Chapter 17 (cont..14) Details of reinforcements in two way slab system

In the previous section, we saw the figs.17.47, 48 and 49, showing the required reinforcements in the slab. Now we will see some main features of those figs:

Division of panels into the three strips:
All the four panels have their longer sides in the horizontal direction, and the shorter sides in the vertical direction (in the plane of paper). So all the panels have their l_y horizontal and l_x vertical on the plane of paper.

In addition to the above, all the four panels have their l_y same, and is equal to 4798mm. So the division in the l_y direction is the same for all the panels. 0.75l_y = 0.75 x 4798 = 3598.5 This can be rounded off to the higher value of 3600mm. So the clear span 4700 can be marked off as 550 +3600 +550 =4700mm

But in the other direction, l_x of the upper two panels is different from those of the lower two panels. So we have to do the calculations separately.

Upper panels 3 & 4:
0.75l_x = 0.75 x 3506mm =2629.5 This can be rounded off to the higher value of 2630mm. So the clear span 3400mm can be marked off as 385 +2630 +385 =3400mm

Lower panels 1 & 2:
0.75l_x = 0.75 x 4006 =3004.5 This can be rounded off to the higher value of 3010mm. So the clear span 3900mm can be marked off as 445 +3010 +445 =3900mm

Curtailment of bottom bars:

Bars in X direction:
Upper panels 3 & 4:
Cut off point from discontinuous edge = 0.15l_x = 0.15 x 3506mm =525.9 = 530mm (from the center of support)
Cut off point from continuous edge = 0.25l_x = 0.25 x 3506mm =876.5 = 880mm (from the center of support)
Lower panels 1 & 2:
Cut off point from continuous edge = 0.25l_x = 0.25 x 4006mm =1001.5 = 1000mm (from the center of support)
Bars in Y direction:
Left side panels 2 & 4:
Cut off point from discontinuous edge = 0.15l_y = 0.15 x 4798mm =719.7 = 720mm
Cut off point from continuous edge = 0.25l_y = 0.25 x 4798mm =1199.5 = 1200mm (from the center of support)
Right side panels 1 & 3:
Cut off point from continuous edge = 0.25l_y = 0.25 x 4798mm =1199.5 = 1200mm (from the center of support)

Curtailment of top bars:
Bars in X direction:
Upper panels 3 & 4:
Cut off point from discontinuous edge = 0.1l_x = 0.1 x 3506mm =350.6 = 350mm (from the face of support)
Cut off point from continuous edge = 0.15l_x = 0.15 x 3506mm =525.9 = 530mm (from the face of support)
Lower panels 1 & 2:
Cut off point from continuous edge = 0.15l_x = 0.15 x 4006mm =600.9 = 600mm (from the face of support)
Bars in Y direction:
Left side panels 2 & 4:
Cut off point from discontinuous edge = 0.1l_y = 0.1 x 4798mm =479.8 = 480mm
Cut off point from continuous edge = 0.15l_y = 0.15 x 4798mm =719.7 = 720mm (from the face of support)
Right side panels 1 & 3:
Cut off point from continuous edge = 0.15l_y = 0.15 x 4798mm =719.7 = 720mm (from the face of support)

This completes all the details of the solved example 17.3

Two way slabs supported on beams

We have seen the analysis and design of two way slabs. The method of using coefficients from table 26 of the code is applicable only if the supports are rigid. So this method can be applied to slabs resting on masonry walls. These masonry walls should be constructed according to the specifications given in the relevant codes.

This method of using the coefficients cannot be used for slabs in a framed structure. This is because, the beams on which the slabs rest, may not be sufficiently rigid. We know that reinforced beams will deflect when loads are applied. Consider the fig.17.50 given below:

Fig.17.50
Part key plan of a framed building

It is a part key plan of a framed building. On an interior beam, there will be slab on either side of the beam. If the slabs on either side have the same loads and same spans, there will not be net moment on the beam. In such a situation, there will only be deflection in the beam.This is shown in fig.17.51(b). But if the spans or loads are different, or if the beam is an edge beam, there will be deflection as well as torsion. This is shown in fig.17.51(c).

Fig.17.51
View of beams under deflection and torsion

In framed structures, the columns that support the beams may also deflect or rotate from their original positions. When such changes occur in  beams or columns, the edge conditions of the slabs will change. In such situations, the basic assumptions that are made in the derivation of the coefficients will not be valid. Thus we can see that the method of coefficients is best suited for slabs resting on rigid masonry walls.

Now we will examine the possibility for extending this method to slabs resting on beams:  Cl.23.2 of the code gives the procedure that should be followed for ensuring the control of deflection. We discussed about it here. When the provisions of this clause are satisfied, we can assume that deflections will be within permissible limits. In addition to this, we must ensure that the beams are adequately stiff. The Canadian code gives a simple guideline for this purpose. In this method we take the following ratio:

Where D_b = Total depth of beam, D_s = total depth of slab, l = clear span of the beam, and b = width of the web of the beam. If this ratio is greater than or equal to 2, then the beam can be considered to be stiff.

Let us use the above method for an actual beam. At the end of a previous section here, a solved example 4.3 was given. It demonstrated the design of a simply supported beam. We will take the final dimensions of that beam. They are as follows: D_b = 450mm, b = 230mm, l = 4770mm. Let the total thickness D_s of the slab be equal to 135 mm. Then applying the above we get the ratio equal to 1.786. This is less than 2. So, if we want to design the slab using the coefficients, we must increase the total depth of the beam so that the ratio will be greater than 2.

cont..