water tank 1Design of Rectangular water tank CASE-1 ( L / B < 2 )Capacity80000Litres(given)MaterialM20Grade Concrete(given)Fe 415Grade HYSD reinforcement(given)Solution :-Provide 6 m x 4 m x 3.5 m tank with free board of 150 mm.Volume = 6 x 4 x 3.35 x 10 3 =80400LitresL / B =6 / 4 =1.5< 2 .The top portion of side walls will be designed as a continuous frame.bottom 1 m or H / 4 whichever is more is designed as cantilever.H / 4 =3.5 / 4 =0.875mbottom 1 m will be designed as cantilever.Water pressure at 3.5 - 1 = 2.5 m height from top =2.5 x YW =2.5 x 9.8=24.5KN / m2where Yw is unit weight of water = 9.8 KN / m3To find moment in side walls, moment distribution or kani's method is used. As the frame is symmetrical about both the axes, only one joint is solved6 m6 mAEBF4 m3.5 m24.5 KN / m2DC34.3 KN / m2ElevationPlanFixed end moments :-MAB =w x l2 / 12 ==24.5 x 62 / 12 =73.5KNmMAD =w x l2 / 12 ==24.5 x 42 / 12 =-32.66KNmKani's Method :-- 32.6673.50-12.25-8.1700-12.25-8.170D0A0B-57.1657.16Rotation factor at Joint AJointMemberRelative Stiffness( K ) KRotation Factor u =(-1/2) k / KAABI / 65 * I / 12- 2 / 10ADI / 4- 3 / 10Sum of FEMMAF =73.5-32.6640.84KNmMAB =MABF + 2 MAB' + MBA'=73.5 + 2 x (- 8.17 ) + 0=57.16MAD =MADF + 2 MAD' + MDA'=(- 32.66 ) + 2 x (- 12.25 ) + 0=-57.16B.M. at centre of long span =w x l 2 / 8 - 57.16=24.5 x 62 / 8 - 57.1653.09KNmB.M. at centre of short span =w x l 2 / 8 - 57.16=24.5 x 42 / 8 - 57.16-8.16KNmDirect tension in long wall =Yw ( H - h ) x B / 2=24.5 x 4 / 2 =49KNDirect tension in short wall =Yw ( H - h ) x L / 2=24.5 x 6 / 2 =73.5KNTABLE 9-6Design of Long Walls :-Balanced Design Factors for members in bendingAt supportFor M20 Grade Concrete MixM =57.16KNmd / DMild steelHYSD barsT =49KNTension on liquid face.Q = M / bD2PtQ = M / bD2PtFrom Table 9-6Q = 0.306Assuming d / D = 0.90.750.30.40.2950.289D =M / Q x b0.80.3050.370.2990.272=57.16 x 10 6 / 0.306 x 10000.850.310.3550.3020.258=432.2mm,Assuming d / D = 0.90.90.3140.3350.3060.246Take D =450 mmd =450 - 25 - 8TABLE 9-5=417 mmMembers in bending ( Cracked condition )From Table 9-5Coefficients for balanced designAst1 for moment =M / st x j x dGrade of concreteGrade of steelcbc N / mm2st N / mm2kjQPt,bal=57.16 x 10 6 / 150 x 0.872 x 417For members less than 225mm thickness and tension on liquid face=1048mm2M20Fe25071150.4450.8511.331.36Fe41571500.3840.8721.170.98Ast2 for direct tension = T / stFor members more than 225mm thickness and tension away from liquid face=49 x 10 3 / 150M20Fe25071250.4270.8581.281.2=327mm2Fe41571900.3290.891.030.61Total Ast1 + Ast2 =1048 + 327=1375mm2Provide 16 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=200.96 x 1000 / 1375=146.1527272727mmProvide 16 mm O bar @ 130 mm C/Cmarked(a)= 1546mm2 / m.Larger steel area is provided to match with the steel of short walls.At centreM =53.09KNmT =49KNtension on remote facee =M / T =53.09 / 49=1.08mLine of action of forces lies outside the sectioni.e.tension is smallE =e + D / 2 - db=1080 + 450 / 2 - 417=888mmD / 2Dmodified moment= 49 x 0.888d=43.51KNme = M / Td'Ast1 for moment =M / st x j x dE = e + D / 2 - d=43.51 x 10 6 / 190 x 0.89 x 417=617mm2TABLE 9-3Minimum Reinforcement for Liquid Retaining StructuresAst2 for direct tension = T / stThickness, mm% of reinforcement=49 x 10 3 / 150Mild SteelHYSD bars=327mm21000.30.24Total Ast1 + Ast2 =617 + 3271500.2860.229=944mm22000.2710.217Provide 16 mm O bar2500.2570.206spacing of bar =Area of one bar x 1000 / required area in m2 / m3000.2430.194=200.96 x 1000 / 9443500.2290.183=212.8813559322mm4000.2140.171Provide 16 mm O bar @ 200 mm C/Cmarked(b)= 1005mm2450 or more0.20.16From Table 9-3minimum reinforcement 0.16 %Distribution steel =0.16 / 100 x 1000 x 450=720mm2On each face =360mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 /360=139.5555555556mmProvide 8 mm O bar @ 130 mm C/Cmarked(d)= 385mm2Vertical Steel ( c)Provide 10 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=78.50 x 1000 /360=218.0555555556mmProvide 10 mm O bar @ 200 mm C/Cmarked(c)= 392mm2Horizontal steel :-Remote face ( b) - Provide 16 mm O @ 200 mm C/C = 1005mm2Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =385mm2Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392mm2Design of short walls :-At supportM =57.16KNmT =73.5KNtension on liquid faceFrom Table 9-5Ast1 for moment =M / st x j x d=57.16 x 10 6 / 150 x 0.872 x 417=1048mm2Ast2 for direct tension = T / st=73.5 x 10 3 / 150=490mm2Total Ast1 + Ast2 =1048 + 490=1538mm2Provide 16 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=200.96 x 1000 / 1538=130.6631989597mmProvide 16 mm O bar @ 130 mm C/Cmarked(b)= 1546mm2 / m.At centreM =8.16KNmT =73.5KNtension on liquid faceFrom Table 9-5Ast1 for moment =M / st x j x d=8.16 x 10 6 / 150 x 0.872 x 417=150mm2Ast2 for direct tension = T / st=73.5 x 10 3 / 150=490mm2Total Ast1 + Ast2 =150 + 490=640mm2Provide 12 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=113.04 x 1000 / 640=176.625mmProvide 12 mm O bar @ 130 mm C/Cmarked(e)= 869mm2 / m.From Table 9-3minimum reinforcement 0.16 %Distribution steel =0.16 / 100 x 1000 x 450=720mm2On each face =360mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 /360=139.5555555556mmProvide 8 mm O bar @ 130 mm C/Cmarked(d)= 385mm2Vertical Steel ( c)Provide 10 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=78.50 x 1000 /360=218.0555555556mmProvide 10 mm O bar @ 200 mm C/Cmarked(c)= 392mm2 .Horizontal steel :-Remote face ( e) - Provide 12 mm O @ 130 mm C/C = 869mm2Liquid face ( d) - Provide 8 mm O @ 130 mm C/C =385mm2Vertical Steel ( c) - Provide 10 mm O @ 200 mm C/C both faces = 392mm2Bottom 1 m will be designed as cantileverCantilever moment : -M =Yw x H x h2 / 6ORYw x H / 6, whichever is greater.=9.8 x 3.5 x 1 / 6=9.8 x 3.5 / 6=5.72KNm=5.72,tension on liquid face.From Table 9-5Ast for moment =M / st x j x d=5.72 x 10 6 / 150 x 0.872 x 417=105mm2From Table 9-3minimum reinforcement 0.16 %Distribution steel =0.16 / 100 x 1000 x 450=720mm2On each face =360mm2Provide 10 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 /m=78.50 x 1000 /360=218mmProvide 10 mm O bar @ 200 mm C/Cmarked(c)= 392mm2 .each faceBase slab :-Base slab is resting on ground. For a water head 3.5 m, provide 150 mm thick slab.3500From table 9-3Minimum steel =0.229%=0.229 / 100 x 1000 x 150150=344mm2,172 mm2 bothway1 : 4 : 8 P.C.C.Provide 8 mm O bar1504506000450150spacing of bar =Area of one bar x 1000 / required area in m2 / mElevation=50.24 x 1000 /1721500=292mm450Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.1000Ast =346mm2Designed section,Elevation etc. are shown in fig.Top slab may be designed as two-way slab as usual for a live load of 1.5 KN / m24000Top slab : -consider 1 m wide strip. Assume 150 mm thick slab.lx =4 + 0.15 = 4.15 say 4.5 m1000ly =6 + 0.15 = 6.15 say 6.5 m450Dead Load : self 0.15 x 25 =3.75KN / m215001500floor finish =1.0KN / m26000Live load =1.5KN / m24504506.25KN / m2Section A-AFor 1 m wide stripPU =1.5 x 6.25=9.38KN / mly / lx =6.5 / 4.5=1.4AS Per IS-456-2000,Four Edges Discontinuous,positive moment at mid-span.Table 6-3Table 26x =0.085Limiting Moment of resistance factor Q lim, N / mm2y =0.056fck N / mm2fy, N / mm2Mx =x x w x lx2My =y x w x lx2250415500550=0.085 x 9.38 x 4.52=0.056 x 9.38 x 4.52152.222.072.001.94=16.15KNm=10.64KNm202.962.762.662.58From Table 6-3 ,Q = 2.76253.703.453.333.23drequired =M / Q x b304.444.143.993.87=16.15 x 10 6 / 2.76 x 1000=76.50mmdshort =150 - 15(cover) - 5=130 > 76.50mm(O.K.)dlong =130 - 10 =120mmLarger depth is provided due to deflection check.Mu / b x d2 (short) = 16.15 x 106 / 1000 x 130 x 130=0.96Pt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fck=50 1-1-(4.6 / 20) x (0.96)415 / 20=50 [(1-0.88) x 20 / 415 ]=0.29%Ast (short) =0.29 x 1000 x 130 / 100=377mm2Provide 10 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 /m=78.50 x 1000 /377=208mmProvide 10 mm O bar @ 210 mm c/c= 374mm2 .Mu / b x d2 (long) = 10.64 x 106 / 1000 x 120 x 120=0.74Pt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fck=50 1-1-(4.6 / 20) x (0.74)415 / 20=50 [(1-0.91) x 20 / 415 ]=0.22%Ast (long) =0.22 x 1000 x 120 / 100=264mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 /264=190mmProvide 8 mm O bar @ 190 mm c/c= 264mm2 .

AA15001500-3/10-2/1040.842.5 m1 m1501508 O @ 290 c/c both ways top and bottom10 O @ 200 c/c10 O @ 200 c/c- shape- shape10 O @ 210 c/c8 O @ 190 c/c150 Free boardvvvv100016 O @ 130 c/c (a)10 O @ 200 c/c both faces (c)12 O @ 130 c/c (e)16 O @ 200 c/c (b)8 O @ 130 c/c (d)( a )( a )( b )( d )( c )( d )

water tank 2Design of Rectangular water tank CASE-2 ( L / B 2 )Size of tank : 3.6 m x 8.0 m x 3.0 m high(given)MaterialM20Grade Concrete(given)Fe 415Grade HYSD reinforcement(given)Solution :-Size of tank : 3.6 m x 8.0 m x 3.0 m highVolume = 3.6 x 8 x 3.0 x 10 3 =86400LitresL / B =8 / 3.6 =2.22> 2 .The long walls are designed as vertical cantilevers from the base.The short walls are designed as supported on long walls.If thickness of long walls is 400 mm, the span of the short wall = 3.6 + 0.4 = 4.0 m.bottom 1 m or H / 4 whichever is more is designed as cantilever.H / 4 =3.0 / 4 =0.75mbottom h = 1 m will be designed as cantilever.Moments and tensions :Maximum B.M. in long walls at the base=(1 / 6 ) x Yw x H3=( 1 / 6 ) x 9.8 x 33=44.1KNm.Maximum ( - ve ) B.M. in short walls at support=Yw x ( H - h ) x B2 / 12=9.8 x ( 3 - 1 ) x 42 / 12=26.13KNm.Maximum ( + ve ) B.M. in short walls at centre=Yw x ( H - h ) x B2 / 16=9.8 x ( 3 - 1 ) x 42 / 16=19.60KNm.For bottom portionM =Yw x H x h2 / 6ORYw x H / 6 , whichever is greater=9.8 x 3.0 x 1 / 6=9.8 x 3.0 / 6=4.90KNm=4.90KNmDirect tension in long wall =Yw x ( H - h ) x B / 2=9.8 x ( 3 - 1 ) x 3.6 / 2=35.28KNDirect tension in short wall=Yw ( H - h ) x 1=9.8 x ( 3 - 1 ) x 1=19.6KNIt is assumed that end one metre width of long wall gives direct tension to short walls.Design of long walls : -M ( - ) =44.1KNm( water face )T =35.28KN( perpendicular to moment steel )From Table 9-6Assume d / D = 0.9Q = 0.306D =M / Q x b=44.1 x 10 6 / 0.306 x 1000=379.6mm,Take D =400 mmd =400 - 25 - 8=367 mmFrom Table 9-5 ,Ast for MomentAst =M / st x j x d=44.1 x 10 6 / 150 x 0.872 x 367=918.68mm2Provide 16 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=200.96 x 1000 / 918.68=218.7486393521mmProvide 16 mm O bar @ 200 mm c/c= 1005mm2 .Note : The design is made at the base. The moment reduces from base to top.For economy, the reinforcement can be curtailed or the thickness of wall can also be reduced as we have done for cantilever retaining wall.From Table 9-3Distribution steel = 0.171 % for 400 mm depthAs =( 0.171 / 100 ) x 1000 x 400=684mm2 .on each face =342mm2 . ( 1 )Steel required for direct tension=T / st=35.28 x 103 / 150=235mm2 . ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 342=146.9005847953mmProvide 8 mm O bar @ 140 mm c/c on each face= 357mm2on each faceDesign of short walls :-At supportM =26.13KNmT =19.6KNFrom Table 9-5Ast1 for moment =M / st x j x d=26.13 x 10 6 / 150 x 0.872 x 367=544mm2Ast2 for direct tension =T / st=19.6 x 10 3 / 150=131mm2Total Ast1 + Ast2 =544 + 131=675mm2Provide 12 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=113.04 x 1000 / 675=167.4666666667mmProvide 12 mm O bar@160 mm c/c= 706mm2.1000203.56x400367163.44checking :modular ratio m =280 / 3 x cbc=13.33x =b x D2 / 2 + Ast ( m - 1 ) x db x D + ( m - 1 ) x Ast=( 1000 x 4002 / 2 ) + ( 706 x ( 13.33 - 1 ) x 367 )( 1000 x 400 ) + ( ( 13.33 - 1 ) x 706 )=203.56mmD - x =196.44mmd - x =163.44mmAT =b x D + ( m - 1 ) x Ast=1000 x 400 + (13.33 - 1 ) x 706=408705mm2Ixx =( 1 / 3 ) x b x ( x3 + ( D - x )3 ) + ( m - 1 ) x Ast x ( d - x )2=( 1 / 3 ) x 1000 x ( 203.563 + 196.443 ) + ( 13.33 - 1 ) x 706 x 163.442=5.34E+09+2.33E+08=5.57E+09mm4fct =T / AT=19.6 x 10 3 / 408705=0.048N / mm2fcbt =M x ( d - x ) / Ixx=26.13 x 106 x 163.44 / 5.57 x 10 9=0.767N / mm2check :( fct / ct ) + ( fcbt / cbt ) 1( 0.048 / 1.2 ) + ( 0.767 / 1.7 ) 1From Table 9-2Table 9-20.04 + 0.4512 1Permissible concrete stresses in calculations relating to resistance to cracking0.4912 1.. ( O. K. )Grade of concretePermissible stresses in N / mm2At centre :Direct tension ctTension due to bending cbtShear stress v = V / b j dM =19.6KNmT =19.6KNM151.11.51.5From Table 9-5M201.21.71.7Ast1 for moment =M / st x j x dM251.31.81.9=19.6 x 10 6 / 150 x 0.872 x 367M301.52.02.2=408mm2M351.62.22.5M401.72.42.7Ast2 for direct tension =T / st=19.6 x 10 3 / 150=131mm2Total Ast1 + Ast2 =408 + 131=539mm2Provide 12 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=113.04 x 1000 / 539=209.7217068646mmProvide 12 mm O bar @ 200 mm c/c= 565mm2.From Table 9-3Distribution steel = 0.171 % for 400 mm depthAs =( 0.171 / 100 ) x 1000 x 400=684mm2 .on each face =342mm2 . ( 1 )Steel required for direct tension=T / st=19.6 x 103 / 150=131mm2 . ( 2 )From ( 1 ) and ( 2 ) , minimum steel is sufficient for resisting direct tension.Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 342=146.9005847953mmProvide 8 mm O bar @ 140 mm c/c on each face= 357mm2on each faceBottom cantileverM =4.9KNm150From Table 9-5Ast =M / st x j x d=4.9 x 10 6 / 150 x 0.872 x 3673000=102mm2Minimum steel = 342 mm2 on each face.Provide 8 mm O bar @ 140 mm c/c on each faces= 357mm2150on each face1 : 4 : 8 P.C.C.1501504008000400Base slab :-Section A-ABase slab is resting on ground. For a water head 3 m, provide 150 mm thick slab.2000From table 9-3400Minimum steel =0.229%900=0.229 / 100 x 1000 x 150=344mm2,172 mm2 bothwayProvide 8 mm O bar3600spacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 /172=292mm900Provide 8 mm O bar @ 290 mm C/C both ways, top and bottom.400Ast =346mm220002000Designed section,Elevation etc. are shown in fig.8000Top slab may be designed as a one-way slab as usual for a live load of 1.5 KN / m2400Sectional plan400Top slab : -consider 1 m wide strip. Assume 150 mm thick slab.lx =3.6 + 0.4 = 4 say 4 m150ly =8 + 0.15 = 8.15 say 8.5 mDead Load : self 0.15 x 25 =3.75KN / m2floor finish =1.0KN / m2Live load =1.5KN / m230006.25KN / m2Base details not shown for clarityFor 1 m wide stripPU =1.5 x 6.25150=9.38KN / m1501504003600 400150Maximum moment =9.38 x 42 / 8Section B- B=18.76KNmMaximum shear =9.38 x 3.6 / 2=16.88KNFrom Table 6-3 ,Q = 2.76drequired =M / Q x b=18.76 x 10 6 / 2.76 x 1000=82.44mmdprovided =150 - 15(cover) - 6=129 > 82.44(O.K.)Larger depth is provided due to deflection check.Design for flexure :Mu / b x d2 = 18.76 x 106 / 1000 x 129 x 129=1.13Pt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fck=50 1-1-(4.6 / 20) x (1.13)415 / 20=50 [(1-0.86) x 20 / 415 ]=0.34%Ast =0.34 x 1000 x 129 / 100=439mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 /m=50.24 x 1000 /439=114mmProvide 8 mm O bar @ 110 mm c/c= 457mm2 .Minimum steel is 0.15 % for mild steel and 0.12 % for HYSD Fe415 reinforcementDistribution steel =( 0.12 / 100 ) x 1000 x 150=180mm2Provide 6 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=28.26 x 1000 /180=157mmProvide 6 mm O bar @ 150 mm c/c= 188mm2 .

8 O @ 140 c/c (chipiya)200020008 O @ 290 c/c both ways top and bottom12 O @ 200 c/c- shape8 O @ 110 c/c6 O @ 150 c/cvvvv90012 O @ 160 c/c(chipiya)16 O @ 200 c/c ( chipiya )1508 O @ 140 mm c/c8 O @ 140 c/c12 O @ 200 c/cAABB8 O @ 140 c/c8 O @ 140 c/c8 O @ 110 c/c6 O @ 150 c/c16 O @ 200 c/c ( chipiya )9009008 O @ 140 c/c

slab ( one way )Design of simply supported one way slabNOTE : -If ly / lx 2 ,called one way slab provided that it is supported on all four edges . Note that , if all four edge is not supported and ly / lx < 2 , then also it is one-way slab,If ly / lx < 2 , called two-way slab.provided that it is supported on all four edges.effective span =4 m supported on masonry wall of 230 mm thickness( given )If clear span = 3.77 m , effective span = 3.77 + 0.23 = 4 m OR effective span = 3.77 + 0.16 ( effective depth ) = 3.93 mLive load =2.5KN / m2( given )Floor finish =1KN / m2( given )materialM15 grade concrete( given )HYSD reinforcement grade Fe415( given )solution : -Assume 0.4 % steel , a trial depth by deflection criteria0.6 % for mild steel reinforcement and 0.3 to 0.4 % for HYSD Fe415 grade reinforcementIS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementIS 456-2000 clause 22.2modification factor = 1.26NOTE : -Effective Spanfor simply supported, basic span / effective depth ratio = 20( a ) Simply Supported Beam or Slab -( span / d ) ratio permissible = 1.26 x 20The effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre to centre of supports , whichever is less.=25.2drequired =4000 / 25.2=158.7mm( b )Continuous Beam or Slab - In the case ofD =158.7 + 15 ( cover ) + 5 ( assume 10 O bar )continuous beam or slab, if the width of the=178.7mmsupport is less than l/12 of the clear span, theAssume an overall depth =180mmeffective span shall be as in (a). If theDL = 0.18 x 25 =4.5KN / m2supports are wider than I/12 of the clear spanFloor finish =1.0KN / m2or 600 mm whichever is less, the effective spanLive load =2.5KN / m2shall be taken as under:Total8.0KN / m21) For end span with one end fixed and theFactored load =1.5 x 8 =12KN / mother continuous or for intermediate spans,Consider 1 m length of slabthe effective span shall be the clear spanMaximum moment =w x l2 / 8between supports;=12 x 42 / 82) For end span with one end free and the other=24KNmcontinuous, the effective span shall be equalMaximum shear =w x l / 2to the clear span plus half the effective depth=12 x 4 / 2of the beam or slab or the clear span plus=24KNhalf the width of the discontinuous support,Design for flexure : -whichever is less;d =180 - 15 - 53) In the case of spans with roller or rocket=160mmbearings, the effective span shall always bethe distance between the centres of bearings.Mu / b x d2 =24 x 10 6 / 1000 x (160)2( c )Cantilever-The effective length of a cantilever=0.94shall be taken as its length to the face of thePt =50 1-1-(4.6 / fck) x (Mu / b x d2)support plus half the effective depth exceptfy / fckwhere it forms the end of a continuous beamwhere the length to the centre of support shall=50 1-1-(4.6 / 15) x (0.94)be taken.415 / 15( d )Frames-In the analysis of a continuous frame,centre to centre distance shall be used.=50 [(1-0.84) x 15 / 415 ]=0.289%Ast =0.289 x 1000 x 160 / 100=462mm2Provide 10 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=78.50 x 1000 /462=170mmProvide 10 mm O bar @ 170 mm c/c= 462mm2 .Half the bars are bent at 0.1 l = 400 mm , and remaining bars provide 231 mm2 area100 x As / ( b x D ) =100 x 231 / ( 1000 x 180 )=0.128> 0.12 % ( minimum steel for Fe415)For mild steel minimum reinforcement 0.15 %i.e. remaining bars provide minimum steel. Thus, half the bars may be bent up.Distribution steel =( 0.12 /100 ) x 1000 x 180=216mm2Maximum spacing =5 x 160 = 800 or 450 mmi.e. 450 mmProvide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 /m=50.24 x 1000 /216=233mmProvide 8 mm O bar @ 230 mm c/c= 218mm2 .Check for shear : -Vu =24KNActual Shear stress =Vu / b x d=24 x 103 / 1000 x 160=0.150N / mm2< ( C ) N / mm2( too small )For bars at supportd =160mmAs =231mm2 .100 x As / b x d =100 x 231 / 1000 x 160=0.144Design shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )6 x =0.8 x fck / 6.89 Pt , but not less than 1.0=0.8 x 15 / 6.89 x 0.144=12.1Design shear strength c =0.85 0.8 x 15 ( 1 + 5 x 12.1 - 1 )6 x 12.1=0.277IS 456-2000 Table 19 from table 7-1for Pt = 0.144 c = 0.28N / mm2IS 456-2000 clause 40.2.1.125 difference-0.05k =1.24for 180 mm slab depth20 difference?-0.04Design shear strength =1.24 x 0.28=0.347N / mm2.( O.K.)Check for development length : -Assuming L0 =8 O(HYSD Fe415 steel ) For continuing barsFor checking development length , l0 may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.As =231mm2Mu1 =0.87 x fy x Ast x d{ 1 - ( fy x Ast / fck x b x d ) }ORPt =100 x As / b x d=0.87 x 415 x 231 x 160{ 1 - (415 x 231 / 15 x 1000 x 160 ) }=100 x 231 / 1000 x 160=13.34{ 1- 0.0399 }=0.14=12.8119661208KNmFrom equationPt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fckwe get ,Mu1 / b x d2 =0.49Mu1 =0.49x 1000 x 1602 x 10-6Vu =24KN=12.54KNmDevelopment length of bars Ld =O s / 4 x bd= 56 O ( from Table 7-6 )1.3 x ( Mu1 / Vu ) + L0 Ld1.3 x ( 12.81 x 106 / 24 x 103 ) + 8 O 56 O693.875+ 8 O 56 O48 O 693.88which givesO 14.46mm.( O.K.)Check for deflection : -Basic ( span / d ) ratio =20160Pt =100 x Ast / b x d =100 x 462 / 1000 x 160180=0.289IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement400400modification factor = 1.42( span / d ) ratio permissible = 1.42 x 204000=28.4Actual (span / d ) ratio =4000 / 160=25.00 v.( O.K.)Span BC :Vu =17.14 KNShear stress v =Vu / b x d=17.14 x 10 3 / 1000 x 99=0.173N / mm2< c.( O.K.)Pt =100 x As / b x d =100 x 563 / 1000 x 99=0.569Design shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )6 x =0.8 x fck / 6.89 Pt , but not less than 1.0=0.8 x 15 / 6.89 x 0.569=3.06Design shear strength c =0.85 0.8 x 15 ( 1 + 5 x 3.06 - 1 )6 x 3.06=0.487IS 456-2000 Table 19 from table 7-1for Pt = 0.569 c = 0.48N / mm2IS 456-2000 clause 40.2.1.10.25 difference0.08k =1.3for 120 mm slab depth0.181 difference?0.05792Design shear strength =1.3 x 0.48=0.624N / mm2> v.( O.K.)Check for deflection : -For span AB :Basic ( span / d ) ratio =20Pt =100 x Ast / b x d =100 x 462 / 1000 x 99=0.46712001200IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 2120125( span / d ) ratio permissible = 2 x 20=40Actual (span / d ) ratio =3000 / 99=30.30 81.23 mm.( O.K.)Mu / b x d2 ( + ) =10.17 x 10 6 / 1000 x (120)2=0.706Pt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fck=50 1-1-(4.6 / 15) x (0.706)415 / 15=50 [(1-0.885) x 15 / 415 ]=0.208%Ast =0.208 x 1000 x 120 / 100=250mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 250=200.96mmProvide 8 mm O bar@ 200 mm c/c= 251mm2.Mu / b x d2 ( - ) =13.66 x 10 6 / 1000 x (130)2=0.81Pt =50 1-1-(4.6 / fck) x (Mu / b x d2)fy / fck=50 1-1-(4.6 / 15) x (0.81)415 / 15=50 [(1-0.867) x 15 / 415 ]=0.240%Ast =0.24 x 1000 x 130 / 100=312mm2Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 312=161mmProvide 8 mm O bar@ 150 mm c/c= 335mm2.For HYSD Fe415Minimum steel =( 0.12 / 100 ) x 1000 x 150=180mm2At discontinuous edges 4 and 5 , 50 % of the positive steel is required at top=( 1 / 2 ) x 251=126mm2This is less than minimum , therefore , use minimum steel at location 4 and 5 .Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 180=279.1111111111mmProvide 8 mm O bar@ 260 mm c/c= 193mm2.More steel is provided to match with the torsion reinforcement.In edge strip , minimum reinforcement is provided equal to 8 mm O @ 260 mm c/c.Torsion steel :-At corner A , steel required = ( 3/4 ) x 250=188mm2.Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 188=267.2340425532mmProvide 8 mm O bar@ 260 mm c/c= 193mm2.This will be provided by minimum steel of edge strip,At corner B , steel required = ( 1/2 ) x 188=94mm2.Provide 8 mm O barspacing of bar =Area of one bar x 1000 / required area in m2 / m=50.24 x 1000 / 180=279.1111111111mmProvide 8 mm O bar@ 260 mm c/c= 193mm2.This will be provided by minimum steel .Note that positive reinforcements are not curtailed because if they are curtailed , the remaining bars do not provide minimum steel.Check for shear : -At point 1 or 3S.F. =w x l / 2 + Moment @ point 1 or 3 in that span=11.625 x 5 / 2 + 13.66 / 5=31.795KN100 x As / b x d =100 x 335 / ( 1000 x 130 )=0.258from table 7-1for Pt = 0.258 , c = 0.354N / mm20.25 difference0.11IS 456-2000 clause 40.2.1.10.242 difference?-0.10648k =1.3for 150 mm slab depthDesign shear strength =1.3 x 0.354=0.460N / mm2.( O.K.)ORDesign shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )6 x =0.8 x fck / 6.89 Pt , but not less than 1.0=6.75Design shear strength c =0.85 0.8 x 15 ( 1 + 5 x 6.75 - 1 )6 x 6.75=0.356IS 456-2000 clause 40.2.1.1k =1.3for 150 mm slab depthDesign shear strength =1.3 x 0.356=0.463N / mm2.( O.K.)Actual shear stress =Vu / b x d=31.795 x 103 / ( 1000 x 130 )=0.245N / mm2< c.( O.K.)At point 4 or 5S.F. =w x l / 2=11.625 x 5 / 2=29.06KN100 x As / b x d =100 x 251 / ( 1000 x 120 )=0.209from table 7-1for Pt = 0.209 , c = 0.321N / mm20.1 difference0.07IS 456-2000 clause 40.2.1.10.041 difference?-0.0287k =1.3for 150 mm slab depthDesign shear strength =1.3 x 0.321=0.417N / mm2ORDesign shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )6 x =0.8 x fck / 6.89 Pt , but not less than 1.0=8.33Design shear strength c =0.85 0.8 x 15 ( 1 + 5 x 8.33 - 1 )6 x 8.33=0.326IS 456-2000 clause 40.2.1.1k =1.3for 150 mm slab depthDesign shear strength =1.3 x 0.326=0.424N / mm2Actual shear stress =Vu / b x d=29.06 x 103 / ( 1000 x 120 )=0.242N / mm2< c.( O.K.)Check for development length : -This is critical at point 4 or 5.At point 4 or 5 ,Vu =29.06KNNo bar is curtailed or bent up.Assuming L0 =8 O(HYSD Fe415 steel )Pt =100 x As / b x d=100 x 251 / 1000 x 120=0.209From equationORPt =50 1-1-(4.6 / fck) x (Mu / b x d2)Mu1 =0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck )fy / fck=0.87 x 415 x 251 x 120 ( 1 - 415 x 251 / 1000 x 120 x 15 ) x 10 -6we get ,Mu1 / b x d2 =0.711=10.874826( 1 - 0.0579 )Mu1 =0.711 x 1000 x 1202 x 10-6=10.246KNm=10.24KNmDevelopment length of bars Ld =O s / 4 x bd=54.3 O(From Table 7-6 )1.3 x ( Mu1 / Vu ) + L0 Ld1.3 x ( 10.246 x 106 / 29.06 x 103 ) + 8 O 54.3 O458.3551273228+ 8 O 54.3 O46.3 O 458.3551273228which givesO 9.90mm.( O.K.)Short spanVu =11.25 x ( 4.5 / 2 )= 25.31KNAssuming L0 =8 O(HYSD Fe415 steel )Pt =100 x As / b x d=100 x 462 / 1000 x 160=0.289From equationORPt =50 1-1-(4.6 / fck) x (Mu / b x d2)Mu1 =0.87 x fy x Ast x d ( 1 - fy x Ast / b x d x fck )fy / fck=0.87 x 415 x 462 x 160 ( 1 - 415 x 462 / 1000 x 160 x 15 ) x 10 -6we get ,Mu1 / b x d2 =0.9595=26.688816( 1 - 0.0799 )Mu1 =0.9595 x 1000 x 1602 x 10-6=24.55388KNm=24.56KNmDevelopment length of bars Ld =O s / 4 x bd=56 O(From Table 7-6 )1.3 x ( Mu1 / Vu ) + L0 Ld1.3 x ( 24.56 x 106 / 25.31 x 103 ) + 8 O 56 O1261.4776768076+ 8 O 56 O48 O 1261.4776768076which givesO 26.28mm.( O.K.)Note that the bond is usually critical along long direction.Check for deflection : -Basic ( span / d ) ratio =26positive moment steel =251mm2.actual d =150 - 15 -8 - 4=123mm.Pt =100 x Ast / b x d =100 x 251 / 1000 x 123240.7=0.204IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementmodification factor = 1.65( span / d ) ratio permissible = 1.65 x 26=42.9Actual (span / d ) ratio =5000 / 123=40.65 0.7 x 110 KNmmember is not less than 70 percent of the moment at that sectionBC ( - ) = 11.6 KNmobtained from an elastic maximum moment diagram covering allBC ( + ) = 51.6 KNm > 0.7 x 58 KNm.appropriate combinations of loads.Note that after redistribution , the design positive moments also have been reduced.( c ) The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30 % of the numerically largest moment given anywhere by the elastic maximum moments diagram for the particular member , covering all appropriate combination of loads.( b ) Design for flexure :Span AB or CDMu ( + ) = 111.6 KNm.The beam acts as a flanged beamFor T-beams , bf = ( l0 / 6 ) + bw + 6 DfIS 456-2000 ,Clause 23.1.2 Effective width of flange -bf = ( 0.7 x 6000 / 6 ) + 230 + 6 x 120( As per IS456-2000 ,Clause 23.1.2 , Note )In the absence of more accurate determination , the effective width of flange may be taken as the following but in no case greater than the breadth of the web plus half the sum of the clear distances to the adjacent beams on either side :=1650 mm > 3000 mmAssuming one layer of 20 mm diameter barsd = 450 + 120 -25 - 10 = 535 mm .Minimum Ast = ( 0.205 / 100 ) x 230 x 535 = 252 mm2( As per IS456-2000 ,Clause 26.5( a ) )( a ) For T-beams , bf = ( l0 / 6 ) + bw + 6 Dfbf / bw = 1650 / 230 = 7.17For bf /bw = 7For bf /bw = 8( b ) For L-beams , bf = ( l0 / 12 ) + bw + 3 DfDf / d = 120 / 535 = 0.2240.01 diff. 0.0140.016Where ,Mu,lim / fck bw d2 = 0.671( As per SP:16 ,Table 58 )0.006 diff. ??bf = effective width of flange ,Mu.lim = 0.671 x 15 x 230 x 5352 x 10-6-0.0084-0.0096l0 = distance between points of zero moments in the beam ,=662.59656375KNm > 111.6 KNmFor 0.2240.65660.743bw = breadth of the web ,1 diff0.0864Df = thickness of flange , and0.83 diff?0.0717120.671b = actual width of the flange.If Mu < Mu,lim : design as under-reinforced section (singly reinforced beam) as explained below.NOTE - For continuous beams and frames , ' l0 ' may be assumed as 0.7 times the effective span.Ast = Mu / 0.87 x fy x lever armwhere lever arm = d - Df / 2= 535 - 120 / 2 = 535 - 60Ast = 111.6 x 106 / 0.87 x 415 x ( 535 - 60 )IS 456-2000 Clause 26.5 Requirements of Reinforcement for=651mm2 .Structural MembersProvide 6 - 12 mm O = 678 mm2 .26.5.1 BeamsSpan BC :26.5.1.1 Tension ReinforcementMu ( + ) = 51.6 KNma ) Minimum reinforcement - The minimum area of tension reinforcementAst = 51.6 x 106 / 0.87 x 415 x ( 535 - 60 )shall not be less than that given by the following :=301mm2 .As / b d = 0.85 / fyProvide 3 - 12 mm O = 339 mm2 .where ,In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ) from A and at 900 mm ( 0.15 ) from B.As = minimum area of tension reinforcement ,Continue 3 - 12 O in span BC as required for flexure.b = breadth of the beam or the breadth of the web of T- beam ,Support B or C :d = effective depth , andMu ( - ) = 110.4 KNmfy = characteristic strength of reinforcement in N / mm2 .Mu / bd2 = 110.4 x 106 / 230 x 5352 = 1.68 < 2.07.( From Table 6-3 )b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.The section is under-reinforced.Pt = 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )Minimum steel %fy / fckFor mild steelPt = 501 - 1 - ( 4.6 / 15 ) x ( 1.68 )100 As / b d = 100 x 0.85 / 250 = 0.34415 / 15For HYSD steel , Fe415 grade=0.549100 As / b d = 100 x 0.85 / 415 = 0.205Ast =( 0.549 / 100 ) x 230 x 535 = 676 mm2For HYSD steel , Fe500 grade20 % of steel should be carried through the span = 0.2 x 676 = 135 mm2100 As / b d = 100 x 0.85 / 500 = 0.17Provide 2- 10 mm O anchor bars = 157 mm2 . At support , provide 3 - 16 mm O extra at top = 603 mm2 ,one of which may be curtailed at 0.15 = 900 mm from centre of support B and remaining 2 - 16 mm OTable 6-3at 0.25 = 1500 mm from B .Limiting Moment of resistance factor Q lim, N / mm2In span AB , curtail 3 - 12 O at 300 mm ( 0.05 ) from A and at 900 mm ( 0.15 ) from B.For singly reinforced rectangular sectionsThe moment of resistance of remaining barsfck N / mm2fy, N / mm2=0.87 fy Ast ( d - Df / 2 )250415500550=0.87 x 415 x 339 ( 535 - 60 ) x 10-6152.222.072.001.94=58.14KNm202.962.762.662.58Note that lever arm assumed here is d - Df / 2253.703.453.333.23Theoretical cut off point can be found out from case ( c ) as304.444.143.993.8758.14 = 89.6 x - 36 x2 / 2i. e. x2 - 4.98 x + 3.23 = 0check for shearx = ( -b + b2 - 4ac ) / 2 aIS 456-2000 , Table 19=4.98 + 4.982 - 4 x 1 x 3.23 / 2or=( 4.98 - 4.982 - 4 x 1 x 3.23 ) / 2Table 7-1=( 4.98 +3.45 ) / 2=( 4.98 - 3.45 ) / 2Design shear strength of concrete , C, N / mm2x =4.22mx =0.765mPt = 100 x As b x dConcrete gradeAccording to curtailment rules, the bars are cut off at 12 O or deff whichever is greater i.e. 535 mmM15M20M25M30M35M40From A , curtail 3 - 12 O bars at 765 - 535 = 230 mm ( i.e. 230 - 115 = 115 mm from face of support say 300 mm) 0.150.280.280.290.290.290.30and at 4220 + 535 = 4755 mm ( i.e. 1245 mm from B say 300 mm ).0.250.350.360.360.370.370.38The curtailment rules are to be satisfied . The third rule that the continuing bars provide double the area0.500.460.480.490.500.500.51required for flexure at cut off point and the shear does not exceed 3 / 4 th that permitted.0.750.540.560.570.590.590.60Moment at 0.230 m from A1.000.600.620.640.660.670.68=89.6 x 0.23 - ( 0.232 / 2 ) x 360.7451.250.640.670.700.710.730.74=19.66KNm < ( 1 / 2 ) x 58.14 KNm.( O.K.)1.500.680.720.740.760.780.79Moment at 4.755 m from A1.750.710.750.780.800.820.84=89.6 x 4.755 - ( 4.7552 / 2 ) x 362.000.710.790.820.840.860.88=19.07KNm < ( 1 / 2 ) x 58.14 KNm.( O.K.)2.250.710.810.850.880.900.92Minimum design shears at cut off points, referring to design shear diagram .2.500.710.820.880.910.930.95Vu =4 / 3 [ 81.32 ] = 108.43 KN at 0.23 m from A< 89.6 KN ,take 81.32 KN2.750.710.820.900.940.960.98Vu =4 / 3 [ 81.56 ] = 108.75 KN at 1.245 m from B.< 126.4 KN3.000.710.820.920.960.991.01Negative moment reinforcementThe above given table is based on the following formula20 % of steel should be carried through the span = 0.2 x 676 = 135 mm2Design shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )Provide 2- 10 mm O anchor bars = 157 mm2 . At support , provide 3 - 16 mm O extra at top = 603 mm2 ,6 x one of which may be curtailed at 0.15 = 900 mm from centre of support B and remaining 2 - 16 mm O =0.8 x fck / 6.89 Pt , but not less than 1.0at 0.25 = 1500 mm from B .For negative moment top is tension and bottom is compression.The theoretical cut-off points and actual positions for curtailing 40 % bars i.e. 2 - 16 O may now be determined .IS 456-2000 , Table 20Remaining bars 2-10 O & 2-16 O provide 157 + 402 = 559 mm2 area.Table 7-2The strength of section is controlled by top reinforcement only and is given approximately byMaximum shear stress , C, N / mm2M.R. = 0.87 fy Ast ( d - Df / 2 )Concrete gradeM15M20M25M30M35M40M.R. = 0.87 x 415 x 559 ( 535 - 60 ) x 10-6( c )max N/mm22.52.83.13.53.74.0=0.20 x 475=95.00KNmIS 456-2000 Clause 26.5.1.6 Minimum shear reinforcementThe theoretical cut off can be given by an equationMinimum shear reinforcement in the form of stirrups shall be provided89.6 x - 36 x2 / 2 = -95 [ case ( a ) ]such that :Asv / b sv 0.4 / 0.87 fyx2 - 4.98 x - 5.28 = 0where,x = ( -b + b2 - 4ac ) / 2 aAsv =total cross-sectional area of stirrup legs effective in shear ,=4.98 + 4.982 + 4 x 1 x 5.28 / 2Sv =stirrup spacing along the length of the member ,=( 4.98 +6.776 ) / 2b =breadth of the beam or breadth of the web of flanged beam , andx =5.88m from Afy =characteristic strength of the stirrup reinforcement in N / mm2i.e. 0.12 m from Bwhich shall not be taken greater than 415 N / mm2 .Cut off 3 - 16 O bars at 120 + 535 ( deff ) = 655 mm say 900 mm from B .At 900 mm from B i.e. at 5.1 m from A , the moment is89.6 x 5.1 - 36 x 5.12 / 2 = -11.22 KNmMoment capacity required at actual cut off point as per curtailment rules = 2 ( -11.22 ) = - 22.44 KNmMoment capacity of the section = - 95 KNm( O.K. )Note that shear capacity will be increased as per positive reinforcement curtailment check.Let us now determine the point of curtailment of 80 % of bars.Remaining bars are 2-10 O , i.e. 157 mm2Moment capacity = 0.87 fy Ast ( d - Df / 2 )check for development length=0.87 x 415 x 157 ( 535 - 60 ) x 10-6IS 456-2000 clause 26.2.1=0.057 x 475Development length of bars Ld =O s / 4 x bd=27.08KNmIS 456-200 clause 26.2.1.1Distance from A for this momentTable 7-589.6 x - 36 x2 / 2 = - 27.08Design bond stress (bd ) for plain bars in tensionx2 - 4.98 x - 1.50 = 0Concrete gradeM15M20M25M30M35M40x = ( -b + b2 - 4ac ) / 2 a( bd N / mm21.01.21.41.51.71.9=4.98 + 4.982 + 4 x 1 x 1.50 / 2Note-1 : bd shall be increased by 25 % for bars in compression=( 4.98 +5.55 ) / 2Note-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of bd shall be increased by 60 %.x =5.27m from Ai.e. at 0.73 m from B .For mild steel Fe250s =0.87 x fyCut off the bars at 730 + 535 ( deff ) = 1265 mm say 1500 mm from B.For Fe415s =0.67 x fy20 % bars shall be continued throughout the span. Note that this is required for case ( c ) in span BCIS 456-2000 clause 26.2.3.3where point of contraflexure lies at 3 m from B.Ld M1 / V + L0Continue 2 no. 10 mm O bars giving Ast = 157 mm2 area . These will be used as anchor bars.L0 = effective depth of the members or 12 O , whichever is greater( c ) design for shear reinforcement :if ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )At A , Vu = 89.6 KN also Vu = 81.32 KN upto 0.23 m from A100 As / bd = 100 x 3 x 113 / 230 x 5350.25 diff. 0.11Table 7-6=0.280.22 diff. ?-0.10Development length for single mild steel bars( From IS 456-2000 , table 19 table 7-1 )fy N / mm2Tension barsCompression barsDesign ( permissible )shear strength of concrete , C, = 0.36 N / mm2M15M20M15M20Actual shear strength v =Vu / bd25055 O26 O44 O37 O=81.32 x 103 / ( 230 x 535 )41556 O47 O45 O38 O=0.661N / mm2> c50069 O58 O54 O46 Oshear design is necessary .At support ,check for deflectionVus = Vu - c b dORVus = ( 0.661 - 0.36 ) 230 x 535 x 10-3Basic values of span to effective depth ratios for spans upto 10 m :=81.32 - 0.36 x 230 x 535 x 10-3=37.03805KNcantilever7=81.32 -44.298simply supported20=37.02KNcontinuous26Assuming 6 mm O two - legged stirrups , Asv = 57 mm2 , fy = 415 N / mm2 .For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.Sv = 0.87 fy Asv d / Vus=0.87 x 250 x 57 x 535 / 37.02 x 103=179mmcheck for crackingFrom IS 456-2000 clause 26.5.1.6IS 456-2000 26.3.3 Maximum distance between bars in tensionSpacing of minimum shear reinforcement using 6 mm O stirrupsTable 15 Clear distance Between Bars ( Clause 26.3.3 )=0.87 Asv fy / 0.4 b=0.87 x 57 x 250 / 0.4 x 230fy% redistribution to or from section considered=134.8mm-30-150+15+30spacing should not exceedClear distance between bars( i ) 450 mmN / mm2mmmmmmmmmm( ii ) 0.75 d = 0.75 x 535 = 401.25 mm250215260300300300( iii ) 134.8 mm ( minimum )415125155180210235( iv ) 179 mm ( designed )500105130150175195Minimum shear reinforcement of 6 mm O @ 130 mm c/c will be used .capacity of section with minimum shear reinforcementVus = 0.87 fy Asv d / Sv=( 0.87 x 250 x 57 x 535 / 130 ) x 10-3=51KN> 37.02 KN.( O.K.)At B , Vu = 126.4 KN also Vu = 108.75 KN upto 1.245 m from B.2-10 O + 3-16 O= 157 + 603 = 760 mm2100 As / bd = 100 x 760 / 230 x 5350.25 diff. 0.08=0.620.13 diff. ?-0.04( From IS 456-2000 , table 19 table 7-1 )Design ( permissible )shear strength of concrete , C, = 0.50 N / mm2Actual shear strength v =Vu / bd=108.75 x 103 / ( 230 x 535 )=0.884N / mm2> cshear design is necessary .At support ,Vus = Vu - c b dORVus = ( 0.884 - 0.50 ) 230 x 535 x 10-3=108.75 - 0.50 x 230 x 535 x 10-3=47.2512KN=108.75 -61.525=47.23KNAssuming 6 mm O two - legged stirrups , Asv = 57 mm2 , fy = 415 N / mm2 .Sv = 0.87 fy Asv d / Vus=0.87 x 250 x 57 x 535 / 47.23 x 103=140mmFrom IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 mm O stirrups=0.87 Asv fy / 0.4 b=0.87 x 57 x 250 / 0.4 x 230=134.8mmspacing should not exceed( i ) 450 mm( ii ) 0.75 d = 0.75 x 535 = 401.25 mm( iii ) 134.8 mm ( minimum )( iv ) 140 mm ( designed )Minimum shear reinforcement of 6 mm O @ 130 mm c/c will be used .capacity of section with minimum shear reinforcementVus = 0.87 fy Asv d / Sv=( 0.87 x 250 x 57 x 535 / 130 ) x 10-3=51KN> 47.25 KN.( O.K.)( d ) Check for development length :At A, Mu1 = 58.14 KNm ( Calculated previously )Vu = 89.6 KNL0 = 12 OLd = 56.4 Ocheck1.3 Mu1 / Vu + L0 Ld1.3 x 58.14 x 106 / 89.6 x 103 + 12 O 56.4 O843.5491071429 44.4 OO 19 mm.( O.K.)At point of contraflexure i.e. 1.245 m from BMu1 = 58.14 KNmVu = 126.4 - 1.245 x 36 = 81.58 KN115.6L0 = 12 O ( L0 is much more than this . It can be taken equal to deff.However to permitthe lapping of bars we shall consider 12 O . )check1.3 Mu1 / Vu + L0 Ld1.3 x 58.14 x 106 / 81.58 x 103 + 12 O 56.4 O926.4770777151 44.4 OO 21 mm.( O.K.)( e ) Check for deflection :Basic span / d ratio = 26100 Ast / b d = 100 x 678 / 1650 x 535 = 0.077modification factor = 2IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementbw / bf = 230 / 1650 = 0.14< 0.3IS 456-2000 clause 23.2.1 fig-6 ,for flanged beamReduction factor = 0.8Span / d permissible = 26 x 2 x 0.8 = 41.6Actual span / d = 6000 / 535 = 11.21( safe )( d ) Check for cracking (spacing of bars ) :Clear distance between bars=( 230 - 50 - 3 x 12 ) / 2 = 72 mmMinimum clear distance permitted=hagg + 5 mm =20 + 5 = 25 mm or 12 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted=180 mm( cracking - table 8-1 , IS 456-200 , table 15 )( safe )

B10B11B126 m6 m6 m6 m3 m3 m36 KN / m36 KN / m22 KN / m6 m6 m6 mABCD36 KN / m16236 KN / m162ABBC36 KN / m16222 KN / m99BCCD36 KN / m36 KN / m22 KN / m6 m6 m6 mABCD1389616216299( - )( + )( - )( + )96( - )( - )( + )( + )( + )935145138851312.36 m508211510122 KN / m36 KN / m22 KN / m6 m6 m6 mABCD22 KN / m9936 KN / m162ABBC36 KN / m16222 KN / m99BCCD22 KN / m36 KN / m22 KN / m6 m6 m6 mABCD1041049916299( - )( - )( + )( - )( - )( + )( + )( + )47475810448.6783.332.21 m48.6783.3310810810436 KN / m22 KN / m36 KN / m6 m6 m6 mABCD36 KN / m16222 KN / m99ABBC22 KN / m9936 KN / m162BCCD36 KN / m22 KN / m36 KN / m6 m6 m6 mABCD10410416299162( - )( + )( - )( + )104( - )( - )( + )( + )1101105.010490.67125.332.52 m90.67125.33666689.6 KN126.4 KN110.4 KN110.4 KN126.4 KN89.6 KN2.49 m3.51 m90090015001500900150015009003003003003003 - 12 O6- 12 O6- 12 O2 - 10 O2 - 10 O + 3 - 16 O ( Extra )2 - 10 O + 2 - 16 O2 - 10 O2 - 10 O600060006000AABBD N SD N SD N SD N S6 mm O @ 130 mm c/c6 mm O @ 130 mm c/c6 mm O @ 130 mm c/c6 - 12 O20 O pin3 - 12 O2 - 10 O2-10 O + 3-16 O Extra at top230230450120450120Section A - ASection B - B120.1510.1520.1510.1520.2510.2520.11At least 20 % of steel placed over support Bto be continued through beamCut off not more than 40 % of steel at BCut off not more than 40 % of steel placed over support BCut off not more than 70 % of steel required at mid span

Cantilever BeamDesign of Cantilever BeamSpan = 3 mCharacteristic U.D.L. = 12 KN /mAssume that sufficient safety against overturning is there( Given )and reinforcement anchorages are also available.MaterialM15 grade concretemild steel reinforcement.Solution : -Assume an initial trial section of 230 mm x 600 mm overall depth to consider self-weight.The self-weight =0.23 x 0.6 x 25 =3.45KN / mTotal load =12 + 3.45 = 15.45KN / mFactored load =1.5 x 15.45 = 23.2 KN / mFactored S.F. =w x = 23.2 x 3 = 69.6 KNFactored B.M. =w x 2 / 2 = 23.2 x 32 / 2 = 104.4 KNmDepth required = M / Q b= ( 104.4 x 106 ) / ( 2.22 x 230 )( From Table 6-3 )= 452mmUsing one layer of 20 mm O bars and overall depth of 550 mmd = 550 - 25 - 10 = 515 mmORMu / b d2 == 104.4 x 106 / ( 230 x 5152)Mu,lim =2.22 x 230 x 5152 x 10-6=1.71< 2.22 ( Table 6-3 )=135.423885KNmThe section is singly reinforced ( under-reinforced )Mu < Mu,limPt = 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )fy / fckPt = 501 - 1 - ( 4.6 / 15 ) x ( 1.71 )250 / 15=0.93Ast =( 0.93 / 100 ) x 230 x 515 = 1102 mm2Provide 4 - 20 mm O giving Ast = 4 x 314 = 1256 mm2 .Check for development length :Ld = 1102 / 1256 = 55 O( From Table 7-6 )IS 456-2000 clause 26.2.1( From Table 7-5 )Development length of bars Ld =O s / 4 x bd=O ( 0.87 x 250 x 1102 ) / ( 1256 x 4 x 1.0 )=47.71O=47.71 x 20=954mmThe bar shall extend into the support for a straight length of 954 mm . Provideanchorage of 1200 mm . If in some case the bars are to be bent e.g. anchored inIS 456-2000 , Table 19column , the bearing stress around the bend has to be checked as discussedTable 7-1in the next example .Design shear strength of concrete , C, N / mm2Check for shear :Pt = 100 x As b x dConcrete gradeVu = 69.6 KNM15M20M25M30M35M40Actual shear strength v =Vu / bd 0.150.280.280.290.290.290.30=69.6 x 103 / ( 230 x 515 )0.250.350.360.360.370.370.38=0.588N / mm20.500.460.480.490.500.500.51100 x As / b d = ( 100 x 1256 ) / ( 230 x 515 )( From IS 456-2000 , table 19 table 7-1 )0.750.540.560.570.590.590.60=1.060.25 difference0.041.000.600.620.640.660.670.68design ( permissible ) shear strength c =0.61N / mm2> v0.19 difference?-0.03041.250.640.670.700.710.730.74Provide minimum shear reinforcement . For 230 mm wide beam ,1.500.680.720.740.760.780.79From IS 456-2000 clause 26.5.1.61.750.710.750.780.800.820.84Spacing of minimum shear reinforcement using 6 O stirrups2.000.710.790.820.840.860.88=0.87 Asv fy / 0.4 b2.250.710.810.850.880.900.92=0.87 x 56 x 250 / 0.4 x 2302.500.710.820.880.910.930.95=132.4mm2.750.710.820.900.940.960.98spacing should not exceed3.000.710.820.920.960.991.01( i ) 450 mmThe above given table is based on the following formula( ii ) 0.75 d = 0.75 x 515 = 386 mmDesign shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )( iii ) 132.4 mm ( minimum )6 x ( iv ) 569.2 mm ( designed ) =0.8 x fck / 6.89 Pt , but not less than 1.0Provide 6 mm O two-legged stirrups @ 130 mm c/cCheck for deflection :Basic span / d ratio = 7100 Ast / b d = 100 x 1256 / 230 x 515 = 1.06modification factor = 1.4IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.4 = 9.8Actual span / d = 3000 / 515 = 5.83< 9.8( safe )Check for cracking (spacing of bars ) :Clear distance between bars=( 230 - 50 - 4 x 20 ) / 3 = 10 mmMaximum clear distance permitted=300 mm( cracking - table 8-1 , IS 456-2000 , table 15 )( safe )The design beam is shown in fig.Design of Cantilever BeamA cantilever rectangular bracket projects from a column of size 230 mm x 500 mmin the direction of 500 mm for a length of 3 mFactored load of 20 KN / m inclusive of self - weight( given )MaterialM15 grade concreteHYSD reinforcement of grade Fe415 .Solution : -Vu =w x = 20 x 3 = 60 KNMu =w x 2 / 2 = 20 x 32 / 2 = 90 KNm( a ) Moment steel :Take size of Beam 230 mm x 550 mm overall.Assuming 16 mm diameter bars in one layerd = 550 - 25 ( Cover ) -8OR=517mmMu,lim =2.07 x 230 x 5172 x 10-6Mu / b d2 =90 x 106 / ( 230 x 5172)=127.2562929KNm=1.46< 2.07 ( Table 6-3 )Mu < Mu,limThe section is singly reinforced ( under-reinforced ) BeamPt = 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )fy / fckPt = 501 - 1 - ( 4.6 / 15 ) x ( 1.46 )415 / 15=0.464Ast =( 0.464 / 100 ) x 230 x 517 = 552 mm2Provide 16 mm O - 3 No. = 603 mm2.Use 2- 10 mm O as bottom anchor bars .( b ) Check for development length :IS 456-2000 clause 26.2.1Development length of bars Ld =O s / 4 x bdStress in bar s =0.87 x fy= 0.87 x 415 x 552 / 603 =330.5N / mm2( bd = 1.6 x 1 = 1.6 N / mm2IS 456-2000 clause 26.2.1( From Table 7-5 )Development length of bars Ld =O s / 4 x bd=O 330.5 / ( 4 x 1.6 )=51.64O=51.64 x 16=826mmThe arrangement of anchoring the bars is shown in fig. and is equivalent to an anchorageof 1127 mm. The bearing stress inside the bend is now checked .O = 16 mma = 82 mm for internal bar ( centre to centre distance between bar )( = ( 230 - 50 - 3 x 16 ) / 2 + 16 = 82 mm )= 25 + 16 = 41 mm for external bar .( = cover + dia of bar )The bearing stress is critical in external bar . Check for this stress for a = 41 mm.IS 456-2000 Clause 26.2.2.5 Bearing stresses at bendsDesign bearing stress = 1.5 x fck / [1 + ( 2O/a ) ]The bearing stress in concrete for bends and hooks described in IS : 2502-1963 need not be checked . The bearing stress inside a bend in any other bend shall be calculated as given below :=1.5 x 15 / [ 1 + ( 2 x 16 / 41 ) ]=22.5 / 1.78=12.6404494382N / mm2Bearing stress = Fbt / r OAt the centre of bend , the anchorage available = 279 + 147 = 426 mmWhere ,Stress in bar at centre of the bend = 330.5 x ( 826 - 426 ) / 826Fbt = tensile force due to design loads in a bar or group of bars,=160N / mm2r = internal radius of the bend , andFbt =160 x 201 x 10-3 = 32.16 KN( area of 16 O bar = 201 mm2 )O = size of the bar or , in bundle , the size of bar of equivalent areaBearing stress = Fbt / r OFor Limit state method of design , this stress shall not exceed 1.5 fck / [ ( 1+ 2O/a ) ]=32.16 x 103 / 180 x 16Where,=11.16 N / mm2 < 12.64 N /mm2..( safe )fck is the characteristic strength of concrete andThe arrangement is thus satisfactory.a is the centre to centre distance between bars or group of bars , for a bar or group of bars adjacent to the face of the member a shall be taken as the cover plus size of bar ( O )( c )Check for shear :Vu = 60 KNActual shear strength v =Vu / bd=60 x 103 / ( 230 x 517 )=0.505N / mm2100 x As / b d = ( 100 x 603 ) / ( 230 x 517 )( From IS 456-2000 , table 19 table 7-1 )=0.5070.25 difference0.08design ( permissible ) shear strength c =0.46N / mm2< v0.243 difference?-0.07776shear design is necessary .At support ,Vus = Vu - c b dORVus = ( 0.505 - 0.46 ) x 230 x 517 x10-3=60 - 0.46 x 230 x 517 x 10-3=5.35KN=60 -54.6986=5.3KNUsing 6 mm O ( mild steel ) Two legged stirrups , Asv = 28 x 2 = 56 mm2 .fy = 250 N / mm2Sv = 0.87 fy Asv d / Vus=0.87 x 250 x 56 x 517 / 5.3 x 103=1188mmProvide minimum shear reinforcement . For 230 mm wide beam ,From IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 6 O stirrups=0.87 Asv fy / 0.4 b=0.87 x 56 x 250 / 0.4 x 230=132.4mmspacing should not exceed( i ) 450 mm( ii ) 0.75 d = 0.75 x 517 = 388 mm( iii ) 132.4 mm ( minimum )( iv ) 1188 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/c( d ) Check for deflection :Basic span / d ratio = 7100 Ast / b d = 100 x 603 / 230 x 517 = 0.507modification factor = 1.2IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementSpan / d permissible = 7 x 1.2 = 8.4Actual span / d = 3000 / 517 = 5.8< 8.4( safe )( e ) Check for cracking (spacing of bars ) :Clear distance between bars=( 230 - 50 - 3 x 16 ) / 2 = 66 mmMinimum clear distance permitted=hagg + 5 mm =20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted=180 mm( cracking - table 8-1 , IS 456-2000 , table 15 )( safe )The design beam is shown in fig.

3 m23.2 KN /m69.6 KN-104.4 KNm6 O @ 130 c/cFixed end support120030002-12 O4 -20 Ovvvvvv2305504-20 O2-12 OStirrups 6 O @ 130 c/c279295295108150internal radiusr = 180rr5003000 mm2-10 O3-16 O505502302-10 O3-16 OStirrups6 mm O @ 130 mm c/c230550333382821625

Simply supp. ( doubly ) beamDesign of Simply supported Doubly reinforced BeamSpan = 5 m( simply supported rectangular beam )super-imposed load = 40 KN / mBeam section = 230 mm x 600 mm( Given )MaterialM15 grade concreteHYSD reinforcement of grade Fe415Solution : -DL of beam 0.23 x 0.60 x 25 = 3.45 KN / msuper-imposed load = 40 KN / mTotal43.45 KN / mFactored load = 1.5 x 43.45 = 65.18 KN / mMu =w x 2 / 8=65.18 x 52 / 8Table 6-6=203.6875KNmSTRESS IN COMPRESSION REINFORCEMENT fsc , N / mm2 IN DOUBLY REINFORCED BEAMSVu =w x / 2=65.18 x 5 / 2fy N / mm2d ' / d=162.95KN0.050.10.150.2( a ) Moment steel :250217217217217Assuming 20 mm diameter bars in two layer415355353342329d = 600 - 25 ( Cover ) - 20 - 10OR500424412395370=545mmMu,lim =2.07 x 230 x 5452 x 10-6550458441419380Mu / b d2 =203.69 x 106 / ( 230 x 5452)=141.4136025KNm=2.98> 2.07 ( Table 6-3 )Mu > Mu,limIf Mu > Mu,lim : design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance Mu2 needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. Mu2 = Mu - Mu,lim as explained below .The section is Doubly reinforced ( over-reinforced )Mu,lim =2.07 x 230 x 5452 x 10-6=141.4136025KNmMu2 =Mu - Mu,lim=203.69 - 141.41=62.28KNmLet the compression reinforcement be provided at an effective cover of 40 mm .Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) )d' / d =40 / 545Asc = Mu2 / ( fsc x ( d - d' ) )=0.07consider d' / d = 0.1 .Ast2 = Asc fsc / 0.87 fyStress in compression steel , fsc = 353 N / mm2 ( refer to table 6-6 )Ast = Ast,lim + Ast2 .Asc = Mu2 / ( fsc x ( d - d' ) )If Xu < Xu,max the section is under-reinforced ( singly reinforced )=62.28 x 106 / 353 ( 545 - 40 )If Xu = Xu,max the section is balanced=349mm2If Xu > Xu,max the section is over-reinforced ( doubly reinforced )Corresponding tension steelwhere ,Ast2 = Asc fsc / 0.87 fyXu,max = 0.53 x d ( for Fe250 mild steel )=349 x 353 / ( 0.87 x 415 )Xu,max = 0.48 x d ( for Fe415 HYSD steel )=341.2mm2Xu,max = 0.46 x d ( for Fe500 HYSD steel )Ast,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) )Xu,max = 0.44 x d ( for Fe550 HYSD steel )=141.41 x 106 / ( 0.87 x 415 ( 545 - 0.42 x 0.48 x 545 ) )Xu =( 0.87 fy Ast ) / ( 0.36 fck b )=141.46 x 106 / 361.05 ( 435.13 )=900mm2Ast = Ast,lim + Ast2 .=900 + 341.2 mm2=1241.2mm2ProvideAsc =2-16 O =402mm2Ast =4-20 O =1256mm2( all straight )( b ) Check for development length :As all the bars are taken into support , Mu1 may be taken as Mu .Assume L0 = 12 OLd = 56 O ( From Table 7-6 )1.3 x M1 / V + L0 Ld1.3 x 203.69 x 106 / 162.95 x 103 +12 O 56 O1625.0244 OO36.93mmOprovided = 20 mm ..( safe )( c ) Check for shearVu =162.95KNActual shear strength v =Vu / bd=162.95 x 103 / ( 230 x 545 )=1.3N / mm2100 x As / b d = ( 100 x 1256 ) / ( 230 x 545 )( From IS 456-2000 , table 19 table 7-1 )=1.0design ( permissible ) shear strength c =0.6N / mm2< vAs the ends are confined with compressive reaction , shear at distance d will be used forchecking shear at support . At 545 mm , shear is equal toVu =162.95 - 0.545 x 43.45 = 139.27 KNVus = Vu - c b d=139.27 - 0.6 x 230 x 545 x 10-3 = 64.06 KNAssuming 8 mm O two - legged stirrups , Asv = 100 mm2 , fy = 415 N / mm2 .Sv = 0.87 fy Asv d / Vus=0.87 x 415 x 100 x 545 / 64.06 x 103=307mmFrom IS 456-2000 clause 26.5.1.6Spacing of minimum shear reinforcement using 8 mm O stirrups=0.87 Asv fy / 0.4 b=0.87 x 100 x 415 / 0.4 x 230=392.4mmspacing should not exceed( i ) 450 mm( ii ) 0.75 d = 0.75 x 545 = 408 mm( iii ) 392.4 mm ( minimum )( iv ) 307 mm ( designed )Minimum shear reinforcement of 8 mm O @ 390 mm c/c will be used .At support provide 8 mm O @ 300 mm c/c .Shear resistance of beam with minimum shear reinforcement=( 0.87 fy Asv d / Sv )+ c b d=(0.87 x 415 x 100 x 545 x 10-3 / 390 ) + 0.6 x 230 x 545 x 10-3=50.45+75.21=125.66KNDesigned shear reinforcement is required from face of the support upto thedistance equal to162.95 - 125.66 / 43.45 = 0.858 mNo. of stirrups = ( 858 / 300 ) + 1 = 3.86 say 4Provide 8 mm O two-legged stirrups @ 300 mm c/c upto 4 no. from face of the supportand 8 mm O @ 390 mm c/c in remaining central portion .( d ) Check for deflection :Basic span / d ratio = 20100 Ast / b d = 100 x 1256 / 230 x 545 = 1.0modification factor = 0.97IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcement100 Asc / b d = 100 x 402 / 230 x 545 = 0.32modification factor = 1.08IS 456-2000 clause 23.2.1 fig-5 ,for compression reinforcementSpan / d permissible = 20 x 0.97 x 1.08 = 20.95Actual span / d = 5000 / 545 = 9.17( safe )( d ) Check for cracking (spacing of bars ) :Clear distance between bars=( 230 - 50 - 4 x 20 ) / 3 = 10 mmMinimum clear distance permitted=hagg + 5 mm =20 + 5 = 25 mm or 20 mm ( O of bar ) i.e. 25 mm .Maximum clear distance permitted=180 mm( cracking - table 8-1 , IS 456-2000 , table 15 )( safe )The design beam is shown in fig.

2306005452-16 O4- 20 O6005000 c/c2-16 O4- 20 O ( straight )600DIANO.SPA.8 O8 O8 O44rest300300390bearing

Simply supp. ( singly ) BeamDesign of Simply supported Singly reinforced BeamConsider width of the beam equal to 230 mm. The depth may be assumedSpan = 6 m( simply supported rectangular beam )as 1 / 10 to 1 / 8 of the span.characteristic load = 20 KN / m inclusive of its self-weightTo find steel areaBeam section = 230 mm x 600 mm( 1 ) For a given ultimate moment ( also known as factored moment )MaterialM15 grade concrete( Given )and assumed width of section , find out d from equationHYSD reinforcement of grade Fe415d = Mu / Qlim bThe beam is resting on R.C.C. columns.This is a balanced section and steel area may be found out from table Pt,lim , SP : 16 ,2.3Solution : -Factored load = 1.5 x 20 = 30 KN /m.Mu =w x 2 / 8( 2 ) For a given factored moment ,width and depth of section .=30 x 62 / 8Obtain Mu,lim = Qlim bd2 .=135KNmIf Mu < Mu,lim : design as under-reinforced section (singly reinforced beam) as explained below.Vu =w x / 2=30 x 6 / 2Pt = 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )=90KNfy / fck( a ) Moment steel :If Mu = Mu,lim : design as balanced section as explained in ( 1 ).Assuming 20 mm diameter bars in one layerIf Mu > Mu,lim : design the section either increasing the dimensions of section or deign as Over reinforced section ( doubly-reinforced beam ).The additional moment of resistance Mu2 needed is obtained by providing compression( top ) reinforcement and additional tensile reinforcement. Mu2 = Mu - Mu,lim as explained below .d = 600 - 25 ( Cover ) -10OR=565mmMu,lim =2.07 x 230 x 5652 x 10-6Mu / b d2 =135 x 106 / ( 230 x 5652)=151.9830225KNm=1.84< 2.07 ( Table 6-3 )Mu < Mu,limAst,lim = Mu,lim / ( 0.87 fy ( d - 0.42 Xu,max ) )The section is singly reinforced ( under-reinforced )Asc = Mu2 / ( fsc x ( d - d' ) )Pt = 501 - 1 - ( 4.6 / fck) x ( Mu / bd2 )Ast2 = Asc fsc / 0.87 fyfy / fckAst = Ast,lim + Ast2 .Pt = 501 - 1 - ( 4.6 / 15 ) x ( 1.84 )If Xu < Xu,max the section is under-reinforced ( singly reinforced )415 / 15If Xu = Xu,max the section is balanced=0.614If Xu > Xu,max the section is over-reinforced ( doubly reinforced )Ast =( 0.614 / 100 ) x 230 x 565 = 798 mm2where ,As per IS 456-2000 clause 26.5.1.1 ( a )Xu,max = 0.53 x d ( for Fe250 mild steel )Minimum steel , As = ( 0.205 / 100 ) x 230 x 565 = 266 mm2Xu,max = 0.48 x d ( for Fe415 HYSD steel )From Table 6-4Xu,max = 0.46 x d ( for Fe500 HYSD steel )Ast,lim = 0.72 / 100 x 230 x 565 = 936 mm2.Xu,max = 0.44 x d ( for Fe550 HYSD steel )Provide 16 mm O - 4 No. = 804 mm2.Xu =( 0.87 fy Ast ) / ( 0.36 fck b )Let 2 bars are bent at 1.25 D = 1.25 x 600 = 750 mm , from the face of the support .( b ) Check for development length :Table 6-2( 1 ) A bar can be bent up at a distance greater than Ld = 56 O ( Table 7-6 )Limiting Moment of Resistance and ReinforcementFrom the centre of the support , i.e. 56 x 16 = 896 mm .Index for Singly Reinforced Rectangular Sectionsin this case , the distance is ( 3000 - 750 ) = 2250 mm .( safe )fy , N / mm2250415500550( 2 ) For the remaining bars , Ast = 402 mm2Mu,lim / fck b d20.1480.1380.1330.129Mu1 = 0.87 fy Ast d ( 1 - [( fy Ast ) / ( fck b d ) ] )Pt,lim fy / fck21.9319.8619.0318.2=0.87 x 415 x 402 x 565 ( 1 - [ ( 415 x 402 ) / ( 15 x 230 x 565 ) ] ) x 10-6=82.01x0.91Table 6-3=74.99KNmLimiting Moment of resistance factor Q lim, N / mm2Vu =90 KN ,L0 = 12 O ( assume )For singly reinforced rectangular sectionsAs the reinforcement is confined by compressive reactionfck N / mm2fy, N / mm21.3 x M1 / V + L0 Ld2504155005501.3 x 74.99 x 106 / 90 x 103 +12 O 56 O152.222.072.001.941083.1944 O202.962.762.662.58O24.62mm253.703.453.333.23Oprovided = 16 mm ..( safe )304.444.143.993.87( c ) Check for shear :At support , Vu = 90 KNTable 6-4As the ends of the reinforcement are confined with compressive reaction , shear at distance dLimiting Percentage of Reinforcement Pt,limwill be used for checking shear at support.For singly reinforced rectangular sectionsVu = 90 - 0.565 x w = 90 - 0.565 x 30 = 73.05 KNfck N / mm2fy, N / mm2Actual shear strength v =Vu / bd250415500550=73.05 x 103 / ( 230 x 565 )151.320.720.570.50=0.562N / mm2201.750.960.760.66100 x As / b d = ( 100 x 402 ) / ( 230 x 565 )( From IS 456-2000 , table 19 table 7-1 )252.191.200.950.83=0.3090.25 difference0.11302.631.441.140.99design ( permissible ) shear strength c =0.376N / mm2< v0.191 difference?-0.08404shear design is necessary .IS 456-2000 Clause 26.5 Requirements of Reinforcement forAt support ,Vus = Vu - c b dORVus = ( 0.562 - 0.376 ) x 230 x 565 x10-3Structural Members=73.05 - 0.376 x 230 x 565 x 10-3=24.17KN26.5.1 Beams=73.05 -48.8626.5.1.1 Tension Reinforcement=24.19KNa ) Minimum reinforcement - The minimum area of tension reinforcementCapacity of bent bars to resist shearshall not be less than that given by the following :=2 x 201 x 0.87 x 415 x sin 45 x 10-3( 0.87 fy Asv sin )As / b d = 0.85 / fy=102.6KNwhere ,Bent bars share 50 % = 12.09 KNAs = minimum area of tension reinforcement ,Stirrups share 50 % = 12.09 KNb = breadth of the beam or the breadth of the web of T- beam ,Using 6 mm O ( mild steel ) Two legged stirrups , Asv = 28 x 2 = 56 mm2 .d = effective depth , andSv = 0.87 fy Asv d / Vusfy = characteristic strength of reinforcement in N / mm2 .=0.87 x 250 x 56 x 565 / 12.09 x 103b ) Maximum reinforcement - The maximum area of tension reinforcement shall not exceed 0.04 b D.=569.2mmFrom IS 456-2000 clause 26.5.1.6Minimum steel %Spacing of minimum shear reinforcement using 6 O stirrupsFor mild steel=0.87 Asv fy / 0.4 b100 As / b d = 100 x 0.85 / 250 = 0.34=0.87 x 56 x 250 / 0.4 x 230For HYSD steel , Fe415 grade=132.4mm100 As / b d = 100 x 0.85 / 415 = 0.205spacing should not exceedFor HYSD steel , Fe500 grade( i ) 450 mm100 As / b d = 100 x 0.85 / 500 = 0.17( ii ) 0.75 d = 0.75 x 565 = 423 mm( iii ) 132.4 mm ( minimum )For checking development length , l0 may be assumed as 8 O for HYSD bars ( usually end anchorage is not provided ) and 12 O for mild steel ( U hook is provided usually whose anchorage length is 16 O.( iv ) 569.2 mm ( designed )Provide 6 mm O two-legged stirrups @ 130 mm c/cAt 1.25 D = 750 mm from face of the support where contribution of bent bars is not availableVu = 90 - 0.75 x w = 90 - 0.75 x 30 = 67.5 KNVus = Vu - c b d=67.5 - 0.376 x 230 x 565 x 10-3=67.5 -48.86=18.64KNcheck for development lengthProvide minimum 6 mm O M.S. two -legged stirrups @ 130 mm c/c throught the beam.IS 456-2000 clause 26.2.1( d ) Check for deflection :Development length of bars Ld =O s / 4 x bdBasic span / d ratio = 20IS 456-200 clause 26.2.1.1100 Ast / b d = 100 x 804 / 230 x 565 = 0.62Table 7-5modification factor = 1.1IS 456-2000 clause 23.2.1 fig-4 ,for tension reinforcementDesign bond stress (bd ) for plain bars in tensionSpan / d permissible = 20 x 1.1 = 22Concrete gradeM15M20M25M30M35M40Actual span / d = 6000 / 565 = 10.62( safe )( bd N / mm21.01.21.41.51.71.9( d ) Check for cracking (spacing of bars ) :Note-1 : bd shall be increased by 25 % for bars in compressionClear distance between barsNote-2 : In case of deformed bars confirming to IS : 1786-1985 , the value of bd shall be increased by 60 %.=230 - 50 - 2 x 16 = 148 mmMinimum clear distance permittedFor mild steel Fe250s =0.87 x fy=hagg + 5 mm =20 + 5 = 25 mm or 16 mm ( O of bar ) i.e. 25 mm .For Fe415s =0.67 x fyMaximum clear distance permittedIS 456-2000 clause 26.2.3.3=180 mm( cracking - table 8-1 , IS 456-200 , table 15 )( safe )Ld M1 / V + L0The design beam is shown in fig.L0 = effective depth of the members or 12 O , whichever is greaterif ends of the reinforcement are confined by a compressive reaction M1 / V increased by 30 % ( 1.3 M1 / V )Table 7-6Development length for single mild steel barsfy N / mm2Tension barsCompression barsM15M20M15M2025055 O26 O44 O37 O41556 O47 O45 O38 O50069 O58 O54 O46 Ocheck for shearIS 456-2000 , Table 19Table 7-1Design shear strength of concrete , C, N / mm2Pt = 100 x As b x dConcrete gradeM15M20M25M30M35M40 0.150.280.280.290.290.290.300.250.350.360.360.370.370.380.500.460.480.490.500.500.510.750.540.560.570.590.590.601.000.600.620.640.660.670.681.250.640.670.700.710.730.741.500.680.720.740.760.780.791.750.710.750.780.800.820.842.000.710.790.820.840.860.882.250.710.810.850.880.900.922.500.710.820.880.910.930.952.750.710.820.900.940.960.983.000.710.820.920.960.991.01The above given table is based on the following formulaDesign shear strength c =0.85 0.8 x fck ( 1 + 5 x - 1 )6 x =0.8 x fck / 6.89 Pt , but not less than 1.0IS 456-2000 , Table 20Table 7-2Maximum shear stress , C, N / mm2Concrete gradeM15M20M25M30M35M40( c )max N/mm22.52.83.13.53.74.0IS 456-2000 Clause 26.5.1.6 Minimum shear reinforcementMinimum shear reinforcement in the form of stirrups shall be providedsuch that :Asv / b sv 0.4 / 0.87 fywhere,Asv =total cross-sectional area of stirrup legs effective in shear ,Sv =stirrup spacing along the length of the member ,b =breadth of the beam or breadth of the web of flanged beam , andfy =characteristic strength of the stirrup reinforcement in N / mm2which shall not be taken greater than 415 N / mm2 .check for deflectionBasic values of span to effective depth ratios for spans upto 10 m :cantilever7simply supported20continuous26For spans above 10 m, the values in (a) may be multiplied by 10 / span in metres , except for cantilever in which case deflection calculations should be made.check for crackingIS 456-2000 26.3.3 Maximum distance between bars in tensionTable 15 Clear distance Between Bars ( Clause 26.3.3 )fy% redistribution to or from section considered-30-150+15+30Clear distance between barsN / mm2mmmmmmmmmm250215260300300300415125155180210235500105130150175195

2306005652-10 O4- 16 O7506000 c/c2-10 O4- 16 O ( 2 straight +2 bent )6 mm O @ 130 mm c/c

Slender ( Long ) ColumnDesign of slender ( Long ) columns ( with biaxial bending )IS 456-2000 clause 25Size of column 400 x 300 mmColumn is restrained against sway.Note :- A column may be considered as short when both slenderness ratios lex / D and ley / b 12 OR lex / ixx < 40 where, For Circular column lex / D 10 .Concrete grade M 30Characteristic strength of reinforcement 415 N/mm2Effective length for bending parallel to larger dimension ex = 6.0 mEffective length for bending parallel to shorter dimension ey = 5.0 mlex = effective length in respect of the major axisUnsupported length = 7.0 m( Given )D = depth in respect of the major axisFactored load 1500kNley = effective length in respect of the minor axisFactored.moment in the direction of larger dimension = 40 kNm at top and 22.5 KNm at bottom.b = width of the memberixx = radius of gyration in respect of the major axis.Factored.moment in the direction of shorter dimension = 30 kNm at top and 20 KNm at bottom.iyy = radius of gyration in respect of the minor axis.Solution : -IS 456 : 2000About x axis :Table 28 Effective Length of Compression Members ( Clause E-3 )1 and 2 are the sameDegree of End Restraint of Compression membersSymbolTheoretical Value of effective lengthRecommended Value of effective lengthFor Beam :bf = l0 / 6 + bw + 6 DfEffectively held in position and restrained against rotation in both ends0.5 0.65 =0.7 x 5000 / 6 + 230 + 6 x 120=1533.3333333333bf / bw = 1533.3 / 230 = 6.67Df / D = 120 / 600 = 0.2Effectively held in position at both ends , restrained against rotation at one ends0.7 0.8 Kt from chart 88 , SP : 16 =2.07Beam stiffnessKb = 1.5 x Ib / l=1.5 x ( 2.07 x ( 1 / 12 ) x 230 x 6203 ) / 5000Effectively held in position at both ends , but not restrained against rotation .1.0 1.0 =2836699mm3Column stiffnessKc = Ic / l = 1/12 x 230 x 4003 / 7000=175238.095238095mm3Effectively held in position and restrained against rotation at one end , and at the other end restrained against rotation but not held in position1.0 1.2 1 = 2 = Kc / ( Kc + Kb )=2 x 175238 / 2 ( 175238 + 2836699 )=0.0582as per IS 456-2000 fig. 27Effectively held in position and restrained against rotation at one end , and at the other partially restrained against rotation but not held in position-1.5 lef / l =1.035 < 1.2.consider 1.2lex =1.2 x unsupported length=1.2 ( 7000 - 620 )=7656mmEffectively held in position at one end but not restrained against rotation and at the other end restrained against rotation but not held in position2.0 2.0 lex / D =7656/400=19.14> 12The column is long about x direction.About Y axis :Effectively held in position and restrained against rotation at one end but not held in position nor restrained against rotation at the other end.2.0 2.0 Beam stiffnessKb = 1.5 x Ib / l=1.5 x ( ( 1 / 12 ) x 230 x 4203 ) / 5000=426006mm3NOTE - is the unsupported length of compression member.Column stiffnessKc = Ic / l = 1/12 x 400 x 2303 / 7000=57938.0952380952mm31 = 2 = Kc / ( Kc + Kb )=2 x 57938.1 / 2 ( 57938.1 + 426006 )=0.1197as per IS 456-2000 fig. 27lef / l =1.06 < 1.2.consider 1.2ley =1.2 x unsupported length=1.2 ( 7000 -420 )=7896mmley / b =7896/300=26.32> 12The column is long about Y direction.The column is bent in double curvature. Reinforcement will be distributed equally on four sides.ex / D =6000 / 400=15.0> 12Effective length of column ( ef ) : It is the distance between the points of zero moment (contraflexure ) along the column height .ey / b =5000 / 300=16.7> 12Therefore the column is slender about both the axes.Additional momentsMax =( Pu D / 2000 ) x ( ex / D )2=( 1500 x 400 / 2000 ) x (15)2 x 10-3=67.5KNmMay =( Pu b / 2000 ) x ( ey / b )2=( 1500 x 300 / 2000 ) x (16.7)2 x 10-3=62.75KNmThe above moments will have to be reduced in accordance with clause 39.7.1.1 of the IS 456-2000but multiplication factors can be evaluated only if the reinforcement is known.For first trial , assume p = 3.0 ( with reinforcement equally on all the four sides ).Ag =400 x 300Puz = 0.45 fck Ac + 0.75 fy Asc=120000mm2Ac =400 x 300 - 120000*3/100From chart 63 , puz / Ag = 22.5 N/mm2OR=116400mm2Puz =22.5 x 120000 x 10-3Puz = 0.45 x 30 x 116400 x 10-3 + 0.75 x 415 x 3600 x 10-3=2700KN=1571.4+1120.5Calculation of Pb :=2692KNAssuming 25 mm dia bars with 40 mm coverSP : 16 ,Table 60d' / D ( about xx-axis ) = 52.5 / 400 = 0.13 use d' / D = 0.15Slender compression members- Values of Pbd' / D ( about yy-axis ) = 52.5 / 300 = 0.18 use d' / D = 0.2Rectangular sections : Pb / fck b D = k1 + k2 . p / fckFrom Table 60 , SP 16Circular sections : Pb / fck D2 = k1 + k2 . p / fckPb ( about xx-axis ) = ( k1 + k2 p / fck ) fck b DValues of k1Pbx = ( 0.196 + 0.203 x 3 /30 ) 30 x 300 x 400 x10-3Sectiond' / D=779KN0.050.100.150.20Pb ( about yy-axis ) = ( k1 + k2 p / fck ) fck b DRectangular0.2190.2070.1960.184Pby = ( 0.184 + 0.028 x 3 /30 ) 30 x 300 x 400 x10-3Circular0.1720.1600.1490.138=672KNValues of k2Kx =( Puz - Pu ) / ( Puz - Pbx )Sectiond' / D=( 2700 - 1500 ) / ( 2700 - 779 )fy N/ mm20.050.100.150.20=1200 / 1921Rectangular ; equal reinforcement on two opposite sides250-0.045-0.045-0.045-0.045=0.6254150.0960.0820.046-0.022Ky =( Puz - Pu ) / ( Puz - Pby )5000.2130.1730.104-0.001=( 2700 - 1500 ) / ( 2700 - 672 )Rectangular ; equal reinforcement on four sides2500.2150.1460.061-0.011=1200 / 20284150.4240.3280.2030.028=0.5925000.5450.4250.2560.040The additional moments calculated earlier , will now be multiplied by the above values of k .Circular2500.1930.1480.077-0.020Max =67.5 x 0.625 = 42.2 KNm4150.4100.3230.2010.036May =62.75 x 0.592 = 37.15 KNm5000.5430.4430.2910.056The additional moments due to slenderness effects should be added to the initial momentsIS 456-2000 clause 39.7.1after modifying the initial moments as follows ( see note 1 under 39.7.1 of the code IS 456-2000 )The additional moments Max and May shall be calculated by the following formulae :Mux =( 0.6 x 40 - 0.4 x 22.5 ) =15.0KNmMax =( Pu D / 2000 ) x ( ex / D )2Muy =( 0.6 x 30 - 0.4 x 20 ) =10.0KNmMay =( Pu b / 2000 ) x ( ey / b )2The above actual moments should be compared with those calculated from minimum eccentricityWhere ,consideration ( see 25.4 of the code ) and greater value is to be taken as the initial moment forPu =axial load on the member,adding the additional moments.ex =effective length in respect of the major axis ,ex =( / 500 ) + ( D / 30 ) =( 7000 / 500 ) + ( 400 / 30) =27.3333333333mm> 20 mmey =effective length in respect of the minor axis ,ey =( / 500 ) + ( b / 30 ) =( 7000 / 500 ) + ( 300 / 30) =24mm> 20 mmD =depth of the cross - section at right angles to the major axis , andMoments due to minimum eccentricity :Mux =1500 x 27.33 x 10-3 = 41.0 KNm > 15.0 KNmb =width of the memberMuy =1500 x 24 x 10-3 = 36.0 KNm > 10.0 KNmNOTES : -Total moments for which the column is to be designed are :1) A column may be considered braced in a given plane if lateralMux =41.0 + 42.2 = 83.2KNmstability to the structure as a whole is provided by walls orMuy =36.0 + 37.15 = 73.15KNmbracing or buttressing designed to resist all lateral forces inThe section is to be checked for biaxial bendingthat plane. It should otherwise be considered unbraced.Pu / fck b D =1500 x 103 / ( 30 x 300 x 400 )2 ) In the case of a braced column without any transverse loads=0.417occurring in its height, the additional moment shall be addedp / fck =3 / 30 = 0.10to an initial moment equal to sum of 0.4 Mu1, and 0.6 Mu2,referring to chart 45 (d' / D = 0.15 ) ,where Mu2 is the larger end moment and Mu1 is the smallerMu / fck b D2 =0.104end moment (assumed negative if the column is bent in doubleMux1 =0.104 x 30 x 300 x 4002curvature). In no case shall the initial moment be less than=149.8KNm0.4 Mu2 nor the total moment including the initial moment bereferring to chart 46 (d' / D = 0.2 ) ,less than Mu2. For unbraced columns, the additional momentMu / fck b D2 =0.096shall be added to the end moments.Muy1 =0.096 x 30 x 400 x 30023 ) Unbraced compression members, at any given level or storey,=103.7KNmsubject to lateral load are usually constrained to deflectMux / Mux1 =83.2 / 149.8 =0.56equally. In such cases slenderness ratio for each column mayMuy / Muy1 =73.15 / 103.7 =0.71be taken as the average for all columns acting in the samePu / Puz = 1500 / 2700 = 0.56direction.referring to chart 64 , the maximum allowable value of Mux / Mux1 corresponding to theIS 456-2000 clause 39.7.1.1above values of Muy / Muy1 and Pu / Puz is 0.58 which is slightly higher than the actualThe values given by equation 39.7.1 may be multiplied by the following factor :value of 0.56 . The assumed reinforcement of 3.0 % is therefore satisfactory.K =( Puz - Pu ) / ( Puz - Pb ) 1for Pu / Puz = 0.56 ,n = 1.602where ,From , IS 456-2000 , Clause 39.6Pu =axial load on compression member,n =1.6020.2 difference0.34Puz =as defined in 39.6,Puz = 0.45 fck Ac + 0.75 fy Asccheck :0.04 difference?-0.068Pb =axial load corresponding to the conditionMux+Muy 1of maximum compressive strain ofMux1Muy10.0035 in concrete and tensile strain of0.002 in outer most layer of tension steel.( 0.56 )+( 0.71 )0.395+0.577= 0.972 1..( O.K.)As =p x b x D / 100 = 3.0 x 300 x 400 / 100=3600mm2Provide 25 mm diameter bar ( 3600 / 491 ) = 8 no. = 3928 mm2Ties : -O min =25 / 4=6.25Provide 8 mm O M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( ii ) 16 x 25 = 400 mm( iii ) 48 x 8 = 384 mmProvide 8 mm O M.S. ties @ 300 mm c/cNote that the distance between corner bars in one face is more than 48 Otr( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.Design of slender ( Long ) columns ( with Uniaxial bending )Size of column 230 x 450 mmColumn of a braced frameConcrete grade M 20Characteristic strength of reinforcement 415 N/mm2 HYSD reinforcementUnsupported length in both the direction = 5.0 mFactored load Pu = 1000 kNFactored.moment in the direction of larger dimension Muxx = 80 kNm at top and 60 KNm at bottom.( Given )Factored.moment in the direction of shorter dimension Muyy = 40 kNm at top and 30 KNm at bottom.The column is bent in double curvature and is slender about both the axis.The slenderness ratios ex / Ixx and ey / Iyy are respectively 13.2 and 15.6Assume that the moments due to minimum eccentricities about both the axes are less than applied moments.Solution : -Assume adjustment factor k = 0.8 for the first trial .Additional momentsMax =( Pu D / 2000 ) x ( ex / D )2=( 1000 x 450 / 2000 ) x (13.2 )2 x 10-3=39.2KNmMay =( Pu b / 2000 ) x ( ey / b )2=( 1000 x 230 / 2000 ) x (15.6 )2 x 10-3=28KNmAbout XXPu =1000KNMuxx =Mi + k x MaxMi =0.6 Mu2 + 0.4 Mu1=0.6 x 80 - 0.4 x 60=48 - 24=24 KNm< 0.4 x 80 = 32 KNmTake ,Mi =32 KNmNote that Mu1 is considered negative as the column bends in double curvature .Mu,xx =32 + 0.8 x 39.2=63.36KNm.< 80 KNmTake ,Mu,xx =80 KNmAbout YYPu =1000KNMuyy =Mi + k x MayMi =0.6 Mu2 + 0.4 Mu1=0.6 x 40 - 0.4 x 30=24 - 12=12 KNm< 0.4 x 40 = 16 KNmTake ,Mi =16 KNmMu,yy =16 + 0.8 x 28=38.4KNm.< 40 KNmTake ,Mu,yy =40 KNmFinally design the column forPu =1000 KNMu,xx =80 KNmMu,yy =40 KNmFor the first trial , assume uniaxial bending about y axis for the following values .P'u =1000 KNM'uy =( 230 / 450 ) ( 80 + 40 ) =61.33KNmd' / D =50 / 230 =0.22Say 0.2P'u / fck b D =1000 x 103 / 20 x 230 x 450 = 0.48M'uy / fck b D2 =61.33 x 106 / 20 x 450 x 2302 = 0.129p / fck =from chart 46 , SP : 16 =0.162p =0.162 x 20=3.24Asc =( 3.24 /100 ) x 230 x 450=3353mm2Assumptions made above for k = 0.8 and for uniaxial moment to find out first trial steel areusually conservative . Let us try 4 - 25 O + 4 - 20 O = 3220 mm2Now check the assumed section as follows :Puz = 0.45 fck Ac + 0.75 fy AscAc =230 x 450 - 3220=100280mm2Puz = 0.45 x 20 x 100280 x 10-3 + 0.75 x 415 x 3220 x 10-3=902.52+1002.225=1904.7KNPb = ( k1 + k2 p / fck ) fck b DAsc =( p / 100 ) x b x Dp / fck =( 3220 x 100 ) / ( 230 x 450 x 20 )=0.156For d' / D = 0.2 , k1 = 0.184 and k2 = -0.022 from table 60 ,SP : 16Pb = ( 0.184 - 0.022 x 0.156 ) x 20 x 230 x 450 x 10-3=0.181x2070=373.8KNk =( Puz - Pu ) / ( Puz - Pb )=( 1904.7 - 1000 ) / ( 1904.7 - 373.8 )=904.7 / 1530.9=0.59Design forPu =1000 KNMux =32 + 0.59 x 39.2 = 55.13 KNm < 80Take , Mux = 80 KNmMuy =16 + 0.59 x 28 = 32.52 KNm < 40Take , Muy = 40 KNmFor p / fck = 0.162 and Pu / fck b D = 0.48 , the reinforcement being equally distributed ,the moment capacities can be found out as follows :About XXd' / D =50 / 450 =0.11 0.15From chart 45 , SP : 16Mux1 / fck b D2 = 0.145Mux1 =0.145 x 20 x 230 x 4502 x 10-6=135.07KNmAbout YYd' / D =50 / 230 =0.22 0.2From chart 46 , SP : 16Muy1 / fck b D2 = 0.13Muy1 =0.13 x 20 x 450 x 2302 x 10-6=61.89KNmCheck :Pu / Puz = 1000 / 1904.7 = 0.525From , IS 456-2000 , Clause 39.6n =1.5420.2 difference0.34check :0.075 difference?-0.1275Mux+Muy 1Mux1Muy180+40135.0761.890.446+0.510.956 1.( O.K.)Provide 4 - 25 O + 4 - 20 O equally distributed.Ties : Minimum diameter Otr = 25 / 4= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 230 mm( least lateral dimension )( ii ) 16 x 25 = 400 mm( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 230 mm c/cNote that the distance between corner bars in one face is less than 48 Otr( 450 - 80 -25 = 345 < 48 x 8 = 384 ). Therefore two sets of Opened ties shall be used .Note that if this distance would be more than ( 48 x 8 ) mm , closed ties for internal bars would be sufficient.

( i + 1 ) th floor( i ) th floorslabBeamnn1.6021.6025008-25 O8 O @300 c/c(two sets )300nn1.5421.5424502304-25 O + 4 -20 O8 O @ 230 mm c/c( three sets )

Short Biaxial bending columnDesign of short eccentrically loaded square columns - Biaxial bending.Size 500 mm x 500 mmAxial factored load 1500 KNFactored moment Mux = 90 KNm( Given )Muy = 120 Knmmoment due to minimum eccentricity is less than the applied moment.Material M15 grade concreteHYSD reinforcement of grade Fe415Solution :-Assume an axial load P'u of 1500 KN and a uniaxial moment M'ux = 90 + 120 = 210 KNm .P'u / ( fck x b x D ) =1500 x 103 / ( 15 x 500 x 500 )=0.4M'ux / ( fck x b x D2 ) =210 x 106 / ( 15 x 500 x 5002 )=0.112d' = 40 + 10 = 50 mmd' / D = 50 / 500 = 0.1From chart 32 , SP-16p / fck = 0.078p = 0.078 x 15 = 1.17As = 1.17 x b x D = 1.17 x 500 x 500 / 100=2925mm2Provide 4-25 mm O + 4-20 mm O = 3220 mm2 , equally distributedp = 3220 x 100 / ( 500 x 500 )p =1.288p / fck = 1.288 / 15=0.086The assumed section is now checked.For p / fck = 0.086 and Pu / ( fck x b x D ) = 0.4 ,39.6 Members Subjected to Combined Axial LoadThe reinforcement being equally distributed , the moment capacities fromand Biaxial Bendingwhere ,Chart - 44 , SP-16Mux , Muy =moments about x and y axesMux1 / ( fck x b x D2 ) = Muy1 / ( fck x b x D2 ) = 0.108Mux+Muy 1due to design loads,Mux1 = Muy1 = 0.108 x 15 x 500 x 5002 x 10-6Mux1Muy1Mux1, Muy1 =maximum uniaxial moment=202.5KNmcapacity for an axial load of Pu,Puz = 0.45 fck Ac + 0.75 fy Ascbending about x and y axesAc =500 x 500 - 3220respectively, and=246780mm2n is related to Pu/PuzPuz = 0.45 x 15 x 246780 x 10-3 + 0.75 x 415 x 3220 x 10-3where Puz = 0.45 fck Ac + 0.75 fy Asc=1665.765+1002.225For values of Pu / Puz = 0.2 to 0.8, ( 0.2 , 0.4 , 0.6 , 0.8 )the values of n vary=2667.99KNlinearly from 1 .0 to 2.0.( 1.0 , 1.33 , 1.67 , 2.0 ) For values less than 0.2, n isPu / Puz =1500 / 26681 .O; for values greater than 0.8, n is 2.0.=0.56From , IS 456-2000 , Clause 39.6n =1.6020.2 difference0.34check :0.04 difference?-0.068Mux+Muy 1Mux1Muy190+120202.5202.50.273+0.4330.706 1Ties : Minimum diameter Otr = 25 / 4= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 500 mm( least lateral dimension )( ii ) 16 x 25 = 400 mm( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 350 mm c/cDesign of short eccentrically loaded rectangle columns - Biaxial bending.Size 300 mm x 500 mmAxial factored load 1500 KNFactored moment Mux = 60 KNm( Given )Muy = 60 KNmmoment due to minimum eccentricity is less than the applied moment.Material M15 grade concreteHYSD reinforcement of grade Fe415Solution :-Assume an axial load P'u of 1500 KN and a uniaxial moment M'uy = 300 / 500 ( 60 + 60 ) = 72 KNm .P'u / ( fck x b x D ) =1500 x 103 / ( 15 x 300 x 500 )=0.67M'u / ( fck x b x D2 ) =72 x 106 / ( 15 x 500 x 3002 )=0.107d' = 40 + 10 = 50 mmd' / D = 50 / 300 = 0.167..use 0.2From chart 34 , SP-16p / fck = 0.167p = 0.167 x 15 = 2.5As = 2.5 x b x D = 2.5 x 300 x 500 / 100=3750mm2Provide 8-25 mm O = 3928 mm2 , equally distributedp = 3928 x 100 / ( 300 x 500 )p =2.62p / fck = 2.62 / 15=0.175The assumed section is now checked.About Xd' / D = 50 / 500 = 0.1 , For p / fck = 0.175 and Pu / ( fck x b x D ) = 0.67 ,Chart - 44 , SP-16Mux1 / ( fck x b x D2 ) = 0.13Mux1 = 0.13 x 15 x 300 x 5002 x 10-6=146.25KNmAbout Yd' / D = 50 / 300 = 0.167 say 0.2 , For p / fck = 0.175 and Pu / ( fck x b x D ) = 0.67 ,Chart - 46 , SP-16Muy1 / ( fck x b x D2 ) = 0.103Muy1 = 0.103 x 15 x 500 x 3002 x 10-6=69.53KNmPure axial load capacityPuz = 0.45 fck Ac + 0.75 fy AscAc =300 x 500 - 3928=146072mm2Puz = 0.45 x 15 x 146072 x 10-3 + 0.75 x 415 x 3928 x 10-3=985.986+1222.59=2208.576KNPu / Puz =1500 / 2208.6=0.68From , IS 456-2000 , Clause 39.6n =1.8020.2 difference0.33check :0.12 difference?-0.198Mux+Muy 1Mux1Muy160+60146.2569.530.2+0.770.97 1Ties : Minimum diameter Otr = 25 / 4= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( least lateral dimension )( ii ) 16 x 25 = 400 mm( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 300 mm c/cNote that the distance between corner bars in one face is more than 48 Otr( 500 - 80 -25 = 395 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.

nn1.6021.6025005004-25 O +4-20 O8 O @350 c/cnn1.8021.8025003008-25 O8 O @300 c/c(two sets )nn

Short uniaxial bending column.Design of short eccentrically loaded columns - uniaxial bending.Size 300 mm x 600 mmAxial factored load 600 KNFactored moment 300 KNm( Given )moment due to minimum eccentricity is less than the applied moment.Material M15 grade concreteHYSD reinforcement of grade Fe415Solution :-Using 25 mm diameter bars with 40 mm clear coverd' = 40 + 12.5 = 52.5 mmd' / D = 52.5 / 600 = 0.088 ,say 0.1Pu / ( fck x b x D ) =600 x 103 / ( 15 x 300 x 600 )=0.222Mu / ( fck x b x D2 ) =300 x 106 / ( 15 x 300 x 6002 )=0.185From chart 32 , SP : 16p / fck = 0.1p = 0.1 x 15 = 1.5As = 1.5 x b x D = 1.5 x 300 x 600 / 100=2700mm2Provide 22 mm O ( 2700 / 380 = 8 no.) = 3041 mm2As the distance between two opposite corner bars is more than 300 mm ,provide 2- 12 mm O at centre of long sides of the column, which may be tied by open ties.Provide 8- 25 O + 2 - 12 O = 4151 mm2Ties : Minimum diameter Otr = 25 / 4= 6.25 mmUse 8 mm diameter M.S. ties.Spacing should not exceed lesser of( i ) 300 mm( least lateral dimension )( ii ) 16 x 25 = 400 mm( 16 times least longitudinal diameter of bar )( iii ) 48 x 8 = 384 mm( 48 times diameter of tie )Provide 8 mm O M.S. ties @ 300 mm c/c ( two sets )Note that the distance between corner bars in one face is more than 48 Otr( 600 - 80 -25 = 495 > 48 x 8 = 384 ). Therefore two sets of closed ties shall be used .Note that if this distance would be less than ( 48 x 8 ) mm , open ties for internal bars would be sufficient.

6003008-25 O+ 2-12 O8 mm O @300 mm c/c( two sets )

Short circular columnDesign of short circular columnWorking load = 1200 KNAssume emin < 0.05 DGiven( a ) lateral ties &( b ) helical reinforcementMaterialM20 grade concreteHYSD steel Fe415For Lateral reinforcement mild steel Fe250Solution : -Factored load =1.5 x 1200=1800KN( a ) lateral ties : -Pu = 0.4 fck Ac + 0.67 fy AscAssume 0.8 % minimum steel.Then ,Asc =0.008 AgAc =Ag - Asc=0.992 AgSubstituting , we have1800 x 103 =0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag1800 x 103 =7.936Ag +2.2244Ag1800 x 103 =10.1604AgAg =177165.354330709mm2If D is the diameter of the column( / 4 ) x D2 =177165.354330709D = 225688D =475 mmUse 475 mm diameter column.Asc =0.008 x 177165=1417mm2Minimum 6 bars shall be used.Provide 16 mm diameter bars ( 1417 / 201 ) = 8 no.Asc =8 x 201=1608mm2Use 6 mm O lateral ties , Spacing shall be lesser of( i ) 475 mm( least lateral dimension )( ii ) 16 x 16 = 256 mm( 16 times least longitudinal diameter of bar )( iii ) 48 x 6 = 288 mm( 48 times diameter of tie )Provide 6 mm O lateral ties @ 250 mm c/c .( b ) Helical reinforcement : -The column with helical reinforcement can support 1.05 times the load of a similarmember with lateral ties. ThereforePu = 1.05 [ 0.4 fck Ac + 0.67 fy Asc ]1800 x 103 =1.05 [0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag ]1800 x 103 =8.3328Ag +2.33562Ag1800 x 103 =10.66842AgAg =168728.908886389mm2If D is the diameter of the column( / 4 ) x D2 =168728.908886389D = 214941D =463 mmUse 450 mm diameter column.Then ,Ag =( / 4 ) x 4502=158962.5mm21800 x 103 =1.05 [0.4 x 20 x (158963 - Asc ) + 0.67 x 415 x Asc ]=1335289.2-8.4Asc+291.9525Asc464711=283.55AscAsc =1638.903191677mm2provide 20 mm diameter bars ( 1638.9 / 314 =)6 No.Asc =6 x 314=1884mm2Assume 8 mm O M.S. bars for helix at 40 mm clear cover .Dc =450 - 40 - 40=370mmasp =( / 4 ) x 82=50mm2Minimum s =0.36 ( (Ag / Acr) - 1 ) fck / fy=0.36 ( ( 4502 / 3702 ) - 1 ) x 20 / 250=0.36 x 0.4792 x 20 / 250=0.01380096Nows =4 x asp / p x Dc0.0138 =4 x 50 / p x 370p =39.17mm.. ( 1 )As per IS 456-2000 clause 26.5.3.2 ( d )The pitch < 75 mm< Dc / 6 ( = 370 / 6 = 61.67 mm )> 25 mm> 3 x dia of helix bar = 3 x 8 = 24 mm( 2 )From ( 1 ) and ( 2 ) , provide 8 mm O helix @ 35 mm pitch.

4506 - 20 O6 - 20 O358 mm O @35 mm c/c

short columnDesign of short columnIS 456-2000 clause 25Factored load = 1500 KNNote :- A column may be considered as short when both slenderness ratios lex / D and ley / b 12 OR lex / ixx < 40 where,Assume emin < 0.05 DGivenMaterialM15 grade concretelex = effective length in respect of the major axismild steel Fe250D = depth in respect of the major axisSolution : -ley = effective length in respect of the minor axisHere , emin < 0.05 D , But emin = 20 mmb = width of the memberTherefore , size of column shall be minimum ( 20 / 0.05 = 400 )400 mm x 400 mmixx = radius of gyration in respect of the major axis.Assume 0.8 % minimum steel.iyy = radius of gyration in respect of the minor axis.Then ,Asc =0.008 AgAll columns shall be designed for minimum eccentricity equal to the unsupported length of column / 500 plus lateral dimension / 30 , subject to a minimum of 20 mm. For bi-axial bending this eccentricity exceeds the minimum about one axis at a time.Ac =Ag - Asc=0.992 AgPu = 0.4 fck Ac + 0.67 fy Asc=0.4 x 15 x 0.992 Ag + 0.67 x 250 x 0.008 AgIS 456-2000 clause 39.31500 x 103 =5.952Ag +1.34AgWhen emin 0.05 D , the column shall be designed by the following equation:1500 x 103 =7.292AgAg =205704.882062534mm2Pu = 0.4 fck Ac + 0.67 fy AscIf the column is to be a square , the side of column =205705 = 453 mmAc = Area of concrete ,Adopt450 mm x 450 mm size column .Asc = Area of longitudinal reinforcement for columns.Then ,If emin > 0.05 D , the column shall be designed for moment also.1500 x 103 =0.4 x 15 x ( 450 x 450 - Asc )+ 0.67 x 250 x AscIS 456-2000 clause 26.5.3.11500 x 103 =1215000- 6 Asc+167.5AscThe cross-sectional area of longitudinal reinforcement , shall be not less than 0.8 % nor more than 6 % of the gross cross-sectional area of the column.285000=161.5AscAsc =1765mm2Provide 20 mm diameter bars = ( 1765 / 314 ) = 6 No.NOTE - The use of 6 percent reinforcement may involvegiving , Asc = 6 x 314 = 1884 mm2practical difficulties in placing and compacting of concrete;Note that the distance between the bars exceeds 300 mm on two parallel sides .hence lower percentage is recommended. Where bars fromthe arrangement of reinforcement should be changed.the columns below have to be lapped with those in theProvide then 4 no. 20 mm diameter bars