Chapter 17 (cont..10)-Details of Corner bars

'L' Type:
The requirements for this type is given in cl.D-1.8 of the code. We know that corner reinforcements should be provided near both top and bottom surfaces of the slab. Both of them are the same. That is., if we design the reinforcements to be provided near the top surface, the exact same can be provided near the bottom surface also. So we need to discuss about any one, say the one at the top only.

The top one consists of a mesh. A mesh, formed by two layers of bars. The first layer consists of bars parallel to l_x. The second layer consists of bars parallel to l_y. This second layer is stacked just below the first layer and the two layers are tied together. Thus the mesh is formed. The length of all the bars in both the layers is the same, and should not be less than 1/5 times the smaller side of the slab panel. The smaller side is l_x. So length of each bar = 1/5(l_x) = 0.2l_x.

Now we need to know the number of bars that have to be provided in each direction of the mesh. According to the code, the total number of bars in the X direction in a single mesh should give an area which is not less than 0.75 times A_st,x. Where A_st,x is the area required to resist the maximum bending moment at the midspan. The subscript 'x' is given in A_st,x because we readily know that the maximum bending moment in the midspan will be that in the X (direction of shorter span according to fig.17.1) direction. So A_st,x is the area of steel required to resist the bending moment M_u,x⁺ =  α_x⁺  w_u l_x² . This same number of bars should be given in the Y direction also.

So now we know the details required to make the mesh near the top surface. Another mesh is to be made, which is exactly identical to this. And this second mesh should be placed near the bottom surface, in line with with the top mesh. These two meshes together will satisfy the requirements of torsion reinforcement at the 'L' Type corner. This is shown in the fig. below:

Fig.17.37
Torsion reinforcement at ‘L’ type corner
 

A View showing the bottom mesh at this type of corner in a slab is given in fig.17.38 below:

Fig.17.38
View of bottom mesh

We can see that one layer consists of bars parallel to one side of the room, and the second layer consists of bars parallel to the wall which is perpendicular to the first wall. These two layers together form the ‘mesh’. There will be an exactly identical mesh near the top surface of the slab. Thus there will be a total of four layers. The top mesh is not shown in the view above for clarity.

The two meshes can be formed by using ‘U’ bars. This was shown earlier in fig.17.17 and 17.18. A view of this type is shown in the fig.17.39 below:

Fig.17.39
Corner reinforcement using ‘U’ bars

'T' Type:
The requirements for this type is given in cl.D-1.9 of the code. The method of making the meshes and placing them in the slab are same as that of the 'L' type. But the required areas of the bars is different. As for the 'L' type, we will discuss all the details:

First we will see the top mesh. This mesh is formed by two layers of bars. The first layer consists of bars parallel to l_x. The second layer consists of bars parallel to l_y. This second layer is stacked just below the first layer. Thus the mesh is formed. The mesh does not have a 'square' shape. So the length of bars parallel to l_x (top layer) is different from the length of the bars parallel to l_y (bottom layer). We have to calculate the lengths carefully. From the fig.17.38 below, we can see that at a 'T' type corner, one wall is common to two panels. The bars parallel to l_x, provided at this wall extends into both the panels. So these reinforcements should satisfy the requirements of both the panels.

Fig.17.38
Torsion reinforcement at ‘T’ type corner

According to the cl.D-1.8, the bars should extend to a distance of 0.2l_x. But in a 'T' type corner, there are two panels, and l_x may be different. So in the fig.17.38, we can see that the bars extend for a distance of 0.20l_x1 into the upper panel, and a distance of 0.20l_x2 into the lower panel. (Both the lengths are measured from the face of the support). This gives the length of the bars in the direction parallel to l_x, and thus the length of the mesh can be fixed up. Now we look at the width of the mesh. Unlike the length, the width of the mesh is common to both the panels. That is., once we fix up the width in the direction parallel to l_y, it will be applicable to both the panels. So we must take the greater of 0.20l_x1 and 0.20l_x2 as the width. This is shown in the fig.17.38 above.

Now we need to know the number of bars that have to be provided in each direction of the mesh. According to the code, the total number of bars in the X direction in a single mesh should give an area which is not less than 0.375 times A_st,x (half of that in a 'L' type = 0.5 x 0.75 = 0.375 According to cl.D-1.9) . Where A_st,x is the area required to resist the 'maximum' bending moment at the midspan. The subscript 'x' is given in A_st,x because we readily know that the maximum bending moment in the midspan will be that in the X direction. So A_st,x is the area of steel required to resist the bending moment M_u,x⁺ =  α_x⁺  w_u l_x² . But just as we fixed up the dimensions of the mesh, in fixing up the number and diameter of bars also, we have to take the 'common' factor into consideration. That is., once we fix up the number of bars parallel to l_x, they will be common to two panels. So we must take the greater of A_st,x1 and A_st,x2. But while fixing up the number of bars parallel to l_y, the 'common' factor does not arise: In the fig.17.38, the bars in the top panel is calculated from 0.375A_st,x1 and the bars in the bottom panel is calculated from 0.375A_st,x2.

So now we know the details required to make the mesh near the top surface. Another mesh is to be made, which is exactly identical to this. And this second mesh should be placed near the bottom surface, in line with with the top mesh. These two meshes together will satisfy the requirements of torsion reinforcement at the 'T' Type corner.

It may be noted that in the figs.17.38, the longer spans of both the upper room and lower room (in the plane of paper) are horizontal, and the shorter sides are vertical. But some times it may so happen that the lower room has it's longer span vertical, and the shorter span horizontal. Part plan of such a situation is shown in fig.17.39 below:

Fig.17.39
Lower room having shorter side horizontal

The notation for the lower room has changed from l_x2 to l_y2. But as far as the torsion reinforcements are concerned, the notations used for specifying the dimensions of the meshes, and the quantities of bars in the meshes will not change because, they all depend on the properties of the panels in the X direction only: Dimensions depend on l_x values and quantities depend on A_st,x values. However we must thoroughly analyse each structure carefully and find the appropriate values.

In the next section we will see some solved examples that will illustrate the details that we have discussed above.

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