## New Concept of Modern Bathrooms

To say that bath spaces have changes would be an understatement. The enormous sophistication in the range of bath products and the change in the way people look at the space, has attracted several international players into the Indian markets, witha slew of technologically advanced and designed products.

The new trendy bathroom designs are making use of lot of wood furniture into the bathroom. Various companies are into the race for designing the furniture specifically for the application area and is resistant to both water and humidity.

Wood surfaces act as a link between the bathroom as a space and its diverse range of elements. When fitted with panelling, drawers or shelves, washbasin, bathtubs and shower trays become items of furniture in their own right. These elements can also be complemented by wall cabinets with valuable storage space, bathrooms are currently in a period of transition.

## Guide to design of Singly reinforced beams | Solved examples

#### For design of “Singly reinforced beam” article series, we have covered the following:

Now we will move on with another solved example where we will calculate the Moment of resistance and determine the position of the Neutral axis. For this we will have to use the formulas that we have derived earlier in our previous articles in the list above.

#### Numerical Problem

Calculate the moment of resistance of an RC beam 250x550mm overall. Reinforcement is 1521mm2 and is placed at a distance of 25mm from the bottom.

cbc = 7N/mm2, σst = 140N/mm2, m = 13.33)

#### Given that:

b = breadth of a rectangular beam = 250mm

d = effective depth of a beam = 550 – 25 = 525mm

x = depth of neutral axis below the compression edge = ?

Ast = cross-sectional area of steel in tension = 1531mm2

σcbc = permissible compressive stress in concrete in bending = 7N/mm2

σst = permissible stress in steel = 140 N/mm2

m =  modular ratio = 13.33

We have to find the value for Moment of resistance. To calculate Mr, we have to first calculate NA(critical) and NA (actual).

#### To find Neutral axis (NA) (critical):

σcbc /(σst/m) = xc/(d – xc)

7/(140/13.33) = xc/(500 – xc)

xc = 209.969mm = 210mm

## Numerical example 2 | Singly reinforced Sections

#### Guide to design of Singly reinforced Sections | Civil Engineering

For “Singly reinforced sections” article series, we have covered the following:

Now we will move on with our next solved example in which we will make use of formulas derived earlier. That is why it is necessary that you go through the entire step by step guide in order to gain complete understanding.

#### Numerical Problem

Calculate the moment of resistance of an RC beam 250mm wide, the depth of the centre of reinforcement being 500mm. Assume σcbc = 5N/mm2, σst = 140 N/mm2, modular ratio = 18.66

#### Given that,

b = width of the beam = 250mm

d = depth of the beam = 500mm

σcbc = permissible compressive stress in concrete in bending = 5N/mm2

σst = permissible stress in steel = 140 N/mm2

m =  modular ratio = 18.66

#### To find Neutral Axis (NA)

σcbc /(σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(500 – xc)

Xc = 199.95mm = 200mm

## Solved numerical examples | Design of Singly reinforced sections

#### Guide to design of Singly reinforced Sections | Building Construction

For the “design of Singly reinforced sections” article series, we have covered the following:

Now we will move on with our next solved numerical example in which we will make use of the formulas that we have derived in our earlier articles.

#### Numerical Problem

Determine the following:

a)      The position of the neutral axis

b)      Lever arm

c)       Moment of resistance

d)      Percentage of steel

(For a rectangular beam section of width b mm and effective depth d mm. Take σcbc = 5 N/mm2, σst = 140 N/mm2, m = 18.66

#### To find Neutral Axis (NA)

σcbc /(σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(d – xc)

Therefore, xc  = 0.399d mm= 0.4dmm

## Moment of Resistance | Design of Singly reinforced Sections

#### Moment of Resistance | Guide to design of Singly reinforced Sections

For the design of Singly reinforced Sections article series, we have covered the following:

Now we will move on with our discussion on “Moment of resistance” and derive the formula for Moment of resistance for balanced section, under-reinforced section and over reinforced section.

The moment of resistance of the concrete section is the moment of couple formed by the total tensile force (T) in the steel acting at the centre of gravity of reinforcement and the total compressive force (C) in the concrete acting at the centre of gravity (c.g.) of the compressive stress diagram. The moment of resistance is denoted by M.

The distance between the resultant compressive force (C) and tensile force (T) is called the lever arm, and is denoted by z.

From the diagram above, it is clear that the intensity of compressive stress varies from maximum at the top to zero at the neutral axis.

Therefore, centre of gravity of the compressive force is at a distance x/3 from the top edge of the section.

Therefore, z = d-x/3

## Solved Numericals for Singly reinforced Sections | Design Method 2

#### Design of Singly reinforced Sections | Method 2

In our article series for Singly reinforced sections, we have covered the following:

#### Numerical Problem

Find the position of the neutral axis of a reinforced concrete beam 150mm wide and 400mm deep (effective). Area of tensile steel is 804mm2. (modular ratio = m = 18.66)

#### Given that:

b = breadth of a rectangular beam = 150mm

d = effective depth of a beam = 400mm

Ast = cross-sectional area of steel in tension = 804mm2

x = depth of neutral axis below the compression edge

m = modular ratio = 18.66

Taking moments of area of compression and tension sides about neutral axis,

bx.x/2 = mAst (d – x)

## Design of Singly reinforced sections | Design Method 2

#### Guide to design of Singly reinforced Sections

For “Singly reinforced sections” article series, we have covered the following:

Now we will move on with our discussion on “2nd Design method” for the design of Singly reinforced Sections.
We will follow a simple three-step procedure for the design of singly reinforced sections.

#### Step One:

Given that:

• Dimensions of section (b and d)
• Area of tensile steel (Ast)
• Modular ratio (m)

From the figure above, we can see that the neutral axis is situated at the centre of gravity of a given section. Therefore, the moments of area on either side are equal.

## Solved numericals for Singly reinforced Sections | Design Method 1

#### Design of Singly reinforced Sections | Method 1

In our article series for Singly reinforced sections, we have covered the following:

#### Numerical Problem

An RC beam 200mm wide has an effective depth of 350mm. The permissible stresses in concrete and steel are 5N/mm2 and 140 N/mm2 respectively. Find the depth of neutral axis, area of steel and percentage of steel. (modular ratio (m) = 18.66)

#### Given that:

b = breadth of a rectangular beam = 200mm

d = effective depth of a beam = 350mm

x = depth of neutral axis below the compression edge = ?

Ast = cross-sectional area of steel in tension = ?

σcbc = permissible compressive stress in concrete in bending = 5N/mm2

σst = permissible stress in steel = 140 N/mm2

m = modular ratio = 18.66

#### From the concrete stress diagram, the formula is given as,

σcbc/(σst/m) = x/(d – x)

5/(140/18.66) = x/(350-x)

Therefore, x = 139.97mm

## Design Methods for Singly reinforced Sections

#### Singly reinforced sections | Design of RCC structures

Earlier we discussed some basic terms in reference to singly reinforced sections design. It is important that you are thorough with the basic definitions and have complete understanding of stresses in concrete and steel. You should also possess the knowledge of reinforcement and terminology of beams which includes understanding singly reinforced beam, doubly reinforced beam, under reinforced beam, over reinforced beam and balanced reinforced beam.

There are two methods for the design of singly reinforced sections. In this article we will discuss the first method of singly reinforced section in a stepwise manner. The discussion will include the method for determining the value of neutral axis followed by a formula for the area of steel calculations.

Let,

b = breadth of a rectangular beam

d = effective depth of a beam

x = depth of neutral axis below the compression edge

Ast = cross-sectional area of steel in tension

σcbc = permissible compressive stress in concrete in bending

σst = permissible stress in steel

m = modular ratio

Neutral axis

Neutral axis is denoted as NA.

There are two methods for determining the neutral axis depending on the data given.

In this article, we will discuss the first method followed by a couple of numericals for your understanding and then move on to the second method.

## Assumptions for Singly reinforced Sections | RCC Structures

#### In our series of articles for singly reinforced sections, we have covered the following:

Now, we will move on with our discussion on “assumptions for singly reinforced sections”.

1. The sections that are plane before bending remain plane after bending, at any cross-section.
2. All tensile stresses are taken up by steel reinforcement and none by concrete.
3. The stress to strain relationship of steel and concrete under working load is a straight line.
4. The modular ratio m has the value 280/3σcbc
5. There is a perfect adhesion between steel and concrete and no slip takes place between steel and concrete.