One-Way Simply Supported Slab Design in Building Construction

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

1. Simply Supported 
2. Continuos 
3. 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

1. M.F. Modifiction factor from— IS: 456, p.38.Fig-4
2. 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

Dead load	4.375	Kn/m
Floor Finsh	1	Kn/m
Live Load	2.5	Kn/m
Total Load	7.875	Kn/m

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:

FAQ: Comprehensive Guide to One-Way Simply Supported Slab Design in Building Construction

What Is a One-Way Spanning Slab?

A one-way spanning slab is a type of slab supported on two opposite sides, transferring loads primarily to these two supports. It bends in one direction, and the main reinforcement is provided along the shorter span.

How Do You Determine If a Slab Is One-Way or Two-Way?

To determine if a slab is one-way or two-way, calculate the ratio of the longer span (ly) to the shorter span (lx). If ly/lx ≥ 2, it is a one-way slab. If ly/lx < 2, it is a two-way slab.

What Are the Key Steps in Designing a One-Way Simply Supported Slab?

The key steps in designing a one-way simply supported slab include:

1. Determining the effective depth (d)
2. Calculating the effective span
3. Assessing reinforcement requirements
4. Checking for cracking
5. Checking for deflection
6. Ensuring adequate development length (Ld)

How Is the Effective Depth (D) of a Slab Calculated?

The effective depth (d) is calculated using the formula L/d = 20 x M.F., where L is the span length and M.F. is the modification factor obtained from IS 456. For example, for Fe-415 steel with a modification factor of 1.15, the formula is adjusted accordingly.

What Is the Importance of Reinforcement in Slab Design?

Reinforcement is crucial in slab design as it helps resist tensile stresses, prevents cracking, and ensures the structural integrity of the slab. The reinforcement requirements are calculated based on the type and amount of load the slab will bear.

How Do You Check for Deflection in a Slab Design?

Deflection is checked by comparing the actual L/d ratio to the allowable L/d ratio, which is calculated as 20 x M.F. If the actual L/d is less than the allowable L/d, the slab design is considered satisfactory.

What Is the Development Length (Ld) and Why Is It Important?

The development length (Ld) is the length of rebar required to develop its full strength in tension or compression. It ensures that the reinforcement is properly anchored in the concrete to prevent slipping and ensure the slab’s structural performance.

Can You Provide an Example of a One-Way Simply Supported Slab Design?

Yes, an example design includes calculating for a slab size of 3.2m x 9.2m, with specific calculations for effective depth, effective span, reinforcement requirements, and checks for cracking, deflection, and development length. Detailed steps and formulas are provided in the guide.

What Are the Minimum Reinforcement Requirements for a One-Way Slab?

Minimum reinforcement requirements vary based on the type of steel used:

1. For Fe-250: 0.15% of the total cross-sectional area
2. For Fe-415: 0.12% of the total cross-sectional area
3. For Fe-500: 0.12% of the total cross-sectional area

Where Can I Find More Resources on Slab Design?

Additional resources on slab design, including detailed calculations, example designs, and video tutorials, can be found in the downloadable Excel sheet and linked video tutorials provided in the comprehensive guide.