# Part-1: Introduction

Can you design and calculate a simple house plan?

I have noticed that so many people are struggling to find the full structure design of a simple house.

There are so many books that are available for reinforced concrete structure design.

I have found that books provide overall detail of structure design but in actual practice, the methods are different. that's why I decided to give you the complete detail of how we can actually design the house in the structural terms.

In the design of the house, we need to calculate all the members. it is the step by step process.

There are so many software are available currently but, we should know how to calculate it manually.

So, a very basic way to lear full design is to practice manually on paper after that practice in an excel sheet after that software. we can use software for the analysis and for other requirements.

If we calculate all the members in a paper it might take so many days or even months because even a simple house could have a minimum of 15 to 20 columns and beams and also so many slabs per floor.

all the members need to be designed and analyzed which is a very lengthy process. I suggest practice only 4 to 5 different beams, columns, and slabs manually and after that do it on the software it saves the time. Architectural plan

I will provide you a stepwise process for house structure design.

The steps include column position and orientation, the calculation for slabs, beams, columns, and footing.

First of all, we need to study the basic parameter and general codel provision. after that, we will start the step by step design calculation.

In G+2 structures, the earthquake, and the wind load should not be considered. snow load should be considered according to the site condition if applicable.

firstly, we will calculate the superstructure and collect the required data and then footing design will be done.

Following are the types of loads that we will require for further consideration in the designing of the members.

The following are the formulas that can be useful in the design of reinforced concrete structures :

### Floor finish :

The floor finish should be assumed as 1 kN/m2.

### Sunk :

Sunk weight depends upon the material to be filled and taken depth. In this example, we will take 300 mm down sunk withe material density of 16 kN/m3.
Hence, the total sunk weight will be 4.8 kN/m2.

Wall :

Thickness of wall  x (height of the wall - beam depth) x Density kN/m.

### Slab Weight :

Thickness x density kN/m2.

Self-weight of beam :

Breadth x depth x density kN/m.

### Self-weight of column :

Breadth x depth x density kN/m.

Live load also known as an imposed load which is taken from the IS 875 (part- 2): 1987.

Some of the basic consideration such as the spacing of reinforcement, minimum, and maximum reinforcement criteria for reinforcement in a beam, slab, and column are given in (IS: 456-2000) which are as follow.

## Spacing in reinforcement

(IS: 456-2000, p.45)

### Horizontal distance

The minimum horizontal distance between bars should not be less than greater of below three values,

• The diameter of the bars are equal then it should not be less than the diameter of the bar,
• If the diameter of the bars are not equal then it should not be less than the diameter of the larger bar,
• Maximum size of the coarse aggregate + 5 mm.

### Vertical distance

The minimum vertical distance of bars should not be less then greater of below three if the bars are arranged in rows.

•     Should not less than 15 mm,
•     Should not less than two-thirds of the maximum size of the aggregate,
•     Should not less than the maximum size of the bar.

## Slab

(IS: 456-2000, P- 48)

### Minimum reinforcement

Mild steel (Fe 250) - 0.15 % of total c/s area

High yield strength deformed bars  (HYSD) or tor steel ( Fe 415) -                             0.12 % of the total c/s area.

### ·  Maximum diameter

The maximum bar diameter should be < 1/8 x slab thickness

### ·  Minimum diameter

Main steel,

Minimum dia. for plain bars         -   10 mm

Minimum dia. for deformed bars -   8 mm

Distribution steel,

Minimum dia. - 6 mm

## Beam

IS : 456-2000, P- 46

### ·  Minimum tension reinforcement

The minimum tension reinforcement can be calculated from the equation

As   = 0.85

bd         fy

In percentage,

For mild steel (fe- 250)     - 0.34 %,

For HYSD steel  (Fe- 415) - 0.20 %.

### ·  Maximum reinforcement

The maximum compression or tension  reinforcement shall not exceed 4 % of the total area.

Max. reinforcement  <  0.04bD

### Shear reinforcement (stirrups)

Max. spacing :

Should not greater than following,

For vertical stirrups = 0.75 d,

For inclined stirrups = d,

300 mm

·  Min. shear reinforcement :

Min. shear reinforcement should be provided from the following equation :

Asv     0.4

bSv      0.87 fy

## Column

IS : 456-2000, P- 46

### ·   Longitudinal reinforcement

1.       the diameter of the bars should not less than 12 mm,

2.      The c/s area of bars should be between 0.8 % to 6 % of the gross cross-sectional area of the column.

3.      For square or rectangular column, minimum 4 numbers of bars and for a circular column, a minimum of 6 numbers of bars should be provided.

4.      for helical reinforcement, a minimum of 6 number of bars should be provided.

5.      The spacing of the longitudinal bars measured along the periphery of the column should not greater than 300 mm.

### ·    Lateral ties

1.       Pitch (vertical distance between two consecutive ties)

Should not greater than following value :

A.       The least lateral dimension of the column,

B.       16 x the diameter of the smallest bar,

C.        300 mm

2.      Diameter

The diameter of the lateral ties should not less than the following values :

A.     1/4 x diameter of the largest bar

B.     6 mm.

After getting the architectural plan and required data, the first step is to place the column according to design and draw a beam grid.
We will work on the following plan,

The next step is the column position and beam grid.

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