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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Chapter 5 - Analysis of One-way slabs

In the previous section we completed the analysis and design of singly reinforced beams for flexure. Now we will see the analysis of 'One way slabs'. The bending (flexure) of a one way slab is similar to the bending of a beam. 

The following presentation shows some basic details about the behaviour of simply supported slabs :



From the above presentation, we get a clear idea of how to distinguish between a one way slab and a two way slab:

We take the following ratio:
5.1: l_y/l_x. If this ratio is greater than 2, then the slab is a one way slab. If it is less than or equal to 2, then it is a two way slab.

This method is to be used only on those slabs which are resting on walls on all the four sides. If there are walls only on the opposite sides, the load transfer will occur in that direction only, and it will be a one way slab.

Now we will see how a one way slab can be compared to a beam: We divide the slab into a number of strips as shown in the fig.5.10 below:

Fig.5.10
One way slab divided into strips

The strips extend from one support to the other. Each strip has a width of 1m (shown in the fig. as 100cm). Each of these strips act as individual beams. So each of these beams will have a width of 1m and a total depth D, where D is the total depth of the slab.

Assuming that only 'uniformly distributed load' is acting on the slab, and that the slab is of uniform thickness, all these strips are identical. We can take any one of them for analysis and design. Let us take one strip as shown in the fig.5.11 below:

Fig.5.11
Loads on an individual strip

For designing this strip, we want to know the maximum bending moment acting on it. We can calculate it by using the same method that we would use if it was an individual beam. That is., we can use the formula:

M_max = wl²/8.

where w is the load per unit length of the strip, and l is the effective span of the strip.

So we want w. For this, in the above fig.5.11, a length of 1m has been marked off on the strip, in a direction parallel to the length of the strip. When we mark off 1m on the strip parallel to it's length, there will be a dimension perpendicular to the length. As we are considering strips of 1m width, this perpendicular dimension will also be 1m. So we are having an area of 1x1m² on the strip. The load acting on this 'area' is the load acting per meter 'length' of the strip. So we can summarize the above discussion as follows:

• To design the one way slab, we take any one strip of 1m width.
• We want the load per unit 'length' of this strip
• This load load per unit 'length' that we require, is same as the load per unit 'area' of the slab.

So we will now see the details of the load per unit area:

Self weight of the slab in the unit area: - - - - (1)
Volume of the concrete block in the unit area = 1m x1m x'D'm  = 'D'm³. ('D' should be in meters.)
So weight of the concrete block in unit area = D x25kN/m³ = 25DkN/m²

Self wt. of finishes: - - - - (2)
This can be obtained from data books or relevant codes. Usually it varies from 0.5 to 1.0 kN/m².

Live loads: - - - - (3)
This can be obtained from IS code 875. This load depends upon the nature of use of the building. (Residential, office, storage purpose etc.,). Usually it is given in kN/m² 

All the relevant clauses of the code must be considered in arriving at the appropriate value of the loads to be used in the design.

Sum of the above three items will give the characteristic load per meter square of the slab. This is equal to the characteristic load per unit length of the strip. We must multiply it by the load factor to obtain the factored load. 

Calculation of Area of steel and spacing of bars

Now that we know how to obtain the bending moment acting on a strip of a one-way slab, we can analyze/design it as a beam, whose width is equal to 1m (or 1000mm), and whose effective depth is equal to the effective depth of the slab.

When we are given a slab for analysis, we will be given it's steel in terms of 'spacing of bars of a particular diameter'. For example: '#8 @ 180 c/c'. This means that 8 mm bars are provided at a spacing of 180mm c/c. But to analyse the slab as a beam, we want the area A_st within a width of the beam (1000mm). Let us see how we can obtain this from the 'spacing': The fig.5.12 below shows a part sectional elevation of a slab. The spacing of bars is denoted as 's' mm.

Fig.15.12
Area of steel in a slab

From the fig. we can see that the number of bars 'n' in a width of 1m will be equal to 1000/s. If the diameter of bar is denoted as Φ, then the area of one bar = πΦ² /4. So the total area of bars in the strip =  nπΦ² /4 = (1000/s)( πΦ² /4) Thus we get 
Eq.5.2:

(Where Φ and s are in mm)

When we are required to analyse a given one-way slab section, Eq.5.2 can be directly used to calculate A_st in a 1000 mm wide strip. Then the strip can be treated as a beam of width 1000 mm and can be analysed to find M_uR.

While designing a beam, we can use the converse of the above: As the result of the design, we will get A_st that has to be provided in one strip of 1000mm width. But we want to express it as 'spacing of bars of a particular diameter'. If Φ is the diameter, area of a single bar = πΦ² /4 . So the number of bars required to make up A_st = n = A_st /( πΦ² /4) =4A_st / ( πΦ²)   Thus we get
5.3:

(Where Φ is in mm and A_st is in mm²)

Transverse moments in one-way slabs

In the above discussions, we considered a 1m wide strip, and it was assumed to act as an independent beam. But there is a difference between the bending of a beam and the bending of a strip of slab. First let us consider the bending of a beam. When the beam is subjected to a sagging moment, the portion above the NA is under compression. Due to this compression, there will be a lateral expansion for the portion above the NA. This is due to the poisson effect. In the same manner, the portion below the NA is under tension, and hence there will be a lateral contraction. So after bending, the cross section of the beam will have a nearly trapezoidal shape as shown in fig. 5.13 below:

Fig.5.13
Lateral deformation of beam section

Now let us consider the strip of slab. In this case, the expansion above the NA and the contraction below the NA is prevented by the strips on the two sides of the design strip. So the portion above the NA will experience a lateral compressive reaction from the adjacent strips, so that the lateral expansion is prevented. The portion below the NA will experience a lateral tensile reaction from the adjacent strips, so that the lateral contraction is prevented. These lateral forces will give rise to secondary moments in the transverse direction as shown in the fig. 5.14 below:

Fig.5.14
Secondary moments in slab

In the above fig., the restraining forces are indicated by the magenta colored arrows. 
• The top arrows shows compressive reaction from the adjacent strips. This compressive reaction prevents the expansion (above the neutral axis) of the design strip. 
• The bottom arrows shows tensile reaction from the adjacent strips. This tensile reaction prevents the expansion (below the neutral axis) of the design strip . 

These two reactions together form a couple. It will try to bend the slab in the transverse direction. Thus we see that the one way slab require reinforcements in the transverse direction to resist the secondary moments. These reinforcements in the transverse direction are called secondary reinforcements.

The secondary reinforcements serve some other purposes also:
• Effects due to concentrated loads:
When a concentrated load is applied on the slab, bending moments in the transverse direction are induced in the slab. So we want secondary reinforcements in the  transverse direction to resist these moments.

• Shrinkage and temperature effects: 
Freshly placed concrete will shrink when it dries. The extent of this shrinkage can be minimized by using appropriate water cement ratio, and by proper moist curing. But even after taking all necessary measures, some shrinkage will always occur. If a slab of usual dimensions rests freely on it's supports, it can freely shrink. But usually the slab is kept in position by beams, walls above supporting walls etc., So it cannot shrink freely. So when restrained slabs shrink, tensile stresses will develop in it. This will give rise to cracks. 

A similar effect is produced due to temperature variations also. There will be thermal expansion and contraction of the slab due to temperature variation. A restrained slab can not expand or contract freely. So this will give rise to cracks.

These cracks can be minimized if we provide steel in a direction perpendicular to the cracks. But shrinkage and temperature effects occur in all direction, and so the cracks can occur in any direction. If bars are provided in two sets, each set perpendicular to the other (in the form of a grid), crack formation in any direction can be resisted. We already have one set which is provided to resist the bending moment. We also learned about the secondary reinforcements which are provided in the transverse direction. So these two sets will resist the formation of cracks due to shrinkage and temperature effects.

It may be noted that during shrinkage, the slab is trying to pull inwards. The bars which try to resist this inward pull will be experiencing compressive stresses. 

Thus we can see that the secondary reinforcements are essential not only for resisting the secondary moments, but also to reduce the formation of cracks due to shrinkage and temperature effects. We will discuss about the 'quantity' of secondary reinforcements to be provided as per the code, when we take up the design of one-way slabs.

The following solved examples demonstrates the process of analysis of a singly reinforced one-way slab:

Solved example 5.1 

We will now see the solved example that we did earlier based on Fig.3.32. There we analysed the beam and calculated the safe load that the beam can carry. Now let us analyse the slab and find the safe load that it can carry. The fig.3.32 is shown below again. But this time the section of the slab is shown:

Analysis steps are shown here. We get M_uR = 14.4kNm

Now we calculate M_u:

M_u = w_ul²/8. where, l= effective span; w_u = load per meter square area of slab.

First we calculate effective span:

• c/c distance between supports = 2950 +250 =3200mm
• clear span + effective depth = 2950 +126 = 3076mm
l = lesser of the above = 3076mm

w_u is the unknown. Equating M_u and M_uR we get:
w_ul²/8 = 14.4 ⇒ w_u x 3.076² x (1/8) = 14.4 ⇒ w_u =12.18 kN/m
But w_u is the load per m² on the slab. This load consists of the following three components:
(1) Self wt = 25D = 25 x .16 = 4kN/m²
(2) Wt. of finishes = 1 kN/m² (assumed)
(3) Wt. of partitions = 1.25 kN/m² (assumed)
So total DL = 6.25 kN/m². But there may be uncertainties in the DL. so we must multiply it by 1.5 Thus we get 6.25 x1.5 = 9.375kN/m. 

So we can write: (9.375 + 1.5 x LL) = 12.18 ⇒LL = 1.87 kN/m²

Thus we can specify that the maximum characteristic LL that can be applied on the slab is 1.87 kN/m²    
    

When we analysed the beam based on fig.3.32, we found that a characteristic LL of 7.8 kN/m² can be applied on the slab, as far as the safety of the beam is concerned. But now we find that such a load can never be applied on the slab. As far as the safety of the slab is concerned, we can specify that a characteristic LL of only 1.87 kN/m² can be applied on the slab.     

This completes the details about the procedure of analysis of a one way slab. In the next chapter we will see the design part.

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