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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.