Chapter 17 (cont..9)-Top bars and corner bars in restrained two way slabs

Here we will discuss the arrangement of top bars in the slab. We know that the top bars are given at the supports. They are calculated from the negative moments M_u,x^- and M_u,y^-. The diameter and spacing calculated from these values should be provided in the middle strips in the respective directions. As usual, we can curtail some of these bars. The guidelines for the curtailment of these top bars are given in cl.D-1.5 of the code. The top bars are essential up to a distance of 0.3l from the face of the support. But all the bars are not required upto this distance. 50% can be curtailed at a distance of 0.15l. This is shown in the figs.17.33 and 17.34

Fig.17.33
Layout of Top bars in the middle strip in the X direction

Fig.17.34
Layout of Top bars in the middle strip in the Y direction
 

Technically, it is sufficient to show two bars belonging to a particular set of bars, in a drawing. But if more bars are shown, a definite pattern and symmetry in the arrangement of bars will become more clear. So 5 top bars (at the continuous support) are shown in the drawings above.

In the above figs., every  bar in one set of alternate bars at the continuous support in a particular direction has the same length (0.15l1 + 0.15l1 + wall thickness + 0.15l2 = 0.3l1 + wall thickness + 0.15l2 ). And every bar in the other set of alternate bars have the same length (0.3l2 + wall thickness + 0.15l1.) Alternate bars are given an extension of 0.15l on opposite sides of the wall. We can see that, on either side of the wall, 100% steel is available upto a distance of 0.15l from the face of the wall. Beyond this distance, and within 0.3l from the face of the wall, 50% is available.

At this time we can see a small topic which is related to our present discussion:

Design negative moment at continuous support

In a continuous slab system supported on walls, we determine the bending moments using the coefficients given in the code. Each slab panel is analysed separately. In the fig.17.33 above, there are two adjacent spans l_x1 and l_x2 . We are considering the panel with span l_x1. We get the factored negative moment on the continuous support as M_u,x1^- =  α_x1^-  w_u l_x1². From this we will get the steel required to resist this bending moment. But we cannot finalize it yet.

As it is a continuous system, the steel at the support will be extending to either side of the wall, and will be resisting the negative moment created at the wall, due to the loads on the span l_x2 also. This negative moment is given by M_u,x2^- =  α_x2^-  w_u l_x2² . This quantity may be different from M_u,x1^- due to one or more of the following reasons:

• the coefficients α_x1^-  and α_x2^-  may not be equal because of unequal spans or different boundary conditions.
• span l_x1 may not be equal to l_x2
• the load w_u may not be equal on both the spans.

So, when the negative bending moments at the support are not equal, we must take the larger of the two for the design. This is written in both the figs.17.33 and 17.44. In this way, the top steel at the support will be able to resist the negative bending moments from either spans.

It may be noted that in the figs.17.33 and 17.34, the spans l_y1 and l_y2 are horizontal and l_x1 and l_x2 are vertical on the plane of the paper. This indicates that the adjacent rooms also have their longer side horizontal and shorter side vertical. But some times it may so happen that an adjacent room has it's longer side vertical, and the shorter side horizontal. Part plan of such a situation is shown in fig.17.35 below:

Fig.17.35
Adjacent room having shorter side horizontal

In such cases, the notations for the edges of the panels will change, and appropriate changes should be made in the calculations. For example, the diameter and spacing of the top bars at the wall which is common to the two rooms should be calculated from the greater of M_u,y1^-  and M_u,x2^- . This is different from what is written in the fig.17.34 at the common wall. So we must thoroughly analyse each structure carefully and find the appropriate values.

At the discontinuous edge, we must provide top steel to resist any negative moment that may develop due to any partial fixity. The area of steel required for this purpose, is half of the quantity that is provided for resisting the midspan moment in the respective directions. These bars have to be given an extension of 0.1l into the slab panel, from the face of the support. 

Corner bars for resisting Torsion

We have seen some basic details about corner bars in fig.17.17. Now we will discuss about the actual area to be provided for these bars, and also the exact position in which they must be provided.

Corners are formed when two or more edges meet at a point. We are considering two-way slab systems which consists of square or rectangular slab panels. So all the corners that we come across in this discussion will be right angled corners. Also all these corners will fall into one of these categories:

• 'L' Type in which a corner is formed by two edges.
• 'T' Type in which a corner is formed by three edges.
• '+' (Plus) Type in which a corner is formed by four edges.

These different types of corners in actual building plans are shown in the fig.17.36 below:

Fig.17.36
Corners in slab panels

We will first see the '+' Type. This is an interior corner. The Code mentions about this type in cl.D-1.10. From fig.17.36 we can see that this type is formed by two edges, and the slab is continuous over both these edges. This type of continuity of the slab panel happens in an interior corner. And according to this clause, torsion reinforcements need not be provided at such corners.

So that leaves the 'L' Type and 'T' type only. If we know the requirements for these two types, we can do a complete design of torsion reinforcements for any two-way slab systems consisting of square or rectangular panels. We will discuss about them in the next section.

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