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.

PREVIOUS      CONTENTS   NEXT

             Copyright ©2015 limitstatelessons.blogspot.com - All Rights Reserved