Content of this article:Learn In singly Reinforced beam Strain distribution,Stress distribution,Stress block parameters,Neutral axis depth,Limiting depth of neutral axis

Table of Contents

## Analysis of a singly Reinforced beam

### singly Reinforced beam

The beam that is longitudinally reinforced only in tension zone, it is known as singly reinforced beam. In Such beams, the ultimate bending moment and the tension due to bending are carried by the reinforcement, while the compression is carried by the concrete.

Practically, it is not possible to provide reinforcement only in the tension zone, because we need to tie the stirrups. Therefore two rebars are utilized in the compression zone to tie the stirrups and the rebars act as false members just for holding the stirrups.

Consider a singly reinforced beam section subjected to bending as shown in fig

### 1.Strain Distribution:

The assumption (1) of the limit state theory gives a linear strain distribution across the cross section as shown in fig (b).

It varies as zero at the neutral axis and maximum at the extreme fibres.

The various silent points of the strain diagram are:

(i) Strain at neutral axis = 0

(ii) Maximum or ultimate strain in concrete at extreme fibre,∈_{cu} = 0.0035

(iii) Strain at constant stress of 0.67 ƒ_{ck},,∈_{c} = 0.002

(iv) Ultimate strain in steel corresponding to maximum stress at failure,

[Latexpage]

\[\epsilon _{su}=\frac{0.87f_{y}}{E_{s}}+0.002\]

### 2. Stress distribution

The stress diagram is shown in fig c.

It has a parabolic shape from A to B and then linear from B to C above the neutral axis.

The various salient points of the stress diagram are:

(i) Stress at neutral axis (Pt. A) = 0

(ii) Stress at 0.002 strain (Pt. B)

\[\frac{0.67f_{ck}}{1.5}= 0.446 f_{ck}\]

(iii) Stress at extreme fibre (Pt. C) = 0.446 f_{ck}

(iv) Below the neutral axis, the concrete is assumed to be cracked and maximum stress in steel

\[\frac{f_{y}}{1.15} = 0.87f_{y}\]

### 3. Stress block parameters

For the stress strain curve of concrete the design stress block parameter are taken as following as per IS 456:2000

Area of stress block = 0.36f_{ck}x_{u}

Depth of centre of compressive force from the extreme fibre in compression = 0.42x_{u}

### 4. Neutral axis depth ( x_{u })

Neutral axis is the axis at which the stresses are zero and it is situated at the centre of gravity of the section. The depth of neutral axis for a singly reinforced beam is calculated by equilibrium of tensile and compressive forces as shown in fig

b = Width of section

d = Effective depth of beam

A_{st} = Area of steel reinforcement

x_{u} = Depth of neutral axis

C = Total compression

T = Total tension

Total tension (T) = 0.87f_{y}.A_{st}

Total compression (C) = 0.36 f_{ck}x_{u}.b

at 0.42x_{u} from the top extreme fibre (as per IS 456:2000)

For equilibrium of forces

Total Tension = Total compression

0.87f_{y}.A_{st} = 0.36 f_{ck}x_{u}.b

\[x_{u} = \frac{0.87f_{y}A_{st}}{0.36f_{ck}bd}\]

which is given in clause G1.1 of IS code

\[x_{u} = \frac{0.87f_{y}A_{st}}{0.36f_{ck}bd}\]

### 4.1 Limiting depth of neutral axis (x_{u} max)

The strain distribution for the singly reinforced beam is shown in fig.

The maximum strain in concrete is 0.0035.

As per code, The strain in steel at failure should not be less than

\[\epsilon _{su}=\frac{0.87f_{y}}{E_{s}}+0.002\]

From the fig.

[latexpage]

\[\frac{0.0035}{x_{umax}}=\frac{\frac{0.87f_{y}}{E_{s}}+0.002}{d-x_{umax}}\]

\[\frac{x_{u max}}{d-x_{u max}}= \frac{0.0035}{\frac{0.87f_{y}}{E_{s}}+0.002}\]

\[x_{u max}\left ( \frac{0.87f_{y}}{E_{s}}+0.002+0.0035 \right )= 0.0035d\]

\[\frac{x_{u max}}{d}= \frac{0.0035}{\frac{0.87f_{y}}{E_{s}}+0.0055}\]

The above equation gives the limiting or maximum values of depth of neutral axis for different grades of steel.

The following table is based upon the above equation and gives limiting values depth of neutral axis for different grades of concrete.

Table Values of x_{umax}/d

Grade of steel | fy(N/mm2) | Xumax/d |
---|---|---|

Mild steel Fe250) | 250 | 0.53 |

Fe 415 | 415 | 0.48 |

Fe 500 | 500 | 0.46 |

The value of E

_{s}or modulus of elasticity is taken as 2×10

^{5}N/mm

^{2}