# Part 5 - One way and two-way slab design

To design the slab, we specify the reinforcement detail according to the span length by either one-way slab or two-way slab.

The major difference between one way and two-way slab is the tendency of a slab to bend in a different direction.

One way slab bend in one direction and two-way slab bend in both the direction.

If the ratio of longer span to a shorter span of the slab is less than 2 then the slab is called a two-way slab. and, if this ratio is higher than two then the slab is called a one-way slab.

For better understanding, every designer or civil engineer should visit the site and check the slab reinforcement arrangement, the following is an image from my site for the slab. On-site, one-way slab

## One way slab

One way slab is mostly used in balcony, chajjas, passage, etc. where there is a huge difference between the shorter and longer span and the ratio is more than two.

One way slab is more economical than the two-way slab and but also the two-way slab transfers the load in both the direction.

The simple one-way slab reinforcement details are as follows. One-way slab, bend tendency

The design of a one-way slab can be done by using the Indian standard code. ( IS: 456-2000 )

There are some basic steps to calculate and design the one-way simply supported slab. which are given below.

## One way slab design steps

• Effective depth,
• Effective span,
• Bending moment calculation,
• Check for effective depth,
• Reinforcement detail,
• Design checks.

## Design calculation

To design the one-way slab, we assume the 1-meter strip. which means in our design calculation the breadth, b will be equal to 1000 mm.

We will calculate the S-6 slab, which is a one-way slab, and the usage will be the passage.

The loads will transfer from S-6 to beams B-7, 8, and 5.

Following is the simplified version of the above plan. Simplified one-way slab plan

The dimension of the above plan is the c/c distance. the slab is resting on the beam with a size of 230 mm in all the directions.

To calculate the S-6, we have the following value,

fck = 20 N/mm2

fy = 415 N/mm2

Shorter span = lx = 0.953 m (c/c distance)

Longer span = ly = 4.76 m (c/c distance)

Floor finish = 1 kN/㎡

### Effective depth

(From IS: 456-2000, P.-37 & 38)

\frac{l}{d} = 20 X M.F

M.F = 1.15 IS: 456-2000, P.- 38, fig. 4 )

fs = 240 N/ mm2

d = \frac{953}{20 X 1.15} = 41.43 mm

Provide, d = 76 mm and assume 8 mm bars

Total depth, D = d + Φ/2 + clear cover

= 76+ 4 + 20

D = 100 mm

### Effective span

( from IS: 456-2000, p- 34, Cl. 22.2 (a) )

(1) lx + d = 953 + 76 = 1029 mm

(2) c/c of support = 953 + 230 = 1183

Whichever is smaller,

Effective span, l = 1029 mm.

We have considered 1 m strip,

(1) dead load    = (0.100 X 1 X 25)   = 2.5 kN/m

(2) floor finish   = ( 1 X 1)                   = 1 kN/m

(3) live load       =  ( 3 X 1)                  = 3 kN/m

Total load = 2.5 + 1 + 3 = 6.5 kN/m

Factored load w = 6.5 X 1.5 = 9.75 kN/m

### Bending moment

M_{u} = \frac{wl^{2}}{8}

M_{u} = \frac{9.75 X 1.02^{2}}{8}

Mu = 1.26 kN⋅m

### Check for effective depth

( SP. 16, P- 10, table - C)

M_{u} = 0.138 f_{ck} b d^{2}

1.26 X 10^{6} = 0.138 X 20 X 1000 X d^{2}

d = 21.37 mm < 76 mm ....... O.K.

### Steel

#### Main steel

p_{t} = 50 \frac{f_{ck}}{f_{y}} [ 1 - \sqrt{1 - \frac{4.6 M_{u}}{f_{ck}b d^{2}}}]

= 50 X \frac{20}{415} [ 1 - \sqrt{1 - \frac{4.6 X 1.26 X 10^{6}}{20 X 1000 X 76^{2}}}]

pt = 0.061 %

According to IS standard, the minimum percentage of steel for Fe 415 is 0.12 %.

pt = 0.12 %

A_{st} = \frac{p_{t}}{100} bd

A_{st} = \frac{0.12}{100} X 1000 X 76

Ast = 91.2 mm2

For 8 mm bars,

a_{st} = \frac{\pi}{4} X d^{2} = 50.24 mm^{2}

Spacing,

\frac{50.24}{91.2} X 1000 = 550.87 mm

According to IS specification, the minimum spacing between bars shall be,

(1) 3 d = 3 x 76 = 228 mm

(2) 300 mm

Whichever is smaller,

Provide, 228 mm c/c.

A_{st} = \frac{50.24}{228} X 1000

Ast = 220.35 mm2

Provide, 8 mm Φ - 230 mm c/c

#### Distribution steel

pt = 0.12 %

Ast = 91.2 mm2

According to IS: 456- 2000 P.- 48, CL. 26.5.2.1,

Minimum spacing for distribution bars is,

(1) 5d = 5 X 76 = 380 mm

(2) 450 mm

Whichever is smaller,

Provide, 380 mm c/c.

Assume 8 mm dia. bars,

A_{st} = \frac{50.24}{380} X 1000

Ast = 132.21 mm2

Provide, 8 mm Φ - 380 mm c/c

(6 mm dia. bars also can be used)

## Design Checks

### Check for cracking

For main steel,

(1) 3 d = 3 x 76 = 228 mm

(2) 300 mm

230 mm provided < 300 mm ..... O.K.

For distribution steel,

(1) 5d = 5 X 76 = 380 mm

(2) 450 mm

380 mm provided < 450 mm ..... O.K.

### Check for deflection

Allowable, l/d = 20 X M.F = 20 X 1.15 = 23

Actual, l/d = 1029/76 = 13.53

13.53 < 23 ..... O.K.

### Check for development length

L_{d} \leq 1.3 \frac{M_{1}}{V} + L_{0}

(1) d = 76 mm

(2) 12 Φ = 12 X 8 = 96 mm

Taking larger of two value,

L0 = 96 mm

Shear force at support,

take, l = clear span

V_{u} = \frac{wl}{2}

V_{u} = \frac{9.75 X 0.953}{2} = 4.64 kN

50 % are bent up bars near the support,

Ast at support = 50 % of Ast at mid-span

= 220.35 X 50/100

= 110.175 mm2

(From IS: 456-2000, P. - 96 )

M_{1} = 0.87 f_{y} A_{st} d (1- \frac{f_{y}A_{st}}{f_{ck}bd})

M_{1} = 0.87 X 415 X 110.18 X 76 (1- \frac{415 X 110.18}{20 X 1000 X 76})

M1 = 2.93 kN⋅m

1.3 \frac{M_{1}}{V} + L_{0} = 1.3 X \frac{2.93 X 10^{6}}{4.64 X 10^{3}} + 96

= 916.90 mm

(From IS: 456-2000, P.-42)

L_{d} = \frac{\phi \sigma_{s}}{4 \tau_{bd}}

\sigma_{s} = stress in bar at the section considered at design load.

\sigma_{s} = 0.87 X f_{y} = 0.87 X 415

= 361.05 N/mm2

\tau_{bd} = 1.2 X \frac{160}{100}

(Increased by 60 % = 100+60 = 160 %)

\tau_{bd} = 1.92 N/ mm2

L_{d} = \frac{\phi \sigma_{s}}{4 \tau_{bd}}

L_{d} = \frac{8 X 361.05}{4 X 1.92} = 376.09 mm

376.09 mm < 916.90 mm ..... O.K.

### Check for shear

\tau_{v} < \tau_{c}'

\tau_{v} = \frac{V_{u}}{bd} = \frac{4.64 X 10^{3}}{1000 X 76} = 0.061 mm^{2}

50 % bent up bars at support,

Ast = 110.175 mm2

p_{t} = \frac{100 A_{st}}{bd} = \frac{100 X 110.18}{1000 X 76}

= 0.145 %

( From IS: 456- 2000, P.- 73, table - 1 )

\tau_{c} = 0.28 N/ mm2

k = 1.30 ( IS: 456- 2000, P.- 72, CL. 40.2.1.1)

\tau_{c}' = k \cdot \tau_{c}

= 1.30 X 0.28

= 0.364 N/ mm2

0.061 < 0.364 ..... O.K.

So, from the above design checks, we can say that our slab is safe.

## Reinforcement details

The reinforcement details for the Slab S-6 is given below. One way slab reinforcement detail

After completing the one-way slab design, we will move on to the two-way slab design.

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