Design Report of Damauli Control Building

BEAM DESIGN OF DAMAULI CONTROL BUILDING

Max moment 175.611 KNmMax shear

force 120.404 KN

Input Data: B DSize of beam= 375 500 mm2Material useConcrete M20 Concrete fck 20 N/mm2Steel Fe415 Steel fy 415 N/mm2

mmClear cover= 20 mm

Factored moment (Hogging) 175.611 kN-mDiameter of bar use (assume)= 20 mmDiameter of Stirrups used (assume) = 8 mm

Assuming 20mm diameter bars with 20mm clear cover and two layers of barsNominal diameter of agg size (assume)= 20 mmd'=dia/2+diastirrups+clear cover= 38 mmd=500-20-dia-agg dia/2 462 mmMlim=0.133*fck*b*d2= 212.91 kN-m > 175.611 kN-mSingly Section DesignMlim= 0.87*fy*Ast(d-Ast*fy/fck*b)=solving quadratic eq for Ast1: 0.87*fy2*Astx2/fck*b-0.87fydAstx+Mlim=0

a=0.87*fy*fy/fck*bb=-0.87*fy*dc=Mlimd1=sqrt(b2-4ac)x1= 6776.80x2= 1572.60Ast1(mm2)= 1572.60 mm2

If,d'/d=(38/462)= 0.082Stress fsc in compression reinforcement for interpolation

a b c d0.05 0.10 0.15 0.20

415 355 353 342 329500 424 412 395 370

fsc=355+(355-353)/(0.05-0.1)*(0.0822510822510823-0.05)=Then,fsc= 353.71 N/mm2 Interpolated values

Now,Mu-Mlim=(fsc-fcc)Asc*(d-d')=

Mu-Mlim= -37.30 kN-mfcc=0.446*fck= 8.92(fsc-fcc)*(d-d')= 146190.94

DESIGN OF BEAMfor 5.5 m span beam

BEAM ID B2/41

19.98-166805.102.13E+08103970.00

The remaining bending moment has to be resisted by couple cosisting of compression steel and the corresponding tension steel.

Grade of steel(fy) N/mm2

d'/d

yf mx 'd 2scA scA


Asc= -255.14 mm2Now, C2 = T2

Corresponding tensile steel (Ast2)0.87fyAst2=fsc*Asc

0.87*fy= 361.05fsc*Asc= -90246.12

Ast2= -249.95 mm2Total Tensile steel=Ast1+Ast2= 1322.65 mm2

Check for Ast minimum

Min Ast= 354.8493976 mm2 Ast

Hence,Tension reinforcement at top (Ast)provide 5-20 bars 5

Ast actual of 5-20 = 1570.796327 mm2Total tension steel provided in two layer= 1570.796327 mm2 > 1322.65 mm2

Compression reinforcement at Bottom(Asc)provide 4-20 bars 4

Ast actual of 4-20 = 1256.637061 mm2Total Compression steel provided in two layer= 1256.637061 mm2 > -255.14 mm2

Check for shear:Ast= 1570.796327 mm2

0.91 %Design Shear Strength of concrete c ( by interpolation ) (From table 19, p.73) IS 456:2000

0.75 1 1 00.56 0.62

Then,c= 0.598 N/mm2 Interpolated valuesVu= 120.404 kN

v=Vu/bd= 0.69 N/mm2cmax= 2.8 N/mm2 (table 20 p.73)

c


= Basic value of span to effective depth ratios for spans up to spans 10m.= 26 for contineous(cl.23.2.1(a),p37,IS 456:2000)

= a factor which accounts for correction in the values of for spans greater than 10 m= 1 (since span= Lx/= 0 mm


Ductility check for shear Spacing of stirrups,Spacing as per IS: 456:2000 adopt 100.00 mmAt distance up to 2d, X=


Max moment 139.511 KNmMax shear

force 109.826 KN

Input Data: B DSize of beam= 375 550 mm2Material useConcrete M20 Concrete fck 20 N/mm2Steel Fe415 Steel fy 415 N/mm2

mmClear cover= 20 mm

Factored moment (Hogging) 139.511 kN-mDiameter of bar use (assume)= 20 mmDiameter of Stirrups used (assume) = 8 mm

Assuming 20mm diameter bars with 20mm clear cover and two layers of barsNominal diameter of agg size (assume)= 20 mmd'=dia/2+diastirrups+clear cover= 38 mmd=550-20-dia-agg dia/2 512 mmMlim=0.133*fck*b*d2= 261.49 kN-m > 139.511 kN-mSingly Section DesignMlim= 0.87*fy*Ast(d-Ast*fy/fck*b)=solving quadratic eq for Ast1: 0.87*fy2*Astx2/fck*b-0.87fydAstx+Mlim=0

a=0.87*fy*fy/fck*bb=-0.87*fy*dc=Mlimd1=sqrt(b2-4ac)x1= 7510.22x2= 1742.80Ast1(mm2)= 1742.80 mm2

If,d'/d=(38/512)= 0.074Stress fsc in compression reinforcement for interpolation

a b c d0.05 0.10 0.15 0.20

415 355 353 342 329500 424 412 395 370

fsc=355+(355-353)/(0.05-0.1)*(0.07421875-0.05)=Then,fsc= 354.03 N/mm2 Interpolated values

Now,Mu-Mlim=(fsc-fcc)Asc*(d-d')=

Mu-Mlim= -121.98 kN-mfcc=0.446*fck= 8.92(fsc-fcc)*(d-d')= 163582.73

DESIGN OF BEAMfor 6.30 m span beam

BEAM ID B2/32

19.98-184857.602.61E+08115222.17

The remaining bending moment has to be resisted by couple cosisting of compression steel and the corresponding tension steel.

Grade of steel(fy) N/mm2

d'/d

yf mx 'd 2scA scA


Asc= -745.66 mm2Now, C2 = T2

Corresponding tensile steel (Ast2)0.87fyAst2=fsc*Asc

0.87*fy= 361.05fsc*Asc= -263988.11

Ast2= -731.17 mm2Total Tensile steel=Ast1+Ast2= 1011.63 mm2

Check for Ast minimum

Min Ast= 393.253012 mm2 Ast

Hence,Tension reinforcement at top (Ast)provide 4-20 bars 4

Ast actual of 4-20 = 1256.637061 mm2Total tension steel provided in two layer= 1256.637061 mm2 > 1011.63 mm2

Compression reinforcement at Bottom(Asc)provide 4-20 bars 4

Ast actual of 4-20 = 1256.637061 mm2Total Compression steel provided in two layer= 1256.637061 mm2 > -745.66 mm2

Check for shear:Ast= 1256.637061 mm2

0.65 %Design Shear Strength of concrete c ( by interpolation ) (From table 19, p.73) IS 456:2000

0.50 0.75 0.75 00.48 0.56

Then,c= 0.529 N/mm2 Interpolated valuesVu= 109.826 kN

v=Vu/bd= 0.57 N/mm2cmax= 2.8 N/mm2 (table 20 p.73)

c


= Basic value of span to effective depth ratios for spans up to spans 10m.= 26 for contineous(cl.23.2.1(a),p37,IS 456:2000)

= a factor which accounts for correction in the values of for spans greater than 10 m= 1 (since span= Lx/= 0 mm


Ductility check for shear Spacing of stirrups,Spacing as per IS: 456:2000 adopt 100.00 mmAt distance up to 2d, X=

Design Of Column of Damauli SS Building

Design of column

A. Column no C1

1 INPUT DATASize of Column= 450 450 mm2Beam Along XX grid = 375 550 mm2Beam Along YY grid = 375 550 mm2Centre to centre length(L)= 4.2 m Effective length (le)=0.65*L= 0.65 2.73 m Both ends fixedMaterial useConcrete M20 Concrete fck 20 N/mm2Steel Fe415 Steel fy 415 N/mm2Diameter of bar use (assume)= 25 mm

Clear cover= 40 mmd'=dia+clear cover= 52.5 mmData from staad analysis

Pu Mux Muy0 -263.5233 8.165 -41.873167

2.1 -255.0233 0.37783333 21.6324.2 -246.5233 -7.40925 85.1371667

Max -263.5233 8.165 85.1371667

2 Slenderness ratiole/LLD=2730/450= 6.07 20mmAlong grid Y-Y,eymin= 22.30 mm >20mmMoments due to the Minimum eccentricity=Pu*emin=

Mux=Pu*exmin= -5.88 kN-mMuy=Pu*eymin= -5.88 kN-m

Moments due to minimum ecentricity are less than the values given above.Reinforcement is distributed equally on four sides.Design Final Moments= Mux 8.17 kN-m

Muy 85.14 kN-m4 Determine the trial section:

First trial:-Let, Assume percentage of reinforcement(p)= 1.5 %

p/fck= 0.075d'/D= 0.117

SP-16 chart for d'/D= 0.1 used.Pu/fckbD= -0.065

Referring to SP-16 chart= 44 Page:- 129Mu/fckbD2= 0.12Now,Mux1= 218.70 kN-m

Since size Column is square so,Mux1=Muy1= 218.70 kN-m

Calculation of Puz:

1.2 (DL+LL+EQ)Load combination

C2/7

Column Dist m

ckf yf el


Puz = 0.45fck*Ac+0.75*As*fy Ref. (SP-16)= 0.45*20*450*450+0.75*0.015*450*450*415= 2767.92 kN

Pu/Puz= 0.10

n= 0.8 2 For intermediate values ,Linear interpolation done.

n= 1.000Mux/Mux1= 0.037Muy/Muy1= 0.389

0.427 OK

Hence safe ,OK, the section & percentage of steel has been carried out.Second trial:-Let, Assume percentage of reinforcement(p)= 1.5 %

p/fck= 0.075d'/D= 0.117

SP-16 chart 44 and 45 for d'/D= 0.15 and 0.1 used.Pu/fckbD= -0.065



Calculation of Puz:Referring to Chart Page:

Puz = 0.45fck*Ac+0.75*As*fy= 0.45*20*450*450+0.75*0.015*450*450*415= 2767.92 kN

Pu/Puz= -0.10


n= 1.000Mux/Mux1= 0.025Muy/Muy1= 0.260

0.284 OK

Hence safe ,OK, the section & percentage of steel has been carried out.Hence provide= 1.500 %

Area of compression steel (Asc)= 3037.5 mm2Diameter of bars= 25 mm Area of steel considered 55 %No of bars= 3.4033693 mm 4Diameter of bars= 20 mm Area of steel considered 45 %No of bars= 4.35089826 mm 4

Total Area of steel provided = 3220.13 mm2

Pu/Puz

Pu/Puz

uyM uzM

nn

MuyMuy

MuxMux

11 +

uyM uzM

nn

MuyMuy

MuxMux

11 +


Provide 4@25 + 4@20 (Asc provided = 3220.14 mm2)

According to IS456:2000,cl.26.5.1(a),p-48,minimum area of reinforcement in column=0.8%Check for minimum reinforcement 0.800 % 1,620.00 mm2

OK

5 Design of diameter and pitch of lateral ties:i) The dia. of lateral ties should not be less than one-forth, dia. of largest longitudinal bar

= 25/4 6.25 mm

6 mmHence provide 8mm dia. 8 mmii) Pitch of ties:Least lateral dimension (LLD) 450 mmnot greater than 16x25mm of main bar 400 mmnot greater than 48x8mm lateral bars 384 mm

Adopt minimum of above three = 384 mmHence provide 384mm c/c

Check for ductility criteriai) The spacing of hoops shall not exceed half the LLD of column.

225 mmii) Special confining reinforcement shall be provided over a length l 0 from each joint face.

a) LLD 450 mmb) 1/6 of clear span of member 608.33 mmc) 450mm 450 mm

Adopt minimum of above three = 450 mmSpacing provided in above mentioned case =450mmc/ciii)Lap splices shall be provided only in the central half of the member.

a) Hoops shall be provided over the entire splices .Spacing shall no be exceeded 150mmc/c at splices.

52.155UZp


Design of column

A. Column no C2



Pu Mux Muy0 -193.2504 5.98766667 -30.706989

2.1 -187.0171 0.27707778 15.86346674.2 -180.7838 -5.43345 62.4339222

Max -193.2504 5.98766667 62.4339222


Mux=Pu*exmin= -4.31 kN-mMuy=Pu*eymin= -4.31 kN-m

Moments due to minimum ecentricity are less than the values given above.Reinforcement is distributed equally on four sides.Design Final Moments= Mux 5.99 kN-m


First trial:-Let, Assume percentage of reinforcement(p)= 1.5 %

p/fck= 0.075d'/D= 0.117

SP-16 chart for d'/D= 0.1 used.Pu/fckbD= -0.048



Calculation of Puz:

Column Dist m Load combination1.2 (DL+LL+EQ)

C2/7

ckf yf el



Pu/Puz= 0.07


n= 1.000Mux/Mux1= 0.027Muy/Muy1= 0.285

0.313 OK


p/fck= 0.075d'/D= 0.117

SP-16 chart 44 and 45 for d'/D= 0.15 and 0.1 used.Pu/fckbD= -0.048




Puz = 0.45fck*Ac+0.75*As*fy= 0.45*20*450*450+0.75*0.015*450*450*415= 2767.92 kN

Pu/Puz= -0.07


n= 1.000Mux/Mux1= 0.018Muy/Muy1= 0.190

0.209 OK


Area of compression steel (Asc)= 3037.5 mm2Diameter of bars= 25 mm Area of steel considered 100 %No of bars= 6.18794419 mm 8Diameter of bars= 20 mm Area of steel considered 0 %No of bars= 0 mm 0


Pu/Puz

Pu/Puz

uyM uzM

nn

MuyMuy

MuxMux

11 +

uyM uzM

nn

MuyMuy

MuxMux

11 +




OK


= 25/4 6.25 mm

6 mmHence provide 8mm dia. 8 mmii) Pitch of ties:Least lateral dimension (LLD) 450 mmnot greater than 16x25mm of main bar 400 mmnot greater than 48x8mm lateral bars 384 mm







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Design of column

A. Column no C3



Pu Mux Muy0 350.45 -2.61 31.136

2.1 -320.4 -0.457 1.471166674.2 -347.76 -0.564 7.68616667

Max 350.45 -2.61 31.136


Mux=Pu*exmin= 7.82 kN-mMuy=Pu*eymin= 7.82 kN-m

FALSEReinforcement is distributed equally on four sides.Design Final Moments= Mux -2.61 kN-m


First trial:-Let, Assume percentage of reinforcement(p)= 2 %

p/fck= 0.100d'/D= 0.117

SP-16 chart for d'/D= 0.1 used.Pu/fckbD= 0.087



Calculation of Puz:

Column Dist m Load combination1.2 (DL+LL+EQ)

C1/15

ckf yf el



Pu/Puz= 0.11


n= 1.000Mux/Mux1= -0.010Muy/Muy1= 0.114

0.104 OK


p/fck= 0.125d'/D= 0.117

SP-16 chart 44 for d'/D= 0.1 used.Pu/fckbD= 0.087




Puz = 0.45fck*Ac+0.75*As*fy= 0.45*20*450*450+0.75*0.02*450*450*415= 3398.20 kN

Pu/Puz= 0.10


n= 1.000Mux/Mux1= -0.008Muy/Muy1= 0.100

0.092 OK


Area of compression steel (Asc)= 5062.5 mm2Diameter of bars= 25 mm Area of steel considered 40 %No of bars= 4.12529612 mm 4Diameter of bars= 20 mm Area of steel considered 60 %No of bars= 9.66866279 mm 8


Pu/Puz

Pu/Puz

uyM uzM

nn

MuyMuy

MuxMux

11 +

uyM uzM

nn

MuyMuy

MuxMux

11 +




OK


= 25/4 6.25 mm6 mm

Hence provide 8mm dia. 8 mmii) Pitch of ties:Least lateral dimension (LLD) 450 mmnot greater than 16x25mm of main bar 400 mmnot greater than 48x8mm lateral bars 384 mm







52.155UZp

DAMAULI CONTROL BUILDING Design of Slab

Slab No. S4 Level: TOPPreliminary Design Data:

Slab effective depth (mm) = 155 mmClear span C/C span Lc+d

Effective span (lesser of two)

Span in shorter direction = 5.500 m 5.50 m 5.66 m 5.50Span in longer direction = 6.300 m 6.30 m 6.46 m 6.30

1 Interior Panel2

One short edge discontineous

3One long edge discontineous

4Two adjacent edge discontineous

Short span 5Two short edge discontineous

B or Lx = 5.50 m 6Two long edge discontineous

7Three edge discontineous (One long edge contineous)

Long Span 8Three edge discontineous (One short edge contineous)

L or Ly = 6.30 m 9 Four edge discontineousInput Data:

Slab size = 6.30 5.50 sq. mSlab is contineour over T-BeamMaterial useConctrete M20 20 N/mm2Steel Fe415 415 N/mm2 xu/d = 0.48Bar dia. At short span 10 mmArea of bars (ax) 78.54 mm2Bar dia. At long span 10 mmArea of bars (ay) 78.54 mm2Distribution bar dia. 10 mmArea of bars (Asty) 78.54 mm2Clear cover 15.00 mm

Design of Two way slab (IS 456:2000) (Limit State Method)

Slab Panel


IS 456: 2000 OutputSlab dataLx = 5.50 mLy = 6.30 mLy/Lx= 1.145454545 Two Way SlabPreliminary Design of SlabMinimum thickness of slab based on serviceability limit state criteria for deflection is determined from the effective spanto effective depth ratio (I.e. based on deflection criteria & depth required to resist the bending moment) (IS 456:2000, 22.2 a)The basic span to effective depth ratio for contineous slab (35 to 40)From deflection criteria, we have,

IS 456:2000clause 23 - 2.1 l/d


Dead load of Cement Punning = 20.4 KN/m3 * 0.003m = 0.0612 KN/m2

iv Cement PlasterThickness 12.5 mm 20.4 KN/m3Dead load of Cement Plaster = 20.4 KN/m3 * 0.0125m = 0.255 KN/m2

v Partition wall (average)Thickness 110 mm 18.5 KN/m3Dead load of Partition wall (average) = 0KN/m2 0 KN/m2

Total of (I + ii + iii + iv + v) 5.72 KN/m2IS 876:1986 (II) I) Live Load (LL)

For single storey without stairs 1.5 KN/m2Total Design Load (DL + LL) 7.22 KN/m2The Factored Load (Wu) = 1.5*Design Load = 10.83 KN/m2 1.5 Factor

IS: 456:2000 Critical Support Condition:Table 26, Page 91 Two short edge discontineous 5

Calculation of Moments:

-ve moment at contineous edge (support) x1 0.050363636 y1 0 +ve moment at mid span x2 0.038363636 y2 0.035

-ve moment at contineous edge (support) -Mx = x1*wu*lx2 16.50 KN-m -My = y1*wu*lx2 0.00 KN-m +ve moment at mid span +Mx = x2*wu*lx2 12.57 KN-m +My = y2*wu*lx2 11.47 KN-mDepth Calculation:Depth Required (d) = sqrt(Mu/(0.36*xumax/d*(1-.42*xumax/d)*b*fck)) 77.33 mm OKHence, for safe design adopt above more depth D= 150 mm 2 % decreaseEffective Depth (main) dx= D-1/2-clear cover 130 mmEffective Depth (secondary) dy= D-clear cover-main steel dia-2/2 120 mmReinforcement Calculation:

Moment

Short Span Long Span


-ve Mx -ve My +ve Mx +ve MyMoment (KN-m) 16.50 0.00 12.57 11.47Solving quadratic equation Mu=0.87fyAstd(1-Astfy/bdfck)a = 0.87*fy^2/(b*fck) 7.49 7.49 7.49 7.49b=-0.87fyd -46936.50 -43326.00 -46936.50 -43326.00c=Mu 16499505.00 0.00 12568215.00 11466262.50x=(-b+-sqrt(b2-4ac))/2ax1=(-b+sqrt(b2-4ac))/2a 5891.23 5783.13 5984.75 5505.12x2=(-b-sqrt(b2-4ac))/2a 373.83 0.00 280.31 278.02Astx (mm2) = 373.83 0.00 280.31 278.02Ast req. (mm2) 373.83 0.00 280.31 278.02

cl. 25.5.1 Min As(0.12%bD) mm2 156 144 156 144 Not OK FALSESpacing =b*area of one bar/Ast 210.09 0.00 280.19 282.50

cl. 26.3.3(b) 1 3d 390 390 390 390 Minimum of 3d or 300300 300 300 300 300

Adopt the spacing (min of above three) 210.09 0.00 280.19 282.50Spacing provided in mm 210 0 280 280Ast provided (mm2) 374.00 0.00 280.50 280.50% of Ast (Provided) 0.25 0.00 0.19 0.19Main Reinforcement 10 @ 210mm c/c 10 @ 280mm c/c 10 @ 280mm c/cDistribution ReinforcementMin As(0.12%bd) mm2 156 mm2Bar size 10 mmBar Spacing calculated 503.46 mm

cl. 26.3.3(b) 2 Maximum bar spacing 5d or 450 5d = 650 & 450 450 mmAdopt the spacing (min of above two) 450.00 mmProvide Distribution reinforcement 10 @ 450mm c/c 450.00Area of steel provided 174.53 mm2% of reinforcement provided 0.13 %Check for Design:

IS 456:2000 Check for deflectioncl 23-2.1 From deflection criteria, we have,Page - 37 l/d


fs=0.58*fy*Ast req./Ast provided 240.59 N/mm2from graph () = 1.65

cl 23-2.1(d) = a factor which depends on the area of compression reinforcement. Page 39, (Fig 5) = 1

No compression reinforcement is provided

cl 23-2.1(e) = a factor for flanged beams which depends on the ratio of web width to the flange widthPage 39, (Fig 6) = 1 No flange beam

lx/d permissible = 42.9lx/d provided = 5500/130 = 42.31 Safe in deflectionCheck for ShearMax Shear Force (V) = w*lx/2 29.7825 KNTv=V/bd 0.229096154 N/mm2% of steel provided 0.25 %

Table 19, Page 73 Shear capacity of section without reinforcement

Tc from table 100As/bd M200.15 0.280.25 0.36 0.36 N/mm2

Shear strength in T' = kTcThe provided Depth of slab (D) = 150.00 mm

D k150 1.3175 1.25 1.30

T' = kTc 0.47 N/mm2 Safe in shearCheck for Development lengthLd


Summary:Overall Depth of slab = 150 mmMain Bar:Astx +ve = 10 @ 280mm c/cAstx -ve = 10 @ 210mm c/cAsty +ve = 10 @ 280mm c/cAsty -ve = Distribution Bar:10 @ 450mm c/c

DESIGN OF ISOLATED RCC FOOTING (IS 456, 2000)Footing ID: F2 COMB 11COLUMN Length (l, dim. || Z axis ) = 450 mmBreadth (b, dim. || X axis) = 450 mmHeight of pedestal = 2 m Breadth 1.8 mWeight of pedestal/column = 10.13 KN

FOOTINGFoot length (L, dim. || Z axis) = 1.8 mFoot Breadth (B, dim. || X axis) = 1.8 mThickness of footing (t) = 350 mmClear cover to Reinforcement = 50 mmMain bar dia of footing = 10 mmEffective depth of footing = 295 mm Length 1.8 mSelfweight of the footing = 28.35 KNArea of Footing(A) = 3.24 m2Sect mod of foot about Z axis (Zz) = 0.97 m3Sec mod of foot about X axis (Zx) = 0.97 m3

MATERIALS OF CONSTRUCTIONGrade of concrete fck = 20 N/mm2Grade of steel fy = 415 N/mm2

CHECK FOR GROSS BEARING PRESSURESafe NET bearing pressure = 200 KN/m2Safe gross bearing pr. = 236.90 KN/m2 (net pr. + depth of foot * soil unit wt)Unfactored load from output (P1) = 292.0417 KN 350.45Moment about Z axis (Mz) = -2.17833 KN-m -2.614Moment about X axis (Mx) = 25.94583 KN-m 31.135Depth of top of foot. from ground = 1.7 mUnit wt of soil = 18 KN/m3Weight of soil retained above foot = 92.95 KNP = (P1+soil+pedestal+foot selfwt) = 423.46 KNMz/Zz= 2.24108368 KN/m2Mx/Zx= 26.6932442 KN/m2Maximum/Minimum bearing pressure = (+,+) 155.15 KN/m2(+,-) 101.76 KN/m2(-,+) 159.63 KN/m2(-,-) 106.25 KN/m2Maximum soil Pressure = 159.63 KN/m2 0 KN/m2Hence footing is safe against max bearing pressure.Moment steel:Net cantilever on x-x or z-z =0.5(L-l) 0.675 m

globalZ

globalX

globalX

globalZ

Footing Dimensions

x

x

y

y

ZM

ZM

AP

CALCULATION FOR BOTTOM STEELM about X1 X1/Z1-Z1 = ( pe max x length2/2)= 36.3664258 KN-m per meterMu 54.55 KN-m per meter

Mulimit = 240.37 KN-m per meterThe section is singly reinforced

Hence, Ast = 671.96 mm2Min Ast = 420.00 mm2 (0.12 % for slab, cl 26.5.2.1)Spacing = 116.88 mm (considering max of above two calculated values of Ast)pt provided = 0.23 %Hence provide 10 mm dia bar @ 116 mm c/c parellel to length of footing

Arrangement of bottom reinforcement as per above design is shown below

10 mm dia bar @ 116 mm c/c


1 1

Footing Length 1800 mm Breadth 1800 mm

Sec 1-1

745 450745

L1 X1 X

a a

Z ZN1 N1

a a

L2 L2

380 X1 XL1 Breadth 1800 mm

450 Footing Length 1800 mm 380

PLAN

bdbdfM

ffA

ck

u

y

ckst

2

6.4115.0

CHECK FOR ONE WAY SHEAR :One way shear at critical sectionsDistance of critical sec. from edge of footing = 0.38 mShear force Vs =pe max x 0.38 x 1m width of footing = 60.66059 KNShear force Vu 90.99 KNShear stress tv = Vs/bd = 0.308 N/mm2tc = 0.35 N/mm2tv < tc hence O.K.

CHECK FOR TWO WAY SHEARRef. cl 34.2.4 and cl.31.6.3 of IS 456 : 2000Allowable shear stress tv allowable = kstcks = ( 0.5 + bc) = 1.5 >1Hence, ks= 1tc = 0.25 (fck)0.5 = 1.11803399 N/mm2tv allowable = ks x c = 1.11803399 N/mm2Shear force Vs = 16.024 ( 1.8 x 1.8 - 0.745 x 0.745) = 428.61 KNLength of critical section = 2 x ( 745 + 745) = 2980 mmArea of the critical section (length of critical sec x eff. d ) = 879100 mm2Hence shear stress v = 0.488 N/mm2tv < allowable hence O.K.

DESIGN OF ISOLATED RCC FOOTING (IS 456, 2000)Footing ID: F3 COMB 11COLUMN Length (l, dim. || Z axis ) = 450 mmBreadth (b, dim. || X axis) = 450 mmHeight of pedestal = 2 m Breadth 1.6 mWeight of pedestal/column = 10.13 KN

FOOTINGFoot length (L, dim. || Z axis) = 1.6 mFoot Breadth (B, dim. || X axis) = 1.6 mThickness of footing (t) = 350 mmClear cover to Reinforcement = 50 mmMain bar dia of footing = 10 mmEffective depth of footing = 295 mm Length 1.6 mSelfweight of the footing = 22.40 KNArea of Footing(A) = 2.56 m2Sect mod of foot about Z axis (Zz) = 0.68 m3Sec mod of foot about X axis (Zx) = 0.68 m3

MATERIALS OF CONSTRUCTIONGrade of concrete fck = 20 N/mm2Grade of steel fy = 415 N/mm2

CHECK FOR GROSS BEARING PRESSURESafe NET bearing pressure = 200 KN/m2Safe gross bearing pr. = 236.90 KN/m2 (net pr. + depth of foot * soil unit wt)Unfactored load from output (P1) = 161.042 KN 193.2504Moment about Z axis (Mz) = 4.989722 KN-m 5.987667Moment about X axis (Mx) = 52.02827 KN-m 62.43392Depth of top of foot. from ground = 1.7 mUnit wt of soil = 18 KN/m3Weight of soil retained above foot = 72.14 KNP = (P1+soil+pedestal+foot selfwt) = 265.71 KNMz/Zz= 7.30916341 KN/m2Mx/Zx= 76.213284 KN/m2Maximum/Minimum bearing pressure = (+,+) 187.31 KN/m2(+,-) 34.89 KN/m2(-,+) 172.70 KN/m2(-,-) 20.27 KN/m2Maximum soil Pressure = 187.31 KN/m2 0 KN/m2Hence footing is safe against max bearing pressure.Moment steel:Net cantilever on x-x or z-z =0.5(L-l) 0.575 m

globalZ

globalX

globalX

globalZ

Footing Dimensions

x

x

y

y

ZM

ZM

AP

CALCULATION FOR BOTTOM STEELM about X1 X1/Z1-Z1 = ( pe max x length2/2)= 30.9653561 KN-m per meterMu 46.45 KN-m per meter

Mulimit = 240.37 KN-m per meterThe section is singly reinforced

Hence, Ast = 526.37 mm2Min Ast = 420.00 mm2 (0.12 % for slab, cl 26.5.2.1)Spacing = 149.21 mm (considering max of above two calculated values of Ast)pt provided = 0.18 %Hence provide 10 mm dia bar @ 149 mm c/c parellel to length of footing

Arrangement of bottom reinforcement as per above design is shown below



1 1

Footing Length 1600 mm Breadth 1600 mm

Sec 1-1

745 450745

L1 X1 X

a a

Z ZN1 N1

a a

L2 L2

280 X1 XL1 Breadth 1600 mm

450 Footing Length 1600 mm 280

PLAN

bdbdfM

ffA

ck

u

y

ckst

2

6.4115.0

CHECK FOR ONE WAY SHEAR :One way shear at critical sectionsDistance of critical sec. from edge of footing = 0.28 mShear force Vs =pe max x 0.28 x 1m width of footing = 52.44794 KNShear force Vu 78.67 KNShear stress tv = Vs/bd = 0.267 N/mm2tc = 0.31 N/mm2tv < tc hence O.K.

CHECK FOR TWO WAY SHEARRef. cl 34.2.4 and cl.31.6.3 of IS 456 : 2000Allowable shear stress tv allowable = kstcks = ( 0.5 + bc) = 1.5 >1Hence, ks= 1tc = 0.25 (fck)0.5 = 1.11803399 N/mm2tv allowable = ks x c = 1.11803399 N/mm2Shear force Vs = -1.669 ( 1.6 x 1.6 - 0.745 x 0.745) = 375.56 KNLength of critical section = 2 x ( 745 + 745) = 2980 mmArea of the critical section (length of critical sec x eff. d ) = 879100 mm2Hence shear stress v = 0.427 N/mm2tv < allowable hence O.K.

Beam Design 5.5MBeam Design 6.3MColumn-1Column-2Column-3SLAB DESIGN ALLF-2F3

Design Report of Damauli Control Building

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