Chapter 17 (cont..3)- Check for One way shear in a Two way slab

Calculation of one way shear at the supports of a two way slab

We have seen the procedure for shear check in one way slabs. In one way slabs, the loads are transferred to two opposite walls. If the slab is subjected to uniformly distributed loads, then one half of the load will be transferred to one wall, and the other half will be transferred to the other opposite wall. But in the case of a two way slab, there are four walls, and the loads are transferred to all the four walls. If all these four walls are of equal length, then we can assume that, one fourth of the total uniformly distributed load will be transferred to each wall. But if the lengths are not equal, then the distribution of loads to each wall is complicated. In this situation, we can use the recommendation given by cl.24.5 of the code. According to this clause, the distribution of load on the short side is triangular, and the distribution of load on the long side is trapezoidal. This is shown in the fig.17.6 below:

Fig.17.6
Distribution of load in a two way slab

The portions that each wall will carry can be demarcated by drawing 45^o lines from the corners. When these 45^o lines are drawn from all the four corners, we will get a triangle (shown in blue color) at each short side, and a trapezium (shown in green color) at each long side. When the area is divided in this manner, the height of the triangles and the height of trapeziums will be the same, and is equal to half of the length of the short side.

We are designing a 1m wide strip of the slab. So the shear force will be maximum when the strips are taken at the apex point of the triangle and the trapezium. We need to consider any one only (triangle or trapezium) because the strip for both the cases are of the same length 0.5 l_xn. where l_xn is the clear span in the short direction. Also, we need not consider this whole length of 0.5 l_xn. Only the load to the right of the critical section (which is at a distance of d, the effective depth of the slab, shown by the red dashed line in plan and section from the face of the support) need to be considered. This is because the shear force at the critical section will be caused by the loads acting on the right side of the section. This is shown in the section XX in the above fig.17.6

The area of the strip to the right side of critical section = (0.5 l_xn – d) x 1 = (0.5 l_xn – d) m²

So total factored load on this area is obtained as:

Eq.17.12
V_u = w_u(0.5 l_xn -d)

The rest of the procedure is same as that for a one way slab: The factored shear force V_u should be less than the shear contribution from concrete. The shear contribution from concrete is calculated as kτ_cbd. Where,

k is the modification factor given in cl.40.2.1.1 of the code, calculation of which we discussed in the case of one way slabs.
τ_c is the design shear strength of concrete, obtained from table 19 of the code,
b is the width of the strip of slab = 1000 mm, and
d is the average effective depth of the slab = (d_x + d_y)/2

Thus the procedure for shear check of a two way slab is complete.

Though we have had a somewhat lengthy discussion about the analysis and design of the simply supported two way slab, it can be summarized into just a very few steps:

(a) Determine the preliminary value of total thickness, and effective spans l_x and l_y as described in the previous section. l_x is the shorter span, and l_y is the longer span.

(b) Determine the factored load w_u [Load factor x (sum of self wt., finishes and LL)]

(c) Calculate r = ly/lx

(d) based on r, calculate α_x and α_y

(e) Calculate the factored moments  M_ux =  α_x w_u l_x²  , and  M_uy =  α_y w_u l_x²   

(f) Provide the steel determined from M_ux in the direction parallel to l_x (in the bottom most layer) , and the steel determined from M_uy in a direction parallel to l_y.

(g) Do the shear check and also all other required checks that we do for a one way slab.

The solved example given below will illustrate all the points that we discussed above:

Solved example 17.1

The reinforcement details according to the above design is shown in the figs.17.7,8 and 9 below:

Fig.17.7
Plan view

Fig.17.8
Section AA

Fig.17.9
Section BB

We will now discuss the main features of the above figs:
The steel A_st,x is calculated from M_ux. From A_st,x we fixed up #8 bars @ 140 mm c/c. This steel is laid parallel to l_x. Consider the set of all the strips parallel to l_x. That is., all the blue strips in fig.17.3. The strips in the middle region will require the steel A_st,x. But those on the sides, which are nearer to the shorter walls will be carrying lesser loads, and so they will not require the same A_st,x. For those strips, the spacing can be increased. But, as mentioned earlier, we assume that all the blue strips carry the same load, and so the same steel is provided for all the blue strips. Thus in the above fig.17.7, we can see that #8 @140 mm c/c is provided from the left short wall to the right short wall. Similarly, in the set of red strips parallel to l_y in fig. 17.3, the strips nearer to the long walls will not require #8 @ 220mm c/c. For those strips, the spacing can be increased. But we assume that all the red strips carry the same load, and so the same steel is provided for all the red strips. Thus in the above fig.17.7, we can see that #8 @220 mm c/c is provided from the bottom long wall to the top long wall.

Now we will discuss another feature that can be seen in the above figs. Let us consider any one of the blue strips. It is simply supported on the two long walls. So the bending moment at the ends will be zero. So, like any simply supported member, the bending moment varies from a maximum at the middle of the strip to zero at the supports. Thus it is possible to curtail the bars. We have seen the details of curtailment here. For a simply supported two way slab we follow simplified rules of curtailment: The main bars are terminated at alternate ends, at a distance of 0.l from the face of the support. So 50% of the main bars are curtailed, and the other 50% continues into the support. We can see from the fig. that, this type of curtailment is done in both x and y directions.

The curtailed bars are not just stopped at 0.1l from the face of the support. They are given a 45^o bend at that point, and thus they become top bars at the support. Thus we will get 50%  A_st as top bars at the supports. These top bars are necessary to resist any hogging moment that may develop at the supports.

In the next section we will see some basic details about torsionally restrained two way slabs.

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