Chapter 8 (cont..2) - Checks for continuous slab

Distribution bars
In the previous section we completed the layout of main bars of the slab ABCD. In this section we will design the distribution bars. We know that the quantity of distribution bars required depends only on A_g, the gross area of cross section of the slab, which is obtained as A_g = b xD = 1000mm x D. So even when our continuous slab has many spans, the area of distribution bars required will be the same for all spans. This is because, all spans in our slab has the same depth D, which is equal to 200mm.

Let us use #8 of Fe415 grade steel. So the area required = 0.0012A_g = 0.0012 x bD = 0.0012 x1000 x200 = 240mm²

Now, spacing required can be determined using Eq.5.3

Where Φ= 8mm and A_st = 240mm²

Thus we get spacing s =209.44mm. Let us provide #8 @200 mm c/c

The actual area provided is given by Eq.5.2 

Where s is the actual spacing provided, which is equal to 200mm. So we get A_st provided =251.33mm²

Maximum spacing allowable between distribution bars of the slab : 
(cl 26.3.3.b(2) of the code)

According to this clause, the spacing should not be more than the smallest of the following: 

1. Five times the effective depth of the slab = 5 x d = 5 x165 =825mm
2. 450mm

So the spacing should not be more than 450mm. Thus the spacing of 200mm is OK

Check whether the minimum required distribution steel is provided:
The spacing of 200mm which is actually provided, is less than the required spacing of 209.44mm. So the area will be greater than the minimum required.

Thus we have completed the design of distribution bars also. We can now do the various checks.

The pdf file given below gives the detailed steps involved in the various checks:
Solved example 8.1 - final checks

Curtailment of bars
Now we can discuss about 'curtailment of bars' in our slab. We know that in a continuous beam or slab, there will be hogging moment at support, and sagging moment at midspan. This can be seen in the BM diagram of a continuous member.

We earlier saw the BM diagram for our present slab in fig.7.5.  On either side of the support, the hogging moment is decreasing progressively. That is., at greater distances away from the support, there will be lesser hogging moments. 

We have designed the top steel at supports for the full design moment. But now we see that at sections away from the support, the BM decreases, and so the full steel is not necessary. So we can reduce the steel. A detailed discussion about 'Development length and curtailment' can be seen here. But for our problem, we will use the recommendations given in SP16. It must be noted that, to use these recommendations, the analysis of the continuous member should be done using the 'method of coefficients'. We have indeed used it in the analysis of our slab, and the results thus obtained were used in the design.

We will learn about the curtailment of top bars by taking support B of our continuous slab as an example. The fig.8.10 given below shows the details.

Fig.8.10
Curtailment of top bars

Here, bar types 'b' from span AB, and bar type 'c' from span BC are working together to resist the hogging moment at support B. These bars are shown separately only for clarity. In reality, they are at the same level, as indicated by the 0mm distance in the fig.

We can see a distance of 0.15l (from the face of the support) marked off on either side. So there is a particular horizontal length equal to 0.15l₁ + 0.15l₂ + width of the support. Within this length, no curtailment is allowed. In other words, within this length, all bars (which are intended to resist the design hogging moment) should be compulsorily present. So the quantity of steel in this length is denoted as A_st,sup.B. Just the availability of a length of '0.15l₁ + 0.15l₂ + width of the support' is not good enough. We must ensure that 0.15 times the respective spans is available on both sides. This is indicated by the '≥' sign. In this length, all of 'b' and 'c' are working together. But beyond this distance, some bars can be allowed to leave. We see that on the left side, 'b' bars are leaving. So after this distance, on the left side, only 'c' is present. The 'b' bars have taken deviation and left. They will become bottom bars at the midspan region. The 'c' bars remain as top bars, and continue to do their work. We can say that, the group does not need 'b' beyond 0.15l because, the moment is of a lower magnitude in that region. 

But there is a restriction on the quantity of bars that can be curtailed in this way: 50% of the bars required to resist the full hogging moment must be compulsorily present beyond the 0.15l length. Thus, in the fig., the quantity of steel beyond 0.15l is denoted as 0.5A_st,sup.B. So all of 'c' must continue. This will ensure the required 50%. There is a restriction on the length also. All the 'c' must compulsorily extend a distance of another 0.15l. So they will have a length of 0.30l from the face of the support.

A mirror image of the above arrangement happens on the right side of the support. There, 'c' bars will be leaving the group of top bars.

The above fig.8.10 is applicable to interior supports. Now we will see the curtailment of bottom bars. We will take the example of span BC, which is an interior span. Fig.8.11 given below shows the details:

Fig.8.11
Curtailment of bottom bars 

Here, the bars 'c' and 'd' are working together to resist the sagging moment at midspan BC. These bars are shown separately only for clarity. In reality, they are at the same level, as indicated by the 0mm distance in the fig.

We can see a distance of 0.25l₂ (from the center of supports) marked near either supports. So there will be a portion (of length l₂ -0.50l₂ =0.50l₂) at the middle of the slab. This is the important portion as far as the 'sagging moment in an interior span' is concerned. All the bars which are intended to resist the sagging moment at midspan should be completely present in this portion. Thus, in the fig., the quantity of steel is denoted as A_st,mid.BC. Beyond this portion, on either sides, the BM is of lower magnitude. So 'c' is allowed to leave on the left side, and 'd' is allowed to leave on the right side. They are bent up and will become top bars on the respective supports.

Like in the case of top bars, here also, there are some restrictions: Half the area of bars at the midspan should be compulsorily present after curtailment on either sides. Thus, in the fig., the quantity of steel after curtailment is denoted as 0.5A_st,mid.BC. There is a restriction on length also: All the bars remaining after curtailment should be compulsorily extended into the supports on either sides.

So we have completed the discussion on the curtailment of bars at interior supports and spans. From the above figs.8.10 and 8.11, one point can be noted: Each bent-up bar will be marked with three lengths. They are 0.15l and 0.30l at the top and 0.25l at the bottom. The lengths at top should be measured from the respective face of support. The length at bottom should be measured from the center line of support.

Now we will see the curtailment at an end span. Together, we will see the curtailment at an end support also. Fig.8.12 below shows the details.

Fig.8.12
Curtailment details in end span

First we will see the details of top bars at the end support A. The length marked here is 0.1l from the face of the support. So we must ensure that the top horizontal portion of all the 'a', which are bent-up bars, have a length of 0.1l from the face of the support. The area of these bars should not be less than half of that provided at midspan for the sagging moment.

Next we will see the bottom bars. This is similar to what we saw in fig.8.11 above, except that the distance from the center line of end support is 0.15l instead of 0.25l. So the 'important' portion in the midspan region has a length of l₁ -0.15l₁ -0.25l₁ =0.60l₂ . All the bars should be present in this region. Beyond this region, the bars should have an area of half of that provided at midspan, and they must extend into supports on either sides.

This completes the details of the arrangement using 'bent-up bars'. In the next section we will see how the same reinforcement requirements of this slab can be satisfied with 'straight bars'.    

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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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