How to Structural Design a Building/House Step by Step Part-1 (One Way Simply Support Slab)

Any construction building design as per the below step. In this article, we design only slab and next article next steps calculation.

(1) Slab Design

(2) Slab Beam Design

(3) Lintel Beam Design

(4) Column Design

(5) Plinth Beam Design

(6) Footing Design

Some Important Point for Slab Design

Type of Slab

A slab is a plate element having a depth (D), Very small as compared to its length and width slab is used as floor or roof in building, carry uniformly distributed load.

Slab may be

• Simply Supported 
• Continuos 
• Cantilever 

Type of Slab Based on Support Conditions Are:

1). One Way Spanning Slab

2). Two Way Spanning Slab

3). Flat Slab Resting Directly on Columns Without Beams

4). Grid Slabs or Waffle Slabs

5).  Circular Slab and Other Shapes

One Way Spanning Slab

If the slab is supported on two opposite sides. it is called a one-way spanning slab. In this type of slab, lads are transferred on two opposite supports as shown below figure.

One Way Slab

If the slab is supported on four sides, and if ly/lx ≥ 3 one way spaning slab.

For any slab, if ly = lx, the slab has a tendency to bend in both the directions Which increase is provided along lx (Short Span)

Two Way Spanning Slab

It the slab is supported on all four edges and if ly / lx < 2,

The tendency of the slab is to bend in both directions. Such slabs are called a two-way slab. ( as shown below figure.)

In a two-way slab, the main reinforcement is provided along with lx as well as ly direction.

Flat Slab

When the slab is directly supported on columns, without beams, it is known as a flat slab.

Flat slabs are provided to increased the floor height and to permitted a large amount of light which might be obstructed by the depth of beams.

Grid Slabs

When the slab is required on beams with columns only on the periphery of the hall, the slab is called grid slab

Sometimes, in a large hall, public places, marriage halls, auditoriums, etc. a large column-free area is required. In these cases, large deep beams may be permitted but the columns are permitted only on the periphery

One Way Simply Support Slab Calculation /Design

One Way Simply Support slab Below Point Calculation Required

1. Effective Depth (d)

2. Effective Span

3. Reinforcement Requirements

4. Check for Cracking

5. Check for Deflection

6. Check for Development Length (Ld)

Effective Depth (d)

For deflection control 

L/d = 20 X M.F

• M.F. Modifiction factor from— IS: 456, p.38.Fig-4
• Assume % steel 0.3 to 0.6%

Fs =  0.58 Fy X (Ast requierd / Ast Provied)

Initially assume that Ast reqierd = Ast Provided 

Fy = 250 N/ Sq.mm —– Fs = 0.58 X 250 = 145 N/ sq.mm.

Fy = 415 N/ Sq.mm —– Fs = 0.58 X 415 = 240 N/ sq.mm.

Fy = 500 N/ Sq.mm —– Fs = 0.58 X 500 = 290 N/ sq.mm.

Effective Span

Clear Span + d

c/c of Supports

Whichever is smaller ——–  as per IS 456-2000 P. 34, CI 22.2.a

Reinforcement Requirements

Minimum reinforcement 

For Fe-250     Pt = 0.15 % of total C/s area (d x D)

For Fe-415     Pt = 0.12 % of total C/s area (d x D)

For Fe-500     Pt = 0.12 % of total C/s area (d x D) ——–  as per IS 456-2000 P. 48, CI 26.5.2.1

Maximum diameter (Sp 34)

For minbar:

• Plain bars———–10 mm Ø min dia
  Deformed bars—–8 mm Ø min dia

For Distribution bars:

• Plain bars———–6 mm Ø min dia
  Deformed bars—–6 mm Ø min dia

Check for Cracking

For Min Steel:

3d ——— Where. d = Effective depth

300 mm

Spacing should not exceed smaller these two values.

For Distribution steel:

5 d

450 mm

Spacing should not exceed smaller these two values. ——- IS: 456-2000, P.46

Check for Deflection

Allowable L/d  = 20 X M.F.

• M.F is Obtained from IS:456-200 P-38 Fig 4

Find actual, L/d

If Actual L/d < allowable  L/d ———- Ok

Check for Development Length (Ld)

IS 456-2000,P.44, Cl. 26.2.3.3 C

Ld should be ≤ 1.3  (M1/V) + L0

Where

Ld = (Ø.σs / 4 τ bd )—————–σs = 0.87 fy          As per IS 456-200, P.42

50 % of steel is bent up near support. Therefore find M.R for 50 % of steel only

M1 = M.R. for 50% steel support

V = Shear force at the support 

L0 = Sum of anchorage beyond the center of support

 d

12  Ø

Take L0 as the smaller of two values.

One Way Simply Support SlabCalculation /Design- Example

Sum Point Consider As below

Slab Size 3.2m X 9.2 m

The slab is resting on 300mm thick wall

Find One Way Slab or Two Way Slab

ly/lx = 9.2 / 3.2 = 2.875 > 2  

As per the type of slab

Here this one-way slab, So we design the slab as one-way simply supported slab

Effective Depth (d)

Here Consider shorter span as l,

l = 3200 mm = 3.2 m

l/d = 20 x M.F

fy – 415 N/sq.mm, fs = 240 N / Sq.mm

M.F = 1.15 | As per IS Code 456, Fig.4

l/d = 20 x 1.15

3200/d = 20 x 1.115

d = 139.13 mm

Here, d = 150 mm , Assume 10 mm Ø bars

Overall Depth, D = 150 + (Ø / 2 ) + Clear Cover

D = 175 mm

Effective Span

1).  3200 + 150 = 3350 mm

2). c/c of Supports = 3200 + 300 = 3500  mm   | IS: 456-2000,P-34, CI. 22.2.a

Whichever is smaller

Effective Span = l = 3350 mm = 3.35 m.

Reinforcement Requirements

Load Calculations

Factored Load = 1.5 x 7.875

w = 11.82 kn/m.

Bending Moment

Mu = (w. l²) / 8 = (11.82 / 3.35²)

Mu = 16.58 Kn.m.

Main Steel

Pt = 50 (fck/fy) [1 – √ ( 1 -{(4.6 Mu)/ (fck x bd²)})]

Pt = 50 x  (0.482) x (0.0945)

Pt = 0.215 %

Ast = ( pt / 100) x 1000 x 150

Ast = 322.5 mm²

For Spacing 

Sapcing = ({[π/4] x d²}/Ast ) x 1000 mm

Sapcing = ( 78.53 / 322.5 ) x 1000 = 243.50 mm

Distrbution Steel 

Provide a minimum of 0.120% of Total C/s Area | As per IS 456-200 P 48, CI. 26.5.2.1

Ast = (0.12/100) x 1000 x 175 = 210 Sq.mm

Check for Cracking

For Main Steel

1). 3 d = 3 x 150 = 450 mm

2). 300 mm | IS 456-2000 P-46

240 mm provided < 300 mm …………….. o.k.

For Distribution Steel 

1). 5 d = 5 x 150 = 750 mm

2). 450 mm

130 mm provided < 450 mm …………….. o.k.

Check for Deflection

Allowable (l/d) = 20 x M.F.

% pt Provided  = 100 Ast / bd = (100 x 327) / (1000 x 150) = 0.218% | IS Code 456-2000 P.38, Fig 4

M.F = 1.6

Allowable l/d = 20 x 1.6 = 3350 / 150 = 22.33

22.33 < 32 …………….. o.k.

Check for Development Length (Ld)

1). d = 150 mm

2). 12 Ø = 12 x 10 = 120 mm

Taking larger of two values L0 = 150 mm

S.F. at support = 50% of East at mid-span = 327 / 2 = 163.5 Sq.mm

M1 = 0.87 x 415 x 163.5 x 150 x [1-(415 x 163.5) / (20 x 1000 x 150)]

M1 = 8.65 x 10⁶ N.mm = 8.65 kN.m.

1.3 [ M1/V] + L0 = 1.3 x (8.65 x 10⁶ ) / (18.91 x 10³ ) + 150

M1 = 744.65 mm

for M 20 , fy = 415 N/mm²

10 mm Ø. bar, tension

Ld = 470 mm

470 mm < 744.65 mm …………….. o.k.

Reinforcment Details 

One Way Simply Support Slab Calculation /Design Excel Sheet – Download 

Video tutorial for better understanding: