Chapter 17 (cont..2) Thickness and Effective span of Two way slabs

Preliminary value for the thickness of a two way slab

When fixing the thickness of slab, deflection criterion is more important than the bending moment criterion. That is., a lower value of thickness may be sufficient to resist the bending moment. But that lower value will not satisfy the deflection check. So we will do the procedure for deflection check in a sort of 'reverse' manner to fix up a preliminary value for 'D', the total thickness of the slab. This can be done as follows:

We know that, for spans less than 10m, to satisfy the deflection check, the following condition should be satisfied:

  - - - - (1)

For simply supported members, (l/d)_basic = 20.
Now we take k_t. This can be calculated by using the expression given in SP24:

  - - - - (2)

Where

  - - - - (3)

But in the above expressions, there are three unknowns. (a) pt, (b) Area of steel required, and (c) Area of steel provided. These three unknowns can be solved by making some assumptions:

From the numerous design problems of One way slabs that has been already done for various buildings and other structures, we can say that the percentage of reinforcement is around 0.45% when Fe415 steel is used. Note that this is for one way slabs. If we assume that the area of steel required is same as the area of steel provided, then from (3):

f_s = 0.58 x 415 x 1 = 240.7 N/mm² - - - - (4)

Substituting this value of f_s and p_t (= .45) in (2) we get k_t = 1.277

Effective spans of a two way slab

Now, to use the l/d ratio, we must have the value for l. For a two way slab, we have two values for l. They are l_x and l_y. It is the value of l_x, (cl.24.1, Note 1) the shorter side that we have to use in the l/d calculation. But the percentage of steel required for resisting M_x in a two way slab is less than the percentage of steel (0.45%) required to resist the moment in a one way slab with the same effective span as l_x. Because of this lower value of p_t, k_t will be higher than 1.277.

It may be noted that instead of using (2) we can use fig.4 of the code. From that fig., it can be seen that when p_t decreases, k_t increases.

So we use a value of k_t = 1.5
Substituting this in (1) we get lx/d provided ≤ 20 x 1.5.
So we get:

Eq.17.9:

lx/d _provided ≤ 30

OR d _provided ≥ lx/30

However, the design using this value should be finalized only after the checks using actual area of steel required and the actual area of steel provided.

There is another method to fix up the total depth D of a two way slab. But to use this method, the slab should satisfy the following conditions:

• l_x is less than or equal to 3.5m, and
• LL is less than or equal to 3kN/m²,

When the above two conditions are satisfied, we can use cl.24.1-Note 2 of the code for calculating the total depth D of the two way slab:

Eq.17.10: 

Fe250 steel:
D ≥ lx/35 for simply supported slabs
D ≥ lx/40 for continuous slabs

For Fe415 steel, 35 and 40 can be multiplied by 0.8. Thus:

For Fe415 steel :
D ≥ lx/28 for simply supported slabs
D ≥ lx/32 for continuous slabs

If the conditions are satisfied, and if D is calculated using the above method, then according to this clause, we do not need to do deflection checks after the design.

So we now know the method to fix up the preliminary value for the total depth or effective depth of a two way slab. For this method, we need to know the value of l_x. But fixing up the value of l_x itself needs a trial and error method:

We know that the effective span of a simply supported beam or slab is equal to the lesser of the following two items: Details here

• Clear span plus effective depth
• c/c distance between the supports 
         - - - - (5)

But we are trying to determine the effective depth. It is a fraction of the effective span. And the effective span can be computed only if the effective depth is known. This forms a loop. So here we do a trial and error method:

(a) Take the clear shorter span (mm)
(b) Add 150 mm to it. (The average value of the effective depths of slabs with different spans generally seen in practice is around 150 mm)
(c) Find the lesser of {the above sum; c/c distance between supports}.Assume this lesser value to be l_x. Use any one of the three equations as appropriate to find d:

▪ For Fe415, d provided ≥lx/30 (Eq.17.9)
▪ For Fe250, D ≥ lx/35 for simply supported slabs (Eq.17.10)
▪ For Fe415, D ≥ lx/28 for simply supported slabs (Eq.17.10)

(d) Round off the value of d, by assuming the diameter of bars and clear cover.
(e) Use this in (5) to calculate l_x

We will do a sample calculation to illustrate the above steps:

The shorter clear span of a simply supported two way slab is 3700 mm, and the longer clear span is 4500 mm. The slab is supported on 230 mm thick brick masonry walls all around. Determine l_x, l_y, d_x and d_y.

Solution:
(a) Clear shorter span = 3700mm. So c/c distance between supports = 3930mm.
(b) 3700 + 150 = 3850 mm
(c) lesser of {3850;3930} = 3850. So d provided should be greater than or equal to 3850/30 = 128.33 (Eq.17.9)

(d) Assume 10 mm diameter bars at a clear cover of 20 mm. Then total depth D = 128.33 +5 +20 = 153.33mm. Let us provide D = 155mm. Then d_x = 155 -20 -5 = 130mm. d_y = 155 -20 -10 -5 = 120 mm.

e) Short direction:
▪ clear span + effective depth = 3700 + 130 = 3830 mm
▪ c/c distance between the supports = 3700 +230 = 3930 mm
So lx = lesser of the above = 3830 mm

Long direction:
clear span + effective depth = 4500 + 120 = 4620 mm
c/c distance between the supports = 4500 +230 = 4730 mm
So lx = lesser of the above = 4620 mm

We will do one more sample calculation with smaller spans:

The shorter clear span of a simply supported two way slab is 3250 mm, and the longer clear span is 3450 mm. The slab is supported on 230 mm thick brick masonry walls all around. Maximum LL on the slab is 2.5 kN/m². Determine l_x, l_y, d_x and d_y.

Solution:
(a) Clear shorter span = 3250mm. So c/c distance between supports = 3250 +230 =3480mm
(b) 3250 + 150 = 3400 mm
(c) Lesser of {3400;3480} = 3400. D provided should be greater than or equal to 3400/28 = 121.43 (Eq.17.10)
(d) Let us provide D = 125 mm. Assume 8 mm diameter bars at a clear cover of 20 mm. Then d_x = 125 -20 -4 = 101 mm. d_y = 125 -20 -8 -4 = 93 mm.
(e) Short direction:
▪ clear span + effective depth = 3250 + 101 = 3351 mm
▪ c/c distance between the supports = 3250 +230 = 3480 mm
So l_x = lesser of the above = 3351 mm

Long direction:
▪ clear span + effective depth = 3450 + 93 = 3543 mm
▪ c/c distance between the supports = 3450 +230 = 3680 mm
So l_x = lesser of the above = 3543 mm

So we are now in a position to determine the Total depth and effective spans. Knowing the total depth, we can determine the self wt. of the slab:

Eq.17.11:
Self wt. per square meter area of the slab = 1 x 1 x D x 25 = 25D kN/m².

We add this to the self wt. of finishes and the LL. (Both will be given in wt. per unit area). The sum of the three items will be w, the total load acting on the slab.

Knowing w, l_x and l_y, we can determine M_x and M_y using Eq.17.7. And thus the analysis of the slab will be complete.

In the next section, we will discuss the procedure to check whether the section of a two way slab can resist the one way shear applied on it.

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