DESIGN OF BI-AXIAL ISOLATED RCC FOOTING (IS 456, 2000)
Building Name Conveyer Supporting Structure
Footing Number: L2,L3,M2,M3,N1,N4,P1,P4
COLUMN
Length (l, dim. || Z axis ) =
Breadth (b, dim. || X axis) =
600 mm
500 mm
global
X
Breadth 1 m
FOOTING
Foot length (L, dim. || Z axis) =
Foot Breadth (B, dim. || X axis) =
Thickness of footing (t) =
Clear cover of footing =
Main bar dia of footing =
Effective depth of footing dz =
Effective depth of footing dx =
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 fck =
Grade of steel fy =
2
1
530
50
12
474
474
26.50
2.00
0.33
0.67
( Pu/Area+ Muz/Zz + Mux/Zx) =
Minimum effective soil pressure pe min
global
Z
global
Z
global
Length 2 m
Footing Dimensions
2
20 N/mm
2
500 N/mm
CHECK FOR GROSS BEARING PRESSURE
Safe NET bearing pressure =
300
Safe gross bearing pr. =
320.60
Unfactored load case number =
1
Axial load from output (P1) =
304.66667
Moment about Z axis (Mz) =
44.666667
Moment about X axis (Mx) =
7.3333333
Depth of top of foot. from ground =
0.5
Unit wt of soil =
20
Weight of soil retained above foot =
17.00
P = (P1+soil+foot self wt) =
348.17
Maximum bearing pressure =
319.08
Minimum bearing pressure =
29.08
Hence footing is safe against max gross bearing pr.
DESIGN FORCES
Factored load comb. no.
Axial load:(Pu) =
Moment about Z axis (Muz) =
Moment about X axis (Mux) =
Maximum effective soil pressure pe max
m
m
mm
mm
mm
mm
mm
KN
m2
m3
m3
KN/m2
KN/m3
(net pr. + depth of foot * soil unit wt)
KN
KN-m
KN-m
m
KN/m3
KN
KN
KN/m2
KN/m2
P
P MMy y M M

  x x
A
A ZZ
ZxZ x
y y
1
457 KN
67 KN-m
11 KN-m
2
446.00 KN/m
2
( Pu/Area - Muz/Zz - Mux/Zx) =
11.00 KN/m
Design of footing is done using above maximum effective soil pressure
CALCULATION FOR BOTTOM STEEL
Mu about X1 X1 = ( pe max x length2/2)=
Ast 
0.5 f ck
fy
109.27 KN-m per meter
Mulimit =
597.75 KN-m per meter
The section is singly reinforced

4 .6 M u 
1  1 
 bd
f ck bd 2 

2
Hence, Ast =
545.933 mm
2
Min Ast =
636.000 mm
Spacing (reqd.) =
177.83 mm
pt required =
0.13 %
Hence required 12 mm dia bar @ 177 mm
(0.12 % for slab, cl 26.5.2.1)
(considering max of above two calculated values of Ast)
Sp (prov.) =
100 mm
Ast (prov.) =
c/c parellel to length of footing ( || to Z)
pt (prov.) =
0.24 %
Mu about N1 N1 = ( pe max x length2/2)=
13.94 KN-m per meter
2
Calc. Ast =
67.872 mm
The section is singly reinforced
2
mm
Min Ast =
636
(0.12 % for slab, cl 26.5.2.1)
Spacing (reqd.) =
177.83 mm
(considering max of above two calculated values of Ast)
pt required =
0.13 %
Sp (prov.) =
100 mm
Ast (prov.) =
Hence required 12 mm dia bar @ 177 mm c/c parellel to breadth of footing ( || to X)
Arrangement of bottom reinforcement as per above design is shown below
pt (prov.) =
0.24 %
12 mm dia bar @ 100 mm c/c
12 mm dia bar @ 100 mm c/c
1
1
Footing Length 2000 mm
Breadth 1000 mm
Sec 1-1
1074
500
974
X1
L1
X
a
a
Z
Z
N1
N1
a
a
L2
L2
X1
226
X
L1
Breadth 1000 mm
600
Footing Length 2000 mm
PLAN
-224
1130.97
mm2
1130.97
mm2
CHECK FOR ONE WAY SHEAR :
One way shear at critical section L1- L1
Distance of critical sec. from edge of footing =
Shear force Vu =pe max x 0.226 x 1m width of footing =
2
tv = Vs/bd =
Shear stress
0.213 N/mm
2
tc =
0.352 N/mm
0.226 m
100.796 KN
tv < tc hence O.K.
One way shear at critical section L2- L2
Distance of critical sec. from edge of footing =
Shear force Vu =pe max x -0.224 x 1m width of footing =
2
tv = Vs/bd =
Shear stress
-0.211 N/mm
2
tc =
0.352 N/mm
-0.224 m
-99.904 KN
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 tv allowable =
kstc
ks = ( 0.5 + bc) =
Hence, ks=
tc = 0.25 (fck)0.5 =
1.33333 >1
1
2
1.11803 N/mm
2
1.11803 N/mm
Shear force Vs = 446 ( 2 x 1 - 1.074 x 0.974) =
Length of critical section = 2 x ( 1074 + 974) =
Area of the critical section (length of critical sec x eff. d ) =
2
Hence shear stress tv =
0.219 N/mm
tv < ks tc hence O.K.
tv allowable = ks x tc =
1.5 tc =
425.45 KN
4096 mm
2
1941504 mm
2
1.677 N/mm