DESIGN OF ISOLATED FOOTING
Caption Calculation Symbol =ValueUnitReference OR Explanation
Design Parameters
Partial Safety Factor for LoadsLimit State of Collapse (DL+LL)gf=1.5ConstantCl.36.4.1 of IS 456-2000
Partial Safety Factor for Steelgs=1.15ConstantCl.36.4.1 of IS 456-2000
Partial Safety Factor for Concretegc=1.5ConstantCl.36.4.1 of IS 456-2000
Elasticity of SteelEs=200000N/mm^2Cl 5.6.3 of IS 456-2000
Elasticity of ConcreteEc=25000N/mm^2Cl 6.2.3.1 of IS 456-2000
Characteristic strength Concretefck=25N/mm^2
Characteristic strength Steelfy=415N/mm^2
Concrete Weightwc=25 kN/Cum
SECTION PROPERTIES
CoverMaindc=50 mm
Main Bar Dia along XXX_bar=16 mm
Main Bar Dia along YYY_bar=20 mm
Footing Length-XLx=2.4mmXX Direction
Footing Length-YLy=2mmYY Direction
Overall Footing ThicknessDg=600mm
Effective Depth=600-50-16/2D=542 mm
Axial LoadPu=825kNIncreased by 10% to arrive at design load
Moment about XX-axisMx=70kN-m
Moment about YY-axisMy=100kN-m
Footing Plan=Nr of Steps: 1; Step offset: 0.3
Footing Elevation=Nr of Steps: 1; Step offset: 0.3
Earth pressure due to biaxial moments
Biaxial Earth Pressure: Max=(825/ (2.4*2))+(70*2/ 2)/(2.4*2^ 3/12)+(100*2.4/ 2) / (2*2.4^ 3 / 12)QmaxXY=267.71kN/m^2(Pg / (Lx * Ly)) + (Mx * Ly / 2) / (Lx * Ly ^ 3 / 12) + (My * Lx / 2) / (Ly * Lx ^ 3 / 12)
Biaxial Earth Pressure=(825/ (2.4*2))+(70*2/ 2)/(2.4*2^ 3/12)+(100*2.4/ 2) / (2*2.4^ 3 / 12)QmaxX=163.54kN/m^2(Pg / (Lx * Ly)) + (Mx * Ly / 2) / (Lx * Ly ^ 3 / 12) - (My * Lx / 2) / (Ly * Lx ^ 3 / 12)
Biaxial Earth Pressure=(825/ (2.4*2))+(70*2/ 2)/(2.4*2^ 3/12)+(100*2.4/ 2) / (2*2.4^ 3 / 12)QmaxY=180.21kN/m^2(Pg / (Lx * Ly)) - (Mx * Ly / 2) / (Lx * Ly ^ 3 / 12) + (My * Lx / 2) / (Ly * Lx ^ 3 / 12)
Biaxial Earth Pressure: Min=(825/ (2.4*2))+(70*2/ 2)/(2.4*2^ 3/12)+(100*2.4/ 2) / (2*2.4^ 3 / 12)Qminn=76.04kN/m^2(Pg / (Lx * Ly)) - (Mx * Ly / 2) / (Lx * Ly ^ 3 / 12) - (My * Lx / 2) / (Ly * Lx ^ 3 / 12)
Pressure DiagramQmaxXY; QmaxX; QmaxY; Qminn=267.71; 163.54; 180.21; 76.04kN/m^2Biaxial bending:(Pg / (Lx * Ly)) (+/-) (Mx * Ly / 2) / (Lx * Ly ^ 3 / 12) (+/-) (My * Lx / 2) / (Ly * Lx ^ 3 / 12)
Pressure on extreme left along column face=180.21+ ((267.71-180.21)*(2/2 -0.4/ 2)/2)q1=232.71kN/m^2QmaxY + ((QmaxXY - QmaxY) * (Ly / 2 + Dy / 2)) / Ly
Pressure on extreme right along column face=76.04+ ((163.54-76.04)*(2/2 +0.4/ 2)/2)q2=128.54kN/m^2Qminn + ((QmaxX - Qminn) * (Ly / 2 + Dy / 2)) / Ly
Pressure on extreme bottom along column face=163.54+ ((267.71-163.54)*(2.4/2 +0.3/ 2)/2.4)q3=222.14kN/m^2QmaxX + ((QmaxXY - QmaxX) * (Lx / 2 + Dx / 2)) / Lx
Pressure on extreme top along column face=76.04+ ((180.21-76.04)*(2.4/2 +0.3/ 2)/2.4)q4=134.64kN/m^2Qminn + ((QmaxY - Qminn) * (Lx / 2 + Dx / 2)) / Lx
Moment at column face XX
Stress area beyond the column face XX=2.4* (2/ 2 - 0.4/ 2)Axx=1.92SqmAxx = Lx * (Ly / 2 - Dy / 2)
Force due to pressure XX=1.92* (267.71+163.54+232.71+128.54) / 4Fxx=380.4kNAxx * (QmaxXY + QmaxX + q1 + q2) / 4
Lever Arm from the column face XX=(2*QmaxXY+232.71)/(267.71+232.71)*((2/2-Dy/2)) / 3CgXX=0.409mCgXX = Lx * (Ly / 2 - Dy / 2)
Max Force * Distance=380.4*0.409Mxx=155.584kN-MMxx = Force * CgXX (Lever Arm)
Factored Moment = 1.5*155.584Mxx=233.38kN-mMxx = gf* Mxx
Moment at column face YY
Stress area beyond the column face YY=2* (2.4/ 2 - 0.3/ 2)Ayy=2.1SqmAyy = Ly * (Lx / 2 - Dx / 2)
Force due to pressure YY=2.1* (267.71+180.21+222.14+134.64) / 4Fyy=422.468kNFxx * (QmaxXY + Qmaxy + q3 + q4) / 4
Lever Arm from the column face YY=(2*QmaxXY+222.14)/(267.71+222.14)*((2.4/2-0.3/2)) / 3CgYY=0.541mCgYY = Ly * (Lx / 2 - Dx / 2)
Max Force * Distance=422.468*0.541Myy=228.555kN-MMyy = Force * CgYY (Lever Arm)
Factored Moment = 1.5*228.555Myy=342.83kN-mMyy = gf* Myy
CHECK for Min Eff DEPTH
Maximum of Factored Moments = Max(1.5*155.584,1.5*228.555)Mxy=342.8325kN-mMax of Mxx & Myy
Ultimate strain in steel=(415/1.15/200000/)+0.002esu=0.0038Constant((fy / gs / Es) + 0.002
Limiting depth of neutral axis=(0.0035 / (0.0035 +0.0038)) *542Xu=259.708mm(0.0035 / (0.0035 + esu)) * D OR 0.48D
Parabolic depth from origin/neutral axis=0.002*259.708/ 0.0035X1=148.4mm0.002 * Xu / 0.0035
Total area constant of the stress block=(((2/3/1.5)*25*259.708*1000)-((2/3/1.5)*25*148.4/3*1000))/25/259.708/1000k1=0.36Constant0.364 * fck * x * b ; this can be taken as 0.36 for all practical purposes -cl 37.1 of IS 456-2000 {(((2 / 3 / gc) * fck * X * B) - ((2 / 3 / gc) * fck * X1 / 3 * B))}
Distance of the centre of compression from the compression edge=(259.708 - (((2/3/1.5) * 25*259.708*1000*(259.708/2) - ((2/3/1.5) *25*1000*148.4^ 2 /12))/(0.36*25*259.708*1000))) /259.708k2=0.416Constant0.416 OR say 0.42 as per IS 456 clause 37.1c ((X - (((2 / 3 / gc) * fck * X * B * (X / 2) - ((2 / 3 / gc) * fck * B * X1 ^ 2 / 12)) / (k1 * fck * X * B))) / X)
Moment of Resistance by Concrete=0.36*25*259.708*1000*(542-(0.416*259.708))/10^6MRc=1014.33kN-m k1 * fck * xu * B * (D - (k2 * xu))/10^6
Coefficient of Moment of Resistance =1014330371.5/(25*1000*293764)k=0.138Constant0.138 for fe 415 (= Mu/ fck * B * D^2)
Minimum Depth of Section Required=Sqrt(342832500/ (0.138*25*1000))Dr=315.233mmUsing formula Mu = k * fck * B * D^2
DEPTH provided is adequate
Steel along XX
Reqd Steel Percentage %=0.24pt=0.24%By trials; pt = 0.22 produces lower Moment of Resistance than 233.376
Moment of Resistance= 1000*(542^2) * ((415/1.15 ) * (0.24/100 ) * (1-(0.416 * (415/1.15) / (0.36*25) * (0.24/100))))/10^6MR=244.24kN-mB * (D ^ 2) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100))))
Steel Required=0.24*1000*542/100Ast=1300.8mm^2Ast = pt * B * D / 100
Spacing Required: #16@160c/c=Round(Round(1000/(1300.8/ (3.142*16^2/4)) ,0) /10,0) *10Spacing=160mmB / (Ast / (PI * dia ^ 2 / 4)) & rounded to nearest 10 mm
Steel along YY
Reqd Steel Percentage %=0.38pt=0.38%By trials; pt = 0.36 produces lower Moment of Resistance than 342.832
Moment of Resistance= 1000*(526^2) * ((415/1.15 ) * (0.38/100 ) * (1-(0.416 * (415/1.15) / (0.36*25) * (0.38/100))))/10^6MR=355.36kN-mB * (D ^ 2) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100))))
Steel Required=0.38*1000*526/100Ast=1998.8mm^2Ast = pt * B * D / 100
Spacing Required: #20@160c/c=Round(Round(1000/(1998.8/ (3.142*20^2/4)) ,0) /10,0) *10Spacing=160mmB / (Ast / (PI * dia ^ 2 / 4)) & rounded to nearest 10 mm
Critical Section along XX = 2-(2 / 2 + 0.4 / 2 + 0.542)CSx=258 mmfrom face of column; Ly - (Ly / 2 + Dy / 2 + D):
Critical Section along YY=2.4-(2.4 / 2 + 0.3 / 2 + 0.542-0.016)CSy=524 mmfrom face of column; Lx - (Lx / 2 + Dx / 2 + D - X_bar)
shear Force at Critical Section XX=1.5*0.258*2.4* ((267.71+163.54+256.42+152.25) / 4)Vux=195.03kNVux = gf * CSx * Lx * ((QmaxXY + QmaxX + q1 + q2) / 4)
shear Force at Critical Section YY=1.5*0.524*2* ((267.71+180.21+244.97+157.47) / 4)Vuy=336.64kNVuy = gf * CSy * Ly * ((QmaxXY + QmaxY + q3 + q4) / 4)
Shear Planq1; q2; q3; q4=256.42; 152.25; 244.97; 157.47kNq1 = 180.21+ (((2-0.258) / 2) * (267.71-180.21)); q2 = 76.04+ (((2-0.258) / 2) * (163.54-76.04)); q3 = 163.54+ (((2.4-0.524) / 2.4) * (267.71-163.54)); q4 = 76.04+ (((2.4-0.258) / 2.4) * (163.54-76.04))
Shear stress long span=195.03* 10 ^ 3 / (2.4*0.484 * 10 ^ 6)tvx=0.168N/mm^2Vux * 10 ^ 3 / (Lx * Dcs * 10 ^ 6)
Shear stress short span=336.64* 10 ^ 3 / (2*0.468 * 10 ^ 6)tvy=0.36N/mm^2Vuy * 10 ^ 3 / (Ly * Dcs * 10 ^ 6)
Beta Constant= 0.8 * 25/ (6.89 * 0.38)beta=7.639Constant8 * fck / (6.89 * pt)
Shear strength of concrete=(0.85 * Sqrt(0.8 *25)* (Sqrt(1 + 5 *7.639) - 1)) / (6 * 7.639)tc=0.436N/mm^20.85 * Sqrt(0.8 * fck * (Sqrt(1 + 5 * beta) - 1)) / (6 * beta)
Shear along XX Pass
Shear along YY Pass
Revised eff DepthD=0.542mm
Steel along XX
Reqd Steel Percentage %=0.24pt=0.24%By trials; pt = 0.22 produces lower Moment of Resistance than 233.376
Moment of Resistance= 1000*(542^2) * ((415/1.15 ) * (0.24/100 ) * (1-(0.416 * (415/1.15) / (0.36*25) * (0.24/100))))/10^6MR=244.24kN-mB * (D ^ 2) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100))))
Steel Required=0.24*1000*542/100Ast=1300.8mm^2Ast = pt * B * D / 100
Spacing Required: #16@160c/c=Round(Round(1000/(1300.8/ (3.142*16^2/4)) ,0) /10,0) *10Spacing=160mmB / (Ast / (PI * dia ^ 2 / 4)) & rounded to nearest 10 mm
Steel along YY
Reqd Steel Percentage %=0.38pt=0.38%By trials; pt = 0.36 produces lower Moment of Resistance than 342.832
Moment of Resistance= 1000*(526^2) * ((415/1.15 ) * (0.38/100 ) * (1-(0.416 * (415/1.15) / (0.36*25) * (0.38/100))))/10^6MR=355.36kN-mB * (D ^ 2) * ((fy / gs) * (pt / 100) * (1 - (k2 * (fy / gs) / (k1 * fck) * (pt / 100))))
Steel Required=0.38*1000*526/100Ast=1998.8mm^2Ast = pt * B * D / 100
Spacing Required: #16@100c/c=Round(Round(1000/(1998.8/ (3.142*16^2/4)) ,0) /10,0) *10Spacing=100mmB / (Ast / (PI * dia ^ 2 / 4)) & rounded to nearest 10 mm
Punching Shear: 0=
Punching Shear Sectional Dimensions
Depth at the sectionDcs=0.842m
D/2 away from Column=2*((0.3+0.842))+2*(0.4+0.842)BO=4.768m2 * ((Dx + D)) + 2 * (Dy + D)
Shear Area=(2.4*2)-((0.3+0.842)) * (0.4+0.842)Area=3.382sqm((Dx + D)) * (Dy + D)
Shear Stress=1.5*3.382*((267.71+163.54+180.21+76.04)/4 *10^3) /(4.768*0.842*10 ^ 6)tv=0.22N/mm^2gf * A * ((QmaxXY + QmaxX + QmaxY + Qminn) / 4 * 10 ^ 3) / (BO * D * 10 ^ 6)
Section Passes for Punching Shear
Punching Shear: 1=
Punching Shear Sectional Dimensions
Depth at the sectionDcs=0.542m
D/2 away from Step/Pedastal: 1=2*((0.6+0.542))+2*(0.7+0.542)BO=4.768m2 * ((Dx + D)) + 2 * (Dy + D)
Shear Area=(2.4*2)-((0.6+0.542)) * (0.7+0.542)Area=3.382sqm((Dx + D)) * (Dy + D)
Shear Stress=1.5*3.382*((267.71+163.54+180.21+76.04)/4 *10^3) /(4.768*0.542*10 ^ 6)tv=0.34N/mm^2gf * A * ((QmaxXY + QmaxX + QmaxY + Qminn) / 4 * 10 ^ 3) / (BO * D * 10 ^ 6)
Section Passes for Punching Shear
Footing Rebar=Steps: 1; Step Offset: 0.3
Rebars on both directions
Bars along XX Direction#16 mm @100=
Bars along YY Direction#20 mm @153=