### WHAT IS ONE WAY SLAB AND HOW TO DESIGN ONE WAY SLAB WITH DESIGN EXAMPLE FOR A BUILDING PLAN ALSO WITH BAR BENDING SCHEDULE

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 )
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.
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 )
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
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
1. Clear span + eff depth = 3.05+0.105 = 3.115 m
2. c/c distance of support=3.05+0.23   = 3.28 m

eff span l = 3.115 m
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

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