C-25 S-300Km Ks

19.00 3.9520.00 3.9621.00 3.9622.00 3.9723.00 3.9824.00 3.9825.00 3.9926.00 4.0027.00 4.0228.00 4.0329.00 4.0430.00 4.0531.00 4.0632.00 4.0733.00 4.0834.00 4.0935.00 4.1036.00 4.1137.00 4.1238.00 4.1539.00 4.1740.00 4.1841.00 4.2042.00 4.2143.00 4.2444.00 4.2645.00 4.2946.00 4.3147.00 4.3448.00 4.3649.00 4.3950.00 4.4151.00 4.4652.00 4.4953.00 4.5154.00 4.5655.00 4.6256.00 4.6557.00 4.6758.00 4.73

Design of strap footing foundation

I) Proportioning of footingsTrial 1

1.53 m

2291.00 m0.5 m0.5 m

2618.00 m0.5 m0.5 m

Xc= 4.4 m

300 xs

e= 1.25 m3.15 m

Note: neglect wt. of strap beam e10.67

3200.13 KN3.56

Dimensions of footing 1= 3 X 3.56

For footing 21708.87 KN

5.701.602 m

3.56 mDimensions of footing 2= 1.6020034 X 3.56

1066.71 KN/m1066.71 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

1.89772727 -854.44

-2024.32 Shear force diagram

-346.76

33.33

-728.21224455

Bending Moment diagram-1887.47 -1500.06

b) Determination of Depths of Footings

b-1) Footing 1

i) Punching shear

25 Mpa

11.1666667 Mpa

Xc

column1 Column2

Assume half length of a1= b1 b2

a1=

P1=

bc1,1= a1 a2

bc1,2=P2=

bc2,1=bc2,2=

P1 Ws P2

sall= KN/m2

XR=

Determine soil reaction, R1 R1 XR R2

Area of F2 A(F2)=R1=b1=

R2=Area of F2 A(F2)= m2

a2=b2=

q1=q2=

xo

xo=

fck=

fcd=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

300 Mpa

0.002

1.167 Mpa

641.67 Kpa

385.00 Kpa

a= 1433.33b= 1187.50c= -2216.00

0.90 0.90 1893.81 1893.81 0.00

-1.72d= 0.90D= 933

ii) Diagonal tension (Wide beam shear)V= 1068.18

Existing shear= 334.75d 0.90 mD 931

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 570.289245v= 212.07945

ok!

b-1) Footing 2

i) Punching shear

Concrete shear resistance=

Net force along the perimeter=Equating eq (1) and (2), we obtain a quadratic equation

a= 2866.67b= 1583.33c= -2543.00

0.71 m 2182.14 2182.14 0.00

-1.26 md= 0.705 mD 742

ii) Diagonal tension (Wide beam shear)V= 1011.16418

Existing shear= 402.6889d 0.738 mD 773

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 396.329691v= 335.337587

ok!D 773

fs=

rmin=

fctd=

vup=

vud=

ax2+bx+c=0

d1

d2

2*(bc2,1+d+bc2,2)*d*vcp…………(1)

P2-(bc2,1+d)(bc2,2+d)sas…………(2)

d1

d2

C) Reinforcementc-1) Footing 1

i) Direction X (TOP)

Maximum moment= 1887.47 KNm -1887.47

Km= 48 Ks= 4.36 MIN. REINF

9181.03357 1793

f 24 452.39 c/c 46.96

ii) Direction Y (BOTTOM)

M= 351.14 Knm/m

Km= 21 Ks= 3.96 MIN. REINF

Asteel = 1552.599289 1793

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

Asteel = 4066.719734 1793

f 20 314.16 c/c 71.71

ii) Direction Y

M= 351.14 KNm/m

Km= 26 Ks= 4.00 MIN. REINF

Asteel = 1904.997944 1793

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

Asteel = 7311.502942

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

1300.9167 KN

Vc= 187.36667 KN

Shearv= 1873.6667

Thus needs to be designed for shear500 mm

Asteel = mm2 mm2

mm2 mm2

mm2 mm2

mm2 mm2

mm2

VRd=

Kn/m2

f6=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

Design of strap footing foundation

I) Proportioning of footingsTrial 1

1.53 m

1472.00 m0.5 m0.5 m

1798.00 m0.5 m0.5 m

Xc= 4.4 m

300 xs

e= 1.25 m3.15 m

Note: neglect wt. of strap beam e6.85

2056.13 KN2.28

Dimensions of footing 1= 3 X 2.28

For footing 2

1213.87 KN

4.051.77109 m

2.28 mDimensions of footing 2= 1.7710949 X 2.28

685.38 KN/m685.38 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

1.897727 -606.94

-1300.66 Shear force diagram

-205.14

21.42

-517.272156

Bending Moment diagram-1212.73 -963.81

Xc

column1 Column2

Assume half length of a1= b1 b2

a1=

P1=

bc1,1= a1 a2

bc1,2=P2=

bc2,1=bc2,2=

P1 Ws P2

sall= KN/m2

XR=

Determine soil reaction, R1 R1 XR R2

Area of F2 A(F2)=R1=b1=

R2=

Area of F2 A(F2)= m2

a2=b2=

q1=q2=

xo

xo=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

b) Determination of Depths of Footings

b-1) Footing 1

i) Punching shear

25 Mpa

11.16667 Mpa

300 Mpa

0.002

1.167 Mpa

641.67 Kpa

385.00 Kpa

a= 1433.33b= 1187.50c= -1397.00

0.66 0.66 1184.69 1184.69 0.00

-1.48d= 0.66D= 693

ii) Diagonal tension (Wide beam shear)V= 850.78

Existing shear= 568.49d 0.97 mD 1004

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 210.2524v= 72.30953

ok!

b-1) Footing 2

i) Punching shear

Concrete shear resistance=

Net force along the perimeter=Equating eq (1) and (2), we obtain a quadratic equation

a= 2866.67b= 1583.33c= -1723.00

0.55 m 1469.25 1469.25 0.00

-1.10 md= 0.547 mD 584

ii) Diagonal tension (Wide beam shear)V= 816.2791

Existing shear= 654.7173d 0.930 mD 965

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 182.3375v= 110.7111

ok!D 965

fck=

fcd=

fs=

rmin=

fctd=

vup=

vud=

ax2+bx+c=0

d1

d2

2*(bc2,1+d+bc2,2)*d*vcp…………(1)

P2-(bc2,1+d)(bc2,2+d)sas…………(2)

d1

d2

C) Reinforcementc-1) Footing 1

i) Direction X (TOP)

Maximum moment= 1212.73 KNm -1212.73

Km= 36 Ks= 4.11 MIN. REINF

5142.5762 1939

f 24 452.39 c/c 80.86

ii) Direction Y (BOTTOM)

M= 118.81 Knm/m

Km= 11 Ks= 3.95 MIN. REINF

Asteel = 484.22158 1939

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

Asteel = 2215.7596 1939

f 20 314.16 c/c 124.18

ii) Direction Y

M= 118.81 KNm/m

Km= 12 Ks= 3.95 MIN. REINF

Asteel = 504.69065 1939

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

Asteel = 4521.3033

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

1300.9167 KN

Vc= 187.36667 KN

Shearv= 1873.6667

Thus needs to be designed for shear500 mm

Asteel = mm2 mm2

mm2 mm2

mm2 mm2

mm2 mm2

mm2

VRd=

Kn/m2

f6=

Bending Moment diagram

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

Design of strap footing foundation

I) Proportioning of footingsTrial 1

0.61.2 m

305.34 KN0.3 m

0.25 m308.70 KN

0.3 m0.25 m

Xc= 4 m

300 xs

e= 0.45 m3.55 m

Note: neglect wt. of strap beam e1.15

344.05 KN0.96

Dimensions of footing 1= 1.2 X 0.96

For footing 2

269.99 KN

0.901.5 m

0.60 mDimensions of footing 2= 1.5 X 0.60

286.70 KN/m180.00 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

0.915 -135.00

-262.33 Shear force diagram

58.35

3.23

-29.0288028

Bending Moment diagram-116.79 -114.18

Xc

column1 Column2

Assume half length of a1= b1 b2

a1=

P1=

bc1,1= a1 a2

bc1,2=P2=

bc2,1=bc2,2=

P1 Ws P2

sall= KN/m2

XR=

Determine soil reaction, R1 R1 XR R2

Area of F2 A(F2)=R1=b1=

R2=

Area of F2 A(F2)= m2

a2=b2=

q1=q2=

xo

xo=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

b) Determination of Depths of Footings

b-1) Footing 1

i) Punching shear

25 Mpa

11.16667 Mpa

300 Mpa

0.002

1.167 Mpa

641.67 Kpa

385.00 Kpa

a= 1433.33b= 672.92c= -282.84

0.27 0.27 237.96 237.96 0.00

-0.74d= 0.27D= 305

ii) Diagonal tension (Wide beam shear)V= 185.59

Existing shear= 722.18d 0.50 mD 537

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 31.43229v= 52.16498

ok!

b-1) Footing 2

i) Punching shear

Concrete shear resistance=

Net force along the perimeter=Equating eq (1) and (2), we obtain a quadratic equation

a= 2866.67b= 870.83c= -286.20

0.20 m 241.57 241.57 0.00

-0.50 md= 0.199 mD 236

ii) Diagonal tension (Wide beam shear)V= 137.9386

Existing shear= 1157.053d 0.597 mD 632

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= -10.66155v= -11.90296

ok!D 632

fck=

fcd=

fs=

rmin=

fctd=

vup=

vud=

ax2+bx+c=0

d1

d2

2*(bc2,1+d+bc2,2)*d*vcp…………(1)

P2-(bc2,1+d)(bc2,2+d)sas…………(2)

d1

d2

C) Reinforcementc-1) Footing 1

i) Direction X (TOP)

Maximum moment= 116.79 KNm -116.79

Km= 22 Ks= 3.97 MIN. REINF

923.40008 1195

f 24 452.39 c/c 274.61

ii) Direction Y (BOTTOM)

M= 18.90 Knm/m

Km= 9 Ks= 3.95 MIN. REINF

Asteel = 148.70628 1195

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

Asteel = 385.94798 1195

f 20 314.16 c/c 208.17

ii) Direction Y

M= 4.59 KNm/m

Km= 4 Ks= 3.95 MIN. REINF

Asteel = 30.387198 1195

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

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

1300.9167 KN

Vc= 187.36667 KN

Shearv= 1873.6667

Thus Provide nominal shear reinforcement!500 mm

Asteel = mm2 mm2

mm2 mm2

mm2 mm2

mm2 mm2

mm2

VRd=

Kn/m2

f6=

Bending Moment diagram

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

Design of strap footing foundation

I) Proportioning of footingsTrial 1

1.53 m

2233.00 m0.5 m0.5 m

2888.00 m0.5 m0.5 m

Xc= 4.4 m

300 xs

e= 1.25 m3.15 m

Note: neglect wt. of strap beam e10.40

3119.11 KN3.47

Dimensions of footing 1= 3 X 3.47

For footing 2

2001.89 KN

6.671.92545 m

3.47 mDimensions of footing 2= 1.9254486 X 3.47

1039.70 KN/m1039.70 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

1887.06

259.93 886.11

1.897727 -1000.94

-1973.07 Shear force diagram

-324.09

32.49

-853.08069

Bending Moment diagram-1839.69 -1462.08

Xc

column1 Column2

Assume half length of a1= b1 b2

a1=

P1=

bc1,1= a1 a2

bc1,2=P2=

bc2,1=bc2,2=

P1 Ws P2

sall= KN/m2

XR=

Determine soil reaction, R1 R1 XR R2

Area of F2 A(F2)=R1=b1=

R2=

Area of F2 A(F2)= m2

a2=b2=

q1=q2=

xo

xo=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

b) Determination of Depths of Footings

b-1) Footing 1

i) Punching shear

25 Mpa

11.16667 Mpa

300 Mpa

0.002

1.167 Mpa

641.67 Kpa

385.00 Kpa

a= 1433.33b= 1187.50c= -2158.00

0.88 0.88 1843.44 1843.44 0.00

-1.71d= 0.88D= 918

ii) Diagonal tension (Wide beam shear)V= 1057.29

Existing shear= 345.92d 0.88 mD 916

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 543.766v= 205.7813

ok!

b-1) Footing 2

i) Punching shear

Concrete shear resistance=

Net force along the perimeter=Equating eq (1) and (2), we obtain a quadratic equation

a= 2866.67b= 1583.33c= -2813.00

0.75 m 2417.59 2417.59 0.00

-1.30 md= 0.752 mD 789

ii) Diagonal tension (Wide beam shear)V= 1104.986

Existing shear= 423.3403d 0.827 mD 862

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 423.2864v= 265.7884

ok!D 862

fck=

fcd=

fs=

rmin=

fctd=

vup=

vud=

ax2+bx+c=0

d1

d2

2*(bc2,1+d+bc2,2)*d*vcp…………(1)

P2-(bc2,1+d)(bc2,2+d)sas…………(2)

d1

d2

C) Reinforcementc-1) Footing 1

i) Direction X (TOP)

Maximum moment= 1839.69 KNm -1839.69

Km= 49 Ks= 4.39 MIN. REINF

9158.5917 1762

f 24 452.39 c/c 47.07

ii) Direction Y (BOTTOM)

M= 330.78 Knm/m

Km= 21 Ks= 3.96 MIN. REINF

Asteel = 1488.4005 1762

f 14 153.94 c/c 80.35

c-1) Footing 2

i) Direction X

Maximum moment= 853.08 KNm

Km= 35 Ks= 4.10 MIN. REINF

Asteel = 4228.7063 1762

f 20 314.16 c/c 69.15

ii) Direction Y

M= 330.78 KNm/m

Km= 23 Ks= 3.98 MIN. REINF

Asteel = 1590.3654 1762

f 16 201.06 c/c 102.42

d) Design of strap

Note : the strap should be designed to withstand the shear foce and

moment on the respective diagrams

V= 886.11 KN

M= -1462.08 KNm

Km= 43 Ks= 4.24

Asteel = 7084.58

f 24 452.39 No. 15.66

Take width of strip, b= 0.50 m

Assome depth, d= 1.00 m

Shear resistance of concrete

1300.9167 KN

Vc= 187.36667 KN

Shearv= 1873.6667

Thus needs to be designed for shear500 mm

Asteel = mm2 mm2

mm2 mm2

mm2 mm2

mm2 mm2

mm2

VRd=

Kn/m2

f6=

Bending Moment diagram

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

Design of strap footing foundation

I) Proportioning of footingsTrial 1

0.61.2 m

305.34 m0.3 m

0.25 m308.70 m

0.3 m0.25 m

Xc= 4 m

300 xs

e= 0.45 m3.55 m

Note: neglect wt. of strap beam e1.15

344.05 KN0.96

Dimensions of footing 1= 1.2 X 0.96

For footing 2

269.99 KN

0.901.5 m

0.60 mDimensions of footing 2= 1.5 X 0.60

286.70 KN/m180.00 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

0.915 -135.00

-262.33 Shear force diagram

58.35

3.23

-29.0288028

Bending Moment diagram-116.79 -114.18

Xc

column1 Column2

Assume half length of a1= b1 b2

a1=

P1=

bc1,1= a1 a2

bc1,2=P2=

bc2,1=bc2,2=

P1 Ws P2

sall= KN/m2

XR=

Determine soil reaction, R1 R1 XR R2

Area of F2 A(F2)=R1=b1=

R2=

Area of F2 A(F2)= m2

a2=b2=

q1=q2=

xo

xo=

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.

b) Determination of Depths of Footings

b-1) Footing 1

i) Punching shear

25 Mpa

11.16667 Mpa

300 Mpa

0.002

1.167 Mpa

641.67 Kpa

385.00 Kpa

a= 1433.33b= 672.92c= -282.84

0.27 0.27 237.96 237.96 0.00

-0.74d= 0.27D= 305

ii) Diagonal tension (Wide beam shear)V= 185.59

Existing shear= 722.18d 0.50 mD 537

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= 31.43229v= 52.16498

ok!

b-1) Footing 2

i) Punching shear

Concrete shear resistance=

Net force along the perimeter=Equating eq (1) and (2), we obtain a quadratic equation

a= 2866.67b= 870.83c= -286.20

0.20 m 241.57 241.57 0.00

-0.50 md= 0.199 mD 300

ii) Diagonal tension (Wide beam shear)V= 137.9386

Existing shear= 1157.053d 0.597 mD 632

Note: the shear force in the y-direction may be esstimated as for the case of an isolated fooring.

V= -10.66155v= -11.90296

ok!D 632

fck=

fcd=

fs=

rmin=

fctd=

vup=

vud=

ax2+bx+c=0

d1

d2

2*(bc2,1+d+bc2,2)*d*vcp…………(1)

P2-(bc2,1+d)(bc2,2+d)sas…………(2)

d1

d2

C) Reinforcementc-1) Footing 1

i) Direction X (TOP)

Maximum moment= 116.79 KNm -116.79

Km= 22 Ks= 3.97 MIN. REINF

923.40008 1195

f 14 153.94 c/c 114.12

ii) Direction Y (BOTTOM)

M= 18.90 Knm/m

Km= 9 Ks= 3.95 MIN. REINF

Asteel = 148.70628 1195

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

Asteel = 385.94798 1195

f 14 153.94 c/c 114.12

ii) Direction Y

M= 4.59 KNm/m

Km= 4 Ks= 3.95 MIN. REINF

Asteel = 30.387198 1195

f 14 153.94 c/c 114.12

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= 28 Ks= 4.03

Asteel = 1229.5716

f 20 314.16 No. 3.91

Take width of strip, b= 0.30 m

Assome depth, d= 0.50 m

Shear resistance of concrete

361.8 KN

Vc= 56.21 KN

Shearv= 562.1

Thus Provide nominal shear reinforcement!250 mm

Asteel = mm2 mm2

mm2 mm2

mm2 mm2

mm2 mm2

mm2

VRd=

Kn/m2

f6=

Bending Moment diagram

Use the "tools-Goal seek" command and feed the value in "D33" to obtain this result. The value in "D30" should be used as a variable to change in order to achieve this result.