## Numericals of flat slab

**Example 1.
**

Design an interior panel of a flat slab with panel size 6 x 6 m supported by columns of size 500 x 500 mm. Provide suitable drop. Take live load as 4 kN/m^{2}. Use M20 concrete and Fe 415 steel.

**Solution
**

Thickness : Since Fe 415 steel is used and drop is provided, maximum span to thickness ratio permitted is 32

**Thickness of flat slab **

Provide 190 mm thickness. Let the cover be 30 mm and Overall thickness D = 220 mm

Let the drop be 50 mm. Hence at column head, *d* = 240 mm and D = 270 mm

**Size of Drop
**

It should not be less than

Let us provide 3 m x 3 m drop so that the width of drop is equal to that of column head.

hence Width of column strip = width of middle strip = 3000 mm.

**Loads
**

For the purpose of design let us take self-weight as that due to thickness at column strip

**= 17.625 x 6 x 5.5**

**= 581.625 kN**

**Design Total Moment
**

Total moment

Total negative moment = 0.65 x 400 = **260 kNm**

Total positive moment = 0.35 x 400 = **140 kNm**

The above moments are to be distributed into column strip and middle strip

Width of column strip = width of middle strip = 3000 mm

**= 476.928 kNm**

Thus M_{u lim} > M_{u}. **Hence thickness selected is sufficient**