Slab Designer as per IS 456-2000

SLAB DESIGNER ver 2.00 Name of author:Mofiz Ahmed Profession:B-Tech Civil Engineer (student IV year II sems) Software used: Slab Designer ver 2.00 Web site: www.mofizahmed.webs.com

An Example (For educational purpose only)

The given value that we have are: Grade of concrete used is Fck=25N/mm2 ( for M25 grade of concrete) Grade of steel used is Fy=415N/mm2 Shorter length of slab i.e slab in X direction lx=3000mm Longer length of slab i.e slab in Y direction ly=4000mm Given Live load is LL=4KN/m2 Given Dead load (beside the selfweight) is DL=0KN/m2 Given Floor finishing Load FL=0.65KN/m2 Now finding the ratio of ly and lx, ly/lx=1.33 Since the ratio (ly/lx)=1.33 is less than 2 so it is a Two way type of slab Selected type of slab i.e whether continuous or discontinuous= D kind of slab Effective depth of slab d=lx/30=100 Taking the value to nearest multiple of 5 or 10 then d=100 mm

Providing a nominal cover of 30 mm,therefor overall depth D=130 mm

/////////////// LOAD CALCULATION /////////////// Self weight per square unit=Volume*Unit weight of concrete=1*1*D*U concrete and is assumed as U=25 KN/m2

where U is the unit weight of

Self weight SL=3.25 KN/m2 Total load acting on the slab is W= DL+FL+LL+SL Total load acting on the slab W=7.90 KN/m2 Total factored load acting on the slab Wu=Factor of safety * W=1.50*7.90=11.85 KN/m2 Since d is less than support width so the effective spans are............................................... Lefx=lx+d=3100 mm Lefy=ly+d=4100 mm Ratio of effective spans in X and Y direction i.e Lefx/Lefy=1.32 For ratio of Lefy/Lefx=1.32,from Table 26 of IS-456,we get -ve alpha x i.e -ve ax=0.07 For ratio of Lefy/Lefx=1.32,from Table 26 of IS-456,we get -ve alpha y i.e -ve ay=0.05 For ratio of Lefy/Lefx=1.32,from Table 26 of IS-456,we get +ve alpha x i.e +ve ax=0.05 For ratio of Lefy/Lefx=1.32,from Table 26 of IS-456,we get +ve alpha y i.e +ve ay=0.03 As per ANNEX D of IS-456 Moment is calculated as : alpha value * load * square of effective length Positive moment in X direction(shorter span) for alpha positive i.e +ve ax M1=5.68 KN-m Negitive moment for alpha x negitive in X direction i.e -ve ax M2=7.56 KN-m Positive moment in Y direction (longer span) for alpha positive i.e +ve ay M3=3.99 KN-m Negitive moment in Y direction (longer span) for alpha negitive i.e -ve ay M4=5.35 KN-m Shear force as per IS-456 Vu=((Wu*Lefx*r)/2)(1+r)=13.84 KN,

where r=(Lefx/Lefy)^4=3.06

Design Moment is the maximum of all the above moment caculated. .'. Design moment=M=7.56 KN-m As per IS-456 for Fy=415 N/mm2 we have Xu=0.48*d=48.00

//////////// AREA OF STEEL REINFORCEMENT ///////////// From IS-456 we have ultimate moment as Mu=0.36*fck*1000*Xu*(d-(0.42*Xu)) Putting all the respective values in the above equation we get Mu=34.49 KN-m Since Mu is less than M therfor we can design the slab as singly reinforced From IS-456 we have M=0.87*Fy*Ast*d(1-((Fy*Ast)/(b*d*Fck))) where b=1000 mm, Ast=area of reinforcement steel required In this equation,we have all the values except Ast .'. Putting all the values we get Ast value Area of steel required for Moment=5.68 KN-m is calculated as = 161.74 mm2 Spacing required for this area of steel is given as..... (area of each bar*b)/total area of steel Spacing calculated or required spacing=485.59 mm Spacing provided=300 mm since it is less than 3d but more than 300 mm Converting it into lower multiple of 5 or 10 Spacing provided=300 mm Area of steel required for Moment=3.99 KN-m is calculated as = 112.49 mm2 Spacing required for this area of steel is given as..... (area of each bar*b)/total area of steel Spacing calculated or required spacing=698.17 mm Spacing provided=300 mm since it is less than 3d but more than 300 mm Converting it into lower multiple of 5 or 10 Spacing provided=300 mm Area of steel required for Moment=7.56 KN-m is calculated as = 217.11 mm2 Spacing required for this area of steel is given as..... (area of each bar*b)/total area of steel

Spacing calculated or required spacing=361.74 mm Spacing provided=300 mm since it is less than 3d but more than 300 mm Converting it into lower multiple of 5 or 10 Spacing provided=300 mm Area of steel required for Moment=5.35 KN-m is calculated as = 152.08 mm2 Spacing required for this area of steel is given as.... (area of each bar*b)/total area of steel Spacing calculated or required spacing=516.43 mm Spacing provided=300 mm since it is less than 3d but more than 300 mm Converting it into lower multiple of 5 or 10 Spacing provided=300 mm Minimum area of steel reinforcement required=0.12 % of b*d=120 mm2 Spacing required for minimum steel reinforcement =418.88 mm Spacing provided for minimum steel reinforcemen =300 mm since it was greater than 300mm and 5d is also more than 300 mm Finalyy spacing provided for minimum steel reinforcement after converting it into nearest multiple of 5 or 10 =300 mm For moment M2=7.56 KN-m we have reinforcement steel 112.49<120 Calculated steel area is less than minimum steel required so providing minimum steel value and its spacing Area of torsional reinforcement if provided at discontinuous corners, then it will be 3/4 of the maximum area of steel reinforcement. Area of torsional reinforcement=162.84 mm2 Dimension of torsional area are................................ In X direction =Lefx/5=620 mm In Y direction =Lefy/5=820 mm Spacing provided between torsional bars =300 mm

//////////CHECK FOR SHEAR///////////// Tv=Vu/(b*d)=0.14N/mm2

where b=1000 mm

Percentage of steel=(Pi*diameter of main steel bar*100)/(spacing between main bars*effective depth) =(3.141592654*dia*dia*100)/(spacing*d*4) =0.26

From Table 19 of IS-456 for Pt=0.26 and Fck = 25 N/mm2 we get Tc=0.37 N/mm2 From clause 40.2.1.1 we get k =1.30 approximately. .'. Now multiplying Tc with k we get Tc=0.48 From Table 20 of IS-456 for Fck-25 we get Tcmax=3.10 N/mm2 Since Tv is less than Tc and Tcmax,therefor the slab is safe against shear

/////////////Check for Deflection///////////// From IS 456-2000,clause 23.2.1 we have fs=0.58*fy*(Area of steel required/area of steel provided) In case of this slab we have........ Area of main mid steel required/Area of main mid steel provided=0.62 So we get fs=0.58*fy*(Ast1/Astp) = 0.58*415*0.62= 148.71 We have percentage of steel as Pt=0.26 Now for the above value of Pt and fs we get modification factor F1 from Fig 4 of IS 456-2000 as F1=1.99 Now Maximum L/d ratio allowed = F1*basic L/d ratio as given in clause 23.2.1 of IS 456-2000 Max L/d = F1*26=51.69 Provided L/d = effective length in shorter direction*effective depth=Lefx/d=31 Since Max L/d ratio is greater than provided L/d ratio so Deflection criteria for this Slab is satisfied

///////////////RESULT/////////////////

Slab is designed as : D type of Slab (see the fig. on the left) Load=11.85 KN Moment=7.56 KN.m Ultimate moment=34.49 KN.m shear force=13.84 KN 10 mm Main short mid bar=161.74 mm2, Spacing=300 mm 10 mm Main long mid bar=120 mm2, Spacing=300 mm 10 mm Support short near suppor bar=217.11 mm2, Spacing=300 mm 10 mm Support long near support bar=152.08 mm2, Spacing=300 mm 8 mm Edge strip bar=120 mm2 ,spacing=300 mm 8 mm bar for Torsional Reinforcement mesh at discontinuous corner having area=162.84 mm2 spacing=300mm with side cover of 30 mm and dimension as : 620x820mm Pt=0.26 % Tv=0.14 N/mm2 Tc=0.48 N/mm2 Tcmax=4.03 N/mm2 k=1.30 d=100 mm D=130 mm Check shear (1 = OK,0 = Redesign) = 1 Check for Deflection=Satisfied

***********END*********** © Mofiz Ahmed 2009 www.mofizahmed.webs.com his design of slab is done by using SLAB DESIGNER ver 2.00 Though thorough analysis has been done while designing the software and the output produce,the user are highly recommended to follow manual designing procedure and to use this software for checking purpose only.Ther author takes “NO RESPONSIBILITY” if any type of damage is caused by using the software or by using or reading or following this manual or example.