Week 12 Challenge
1.
- Click on YES to start the Design Process.
- Design of the Footing has been done succesfully.
Following design result has been found out:
- Dimension of footing:3 mx 3 m
- Over all depth of footing D = 650 m
- Depth of footing is checked for BM, Single shear & Double shear and found safe
- No tension steel is needed as there is no negative pressure.
- Development length = 648 mm
Now,for validation of design of footing using hand calculation.
VALIDATION OF DESIGN THROUGH HAND CALCULATION
Footing should be symmetric with respect to column as the moment is reversible( Mx&MyMx&My are same ,that is square footing is to be designed.
Service load = 1500KN
Considering Self weight of footing & weight of backfill soil = 10% of service load
P = 1.1x 1500= 1650 KN
PuPu= 1.5x1650KN = 2475 KN
Area required = 1.1PSBCPSBC = 1.1x1500/200 = 8.25 sqm`
Acal=8.25m2Acal=8.25m2
Therefore L=B=√A=√8.25=2.87mA=8.25=2.87m
`provide square footing of size 3.0 m x 3.0 m
Aprovided=9m2Aprovided=9m2 > Area required which is 8.25 sqm (OK)
STEP-2 : NET UPWARD SOIL PRESSURE
σ(max,min)=PdAprovided+MuZσ(max,min)=PdAprovided+MuZ
= (1.5x 1.1x1500/9) ±±( 1.5x150/(B×L2/6)(B×L2/6) ( B=L=3 M)
= 275±±50
σ(max)σ(max)= 325 KN/ sqm < 370 KN/sqm ( Gross factored Bearing capacity) ( ok)
σ(min)σ(min)= 225 KN-m < 340 KN/sqm
Eccentricity e = M/P = 150/1500 = 0.1M = 100 mm
L/6= 3000/6 =500 mm
e < L/6 ,therefore no negative soil pressure will develop hence compression steel is not required.
STEP-3 DEPTH OF FOOTING ON THE BASIS OF B.M.
The maximum bending moment at the face of column (M)
= P0×(B−b)2/8P0×(B-b)2/8 = 275×3(3−0.3)2/8275×3(3-0.3)2/8= 751.78 KNm
Mu=0.138fckbd2Mu=0.138fckbd2
751.78×106×106 = 0.138x20x3000xd2d2
d = 301 mm Say d = 305 mm
Adopt d = 600 mm due to shear consideration
Add effective cover = 50 mm
D = 650 mm
Depth of footing from one way shear-
d= Pu(L−a)2(P+τc×L2)Pu(L-a)2(P+τc×L2)
= 1.5X1.1X1500( 3-0.3)/2(1.5x1.1x1500+350x9)
= 6682.5/11250 =0.594 m = 594 mm < 600 ( OK)
Depth required for two way shear ( punching shear)
PL2[L2−(a+d)2]=4(a+d)d×τpPL2[L2-(a+d)2]=4(a+d)d×τp
[1.5x1.1x15009][(9−(0.3+0.6)2]1.5x1.1x15009][(9-(0.3+0.6)2]= 4x0.9 x0.6τpτp
τpτp= 1043.98 KN/m^2 = 1.044 N/mm2mm2
Permissible τpτp= 0.25√fckfck
=0.25 √2020
= 0.25x4.472 = 1.118 N/mm2mm2 > 1.044N/mm2mm2 ( OK)
STEP-4: Reinforcement calculation
Ast=0.5fckfy[1−{1−(4.6×Mu/fckBd2)Ast=0.5fckfy[1-{1-(4.6×Mu/fckBd2)
AstAst= 5081 sqmm
p = AstB×d×100AstB×d×100
= (5081/3000x650)x100 = 0.26% > 0.12% which is min reqr , hence OK
xud=1.2−[1.44−6.6Mufck×b×d2]12xud=1.2-[1.44-6.6Mufck×b×d2]12
=1.2−[1.44−(6.6x751780x100020x3000x600x600)0.5=1.2-[1.44-(6.6x751780x100020x3000x600x600)0.5
= 0.0998
Lever Arm Z = d(1-0.416xud)xud)
= 600( 1-0.416x0.0998)
= 575.089 mm
Ast=Mu0.87×fy×ZAst=Mu0.87×fy×Z
= 751780X1000/0.87X415X575.089 = 3620.66 mm2mm2
STEP-5 : DEVELOPMENT LENGTH
As per IS 456: 2000,Cl 26.2.1.1
σs=415Nmm2σs=415Nmm2
τbdτbdfor deformed bar and M20 Grade of concrete = 1.92 N/mm2mm2
Ld=ϕ×σA4τbdLd=ϕ×σA4τbd
415ϕϕ/4x1.92 = 54ϕϕ = 54x12 = 648 mm for 12 mm dia bar.
Result: Isolated Footing for the given data has been designed succesfully using STAAD Foundation and summary of the design is as follow:
- Dimension of footing: 3 mx 3 m
- Over all depth of footing D = 650 mm
- Depth of footing is checked for BM, Single shear & Double shear and found safe
- No tension steel is needed as there is no negative pressure.
- Development length = 648 mm.
2.