Chapter 4 (cont..3) - Minimum and Maximum spacing of bars in a beam

In the previous section we saw the steps required to complete the design part. In this section we will see the rules that we have to follow while arranging the steel inside the beam.

Minimum spacing that should be provided between the bars of a beam

The clear space provided between parallel reinforcing bars should not be less than the minimum value specified in cl 26.3.2 of the code. For a beam, the minimum spacing specified by the code is shown in fig 4.10 below:

Fig.4.10
Minimum horizontal spacing between bars

Fig.4.10(a) shows the case when the all the bottom bars of the beam are of the same diameter. Fig.4.10(b) shows the case when bars of different diameters are used in the bottom layer.

Based on fig.4.10(a), we can write an equation to quickly calculate S_h:
Eq.4.12
S_h = [b-(2C_c +2Φ_l +3Φ)] /2

Where Φ is the diameter of the main bar and Φ_l is the diameter of the links. This equation is applicable when the bottom bars contain only 3 bars and all of them are of the same diameter.

If in the fig., the middle bar is of a different diameter, the eq. can be written as:
Eq.4.13
S_h = [b-(2C_c +2Φ_l +2Φ₁ +Φ₂)] /2
Where Φ₁ is the diameter of the two edge bars and Φ₂ is the diameter of the middle bar.

By providing a spacing greater than this minimum spacing, we can ensure that concrete is placed uniformly in between and around the bars and can be compacted well during the placement of fresh concrete.

In general, the slabs require a low percentage of flexural reinforcement. That is., when we take a 1m width of a slab of thickness Dm, the area of cross section of the slab will be 1 xD =D m², and only a lesser percentage of this D m² is required as the area of tension steel A_st. So there will be sufficient space to provide the required steel in one layer. And there will be no need to reduce the spacing between the bars to values which are below the minimum required values specified. But the beams will be having a limited width, and require relatively higher percentage of flexural reinforcement. So when we arrange the bars by giving the required minimum spacing S_h between the bars and the clear cover C_c required on the two sides, the total space required may exceed the width of the beam. Fig.4.11 shows such an example:

Fig.4.11
Insufficient width of a beam

In the fig.4.11, let  Dia. Φ of all the four bars in the bottom layer = 20mm; Dia of links Φ_l = 10mm; C_c = 30mm; Size of aggegate = 20mm. Then S_h = larger of {20} and {20 +5} = 25mm. So the total width of beam required = 

2C_c + 2Φ_l + 4Φ +3S_h = 235mm

But the total width available is only 200mm. In such a situation, the following options can be considered:

• Increase the beam width

• Place the bars in two or more layers with the minimum required spacing between layers as shown in the fig.4.12
• Bundle groups of parallel bars as shown in fig.4.13. Each bundle can have two, three or four bars

Fig.4.12
Bars of a beam placed in layers

Fig.4.13
Bundles of bars in a beam

A brief description about bundled bars and 'Equivalent diameter' can be seen here.

Maximum Spacing allowable between bars of a beam.

The clear space provided between parallel reinforcing bars should not be greater than the maximum value specified by the code. (cl 26.3.3). In the case of beams, we get the maximum allowable values from table 15 of the code. This table gives the clear distance between the parallel reinforcement bars or groups near the tension face of the beam. The clear distance that we provide between the bars should not exceed these values. By providing a clear distance lesser than the allowable values, we can ensure that the crack widths will be minimum and also that there is sufficient bond between the bars and concrete. 

In the next section we will discuss about the minimum and maximum areas of tension steel in beams.

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