For design & analysis of slabs the following methods have been used:

Coefficient method for two way slabs(Method 3.)

Cantilever method for one way slabs.Using:

Fc = 21 mpa′Fy = 400 mpa

By (Ary Hamafaraj Ahmed, HalmatTahir, and Three other collegues ) (Graduation project

(2008))

Analysis And Design of Slab by using moment coefficient method

Slab numbering

• Step 1 : • Finding thickness of slab:

• Min. t = perimeter/180

• = (2(7.6+4.6))/180 = 0.135m • Assume slab thickness= 0.17 m = 17 cm

Step 2:

• Finding Total factored load• Assume ceiling finishing = 0.36 KN/m² • D.L of slab = 0.17 * 1 * 24 = 4.08 KN/m² • Factored D.L = 1.4*(4.08+0.36)= 6.216 KN/m²• • Assume Roof Live load = 1.9 KN/m²• Factored L.L = 1.7*1.9= 3.23KN/m²• • Total Factored Load = 6.216+3.23= 9.446 KN/m²

Step3 : Finding Moments:

• Slab 1:______ Case(9)• La=4.6 m , Lb=4.6m • m = La/Lb = 4.6/4.6= 1 • Ma = Ca * Wu * La²• Mb= Cb * Wu * Lb²• -ve moments• -Ma= 0.061*9.446*(4.6²) = 12.19 KN.m• -Mb=0.033*9.446*(4.6²) =6.6 KN.m • +ve moments• +Ma D.L = 0.023*6.216*(4.6²)=3.02KN.m• +Ma L.L = 0.030*3.23*(4.6²) = 2.05 KN.m• +Ma= 3.02+2.05 = 5.07 KN.m• • +Mb D.L= 0.020*6.216*(4.6²)=2.63 KN.m• +Mb L.L = 0.028*3.23*(4.6²) = 1.91 KN.m• +Mb = 2.63+1.91= 4.54KN.m

•  • S2___________ Case (3):•  • La=4.6 m , Lb= 5.6m• m = 4.6/5.6= 0.82 •  • -ve moments• -Ma=0 • -Mb= 0.0626*9.446*(5.6²)=18.54 KN.m•  • +Ve moments • +Ma D.L = 0.0308*6.216*(4.6²)=4.051KN.m• +Ma L.L = 0.043 *3.23*(4.6²)=2.94KN.m• +Ma=4.051+2.94= 6.99 KN.m•  • +Mb D.L = 0.0208 *6.216*(5.6²)=4.05 KN.m• +Mb L.L = 0.0228 *3.23*(5.6²)=2.31 KN.m• +Mb=4.05+2.31= 6.36KN.m

• S3_____________Case (2)

• La=4.6m , Lb=5.6 • m = 4.6/5.6 =0.82

• -Ve moments• -Ma=0.063*9.446*(4.6²)= 12.59 KN.m• -Mb=0.0286*9.446*(5.6²)=8.47KN.m

• +Ve moments• +Ma D.L = 0.0252 *6.216*(4.6²)=3.36 KN.m• +Ma L.L = 0.0394*3.23*(4.6²)=2.69KN.m• +Ma=3.36+2 .69= 6.05KN.m

• +Mb D.L = 0.0114*6.216*(5.6²)=2.22 KN.m• +Mb L.L=0.0178*3.23*(5.6²)=1.8 KN.m• +Mb=2.22+1.8=4.02KN.m

Note: The remain slabs should be determined as the other Slabs (the cases Note: The remain slabs should be determined as the other Slabs (the cases are the same)are the same)

S8_________- Cantilever slab• S8 ( 4.6 m X 1m )

• - Ve moment= (W L²)/2 + P*L • • Min. thickness of way slab ( cantilever )

• t = L/10 = 1/10 = 0.10 m = 10 cm • Assume t = 0.17 m = 17 cm• Assume using light weight concrete block ;• With density = 12 KN/m³• D.L1= (0.17 *1*24)= 4.08 KN/m²• P=0.2*1*1*12=2.4KN• Factored P=1.4*2.4=3.36 KN• Factored D.L=1.4*4.08= 5.712 KN/m²• Factored L.L=1.7*1.9=3.23 KN/m²• Wu= 5.712+3.23=8.94KN/m²

• -M=(Wu*L²)/2+ p*L • -M=8.94*1²/2 + 3.36* 1 =7.83 KN.m

• For slab 9,10,11 have the same procedure and moment(7.83KN.m)

• For slab 12• (4.6m X 1.6m)• • P=0.2*1*1*12=2.4KN• factored P= 1.4*2.4=3.36 KN• • Wu=8.94KN/m²• • -M=8.94*(1.6²)/2 + 3.36* 1.6=16.82KN.m• • For slab 13 have the same moment(16.82KN.m)

• For Slab 14• h = 1.4/10=0.14 m = 14 cm • Assume t=17 cm • Wu=1.4*4.08+1.7*1.9 = 8.94 KN/m²

• -M= (8.94 * (1.4²))/2 = 8.76 KN.m

• For slab 15 have the same moment of S14

• For slab 16

• h = 1.8/10= 0.18 m = 18 cm• Assume t = 18 cm• D.L of slab = 0.175 * 1 *24 = 4.2 KN/m² • Factored D.L = 1.4* 4.2 = 5.88 KN/m²• Wu=5.88+3.23=9.11KN/m²• -M=(Wu * L²)/2 = (9.11* (1.8²))/2= 14.75KN.m

• Step (4)

• Checking for d • d available = h – cover – Ø/2• Assume using Ø 12mm• d= 170-20-12/2 = 144 mm

• Using maximum moment for checking d• M= 18.54KN.m• M=Ø ρ b d² fy(1- 0.59 ρ fy/fc )′• Assume ρact. = ρmax. = 0.75 ρbalance• ρb=0.85 * ß * fc /fy(600/(600+fy))′• ß=0.85 • ρb=0.85*0.85*(21/400)*(600/(600+400)) = 0.0227• ρmax.=0.75*0.0227=0.017 • Ø=0.9• 18.54*1000000=0.9*0.017*1000*d²*400(1- 0.59*0.017*(400/21))• d = 61.195mm < 144 mm O.K

For Two way slabMoment KN.m

Short directoinlong direction

- Ve+Ve- Ve+Ve

S112.195.076.64.54

S206.9918.546.36

S312.596.058.474.02

S48.1533.944.992.62

S507.2112.263.96

S616.198.435.452.73

S713.336.2413.336.24

For ( cantilever slab) KN.m 

 

S8,9,10,117.83 

S12,1316.8192 

S14,158.76 

S1614.75 

Table of moments

Table 1.1 Designed moment values for slabs

• Step (5)• • Finding the reinforcement , For example :• • M=Ø As fy ( 0.9d)• For slab 1 for moment= 12.19KN.m• 12.19*1000000=0.9*As*400(0.9*144)• As= 261.27mm²• • From ACI code• Code 13.3.1 — Area of reinforcement in each direction for• two-way slab systems shall be determined from• moments at critical sections, but shall not be less than• required by 7.12.

For two way slabsAs mm²

Short directionLong direction

- ve+ve-ve+ve

S1261.27494.307141.46097.307

S20149.819397.376136.316

S2269.847129.672181.54186.162

S4174.74784.447106.9556.155

S50154.535262.77484.876

S6347.007180.684116.81258.513

S7285.708133.744285.708133.744

for one way slab ( cantilever)As mm²

S8,9,10,11167.82

S12,13360.511

S14,15187.757

S16316.143

 Table 1.2 Design As values for slabs 

Table of ( As ) of slabs

• As minimum for slab

• From ACI code • Code:

• 7.12.1 — Reinforcement for shrinkage and temperature• stresses normal to flexural reinforcement shall be• provided in structural slabs where the flexural

reinforcement• extends in one direction only.• As=ρ* b * h • ρ=0.002 , b=1000 mm , h=170 mm• Asmin.=0.002*1000*170=340 mm²

• Note : For slabs that As required < As min. we should use As min. as a design

• IN theoretical • For slab 1 finding arrangement• • in middle strip(span/2=4.6/2=2.3 m) • +As = As min. = 340 mm²/m• +As=340*2.3=782 mm²/2.3m• Using Ø12mm ______ As for one bar=113.1mm²• No. of bars=As total/As of one bar = 782/113.1=6.9= 7bars• Spacing = 7-1= 6 • (2.3*100)/6 = 38.33= 39 cm • From ACI code• Code 13.3.2 — Spacing of reinforcement at critical sections• shall not exceed two times the slab thickness, except• for portions of slab area of cellular or ribbed construcedges• • 2h=2*17=34 cm < 39cm using 34 cm for spacing between bars • Using Ø12mm @ 68cm c/c straight • Using Ø12mm @ 68cm c/c bent up .

• In edge strip of short direction (span/4=4.6/4=1.15m)• M=2/3 of middle strip moment = 2/3*5.07=3.38 kN.m• As=72.44 mm² < Asmin. Using Asmin.• +As=340mm²/m=340*1.15=391mm²/1.15m• No. of bars=391/113.1=3.45=4 bars• No. of spacing=4-1=3 • 1.15*100/3=38.33cm= 39cm >2h • Then using 2h=34cm spacing between bars• Using Ø12mm @ 68cm c/c straight • Using Ø12mm @ 68cm c/c bent up•

• *at supports middle strip • -As= As min.=340mm²/m=782mm²/2.3m• (2.3*100/34)=6.7= 7 spacing • No. of bars=spacing+1=8 bars• As from bent up= 1/2 * (8*113.1)=452.4mm²/2.3m• -As at support = 340mm²/m=340*2.3=782mm²/2.3m• Required as using as additional = 782-452.4=329.6mm²/2.3m• No. of additional bars=329.6/113.1=2.91= 3 bars• No. of spacing = 2 • 2.3*100/2=115 cm >68 cm • Then using Ø12mm @ 68 cm c/c Additional

• *at supports middle strip • -As= As min.=340mm²/m=782mm²/2.3m• (2.3*100/34)=6.7= 7 spacing • No. of bars=spacing+1=8 bars• As from bent up= 1/2 * (8*113.1)=452.4mm²/2.3m• -As at support = 340mm²/m=340*2.3=782mm²/2.3m• Required as using as additional = 782-452.4=329.6mm²/2.3m• No. of additional bars=329.6/113.1=2.91= 3 bars• No. of spacing = 2 • 2.3*100/2=115 cm >68 cm • Then using Ø12mm @ 68 cm c/c Additional

• In edge strip of short direction (span/4=4.6/4=1.15m)• M=2/3 of middle strip moment = 2/3*12.19=8.12 kN.m• As=174.04mm² < Asmin. Using Asmin.• -As=340mm²/m=340*1.15=391mm²/1.15m• 1.15*100/34=3.38=4 spacing • No . of bars=4+1=5 bars• As from bent = 1/2* (5*113.1)=282.75mm²/1.15m• Required –As= 391-282.75=108.25mm²/1.15m• No. of additional bars=108.25/113.1= 1 bar • Then using Ø12mm @ 68 cm c/c Additional

• Note: generally In practice there is no divisions in the distribution of steel reinforcement concrete

• Arrangement for +As :• for all spans that +As < As min. • Then using As min. as a design = 340mm²/m• No. of bars=340/113.1=3.006 = 4 bars• No. of spacing= 4-1=3• 1*100/3=33.3=33 cm <2h=34cm• Then using Ø12mm@66cm c/c straight • Using Ø12mm@66cm c/c bent up.• -As for supports that As required < As min. • Then using As min. = 340mm²/m• We have –As from bent=1/2 * (4*113.1)=226.2 mm²/m• -As required = 340-226.2=113.8mm²/m• No. of additional bras=113.8/113.1=1.006=2 bars• No. of spacing=2-1=1 • 1*100/1=100 cm > 66 cm• Then using Ø12mm@66cm c/c Additional bras

• For long direction of ( slab 2 )support that –As= 397.376mm²/m

• We have –As from bent = 1/2*(4*113.1)=226.2mm²/m• -As required = 397.376-226.2=171.176mm²/m• No. of additional bars= 171.176/113.1=1.51=2 bars • No. of spacing=2-1=1• 1*100/1=100cm > 66cm• Then using Ø12mm @ 66cm c/c additional

• For short direction of ( slab 6 )support that –As= 347.007mm²/m

• We have –As from bent = 1/2*(4*113.1)=226.2mm²/m• -As required = 347.007-226.2=120.807 mm²/m• No. of additional bars= 120.807/113.1=1.068=2 bars • No. of spacing=2-1=1• 1*100/1=100cm > 66cm• Then using Ø12mm @ 66cm c/c additional

• For cantilever slabs that –As < Asmin • Then using Asmin. =340mm²/m• –As from bent of two way slabs =1/2 *

(4*113.1)=226.2 mm²/m• -As required = 340-226.2=113.8mm²/m• No. of additional bras=113.8/113.1=1.006=2 bars• No. of spacing=2-1=1 • 1*100/1=100 cm > 66 cm• Then using Ø12mm@66cm c/c Additional bras• Finding the spacing between steel bars

• For + As using As min for all panels • At mid span in • +As=340 mm² per one meter

• Using Ø12 mm _________ As=113.1 mm²• No. of bars= Astotal/As one bar. = 340/113.1=3.006 , say = 4 bars• No. of spacing = 3• Spacing = 100/no of spacing

• Spacing = 100/3=33.3 cm = 33 cm• 2h= 2*17 = 34 cm , 33 cm < 2h using 33 cm as spacing • Using: • Ø12mm @ 66 cm C/C straight • Ø12mm@66 cm C/C bent up.

• At support As=340 mm²• From bent we have 1/2*1564=782 mm²/4.6m

• As at support = 1564 mm²/4.6m – 782mm²/4.6m=782mm²/4.6m• No. of bars= 782/113.1=6.91 = 7 bars• No. of spacing= 7-1=6 • Spacing=4.548*100/6=75 cm • Using:• Ø12mm @ 75 cm C/C Additional