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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Chapter 6 - Design of One-way slabs

In the previous section we completed the analysis of one way slabs. Now we will see the design process. As in the case of beams, here also, we will first see the procedure for fixing up the 'preliminary dimensions'. We have seen that for analysis and design, the slabs are considered as strips, and we can take any one strip. As we are dealing with 1m wide strips, the width of the beams that we have to design is fixed at 1000mm. We have to fix up only the preliminary depth. Let us see how this is done:

We know that the depth provided to a beam must be sufficient for it to resist the bending moment acting on it. In addition to this, the depth must be sufficient to control the deflection also. So the depth depends on two criteria:
• To resist the bending moment, and
• To control deflection. 

In the case of slabs, where the depth is relatively low, the criterion for deflection control becomes more critical. That is., a slab with a certain depth may effectively resist the bending moment acting on it. But this same depth may not be sufficient to control the deflection. In other words, the slab generally requires more depth to control it's deflection than it requires to resist bending. 


So to fix up the preliminary depth, we must use the deflection control criterion. Let us take the case of a simply supported slab (whose effective span is less than 10m). We have already discussed the procedure for deflection control here. The effective depth that the slab actually have in the final structure should satisfy the following relation:

(l/d)_actual  ≤  [(l/d)_basic] α k_t  - - - (1) 

• For simply supported spans, (l/d)basic = 20
• As we are considering spans less than 10m, α need not be taken into account. 
Thus we get :
(l/d)_actual  ≤  20 x k_t - - - (2)

From this expression, we want to calculate 'd'. But k_t is also an unknown. So we assume a value for k_t. The assumed value should be as accurate as possible. For this we look at the numerous design examples of simply supported slabs that are done in the past by different designers. By examining them, we can find that the percentage of reinforcement p_t generally falls in the range 0.4–0.5. 

Sample calculation:
Let a slab have the following properties:
D =150mm; d =115mm; Ф =10mm; s =150mm
Then from Eq.5.2,

= 490.87mm². So p_t = 100A_st /bd = [490.87 x100]/[1000 x115] = 0.427 

Based on this information, we adopt the following procedure:
• Assume that the area of steel provided is equal to the area of steel required,
• Then f_st will be equal to 0.58f_y. 
Take the value of f_y that is going to be used. If it is 415, then f_st = 240.7
So we have the values required to find k_t: They are:
(a) p_t in the range 0.4–0.5, and
(b) f_st = 240.7

Corresponding to these values, we will get k_t = 1.25 from Fig.4 of the code.

So substituting in (2) we get
6.1: (l/d)_actual  ≤  20 x 1.25 ⇒ (l/d)_actual  ≤  25. So d that we actually provide in the final slab should be greater than or equal to l/25. This is for simply supported slabs. 

For continuous slabs, we can write:
6.2: (l/d)_actual  ≤  26 x 1.25 ⇒ (l/d)_actual  ≤  32.5. So d that we actually provide in the final slab should be greater than or equal to l/32.5.≈ l/32

From the above expressions, what we get is the effective depth d. To this we must add C_c and half of Ф to get the total depth D. Ф can be assumed to be equal to 10mm. 

As in the case of beams, we must do all the design checks to ensure that the final section satisfies all requirements. If the section fails to do so, the whole process should be repeated with improved dimensions.

So we have fixed the preliminary dimensions. The next step is to find the effective depth required from bending moment considerations. Once this is calculated, we must compare it with the d that we have from the preliminary depth. If the required value is less, we can proceed to find the steel required to resist this bending moment. These two steps are same as that for a beam.


Now we must know the 'rules' for distributing the calculated steel into the slab.

Concrete cover and grade of concrete:
The details about the concrete cover that has to be provided to the bars of a slab are same as those of a beam. They were discussed in the section about the design of beams, and we saw table 4.1.


So, if we are to design a slab which will be subjected to 'moderate' exposure conditions, from table 4.1,we get the value of C_c to be provided as 30 mm , and the minimum grade of concrete to be used as M25

Minimum spacing to be provided between the bars of a slab:
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. We have seen the details of this clause when we discussed about the minimum spacing between bars of beams here. The same is applicable to slabs also. So we can show the application of this clause to slabs as shown in the fig.6.1 below:

Fig.6.1
Minimum spacing between bars of slab

Maximum spacing allowable between bars of a slab:
When we design the slab we will get the area of steel that should be provided. Sometimes, when we convert it into 'spacing of bars of a particular diameter', we may get a large spacing s. This often happens when the bending moment that the slab has to resist, is low. Large values of s may be obtained on other occasions also: When we choose to provide large diameter bars. This can be explained as follows: From 5.3 we know that

In the above equation, for the same value of A_st, we will get a larger s for a larger Ф.  

It may seem to be economical to provide bars at large spacing. But bars at large spacing will not be able to control cracks effectively. Also the bond between steel and concrete will become lesser when bars are at a large distance apart.

So the code specifies some upper limits to  the spacing. The c/c spacing provided between parallel reinforcing bars of a slab should not be greater than that specified in cl 26.3.3 (b) of the code. The requirements stated in this clause can be shown as in fig 6.2 given below:

Fig.6.2
Maximum spacing between bars of slab

In the above fig., it is indicated that the diameter of the main bars should not exceed D/8. This is because 'small diameter bars spaced closer together' is effective in reducing cracks and in improving bond than 'large diameter bars spaced at larger distances'. However, bars smaller than 8mm in diameter are generally not used as main bars of slabs. 

Minimum Area of reinforcement for slabs:
The area of reinforcement provided in slabs should not be less than that specified in cl 26.5.2 of the code. The requirements stated in this clause can be written as follows:

6.3: For Fe 250 steel:
A_st ≮ 0.0015A_g


6.4: For Fe 415 steel:
A_st ≮ 0.0012A_g

Where A_st is the area of steel provided in the slab, and A_g is the gross area of cross section of the slab. A_g is equal to bD where b is the width of the slab and D is the total depth of the slab. When we consider a strip of slab for the design, b is equal to the width of the strip.

Transverse reinforcement for one way slabs :
We have seen in the previous chapter that transverse reinforcement is required for one-way slabs. These bars are also called distribution bars of a One-way slab. The quantity of this reinforcement that we have to provide, is given by the code, and is same as that given by 6.3 and 6.4 above.

The c/c spacing between the distribution bars should not exceed the smallest of the following (cl 26.3.3.b (2)):

• 5 times the effective depth d of the slab
• 450 mm 

This is shown in fig. 6.1 above

Deflection control for one-way slabs
We have seen some details about deflection control checks in one way slabs when we discussed about the 'preliminary depth' at the beginning of this chapter. There we assumed a value for p_t. But now we are discussing the actual 'check'. This is the 'last but one check' that we have to perform after the design of a one way slab (The last one being the check for M_uR). So here we will be using the actual p_t that is provided. The procedure is same as that for a singly reinforced beam. So we can write the same expressions that we wrote for the beam:

4.19
For singly reinforced rectangular slabs with span less than 10m,

(l/d)_actual  ≤  [(l/d)_basic] k_t

4.20
For singly reinforced rectangular slabs with span greater than 10m,

(l/d)_actual  ≤  [(l/d)_basic] α k_t

Partial fixity at supports:
Another topic that we must consider, is the design at supports. If the slab is simply supported, the bending moment will be zero at the supports. The details to be considered in such a situation is demonstrated in the following presentation:


(The first five slides are same as the ones we saw in an earlier presentation about the behaviour of slabs in chapter 5)




This completes the discussion about the design of a simply supported one way slab. We will now see a solved example which demonstrates the actual process of design.

Solved example 6.1

In the above solved example, we have determined the reinforcement requirements of a simply supported one way slab. Based on that design we must draw a plan and a sectional view, showing the layout of bars. This is given in the fig.6.3 and 6.4 below:

Fig.6.3
Plan view showing reinforcement details of slab

Fig.6.4
Sectional elevation

In the plan view, we can see that alternate bars are bent up at supports. The reason for this can be explained as follows: We have designed the slab as 'simply supported'. So the bending moments at the support will be equal to zero. But if in the future, partial fixity (shown in the presentation above) is introduced at any of the supports, bending moments will develop at that support. These bending moments will be hogging in nature. So tension will develop near the top surface of the slab at such supports. Thus we have to provide top steel. This can be achieved by bending up the bars. The requirements for this top reinforcement is given by cl.D-1.6 of the code. According to this clause, the bars must extend a distance of 0.1l from the support into the slab. Also, these bars should have an area equal to 50% of that provided at midspan. Thus, when we bend up alternate bars, we will get 50% at the top.

Technically, we need to show only two bars of a particular set. But if we show more bars, the symmetry and pattern will become more clear. Such a plan view is given here. 

In the sectional view in fig.6.4 above, the two bars are shown separately only for clarity. The two bars are in fact provided in the same layer as indicated by the '0'mm dimension.

This completes the design and detailing of a 'simply supported' one way slab. Next we will see 'continuous' beams and one way slabs. But to analyse continuous beams and slabs, we must know how to calculate the 'effective spans' in continuous members. So in the next chapter we will see the details about this effective span also.

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