01 Footing With Moment

ISOLATED FOOTING WITH BIAXIAL MOMENT

DESIGN OF BI-AXIAL ISOLATED RCC FOOTING (IS 456, 2000) Building Name Footing Number: Node number  COLUMN W IDTH (Breadth) = 600 mm DEPTH (Length) = 600 mm  Axial load from output (P1) = 1000 KN Moment about Z axis (M z) = 20 KN-m KN-m Moment about X axis (M x) = Safe bearing pressure =  AREA=

20 KN-m KN-m 2

400 KN/m 2.75 mm

2

1 x2 Required length of the footing Required width of the footing Depth required 124.090897 Ov er All Depth 180.090897

PEDASTAL Breath = Length =

0 x

-2.75 L w

1.658312 -1.65831

900 mm 900 mm Breadth 1.8 m

FOOTING Revised Footing length (L, dim. ) = Revised Footing Breadth (W , ) = Depth of footing (t) = Clear cover of footing = Main bar dia of footing = Effective depth of footing = Selfweight of the footing =  Area of Footing(A) = Sect mod of foot about Z axis (Zz) = Sec mod of foot about X axis (Zx) = MATERIALS OF CONSTRUCTION Grade of concrete f ck ck = Grade of steel f y = Unit wt of soil = CHECK FOR GROSS BEARING PRESSURE Safe NET bearing pressure = Safe gross bearing pr. =  Axial load from output (P1) = Moment about Z axis (M z) =

1.8 1.8 350 50 12 294 28.35 3.24 0.97 0.97

m m mm mm mm mm KN 2 m 3 m 3 m

global X Length 1.8 m Footing Dimensions

2

25 N/mm 415 N/mm

3

18 KN/m

2

KN/m 3 KN/m KN KN-m KN-m

20 1.15 18 50.30 1078.65 374.07 291.76 Hence footing is safe against max gross bearing pr.

KN-m KN-m m 3 KN/m KN KN 2 KN/m 2 KN/m

Moment about X axis (M x) =

DESIGN FORCES  Axial load:(Pu) =

1500 KN

Moment about Z axis (M uz) =

30 KN KN-m -m

Moment about X axis (M ux) =

30 KN KN-m -m

Maximum effective soil pressure pe max ( Pu/Area+ Muz/Zz + Mux/Zx) = Minimum effective soil pressure pe min

global Z

global Z

400 427.00 1000 20

Depth of top of foot. from ground = Unit wt of soil = Weight of soil retained above foot = P = (P1+soil+foot self wt) = Maximum bearing pressure = Minimum bearing pressure =

global X

524.69 KN/m

2

2

( Pu/Area - Muz/Zz - Mux/Zx) = 401.23 KN/m Design of footing is done using above maximum effective soil pressure

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(net pr. + depth of foot * soil unit wt)

P  A

±

 M   y  Z  y

±

 M   x  Z  x

ISOLATED FOOTING WITH BIAXIAL MOMENT

CALCULATION FOR BOTTOM STEEL 2 Mu about X1 X1 X1 = ( pe max x length /2)=

 Ast  =

0.5  f  ck 

  f   y

é ê1 ê ë

1-

53.13 KN-m per meter  Mulimit = 298.43 KN-m per meter  The section is singly reinforced

ù ú bd    f  ck  bd  û ú 4.6 M  u

2

2 Hence, Ast = 515.746 mm 2 Min Ast = (0.12 % for slab, cl 26.5.2.1) 352.800 mm Spacing required = (considering max of above two calculated values of Ast) 219.29 mm Spacing provided = 150 mm pt provided = 0.26 % pt required = 0.18 % provide a maximum 0.26 % Hence provide 12 mm dia bar @ 150 mm c/c parellel to length of footing ( || to Z) 2

Mu about N1 N1 = ( pe max x length /2)= 53.13 KN-m per meter  2 Calc. Ast = The section is singly reinforced 515.746 mm 2 Min Ast = (0.12 % for slab, cl 26.5.2.1) 352.8 mm Spacing = (considering max of above two calculated values of Ast) 219.29 mm Spacing provided = 150 mm pt provided = 0.26 % pt required = 0.18 % provide a maximum 0.26 % Hence provide 12 mm dia bar @ 150 mm c/c parellel to breadth of footing ( || to X) Arrangement of bottom reinforcement as per above design is shown below 12 mm dia bar @ 150 mm c/c

12 mm dia bar @ 150 mm c/c

1

1

Footing Length 18 1800 mm

Breadth 18 1800 mm

Sec 1-1 600

600 1194

X1

L1

X

a

a

Z

Z N1

N1 a

a

L2

L2 X1

156

X

L1

Breadth 1800 mm

1194 Footing Length 1800 mm PLAN

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156

ISOLATED FOOTING WITH BIAXIAL MOMENT

CHECK FOR ONE WAY SHEAR : One way shear at critical section L1- L1 Distance of critical sec. from edge of footing = Shear force Vs =pe max x 0.156 x 1m width of footing = tv = Vs/bd = 0.278 N/mm Shear stress 2 N/mm tc = 0.37

0.156 m 81.852 KN

tv < tc hence O.K. One way shear at critical section L2- L2 Distance of critical sec. from edge of footing = Shear force Vs =pe max x 0.156 x 1m width of footing = 2 tv = Vs/bd = 0.278 N/mm Shear stress tc = 0.37 N/mm tv < tc hence O.K. CHECK FOR TWO WAY SHEAR Ref. cl 34.2.4 and cl.31.6.3 of IS 456 : 2000  Allowable shear stress stress t v allowable = ks = ( 0.5 + bc) = Hence, ks= tc = 0.25 (fck)

0.156 m 81.852 KN

kstc

1.5 >1 1

0.5

=

tv allowable = ks x tc =

2

1.25 N/mm 1.25 N/mm

951.98 KN 4776 mm 2 1404144 mm

Shear force Vs = 524.691 ( 1.8 x 1.8 - 1.194 x 1.194) = Length of critical section = 2 x ( 1194 + 1194) =  Area of the critical section (length of critical sec x eff. d ) = Hence shear stress tv = 0.678 N/mm tv < allowable hence O.K.

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