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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.
D34
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
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
J76
DESALEGN: The difference between cells H76 and I76
J100
DESALEGN: The difference of cells H99 and I99
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=
H118
DESALEGN: For this particular case the largest is the negative moment. Thus for other cases this should be checked.
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.
D34
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
J76
DESALEGN: The difference between cells H76 and I76
J100
DESALEGN: The difference of cells H99 and I99
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=
H118
DESALEGN: For this particular case the largest is the negative moment. Thus for other cases this should be checked.
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.
D34
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
J76
DESALEGN: The difference between cells H76 and I76
J100
DESALEGN: The difference of cells H99 and I99
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=
H118
DESALEGN: For this particular case the largest is the negative moment. Thus for other cases this should be checked.
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.
D34
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
J76
DESALEGN: The difference between cells H76 and I76
J100
DESALEGN: The difference of cells H99 and I99
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=
H118
DESALEGN: For this particular case the largest is the negative moment. Thus for other cases this should be checked.
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.
D34
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
J76
DESALEGN: The difference between cells H76 and I76
J100
DESALEGN: The difference of cells H99 and I99
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=
H118
DESALEGN: For this particular case the largest is the negative moment. Thus for other cases this should be checked.
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.