 151

    159

    Load per metre run of slab = Avg pressure x 1 m

    ( 158 x 1.00 ) = 158 kN/m

    Load per metre run of slab due to ultimate pressure =

    ( 158 x 1.50 ) = 237 kN/m

    Cantilever projection of slab @ face of beam = = 0.15 m

    Maximum ultimate moment = ( 237 x 0.15 x 0.15 ) = 32

    d required = ( 3 x 10^6 ) = 31 mm

    ( 1000 x 2.76 )

    Try overall depth = 230 mm Width = 1000

    Effective depth d = 184 mm End depth= 175

    Effective end depth d =

    Ast= 40 mm2

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    Required is Y 8 @ 1247 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= -0.034 m

    Shear = -8 kN

    v = -0.06 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    307.5 303

    1.62

    0.115 3.27 0.115

    170 170

    174

    179

    179

    Load Diagram

    1.62 283.43

    20.54 0.541

    19.57

    170.76

    286.96 S.F.Diagram

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    -231.2201

    1.18 1.13

    B.M.Diagram

    As in the central portion of beam, the moment is hogging, i.e. the tension is on th

    top portion of beam, hence the beam at the central portion will be designed as th

    isolated T- beam.

    Reinforcement at the central portion:

    bf= = 0.750 m

    = 750 mm

    bw= = 450 mm

    Let provide depth of beam = 605 mm

    Effective depth of beam = 541 mm

    Ast= 1266 mm2

    Provide 6 Nos. Y 16 mm 1206 mm2

    0.30 %

    Shear at d from face of column = 0.964 m

    Shear = 171 kN

    v = 0.70 N/mm2

    c = 0.330 N/mm2

    (From SP16)

    Section for depth is PROVIDE STIRRUPS

    Shear to be resisted by stirrups = 90.42 kN

    Try stirrups of 2-legged Y 8 mm @ 217 mm

    Provide Y 8 @ 200 mm = 251

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.75 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    27

    685.8

    mm2

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    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=3 Col B=4

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 385.5 kN 2

    Ultimate load carrrid by column B = 372 kN 2

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 257 kNWorking load carrrid by column B = 248 kN

    Self wt of footing (10% of column load)= 51 kN

    Total working load = 556 kN

    Length of footing = 3.50 m

    Requried area of footing = 2.78 m2

    Width of footing = 0.79 m

    Provide width of footing = 0.85 m

    Provide Footing size of 3.50 m x 0.85 m = 2.98 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 386 x 0.115 ) + ( 372 x 3.385 )

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    ( 386 + 372 )

    = 1.72 m

    C.G. of footing = 1.75 m

    Eccentricity of load w.r.t c.g. of footing = 0.03 m

    Moment due to eccentricity = ( 505 x 0.03 ) = 15 kN m

    Total Moment = 13 kN m

    Pressure calculation :

    Intensity of pressure due to Axial load = P / A =

    = ( 257 + 248 ) / 2.98 = 170 kN/m2

    Intensity of pressure due to Moment = M / Z =

    = 13 / ( 0.85 x 3.50 x 3.50 ) / 6

    = = 8 kN/m2

    Pmax= ( 170 + 8 ) = 177 kN/m2

    OK

    Pmin= ( 170 - 8 ) = 162 kN/m2

    OK

    1

    0.115 0.115

    163

    177 173 162

    177

    Load per metre run of slab = Avg pressure x 1 m

    ( 175 x 1.00 ) = 175 kN/m

    Load per metre run of slab due to ultimate pressure =

    ( 175 x 1.50 ) = 263 kN/m

    Cantilever projection of slab @ face of beam = = 0.20 m

    Maximum ultimate moment = ( 263 x 0.20 x 0.20 ) = 52

    d required = ( 5 x 10^6 ) = 44 mm

    ( 1000 x 2.76 )

    Try overall depth = 230 mm Width = 1000

    Effective depth d = 184 mm End depth= 175

    Effective end depth d =

    Ast= 80 mm2

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    Required is Y 8 @ 629 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= 0.016 m

    Shear = 4 kN

    v = 0.03 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    385.5 372

    1.61

    0.115 3.27 0.115

    207 207

    217

    225

    226

    Load Diagram

    1.61 348.18

    25.96 0.692

    23.82

    179.32

    359.54 S.F.Diagram

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.85 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    mm2

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    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=3 Col B=4

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 417 kN 2

    Ultimate load carrrid by column B = 415.5 kN 2

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 278 kNWorking load carrrid by column B = 277 kN

    Self wt of footing (10% of column load)= 56 kN

    Total working load = 611 kN

    Length of footing = 3.50 m

    Requried area of footing = 3.05 m2

    Width of footing = 0.87 m

    Provide width of footing = 0.85 m

    Provide Footing size of 3.50 m x 0.85 m = 2.98 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 417 x 0.115 ) + ( 416 x 3.385 )

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    ( 417 + 416 )

    = 1.75 m

    C.G. of footing = 1.75 m

    Eccentricity of load w.r.t c.g. of footing = 0.00 m

    Moment due to eccentricity = ( 555 x 0.00 ) = 2 kN m

    Total Moment = 4 kN m

    Pressure calculation :

    Intensity of pressure due to Axial load = P / A =

    = ( 278 + 277 ) / 2.98 = 187 kN/m2

    Intensity of pressure due to Moment = M / Z =

    = 4 / ( 0.85 x 3.50 x 3.50 ) / 6

    = = 3 kN/m2

    Pmax= ( 187 + 3 ) = 189 kN/m2

    OK

    Pmin= ( 187 - 3 ) = 184 kN/m2

    OK

    1

    0.115 0.115

    184

    189 188 184

    189

    Load per metre run of slab = Avg pressure x 1 m

    ( 188 x 1.00 ) = 188 kN/m

    Load per metre run of slab due to ultimate pressure =

    ( 188 x 1.50 ) = 283 kN/m

    Cantilever projection of slab @ face of beam = = 0.20 m

    Maximum ultimate moment = ( 283 x 0.20 x 0.20 ) = 62

    d required = ( 6 x 10^6 ) = 45 mm

    ( 1000 x 2.76 )

    Try overall depth = 230 mm Width = 1000

    Effective depth d = 184 mm End depth= 175

    Effective end depth d =

    Ast= 86 mm2

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    Required is Y 8 @ 585 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= 0.016 m

    Shear = 5 kN

    v = 0.04 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    417 415.5

    1.63

    0.115 3.27 0.115

    235 235

    238

    241

    241

    Load Diagram

    1.63 388.51

    27.72 0.692

    26.99

    196.55

    389.28 S.F.Diagram

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    -314.2662

    1.59 1.55

    B.M.Diagram

    As in the central portion of beam, the moment is hogging, i.e. the tension is on th

    top portion of beam, hence the beam at the central portion will be designed as th

    isolated T- beam.

    Reinforcement at the central portion:

    bf= = 0.850 m

    = 850 mm

    bw= = 450 mm

    Let provide depth of beam = 750 mm

    Effective depth of beam = 692 mm

    Ast= 1320 mm2

    Provide 7 Nos. Y 16 mm 1407 mm2

    0.24 %

    Shear at d from face of column = 0.823 m

    Shear = 197 kN

    v = 0.63 N/mm2

    c = 0.330 N/mm2

    (From SP16)

    Section for depth is PROVIDE STIRRUPS

    Shear to be resisted by stirrups = 93.79 kN

    Try stirrups of 2-legged Y 8 mm @ 268 mm

    Provide Y 8 @ 250 mm = 201

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.85 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    mm2

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    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=3 Col B=4

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 441 kN 1

    Ultimate load carrrid by column B = 441 kN 1

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 294 kNWorking load carrrid by column B = 294 kN

    Self wt of footing (10% of column load)= 59 kN

    Total working load = 647 kN

    Length of footing = 3.50 m

    Requried area of footing = 3.23 m2

    Width of footing = 0.92 m

    Provide width of footing = 0.95 m

    Provide Footing size of 3.50 m x 0.95 m = 3.33 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 441 x 0.115 ) + ( 441 x 3.385 )

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    ( 441 + 441 )

    = 1.75 m

    C.G. of footing = 1.75 m

    Eccentricity of load w.r.t c.g. of footing = 0.00 m

    Moment due to eccentricity = ( 588 x 0.00 ) = 0 kN m

    Total Moment = 2 kN m

    Pressure calculation :

    Intensity of pressure due to Axial load = P / A =

    = ( 294 + 294 ) / 3.33 = 177 kN/m2

    Intensity of pressure due to Moment = M / Z =

    = 2 / ( 0.95 x 3.50 x 3.50 ) / 6

    = = 1 kN/m2

    Pmax= ( 177 + 1 ) = 178 kN/m2

    OK

    Pmin= ( 177 - 1 ) = 176 kN/m2

    OK

    1

    0.115 0.115

    176

    178 177 176

    178

    Load per metre run of slab = Avg pressure x 1 m

    ( 177 x 1.00 ) = 177 kN/m

    Load per metre run of slab due to ultimate pressure =

    ( 177 x 1.50 ) = 266 kN/m

    Cantilever projection of slab @ face of beam = = 0.25 m

    Maximum ultimate moment = ( 266 x 0.25 x 0.25 ) = 82

    d required = ( 8 x 10^6 ) = 55 mm

    ( 1000 x 2.76 )

    Try overall depth = 230 mm Width = 1000

    Effective depth d = 184 mm End depth= 175

    Effective end depth d =

    Ast= 127 mm2

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    Required is Y 8 @ 395 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= 0.066 m

    Shear = 18 kN

    v = 0.14 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    441 441

    1.64

    0.115 3.27 0.115

    251 251

    252

    253

    253

    Load Diagram

    1.64 412.16

    29.12 0.692

    28.84

    209.21

    411.88 S.F.Diagram

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    -333.9015

    1.67 1.66

    B.M.Diagram

    As in the central portion of beam, the moment is hogging, i.e. the tension is on th

    top portion of beam, hence the beam at the central portion will be designed as th

    isolated T- beam.

    Reinforcement at the central portion:

    bf= = 0.889 m

    = 889 mm

    bw= = 450 mm

    Let provide depth of beam = 750 mm

    Effective depth of beam = 692 mm

    Ast= 1404 mm2

    Provide 7 Nos. Y 16 mm 1407 mm2

    0.23 %

    Shear at d from face of column = 0.833 m

    Shear = 209 kN

    v = 0.67 N/mm2

    c = 0.330 N/mm2

    (From SP16)

    Section for depth is PROVIDE STIRRUPS

    Shear to be resisted by stirrups = 106.44 kN

    Try stirrups of 2-legged Y 8 mm @ 236 mm

    Provide Y 8 @ 225 mm = 223

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.95 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    mm2

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    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=3 Col B=4

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 414 kN 2

    Ultimate load carrrid by column B = 408 kN 1

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 276 kNWorking load carrrid by column B = 272 kN

    Self wt of footing (10% of column load)= 55 kN

    Total working load = 603 kN

    Length of footing = 3.50 m

    Requried area of footing = 3.01 m2

    Width of footing = 0.86 m

    Provide width of footing = 0.95 m

    Provide Footing size of 3.50 m x 0.95 m = 3.33 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 414 x 0.115 ) + ( 408 x 3.385 )

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    Required is Y 8 @ 419 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= 0.066 m

    Shear = 17 kN

    v = 0.13 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    414 408

    1.62

    0.115 3.27 0.115

    230 230

    235

    240

    240

    Load Diagram

    1.62 381.57

    27.58 0.692

    26.43

    193.92

    386.42 S.F.Diagram

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    -311.9913

    1.59 1.52

    B.M.Diagram

    As in the central portion of beam, the moment is hogging, i.e. the tension is on th

    top portion of beam, hence the beam at the central portion will be designed as th

    isolated T- beam.

    Reinforcement at the central portion:

    bf= = 0.889 m

    = 889 mm

    bw= = 450 mm

    Let provide depth of beam = 750 mm

    Effective depth of beam = 692 mm

    Ast= 1307 mm2

    Provide 7 Nos. Y 16 mm 1407 mm2

    0.23 %

    Shear at d from face of column = 0.813 m

    Shear = 194 kN

    v = 0.62 N/mm2

    c = 0.330 N/mm2

    (From SP16)

    Section for depth is PROVIDE STIRRUPS

    Shear to be resisted by stirrups = 91.16 kN

    Try stirrups of 2-legged Y 8 mm @ 275 mm

    Provide Y 8 @ 225 mm = 223

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.95 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    mm2

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    3.5

    Y dir

    Col A Col B

    0.45 X dir

    ETABS NODE NO. Col A=3 Col B=4

    Conc grade = M20

    Steel grade = Fe415

    A) Proportioning of base size: -

    Y - dir X - dir

    Size of column A = 300 mm x 230 mm Additional

    Size of column B = 300 mm x 230 mm MY

    Ultimate load carrrid by column A = 316.5 kN 2

    Ultimate load carrrid by column B = 237 kN 3

    SBC of the soil = 200 kN/m2

    Working load carrrid by column A = 211 kNWorking load carrrid by column B = 158 kN

    Self wt of footing (10% of column load)= 37 kN

    Total working load = 406 kN

    Length of footing = 3.50 m

    Requried area of footing = 2.03 m2

    Width of footing = 0.58 m

    Provide width of footing = 0.75 m

    Provide Footing size of 3.50 m x 0.75 m = 2.63 m

    2

    As in this case, the property line is present on both end of column, hence there

    is no possibility of projection, so the pressure will not be uniform if the c.g. of

    footing and the c.g. of load does not coincide.

    In such case fooring will become eccentric and hence the pressure will be non-uni

    C.G of load system from end face of col A=

    = ( 317 x 0.115 ) + ( 237 x 3.385 )

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    ( 317 + 237 )

    = 1.52 m

    C.G. of footing = 1.75 m

    Eccentricity of load w.r.t c.g. of footing = 0.23 m

    Moment due to eccentricity = ( 369 x 0.23 ) = 87 kN m

    Total Moment = 62 kN m

    Pressure calculation :

    Intensity of pressure due to Axial load = P / A =

    = ( 211 + 158 ) / 2.63 = 141 kN/m2

    Intensity of pressure due to Moment = M / Z =

    = 62 / ( 0.75 x 3.50 x 3.50 ) / 6

    = = 40 kN/m2

    Pmax= ( 141 + 40 ) = 181 kN/m2

    OK

    Pmin= ( 141 - 40 ) = 100 kN/m2

    OK

    1

    0.115 0.115

    103

    178 158 100

    181

    Load per metre run of slab = Avg pressure x 1 m

    ( 169 x 1.00 ) = 169 kN/m

    Load per metre run of slab due to ultimate pressure =

    ( 169 x 1.50 ) = 254 kN/m

    Cantilever projection of slab @ face of beam = = 0.15 m

    Maximum ultimate moment = ( 254 x 0.15 x 0.15 ) = 32

    d required = ( 3 x 10^6 ) = 32 mm

    ( 1000 x 2.76 )

    Try overall depth = 230 mm Width = 1000

    Effective depth d = 184 mm End depth= 175

    Effective end depth d =

    Ast= 43 mm2

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    Required is Y 8 @ 1161 mm

    Provide Y 8 @ 200 mm = 251

    Distribution steel = ( 0.12 % x 1000 x 203 ) = 243

    Provide Y 8 @ 225 mm = 223

    Shear @ d= -0.034 m

    Shear = -9 kN

    v = -0.07 N/mm2

    Enter depth at d location

    c = 0.355 N/mm2

    (From SP16)

    Section for depth is OK

    Design of Longitudinal beam : -

    316.5 237

    1.42

    0.115 3.27 0.115

    116 113

    164

    201

    204

    Load Diagram

    1.42 223.86

    23.23 0.547

    13.14

    156.55

    293.27 S.F.Diagram

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    -225.287

    1.34 0.75

    B.M.Diagram

    As in the central portion of beam, the moment is hogging, i.e. the tension is on th

    top portion of beam, hence the beam at the central portion will be designed as th

    isolated T- beam.

    Reinforcement at the central portion:

    bf= = 0.750 m

    = 750 mm

    bw= = 450 mm

    Let provide depth of beam = 605 mm

    Effective depth of beam = 547 mm

    Ast= 1216 mm2

    Provide 6 Nos. Y 16 mm 1206 mm2

    0.29 %

    Shear at d from face of column = 0.758 m

    Shear = 157 kN

    v = 0.64 N/mm2

    c = 0.330 N/mm2

    (From SP16)

    Section for depth is PROVIDE STIRRUPS

    Shear to be resisted by stirrups = 75.32 kN

    Try stirrups of 2-legged Y 8 mm @ 263 mm

    Provide Y 8 @ 250 mm = 201

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    SUMMARY: -

    Provide Footing size of 3.50 m x 0.75 m

    SLAB RENFORCEMENT: -

    Provide Y 8 @ 200 mm

    Provide Y 8 @ 300 mm

    BEAM REINFORCEMENT: -TOP: - 9 Nos. Y 12 mm

    0.459375

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    oment

    MX

    1

    2

    orm.

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    kN m / metre

    mm

    mm

    129 mm

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    mm2

    0.16 %

    mm2

    mm2

    0.17 %

    Load per metre

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    mm2

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    Project Comments:

    User ARIF Date Time 03:13

    Footing Identifier =

    Safe Bearing Capacity of Soil = 20 T/m2

    Depth of Founding Level below Ground (Df) = 2 m

    m

    Weight Density of Soil & Backfill togethe = 1.8 T/m3

    Load Factor for Limit State Method (LF) = 1.5 Factor

    Concrete Grade (Fck) = 20 N/mm2

    Steel Grade (fy) = 415 N/mm2

    Column Dimensions: E_W (L1) = 0.3 m WidthColumn Dimensions: N_S (B1) = 0.23 m Width

    Offset from face of column = 75 mm

    Crack width = 0.3 m

    LOAD CASES

    Case Load (T) Soil over

    MZ( @Z ) MX( @X ) Stress Actual /

    P M_E-W M_N-S Factor Allowable

    I 14 0 0 1 0.97

    II 14 0 0 1.25 0.92

    III 13 0 0 1.25 0.83

    IVV

    VI

    VII

    VIII

    For SBC

    Punching

    L / B 0.88 Stress (E

    Length - L 0.91 M E_W AREA 0.728 m Stress (NS

    Width - B 0.80 M N_S Depth (be

    Z_NS 0.1 m Reinf. (Be

    if (P > Pp) then 'Revise Footing Size' Z_EW 0.1 m Bearing pr

    Depth of Footing at Centre 550 mm Depth of Footing at Ed

    Eff. Cover to Bott. Reinf. d' 75 mm de=D-d'= 475

    Distances from CL of to a) Column Face, b) De from & its Distance from Edge,

    Perimeter & Punching Area for Shear ECT,.

    E-W N-S perimeter area, Ap

    L1 (E-W) 0.3 Xf 0.15 0.115

    L (E-W) 0.91 Lf 0.305 0.285

    D E S I G N O F I S O L A T E D SLOPED

    18-Aug-14

    Moments (T.M)

    ETABS NO. 10 DWG NO. C16

    0.8

    0

    Trial Footing Size

    Section Modulus

    For Moment For punching shear

    B=

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    R = Mu / b * de

    - N/mm

    Pt (Req)= 0.5*Fck/Fy{[1-(1-4.6*Mu/B*de^2)/Fck]^0.5}*b*de

    Pt (Req) Min = 0.12%

    Ast - Reinforcement to be required = Pt (req) * A * d

    Ast - Reinforcement Provided

    Pt (Provided) @ Efffective depth d from face of column

    Allowable Shear Stress (t/m2) =0.85*sqrt(0.8*Fck)*(sqrt(1+5*b)-1)/6*b

    b =0.8 * Fck / 6.89 * pt 1.00 for E_W 1.00 for N_S

    Actual Shear stress (t/m2)

    Bearing pressure = Pu/bD in t/m2 0.30

    Permissible bearing pressure = 0.45 fck (sqrt(A1/A2)) 1.80

    A1 = (min of (Lf x Bf or ( b + 4Df )x ( D + 4 Df ) 728000

    A2 = b x D 69000

    where sqrt(A1/A2) should not be greater than 2

    Footing Size

    Pedestal Dimensions: E_W = 0.30 m

    Pedestal Dimensions: N_S = 0.23 m

    Length - L: E_W = 0.91 m

    Width - B: N_S = 0.80 m

    Depth = Column face 550 mm

    Footing Edge 230 (E_W)

    Ast =

    1 Excavation 3.10 m3

    2 PCC 0.13 m3

    3 RCC 0.28 m3 1 Concrete

    4 Formwork 0.8 m2 2 Formwork

    5 Reinforcement 8 Kgs

    Total reinforcement per cft = 8 0.28 35.314 0.819724

    Long Side

    Bottom Reinf.

    Quantities

    F o o t i n g P e d e s t a

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    L (E-W)= 0.91 m & B (N-S) = 0.80 m

    Column offset+2xEffecti

    = -0.11 m2

    (Area of trapaezoid) Footing base dimension

    = -0.16 m2

    (Area of trapaezoid)

    a = 1-(1/(1+2/3*SQRT(Lpu/Bpu))) 1-(1/(1+2/3*SQRT(Bpu/Lpu)))

    J=

    C= Lpu/2 Bpu/2

    M= a/ (0.85*J_E-W) a/ (0.85*J_N-S)

    e) x 1.8

    2.76

    ) P-edge=Ptot*(1+6*El/L)

    B^2*Xf) P-face=Ptot*(1+12*El/L^2*Xf)

    ^2*Xd) P-d =Ptot*(1+12*El/L^2*Xd)

    3+P-face/6-Pob/2}L TM

    d)*0.5-Pob}L T/m

    P-d M-face V@De Punch.sh P-edge P-face P-d M-face

    t/m2 tm t strs t/m2 t/m2 t/m2 t/m2 tm

    23.14 0.71 (2.61) 47.83 22.91 22.81 22.97 0.70

    28.26 0.83 (3.22) 47.92 23.27 23.21 23.31 0.72

    22.72 0.70 (2.56) 47.75 24.45 22.95 25.45 0.74

    0.8 -2.56 47.92 0.74

    1.2 -3.84 71.87 1.11

    7.49 6.66

    111.80

    Depth OK

    E - W N - S

    [2*(De*Lpu^3)/12]+[2*(Lpu*De^3)/12]+[De*(Bp

    u*Lpu^2/2))]

    [2*(De*Lpu^3)/12]+[2*(Lpu*De^3)/

    [(De*Lpu*Bpu^2)/2)]

    W I T H N O T E N S I O N

    FOR - M_E-W only FOR - M_N-S only

    Depth OK

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    0.07 0.05

    0.019 0.015

    0.12 0.12

    456 519

    550 628

    -0.48 -0.39

    82.14 .

    58.81

    Kgs Nos. Dia Spacing (N_S) Kgs Nos. Dia

    4 7 10 150 4 8 10

    0 m3

    2 m2

    Depth OK Depth OK

    OK

    S u m m a r y

    Short Side

    l

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    Pedge

    Pd

    ce

    TensionAllowed

    CaseII

    ce

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    e depth

    D_os

    D min

    V@De

    t

    (3.30)

    (3.36)

    (3.65)

    -3.30

    -4.95

    12]+

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    82.14

    40.41

    Spacing

    150

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    Story Point Load FX FY FZ MX MY MZ

    BASE 69 COMB1 1.26 -0.1 202.33 0.196 0.429 0.012

    BASE 69 COMB2 -0.82 -0.04 165.17 0.101 -4.63 -0.03

    BASE 69 COMB3 2.84 -0.12 158.55 0.213 5.317 0.049

    BASE 69 COMB4 0.66 -0.96 165.68 2.329 0.389 0.046

    BASE 69 COMB5 1.36 0.79 158.04 -2.015 0.298 -0.027

    BASE 69 COMB6 -1.04 -0.03 197.6 0.11 -5.832 -0.034BASE 69 COMB7 3.52 -0.14 189.33 0.25 6.602 0.065

    BASE 69 COMB8 0.8 -1.18 198.24 2.895 0.442 0.061

    BASE 69 COMB9 1.68 1.01 188.69 -2.535 0.327 -0.03

    BASE 69 COMB10 -1.54 0 120.21 0.038 -5.986 -0.04

    BASE 69 COMB11 3.03 -0.11 111.95 0.178 6.448 0.059

    BASE 69 COMB12 0.31 -1.14 120.86 2.823 0.288 0.055

    BASE 69 COMB13 1.18 1.04 111.3 -2.607 0.174 -0.036

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    135 0.29 0.13

    138 -3.86 0.08

    132 4 0

    138 0 2

    132 0 -2

    132 -4 0126 4 0

    132 0.29 1.93

    126 0 -2