Reinforced Concrete Design: 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_stx2A_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