**Isolated Footing Design Example:**

Let us consider an isolated footing for an RCC column of size 450mm x 450mm. Loads from this column to the foundation are:
Vertical Load: 1000 kN
Uniaxial Moment: 100 kNm
The safe bearing capacity (SBC) of soil is 300 kN/m2. The grade of concrete to be used is M30 and grade of steel is Fe415.
**Step by Step Procedure of Isolated Footing Design:**

**Step -1: Determining size of footing:**Loads on footing consists of load from column, self weight of footing and weight of soil above footing. For simplicity, self weight of footing and weight of soil on footing is considered as 10 to 15% of the vertical load. Load on column = 1000 kN Extra load at 10% of load due to self weight of soil = 1000 x 10% = 100kN Therefore, total load P = 1100 kN. Size of footing to be designed can be square, rectangular or circular in plan. Here we will consider square isolated footing. Therefore, length of footing (L) = Width of footing (B) Therefore area of footing required =

^{2}Provide Length and width of footing = 2m Area of footing = 2 x 2 = 4m

^{2}Now the pressure on isolated footing is calculated as

When calculated, pmax = 325 kN/m^{2}

pmin = 175 kN/m^{2}

^{2}. Consider width and length of footing = L =B =2.25m Now, pmax = 250.21 kN/m2 (<300 kN/m2 -> OK) and pmin = 144.86 kN/m2 > 0 (OK) Hence, factored upward pressure of soil = pumax = 375.315 kN/m2 pumin = 217.29 kN/m2 Further, average pressure at the center of the footing is given by Pu,avg= 296.3 kN/m2 and, factored load, Pu= 1500 kN, factored uniaxial moment, Mu= 150 kN-m.

**Step 2: Two way shear**Assume an uniform overall thickness of footing, D =500 mm Assuming 16 mm diameter bars for main steel, effective depth of footing ādā is d = 500 ā 50 ā 8 = 452 mm The critical section for the two way shear or punching shear occurs at a distance of d/2 from the face of the column (Fig. 1), where a and b are the dimensions of the column.

**Fig 1: Critical section for Two Way Shear (Punching Shear)**Hence, punching area of footing = (a + d)

^{2}= (0.45 + 0.442)

^{2}= 0.796 m

^{2}where a = b = side of column Punching shear force = Factored load ā (Factored average pressure x punching area of footing)

= 1500 ā (296.3 x 0.0.796)

= 1264.245 kN

Perimeter along the critical section = 4 (a+d) = 4 (450+ 442) = 3568 mm Therefore, nominal shear stress in punching or punching shear stress= 1264.245 x 1000/(3568x442) = 0.802 N/mm2

Allowable shear stress =

where

^{2}Since the punching shear stress (0.802 N/mm2) is less than the allowable shear stress (1.369 N/mm2), the assumed thickness is sufficient to resist the punching shear force. Hence, the assumed thickness of footing D = 500 mm is sufficient. Please note, there is much difference between allowable and actual values of shear stress, so depth of footing can be revised and reduced. For our example, we will continue to use D = 500mm.

**Step 3: Design for flexure:**The critical section for flexure occurs at the face of the column (Fig. 2).

**Fig. 2 Critical section for flexure**

^{2}

Ast = pt x bxd

considering 1m wide footing, Ast required = 1171.1 mm^{2}/ m width Provide 16 dia bar @ 140mm c/c Repeat this exercise for other direction as well. Since, uniform base pressure is assumed, and it is a square footing, Mu and Ast for other direction will be same.

**Step 4: Check for One-Way Shear:**The critical section for one way shear occurs at a distance of ādā from the face of the column. Factored maximum upward pressure of soil, pu,max= 375.315 kN/m2 Factored upward pressure of soil at critical section, pu= 375.315 kN/m2 For the cantilever slab, total Shear Force along critical section considering the entire width B is

Vu = Total Force X (l - d) X B

= 375.315 X (0.9 - 0.442) X 2 = 343.8 kN

Nominal shear stress = Vu/(Bxd) = 0.346 N/mm^{2}

^{2}Therefore, the foundation is safe in one-way shear.

**Step 5: Check for development length**Sufficient development length should be available for the reinforcement from the critical section. Here, the critical section considered for Ld is that of flexure. The development length for 16 mm diameter bars is given by Ld= 47xdiameter of bar = 47 x 16 = 752 mm. Providing 60 mm side cover, the total length available from the critical section is 0.5x( L - a) - 60 = 0.5x(2250 - 450) - 60 = 840 > Ld, Hence O.K.

**Download**

**Isolated Footing Design Excel Calculation**

**Read More:**

**What is Punching Shear? Punching Shear in Slabs and Foundations**

**Combined Footing Design with Example and Types of Combined Footing**