In a cross-section of concrete beam, if the reinforcements are provided in both the compression and tension zones, it is called a “**Doubly reinforced beam**”.

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## Why to provide Doubly reinforced beam?

This type of beam is mostly provided when the depth of the beam is restricted. and by increasing the steel in the tension zone, the moment of resistance cannot be increased.

This type of beam is provided to increase the moment of resistance of a beam having limited dimensions.

## Analysis of Doubly reinforced beam

Analysis steps for doubly reinforced rectangular beams are summarized below:

**Step(1)** Find the limiting value of depth of neutral axis (Xu,max) by using the clause 38.1 of code IS 456:2000.

**Step(2)** Assuming fsc = fst = 0.87fy and considering force equilibrium, find Xu;

Where, fcc = compressive stress in concrete at the level of compressive steel

So, we can take

Fcc = 0.446fck

**Step(3)** Compare Xu with Xu,max, if Xu

If Xu>Xu,max, then the section is over reinforced section and hence use strain compatibility method to find fst.

**Step(4)** Find the strain in compressive steel

**Step(5)** Find the moment of resistance (Mu) by using equation;

### Example:

**Q) Find the moment of resistance (Mu) of the beam for the following data given below:**

**Width of beam (b) = 250 mm**

**Effective depth of beam (d) = 360 mm**

**Effective cover (d’) = 40 mm**

**Grade of concrete is M _{20}**

**Grade of steel is fe 415**

**Steel used is 3-16 mm ****ɸ bars**

Solution:

From given,

B = 250 mm

d = 360 mm

d’ = 40 mm

f_{ck} = 20 N/mm^{2}

f_{y} = 415 N/mm^{2}

3-16 mm ɸ bars are used.

Since steels bars provided in only tension side, so it is singly reinforced rectangular beam.

Now, Analysis steps:

**Step(1) Calculation of area of reinforcement in beam:**

**Step(2) Calculation of depth of neutral axis (xu):**

By using horizontal equilibrium:

C = T

- 0.36 f
_{ck}bxu = f_{s}Ast

Assume, it is under reinforced section, then, we have for under reinforced section

**Step(3) Calculation of limiting value of depth and neutral axis (xu,lim):**

From clause 38.1 of code IS 456:2000

- Xu,max = 0.48*360

= 172.8 mm

**Comparison of xu and xu,max:**

Since, xu (=120.98)<xu,max(=172.8 mm)

Hence, section is under reinforced section

Then, assumption in step(2) is OK.

**Step(5) Calculation of moment of resistance (Mu):**

Since, for under reinforced section,

= 67497298.35 N-mm

Mu = 67.49 KN-m

Hence, Moment of resistance of beam is

** Mu = 67.49 KN-m**

## Design of Doubly reinforced beam

**1) A doubly reinforced beam of size 230 mm x 500 mm effective is subjected to a factored moment of 200 KNm. Use M20 concrete and Fe 415 steel.**

Solution;

Given:

Breadth ( b )= 230 mm

Depth( d )= 500 mm

M_{u} = 200 KNm = 200 x 10^{6} mm

F_{ck} = 20 N/mm^{2}

f_{y} = 415 N/mm^{2}

**Step-1**

**Step-**2

**Limiting moment of resistance (M _{ulim})**

= 0.48 x 500 = 240 mm

M_{u lim} = 0.36x 20x 230x 240 (500- 0.42 x 240 )

= 158658048 Nmm

M_{u2} = M_{u }– M_{u lim}

= 200 x 10^{6} – 158658048

M_{u2} = 41341952 Nmm

**Step-**3

**Area of tension steel (A _{st})**

A_{st } = A_{st1} + A_{st2}

Where,

Total area of tension steel = A_{st1} + A_{st2}

= 1100.7 + 254.2 = 1354.9 mm^{2}

** provide 5 – 20 mm diameter bars as tension steel.**

**Step-**3

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