Civil Engineering Projects

New Concept of Modern Bathrooms

BenzuJK — Sun, 13 May 2012 14:35:57 +0000

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

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.

New Concept of Modern Bathrooms

Today people expect more from them then the traditional showering or cleaning their teeth. The bathroom is becoming a living room space.

And the winner is ….. Technology. “In the fast paced urban living of today, when technology and gizmos are a way of life, homes have become compact and multi-functional and the transition of spaces uncomplicated. A typical living room would combine an entertainment console which could pull out into a dining table, doubling as an office desk.”

New Concept of Modern Bathrooms

At the end of the trend analysis, the last man standing is technology and the ways in which it is used to create both product and ambiance, generating the look of the 21st century. New realities have engendered the need for new solutions and technology has made them happen.

From ergonomic seating solutions to offices, to massage chairs to fire retardant materials and low emission wood conglomerates, and an entire gamut of lighting products technology has helped change the way interiors look all over the world.

Guide to design of Singly reinforced beams | Solved examples

BenzuJK — Fri, 11 May 2012 19:03:10 +0000

Guide to design of Singly reinforced beams | Solved examples

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.

(σ_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 = ?

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

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

σ_st = 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):

σ_cbc /(σ_st/m) = x_c/(d – x_c)

7/(140/13.33) = x_c/(500 – x_c)

x_c = 209.969mm = 210mm

To find N.A (actual):

Taking moment of area of compression and tension side about N.A., we get

bxx/2 = mA_st (d – x)

250x²/2 = 13.33 x 1521 (525 – x)

x² + 162.19x – 85154 = 0

On solving the above equation, we get two values out of which one is positive and the other negative.

x = 221.77mm = 222mm

Therefore, x > x_c

Since x is greater than x_c, it is clear that the actual N.A. is below the critical N.A. Hence the beam is over-reinforced.

To find Moment of resistance:

M_r = C x z

= bx_c (σ_cbc/2)z

= 250 x 222 x (7/2) x (525 – 222/3)

= 87606750 N-mm

= 87.6 kN-m

Numerical example 2 | Singly reinforced Sections

BenzuJK — Thu, 10 May 2012 19:29:18 +0000

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

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

σ_st = permissible stress in steel = 140 N/mm²

m = modular ratio = 18.66

To find Neutral Axis (NA)

σ_cbc /(σ_st/m) = x_c/(d – x_c)

5/(140/18.66) = x_c/(500 – x_c)

X_c = 199.95mm = 200mm

To find lever arm

z = d – x_c/3

= 500 – 200/3

= 433.33mm

To find Moment of resistance

M_r = C x z

= bx_c (σ_cbc/2)z

= 250 x 200 x 5/2 x 433.33

= 54166250 N-mm

= 54.166 kN-m

M_r = T x z

= A_st. σ_st.z —————————equation 1

To find A_st

Equating, C = T

bx_c.σ_cbc/2 = A_st. σ_st

Therefore, A_st = bx_c.σ_cbc/2 σ_st

A_st = (250 x 200 x 5/2)/140

A_st= 892.85 mm2

Substituting the value of A_st in equation 1;

M_r = T x z

= A_st. σ_st.z

= 892.85 x 140 x 433.33

= 54165817 N-mm

= 54.1658 kN-m

From the above example, it is clear that in case of a balanced section, the Mr can be calculated either as Mr = C x z or as Mr = T x z. The values obtained for moment of resistance are the same for both the formulas.

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Civil Engineering Projects

Solved numerical examples | Design of Singly reinforced sections

BenzuJK — Wed, 09 May 2012 18:21:30 +0000

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 σ_cbc = 5 N/mm², σ_st = 140 N/mm², m = 18.66

To find Neutral Axis (NA)

σ_cbc /(σ_st/m) = x_c/(d – x_c)

5/(140/18.66) = x_c/(d – x_c)

Therefore, x_c = 0.399d mm= 0.4dmm

To find lever arm

z = d – x_c/3

= d – 0.4d/3

= 0.867dmm

= 0.87d mm

To find Moment of resistance

M_r = C x z

= bx_c (σ_cbc/2)z

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

= 0.87 bd² N-mm

To find the percentage of steel

Equating, C = T

bx_c.σ_cbc/2 = A_st. σ_st

Therefore, Ast = bx_c.σ_cbc/2 σ_st

= [b (0.4d)(5/2)]/140

= bd/140 mm²

P_t = Ast. 100/bd

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

= 0.71

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Civil Engineering Projects

Moment of Resistance | Design of Singly reinforced Sections

BenzuJK — Tue, 08 May 2012 18:08:56 +0000

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.

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

Moment of resistance is given by,

Mr = C x z

= bx (σcbc/2)(d-x/3)

Mr = T x z

= Ast σst(d – x/3)

For balanced section, the formula is as follows,

Mr = bxc σcbc (d – xc/3)

= Ast σst(d – xc/3)

For under-reinforced section, the formula is as follows,

Mr = T x z

= Ast.σst (d – x/3)

For over-reinforced section, the formula is given as,

Mr = C x z

M_r = bx( σ_cbc /2) (d – x/3)

Solved Numericals for Singly reinforced Sections | Design Method 2

BenzuJK — Mon, 07 May 2012 17:13:24 +0000

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)

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)

150×2/2 = 18.66 x 804 (400 – x)

75×2 = 15002.64(400-x)

75 x2 = 6001056 – 15002.64x

x2 + 200x – 80014 = 0

After solving the quadratic equation, we will get two values (a negative value and a positive value)

x = 200mm

Design of Singly reinforced sections | Design Method 2

BenzuJK — Sun, 06 May 2012 17:11:35 +0000

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)

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.

Therefore, moment of area on compression side = moment of area on tension side

Moment of area on compression side = Area of compression side x distance of c.g. of compression area from N.A.

= (bx).(x/2)

= bx.x/2

Step Two:

Moment of area on tension side = equivalent area of concrete x distance of c.g. of tensile steel from N.A.

= (m Ast) x (d-x)

= mAst(d-x)

Step three:

Where, mAst = equivalent area of concrete

Therefore, bx.x/2 = mAst (d – x)

Solved numericals for Singly reinforced Sections | Design Method 1

BenzuJK — Sat, 05 May 2012 12:55:15 +0000

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)

Step One:

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

Step two:

To find area of steel

Equating total compressive force (C) to total tensile force (T)

C = T

C = area x average compressive stress

= (b.x) X (σcbc + 0)/2

= bx (σcbc/2)

T = area x tensile stress

= Ast x σst

Therefore, bx (σcbc/2) = Ast x σst

200 x 139.97 x 5/2 = Ast x 140

Therefore, Ast = 499.89 mm2

Calculating area of Steel (pt)

Area of steel is expressed as a percentage. The formula for percentage of steel is as follows;

pt = Ast x 100/ bd

= 499.89 x 100/(200×350)

= 0.714

Design Methods for Singly reinforced Sections

BenzuJK — Fri, 04 May 2012 12:46:50 +0000

Design Methods for Singly reinforced Sections

Singly reinforced sections | Design of RCC structures

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

Now we will move on with our step by step discussion on the first method of designing Singly reinforced sections.

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.

Stress strain diagram

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.

We will follow a simple two step procedure.

Step One:

Given that:

Dimensions of the section (b and d)
Permissible stresses in concrete and steel (σcbc and σst)
Modular ratio (m)

From the above diagram, the formula is as follows:

σcbc/(σst/m) = x/(d – x) ——————————– equation 1

From the above equation 1, the value of x can be determined.

Step two:

To find area of steel

Equating total compressive force (C) to total tensile force (T)

C = T

C = area x average compressive stress

= (b.x) X (σcbc + 0)/2

= bx (σcbc/2)

T = area x tensile stress

= Ast x σst

Therefore, bx (σcbc/2) = Ast x σst ———————-equation 2

Calculation of neutral axis can be done from equation 1 and the area of steel from equation 2.

The area of tensile steel is expressed as a percentage (pt) of the effective section.

pt = Ast x 100/bd

Assumptions for Singly reinforced Sections | RCC Structures

BenzuJK — Thu, 03 May 2012 12:17:23 +0000

Assumptions for Singly reinforced Sections | RCC Structures

Singly reinforced Sections | Design of 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”.

The equivalent stress-strain diagram is developed with respect to the mentioned assumptions in the post.

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

In our next article, we will discuss the design methods of Singly reinforced sections.

No related posts.

Civil Engineering Projects

New Concept of Modern Bathrooms

Guide to design of Singly reinforced beams | Solved examples

Singly reinforced beams | Building Construction

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

Numerical Problem

Given that:

To find Neutral axis (NA) (critical):

To find N.A (actual):

To find Moment of resistance:

Numerical example 2 | Singly reinforced Sections

Guide to design of Singly reinforced Sections | Civil Engineering

Numerical Problem

Given that,

To find Neutral Axis (NA)

To find lever arm

To find Moment of resistance

To find Ast

Solved numerical examples | Design of Singly reinforced sections

Guide to design of Singly reinforced Sections | Building Construction

Numerical Problem

To find Neutral Axis (NA)

To find lever arm

To find Moment of resistance

To find the percentage of steel

Moment of Resistance | Design of Singly reinforced Sections

Moment of Resistance | Guide to design of Singly reinforced Sections

For balanced section, the formula is as follows,

For under-reinforced section, the formula is as follows,

For over-reinforced section, the formula is given as,

Solved Numericals for Singly reinforced Sections | Design Method 2

Design of Singly reinforced Sections | Method 2

Numerical Problem

Step One:

Given that:

Design of Singly reinforced sections | Design Method 2

Guide to design of Singly reinforced Sections

Step One:

Step Two:

Step three:

Solved numericals for Singly reinforced Sections | Design Method 1

Design of Singly reinforced Sections | Method 1

Numerical Problem

Step One:

Given that:

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

Step two:

To find area of steel

Calculating area of Steel (pt)

Design Methods for Singly reinforced Sections

Singly reinforced sections | Design of RCC structures

We will follow a simple two step procedure.

Step One:

Step two:

Assumptions for Singly reinforced Sections | RCC Structures

Singly reinforced Sections | Design of RCC Structures

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

To find A_st