Design of one way slab

Classification of slab:-

a) On the basis of shape –square, circular, triangular etc.b) On the basis of supporting condition –simply supported along

it edge continues slab running over number of support, cantilever slab fixed at one end and free at other end, flat slab directly supported by column

c) On the basis of spanning direction- one way slab (when the main reinforcement is in one direction), two way (when the main steel is provided in two or orthogonal directions.

1. Effective Span (clause 22.2):-

a) simply supported slab

The effective span of member that is not built integrally with it support shall be the least of the following.

i. Clear span + d (effective depth of slab).

ii. Center to center of support.

b) For continues slab.If the width of support is less than 1/12 of clear span the effective span shall be as a).if the support are wider than 1/12 of the clear span or 600mm which ever is less the effective span should be taken as under.

i. For end span with one end fixed and other continues or for intermediate span.

Effective span =clear span between two support

ii. For end span with one end fixed and other continues Effective span =clear span +half the effective depth of slab or beam.OrEffective span =clear span +half the width of discontinues support.Which ever is less.

2. Trial Depth

Depth of slab is governed by serviceability requirement of deflections.Calculate the depth of slab based on L/d ratio.

Cover = min. 20mmTable 16 clause 26.4.2 IS456-2000

Allowable L/d ratio = ra = basic L/d ratio X modification factor (α1)Basic L/d ratio (IS456-2000, C-23.2.1)Basic values of span for effective depth ratio for span up to 10mCantilever – 7mSimply supported – 20mContinues – 26mModification factors (α) depends upon Pt% [c-23.2.1 (e)]

For span above 10m, the value in (a) may be multiplied by 10/span in meters, except for cantilever in which case defection should be made.

Ast % the in (a) & (b) shall be modified by multiplying with the modification factor as per fig. given below.

3. Loads:- Calculate load in KN/m on 1m wide strip of the slab.

Dead load = self weight of slab + floor finish.

Live load = 2 to 2.5 (use IS875-part II)

Total working load = DL+LL

Total ultimate load = Wu = 1.5 (DL+LL)

4. Design Moments:-

For simply supported slab Mu = Wu

For continous slab Mu =

Where = 1.5 DL

= 1.5 LL

Moments and shear coefficient for continuous beam. ( Table 12

and 13 )

For moments at support where two unequal span meet or in

case where the span are not equally loaded, the average of

two values for the negative moment at the support may be

taken for the design.

Clause 22.5.2

Check for concrete depth:-

Since the depth of the slab is obtained from serviceability

consideration it is required to be checked from bending

moment requirements.

Calculate maximum moment carrying capacity of the section

For slab b = 1000 mm

( = for fe 415 =0.138 fck )

( = moment of resistance factor)

If >

Then the section is adequate from bending moment

requirements.

Which shall be less than the

provided

If the above condition is not satisfied provide the depth

required from bending moment consideration.

Main steel:-

Ast =

Where b = 1000 mm > Ast minAst min = 0.12% bd for (1+YSD bar)

(Fe 415 & Fe 500)

= 0.15% bd for Fe 250

Required spacing s =

ast = area of c/s of one barAst = total area of steel required

The spacing shall not exceed 3d or 300 mm whichever is less.

Check for deflection:-

Calculate Pt % = 100 Ast / bdWhere Ast is the maximum area of steel required at mid span assumed.

Check that D >

8. Distribution of steel:-

Required Ast = for Fe 415

= for Fe 250

Where b =1000mm and D is the overall depthMaximum spacing csd or 450 m whichever is less

9. Check for shear:- Calculate maximum shear Vu max as per table 13 (IS456)

& =Vu max / area of cross sectionObtain design shear stress corresponding to Pt = 100 Ast / bd (from table 19)Calculate shear resistance of slab Internal stress = K K – factor to increase in the resistance to shear due to membrane action of slab and is given in table.

Overall depth of slab in mm

300 or more

275 250 225 200 175 150 or less

K 1.00 1.05 1.1 1.15 1.20 1.50 1.30

If > the safe else increase the thickness of slab.

9. Check for development length:-

The length of the bars provided for resisting –ve moment should not be less than +ve development length given by

Ld = K =

Example: - design a simply supported rcc slab for a roof of hall 4m x 10m (inside dimension) with 230mm wall all around. Assume a live load of 4 KN/ and finish 1 KN/ . Use M25 and Fe 415

Solution: - data given Room size = 4 x 10 mWall thickness = 230LL = 4 KN/FF = 1 KN/Required – find depth, Ast.

Step 1. Calculation of load.

Assume d = = = 142.8 143 mm

(28=20x1.4=28)Total depth = 143+5+20=168 170Dead load = 0.17 x 25 = 4.25 KN/FF = 1.0 KN/Total D.L = 5.25 KN/LL = 4.0 KN/Factored design load = 1.5(5.25+4) = 13.875 KN/

Span length of slab

Span = effective span + d = 4+0.143 =4.143 m

Ultimate moment and shear:-

3. Check the depth for the bending:-

M=0.138 Fck b

OK

4. rough calculation for shear:-

For grade of concrete M25 is 0.36

Hence OK for shear.

5. Calculation of steel area:-

29.77 x

= 51630.15 Ast -5.99343

- 8614.4578 Ast + 4967105.647=0

5. main steel:- Using 10 bar.

Spacing =

Ast provided = 628

6. check for central of crack:-

Min pt = 0.12

Hence ok for crack control

Dia = =21.25>10mm provided ok.

Max spacing not more than 300. ok

7. recheck for shear

p= %

Reference table 19

8. check for deflection

Basic span to depth ratio =20Multiplying factor for Ast =0.42% =1.40Allowable L/d =1.40 x 20 =28Assumed is also 28 hence okHence safe in deflection

Secondary steel

As =

Spacing less than = 5d or 450 mm =715 or 450 mm

Hence ok

DESIGN OF CONTINUOUS SLABS

Design the interior span of a continuous one way slab for an office floor continuous over T beams spaced at 4m centers.Fck = 25N/ and Fe = 415 steel

1. Calculate factored load

Assume d =

Total depth =135+5+20=160mmDead load =0.16 x 25=4.0 KN/Floor finish = 1.0 KN/ Total = 5.0 KN/

Live load for office floor = 3.0 KN/Ratio of LL/DL =0.6 less than 0.75(Separate analysis of DL+LL not needed)Design factored load = 1.5(5+3) =12 KN/

2. Ultimate moments

At interior support =

At interior span =

3. Check depth for moment

Mu= 0.138 fck b d = 68 mm adopt d= 135mm

4. Rough check for shear

V=

For M25 OK

5. Calculation of steel areas Adopted depth is greater than required.Hence section is under reinforced.

At support

=48741.75 Ast – 5.9934 x

- 8133.11 Ast

Min area of steel

= OK

At interior span

Hence provide minimum steel 192 Main steel at support = 8 of e 130 c/c (Ast=384) (0.284%) Ast interior =8 of 260 c/c (Ast=192) (0.142%)

6. Check for deflection at middle of slab

Basic L/d ratio = 26

F1= 0.58 x fy

= 240.7Modification factor = 2So allowable ratio is 2 x 26 =52Given are 30 hence OK

7. Check for cracking

Steel are greater than 0.12% OK

Spacing less than 3d = 3 x 135 = 405Spacing less than 300 OK Diameter of rod < D/8 (160/8 = 20mm) OK

8. Check for shear OK

9. Check for top steel

For T- beam action

the detailing arrangement provide more than 60% of main steel in mid span of the slab as transverse steel OK

9. development length

Ld =

Minimum embedment length into the support = Ld/3

Length of bar embedment into the support = width of support- clear side cover

Example:-Design a cantilever porch of size 2500 mm wide and 5000 mm long is to be provided at a height of 3 m from floor level. The porch slab which overhangs 2500 mm beyond the face of the beam into be cast in flush with the top face of the beam

Assume live load = 0.75 Floor finish = 0.8 Concrete M20 and Fe 415

Solution:-

1. effective span = = 2565

The trial depth L/d =7 Modification factor is 2For Fe250, fs=0.58 x fy =0.58 x 250 =145

Pt % is 0.4Hence allowable L/d is 7 x 2 = 14Required d = 2565/4 = 180Assuming 10 bar effective cover = 15 + 10/2 = 20 mmProvide total depth of 200 mm d = 200-20=180 mmLet the overall depth of the slab be reduced to 100 mm at the cantilever and where bending moment is zero.

2. Loads

Consider 1 m width of slabSelf weight of slab = (0.2+0.1)/2 x 25 =3.75 KN/mWeight due to finish = 0.80 KN/mLive load =0.75 KN/m Total = 5.30 KN/mUltimate load per meter Wu = 5.3 x 1.5 = 7.95 KN/mMaximum bending moment (-ve) at the face of support

Mu = Wu x

4. Depth from bending moment consideration

Or

d = 91.45<< d consider (180)

Hence ok

5. Area of steel

Using bar

Provide @ 110 c/cArea provided=1000 x 78.5/110 =713

Curtailment of steel It is proposed to curtail 50% of the steel required at the support since the depth of the slab is tapering and bending moment

variations parabolic the area of reinforcement will get reduced to half at a distance greater than half the span from the free end.Bending moment at 1.6 m

Total depth of slab at 1.6 m from free end = 100 + 1.6(200-100)/2.5= 164 mmd = 164 -20 =144 mm

= 335 < (713/2) provided at supportDistance from support 2800-1600=900

Curtail 50% of the bars at a distance greater of the followinga) 900+12 = 900+120 =1020 mmb) 900+ d = 900+144= 1044 mm

So curtail 50% of steel at a distance 1050 mm from support

6. Distribution steel

Area required = 1.5 D = 1.5(200+100)/2 =225 Spacing = 1000 x 28/225 = 124 mmProvide 6 @ 120 mm c/c

7. check for deflection

Pt (say) = 100 x 666/ (1000 x 180) = 0.37 % < 0.4% OK

Or

Pt provided = 100 x 113/ (1000 x 180) = 0.4% OKRequired d = 2565/ 7 x 2 = 180

8. check for development length

Ld =

Available length = 900 mm > Ld

Continuous one way slab

u = 2.5 KN/

Fck = 15 N/

Fe- 415

Factored load

1) Assume d D =120

Dead load = 0.12 x 25 = 3.00 KN/Floor finish 1.00 KN/

Total 4.00 KN/

Live load = 2.5 KN/

Rating LL/DL = 2.5/4.00 = 0.625 < 0.75

2) moments

Middle of the end part (at centre)

Ast support next to end support

Mid of mid span At support of intermediate span

DL LL DL LL DL LL DL LL

=3.375 =5.4 =3.75 =3.375 =2.8125 =4.5 =3.75

Total =7.875 =9.15 =6.1875 =8.25

Mu = 0.138 fck

d =

d = 120-15-5=100mm

3) rough check for shear

v =

OK

4) calculation of steel area

Ast of middle leg and span

Check for deflection of middle span

7) Check for cracking

Ast greater than 0.12%Spacing less than 3d or 300 mmDia of bar < D/8

8) Check for shear

9) Check for top steel for T-beam action

One way slab

Syllabus- assign of simply supported, cantilever, continuous over beam with IS coefficient.

Que- design a one way slab continuous on four support subjected to udl of 2.5 KN/ . Take floor finish as 0.75 KN/ . The c/c distance between two successive supports is 3m. And it is constant for other spans also use m115 grade of concrete and steel grade concrete fe415

Que- a cantilever beam projecting out 2 m from the face of the support carries an udl of 25 KN/m over its entire length and a concentrated load of 5 KN at its free end. Design the beam, using concrete M15 and steel fe415. Assume width of beam = width of support = 300 mm. design also the shear reinforcement (S-99)

Que- a hall of dimension 21m x 8 m effective is provided with monolithic slab and beam floor with beams provided at 3.5 m c/c. the slab is 120 mm thick and carries a live load of 4 KN/ and finishing load of 1 KN/ . Design an intermediate beam if web width is 230 mm. also design the shear reinforcement for the beam. Sketch the reinforcement details for the beam. Use M20 concrete and fe415 steel.

Que- a simply supported slab having an effective span of 3.5 m. the slab is 125 mm thick and carries a live load of 4

KN/ . The tension reinforcement is provided in the form of 12 mm diameter bars at 175 mm c/c at effective cover of 25 mm. calculate the deflections for slab using IS code method IS 456-1978 specifications. Use M15 concrete and fe250 steel.

Que- a hall of effective dimensions of 7 m x 18 m is provided with monolithic slab and beam floor with beams provided 3 m c/c. the walls and beams are 230 mm thick. Design the slab as continuous slab as per IS 456-1978. if it has to carry live load of 3 KN/ and finishing load of 1 KN/

. Use M20 grade of concrete and fe415 steel. Sketch reinforcement.

Que- a hall of effective dimensions of 7 m x 18 m is provided with monolithic slab and beam floor with beams provided 3 m c/c. the walls and beams are 230 mm thick. Design the slab as continuous slab as per IS 456-1978. If it has to carry live load of 4 KN/ and finishing load of 1 KN/

. Use M20 grade of concrete and fe415 steel. Sketch reinforcement.

Que- design a cantilever slab projecting out of distance of 1.6 m from a brick wall of 230 mm thick. The slab is subjected to a working live load of 5 KN/ and it also carries a parapet wall of 100 mm thick with 0.75 m height. Use M20 mix and fe415. Design the slab for flexure and check for development length only.

Que- design a roof slab for a room 6 m x 3.5 m restrained at all the degrees for a service load 4 KN/ . Thickness of wall is 300 mm. use M20 and fe415.

Que- a hall of dimension 5 m x 16.25 m is provided with monolithic slab beam floor with beams at 3.25 m c/c. the thickness of walls and beams is 230 mm. design the slab as continuous slab as per IS456-1978. if it has to carry a live load of 3 KN/ and finishing load of 1.5 KN/ . Use M15 concrete and fe415 steel. Sketch the reinforcement details of slab.

Que- Design a rectangular slab panel having an effective size of 5.5 m x 3.5 m. the panel is continuous over two long edges and carries superimposed load of 3 KN/ . Use M15 grade concrete and fe415 grade of steel.

Fifth semester B.E civilSubject R.C.C structure (limit state)

Practical assignmentDate of submission

R.N Span in m Live load

KN/

F.F load KN/

Grade of concrete N/

Grade of steel N/

Support condition

Lx Ly

1 2 5 2.5 1 20 415 S.S2 3 6.5 5 0.75 20 250 S.S3 3 6.5 3.5 0.75 25 415 End panel of

continous slab

4 2 5 3.5 0.5 25 415 -“-5 3.5 7 2 1.0 20 415 -“-6 2 6 3.5 1.5 15 415 -“-7 3 5 1.5 0.5 20 250 Cantilever slab8 2.5 5 1.5 0.5 20 415 -“-9 2 5 0.75 0.5 20 415 -“-10 2 6 1 0.5 15 415 -“-11 2 6.5 1 0.5 20 250 -“-12 2.5 5 0.75 0.75 15 415 -“-13 3 5.5 0.75 0.5 20 415 -“-14 3.5 4 0.75 0.5 25 415 -“-15 2.8 6 1 0.5 20 500 -“-16 2.6 3 1 0.5 20 415 -“-17 2.25 4 1 0.25 20 250 -“-18 2.5 6 3 1 Intermediate

panel of continous slab

19 2.25 6 3.5 1.25 20 415 -“-20 3 6 2.5 0.75 15 250 -“-21 3.25 6 2 1 20 250 -“-22 3.38 6.46 3.5 1.25 25 250 -“-23 2.5 6 2.5 0.75 20 250 -“-24 3.5 8 3 0.75 15 250 -“-25 4 9 3.5 1 25 415 -“-26 3 7 3 1 20 415 -“-27 4 9 3.5 1.25 20 415 -“-29 2 6 2 0.5 20 250 -“-30 2.5 6 2.5 0.75 20 250 -“-31 3 7.5 3 0.5 20 250 -“-32 3.5 8 3.25 1 20 415 -“-33 4 8.5 4 1 20 415 S.S34 3 7 3.5 1.5 20 415 -“-35 5 11 3.5 1 20 415 -“-36 3.5 7 2.5 1.5 25 415 -“-

37 3 6 1.5 0.75 25 415 Cantilever slab38 3.5 6 1 0.75 20 415 -“-39 2.8 5 2 0.5 25 415 -“-40 2.5 5 2 0.5 20 415 -“-41 4.5 10 3.5 1 25 415 S.S42 4 10 3 1.5 20 415 -“-

NOTE- in cantilever slab Lx is the span of cantilever slab.

(prof. R.G BAIS)Date-

Example on column design

Que:-Explain in detail the interaction diagram used for design of column. If such interaction diagram is not available, set up the

procedure for design of column subjected to an axial load and bending.

Que:- A R.C.C column 300 x 600 mm is reinforced by 3 no.s of 20mm dia bars on each short side cover of 50mm to the centre of steel. Concrete used is M20 & steel is of grade Fe415. Calculate the ultimate axial force & the corresponding ultimate moment when the neutral axis is at 0.48D from the compression face & is parallel to the shorter side.

Que- a rcc column 300 x 500 mm is equally reinforced on two short faces with 5 bars of 20 mm diameter of steel grade fe250 on each face. Concrete used is M20.Calculate the ultimate load and ultimate moment resisted by the section if the column is just on the verge of cracking.(ans pu = 1513 KN and Mu = 90 KNm)

Que- a R.C.C column 300 x 660 mm is reinforced by 5 no of 25 mm diameter bar of grade Fe415 on either short side with a cover of 60 mm to the center of steel. Concrete grade M20 is used. Calculate ultimate load and ultimate moment corresponding to the coeditors of maximum / compressive strain of 0.0035 in concrete and tensile strength of 0.002 in the outermost layer of tension steel.(ans- Pu = 576.7 KN and Mu = 597.4 KNm)

Que- a R.C.C column of 300 x 500 mm is reinforced with 4 no bar of 20mm diameter of grade Fe415 on either short side with a cover of 50 mm to the center of steel. Concrete used is M20. calculate the ultimate load and its eccentricity from the center of column for an untracked section with neutral axis 1.5 D from the highly comopressed edge. The area of stress block is 0.422 fck D. and its CG acting at 0.48 D from the highly compressed edge.(ans Pu = 1964.7 KN and eccentricity = 22.10 mm)

Que- a short R.C column 250 x 400 mm carries an ultimate load of 310 KN. The area of steel consist of 8 bars of 20 mm diameter placed symmetrically along two short edges of column. The concrete M15 grade and steel Fe415 grade is used. Calculate the maximum ultimate moment about x axis, Mux (dividing the depth of the column.) the column can carry when it is subjected to ultimate moment about y axis equal to 41 KNm(ans- Mux = 77.8 KNm)

Que- a axially loaded short column of width 230 mm is subjected to ultimate load of 622 KN and ultimate moment of 184 KNm bisecting the depth of column. Design the column using concrete grade M15 and steel grade Fe415.(ans- 230 mm x 700 mm, 8 16 mm bars in 4 rows with 2 bars in each row.)

Que- design the column for the following data: ultimate load = 1090 KNm. Ultimate moment bisecting the depth = 19 KNm. Concrete grade M20. Steel grade fe415. Width of column=230 mm(ans. Size =230 mm x 400 mm, 6mm 16 mm in 3 rows with 2 bars in each row.)

Que- the corner column of a building of size 230 mm x 300 mm is reinforced with four bars of 20 mm diameter. It is subjected to an ultimate axial load of 341 KN and ultimate moments of 30 KNm and 16 KNm bisecting the depth and width of column respectively. For concrete grade M15 and steel fe415, check the safety of the column.(ans- the value of interaction equation is 0.89 < 1…….safe)

Que - the circular diameter 300 mm is reinforced with 8 bars of 20 mm diameter of grade fe415. The column braced and hinged at both ends and carries an axial ultimate load of 80 KN. The length of the column is 6 m. the concrete grade used is M20. Check the safety of the column. Assume effective length = unsupported length of 6 m.(ans- the slender column is safe)

Que- solves ex. 12.6.1 by taking effective length about x axis equal to 4.9 m other data remaining the same. Check the safety of the column of size 250 x 400 mm reinforced with 6 bars of 20 mm placed 3 bars on each face along the depth of column.(ans- the interaction equation gives value of 1.05>1…….unsafe)