How do I design isolated rectangular footing as per practical considerations?

How do I design isolated rectangular footing as per practical considerations?

footing

 

DEFINITION:- 

A concrete support under a foundation that rests in solid ground and is wider than the structure supported. Footings distribute the weight of the structure over the ground. A footing or a shallow foundation is placed immediately below the lowest part of the superstructure.Foundation may be broadly classified under two heads :shallow foundation and Deep foundation .According to Terzaghi, a foundation is shallow if it is depth is equal to or less than its width. In the case of deep foundations, the depth is equal to or greater than the width. Apart from deep strip, rectangular or square foundations, other common forms of deep foundations are : pier foundation,pile foundation and well foundation. The shallow foundations are of the following types: spread footing (or simply, footing), strap footing,combined footing, and mat or raft footing.

Types of Isolated footing:-

 (i) Pad footing

(ii) Stepped footing

(iii) Sloped footing

 Shapes of Isolated footing:-

(i) Square footing

(ii) Rectangular footing

(iii) Circular footing.

Image:- www.chegg.com

CALCULATION OF FOOTING SIZE:-

Ø Footing Size Calculation of Column :

Axial load of column = 635.74 kn
Considering 10% additional load including in the footing = 635.747 x 1.1 = 699.321 kn.
Bearing capacity of soil = 120 kn/m2
So, Area of footing = 699.321 / 120 = 5.827 m2
Let, The ratio of B to L = 250/400 = 5/8
Now, (5/8L) x L = 5.827 or, L=3m & B = 2m

1.DESIGN CONSTANT:

For M20 concrete and Fe 415 steel,
xu,max/d = 0.479 ; Ru = 2.716

2. CALCULATION OF FOOTING SIZE:

Load in column W=635.74kN Let would be equal to 10% w=63.574Kn
Taking bearing capacity of soil 120 kn/m2
So, area of footing=(635.74+63.574)/120 m2=5.83 m2
Let ratio of B to L=25/40=5/8
So, 5/8L*L=5.83 or L=3m and B=2m
Now net upward pressure Po=635.74/(2*3)=106kN/m2

3. DESIGN OF SECTION:

Bending moment M1 about section X-X is
M1 = {(P0*B)*(L-a) 2}/8 = {(106*2)*(3.0-0.4)2}/8 =179.14 kn-m=179.14*106 N-mm
M1u = 1.5M1 = 1.5*179.14*106 = 268.71*106 N-mm
So, d=√ (M1u/Ru*B) =√ {268.71*106/ (2.716*2000)} =222.41mm =225 mm
Keep d=225 mm and total depth = 291 mm,
Providing uniform thickness for the entire footing.
Now bending moment about section Y-Y is
M2 = {(P0*L)*(B-b)2}/8 ={(106*3)*(2.0-0.25)2}/8 *106 N-mm =121.73*106 N-mm
M2u = 1.5*M2 m= 1.5*121.73*106 = 182.6*106 N-mm
Thus, M2u<M1u so effective depth found above has to be checked for shear.

4. CHECK FOR SHEAR:

Ø Depth on the basis of one way shear
Shear force along section AB is
V=P0B {(L-a)/2-d} =106*2{(3.0-0.4)/2-0.001d}*103 = 212000(1.3-0.001d) N/m
Vu =1.5v = 318000(1.3 – 0.001d) N/m
τ
v
=Vu/Bd =318000(1.3 – 0.001d) /2000d =159(1.3-0.001d)/d N/mm2
Assuming under reinforced section with P=0.3%, we get τ
c
=0.384 N/mm2 and for M20 concrete also k=1 So Permissible shear stress = 1* 0.384 = 0.384 N/mm2
Equating this to the equation we get
0.384 = 159(1.3-0.001d)/d from which d = 380 mm
For the two way action or punching shear action along ABCD
Perimeter ABCD =2{(a + d) + (b + d)} = 2{(0.4+0.38) + (0.25+0.38)} = 2.82m = 2820 mm
Area ABCD = {(a + d) + (b + d)} = {(0.4+0.38)*(0.25+0.38)} = 0.4914 m2
So, punching shear = 1.5*106*[(2*3)-0.4914] = 875.87 kn
τ
v
={(875.87*1000)/(2820*380)} = 0.817 N/mm2
Allowable shear stress τ
c
is
τ
c
= 0.25√fck = 0.25√20 = 1.118 N/mm2
Ks = {0.5+ (o.25/0.4)} = 1.125 however adopt maximum Ks =1
So, ksτ
c
= 1*1.118 = 1.118 N/mm2
This is more than τ
v
= 0.817 N/mm2 Hence safe from punching shear
Now taking depth = 450 mm so, d= (450-60-6) = 384mm providing effective cover = 60 mm in one direction and other direction d = (384-12) =372 mm using 12 mm Ф bar.

5. DESIGN FOR REINFORCEMENT:

Area Ast1 of long bars calculated for moment M1u is
Ast1 = (0.5fck/fy)[1-√{1-(4.6M1u/fckBd12)}]Bd1
= (0.5*20/415)[1-√{1-(4.6*268.71*106/20*2000*3842)}]*2000*384
= 2052.98 mm2
Hence number of 12mm Ф bars = [2052.98/ (π/4)122] = 18 no.
The area Ast2 of short bars calculated for M2u is
Ast2 =(0.5fck/fy)[1-√{1-(4.6M2u/fckLd22)}]Ld2
=(0.5*20/415)[1-√{1-(4.6*182.6*106/20*2000*3722)}]*2000*372
=1396.47 mm2
Ø Hence number of 12mm Ф bars = [1396.47/ (π/4)122] = 12 no.

6. CHECK FOR DEVELOPMENT LENGTH:

Development length Ld = 47 = 564 mm
Providing 60 mm side cover, length available ½[B-b]-60 = ½[2000-250] =815 mm
This is greater than Ld.
Hence it is OK.

Design of isolated footing