Chapter 17 (cont..13)-Corner bars in the slab system

In this section we design the corner reinforcements of the slab system. Details here. From fig.17.46 of the previous section we can see that we have to design
• one 'L' Type corner (Exterior corner of Panel 4)
• one 'T' Type corner (Corner common to Panels 3 and 4)
• one 'T' Type corner (Corner common to Panels 2 and 4)

Design of 'L' Type corner of Panel 4:
Size of the mesh = 0.2l_x = 0.2 x 3506 = 701.2mm
provide a distance of 710mm from the face of the supports

A_st,x in panel 4 (Table 17.10) = #8 @ 220c/c = 228.48 mm²
0.75 of above = 0.75 x 228.48 = 171.36 mm²
So provide 4-#8, giving an area of 201.06 mm²

The top layer of the mesh will have 4-#8 parallel to l_x, distributed evenly in a distance of 710mm. The bottom layer of the mesh will have the same 4-#8 parallel to l_y, distributed evenly in a distance of 710mm. Thus one complete mesh is formed. This mesh is placed near the top surface. An identical mesh is made and it is placed near the bottom surface. This completes the design of the 'L' type corner. This is shown in fig.17.49 below.

Design of 'T' Type corner common to Panels 3 and 4
Width of the mesh = larger of 0.2l_x1 and 0.2l_x2.
But l_x of panels 3 and 4 are the same and is equal to 3506mm
So width of the mesh = 0.2 x 3506 = 701.2mm
provide a distance of 710mm from the face of the supports
Length of the mesh = 0.2l_x1 +width of support +0.2l_x2 = 710 +230 +710 = 1650mm

No. of Longer bars of the mesh:
This is calculated from the larger of 0.375A_stx1 and 0.375A_stx2
Astx1 = #8 @ 250c/c (Table 17.9) = 201.06 mm²
0.375A_stx1 = 0.375 x 201.06 = 75.4 mm²

A_stx2 = #8 @ 220c/c (Table 17.10) = 228.48 mm²
0.375A_stx2 = 0.375 x 228.48 = 85.68 mm²
Larger value = 85.68 mm².
2No. 8mm is sufficient to give this area. But when 2 No. bars is distributed evenly in 710mm, the distance between the bars will be very large. So provide 3-#8.

No. of shorter bars in the mesh:
This is calculated separately for each of the two panels. But we saw that even for the larger value of 0.375A_stx, only 2 bars are required. So provide 3-#8 in each panel.

Thus the mesh is formed with 3-#8 longer bars and 6-#8 (Three in each of the two panels) shorter bars. This mesh is placed near the top surface. An identical mesh is formed and it is placed near the bottom surface.

Design of 'T' Type corner common to Panels 2 and 4
Width of the mesh = larger of 0.2l_x1 and 0.2l_x2.
l_x1 (panel 2) = 4006; 0.2l_x1 = 0.2 x 4006 = 801.2mm
l_x2 (panel 4) = 3506; 0.2l_x2 = 0.2 x 3506 = 701.2mm
So provide a width of 810mm
Length of the mesh = 810 +230 +710 = 1750mm

No. of Longer bars of the mesh:
This is calculated from the larger of 0.375A_stx1 and 0.375A_stx2 A_stx1 = #8 @ 250c/c (Table 17.8) = 201.06 mm²
0.375A_stx1 = 0.375 x 201.06 = 75.4 mm²

A_stx2 = #8 @ 220c/c (Table17.10) = 228.48 mm²
0.375A_stx2 = 0.375 x 228.48 = 85.68 mm²

Larger value = 85.68 mm². 2No. 8mm is sufficient to give this area. But when 2 No. bars is distributed evenly in 710mm, the distance between the bars will be very large. So provide 3-#8.

No. of shorter bars in the mesh:
This is calculated separately for each of the two panels. But we saw that even for the larger value of 0.375Astx, only 2 bars are required. So provide 3-#8 in each panel.

Thus the mesh is formed with 3-#8 longer bars and 6-#8 shorter bars. This mesh is placed near the top surface. An identical mesh is formed and it is placed near the bottom surface.

Thus we have completed the design of the two way slab system. The figs.17.47, 17.48 and 17.49 below show the reinforcement details:

Fig.17.47
Bottom bars
 

Fig.17.48
Top bars

Fig.17.49
Corner bars

In the next section, we will discuss some features of the above three figs.

PREVIOUS      CONTENTS   NEXT

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

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