Learn : Analysis of a doubly reinforced beam working stress method : modular ratio, Equivalent section,critical neutral axis, actual neutral axis, stresses in the section, moment of resistance.

Table of Contents

## Analysis of a doubly reinforced beam working stress method

** Modular Ratio**

(a) Modular ratio for tensile steel is taken as m where \[m = \frac{280}{3\sigma_{cbc}}\].

(b) Modular ratio for compressive steel is denoted by m_{c} and is taken as

m_{c}=1.5m

As per IS 456 (Table 2.2) the compressive stress in steel in the compression zone is calculated by multiplying the stress in surrounding concrete (σ_{c}) by 1.5 m but this value should not exceed the permissible stress in steel bars in compression i.e. σ_{sc}} as given Table 2.2.

** Equivalent Section**

The equivalent or transformed section of the given doubly reinforced beam in terms of concrete is shown in Fig. 2.8(b). In this equivalent section

(a) The tensile steel area (A_{st}) is replaced by an equivalent concrete area i.e., m.A_{st} and the stress in this equivalent concrete area is\[ \frac{\sigma_{st}}{m}\].

(b) The compression area consists of :

- Area of concrete above the neutral axis excluding the area of compressive steel i.e., b.n – A
_{sc} - The area of compression steel (A
_{sc}) is replaced by an equivalent concrete area i.e., m_{c}.A_{sc}

\therefore Net compression area =bn-A_{sc}+m_{c}.A_{sc}

=b.n+(m_{c}-1)A_{sc}

The stress in equivalent concrete area (m_{c}.A_{sc}) is σ_{c} and stress in compression steel is found as

σ_{sc}=m_{c}. σ_{c}

σ_{sc} =1.5m σ_{c}

** Critical Neutral Axis (n**_{c})

_{c})

The depth of critical neutral critical axis is obtained by the same method as that for singly reinforced section i.e., using the permissible stress values.

From the stress diagram

\[\frac{\sigma_{cbc}}{\frac{\sigma_{st}}{m}}=\frac{n_{c}}{d-n_{c}}\]

which is same as that for a singly reinforced section.

**Actual Neutral Axis (n)**

The actual neutral axis of a doubly reinforced section is calculated by taking the moment of the tension and compression area about the neutral axis.

Moment of compression area about neutral axis

= Moment of tensile area about neutral axis

Moment of compression area =\[b.n\frac{n}{2}-A_{sc}(n-d’)+m_{c}.A_{sc}(n-d’)\]

\[=\frac{bn^{2}}{2}+(m_{c}-1)A_{sc}(n-d’)\]

Moment of tensile area about neutral axis

\[=m.A_{st}(d-n)\]

Now equating them, we get

\[\frac{b.n^{2}}{2}+(m_{c}-1)A_{sc}(n-d’)=m.A_{st}(d-n)\]

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

The above equation can be solved and actual neutral axis depth is obtained.

** Stresses in the Section**

The stresses diagram of a doubly reinforced section is shown in Fig. 2.8(c).The stresses developed in steel and concrete are as follows :

(i) Maximum compressive stress in concrete =σ_{cbc}

(ii) Stress in equivalent concrete at the level of compression steel = σ_{c}

(iii) Stress in equivalent concrete at the level of steel tensile steel \[=\frac{\sigma_{st}}{m}\]

(iv) Stress in compression steel

\[m_{c}\sigma’_{c}=1.5m\sigma’_{c}=\sigma_{sc}\]

(v) Stress in tensile steel = σ_{st}

** Moment of Resistance **

The moment of resistance of a doubly reinforced section is calculated by taking the moment of the compressive forces about the centroid of tensile reinforcement. The compressive forces are shown in Fig. 2.8(c).

The moment of resistance of the doubly reinforced beam (M) is written as

\[M_{r}=M_{1}+M_{2}\]

where M_{1} is the moment of resistance of the similar balance section without compression steel.

M_{2} is the additional moment of resistance provided by the compression steel.

M_{1} = Moment of the compressive force is concrete about the center of tensile steel.

\[=C_{1}\times a\]

where C_{1} is the compressive force carried by concrete

\[M_{1}=\frac{1}{2}\sigma_{cbc}b.n\left ( d-\frac{n}{3} \right )\] or Rbd^{2}

\[M_{2}=C_{2}\times(d-d’)\]

where C_{2} is the compressive force carried by compressive steel.

C_{2}= Equivalent area in terms of concrete×Compressive stress

\[C_{2}=(m_{c}-1)A_{sc}\times \sigma’_{c}\]

\therefore \[M_{2}=(m_{c}-1)A_{sc}.\sigma’_{c}(d-d’)\]

\[=(1.5m-1)A_{sc}.\sigma’_{c}(d-d’)\]

\[M_{r}=M_{1}+M_{2}\]

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

In this expression σ_{c }is calculated from the stress diagram.

\[\frac{\sigma_{cbc}}{n}=\frac{\sigma’_{c}}{n-d’}\] [From similar triangles]

\[\sigma’_{c}=\sigma_{cbc}\left ( \frac{n-d’}{n} \right )\]