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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 )
8/11/2019 Combined Isolated Footing
2/65
( 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
8/11/2019 Combined Isolated Footing
3/65
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
8/11/2019 Combined Isolated Footing
4/65
-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
8/11/2019 Combined Isolated Footing
5/65
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
8/11/2019 Combined Isolated Footing
6/65
oment
MX
2
2
orm.
8/11/2019 Combined Isolated Footing
7/65
kN m / metre
mm
mm
129 mm
8/11/2019 Combined Isolated Footing
8/65
mm2
0.16 %
mm2
mm2
0.17 %
Load per metre
8/11/2019 Combined Isolated Footing
9/65
27
685.8
mm2
8/11/2019 Combined Isolated Footing
10/65
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 )
8/11/2019 Combined Isolated Footing
11/65
( 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
8/11/2019 Combined Isolated Footing
12/65
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
8/11/2019 Combined Isolated Footing
13/65
8/11/2019 Combined Isolated Footing
14/65
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
8/11/2019 Combined Isolated Footing
15/65
oment
MX
2
1
orm.
8/11/2019 Combined Isolated Footing
16/65
kN m / metre
mm
mm
129 mm
8/11/2019 Combined Isolated Footing
17/65
mm2
0.16 %
mm2
mm2
0.17 %
Load per metre
8/11/2019 Combined Isolated Footing
18/65
mm2
8/11/2019 Combined Isolated Footing
19/65
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 )
8/11/2019 Combined Isolated Footing
20/65
( 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
8/11/2019 Combined Isolated Footing
21/65
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
8/11/2019 Combined Isolated Footing
22/65
-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
8/11/2019 Combined Isolated Footing
23/65
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
8/11/2019 Combined Isolated Footing
24/65
oment
MX
2
1
orm.
8/11/2019 Combined Isolated Footing
25/65
kN m / metre
mm
mm
129 mm
8/11/2019 Combined Isolated Footing
26/65
mm2
0.16 %
mm2
mm2
0.17 %
Load per metre
8/11/2019 Combined Isolated Footing
27/65
mm2
8/11/2019 Combined Isolated Footing
28/65
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 )
8/11/2019 Combined Isolated Footing
29/65
( 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
8/11/2019 Combined Isolated Footing
30/65
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
8/11/2019 Combined Isolated Footing
31/65
-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
8/11/2019 Combined Isolated Footing
32/65
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
8/11/2019 Combined Isolated Footing
33/65
oment
MX
1
1
orm.
8/11/2019 Combined Isolated Footing
34/65
kN m / metre
mm
mm
129 mm
8/11/2019 Combined Isolated Footing
35/65
mm2
0.16 %
mm2
mm2
0.17 %
Load per metre
8/11/2019 Combined Isolated Footing
36/65
mm2
8/11/2019 Combined Isolated Footing
37/65
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 )
8/11/2019 Combined Isolated Footing
38/65
8/11/2019 Combined Isolated Footing
39/65
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
8/11/2019 Combined Isolated Footing
40/65
-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
8/11/2019 Combined Isolated Footing
41/65
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
8/11/2019 Combined Isolated Footing
42/65
oment
MX
2
2
orm.
8/11/2019 Combined Isolated Footing
43/65
kN m / metre
mm
mm
129 mm
8/11/2019 Combined Isolated Footing
44/65
mm2
0.16 %
mm2
mm2
0.17 %
Load per metre
8/11/2019 Combined Isolated Footing
45/65
mm2
8/11/2019 Combined Isolated Footing
46/65
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 )
8/11/2019 Combined Isolated Footing
47/65
( 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
8/11/2019 Combined Isolated Footing
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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
8/11/2019 Combined Isolated Footing
49/65
-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