# Types of problem in doubly reinforced beams working stress method

Types of problem in doubly reinforced beams working stress method: Determination of moment of resistance of the given section,Determination of actual stresses in concrete and steel,Design of the section.

## Types of problem in doubly reinforced beams working stress method

1. Determination of moment of resistance of the given section.
2. Determination of actual stresses in concrete and steel.
3. Design of the section.

### Determination of Moment of Resistance

Given :

(i)        Dimension of the beam section (b and d)

(ii)       Area of tensile steel Ast and area of compressive steel Asc

(iii)     Permissible stress in concrete {σcbc) and permissible stress in steel (σst)

#### Procedure :

1. Calculate $m=\frac{280}{3\sigma_{cbc}}$
2. Calculate critical neutral axis (nc)

$\frac{n_{c}}{d-n_{c}}=\frac{m.\sigma_{cbc}}{\sigma_{st}}$

1. Calculate actual neutral axis depth (nc)

$\frac{b.n^{2}}{2}+(1.5m-1)A_{sc}(n-d_{c})=m.A_{st}(d-n)$

1. Compare n and nc

(a)       If n>nc the section is under reinforced      (fully stressed)

Maximum tensile stress developed in steel = σst Maximum compressive stress developed in concrete

$\sigma_{cbc}(where \sigma’_{cbc})is less than \sigma_{cbc}$

$\frac{\sigma’_{cbc}}{\sigma_{st}/m}=\frac{n}{d-n}$                                 (from stress diagram)

$\therefore\sigma’_{cbc}=\frac{\sigma _{st}}{m}\left ( \frac{n}{d-n} \right )$

The stress in concrete at the level of compression steel (σc) can be obtained as

$\sigma’_{c}=\frac{\sigma _{cbc}}{n}(n-d’)$

The moment of resistance of the (under reinforced) doubly reinforced section is calculated as :

$M_{r}=\frac{1}{2}\sigma_{cbc}.bn\left ( d-\frac{n}{3} \right )+(1.5m-1)\sigma’_{c}.A_{sc}(d-d’)$

(b)       If n>nc, then section is over reinforced and max compressive stress in concrete is  σcbc. The moment of resistance is calculated as

$\therefore M_{r}=\frac{1}{2}\sigma_{cbc}.b.n\left ( d-\frac{n}{3} \right )+(1.5m-1)\sigma’_{c}.A_{sc}(d-d’)$

where             $\sigma’_{c}=\frac{\sigma_{cbc}}{n}(n-d’)$

### Determination of Stress in Steel and Concrete

Given :

(i)        Dimensions of beam (b and d)

(ii)       Area of tensile and compressive reinforcement i.e., Ast and Asc

(iii)     Material used grade of concrete and steel.

#### Procedure :

1. Determine actual neutral axis of the section

$\frac{bn^{2}}{2}+(1.5m-1)A_{sc}.(n-d’)=m.A_{st}(d-n)$

1. Determine the value of σc in terms of σcbc

$\sigma’_{c}=\frac{\sigma_{cbc}}{n}(n-d’)$

1. Determine the maximum bending moment (M) on the section due to loads and equate it to moment of resistance of the section (Mr)

$M=M_{r}=\frac{1}{2}\sigma_{cbc}b.n.\left ( d-\frac{n}{3} \right )+(1.5m-1)\sigma’_{c}.A_{sc}(d-d’)$

Putting the value of σc in this equation

$M=\frac{1}{2}\sigma_{cbc}b.n\left ( d-\frac{n}{3} \right )+(1.5m-1)A_{sc}\frac{\sigma _{cbc}}{n}(n-d’)(d-d’)$

In the above equation, only σc as unknown and it can be calculated easily.

1. Knowing σcbc the stress in tensile steel (σst) and stress in compressions steel (σsc) are calculated as under

$\sigma_{st}=\frac{m\sigma_{cbc}}{n}(d-n)$

and                  $\sigma’_{st}=\frac{\sigma_{cbc}}{n}(n-d’)$

$\sigma_{sc}=m_{c}.\sigma’_{c}$

### Design of the Section

Given :

(i)        Span of the beam (l) and its dimensions (b and d).

(iii)     Material used-grade of concretes and steel i.e., σcbc and σst.

#### Procedure :

1. Determine maximum bending moment (M) coming on the section due to loads (including self weight of the beam).
2. Calculate the design constants k, j and R for the given materials.
3. Calculate M1=Rbd2.
4. Calculate the area of tensile reinforcement (Ast1) corresponding to M1.

$A_{st}=\frac{M_{1}}{\sigma_{st}jd}$

1. Calculate M2

M2 =M-M1

1. Calculate the additional area of the tensile reinforcement (Ast2) needed to resist M2.

$M _{2}=\sigma _{st}.A _{st _{2}}(d-d’)$

$\therefore A_{st^{2}}=\frac{M_{2}}{\sigma_{st}(d-d’)}$

1. Determine total area of tensile steel (Ast)

$A _{st} =A _{st _{1}}+A _{st _{2}}$

Selecting suitable diameter of the bar, provide Ast.

1. Determine the area of compressive steel (Asc) by equating the moment area of compressive steel (Asc) to the moment of the area of additional tensile steel (Ast2) about neutral axis.

Moment of the area of compressive steel (Asc) about neutral axis.

$= (m _{c}-1) A _{sc} (n-d’) = (1.5m-1) A _{sc} (n-d’)$

Moment of the additional area of tensile steel (Ast2) about neutral axis

$= m.A _{st _{2}}(d-n)$

Equating them and calculating ASC.

$1.5(m-1)A _{sc} (n-d’)=m.A _{st _{2}}(d-n)$

$A_{sc}=\frac{m.A_{st_{2}}(d-n)}{1.5(m-1)(n-d_{c})}$

Calculate the number of bars required for providing Asc.

1. Design for shear and bond is same as that of singly reinforced beam.
2. Draw a neat sketch and give summary of design.

Note : Design consideration for a doubly reinforcement beam as per IS 456:2000.

(i)        Maximum Compression Reinforcement (Asc): The maximum compression reinforcement in a beam cannot be more than 0.04 bD. (4%) of the gross cross-sectional area).

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