Strap Footing

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C-25 Km 19.00 20.00 21.00 22.00 23.00 24.00 25.00 26.00 27.00 28.00 29.00 30.00 31.00 32.00 33.00 34.00 35.00 36.00 37.00 38.00 39.00 40.00 41.00 42.00 43.00 44.00 45.00 46.00 47.00 48.00 49.00 50.00 51.00 52.00 53.00 54.00 55.00 56.00 57.00 58.00

S-300 Ks 3.95 3.96 3.96 3.97 3.98 3.98 3.99 4.00 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10 4.11 4.12 4.15 4.17 4.18 4.20 4.21 4.24 4.26 4.29 4.31 4.34 4.36 4.39 4.41 4.46 4.49 4.51 4.56 4.62 4.65 4.67 4.73

Design of strap footing foundation I) Proportioning of footingsTrial 1

Xc

column1 1.5 m m m m m m m m KN/m2 m m b1

Assume half length of a1= a1= 3 P1= 2291.00 bc1,1= 0.5 bc1,2= 0.5 P2= 2618.00 bc2,1= 0.5 bc2,2= 0.5 Xc= 4.4 sall= 300 e= 1.25 XR= 3.15

Column2 b2

a1

a2

P1

Ws xs

P2

Determine soil reaction, R 1 Note: neglect wt. of strap beam Area of F2 A(F2)= 10.67 R1= 3200.13 KN b 1= 3.56 Dimensions of footing 1= For footing 2 3 X 3.56

R1 e

XR

R2

R2= 1708.87 KN 2 Area of F2 A(F2)= 5.70 m a 2= 1.602 m b 2= 3.56 m Dimensions of footing 2= 1.602003 X 3.56 q1= 1066.71 KN/m q2= 1066.71 KN/m Note: The Distribution per meter run under the pad is the same for each pad. II) Structural Design a) Calculation of shear force and bending moment diagram

1763.56 266.68 909.13

xo xo= 1.8977273-2024.32

-854.44

Shear force diagram-346.76

33.33 -728.2122445

Bending Moment diagram-1887.47 -1500.06

b) Determination of Depths of Footingsb-1) Footing 1

i) Punching shear fck= fcd= fs= 25 Mpa 11.166667 Mpa 300 Mpa

rmin=fctd= vup= vud= ax +bx+c=0 a= b= c= d1 d2 d= D=2

0.002 1.167 Mpa 641.67 Kpa 385.00 Kpa 1433.33 1187.50 -2216.00 0.90 -1.72 0.90 933

0.90

1893.81

1893.81

0.00

ii) Diagonal tension (Wide beam shear) V= 1068.18 Existing shear= 334.75 d 0.90 m D 931 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= 570.28924 v= 212.07945

ok!b-1) Footing 2

i) Punching shear Concrete shear resistance=

2*(bc2,1+d+bc2,2)*d*vcp(1)

Net force along the perimeter= P2-(bc2,1+d)(bc2,2+d)sas(2) Equating eq (1) and (2), we obtain a quadratic equation a= 2866.67 b= 1583.33 c= -2543.00 d1 0.71 m 2182.14 2182.14 d2 -1.26 m d= 0.705 m D 742 ii) Diagonal tension (Wide beam shear) V= 1011.1642 Existing shear= 402.6889 d 0.738 m D 773 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= 396.32969 v= 335.33759

0.00

ok!D 773

C) Reinforcementc-1) Footing 1 i) Direction X (TOP) Maximum moment= 1887.47 KNm -1887.47 Km= 48 Ks= 4.36 MIN. REINF 2 2 Asteel = 9181.03357 mm 1793 mm f 24 452.39 c/c 46.96 ii) Direction Y (BOTTOM) M= 351.14 Knm/m Km= 21 Ks= 3.96 MIN. REINF 2 2 Asteel = 1552.59929 mm 1793 mm f 16 201.06 c/c 100.83 c-1) Footing 2 i) Direction X Maximum moment= 728.21 KNm Km= 37 Ks= 4.12 MIN. REINF 2 2 Asteel = 4066.71973 mm 1793 mm f 20 314.16 c/c 71.71 ii) Direction Y M= 351.14 KNm/m Km= 26 Ks= 4.00 MIN. REINF 2 2 Asteel = 1904.99794 mm 1793 mm f 20 314.16 c/c 141.57 d) Design of strap Note : the strap should be designed to withstand the shear foce and moment on the respective diagrams V= 909.13 KN M= -1500.06 KNm Km= 44 Ks= 4.26 2 Asteel = 7311.50294 mm f 24 452.39 No. 16.16 Take width of strip, b= 0.50 m Assome depth, d= 1.00 m Shear resistance of concrete VRd= 1300.9167 KN Vc= 187.36667 KN 2 Shearv= 1873.6667 Kn/m Thus needs to be designed for shear

f6= 500

mm

Design of strap footing foundation I) Proportioning of footingsTrial 1

Xc

column1 b1

Assume half length of a1= 1.5 a1= 3m P1= 1472.00 m bc1,1= 0.5 m bc1,2= 0.5 m P2= 1798.00 m bc2,1= 0.5 m bc2,2= 0.5 m Xc= 4.4 m 2 sall= 300 KN/m e= 1.25 m XR= 3.15 m

Column2 b2

a1

a2

P1

Ws xs

P2

Determine soil reaction, R 1 Note: neglect wt. of strap beam Area of F2 A(F2)= 6.85 R1= 2056.13 KN b1= 2.28 Dimensions of footing 1= For footing 2 3 X 2.28

R1 e

XR

R2

R2= 1213.87 KN 2 Area of F2 A(F2)= 4.05 m a2= 1.7711 m b2= 2.28 m Dimensions of footing 2= 1.771095 X 2.28 q1= 685.38 KN/m q2= 685.38 KN/m Note: The Distribution per meter run under the pad is the same for each pad. II) Structural Design a) Calculation of shear force and bending moment diagram

1191.06 171.34 584.13

xo xo= 1.897727-1300.66

-606.94

Shear force diagram-205.14

21.42 -517.27216

Bending Moment diagram-1212.73 -963.81

b) Determination of Depths of Footingsb-1) Footing 1

i) Punching shear 25 Mpa fcd= 11.16667 Mpa fs= 300 Mpa rmin= 0.002 fctd= vup= vud= ax2+bx+c=0 a= b= c= d1 d2 d= D= 1.167 Mpa 641.67 Kpa 385.00 Kpa 1433.33 1187.50 -1397.00 0.66 -1.48 0.66 693 fck=

0.66

1184.69

1184.69

0.00

ii) Diagonal tension (Wide beam shear) V= 850.78 Existing shear= 568.49 d 0.97 m D 1004 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= 210.2524 v= 72.30953

ok!b-1) Footing 2

i) Punching shear Concrete shear resistance=

2*(bc2,1+d+bc2,2)*d*vcp(1)

Net force along the perimeter= P2-(bc2,1+d)(bc2,2+d)sas(2) Equating eq (1) and (2), we obtain a quadratic equation a= 2866.67 b= 1583.33 c= -1723.00 d1 0.55 m 1469.25 1469.25 d2 -1.10 m d= 0.547 m D 584 ii) Diagonal tension (Wide beam shear) V= 816.2791 Existing shear= 654.7173 d 0.930 m D 965 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= 182.3375 v= 110.7111

0.00

ok!D 965

C) Reinforcementc-1) Footing 1 i) Direction X (TOP) Maximum moment= 1212.73 KNm -1212.73 Km= 36 Ks= 4.11 MIN. REINF 2 2 Asteel = 5142.576 mm 1939 mm f 24 452.39 c/c 80.86 ii) Direction Y (BOTTOM) M= 118.82 Knm/m Km= 11 Ks= 3.95 MIN. REINF 2 2 Asteel = 484.2216 mm 1939 mm f 16 201.06 c/c 93.95 c-1) Footing 2 i) Direction X Maximum moment= 517.27 KNm Km= 24 Ks= 3.98 MIN. REINF 2 2 Asteel = 2215.76 mm 1939 mm f 20 314.16 c/c 124.18 ii) Direction Y M= 118.82 KNm/m Km= 12 Ks= 3.95 MIN. REINF 2 2 Asteel = 504.6906 mm 1939 mm f 16 201.06 c/c 93.95 d) Design of strap Note : the strap should be designed to withstand the shear foce and moment on the respective diagrams V= 584.13 KN M= -963.81 KNm Km= 35 Ks= 4.10 2 Asteel = 4521.303 mm f 24 452.39 No. 9.99 Take width of strip, b= 0.50 m Assome depth, d= 1.00 m Shear resistance of concrete VRd= 1300.9167 KN Vc= 187.36667 KN 2 Shearv= 1873.6667 Kn/m Thus needs to be designed for shear

f6= 500

mm

Design of strap footing foundation I) Proportioning of footingsTrial 1

Xc

column1 0.6 m KN m m KN m m m KN/m2 m m b1

Assume half length of a1= a1= 1.2 P1= 305.34 bc1,1= 0.3 bc1,2= 0.25 P2= 308.70 bc2,1= 0.3 bc2,2= 0.25 Xc= 4 sall= 300 e= 0.45 XR= 3.55

Column2 b2

a1

a2

P1

Ws xs

P2

Determine soil reaction, R 1 Note: neglect wt. of strap beam Area of F2 A(F2)= 1.15 R1= 344.05 KN b1= 0.96 Dimensions of footing 1= For footing 2 1.2 X 0.96

R1 e

XR

R2

R2= 269.99 KN 2 Area of F2 A(F2)= 0.90 m a2= 1.5 m b2= 0.60 m Dimensions of footing 2= 1.5 X 0.60 q1= 286.70 KN/m q2= 180.00 KN/m Note: The Distribution per meter run under the pad is the same for each pad. II) Structural Design a) Calculation of shear force and bending moment diagram

173.70 43.01 38.71

xo xo= 0.915-262.33

-135.00

Shear force diagram58.35

3.23 -29.028803

Bending Moment diagram-116.79 -114.18

b) Determination of Depths of Footingsb-1) Footing 1

i) Punching shear 25 Mpa fcd= 11.16667 Mpa fs= 300 Mpa rmin= 0.002 fctd= vup= vud= ax2+bx+c=0 a= b= c= d1 d2 d= D= 1.167 Mpa 641.67 Kpa 385.00 Kpa 1433.33 672.92 -282.84 0.27 -0.74 0.27 305 fck=

0.27

237.96

237.96

0.00

ii) Diagonal tension (Wide beam shear) V= 185.59 Existing shear= 722.18 d 0.50 m D 537 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= 31.43229 v= 52.16498

ok!b-1) Footing 2

i) Punching shear Concrete shear resistance=

2*(bc2,1+d+bc2,2)*d*vcp(1)

Net force along the perimeter= P2-(bc2,1+d)(bc2,2+d)sas(2) Equating eq (1) and (2), we obtain a quadratic equation a= 2866.67 b= 870.83 c= -286.20 d1 0.20 m 241.57 241.57 d2 -0.50 m d= 0.199 m D 236 ii) Diagonal tension (Wide beam shear) V= 137.9386 Existing shear= 1157.053 d 0.597 m D 632 Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring. V= -10.6615 v= -11.903

0.00

ok!D 632

C) Reinforcementc-1) Footing 1 i) Direction X (TOP) Maximum moment= 116.79 KNm -116.79 Km= 22 Ks= 3.97 MIN. REINF 2 2 Asteel = 923.4001 mm 1195 mm f 24 452.39 c/c 274.61 ii) Direction Y (BOTTOM) M= 18.90 Knm/m Km= 9 Ks= 3.95 MIN. REINF 2 2 Asteel = 148.7063 mm 1195 mm f 14 153.94 c/c 114.12 c-1) Footing 2 i) Direction X Maximum moment= 58.35 KNm Km= 13 Ks= 3.95 MIN. REINF 2 2 Asteel = 385.948 mm 1195 mm f 20 314.16 c/c 208.17 ii) Direction Y M= 4.59 KNm/m Km= 4 Ks= 3.95 MIN. REINF 2 2 Asteel = 30.3872 mm 1195 mm f 16 201.06 c/c 144.02 d) Design of strap Note : the strap should be designed to withstand the shear foce and moment on the respective diagrams V= 38.71 KN M= -114.18 KNm Km= 12 Ks= #N/A mm2 Asteel = #N/A f 24 452.39 No. #N/A Take width of strip, b= 0.50 m Assome depth, d= 1.00 m Shear resistance of concrete VRd= 1300.9167 KN Vc= 187.36667 KN 2 Shearv= 1873.6667 Kn/m Thus Provide nominal shear reinforcement!

f6= 500

mm

Design of strap footing foundation I) Proportioning of footingsTrial 1

Xc

column1 b1

Assume half l