#### Moment of Resistance calculations | Doubly reinforced sections

**In our article series for “Doubly reinforced sections”, 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 move on with a solved example. This will help you understand the methods in a better way. I suggest that you do them yourselves too. Practice will help you make your concepts more concrete and clear.**

#### Numerical Example:

**An reinforced concrete 300mm x 600mm effective dimensions is provided with tensile and compressive reinforcement of 1256mm2 each. The compressive steel is placed 30mm from the top edge of the beam. If σ_{cbc} = 7N/mm^{2}, σ_{st} = 190N/mm^{2} and m = 13.33, find the moment of resistance of beam by following two methods:**

**1) Elastic theory method**

**2) Steel beam theory method**

**Given that:**

Width of the beam = b = 300mm

Effective depth of the beam = d = 600mm

Ast = Asc = 1256 mm2

Effective depth of the beam = 400 – 30 = 370mm

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

Permissible stress in concrete = σ_{cbc} = 7N/mm^{2}

Permissible stress in steel = σ_{st} = 190N/mm^{2}

Modular ratio = m = 13.33

σ_{sc} = 130N/mm^{2}

**To find Mr by Elastic Theory Method**

**To find x _{c}**

σ_{cbc}/( σ_{st}/m) = x_{c}/(d – x_{c})

7/(190/13.33) = x_{c}/(d – x_{c})

x_{c}= 197.61mm

**To find x**

bxx/2 + (1.5m – 1)A_{sc}(x – d’) = mA_{st }(d – x)

300×2/2 + (1.5 x 13.33 -1)1256 (x – 30)

= 13.33 x 1256 (600 – x)

x^{2} + 270.6x – 71741 = 0

**Solving the above equation, we get,**

x = 164.778 = 165mm

therefore, **x < x _{c}**

Hence, the beam is under-reinforced.

Therefore, σ_{st} = 190N/mm^{2}

**To find σ _{cbc}**

σ_{cbc}/( σ_{st}/m) = x/(d – x)

σ_{cbc}/( 190/13.33) = 165/(600 – 165)

σ_{cbc }= 5.4 N/mm^{2}

**To find σ _{cbc}**

σ_{cbc}= σ_{cbc }[(x – d’)/x)]

= 5.4 [(165-30)/165]

= 4.418 = 4.42 N/mm^{2}

**To find σ _{sc}**

σ_{sc} = 1.5m x σ_{cbc}

= 1.5 x 13.33 x 4.42

= 88.377 = 88.38 N/mm^{2}

**<130 N/mm ^{2}**

Hence, the design is all right.

**To find Mr**

**Taking moments about tensile steel, we get,**

M_{r}= bx. σ_{cbc}/2 (d – x/3)+(1.5m – 1)A_{sc}. σ_{cbc}(d – d’)

= 300 x 165 x (5.4/2)(600-165/3) + (1.5 x 13.33 – 1)1256 x 4.42 (600 – 30)

= 132946390 N-mm

= 132.946 kN-m

**To find Moment of resistance using Steel beam theory Method**

Mr = Ast.σst (d – d’)

= 1256 x 190 (600 – 30)

= 136024000 N-mm

Mr= 136.024 kN-m

You can leave a response, or trackback from your own site.

#1 by

Arun Shankaron August 14, 2012 - 8:24 amIs this procedure as per IS 456 ?