Chap 17 (cont..1)

Let us consider the red strip first. The red strip is the longer strip, parallel to l_y, and having effective span l_y. From 17.1 in the previous section, the share of load acting on it is w_y. So the deflection of the yellow portion can be written as:

Similarly, the deflection of the yellow portion while considering the blue strip can be written as:

As the deflection of the yellow portion is same for both the strips, we can equate the above two as shown below:

From this, we get a simple relation between w_x and w_y:

Eq.17.2:

Based on the above relation, we can infer that w_x will be always greater than w_y, because the ratio ly/lx will always be greater than 1. We know that w_x is the share that makes the cylinder whose axis is perpendicular to l_x. So this is the load that gets transferred to the long walls. So we can say that the load transferred to the long walls is greater than the load transferred to the short walls.

If we add the two shares, we will get the total load w acting on the slab. That is:

Eq.17.3:

From Eqs.17.2 and 17.3 we will get

Eq.17.4:

Where r = l_y / l_x

Steps leading to the derivation can be seen here

w_x and w_y are the UDL acting per one meter square area on the slab. The width of the strips that we are considering is 1m. So w_x and w_y will be the load acting per one meter length of the strips. When we know the load per meter length on the strip, we can calculate the maximum bending moment on that strip. Thus we can write:

Eq.17.5:

We must understand one more concept at this stage. w_x and w_y are loads per one square meter area, and they are the shares transferred in particular directions. So all the strips parallel to a particular direction will be experiencing the same load per one square meter area. That is., all the strips parallel to l_x will be experiencing w_x per unit area. And all the strips parallel to l_y will be experiencing w_y per unit area. So we can say that each of the strips parallel to l_x will experience a maximum bending moment of w_x l_x²/8. And each of the strips parallel to l_y will experience a maximum bending moment of w_y l_y²/8.  So, the above Eq.17.5 which we derived based on a single strip is applicable to the whole slab.

Again, as mentioned before, in an actual two way slab, the load on each strip will be different from the adjacent strip. So the bending moment experienced by each strip will also be different from the adjacent strip. But we are assuming that all the strips in a particular set have the same load and bending moment. This assumption is for the ease of analysis. It will give conservative results for this type of a simply supported two way slab.

Now we will proceed to calculate the actual quantities in Eq.17.5. For this we must substitute for w_x and w_y from Eq.17.4. Thus we get:

Eq.17.6:

We can write the above equation in a form that is given in the code:

Eq.17.7:

In the above Eq.17.7, we must take a special note of the expression for M_y. We can see that it is given in terms of l_x, the smaller span. Also compare the expression for α_y with that of M_y in Eq.17.6. The numerator has become r² instead of 1.

We can make an important inference from Eq.17.7. The difference between M_x and M_y is α_x and α_y. The other quantities w and l_x² are the same. Now, α_x has r⁴ in the numerator while α_y has r². The denominators are the same. As r will always be greater than 1, r⁴ will always be greater than r². So α_x will be greater than α_y. Thus M_x will always be greater than M_y. So the steel required to resist M_x should be in the bottom most layer. This will give more effective depth, and thus greater Moment of resistance to the section. From 17.1, we know that M_x is acting parallel to l_x. So we can write:

17.8:
The steel calculated from M_x is to be laid parallel to l_x, and it should be in the bottom most layer.

Eq.17.7 is the final form given in cl.D-1.1 of the code. So we will follow it. The detailed steps leading to it’s derivation can be seen here.

The values of α_x and α_y are given in the table 27 of the code. A sample calculation can be done as follows:

Let r = ly/lx = 1.3. Then α_x = r⁴ / 1 + r⁴ = 0.0926 and α_y = r² / 1+ r⁴ = 0.0548 The same values given in the table 27.

For using this table, first we must calculate r = ly/lx. then we take the value of α_x and α_y from the table. If r is an intermediate value, we can use linear interpolation. If we are using a computer, we can use Eq.17.7 directly. There is no need to obtain values from the table, and do the interpolation.

So each of the strip parallel to l_x will experience a maximum bending moment of α_xw l_x². And each of the strip parallel to l_y will experience a maximum bending moment of α_yw l_y² . Now the situation has become similar to a simple one way slab. In the case of a one way slab, we divide it into a number of strips. Then we find the maximum bending moment in any one strip, and the steel for resisting this BM is applicable to all the strips. The same happens in two way slab also. Only difference is that, we have to do the design two times, one for each direction. From the chart 17.1, M_x is bending the slab into a cylinder whose axis is perpendicular to l_x. So we provide the steel required to resist M_x in a direction parallel to l_x, as if to straighten the cylinder back to a plane surface. Similarly, we provide the steel required to resist M_y in a direction parallel to l_y.

So now we have the equations for calculating the bending moments in a simply supported two way slab. But to use them, we need to find the load w. For this, we want to know the thickness of the slab to calculate the self wt.  Also, we want the values of the effective spans l_x and l_y. In the next section we will see how these quantities are determined.

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Chapter 17 (cont..2) Thickness and Effective span of Two way slabs

Preliminary value for the thickness of a two way slab

When fixing the thickness of slab, deflection criterion is more important than the bending moment criterion. That is., a lower value of thickness may be sufficient to resist the bending moment. But that lower value will not satisfy the deflection check. So we will do the procedure for deflection check in a sort of 'reverse' manner to fix up a preliminary value for 'D', the total thickness of the slab. This can be done as follows:

We know that, for spans less than 10m, to satisfy the deflection check, the following condition should be satisfied:

  - - - - (1)

For simply supported members, (l/d)_basic = 20.
Now we take k_t. This can be calculated by using the expression given in SP24:

  - - - - (2)

Where

  - - - - (3)

But in the above expressions, there are three unknowns. (a) pt, (b) Area of steel required, and (c) Area of steel provided. These three unknowns can be solved by making some assumptions:

From the numerous design problems of One way slabs that has been already done for various buildings and other structures, we can say that the percentage of reinforcement is around 0.45% when Fe415 steel is used. Note that this is for one way slabs. If we assume that the area of steel required is same as the area of steel provided, then from (3):

f_s = 0.58 x 415 x 1 = 240.7 N/mm² - - - - (4)

Substituting this value of f_s and p_t (= .45) in (2) we get k_t = 1.277

Effective spans of a two way slab

Now, to use the l/d ratio, we must have the value for l. For a two way slab, we have two values for l. They are l_x and l_y. It is the value of l_x, (cl.24.1, Note 1) the shorter side that we have to use in the l/d calculation. But the percentage of steel required for resisting M_x in a two way slab is less than the percentage of steel (0.45%) required to resist the moment in a one way slab with the same effective span as l_x. Because of this lower value of p_t, k_t will be higher than 1.277.

It may be noted that instead of using (2) we can use fig.4 of the code. From that fig., it can be seen that when p_t decreases, k_t increases.

So we use a value of k_t = 1.5
Substituting this in (1) we get lx/d provided ≤ 20 x 1.5.
So we get:

Eq.17.9:

lx/d _provided ≤ 30

OR d _provided ≥ lx/30

However, the design using this value should be finalized only after the checks using actual area of steel required and the actual area of steel provided.

There is another method to fix up the total depth D of a two way slab. But to use this method, the slab should satisfy the following conditions:

• l_x is less than or equal to 3.5m, and
• LL is less than or equal to 3kN/m²,

When the above two conditions are satisfied, we can use cl.24.1-Note 2 of the code for calculating the total depth D of the two way slab:

Eq.17.10: 

Fe250 steel:
D ≥ lx/35 for simply supported slabs
D ≥ lx/40 for continuous slabs

For Fe415 steel, 35 and 40 can be multiplied by 0.8. Thus:

For Fe415 steel :
D ≥ lx/28 for simply supported slabs
D ≥ lx/32 for continuous slabs

If the conditions are satisfied, and if D is calculated using the above method, then according to this clause, we do not need to do deflection checks after the design.

So we now know the method to fix up the preliminary value for the total depth or effective depth of a two way slab. For this method, we need to know the value of l_x. But fixing up the value of l_x itself needs a trial and error method:

We know that the effective span of a simply supported beam or slab is equal to the lesser of the following two items: Details here

• Clear span plus effective depth
• c/c distance between the supports 
         - - - - (5)

But we are trying to determine the effective depth. It is a fraction of the effective span. And the effective span can be computed only if the effective depth is known. This forms a loop. So here we do a trial and error method:

(a) Take the clear shorter span (mm)
(b) Add 150 mm to it. (The average value of the effective depths of slabs with different spans generally seen in practice is around 150 mm)
(c) Find the lesser of {the above sum; c/c distance between supports}.Assume this lesser value to be l_x. Use any one of the three equations as appropriate to find d:

▪ For Fe415, d provided ≥lx/30 (Eq.17.9)
▪ For Fe250, D ≥ lx/35 for simply supported slabs (Eq.17.10)
▪ For Fe415, D ≥ lx/28 for simply supported slabs (Eq.17.10)

(d) Round off the value of d, by assuming the diameter of bars and clear cover.
(e) Use this in (5) to calculate l_x

We will do a sample calculation to illustrate the above steps:

The shorter clear span of a simply supported two way slab is 3700 mm, and the longer clear span is 4500 mm. The slab is supported on 230 mm thick brick masonry walls all around. Determine l_x, l_y, d_x and d_y.

Solution:
(a) Clear shorter span = 3700mm. So c/c distance between supports = 3930mm.
(b) 3700 + 150 = 3850 mm
(c) lesser of {3850;3930} = 3850. So d provided should be greater than or equal to 3850/30 = 128.33 (Eq.17.9)

(d) Assume 10 mm diameter bars at a clear cover of 20 mm. Then total depth D = 128.33 +5 +20 = 153.33mm. Let us provide D = 155mm. Then d_x = 155 -20 -5 = 130mm. d_y = 155 -20 -10 -5 = 120 mm.

e) Short direction:
▪ clear span + effective depth = 3700 + 130 = 3830 mm
▪ c/c distance between the supports = 3700 +230 = 3930 mm
So lx = lesser of the above = 3830 mm

Long direction:
clear span + effective depth = 4500 + 120 = 4620 mm
c/c distance between the supports = 4500 +230 = 4730 mm
So lx = lesser of the above = 4620 mm

We will do one more sample calculation with smaller spans:

The shorter clear span of a simply supported two way slab is 3250 mm, and the longer clear span is 3450 mm. The slab is supported on 230 mm thick brick masonry walls all around. Maximum LL on the slab is 2.5 kN/m². Determine l_x, l_y, d_x and d_y.

Solution:
(a) Clear shorter span = 3250mm. So c/c distance between supports = 3250 +230 =3480mm
(b) 3250 + 150 = 3400 mm
(c) Lesser of {3400;3480} = 3400. D provided should be greater than or equal to 3400/28 = 121.43 (Eq.17.10)
(d) Let us provide D = 125 mm. Assume 8 mm diameter bars at a clear cover of 20 mm. Then d_x = 125 -20 -4 = 101 mm. d_y = 125 -20 -8 -4 = 93 mm.
(e) Short direction:
▪ clear span + effective depth = 3250 + 101 = 3351 mm
▪ c/c distance between the supports = 3250 +230 = 3480 mm
So l_x = lesser of the above = 3351 mm

Long direction:
▪ clear span + effective depth = 3450 + 93 = 3543 mm
▪ c/c distance between the supports = 3450 +230 = 3680 mm
So l_x = lesser of the above = 3543 mm

So we are now in a position to determine the Total depth and effective spans. Knowing the total depth, we can determine the self wt. of the slab:

Eq.17.11:
Self wt. per square meter area of the slab = 1 x 1 x D x 25 = 25D kN/m².

We add this to the self wt. of finishes and the LL. (Both will be given in wt. per unit area). The sum of the three items will be w, the total load acting on the slab.

Knowing w, l_x and l_y, we can determine M_x and M_y using Eq.17.7. And thus the analysis of the slab will be complete.

In the next section, we will discuss the procedure to check whether the section of a two way slab can resist the one way shear applied on it.

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Chapter 17 (cont..4)- Torsionally restrained two way slabs

In the previous sections, we discussed about the analysis and design of Simply supported two way slabs. The method of analysis that we used is called the Rankine-Grashoff method. The values of α_x and α_y that are given in the table 27 of the code are known as the Rankine-Grashoff coefficients. Simply supported two way slabs are usually designed using this method. But there are other types of two-way slabs which are not simply supported. For example, the two-way slab may be continuous over one or more supports. Fig.17.1 that we saw earlier is the plan view of such an example. Another example is when the two-way slab is built monolithically into a supporting beam. Fig.17.10(a) below shows the sectional view of such an example. Yet another example is when a wall is present above the supporting wall as shown in the sectional view in fig.17.10(b). In these cases, the corners of the slab are not free to lift up, and so, Rankine-Grashoff theory is not applicable.

Fig.17.10
Examples of restraints applied at supports

Fig.17.11 below shows the view of a two way slab. The yellow arrows represent the uniformly distributed load acting on the slab. The red arrows represent the restraining force acting at the supports. This view is only a schematic representation. The restraining force may be due to a wall built above the support, or the slab being built monolithically into a beam, or the slab being continuous. In any case, a restraining force is acting at the supports of the slab, and so the corners are being prevented from lifting up. When such a restraint is applied to a slab subjected to uniformly distributed load, it deforms into the shape of a 'dish'.

Fig.17.11
View of a two way slab whose corners are held down

Let us divide the slab into strips. The fig.17.12 below shows the slab divided into a set of strips parallel to l_y.

Fig.17.12
Strips parallel to ly

Let us take one particular strip from this set. The second one from the longer support. It is shown in the fig.17.13 below:

Fig.17.13
Analysis of a single strip

We can see that, at the middle portion of the strip, the inner edge is at a lower level than the outer edge. And towards the ends of the strip, both the edges are at the same level. So the strip is twisted. In other words, the strip is experiencing torsion. This torsional effect is more towards the ends of the strip. When we take the whole set of red strips, the strips nearer to the two long walls experience this torsion to a greater extent than the strips near the middle of the whole slab. In the same way, in the other set of blue strips which are perpendicular to the red strips, the torsional effect will be more at the ends of those blue strips which are nearer to the short walls. So when we take the slab as a whole, we can see that the torsional effect will be more pronounced at the four corners of the slab.

Thus we can see that the slab is resisting the external loads applied on it, not only by bending, but also by torsion. But we have to provide special steel reinforcement to resist this torsion. Let us now see how this steel is to be provided:

Fig.17.14 below shows a corner of the slab separated from the rest. We can see that, due to the restraining effect, cracks will be formed on the top surface of the slab. These cracks are in a direction perpendicular to the diagonal of the slab.

Fig.17.14
Top surface at corner portion

We have to provide steel reinforcement at the top of the slab so that these cracks are intercepted. This means that we have to provide steel bars in a direction parallel to the diagonal of the slab. This is shown in the plan view given below in fig.17.16(a)

Now we will see the under side of the slab. That is., the bottom surface. This is shown in the fig.17.15 below:

Fig.17.15
Bottom surface at corner portion

At the bottom surface, cracks will be formed in a direction parallel to the diagonal of the slab. So here we have to provide steel bars in a direction perpendicular to the diagonal of the slab so that these cracks are intercepted. This is shown in the fig.17.16(b) below:

Fig.17.16
Arrangement of corner bars

There is another method to provide these corner bars. In this method bars are not provided parallel or perpendicular to the diagonal. Instead, they are provided parallel to the sides of the slab. This is shown in the fig.17.17 below:

Fig.17.17
Alternative arrangement for corner bars

From the fig.17.17(a), we can see that near the top surface, there are two sets of bars. One set consists of bars parallel to l_x and the other set consists of bars parallel to l_y. These two sets are provided in two layers. The bottom layer is just below the top layer. Both these sets act together to prevent the formation of the cracks at the top surface.

Similarly (fig.17.17(b)), near the bottom surface also, there are two sets of bars. One set consists of bars parallel to l_x and the other set consists of bars parallel to l_y. These two sets are provided in two layers. The top layer is just above the bottom layer. Both these sets act together to prevent the formation of the cracks at the bottom surface. Thus there will be a total of 4 layers.

It is better to provide 'U' shaped bars (section XX in fig.17.18) so that all the bars at the corners will act as a single unit, and also chances of displacement of the bars at the time of pouring concrete is minimum.

Fig.17.18
U-shaped bars at corner

In later sections, we will see the length required for these corner bars, and also the diameter and spacing.

In the next section we will discuss the methods for obtaining the bending moments in a restrained two way slab.

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