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Vertical MomentNode Fy kN Mx kNm My kNm Mz kNm
C1 1 356 4.522 0.431 -1.419C2 3 597 2.989 0.296 0.632C3 5 1073 10.289 -0.977 -0.057C4 7 1292 39.396 -0.101 4.64C5 9 772 4.81 0.161 5.503C6 11 668 -1.838 0.202 -25.524C1 13 225 -2.706 0.128 -2.116C2 15 639 3.886 0.203 1.099C6 17 625 1.506 -0.128 -2.736C1 19 225 -3.875 -0.034 -3.355C2 21 565 2.765 -0.3 3.342C4 23 1314 -7.703 -1.417 -2.205C4 25 865 -28.873 1.942 0.455C2 27 671 -0.038 0.174 8.419C2 30 605 9.814 0.159 -6.216C4 32 838 0.28 0.177 -22.563C3 34 935 -6.469 0.018 24.37C6 36 776 -5.157 -0.184 -11.478C6 38 937 -3.1 0.454 14.582C2 44 540 0.695 0.119 -9.763C2 46 651 1.093 0.022 7.289C6 50 614 -0.024 -0.216 5.543C1 52 286 -0.135 -0.043 -1.087C6 56 834 3.242 -0.071 3.597C4A 60 935 2.392 -0.464 11.502C4 62 1168 0.11 0.624 -26.429C3 64 992 -1.707 0.493 39.274C3 66 1008 -2.059 -0.153 4.936C4 68 878 -2.455 0.602 -4.452C3 70 1028 -4.049 0.053 8.289C4 72 907 1.873 -0.001 -6.647C9 74 2019 -1.117 0.057 1.808C4A 76 1079 0.144 0.085 -1.63C7A 78 1422 -33.769 -0.134 0.584C6 80 732 -14.246 -0.278 6.812C7 82 1409 7.711 -0.105 3.732C3A 84 948 -16.307 0.197 -7.8C2 89 497 -10.973 -0.109 -3.511C8 90 794 -1.34 0.175 2.429C8 93 748 -6.375 -0.248 -3.37C10 97 263 8.205 0.013 0.618
99 616 0.624 0.012 0.07C10 101 -29 7.564 0.01 7.09C10 103 266 0.799 0.021 7.732C8A 105 305 1.588 -0.001 0.561C8A 107 303 0.151 0.069 0.152C8A 109 150 2.452 -0.005 0.35C8A 111 160 1.439 0.05 0.92C1 325 296 -0.177 0.083 0.957
328 78 0 0 0329 80 0 0 0330 85 0 0 0
331 70 0 0 0332 29 0 0 0333 56 0 0 0
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 356 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.3 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F1
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 1.35 m
Area of the Footing Required = 1.96 Sqm.Footing Size Required = 1.40 m each side Cx
However ProvideFx = 1.35 mFy = 1.45 m
Area Provided = 1.96 Sqm. Cy Fy = 1.45X X m
Zx = 0.47 CumZy = 0.44 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 200.05 Fmin= 200.05 0.560 0.575 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 181.86 KN/Sqm
D = 0.35 md = 0.295 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.25 N/Sqmm 0.35m
Shear Stress (Tvy) in Y = 0.26 N/Sqmm
0.56DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.56 m CantileverMoment = 28.52 KN-m
Ultimate Bending Moment => Mu = 42.77
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.12
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 104.82 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 103.12 kN
Nominal Shear Stress (qv)
0.25 N/sqmm < SAFE
0.26 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.15 m(b+a) = 0.53 m a = 0.295 m(d+a) = 0.60 m
2240.00 mm
Shear Force at this critical section = 299190.04 N
Nominal Shear Stress (qv)
0.68 N/sqmm < SAFE
short side of the column= 0.77
long side of the column
0.5+βc = 1.27But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 0.56 m Cantilever
Moment = 28.52 KN-m
0.49 AREA OF BARS ast1= 78.525Y 8 = 50.272 Sqmm N= 6.66667
Pt = 0.14 % Y 10 = 78.55 Sqmm ast= 523.5Y 12 = ###
413.85 Sqmm Y 16 = ###Provide Y 10 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 10 mm @ 150 mm c/c Y 25 = ###
524 Sqmm Y 32 = ###
For 0.575 m Cantilever Pt , provided= 0.1775 %
Moment = 30.06 KN-m 0.3096 N/sq.mm
0.52Pt = 0.15 %
Ast = 437.05 Sqmm ast1= 78.525N= 6.66667
Provide Y 10 mm @ ### ast= 523.5However provide Y 10 mm @ 150 mm c/c Pt, provided= ### %
524 Sqmm 0.3096 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 671 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.45 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F2
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 1.85 m
Area of the Footing Required = 3.69 Sqm.Footing Size Required = 1.92 m each side Cx
However ProvideFx = 1.85 mFy = 2 m
Area Provided = 3.70 Sqm. Cy Fy = 2X X m
Zx = 1.23 CumZy = 1.14 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 199.49 Fmin= 199.49 0.810 0.775 m
< 200SAFE SAFE
Y
Net Upward Pressure = 181.35 KN/Sqm
D = 0.5 md = 0.445 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.22 N/Sqmm 0.5m
Shear Stress (Tvy) in Y = 0.20 N/Sqmm
0.81DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.81 m CantileverMoment = 59.49 KN-m
Ultimate Bending Moment => Mu = 89.24
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.18
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 198.58 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 166.07 kN
Nominal Shear Stress (qv)
0.22 N/sqmm < SAFE
0.20 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.22 m(b+a) = 0.68 m a = 0.445 m(d+a) = 0.90 m
3140.00 mm
Shear Force at this critical section = 561441.11 N
Nominal Shear Stress (qv)
0.60 N/sqmm < SAFE
short side of the column= 0.51
long side of the column
0.5+βc = 1.01But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 0.81 m Cantilever
Moment = 59.49 KN-m
0.45 AREA OF BARS ast1= 78.525Y 8 = 50.272 Sqmm N= 6.66667
Pt = 0.13 % Y 10 = 78.55 Sqmm ast= 523.5Y 12 = ###
570.90 Sqmm Y 16 = ###Provide Y 10 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 10 mm @ 150 mm c/c Y 25 = ###
524 Sqmm Y 32 = ###
For 0.775 m Cantilever Pt , provided= 0.1176 %
Moment = 54.46 KN-m 0.2579 N/sq.mm
0.41Pt = 0.12 %
Ast = 521.39 Sqmm ast1= 78.525N= 6.66667
Provide Y 10 mm @ ### ast= 523.5However provide Y 10 mm @ 150 mm c/c Pt, provided= ### %
524 Sqmm 0.2579 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1073 KN Cx = 0.3 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F2
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.35 m
Area of the Footing Required = 5.90 Sqm.Footing Size Required = 2.43 m each side Cx
However ProvideFx = 2.35 mFy = 2.5 m
Area Provided = 5.88 Sqm. Cy Fy = 2.5X X m
Zx = 2.45 CumZy = 2.30 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 200.90 Fmin= 200.90 1.025 0.950 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 182.64 KN/Sqm
D = 0.6 md = 0.544 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.24 N/Sqmm 0.6m
Shear Stress (Tvy) in Y = 0.20 N/Sqmm
1.025DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.025 m CantileverMoment = 95.94 KN-m
Ultimate Bending Moment => Mu = 143.91
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.23
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 329.43 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 261.38 kN
Nominal Shear Stress (qv)
0.24 N/sqmm < SAFE
0.20 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.27 m(b+a) = 0.84 m a = 0.544 m(d+a) = 1.14 m
3976.00 mm
Shear Force at this critical section = 896656.15 N
Nominal Shear Stress (qv)
0.62 N/sqmm < SAFE
short side of the column= 0.50
long side of the column
0.5+βc = 1.00But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 1.025 m Cantilever
Moment = 95.94 KN-m
0.49 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 6.66667
Pt = 0.14 % Y 10 = 78.55 Sqmm ast= 753.84Y 12 = ###
754.81 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 150 mm c/c Y 25 = ###
754 Sqmm Y 32 = ###
For 0.95 m Cantilever Pt , provided= 0.1386 %
Moment = 82.42 KN-m 0.2776 N/sq.mm
0.42Pt = 0.12 %
Ast = 645.62 Sqmm ast1= 113.076N= 6.66667
Provide Y 12 mm @ ### ast= 753.84However provide Y 12 mm @ 150 mm c/c Pt, provided= ### %
754 Sqmm 0.2776 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 948 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F3A
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.25 m
Area of the Footing Required = 5.21 Sqm.Footing Size Required = 2.28 m each side Cx
However ProvideFx = 2.25 mFy = 2.35 m
Area Provided = 5.29 Sqm. Cy Fy = 2.35X X m
Zx = 2.07 CumZy = 1.98 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 197.22 Fmin= 197.22 1.010 0.875 m
< 200SAFE SAFE
Y
Net Upward Pressure = 179.29 KN/Sqm
D = 0.475 md = 0.419 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.38 N/Sqmm 0.475m
Shear Stress (Tvy) in Y = 0.29 N/Sqmm
1.01DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.01 m CantileverMoment = 91.45 KN-m
Ultimate Bending Moment => Mu = 137.17
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.22
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 373.51 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 275.93 kN
Nominal Shear Stress (qv)
0.38 N/sqmm < UNSAFE
0.29 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.21 m(b+a) = 0.65 m a = 0.419 m(d+a) = 1.02 m
3336.00 mm
Shear Force at this critical section = 829429.45 N
Nominal Shear Stress (qv)
0.89 N/sqmm < SAFE
short side of the column= 0.38
long side of the column
0.5+βc = 0.88But Ks is limited to 1.00 0.88333333
Permissible Nominal Shear Stress (qc)
0.987597 N/sqmm
DESIGN OF REINFORCEMENTFor 1.01 m Cantilever
Moment = 91.45 KN-m
0.78 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 9.09091
Pt = 0.23 % Y 10 = 78.55 Sqmm ast= 1027.96Y 12 = ###
952.08 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 110 mm c/c Y 25 = ###
1028 Sqmm Y 32 = ###
For 0.875 m Cantilever Pt , provided= 0.2453 %
Moment = 68.63 KN-m 0.3563 N/sq.mm
0.59Pt = 0.17 %
Ast = 705.53 Sqmm ast1= 113.076N= 6.66667
Provide Y 12 mm @ 160.32159622452 mm c/c ast= 753.84However provide Y 12 mm @ 150 mm c/c Pt, provided= ### %
754 Sqmm 0.3115 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1292 KN Cx = 0.3 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F4
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.55 m
Area of the Footing Required = 7.11 Sqm.Footing Size Required = 2.67 m each side Cx
However ProvideFx = 2.55 mFy = 2.8 m
Area Provided = 7.14 Sqm. Cy Fy = 2.8X X m
Zx = 3.33 CumZy = 3.03 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 199.05 Fmin= 199.05 1.125 1.100 m
< 200SAFE SAFE
Y
Net Upward Pressure = 180.95 KN/Sqm
D = 0.6 md = 0.544 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.29 N/Sqmm 0.6m
Shear Stress (Tvy) in Y = 0.28 N/Sqmm
1.125DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.125 m CantileverMoment = 114.51 KN-m
Ultimate Bending Moment => Mu = 171.76
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.25
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 441.56 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 384.83 kN
Nominal Shear Stress (qv)
0.29 N/sqmm < SAFE
0.28 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.27 m(b+a) = 0.84 m a = 0.544 m(d+a) = 1.14 m
3976.00 mm
Shear Force at this critical section = 1117283.96 N
Nominal Shear Stress (qv)
0.77 N/sqmm < SAFE
short side of the column= 0.50
long side of the column
0.5+βc = 1.00But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 1.125 m Cantilever
Moment = 114.51 KN-m
0.58 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.17 % Y 10 = 78.55 Sqmm ast= 942.3Y 12 = ###
906.27 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 120 mm c/c Y 25 = ###
942 Sqmm Y 32 = ###
For 1.1 m Cantilever Pt , provided= 0.1732 %
Moment = 109.48 KN-m 0.3064 N/sq.mm
0.55Pt = 0.16 %
Ast = 865.03 Sqmm ast1= 113.076N= 7.69231
Provide Y 12 mm @ 130.76016976329 mm c/c ast= 869.815However provide Y 12 mm @ 130 mm c/c Pt, provided= ### %
870 Sqmm 0.2958 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1079 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F4
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.35 m
Area of the Footing Required = 5.93 Sqm.Footing Size Required = 2.44 m each side Cx
However ProvideFx = 2.35 mFy = 2.65 m
Area Provided = 6.23 Sqm. Cy Fy = 2.65X X m
Zx = 2.75 CumZy = 2.44 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 190.59 Fmin= 190.59 1.060 1.025 m
< 200SAFE SAFE
Y
Net Upward Pressure = 173.26 KN/Sqm
D = 0.6 md = 0.544 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.25 N/Sqmm 0.6m
Shear Stress (Tvy) in Y = 0.23 N/Sqmm
1.06DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.06 m CantileverMoment = 97.34 KN-m
Ultimate Bending Moment => Mu = 146.01
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.23
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 355.38 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 293.77 kN
Nominal Shear Stress (qv)
0.25 N/sqmm < SAFE
0.23 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.27 m(b+a) = 0.77 m a = 0.544 m(d+a) = 1.14 m
3836.00 mm
Shear Force at this critical section = 925582.57 N
Nominal Shear Stress (qv)
0.67 N/sqmm < SAFE
short side of the column= 0.38
long side of the column
0.5+βc = 0.88But Ks is limited to 1.00 0.88333333
Permissible Nominal Shear Stress (qc)
0.987597 N/sqmm
DESIGN OF REINFORCEMENTFor 1.06 m Cantilever
Moment = 97.34 KN-m
0.49 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.14 % Y 10 = 78.55 Sqmm ast= 942.3Y 12 = ###
766.15 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 120 mm c/c Y 25 = ###
942 Sqmm Y 32 = ###
For 1.025 m Cantilever Pt , provided= 0.1732 %
Moment = 91.02 KN-m 0.3064 N/sq.mm
0.46Pt = 0.13 %
Ast = 714.95 Sqmm ast1= 113.076N= 7.69231
Provide Y 12 mm @ ### ast= 869.815However provide Y 12 mm @ 130 mm c/c Pt, provided= ### %
870 Sqmm 0.2958 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 772 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.45 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F4
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2 m
Area of the Footing Required = 4.25 Sqm.Footing Size Required = 2.06 m each side Cx
However ProvideFx = 2 mFy = 2.2 m
Area Provided = 4.40 Sqm. Cy Fy = 2.2X X m
Zx = 1.61 CumZy = 1.47 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 193.00 Fmin= 193.00 0.885 0.875 m
< 200SAFE SAFE
Y
Net Upward Pressure = 175.45 KN/Sqm
D = 0.45 md = 0.394 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.33 N/Sqmm 0.45m
Shear Stress (Tvy) in Y = 0.32 N/Sqmm
0.885DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.885 m CantileverMoment = 68.71 KN-m
Ultimate Bending Moment => Mu = 103.07
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.19
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 284.29 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 253.18 kN
Nominal Shear Stress (qv)
0.33 N/sqmm < SAFE
0.32 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.20 m(b+a) = 0.62 m a = 0.394 m(d+a) = 0.84 m
2936.00 mm
Shear Force at this critical section = 679595.81 N
Nominal Shear Stress (qv)
0.88 N/sqmm < SAFE
short side of the column= 0.51
long side of the column
0.5+βc = 1.01But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 0.885 m Cantilever
Moment = 68.71 KN-m
0.66 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.19 % Y 10 = 78.55 Sqmm ast= 942.3Y 12 = ###
754.89 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 120 mm c/c Y 25 = ###
942 Sqmm Y 32 = ###
For 0.875 m Cantilever Pt , provided= 0.2392 %
Moment = 67.17 KN-m 0.3524 N/sq.mm
0.65Pt = 0.19 %
Ast = 737.21 Sqmm ast1= 113.076N= 7.69231
Provide Y 12 mm @ ### ast= 869.815However provide Y 12 mm @ 130 mm c/c Pt, provided= ### %
870 Sqmm 0.3405 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 834 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F6
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2 m
Area of the Footing Required = 4.59 Sqm.Footing Size Required = 2.14 m each side Cx
However ProvideFx = 2 mFy = 2.25 m
Area Provided = 4.50 Sqm. Cy Fy = 2.25X X m
Zx = 1.69 CumZy = 1.50 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 203.87 Fmin= 203.87 0.885 0.825 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 185.33 KN/Sqm
D = 0.45 md = 0.394 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.35 N/Sqmm 0.45m
Shear Stress (Tvy) in Y = 0.30 N/Sqmm
0.885DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.885 m CantileverMoment = 72.58 KN-m
Ultimate Bending Moment => Mu = 108.87
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.20
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 307.12 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 239.64 kN
Nominal Shear Stress (qv)
0.35 N/sqmm < SAFE
0.30 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.20 m(b+a) = 0.62 m a = 0.394 m(d+a) = 0.99 m
3236.00 mm
Shear Force at this critical section = 719045.89 N
Nominal Shear Stress (qv)
0.85 N/sqmm < SAFE
short side of the column= 0.38
long side of the column
0.5+βc = 0.88But Ks is limited to 1.00 0.88333333
Permissible Nominal Shear Stress (qc)
0.987597 N/sqmm
DESIGN OF REINFORCEMENTFor 0.885 m Cantilever
Moment = 72.58 KN-m
0.70 AREA OF BARS ast1= 113.076Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.20 % Y 10 = 78.55 Sqmm ast= 942.3Y 12 = ###
799.34 Sqmm Y 16 = ###Provide Y 12 mm @ 141.50610097768 mm c/c Y 20 = 314.2 Sqmm
However Provide Y 12 mm @ 120 mm c/c Y 25 = ###
942 Sqmm Y 32 = ###
For 0.825 m Cantilever Pt , provided= 0.2392 %
Moment = 63.07 KN-m 0.3524 N/sq.mm
0.61Pt = 0.18 %
Ast = 690.50 Sqmm ast1= 113.076N= 7.69231
Provide Y 12 mm @ ### ast= 869.815However provide Y 12 mm @ 130 mm c/c Pt, provided= ### %
870 Sqmm 0.3405 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1409 KN Cx = 0.23 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F7
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.65 m
Area of the Footing Required = 7.75 Sqm.Footing Size Required = 2.78 m each side Cx
However ProvideFx = 2.65 mFy = 2.9 m
Area Provided = 7.69 Sqm. Cy Fy = 2.9X X m
Zx = 3.71 CumZy = 3.39 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 201.68 Fmin= 201.68 1.210 1.150 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 183.34 KN/Sqm
D = 0.6 md = 0.542 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.34 N/Sqmm 0.6m
Shear Stress (Tvy) in Y = 0.31 N/Sqmm
1.21DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.21 m CantileverMoment = 134.22 KN-m
Ultimate Bending Moment => Mu = 201.33
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.27
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 532.76 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 443.11 kN
Nominal Shear Stress (qv)
0.34 N/sqmm < SAFE
0.31 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.27 m(b+a) = 0.77 m a = 0.542 m(d+a) = 1.14 m
3828.00 mm
Shear Force at this critical section = 1247359.37 N
Nominal Shear Stress (qv)
0.90 N/sqmm < SAFE
short side of the column= 0.38
long side of the column
0.5+βc = 0.88But Ks is limited to 1.00 0.88333333
Permissible Nominal Shear Stress (qc)
0.987597 N/sqmm
DESIGN OF REINFORCEMENTFor 1.21 m Cantilever
Moment = 134.22 KN-m
0.69 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.20 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
1073.43 Sqmm Y 16 = ###Provide Y 16 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 1.15 m Cantilever Pt , provided= 0.3091 %
Moment = 121.24 KN-m 0.3930 N/sq.mm
0.62Pt = 0.18 %
Ast = 965.45 Sqmm ast1= 201.024N= 7.69231
Provide Y 16 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.3799 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1422 KN Cx = 0.3 mUnfactored MOM-Mx = 0 KN-m Cy = 0.6 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F7A
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.65 m
Area of the Footing Required = 7.82 Sqm.Footing Size Required = 2.80 m each side Cx
However ProvideFx = 2.65 mFy = 2.9 m
Area Provided = 7.69 Sqm. Cy Fy = 2.9X X m
Zx = 3.71 CumZy = 3.39 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 203.54 Fmin= 203.54 1.175 1.150 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 185.04 KN/Sqm
D = 0.6 md = 0.542 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.32 N/Sqmm 0.6m
Shear Stress (Tvy) in Y = 0.31 N/Sqmm
1.175DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.175 m CantileverMoment = 127.73 KN-m
Ultimate Bending Moment => Mu = 191.60
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.26
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 509.51 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 447.19 kN
Nominal Shear Stress (qv)
0.32 N/sqmm < SAFE
0.31 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.27 m(b+a) = 0.84 m a = 0.542 m(d+a) = 1.14 m
3968.00 mm
Shear Force at this critical section = 1244076.25 N
Nominal Shear Stress (qv)
0.87 N/sqmm < SAFE
short side of the column= 0.50
long side of the column
0.5+βc = 1.00But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 1.175 m Cantilever
Moment = 127.73 KN-m
0.65 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.19 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
1019.37 Sqmm Y 16 = ###Provide Y 16 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 1.15 m Cantilever Pt , provided= 0.3091 %
Moment = 122.35 KN-m 0.3930 N/sq.mm
0.62Pt = 0.18 %
Ast = 974.72 Sqmm ast1= 201.024N= 7.69231
Provide Y 16 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.3799 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 794 KN Cx = 0.38 mUnfactored MOM-Mx = 0 KN-m Cy = 0.38 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F8
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 2.05 m
Area of the Footing Required = 4.37 Sqm.Footing Size Required = 2.09 m each side Cx
However ProvideFx = 2.05 mFy = 2.05 m
Area Provided = 4.20 Sqm. Cy Fy = 2.05X X m
Zx = 1.44 CumZy = 1.44 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 207.83 Fmin= 207.83 0.835 0.835 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 188.94 KN/Sqm
D = 0.45 md = 0.394 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.32 N/Sqmm 0.45m
Shear Stress (Tvy) in Y = 0.32 N/Sqmm
0.835DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.835 m CantileverMoment = 65.87 KN-m
Ultimate Bending Moment => Mu = 98.80
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.19
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 256.21 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 256.21 kN
Nominal Shear Stress (qv)
0.32 N/sqmm < SAFE
0.32 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.20 m(b+a) = 0.77 m a = 0.394 m(d+a) = 0.77 m
3096.00 mm
Shear Force at this critical section = 680813.48 N
Nominal Shear Stress (qv)
0.84 N/sqmm < SAFE
short side of the column= 1.00
long side of the column
0.5+βc = 1.50But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 0.835 m Cantilever
Moment = 65.87 KN-m
0.64 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.18 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
722.34 Sqmm Y 16 = ###Provide Y 12 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 0.835 m Cantilever Pt , provided= 0.4252 %
Moment = 65.87 KN-m 0.4483 N/sq.mm
0.64Pt = 0.18 %
Ast = 722.34 Sqmm ast1= 201.024N= 7.69231
Provide Y 12 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.4339 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 305 KN Cx = 0.38 mUnfactored MOM-Mx = 0 KN-m Cy = 0.38 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F8A
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 1.35 m
Area of the Footing Required = 1.68 Sqm.Footing Size Required = 1.30 m each side Cx
However ProvideFx = 1.35 mFy = 1.35 m
Area Provided = 1.82 Sqm. Cy Fy = 1.35X X m
Zx = 0.41 CumZy = 0.41 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 184.09 Fmin= 184.09 0.485 0.485 m
< 200SAFE SAFE
Y
Net Upward Pressure = 167.35 KN/Sqm
D = 0.325 md = 0.27 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.20 N/Sqmm 0.325m
Shear Stress (Tvy) in Y = 0.20 N/Sqmm
0.485DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 0.485 m CantileverMoment = 19.68 KN-m
Ultimate Bending Moment => Mu = 29.52
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.10
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 72.86 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 72.86 kN
Nominal Shear Stress (qv)
0.20 N/sqmm < SAFE
0.20 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.14 m(b+a) = 0.65 m a = 0.27 m(d+a) = 0.65 m
2600.00 mm
Shear Force at this critical section = 234293.55 N
Nominal Shear Stress (qv)
0.50 N/sqmm < SAFE
short side of the column= 1.00
long side of the column
0.5+βc = 1.50But Ks is limited to 1.00 1
Permissible Nominal Shear Stress (qc)
1.118034 N/sqmm
DESIGN OF REINFORCEMENTFor 0.485 m Cantilever
Moment = 19.68 KN-m
0.40 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.11 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
310.42 Sqmm Y 16 = ###Provide Y 10 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 0.485 m Cantilever Pt , provided= 0.6204 %
Moment = 19.68 KN-m 0.5208 N/sq.mm
0.40Pt = 0.11 %
Ast = 310.42 Sqmm ast1= 201.024N= 7.69231
Provide Y 10 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.5048 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 2019 KN Cx = 0.3 mUnfactored MOM-Mx = 0 KN-m Cy = 0.75 mUnfactored MOM-My = 0 KN-mSBC = 200 Fck = 20 N/Sqmm FOOTING NO. F9
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 3.15 m
Area of the Footing Required = 11.10 Sqm.Footing Size Required = 3.33 m each side Cx
However ProvideFx = 3.15 mFy = 3.5 m
Area Provided = 11.03 Sqm. Cy Fy = 3.5X X m
Zx = 6.43 CumZy = 5.79 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 201.44 Fmin= 201.44 1.425 1.375 m
< 200UNSAFE SAFE
Y
Net Upward Pressure = 183.13 KN/Sqm
D = 0.7 md = 0.642 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.34 N/Sqmm 0.7m
Shear Stress (Tvy) in Y = 0.31 N/Sqmm
1.425DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.425 m CantileverMoment = 185.93 KN-m
Ultimate Bending Moment => Mu = 278.90
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.32
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 752.80 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 634.25 kN
Nominal Shear Stress (qv)
0.34 N/sqmm < SAFE
0.31 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.32 m(b+a) = 0.94 m a = 0.642 m(d+a) = 1.39 m
4668.00 mm
Shear Force at this critical section = 1778869.20 N
Nominal Shear Stress (qv)
0.89 N/sqmm < SAFE
short side of the column= 0.40
long side of the column
0.5+βc = 0.90But Ks is limited to 1.00 0.9
Permissible Nominal Shear Stress (qc)
1.006231 N/sqmm
DESIGN OF REINFORCEMENTFor 1.425 m Cantilever
Moment = 185.93 KN-m
0.68 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.20 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
1254.71 Sqmm Y 16 = ###Provide Y 16 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 1.375 m Cantilever Pt , provided= 0.2609 %
Moment = 173.11 KN-m 0.3658 N/sq.mm
0.63Pt = 0.18 %
Ast = 1164.67 Sqmm ast1= 201.024N= 7.69231
Provide Y 16 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.3535 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
DESIGN OF FLAT FOOTINGLOAD DATA Dimensions of columnUnfactored Load P = 1400 KN Cx = 0.23 mUnfactored MOM-Mx = 114 KN-m Cy = 0.6 mUnfactored MOM-My = 91 KN-mSBC = 180 Fck = 30 N/Sqmm FOOTING NO. F10
Fy= 415 N/Sqmm YSTABILITY CHECK - Check for Stresses Fx = 3.2 m
Area of the Footing Required = 8.56 Sqm.Footing Size Required = 2.92 m each side Cx
However ProvideFx = 3.2 mFy = 3.5 m
Area Provided = 11.20 Sqm. Cy Fy = 3.5X X m
Zx = 6.53 CumZy = 5.97 Cum
f = P/A + Mx/Zx + My/Zy
Fmax = 170.18 Fmin= 104.82 1.485 1.450 m
< 180SAFE SAFE
Y
Net Upward Pressure = 157.68 KN/Sqm
D = 0.7 md = 0.642 m
S F in X dir @ d = ### KNS F in Y dir @ d = ### KN
Shear Stress (Tvx) in X = 0.31 N/Sqmm 0.7m
Shear Stress (Tvy) in Y = 0.30 N/Sqmm
1.485DEPTH OF FOOTING REQUIRED AS PER MOMENT
LIMITING MOMENT OF RESISTANCE
Mulim =
For 1.485 m CantileverMoment = 173.86 KN-m
Ultimate Bending Moment => Mu = 260.80
b (UNIT WIDTH) = 1000
Equating the limiting moment of resistance to the ultimate bending moment
d = 0.25
CHECK FOR SHEAR STRESSES
1) CRITICAL SECTION FOR ONE WAY SHEARThe critical section for one way shear is taken at a distance equal to the effective depth from the column face.
Shear Force at critical section along X directionSFu X = 697.87 kN
KN/m2 KN/m2
KN/m2
0.138fckbd2
d=(SQRT(Mu/0.138fckb))
Shear Force at critical section along Z directionSFu Z = 611.56 kN
Nominal Shear Stress (qv)
0.31 N/sqmm < SAFE
0.30 N/sqmm SAFE
2) CRITICAL SECTION FOR TWO WAY SHEARThe critical section for two way shear is taken at a distance equal to half the effective depth from the column face alround.
d/2 = 0.32 m(b+a) = 0.87 m a = 0.642 m(d+a) = 1.24 m
4228.00 mm
Shear Force at this critical section = 1595278.71 N
Nominal Shear Stress (qv)
0.88 N/sqmm < SAFE
short side of the column= 0.38
long side of the column
0.5+βc = 0.88But Ks is limited to 1.00 0.88333333
Permissible Nominal Shear Stress (qc)
1.209554 N/sqmm
DESIGN OF REINFORCEMENTFor 1.485 m Cantilever
Moment = 173.86 KN-m
0.63 AREA OF BARS ast1= 201.024Y 8 = 50.272 Sqmm N= 8.33333
Pt = 0.18 % Y 10 = 78.55 Sqmm ast= 1675.2Y 12 = ###
1154.39 Sqmm Y 16 = ###Provide Y 16 mm @ ### Y 20 = 314.2 Sqmm
However Provide Y 16 mm @ 120 mm c/c Y 25 = ###
1676 Sqmm Y 32 = ###
For 1.45 m Cantilever Pt , provided= 0.2609 %
Moment = 165.76 KN-m 0.3759 N/sq.mm
0.60Pt = 0.17 %
Ast = 1099.28 Sqmm ast1= 201.024N= 7.69231
Provide Y 16 mm @ ### ast= 1546.34However provide Y 16 mm @ 130 mm c/c Pt, provided= ### %
1547 Sqmm 0.3629 N/sq.mm
qvx = qc
qvz =
b0 =
Su =
qv = qc
βc =
Ks =
qc = Ksx0.25(sqrt(fck)) =
Mu/bD2 =
Ast.reqd. =
Ast.provd. =
qc =
Mu/bD2 =
Ast.provd. = qc =
