Chapter 17 (cont..7)- Coefficients from code for bending moments in two way slabs

So we have seen the nine different possibilities. As we saw before based on fig.17.21, we need four coefficients for any slab panel. So there can be nine sets of coefficients, with each set having four unique coefficients. So in general, there will be a total of 9 x 4 = 36 coefficients. If we are given a two-way slab panel, depending on it's position in the building, and also depending on it's edge supports, it can fall into any of the nine categories. We must be able to determine the appropriate 4 coefficients for the given slab, from among these 36.

At the discontinuous (ie., simply supported) edges, the negative moments will be zero. So, with out any calculations, we can say that the coefficients for the negative moments at such supports will be equal to zero. (This is because, in the equation M_u,x = α w_u l_x² = 0, w_u and l_x cannot be zero. So the only option is that the coefficient α is equal to zero.) Thus some of the coefficients among these 36 will be zero. We will see more details about it later.

We can derive expressions for calculating these coefficients. The derivations are based on the ‘classical theory of plates’. In this method, the slabs are considered as plates. In the derivation, the various corrections that we saw before are taken into account. There will be variables in these expressions by which we can 'input' the type of support at each edge (whether simply supported or fixed/continuous). By giving the appropriate value for the variables, we can obtain any of the 36 coefficients.

Another method is to use the 'Approximate solutions' based on Rankine-Grashoff theory. Corrections to these solutions were proposed by Marcus. This is known as the 'Marcus correction'.

Even as there are different methods such as those mentioned above, we must use the method (Cl. D-1) given by the code. In the Table 26, the values of the coefficients are given directly. The coefficients in this Table are based on 'Yield line analysis'. We do not have to learn the derivation of these coefficients at this stage. Their values are sufficient for carrying out an analysis and design of the slabs.

Let us now see a sample from the Table 26. Suppose that we want to design a slab in which
• both the short edges are continuous
• one long edge is continuous and
• one long edge is discontinuous.

Mentioning all the above three boundary conditions together is equivalent to mentioning just the last one only:
• one long edge is discontinuous.
With the mention of this last condition, the other two are automatically implied.

This slab falls under the category of panel 3 in fig.17.22. This is case no.3 of Table 26. Let ly/lx of our slab be equal to 1.2. So from the Table, we have
• Coefficient for the Negative moment (in the X direction) at the continuous edge (column 5 of Table 26) = α_x^- = 0.052
• Coefficient for the Positive moment (in the X direction) at the midspan (column 5) = α_x⁺ = 0.039
• Coefficient for the Negative moment (in the Y direction) at the continuous edge (column 11) = α_y^- = 0.037
• Coefficient for the Positive moment (in the y direction) at the midspan (column 11) = α_y⁺ = 0.028

We must note one point here: If the value of ly/lx falls between 1.2 and 1.3 (say 1.23 , 1.27 etc.,), then we must use linear interpolation to find α_x^- and α_x⁺ . But no interpolation is required for α_y^- and α_y⁺ . This is because, for case 3, whatever be the value of ly /lx, the code gives constant values of 0.037 and 0.028 for α_y^- and α_y⁺. In fact, each of the nine cases has it’s own constant values for α_y^- and α_y⁺ (column 11), which do not change with the value of ly/lx.

The application of the four coefficients are shown in the fig.17.23 below:

Fig.17.23
Bending moments in the slab

Let us consider one more sample. Suppose that we want to design a slab in which
• both the short edges are continuous
• both the long edges are discontinuous
Mentioning the above two boundary conditions together is equivalent to mentioning just the last one only:
• both the long edges are discontinuous

This slab falls under the category of panel 6 in fig.17.22. This is case no.6 of Table 26. Let ly/lx of our slab be equal to 1.3. So from the table, we have

• Coefficient for the Negative moment (in the X direction) at the continuous edge (column 6):
Here we do not find any value. Why is this so? This can be explained based on the fig.17.24 below:

Fig.17.24
Bending moments in the slab

Perpendicular to the X direction, both the supports (the longer supports) are discontinuous. In other words simply supported. So the bending moment at those supports will be zero. This means that the coefficient will be equal to zero. If any one of these longer supports were continuous, the coefficient would have been present.

• Coefficient for the Positive moment (in the X direction) at the midspan (column 6) = α_x⁺ = 0.057
• Coefficient for the Negative moment (in the Y direction) at the continuous edge (column 11) = α_y^- = 0.045
• Coefficient for the Positive moment (in the y direction) at the midspan (column 11) = α_y⁺ = 0.035

As mentioned earlier we must note one point here also: If the value of ly/lx falls between 1.2 and 1.3 (say 1.23 , 1.27 etc.,), then we must use linear interpolation to find α_x^- and α_x⁺ . But no interpolation is required for α_y^- and α_y⁺ . This is because, for case 6, whatever be the value of ly /lx, the code gives constant values of 0.045 and 0.035 for α_y^- and α_y⁺. As mentioned above, each of the nine cases has it’s own constant values for α_y^- and α_y⁺ (column 11), which do not change with the value of ly/lx.

If we look at any of the columns from (3) to (10), we can see that for cases 6 and 8, there are no coefficients for negative moment at supports in the X direction. Similarly in column (11), we can see that for cases 5 and 7, there are no coefficients for negative moment at supports. So for the four cases 6,8,5 and 7, there are only 3 coefficients. The fourth one is zero because these cases have a pair of opposite edges discontinuous.

In the last case 9, both the pairs of opposite edges are discontinuous. there will not be a negative moment at any of the four supports. So there are only two coefficients. So in general, the total number of coefficients can be calculated as:

4 cases x 4 coefficients = 16
4 cases x 3 coefficients = 12
1 case x 2 coefficients  = 2
Total                          = 30

This is less than 36. It may be noted that all the cases 6,8,5 and 7, having opposite edges discontinuous, belong to the floor plans 2 and 3 in fig.17.22

So we have seen the method for determining the bending moments. In the next section, we will discuss about the arrangement of bars in the two way slabs.

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