Theory Ofoneway Slab

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THEORY OF SLAB

Slabs are, in general, divided in to two categories depending upon the ratio of long to short span.

ONE WAY SLAB

When Ly/Lx is greater than or equal to 2, the slab is to be designed as slab

spanning in one direction. When the slab is supported on two opposite parallel

edges, it spans only in the direction perpendicular to the supporting edges and

bends in one direction whose main steel is provided in the direction of the span

such a slab is known as one way slab. One way slab is followed when two

supports are available on opposite sides or on all four sides with ratio of long to

short span exceeding 2. The distance between two opposite supports is known as

span.In one way slab main reinforcement is provided along the short span and

distribution steel is provided along longer direction, to distribute any

unevenness that may occur in loading and for temperature and shrinkage effects

in that direction.

The main steel is calculated from the bending moment consideration and under

no circumstances should it be less than the minimum specified by the code. The

secondary reinforcement (distributors) provided is usually the minimum

specified by the code for such reinforcement.

A common example of one way slab is the veranda slab spanning in the short

direction with main reinforcements and distribution reinforcements in the

transverse direction.

Types of one way slab :

One way slab is classified in to following categories:

(i) Simply supported slab

(ii) Continuous slab

(iii) Cantilever slab

ONE WAY SIMPLY SUPPORTED SLAB

Purely simply supported slabs are not possible in practice. As such partial fixity

is assumed at supports and alternate main bars are bent up at l/7 from the

centre of supports. For simply supported slabs maximum bending moment at the



centre is wl2 /8. For fixed slabs positive and negative moments may be calculated

by classical methods like moment distribution, etc.

Slab thickness :

Based on L/d ratio: 25

Based on thumb rule D = 40 to 45 mm / metre of span

Overall depth of slab D = effective depth (d) + clear cover (15 to 20mm ) + Φ/2Assumed steel percentage for mf

Grade of Steel Fe 250 Fe 415 Fe 500

% Ptassumed 0.70 0.30 0.25

Modification Factor 1.67 1.48 1.33

Problem 1 :

Data :

Clear span 3.50 m x 8.0 m

Roof slab

Live load (with Access provided ) 1.50 KN/ m2

Edge condition : Simply supported slab by 300 mm wa;; thickness alround

Concrete Mix : M20

Steel Grade : F415

Design:

(1) Type of slab:Ly/Lx = 8/3.5 =2.29 > 2

Hence the slab has to be designed as a one way with simply supported edge

condition.

(2) Effective span :

Effective span = C/C distance of supports ( Clause 22.2 P34)

Effective span in shorter direction lx = 3.5 +0.30 + 0.302 2

= 3.80 m

Effective span in longer direction ly =8.0 + 0.30 + 0.302 2

=8.30 m



(3) Depth of slab :

Depth of slab is governed by serviceability criteria (Clause 23.2.1 P37)

Effective depth based on deflection criteria d = lx / (L/d x mf )

L/d for simply supported slab =20 mf = 1.48 for Pt = 0.30%

d = 3800 = 128.38 mm

20 x 1.48

Overall depth of slab D = 128.38 + 15 + 10/2 = 148.38 mm say 150 mm

OR d = 3300/25 = 140 mm

D= 140 + 15 + 10/2 = 160 mmAs per thumb rule D = 3.50 x 40 = 140 mm to 3.50 x 45 = 157.5 mm say 150 mm

Effective depth of slab = d = 150 – 15- 10/2 =130 mm

According to IS 456 -2000 (Clause 23.0 P 36)Available effective depth in short direction dx = 150-15-10/2 = 130 mm

Available effective depth in long direction dy = 130-10 = 120 mm

Effective span in short span lx = Clear span + effective depth= 3.5 + 0.13 =3.63 m

Effective span in long span ly = Clear span + effective depth= 8.0 + 0.12 =8.12 m

Aspect ratio = Ly/Lx = 8.12/3.63 = 2.24 > 2

Hence it is designed as one way slab.

(4) Loads :

Live load(with access provided ) = 1.50 KN/m2

Dead load :

Self weight of slab

0.15 x 1 x 25 = 3.75 KN/m2

Weathering coarse 2.00 KN/m2

---------------Total dead load 5.75 KN/m

2

----------------

Factored load = 1.50 x( 5.75 + 1.50) = 10.88 KN/m2



(5) Bending moment :

Maximum B.M at mid span = wlx2/ 8 = 10.88 x 3.63

2/8 = 17.92 KN.m

(6) Check for effective depth:

Mulimit for M20 concrete & Fe 415 steel =2.76 b d 2

Mu = 2.76 b d 2 = 17.92 x 106

d required = √ 17.92 x 106 / (2.76 x 1000) =80.58 mm < 130mm( d available)

(7) Reinforcement :Main steel :

Method 1 : First Principle

Ast = Mu / 0.87 fy x z where z = lever arm

For under reinforced section

x = 1.2 – (1.44 – 6.68 Mu )

d fck b d2

= 1.2 – (1.44 – 6.68 x 17.92 x 106 )

20 x 1000 x 1302

= 0.114

z= d( 1 – 0.416 x/d)= 130 ( 1 – 0.416 x 0.114) =123.83 mm

Ast = Mu / 0.87 fy x z

= 17.92 x 106

= 400.82 mm2

0.87x 415 x 123.83

Method 2 : Using Equation

K = Mu / b d

2

= 17.92 x 10

6

/ 1000 x 130

2

= 1.06 4.59 xK

Astx required = 0.5 f ck [1- √ 1- ------------- ]

bd f y f ck



4.59 x 1.06

Astx required = 0.5 x20 [1- √ 1- ------------- ]bd 415 20

Astx required = 0.00313 x 1000 x 130 = 407.53 mm

Method 3: As per Table 2 of IS 456 – 1978 (SP-16)

For K = Mu / b d2

= 17.92 x 106

/ 1000 x 1302

= 1.06

100 Pt from Table 2 of SP- 16 = 0.314

Ast = 0.314 x 1000 x 130 = 408.2 mm2

100

Method 4 : Chart 13 of SP-16

For Mu =17.92 KN.m & d = 130 mm

100 As = 0. 32

bd

Ast = 0.32 x 1000 x 130 = 416 mm2

100

Method 5 : Direct from Ready made Table

(8) Check for deflection :

For Pt required = 0.314% Modification Factor = 1.46

Maximum permissible span/d = 20 x 1.46 =29.2

Actual span /d ratio = 3630 / 130 = 27.92 < 29.2OR d required = 3630/ 29.2 = 124.32 mm < 130 mm (d available)



(9) Distribution Steel :For Fe415 Ast = 0.12% of gross sectional area.

= 0.12 x 1000 x 150100

= 180 mm2

Using Table 96 of Design Aid

Provide 8 Φ RTS straight bars @ 200 mm C/c (Ast = 251 mm2

> 180 mm2)

(10) Check foe shear :Shear force will be maximum at the support.

Max. Shear = wu l /2 = 10.88 x 3.63/2 = 19.75 KN

Since 50% of bars are bent up at support.

Pt % available at bottom face of support = ½ x 100 Asb d

= ½ x 100 x 436 = 0.17%1000 x 130

The design shear stress ζ c from Table = 0.303Permissible shear stress = k ζ cK = 1.3 for slab having overall depth 150 mm or less.

ζ c’ = 1.30 x 0.303 = 0.394 N/mm2

Permissible shear stress = ζ c’ x b d = 0.394 x 1000 x 130 x 10-3

= 51.22 KN > 19.75 KN

(11) Check for development length :Provide L type hook

Lo = bx - x1 +3Φ2

= 300 - 25 +3x 102

M1 = Mu/2 = 17.92/2 = 8.96 KN.m

(50% of bars are available at support = Mu/2)

Vu = 19.75 KN

Ld at discontinuous end ≤ 1.3 M1 + Lo

V

Ld = [0.87 x 415 ] Φ4 x 1.6

= 0.87 x 415 Φ = 57 Φ = 57 x 10 =570 mm

4 x 1.6

Ld ≤ 1.3 M1 + Lo

V



570 ≤ 1.3 x 8.96 x 103

+ 15519.75

570 < 744.77 mm

Hence safe.

(12) Check for control of cracks :

As per clause 25.5.2.1 of IS 456 – 2000

(a) Mini. Pt = 0.12%

Actual Pt provided = 0.323 %

(b) As per clause 25.5.2.2 of IS 456 – 2000

Max. Dia of bar > 1/8 slab thickness.

10 < 150/8 = =18.75 mm

Diameter provided is o.k.

(b) Maximum spacing for main bars shall not be more than 3 timeseffective depth = 3 x 130 = 390 mm (As per clause 25.3.2 (b) of IS

456 -2000)

Actual spacing provided = 180 mm < 390 mm

(13) Detailing :

Theory Ofoneway Slab

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