WHAT IS ONE WAY SLAB

A slab is supported only one direction that is one way slab now tha loads are transferred only this two supports .therefore the main reinforcements are provided perpendicular to supported direction.

Also the slab is supported four sides but the length breadth ( ly/lx)ration is greater than or equal to two , the slab is designed as one way slab. Therefore assumed the loads are transferred only one direction and designed that slab is one way slab.

One way slab analysis and design procedure is same as singly reinforced rectangular beam design.

The slab is designed 1m width similarly to rectangular beam then same details are followed to remaining width of slab.

For beam design first area of steel is find out then to find number of bars depends upon diameter of the bar ,but slab design is differ to find out spacing b/w bars depends upon diameter of the bar.

DESIGN PROCEDURE OF ONE WAY SLAB

1.FIND OUT THE SLAB IS ONE WAY OR TWO WAY SLAB

A slab is supported four direction however to taken the ratio b/w length to breadth than the ratio is greater than or equal two ( ly/lx >2) the slab is designed as one way slab . for other hand the length to breadth ratio is then two the slab is designed as two way slab (ly/lx <2 )

2 SELECTION OF DEPTH OF SLAB

We have two conditions to assume the depth of the slab

(i) d= le/30 ( for simply supported slab )

(i) d= le/30 ( for simply supported slab )

d= le/15 ( for cantilever beam )

(II) Depth required for stiffness

Dreq = lef/B.V x M F

Lef = clear span

B.V = basic value from IS 456-2000 in section 23.2.1

MF = modification factor from drawing 4 of IS 456-2000

Depth of slab is taken maximum value from above both conditions. However the minimum depth is taken 100 mm for building roof and floor slabs.

3.EFFECTIVE SPAN

Eff dpan of simply supported slab will be the least of (i) clear span + eff depth (ii) c/c distance of supports.

4. LOADS CALCULATION

There are two types of loads are considered imposed load and dead load.

Imposed loads are taken form table 2 of IS 875 ( part 2 ) -1987

Dead loads are weight of weathering course and self weight of slab

5. BENDING MOMENT CALCULATION

Maximum bending moment at mid span

Mu = Wul2/8 ( simply supported slab )

Mu = Wul2/2 ( cantilever slab )

Wu = design load

L = eff length

6.DEPTH REQUIRED FOR STRENGTH

Equating the moment of resistance of the section to the maximum bending moment.

(Qbd2 = Mu) from this

..d=√Mu/Qb

7.MAIN REINFORCEMENT

Main reinforcements are find out for using below formula

Mu = 0.87 fy Ast [d-(fy Ast/fck b)]

Mu = design bending moment

Fy = tensile strength of steel

Ast = area of steel

D = eff. Depth

B = breadth as 1m

Fck = compressive strength of concrete

Fy = tensile strength of steel

MINIMUM REINFORCEMENTS IN SLAB

for IS 456-2000 section 26.5.2.1 using at mild steel in two direction (fe 250 ) the area of steel is not less than 0.15 % if cross sectional area. Using of high yield strength deformed bar ( fe 415 ,fe 500 ) area of steel is not less than 0.12 % of cross section.

MAXIMUM SIZE OF BAR

From IS 456-2000 section 26.5.2.2 the diameter of reinforcement of the slab is not more than 1/8 of overall depth of the slab (D/8)

SPACING OF MAIN REINFORCEMENT

From is 456-2000 section 26.3.3 (b) (i) the maximum spacing of main reinforcements in slab is does not exceed 3d and 300 mm which one is maximum .there d is eff. Depth of slab

SPACING OF SECONDARY REINFORCEMENTS

For is 456-2000 section 26.3.3 (b) (2) the maximum spacing of distributor does not exceed 5d and 450 mm which one is maximum.

CURTAILMENT OF MAIN REINFORCEMENTS

Alternate main reinforcements are cranked at distance equal to 0.1l

CHECK FOR SHEAR

Nominal stress and permissible shear stress are find out may be the permissible shear stress of concrete is more than the ktc or 0.5 tcc max .so the depth of slab is increased .

DESIGN EXAMPLE WITH BUILDING PLAN

In this section we are going to learn how to design one way slab with below given building plan so lets go how to solve this problem.

GIVEN DATA FROM THE DRAWING

We are going to design hall room of the building plan

Room size = 6.10 X 3.05 m ( inner dimension )

Wall thick = 0.23 m

Assume grade of concrete fck = M20

Assume grade of steel = fe 415

SOLUTION

1)SIDE RATIO OF THE SLAB

Length breadth ratio of the slab ly/lx =6.10/3.05

= 2

So this is one way slab

2.PROPERTIES OF THE SECTION

For M20 grade concrete and fe 415 steel are used therefore moment of resistance MR=2.76 bd2

3.DEPTH REQUIRED FOR STIFFNESS

Basic value of l/d ratio for simply supported slab BV=20

Assume modification factor (Mf) = 1.5

Clear span of the slab (lef) = 3.05m

Approximate eff depth required for stiffness dreq = lef/ B.v x M.f

= 3050/(20x 1.5)

= 101 mm

So take d = 105 mm

Minimum nominal cover for reinforcement for mild exposure condition in slab =15 mm

Assume that 10 mm dia bars are provided as main tension reinforcement.

Total depth D = 105+15+10/2

D = 125 mm

4.EFF SPAN

Eff span is the simply supported slab is least of

- Clear span + eff depth = 3.05+0.105 = 3.115 m
- c/c distance of support=3.05+0.23 = 3.28 m

eff span l = 3.115 m

5.LOAD CALCULATION

from table 2 of IS 875 (part 2 )-1987

imposed load on roof slab = 1.50 KN/m2

assume weight of weathering coarse = 1.000 KN/M2

self weight of slab = 0.125 x 1 x 25 x 1

= 3.125 KN/M2

Total weight of slab = 1.500+1.000+3.125

= 5.625 KN/M2

Total load on slab per metre length w = 5.625 x 1

Design load Wu = 5.625 X 1.5 ( in limit state design modification fector is 1.5 )

6.DESIGN BENDING MOMENT

Maximum bending moment at mid span Mu = WUL2/8

= 8.438 X3.1552/8

=10.49 KN.M

MU = 10.49 X 106

7.TO CHECK DEPTH ( TO RESIST BENDING MOMENT )

Depth required for strength for equating the moment of resistance of the section to the maximum bending moment.

MU = 2.76bd2 from this equation we want depth “d” so

..d=√Mu/2.23b

..d=√((10.49x1000000))/(2.23x1000)

d = 68.58 mm6

provided eff depth = 105 mm > 68.58 mm

hence the section is safe so we are going proceed further steps

8.TO FIND MAIN REINFORCEMENT

Mu = 0.87 fy Ast [d-(fy Ast/fck b)]

Design bending moment (MU) = 10.46X106 N.MM

Tensile strength of steel (fy) = 415 N/mm2

Compressive strength of concrete (fck) = 20 N/MM2

Eff depth d = 105 mm

10.49 X 106 = 0.87 X 415 X Ast [ 105-((415 Ast )/20 x 1000 ))

Ast = 292.80 mm2

Minimum steel requirement = 1.2 D

= 1.2 x 125

= 150 mm

Provided 10 mm dia bars so spacing S = (astx1000)/Ast

S = (78.5 X 1000)/292.8 ast = area of 10 mm dia bar 78.5 mm2

S = 268.10 mm

Maximum permitted spacing of main reinforcement is least of 3d and 300 mm

=3 x 105 = 315 mm

So provide spacing = 265 mm c/c

9.TO FIND DISTRIBUTORS

Minimum area of steel for distributors =0.12 %

= (0.12/100) x ( 1000 x 125 )

= 150 mm2

Provide 8 mm dia bars for distributors

Spacing S = (astx1000)/Ast

S = 335 mm

Maximum permitted spacing for distributors is least of 5d and 450 mm

5d = 5 x 105 = 525 > 450 mm

So spacing provided = 330 mm

NEXT STEPS ARE CHECK FOR SHEAR AND CHECK FOR STIFFNESS

It is also explained above design procedure so kindly refer that section

ALSO READ--

BAR BENDING SCHEDULE OF THIS SLAB

HOW TO DESIGN SINGLY REINFORCED BEAM

HOW TO DESIGN SEPTIC TANK FOR RESIDENTIAL COLONY

HOW TO CALCULATE WATER TANK SIZE AND CAPACITY

HOW TO CALCULATE RENT OF BUILDING

HOW TO FIND UNIT WEIGHT OF STEEL BARS

HOW TO APPROXIMATELY CALCULATE STEEL BAR IN BEAMS,COLUMNS AND SLABS

BAR BENDING SCHEDULE OF THIS SLAB

HOW TO DESIGN SINGLY REINFORCED BEAM

HOW TO DESIGN SEPTIC TANK FOR RESIDENTIAL COLONY

HOW TO CALCULATE WATER TANK SIZE AND CAPACITY

HOW TO CALCULATE RENT OF BUILDING

HOW TO FIND UNIT WEIGHT OF STEEL BARS

HOW TO APPROXIMATELY CALCULATE STEEL BAR IN BEAMS,COLUMNS AND SLABS

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