Chapter 17 (cont..12)- Solved example of a Two way slab system

Solved example 17.3
The plan of a floor slab is shown in the fig.17.44 below. It is supported on load bearing masonry walls 23 cm thick. The load of floor finishes is 1 kN/m² and the LL is 4 kN/m². Assuming mild exposure conditions and Fe 415 steel, design the slab system.

Fig.17.44
Plan of Two way slab system

Solution:
In the fig.17.45 below, the slab panels are named based on the cases given in table 26 of the code. Details here

Fig.17.45
Plan showing effective spans and type of panels

From the fig., we can see that there are two axes of symmetry. The loads acting on all the spans are the same. So we need to do the design of any one quadrant only. Let us take the top left quadrant.

Fig.17.46
Top left quadrant

The first step is to determine the effective spans and total depths of the panels in this quadrant. The calculations are given here.

The values can be tabulated as below for easy reference:

Table 17.4Properties of Panels 1 & 2

lx = 4006

ly = 4798

dx = 106

dy = 98

D = 130

Table 17.5Properties of Panels 3 & 4

lx = 3506

ly = 4798

dx = 106

dy = 98

D = 130

We can see that except for the value of l_x, all others in the two tables are equal.

Now we can do the load calculations:Table 17.6Load calculation details

Self wt. of slab =25D	3.25	kN/m2
Finishes	1.00	”
LL	4.00	”
Total	8.25	”
Factored load (load factor = 1.5)	12.4	”

The thickness of all the panels are the same. The LL are also same. So all the panels will have the same factored load of 12.4 kN/m² .

So now we know the Effective spans and the factored loads. We can start the analysis and design of each panel. The links to the detailed steps for each panel are given below:

1. Analysis and design of Panel 1

Based on the above calculations, the following table can be prepared for Panel 1:

Table 17.7Main reinforcements for Panel 1

	X Direction	Y Direction
Bottom steel at Midspan	#8 @ 250c/c	#8 @ 250c/c
Top steel at support	#8 @ 210c/c	#8 @ 250c/c

2. Analysis and design of Panel 2

Based on the above calculations, the following table can be prepared for Panel 2:

Table 17.8Main reinforcements for Panel 2

	X Direction	Y Direction
Bottom steel at Midspan	#8 @ 250c/c	#8 @ 250c/c
Top steel at support	#8 @ 190c/c	#8 @ 220c/c

3. Analysis and design of Panel 3

Based on the above calculations, the following table can be prepared for Panel 3:

Table 17.9Main reinforcements for Panel 3

	X Direction	Y Direction
Bottom steel at Midspan	#8 @ 250c/c	#8 @ 250c/c
Top steel at support	#8 @ 190c/c	#8 @ 250c/c

4. Analysis and design of Panel 4

Based on the above calculations, the following table can be prepared for Panel 4:

Table 17.10Main reinforcements for Panel 4

	X Direction	Y Direction
Bottom steel at Midspan	#8 @ 220c/c	#8 @ 250c/c
Top steel at support	#8 @ 170c/c	#8 @ 220c/c

So we obtained the bottom steel at midspans and the top steel at supports for all the panels. But at the supports which are common to two panels, we have to take the greater value. Details here. For this, we can form a table as shown below:

Table 17.11
Top steel at common supports

1st Panel	2nd Panel	Direction in 1st Panel	Direction in 2nd Panel	Steel from 1st Panel	Steel from 2nd Panel	Greater of the Two
1	2	Y	Y	#8 @250c/c	#8 @220c/c	#8 @220c/c
3	4	Y	Y	#8 @250c/c	#8 @220c/c	#8 @220c/c
1	3	X	X	#8 @210c/c	#8 @190c/c	#8 @190c/c
2	4	X	X	#8 @190c/c	#8 @170c/c	#8 @170c/c

Let us see a sample from the above table. Let us take the third case. In this case, Panel 1 is taken as the 1^st panel, and Panel 3 is taken as the 2^nd panel. These two have a common support (fig.17.46). The steel at this common support comes from the Bending moment in the X (the shorter side) direction of Panel 1. This is indicated in the third column of the table. Again, the steel at this common support comes from the Bending moment in the X (the shorter side) direction of Panel 3 also. This is indicated in the fourth column of the table. The quantities of these steels are given in the fifth (Table 17.7) and sixth columns (Table 17.9) respectively. The final seventh column gives the greater of the two quantities (lesser spacing). This greater quantity of steel is the one which should be given at the common support in each case.

Thus we obtained the top steel at all the supports which are common to any two panels. The only top steel remaining is at the outer edges. These outer edges are discontinuous supports. We know that the top steel at such supports are equal to half of that at the midspan in the respective directions, in the respective panels. By inspecting the tables from 17.7 to 17.10, we find that the least spacing of the midspan steel in any direction, in any panel, is equal to 220mm c/c. For 220mmc/c, we have to provide #8 @ 440mm c/c. Even this spacing of 440mmc/c, which is obtained from the least value of 220mm is larger than the largest spacing permissible. So we will provide a spacing of 300mm. That is., #8 @ 300mm c/c at all outer edges.

Now we can design the corner steel. We will see the details in the next section.

PREVIOUS      CONTENTS   NEXT

             Copyright ©2015 limitstatelessons.blogspot.com - All Rights Reserved