Month: September 2012

Design of Doubly reinforced sections | Numerical example

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7 step design procedure for Doubly reinforced sections

In our article series for “Design of Doubly reinforced sections”, we covered the following:

What are doubly reinforced sections?

Methods for determining Neutral Axis?

Solved numerical examples for determining Neutral Axis

Numerical examples for practice (Find Neutral axis)

Methods for calculating Moment of Resistance

Numerical example for calculating Moment of resistance

Types of problems in Doubly reinforced sections

6 step prodecure for determining stresses in steel and concrete 

Numerical example | Stresses in steel and concrete

7 step procedure for designing doubly reinforced sections

We shall now proceed with a numerical example “Design of Doubly reinforced sections” using the 7-step procedure we discussed in the previous article.

Numerical problem:

A doubly reinforced concrete beam 250mm wide and 600mm deep overall has to resist an external bending moment of 95kN-m. Find the amount of tensile and compressive steel required, if cover to the centre of steel on both sides is 50mm. σcbc = 5 N/mm2, σst = 140 N/mm2, m = 18.66

Given that:

Breadth of the beam = b = 250mm

Effective depth of the beam = d = 600 – 50 = 550mm

Distance of compressive steel from the top edge of the beam to the centre of the steel = d’ = 50mm

Permissible stress in concrete = σcbc = 5 N/mm2

Permissible stress in steel = σst = 140 N/mm2

Modular ratio = m = 18.66

Bending moment = M = 95 kN-m

Step one:

To find xc

σcbc/ (σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(550 – xc)

xc = 219.95 =  220mm

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Design procedure for designing doubly reinforced section

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7 step procedure for “Design of Doubly reinforced sections”

In our article series for “Design of Doubly reinforced sections”, we covered the following:

What are doubly reinforced sections?

Methods for determining Neutral Axis?

Solved numerical examples for determining Neutral Axis

Numerical examples for practice (Find Neutral axis)

Methods for calculating Moment of Resistance

Numerical example for calculating Moment of resistance

Types of problems in Doubly reinforced sections

6 step prodecure for determining stresses in steel and concrete 

Numerical example | Stresses in steel and concrete

In our previous article, we discussed a detailed 6 step procedure for determining stresses in steel and concrete followed by a numerical example. Now we shall move on with the “design procedure for doubly reinforced sections”.

Generally the following data are given:

Breadth of the beam = b

Effective depth of the beam = d

Permissible stress in concrete = σcbc

Permissible stress in steel = σst

Modular ratio = m

Bending moment = M

To solve a problem, the following procedure may be followed.

Design the beam as a singly reinforced one (balanced section)

Step One:

Find xc by

σcbc/ (σst/m) = xc/(d – xc)

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6 step procedure for determining stresses in steel and concrete | Doubly reinforced sections

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Numerical example for determining stresses in steel and concrete

In our article series for “Design of Doubly reinforced sections”, we covered the following:

What are doubly reinforced sections?

Methods for determining Neutral Axis?

Solved numerical examples for determining Neutral Axis

Numerical examples for practice (Find Neutral axis)

Methods for calculating Moment of Resistance

Numerical example for calculating Moment of resistance

Types of problems in Doubly reinforced sections

Determining stresses in steel and concrete 

Numerical example | Stresses in steel and concrete

In our previous article, we discussed a detailed 6 step procedure for determining stresses in steel and concrete. Now we shall move on with a numerical example in which we will use the 6 step procedure to solve the problem.

Problem Type two: Determining stresses in steel and concrete using the 6 step procedure

A rectangular beam is 200mm wide and 480mm deep. It has to resist a bending moment of 100 kN-m. The reinforcedment consists of four 25mm ⏀ bars on tension side and three 22mm⏀bars on compression side. The centres of bars being 30mm from the top and bottom edges of the beam. Find the stresses set up in steel and concrete. m=18.66

Given data is as follows:

Breadth of the beam = b = 200mm

Effective depth of the beam = d = 480 – 30 = 450mm

Distance of compressive steel from the top edge of the beam to the centre of the steel = d’ = 30mm

Bending moment = M = 100kN-m

Modular ratio = m = 18.66

Area of tensile steel = Ast = 4 π/4 x 25 x 25 = 1964 mm2

Area of compressive steel = Asc = 4 π/4 x 22 x 22 = 1140 mm2

Step one:

Find x:

bx.x/2 + (1.5m – 1)Asc (x – d’) = mAst(d-x)

200x2/2 + (1.5×18.66 – 1) 1140 (x – 30)

= 18.66 x 1964 x (450 – x)

Therefore, x2 + 674.17x – 174147 = 0

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Determining stresses in Steel and Concrete | Doubly reinforced Sections

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Six step procedure for determining stresses in steel and concrete

In our article series for “Doubly reinforced Sections design guide”, we have covered the following:

What are doubly reinforced sections?

Methods for determining Neutral Axis?

Solved numerical examples for determining Neutral Axis

Numerical examples for practice (Find Neutral axis)

Methods for calculating Moment of Resistance

Numerical example for calculating Moment of resistance

Types of problems in Doubly reinforced sections

Determining stresses in steel and concrete 

Numerical example | Stresses in steel and concrete

Now we shall proceed with a simple 6 step procedure for determining compressive stresses in steel and concrete. Further in our next article, we shall also solve a numerical using the same method.

Generally, the following data is given for reference with the help of which we can determine the stresses in steel and concrete

Breadth of the beam = b

Effective depth of the beam = d

Area of tensile steel = Ast

Area of compressive steel = Asc

Modular raito = m

Bending moment = M

Six step procedure for determining the compressive stresses in steel and concrete:

Step One:

Find x by using the following formula:

bx.x/2 + (1.5m – 1)Asc (x – d’) = mAst(d-x)

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