This document describes flat slab construction, which is a reinforced concrete slab supported directly on columns without beams. It defines the key components of flat slabs and describes the four main types. An example is provided calculating the loads, moments, and shear for a sample interior panel with dimensions of 5m x 5m. The calculations show selecting a slab thickness of 150mm is adequate. Exterior panel calculations for edge columns are also demonstrated.

1. FLAT SLAB

2. What is flat slab ?
Flat slab is a two way reinforced concrete slab with or without drop, supported generally
without beams directly supported on column.
Main components of flat slab are:
• Drop
• Panel
• Column Head
• Column strip
• Middle strip
INTRODUCTION to flat slab 9

3. TYPES OF FLAT SLAB
Flat slabs have four types:
1. Flat slab without Drop and column head.
2. Flat slab with drop but without column head.
3. Flat slab without drop but with column head.
4. Flat slab with drop and column head.
INTRODUCTION to flat slab 10

4. Flat plate example
 Interior panel → 5m × 5m
Size of column → 400mm × 400mm ( IS 456 : 2000 )
Live load → 4 kN/m2
Floor finish → 1 kN/m2
→ Depth of slab :
𝑙
𝑑
= 26 × 𝑀. 𝑓Pt = 0.4 %
5000
𝑑
= 26 × 1.3 M.f =1.3
d = 147.92 ≅ 150 mm
D = 175 mm

5. Loads
Dead load = 0.175 × 25 = 4.375
Floar finish = 1 KN/m2
Live load = 4 KN/m2
Total load = 9.375 KN/m2
Factored load = 1.5 × 9.375
= 14.062 KN/m2
Ln = 5 - 0.4 = 4.6 m
Total design load in panel W = 14.062 L2Ln
= 14.062 × 5 × 4.6
= 323.42 KN

6. Moments
Panel moments Mo =
𝑊𝐿𝑛
8
= 323.42 × 4.6/8
= 185.96 KN.m
Negative moment = 0.65 × 185.96 = 120.87 KN.m
Positive moment = 0.35 × 185.96 = 65.086 KN.m
Column strip Middle strip
Negative moment 0.75×120.87= 90.65 30.22
Positive moment 0.60×65.086 = 39.051 26.03
Distribution of moment

7. Checking for thickness selected :
Fe 415,
Mu lim = 0.138 fck bd2
Width of column strip = 0.5 × 5000
= 2500 mm
Mu lim = 0.138 × 20 × 2500 × 1502
= 155.25 × 106 N.mm
= 155.25 KN.m
Singly reinforcement section may provided .

8. Check for shear :
Periphery of article section = 400 + d/2+ d/2 = 550 mm
Shear , V = ( 14.062 × 5 × 5 )-( 14.062 × 0.55×0.55)
= 347.29 KN
𝜏𝑣 =
𝑉𝑢
𝑏. 𝑑
= 347.29 ×
103
4 × 550 × 150
= 1.052 N/mm2
Bc=
𝐿1
𝐿2
=
5
5
= 1Ks =1+Bc subjected to maximum of 1.
Ks= 1+1 =2 ˃1 Ks = 1.
τc= 0.25 𝑓𝑐𝑘
= 0.25 20
= 1.118 N/mm2 If , τv˂1.5 τcthen shear reinforcement shall provide
550
550
400
400

9. Exterior panel (Edge column) :
Calculation αc ,
For column , kc =
4𝐸𝐼
𝑙
Assume height of column =3500mm
For column : For slab :
Kc =
4×E×400×400×400×400
3500×12
Ks =
4𝐸𝐼
𝑙
= 2.43×106 E =
4×E×5000×175×175×175
5000×12
=1.786×106 E

10. Now, αc =
𝛴𝑘𝑐
𝛴𝑘𝑠
=
2.43×106
𝐸
1.786×106
𝐸
= 1.360 αcmin, = 0.333
αc˃ αcmin.
No correction is required.
End span
• Interior negative design moment:- Positive design moment:-
= 0.75 −
0.10
1+
1
α0
𝑀𝑜 = 0.63 −
0.28
1+
1
αc
𝑀𝑜
= 0.75 −
0.10
1+
1
1.360
× 185.96 = 0.63 −
0.28
1+
1
1.360
× 185.96
=128.75 KN.m =87.149 KN.m

11. 3.exterior negative design moment:-
=
0.65
1+
1
αc
𝑀𝑜
=
0.65
1+
1
1.360
× 185.96
=69.65KN.m
Check for shear:-
b0= 475+475+550 =1500 mm
V=(14.602×5×5)-(14.062×0.475×0.550) =347.87 KN
τv =
𝑉𝑢
𝑏𝑑
Ks=1 , τ v ˃ τc
=
347.87 ×103
1500×150
τc =0.25 𝑓𝑐𝑘shear reinforcement is required.
=1.546 N/mm2 =1.118 N/mm2
475
475 400
400

12. Excel Sheet
• flat slab.xlsx

13. THANK YOU