Chapter 17 (cont..6)- Combinations of edge supports in restrained two way slabs

In the case of simply supported two-way slabs, we used two center strips in perpendicular directions for the derivation of the Rankine-Grashoff coefficients. In a similar way, we can use the two center strips shown in fig.17.19 in the previous section, to determine the coefficients for restrained two-way slabs. In the fig.17.19, we have a central red strip (the long strip) and a central blue strip (the short strip). Both these strips have their ends tied down to the support. To determine the coefficients, first we find the general form of the maximum bending moments in the two strips, considering them as independent beams. Then we analyse the various corrections to be applied, and determine the general form for those corrections. Finally, we obtain the general form for the coefficients for the bending moments. The coefficients so obtained will be applicable to any slab which has all the four edges tied down to the support. That means, if those coefficients are to be applicable to a slab, each of it's four sides should be either continuous, or built into a beam, or there should be a wall above the supporting wall.

But what if we have a slab which has one edge, say one short edge, simply supported? If one short edge is simply supported, then one end of the red strip will be simply supported. In that case, the process of analysis will be different. So we have to do the whole process again to obtain the general form of the coefficients. Those coefficients will be applicable to any slab which has one short edge simply supported, and all other edges fixed or continuous.

Again, the coefficients mentioned just above will not be applicable to a slab with one long edge simply supported, and all other edges fixed/continuous. In this case, one end of the blue strip will be simply supported, and all other ends will be fixed. We have to do the whole process again and find the general form of the coefficients. These coefficients will be applicable to any slab with one long edge simply supported and all other edges fixed/continuous.

There are several combinations possible:
[One long edge and one short edge simply supported and the other two edges fixed/continuous], [Two long edges and one short edge simply supported and the other edge fixed/continuous], etc., are some of the possible combinations. Let us find out the number of all the possible combinations and their details.

We will use the term 'Discontinuous' (short form: DC) edge to denote a simply supported edge. A fixed/continuous edge will be denoted as 'Continuous' (short form:C) edge. There will be two short edges in a slab and they will be denoted as SE1 and SE2. The long edges will be denoted as LE1 and LE2. Now we can form a table given below which shows all the possible combinations.

Table 17.1
Possible combinations of edges

We can see that there are 16 possible combinations. Let us see the details of any one sample row in the above Table, and see it's details. Take the combination No.3. It has both the short edges discontinuous, and both the long edges continuous. We have to cite an example of such a slab in a real building. For this we look at the fig.17.22 given below:

Fig.17.22
Examples of different combinations

It shows four floor plans. Plan 1 is the plan of a building with nine rooms. Plan 2 and plan 3 are those of buildings with 3 rooms. Plan 4 is that of a building with a single room. The combination that we are considering now, has two short edges discontinuous and two long edges continuous. Such a slab can be seen in plan 3. The middle panel which is marked as 5, belongs to this category. So 5 is entered in the 'Example' column of the Table, in the row for combination No.3.

Let us see one more example. Take combination No.6. It has two short edges and one long edge continuous, and the remaining long edge discontinuous. Such a slab panel can be seen in Plan 1. The panels marked as 3 belongs to this category. So 3 is entered in the 'Example' column of the Table, in the row for combination No.6.

We are considering square or rectangular slabs in our present discussion about two-way slabs. So opposite sides will be equal. That is., SE1 = SE2 and LE1 = LE2. So when we make all the possible combinations, there will be some duplications. For example, the combination with one short edge and one long edge C and the other Short edge and Long edge DC can have four possible combinations, all of them giving the same result of panel 4. This is indicated by the four red blocks in the Table. As the four combinations give the same result, we need to consider only one, and the rest three can be discarded. Similarly there are four more cases, indicated by the green, yellow, cyan and magenta colors. Each have two combinations which give the same results. We need to take one only from each. Thus the net number of combinations = 16 -3 -1 -1 -1 -1 = 16 -7 = 9. We will modify the Table by avoiding duplications and thus showing only the nine cases. We will also change the order so that they are in the same order as given in the code. This modified table is given below:

Table 17.2
Modified Table showing the nine combinations

It may be noted that the numbering of the panels in fig.17.22 above was done based on the order of the cases given in the code. That is how we get sequential order in the ‘Example’ column in the above Table 17.2.

In the next section we will obtain the actual values of the coefficients for each of the above nine combinations.

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