New Concept of Modern Bathrooms

Posted by BenzuJK in Building Construction on May 13, 2012

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.

New concept of Modern Bathrooms

New concept of Modern Bathrooms

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.


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Guide to design of Singly reinforced beams | Solved examples

Posted by BenzuJK in Building Construction on May 11, 2012

Singly reinforced beams | Building Construction

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.

(σcbc = 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


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Numerical example 2 | Singly reinforced Sections

Posted by BenzuJK in Building Construction on May 10, 2012

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


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Solved numerical examples | Design of Singly reinforced sections

Posted by BenzuJK in Building Construction on May 9, 2012

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

To find lever arm

z = d – xc/3

= d – 0.4d/3

= 0.867dmm

= 0.87d mm

To find Moment of resistance

Mr = C x z

= bxc (σcbc/2)z

= b (0.4d)(5/2)(0.87d)

= 0.87 bd2 N-mm

To find the percentage of steel

Equating, C = T

bxc.σcbc/2 = Ast. σst

Therefore, Ast = bxc.σcbc/2 σst

 = [b (0.4d)(5/2)]/140

= bd/140 mm2

Pt = Ast. 100/bd

= (bd/140) x (100/bd)

= 0.71

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Moment of Resistance | Design of Singly reinforced Sections

Posted by BenzuJK in Building Construction on May 8, 2012

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.

Moment of resistance | Singly reinforced Section

Moment of resistance | Singly reinforced Section

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


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Solved Numericals for Singly reinforced Sections | Design Method 2

Posted by BenzuJK in Building Construction on May 7, 2012

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)

Step One:

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)


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Design of Singly reinforced sections | Design Method 2

Posted by BenzuJK in Building Construction on May 6, 2012

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)
Stress-strain diagram

Stress-strain diagram

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.


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