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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Chapter 4 (cont..2) - Calculation of required Effective depth and Area of steel

In the previous section we saw how to fix up the preliminary dimensions of the beam section. We also saw the checks that have to be done immediately after fixing up these dimensions.

We can now take up the actual design of the beam. After fixing up the preliminary cross sectional dimensions, we can do the structural analysis, and calculate the total factored BM that the beam will have to resist. This factored BM is denoted as M_u. The beam section that we provide will have a unique value of 'resisting moment' that it can offer at ultimate state. We have learned about it in the previous section. It is called the 'Ultimate Moment of Resistance' M_uR of the section. So M_uR should be greater than or equal to M_u. Thus we can write one condition that the design should satisfy:

4.7: M_uR ≥ M_u

Another condition that the design should satisfy is that, the beam that we design should be under reinforced. This is to ensure that , if the beam is over loaded and fails, the failure will be a 'tension failure'. So there will be enough warning about the failure. We know that in an under reinforced section, x_u, the depth of NA will be less than x_u,max. So we can write the second condition:

4.8: x_u < x_u,max

We can write the steps for designing a beam which satisfies the above conditions:

Step 1: We want a beam section which will 'work' at it's 'full potential' to resist M_u. We know how to calculate the 'full potential' of a section. It is called the 'Limiting Moment of Resistance' M_u,lim. We already have a section with preliminary cross sectional dimensions. Let us forget the depth D of that section for a while. We will need D in Step 2. But for now, we are discarding it. But we need it's b. We want a section with the same width b, and whose M_u,lim is equal to M_u. So we write the M_u,lim of a section with width b:

Eq.3.26:

Let us put the quantity on the right side equal to 'K'. So we get
M_u,lim = Kbd² 
Now we equate M_u,lim to M_u:
M_u,lim = M_u = Kbd² 

In this equation, the only unknown is d. We can bring it to the left side and write:

Eq.4.9:

So Step 1 can be summarized as :

• Write Eq.4.9 and get the effective depth d of the section whose width = b and M_u,lim = M_u.

Step 2: If we make a section with width = b and effective depth =d obtained from eq.4.9, we will be having a section whose M_u,lim = M_u.  Effective depth can never be less than the d that is obtained from eq.4.9, because, then the M_u,lim will be less than M_u. So in this step we give a 'final' form to our section. For this we assume a diameter Φ for the reinforcing bars, and a diameter Φ_l for the links. Then we will get 
Eq.4.5 (based on fig.4.8)

D = d + ^Φ/₂ + Φ_l + C_c

Another easier method is to use Table 4.2. By using that table, we are assuming that Φ = 20mm and  Φ_l = 10mm. Then
D = d + effective cover

This D should be less than the preliminary value of D that we have used earlier. This is shown in the fig.4.9 below:

Fig.4.9
Calculated required depth should be less than the preliminary depth

If the calculated value is larger, then it is obvious that the section with the preliminary dimensions will not be capable to resist M_u.  Then we must go back and do the structural analysis again with new improved preliminary values of cross sectional dimensions. If the required value is lesser as shown in the fig.4.9 above, then we can discard the required value and finalize the section with the preliminary values. In this situation, the 'preliminary dimensions' will hereafter be called the 'Final dimensions' of the beam section

When we finalize the section in this way, one important point should be noted. The finalized D is greater than the required D obtained from eq.4.5 above. So the final d will also be greater than that calculated using.4.9. So we must calculate the final d as:

Eq.4.6:
d = D - (^Φ/₂ + Φ_l + C_c)

It is very important to make sure that the final d calculated using eq.4.6 above is greater than d calculated in step 1. Because if it is less, it would mean that the final section will have a M_u,lim less than M_u.

So step 2 can be summarized as:

• Using d calculated in step 1, assumed diameters Φ and Φ_l , find D using eq.4.5

• If this D is greater than the preliminary value, go back and do the structural analysis again with new improved values.

• If this D is less than the preliminary value, finalize the section with the preliminary values, and calculate final d using eq.4.6

• Make sure that new d is greater than that calculated in step 1

Step 3: We started off with the calculation of d in step 1. If we give this exact d, then we will be having a section whose M_u,lim is exactly equal to M_u. Then the area of steel will be equal to A_st,lim which corresponds to M_u,lim. But dimensions have changed from those in step1. 

We now have a new d, and a section with finalized values for it's cross section. We have to complete the design by giving it the required steel. How can we know the area of this required steel?

We can think in this way: Suppose we give a certain area of steel. Corresponding to this area of steel and the final values of b and d, the beam will have a certain value of M_uR. This M_uR should be greater than or equal to M_u. So one condition is that: the area of steel which we are going to provide should enable the section to attain an M_uR value which is greater than or equal to M_u. (condition written in 4.7 above)

We have one more condition. The final section should be under reinforced. This is because, if over loading occurs, and the beam fails, the failure should be a 'tension failure'. So we have a second condition: x_u of the section should be less than x_u,max. (condition written in 4.8 above)

With these two conditions, we can calculate the required area of steel. Let us see how this is done:

If the depth of NA is denoted as xu, then the Compressive force (Eq.3.7) =
C_u = 0.362 f_ck b x_u

Lever arm (Eq.3.8) =  z= d - 0.416 x_u

So the moment that the section will offer at ultimate state =
M_uR = 0.362 f_ck b x_u (d - 0.416 x_u)

This moment should be equal to M_u.

So we can write:

M_u = 0.362 f_ck b x_u (d - 0.416 x_u) 

Let us rearrange this so that it takes a 'familiar' form:

Let us simplify it further:

We can see that it is a quadratic equation. We can write it as:

Eq.4.10:
A X² + B X + C = 0

Where A = 0.1506 bd²f_ck ; B = -0.362 bd²f_ck  ; C = M_u and X = x_u/d

A, B and C contain only the known values b, d, f_ck and M_u. So if we are using a spread sheet program, it will calculate A, B and C just when the input data are given. The program will then directly give the two solutions of the quadratic equation. That is., we will get two values for x_u/d. Out of these, one will be greater than 1, and the other will be less than 1. We must choose the lesser value because x_u is always less than d. If we choose the value which is greater than 1, when we multiply it with d, we will get a value for x_u which is greater than d. 

This step comes to an end with the calculation of x_u . So step 3 can be summarized as:
• Calculate the values of A, B and C and then solve x_u

Step 4: With the calculation of x_u, we can split the section into an upper part and a lower part. The compressive force in the upper part is given by the same eq.3.7 that we saw in step 3 above:
Eq.3.7:
C_u = 0.362 f_ck b x_u

The tensile force in the steel steel in the lower part is given by :

T_u = 0.87 f_y A_st

Where A_st is the area of steel. The stress in steel is equal to 0.87 f_y because, it is an under reinforced section, and so the steel would have yielded at the ultimate state.

For equilibrium, C_u = T_u. So, equating the two, we get

Eq.4.11:

So step 4 can be summarized as:
• Calculate A_st the area of steel by using Eq.4.11

Fixing the number of bars and their diameters

The area of steel calculated above is the "Area of steel required". For beams, this has to be expressed in the form of Diameter of bars and number of bars of that particular diameter.

The following table 4.3 gives the areas of standard bar sizes:

Table 4.3

For any diameter Ф, the area A_b of the bar can be obtained as
A_b = πФ²/4

So for a chosen bar diameter, the number of bars required to provide the area of tension steel will be given by A_st / A_b . This ratio must be rounded off to the nearest higher whole number. In some cases, it may be economical to make a combination of two different bar diameters (close to each other) in order to get an area which is just higher than the area A_st.

After fixing the number and diameter of bars, we must calculate A_st,p which is the actual area of tension steel provided in the beam. A_st,p should always be greater than A_st.  

This completes the design part. Now we have to learn some details like the 'rules for distributing the calculated steel in the beam'. Also we have to see the details about the checks that have to be done after completing the design part. This includes 'minimum steel that has to be provided in a beam section', 'Check for deflection' etc.,. We will learn about them in the next section.

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Chap 4 (cont..1) Concrete cover for beams

In the previous section we saw the method to fix up the depth of the beam. Now we will see how to fix up the width b. For this, we can consider the following aspects: If the beam has to support a masonry wall, it is better to give a width in such a way that the sides of the beam are flush with the finished surface of the wall.  This will generally lead us to give a width of 200 mm or 230mm for the beam. This is because masonry walls are generally 200 or 230 mm thick. An example for this is shown in the fig. 4.3 below:

Fig.4.3
Beam supporting a masonry wall

In the above fig., the finished surface of the walls is shown. At the time of making the formworks for the beam, we must consider the the thickness of finishes which will be applied on the masonry work and the concrete work. The width of the form work should be in such a way that, when all the finishes are applied, the side surfaces of the beam will be flush with the side surfaces of the wall.

If the beam that we are considering is part of a framed structure, its width should be less than or equal to the lateral dimension of the column into which it frames. This generally leads us to give widths of 200 mm,250 mm, 300 mm etc., An example is shown in the fig.4.4 below:

Fig.4.4
Beam framing into a column

Also we have to consider the availability of formworks. Usually, formworks are made in such a way that their widths are kept equal to the width of beams commonly adopted in the locality. This will fecilitate their re-usability.

Thus we can fix up the preliminary depth and width of a beam. So the self wt. can be calculated and the analysis and design can be carried out. After the design is complete, various checks have to be done to make sure that the designed beam is capable to resist all the load effects acting upon it. If it is found that the beam designed with these preliminary dimensions do not have adequate strength, new improved dimensions should be given, and the analysis and design should be done again.

After designing the beam, when the checks are being done, the following scenario can arise: The dimensions are found to be unsatisfactory. Either the width or depth or both has to be increased. In such a case, if we are free to increase width and depth, we must opt to increase the depth. This will increase the moment resisting capacity and also will increase the stiffness against bending. So there will be lesser deflections, curvatures and crack widths. But very deep beams are undesirable because they lead to loss of headroom or an increase in the overall height of the building.

4.2: In general, the recommended ratio of overall depth to width (D/b) is in the range 1.5 to 2. But it can go up to 3 or even more for beams carrying very heavy loads such as bridge girders.

Widths and depths are also governed by the shear force at the section. We will discuss about it when we take up the topic of shear design.

Sometimes architectural requirements will have to be considered when deciding upon the size of beams. It may so happen that the depth of the beam has to be kept within a certain limit. In such a situation, we can design it as a doubly reinforced section, and thus achieve the desired strength. 

High strength concrete and steel can also be used for increasing the strength of a beam section.

If the slab which is to be supported over a beam is cast integrally with the beam, it can be modelled as a flanged beam. This will also give a higher strength to the beam, details of which will be discussed in the section on T- beams and L-beams.

So now we know how to fix up the preliminary dimensions of the cross section of the beam. Before proceeding to the next step, we must do some checks related to these cross sectional dimensions.

Check whether the beam is a deep beam.

4.3: For this, we must calculate the ratio l/D (span to overall depth of the beam). Then we must compare it with the values given in cl.29.1 of the code:

• If l/D is less than 2 for a simply supported beam, it should be considered as a deep beam

• If l/D is less than 2.5 for a continuous beam, it should be considered as a deep beam

Note that the overall depth D is being used in the ratio. Deep beams are discussed in a later chapter.

Check whether the beam is a slender beam

4.4: We must also check whether the beam is slender or not. If the beam is too long when compared to it's lateral dimensions, chances are that the beam will fall under the category of 'slender beams'. Slender beams will have lateral instability. To check whether the beam is slender or not, we must refer the cl 23.3 of the code

According to this clause:
For a simply supported or continuous beam, the clear distance between the lateral restraints should not be greater than the lesser of the following two values:

• 60b
• ^250b2⁄_d

For a cantilever beam, the clear distance between free end of the cantilever to the lateral restraint should not be greater than the lesser of the following two values:

• 25b
• ^100b2⁄_d

In the above checks b is the width of the beam, and d is the effective depth.

So now we have three checks (4.2, 4.3 and 4.4) that should be done just after fixing up the preliminary dimensions. These checks should be done again if these preliminary dimensions are changed.

In the above check for slenderness, we find that the effective depth d is coming in the calculation. We know that when the effective depth increases, the lever arm z also increases, and so the beam will be able to give more resisting moment. So we must get the maximum d possible. We have fixed up the preliminary value for D. In this situation, the concrete cover given to the bottom bars will decide the value of d, as shown in the fig.4.5 below. In the fig., Φ is the diameter of the bottom bars. 

Fig.4.5
Concrete cover and effective depth

So the knowledge about concrete cover is vital for calculating d. We will soon see that, the value of d is important not only for the above check for slenderness, but also for the design of the beam itself. So we have to learn how the thickness of this concrete cover can be fixed:

Concrete cover

When steel bars are placed in a structural member, it will be embedded completely in concrete. There will naturally be sufficient embedment on the inner sides of the member. But we have to take special care about the outer or exposed surface. For example, the bottom bars of a beam will have enough concrete on it's top side. But there will be a lesser concrete near the bottom and the sides. This is shown in the fig.4.6 below. We call this thickness of concrete which covers the steel as the 'concrete cover'.

Fig.4.6

Concrete cover

The concrete cover serves the following purposes:

• It gives sufficient protection to the bars from corrosion and fire.

• When the bars are stressed by a tensile or compressive force, they tend to move from their original positions. But this movement should be prevented to enable the effective transfer of stresses between the concrete and steel. By providing the required cover on all the sides of the bar, they will be sufficiently embedded inside concrete, and so, this movement will not occur.

The concrete cover, usually denoted as C_c, should not fall below the values given by the code. In the code, it is given as 'Nominal cover' in the cl. 26.4.1. We can obtain the values of C_c by referring clauses 26.4.1 and 26.4.2 of the code. These specified values should be given from the outer surface of the links. This is shown in the fig.4.7 below:

Fig.4.7

Specified cover provided from outer surface of links

Based on the requirements given in the code, we can prepare Table 4.1 below which gives the value of C_c for various conditions.

Table 4.1

For example,if we are to design a beam which will be subjected to 'severe' exposure conditions, the value of C_c to be provided is 45 mm , and the minimum grade of concrete to be used is M 30 .

The code also gives some guidelines about the changes from the specified cover when the actual work is carried out at the site. In actual construction, the cover provided for the bar may become different from the values specified by the designer. The allowable tolerance is given as Note no. 2 in Table 16 of the code. From this note, it is clear that the cover can increase by a maximum value of 10 mm from the value specified by the designer. But it cannot decrease (-0 mm tolerance) from that value.

The code also specifies concrete cover requirements based on fire resistance. This is given in cl 26.4.3 and table 16.

Thus now we know how to fix the concrete cover C_c . So we can write the equations for calculating the total depth D and effective depth d. We can write them based on the fig.4.8 below:

Fig.4.8
Details of concrete cover and effective depth

From the fig.4.8 we get 
Eq.4.5: D = d + ^Φ/₂ + Φ_l + C_c

and  Eq.4.6: d = D - ( ^Φ/₂ + Φ_l + C_c )  

Where Φ is the diameter of main bars and Φ_l is the diameter of bars used for making the links.

Some examples demonstrating the calculation of d is given here 

So now we know how to calculate the effective depth d of any beam section. But there is one more obstacle that we have to overcome in this calculation: The values of Φ and Φ_l are not known at the time when the preliminary dimensions are fixed. So we cannot do the check 4.4.

In this situation, we assume a value for Φ and another for Φ_l. The values usually assumed are 20mm for Φ and 10mm for Φ_l. We know the five possible values for C_c from table 4.1 above. Corresponding to each of these five values, there will be a value for the 'effective cover'. This is the cover measured from the nearest outer surface of concrete upto the center of the steel bars. So the table 4.2 can be prepared which will directly give the effective cover for each of the exposure condition. In the table, all values are in mm

Table 4.2

Exposure condition	Cc	Φ	Φl	Effective cover
Mild	20	20	10	40
Moderate	30	20	10	50
Severe	45	20	10	65
Very Severe	50	20	10	70
Extreme	75	20	10	95

Sample calculation:
Let us take 'Severe' exposure condition. Then 'effective cover' = 45 +10 +20/2 = 65mm

For all the values of 'effective cover' in the above table, 20mm and 10mm are assumed. So after finalizing the beam section, the check should be done again with the actual diameters provided in the beam.

So we have the basic information required to start the design of a beam. In the next section we will see the design process.

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