#### 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. σ_{cbc }= 5 N/mm^{2}, σ_{st }= 140 N/mm^{2}, 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 = σ_{cbc }= 5 N/mm^{2}

Permissible stress in steel = σ_{st} = 140 N/mm^{2}

Modular ratio = m = 18.66

Bending moment = M = 95 kN-m

#### Step one:

To find x_{c}

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

5/(140/18.66) = x_{c}/(550 – x_{c})

x_{c} = 219.95 = 220mm

#### Step two:

Find A_{st} by:

C = T

b x_{c}σ_{cbc }/2 = σ_{st}.A_{st}1

250 x 220×5/2 = 140 x A_{st}1

A_{st}1 = 982mm^{2}

#### Step three:

Find the Mr of singlt reinforced balanced beam

Mr = b x_{c}σ_{cbc }/2[d – (x_{c}/3)]

= 250 x 200 x 5/2 x [550 – (220/3)]

= 65541667 N-mm

= 65.54 kN-m

#### Step four:

Find the remaining bending moment ‘M1’

M1 = M – M_{r}

= 95 – 65.54 = 29.46kN-m = 29.46 x 10^{6 }N-mm

#### Step five:

Find A_{st2 }for M1

M1 = T x lever arm

= A_{st2.}σ_{st }x (d – d’)

= M1 / σ_{st }x (d – d’)

= 420.85 = 421 mm^{2}

#### Step six:

Ast = Ast1 + Ast2

= 982 + 421 = 1403 mm^{2}

#### Step seven:

Find Asc

Equating moments of equivalent area of tensile and compressive steel about Neutral axis(N.A)

mAst (d – x_{c}) = (1.5m – 1) Asc (x_{c}– d’)

A_{sc}= mAst (d – x_{c})/ (1.5m – 1)(x_{c}– d’)

18.66 x 421 x (550 – 220) = (1.5 x 18.66 – 1)Asc (220 – 50)

Asc = 565 mm^{2}

#### Another similar problem for practice:

A rectangular reinforced beam is 360 x 750mm effective. The beam has to resist a bending moment of 300 kN-m. Find the tensile and compressive steel required for the beam. Take σ_{cbc }= 7 N/mm^{2}, σ_{st }= 190 N/mm^{2}, m = 13.33 and d’ = 50mm

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