of 45
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JOB No. 0-3850-20 DOC No. C-667-1316-001
SHEET 5 OF 45 .
3- GEOMETRY & OUTLINE
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JOB No. 0-3850-20 DOC No. C-667-1316-001 22.79 KN O.K Sat isf ied
@ 200 mm
@ 250 mm
450.00
Mx2 (+ve)
Qx1
4 250.0
My1 (-ve)
0.00114 0.00333 0.00152
1 -2 x Mu
Mx1 (-ve)
My2 (+ve)
Qy1
125.19
0.00152
16
1 -
804.25
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JOB No. 0-3850-20 DOC No. C-667-1316-001 So use :-
ry = < So use :-
1.33rx =1.33ry =
= rx b x d1 = 2056 mm2/m
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JOB No. 0-3850-20 DOC No. C-667-1316-0
SHEET 22 O
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Therefore
Providing D25 @ 200 C/C In Vertical Direction Both Faces
Providing D25 @ 250 C/C In Horizontal Direction Both Faces
Reinforcement Used in drawing:-
D25 C/C In Vert ical Direct ion Both Faces
D25 C/C In Horizontal Direct ion Both Faces
Shear Check:Vu(max) = 337.1 KN
f Vc = .85 x (f c')1/2
x b x d
6
= KN > 337.1 KN O.K Satisfied
REINFORCEMENT PROVIDED IN X- DIRECTION= VERTICAL
200
250
384.19
1963.50 1625.00
4
REINFORCEMENT PROVIDED IN Y- DIRECTION =HORIZONTAL
25
2454.37 2055.78
25 5
@ 200 mm
@ 200 mm
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JOB No. 0-3850-20 DOC No. C-667-131
SHEET 23
4.5.2- Design Of Walls (W-3 & W-4)
gsoil = Soil Density =gwater = Water Density =
Surcharge =
q = Angle of internal Friction = Degrees
H = Height of Wall Iy = m Top of Wall m
h = Thickness of wall = mm Bot of Wall m
Ground m
cc = Clear Cover = mm
d1 for ver-dir = h-cc-dia bar/2 = mm Soil Height Hs = 6.6 m
d2 for hor-dir = h-cc-dia bar-dia bar/2 = mm
fy = Yield strength of steel = MPafc' = Conc. cylinder strength = MPa
Ko = = 0.40
Ka = = 0.25
rm in = 1.4 / fy = rmax =
rmin vert-dir =
rmin hor-d i r =
ly / lx = b = 1000 mm
a- Empty Condition ( Only submerged density and ground water pressure.)
lx= 3.40 m
Soil
ly
qs qq
qs = Ka x gs x Hs = 31.2
qq = Ka x Surcharge = 2.49
Wv Total Factored Triangular Load = (1.6 x qs) 1.6 x 31.2 = 49.9 KN/m2
Wu Total Factored Uniform Load = (1.6 x qq) 1.6 x 2.49 = 3.98
tan2
( 45o
- f / 2)
KN/m2
0.00150
0.00250
KN/m2
Surcharge
KN/m2
1 - sin q
2.06
28
0.00333
400
75
312.5
420
37
7.00
0.02125
KN/m3
KN/m3
KN/m2
1910
10.00
100.4
93.4
100
287.5
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JOB No. 0-3850-20 DOC No. C-667-1316
SHEET 24
For Triangular Loading:- ( Due to soil)
Where
CF = Moment coefficient taken from the wall with 4 sides fixed graphCF = Shear coefficient taken from the wall with 4 sides fixed graph
Wv= Total factored trinagular load in KN
= 0.042 x 49.9 x 11.6 = 24.2 KN-m
= 0.020 x 49.9 x 11.6 = 11.5 KN-m
= 0.360 x 49.9 x 3.4 = 61.0 KN
= 0.046 x 49.9 x 11.6 = 26.5 KN-m
= 0.012 x 49.9 x 11.6 = 6.9 KN-m
= 0.400 x 49.9 x 3.4 = 67.8 KN
For Rectangular Loading :- ( Due to surcharge)
Where
CF = Moment coefficient taken from the wall with 4 sides fixed graph
CF = Shear coefficient taken from the wall with 4 sides fixed graphWu= Total factored Rectangular load in KN
= 0.083 x 4.0 x 11.6 = 3.8 KN-m
= 0.040 x 4.0 x 11.6 = 1.8 KN-m
= 0.520 x 4.0 x 3.4 = 7.0 KN
= 0.057 x 4.0 x 11.6 = 2.6 KN-m
= 0.010 x 4.0 x 11.6 = 0.5 KN-m
= 0.460 x 4.0 x 3.4 = 6.2 KN
Total Net Moments:
= 24.2 + 3.82 = 28.0 KN-m
= 11.5 + 1.84 = 13.4 KN-m
= 61.0 + 7.03 = 68.1 KN= 26.5 + 2.62 = 29.1 KN-m
= 6.9 + 0.46 = 7.4 KN-m
= 67.8 + 6.22 = 74.1 KN
b- Test Condition ( Pit filled with water and no back fill)
lx = 3.4 m
ly
qw
qw = gwater x H = 70
Wv Total Factored Triangular Load = 1.6 x qw = 112 KN/m2
My1 (-ve)
Qy1
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
M= CF x Wv x H2
Q= CF x Wv x H
Water
Q= CF x Wu x H
Mx1 (-ve)
Qy1
Mx1 (-ve)
Mx2 (+ve)
Qx1
My2 (+ve)
M= CF x Wu x H2
KN/m2
Mx1 (-ve)
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
Qy1
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JOB No. 0-3850-20 DOC No. C-667-131
SHEET 25
For Triangular Loading:- ( Due to water )
Where
CF = Moment coefficient taken from the wall with 4 sides fixed graphCF = Shear coefficient taken from the wall with 4 sides fixed graph
Wv= Total factored triangular load in KN
= 0.042 x 112.0 x 11.6 = 54.4 KN-m
= 0.020 x 112.0 x 11.6 = 25.9 KN-m
= 0.360 x 112.0 x 3.4 = 137.1 KN
= 0.046 x 112.0 x 11.6 = 59.6 KN-m
= 0.012 x 112.0 x 11.6 = 15.5 KN-m
= 0.400 x 112.0 x 3.4 = 152.3 KN
Total Net Moments:
= 54.4 = 54.4 KN-m= 25.9 = 25.9 KN-m
= 137.1 = 137.1 KN
= 59.6 = 59.6 KN-m
= 15.5 = 15.5 KN-m
= 152.3 = 152.3 KN
Controlling Mx Moment = 54.4 KN-m = N-mm
Controlling My Moment 59.6 KN-m = N-mm
Controll ing shear = 152.3 KN = 152.3 KN
Consider 1m strip:-
r = 0.85 x fc'
fy 0.85 x 0.9 x fc' x b x d
rx = < So use :-
ry = < So use :-
1.33rx =1.33ry =
= rx b x d1 = 620 mm2/m
= rminx b x h = 1000 mm2/m
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JOB No. 0-3850-20 DOC No. C-667-1316-
SHEET 26
Bar Dia = No. of Bars = Pitch =
As (provided) = mm2
> mm2
OK
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Therefore
Providing D25 @ 250 C/C In Horizontal Direction Both Faces
Providing D25 @ 250 C/C In Vertical Direction Both Faces
Reinforcement Used in drawing:-
D25 C/C In Horizontal Direct ion Both Faces
D25 C/C In Vert ical Direct ion Both Faces
Shear Check:Vu(max) = 152.3 KN
f Vc = 0.85 x (f c')1/2
x b x d
6
= KN > 152.3 KN O.K Satisfi ed
250
234.26
1963.50 625.97
250
REINFORCEMENT PROVIDED IN Y- DIRECTION =VERTICAL
25 4
1963.50 1000.00
@ 200 mm
@ 200 mm
REINFORCEMENT PROVIDED IN X- DIRECTION= HORIZONTAL
25 4
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JOB No. 0-3850-20 DOC No. C-667-1316
SHEET 27
4.5.3- Design Of Walls (W-5 , W-6 & W-7)
gsoil = Soil Density =gwater = Water Density =
Surcharge =
q = Angle of internal Friction = Degrees
H = Height of Wall Iy = m Top of Wall m
h = Thickness of wall = mm Bot of Wall m
Ground m
cc = Clear Cover = mm
d1 for ver-dir = h-cc-dia bar/2 = mm Soil Height Hs = 6.6 m
d2 for hor-dir = h-cc-dia bar-dia bar/2 = mm
fy = Yield strength of steel = MPafc' = Conc. cylinder strength = MPa
Ko = = 0.40
Ka = = 0.25
rm in = 1.4 / fy = rmax =
rmin vert-dir =
rmin hor-d i r =
ly / lx = b = 1000 mm
a- Empty Condition ( Only submerged density and ground water pressure.)
lx = 5.80 m
Soil
ly
qs qq
qs = Ka x gs x Hs = 31.2
qq = Ka x Surcharge = 2.49
Wv Total Factored Triangular Load = (1.6 x qs) 1.6 x 31.2 = 49.9 KN/m2
Wu Total Factored Uniform Load = (1.6 x qq) 1.6 x 2.49 = 3.98
tan2
( 45o
- f / 2)
KN/m2
0.00150
0.00250
KN/m2
Surcharge
1 - sin q
KN/m2
1.21
28
0.00333
400
75
315
420
295
KN/m3
KN/m3
KN/m2
1910
10.00
100.4
93.4
100
7.00
37
0.02125
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JOB No. 0-3850-20 DOC No. C-667-1316-
SHEET 28
For Triangular Loading :- ( Due to soil)
Where
CF = Moment coeffic ient taken from the wall with 3 sides fixed graphCF = Shear coeffic ient taken from the wall with 3 sides fixed graph
Wv= Total factored trinagular load in KN
= 0.015 x 49.9 x 33.6 = 25.2 KN-m
= 0.008 x 49.9 x 33.6 = 13.4 KN-m
= 0.065 x 49.9 x 5.8 = 18.8 KN
= 0.036 x 49.9 x 33.6 = 60.4 KN-m
= 0.009 x 49.9 x 33.6 = 15.1 KN-m
= 0.050 x 49.9 x 5.8 = 14.5 KN
For Rectangular Loading: - ( Due to surcharge)
Where
CF = Moment coeffic ient taken from the wall with 3 sides fixed graph
CF = Shear coeffic ient taken from the wall with 3 sides fixed graphWu= Total factored Rectangular load in KN
= 0.085 x 4.0 x 33.6 = 11.4 KN-m
= 0.041 x 4.0 x 33.6 = 5.5 KN-m
= 0.470 x 4.0 x 5.8 = 10.8 KN
= 0.057 x 4.0 x 33.6 = 7.6 KN-m
= 0.010 x 4.0 x 33.6 = 1.3 KN-m
= 0.042 x 4.0 x 5.8 = 1.0 KN
Total Net Moments:
= 25.2 + 11.4 = 36.5 KN-m
= 13.4 + 5.49 = 18.9 KN-m
= 18.8 + 10.8 = 29.6 KN= 60.4 + 7.63 = 68.0 KN-m
= 15.1 + 1.34 = 16.4 KN-m
= 14.5 + 0.97 = 15.4 KN
b- Test Condition ( Pit filled with water and no back fill)
lx = 5.80 m
ly
qw
qw = gwater x H = 70
Wv Total Factored Triangular Load = 1.6 x qw = 112 KN/m2
Mx1 (-ve)
Mx2 (+ve)
Qx1
My2 (+ve)
My1 (-ve)
My2 (+ve)
M= CF x Wv x H2
Water
Q= CF x Wv x H
Mx1 (-ve)
Qy1
Mx2 (+ve)
Qx1My1 (-ve)
Qy1
M= CF x Wu x H2
Q= CF x Wu x H
KN/m2
Mx1 (-ve)
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
Qy1
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JOB No. 0-3850-20 DOC No. C-667-1316
SHEET 29
For Triangular Loading:- ( Due to water )
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graphCF = Shear coefficient taken from the wall with 3 sides fixed graph
Wv= Total factored triangular load in KN
= 0.036 x 112.0 x 33.6 = 135.6 KN-m
= 0.008 x 112.0 x 33.6 = 30.1 KN-m
= 0.065 x 112.0 x 5.8 = 42.2 KN
= 0.015 x 112.0 x 33.6 = 56.5 KN-m
= 0.009 x 112.0 x 33.6 = 33.9 KN-m
= 0.050 x 112.0 x 5.8 = 32.5 KN
Total Net Moments:
= 135.6 = 135.6 KN-m= 30.1 = 30.1 KN-m
= 42.2 = 42.2 KN
= 68.0 = 68.0 KN-m
= 33.9 = 33.9 KN-m
= 32.5 = 32.5 KN
Controlling Mx Moment = 135.6 KN-m = N-mm
Controlling My Moment 68.0 KN-m = N-mm
Controll ing sh ear = 42.2 KN = 42.2 KN
Consider 1m strip:-
r = 0.85 x fc'
fy 0.85 x 0.9 x fc' x b x d
rx = > So use :-
ry = < So use :-
1.33rx =1.33ry
= rx b x d1 = 1178 mm2/m
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JOB No. 0-3850-20 DOC No. C-667-1316-0
SHEET 30 O
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Therefore
Providing D20 @ 250 C/C In Horizontal Direction Both Faces
Providing D20 @ 250 C/C In Vertical Direction Both Faces
Reinforcement Used in drawing:-
D20 C/C In Horizontal Direct ion Both Faces
D20 C/C In Vert ical Direct ion Both Faces
Shear Check:Vu(max) = 42.2 KN
f Vc = 0.85 x (f c')1/2
x b x d
6
= KN > 42.2 KN O.K Satisfied
250
@ 200 mm
1256.64 1178.00
REINFORCEMENT PROVIDED IN Y- DIRECTION =VERTICAL
20
@ 200 mm
20 4
250
236.13
1256.64 723.39
4
REINFORCEMENT PROVIDED IN X- DIRECTION= HORIZONTAL
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JOB No. 0-3850-20 DOC No. C-667-131
SHEET 31
4.5.4- Design Of W all (W-8)
gsoil = Soil Density =gwater = Water Density =
Surcharge =
q = Angle of internal Friction = Degrees
H = Height of Wall Ix = m Top of Wall m
h = Thickness of wall = mm Bot of Wall m
Ground m
cc = Clear Cover = mm
d1 for ver-dir = h-cc-dia bar/2 = mm Soil Height Hs = 4.85 m
d2 for hor-dir = h-cc-dia bar-dia bar/2 = mm
x
fy = Yield strength of steel = MPa
fc' = Conc. cylinder strength = MPa
Ko = = 0.40
Ka = = 0.25
rm in = 1.4 / fy = rmax =
rmin vert-dir =
rmin hor-d i r =
ly / lx = b = mm
a- Empty Condition ( Only submerged density and ground water pressure.)
ly = 5.50 m
Soil
qs qq
qs = Ka x gs x Hs = 22.9
q = Ka x Surcharge = 2.49
Wv Total Factored Triangular Load = (1.6 x qs) 1.6 x 0 = 36.7 KN/m2
Wu Total Factored Uniform Load = (1.6 x qq) 1.6 x 2.486 = 3.98
1 - sin qtan
2( 45
o- f / 2)
0.00150
0.00250
KN/m2
KN/m2
lx
0.02125
1.10
28
1000
0.00333
400
75
315
420
37
5.00
KN/m3
KN/m3
KN/m2
1910
10.00
KN/m2
Surcharge
100.15
95.15
100
295
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JOB No. 0-3850-20 DOC No. C-667-1316-
SHEET 32
For Triangular Loading:- ( Due to soil)
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graphCF = Shear coefficient taken from the wall with 3 sides fixed graph
Wv= Total factored trinagular load in KN
= 0.014 x 36.7 x 25.0 = 12.8 KN-m
= 0.008 x 36.7 x 25.0 = 7.3 KN-m
= 0.050 x 36.7 x 5.0 = 9.2 KN
= 0.036 x 36.7 x 25.0 = 33.0 KN-m
= 0.008 x 36.7 x 25.0 = 7.3 KN-m
= 0.330 x 36.7 x 5.0 = 60.5 KN
For Rectangular Loading :- ( Due to surcharge)
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graph
CF = Shear coefficient taken from the wall with 3 sides fixed graph
Wu= Total factored Rectangular load in KN
= 0.085 x 4.0 x 25.0 = 8.5 KN-m
= 0.042 x 4.0 x 25.0 = 4.2 KN-m
= 0.510 x 4.0 x 5.0 = 10.1 KN
= 0.057 x 4.0 x 25.0 = 5.7 KN-m
= 0.010 x 4.0 x 25.0 = 1.0 KN-m
= 0.400 x 4.0 x 5.0 = 8.0 KN
Total Net Moments:
= 12.8 + 8.45 = 21.3 KN-m
= 7.3 + 4.18 = 11.5 KN-m
= 9.2 + 10.1 = 19.3 KN= 33.0 + 5.67 = 38.7 KN-m
= 7.3 + 0.99 = 8.3 KN-m
= 60.5 + 7.95 = 68.4 KN
b- Test Condition ( Pit filled with water and no back fill)
ly = 5.50 m
qw
qw = gwater x H = 50
Wv Total Factored Triangular Load = 1.6 x qw = 80 KN/m2
Water
Qy1
Mx1 (-ve)
Mx2 (+ve)
Qx1
My2 (+ve)
My1 (-ve)
M= CF x Wv x H2
Q= CF x Wv x H
M= CF x Wu x H2
Q= CF x Wu x H
Mx1 (-ve)
Qy1
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
lx
KN/m2
Mx1 (-ve)
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
Qy1
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JOB No. 0-3850-20 DOC No. C-667-1316
SHEET 33
For Triangular Loading:- ( Due to water )
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graphCF = Shear coefficient taken from the wall with 3 sides fixed graph
Wv= Total factored triangular load in KN
= 0.014 x 80.0 x 25.0 = 28.0 KN-m
= 0.008 x 80.0 x 25.0 = 16.0 KN-m
= 0.050 x 80.0 x 5.0 = 20.0 KN
= 0.036 x 80.0 x 25.0 = 72.0 KN-m
= 0.008 x 80.0 x 25.0 = 16.0 KN-m
= 0.330 x 80.0 x 5.0 = 132.0 KN
Total Net Moments:
= 28.0 = 28.0 KN-m= 16.0 = 16.0 KN-m
= 20.0 = 20.0 KN
= 72.0 = 72.0 KN-m
= 16.0 = 16.0 KN-m
= 132.0 = 132.0 KN
Controlling Mx Moment = 28.0 KN-m = N-mm
Controlling My Moment 72.0 KN-m = N-mm
Controll ing shear = 132.0 KN = 132.0 KN
Consider 1m strip:-
r = 0.85 x fc'
fy 0.85 x 0.9 x fc' x b x d2
rx = < So use :-
ry = < So use :-
1.33rx =1.33ry
= rx b x d2 = 295 mm2/m
= rminx b x h = 600mm /m
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JOB No. 0-3850-20 DOC No. C-667-1316-0
SHEET 34 O
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Therefore
Providing D20 @ 250 C/C In Vertical Direction Both Faces
Providing D20 @ 250 C/C In Horizontal Direction Both Faces
Reinforcement Used in drawing:-
D20 C/C In Vert ical Direct ion Both Faces
D20 C/C In Horizontal Direct ion Both Faces
Shear Check:Vu(max) = 132.0 KN
f Vc = 0.85 x (f c')1/2
x b x d
6
= KN > 132.0 KN O.K Satisfied
@ 200 mm
1256.64 600.00
250
236.13
1256.64 1000.00
REINFORCEMENT PROVIDED IN Y- DIRECTION =HORIZONTAL
20 4
@ 200 mm
25020 4
REINFORCEMENT PROVIDED IN X- DIRECTION= VERTICAL
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JOB No. 0-3850-20 DOC No. C-667-1316
SHEET 35
4.5.5- Design Of Walls (W-9 & W-10)
(Designing the most critical wall among these three which is wall W-7)
gsoil = Soil Density =gwater = Water Density =
Surcharge =
q = Angle of internal Friction = Degrees
H = Height of Wall Iy = m Top of Wall m
h = Thickness of wall = mm Bot of Wall m
Ground m
cc = Clear Cover = mm
d1 for ver-dir = h-cc-dia bar/2 = mm Soil Height Hs = 4.85 m
d2 for hor-dir = h-cc-dia bar-dia bar/2 = mm
fy = Yield strength of steel = MPafc' = Conc. cylinder strength = MPa
Ko = = 0.40
Ka = = 0.25
rmi n = 1.4 / fy = rmax =
rmin vert-dir =
rmin hor-d i r =
ly / lx = b = mm
a- Empty Condition ( Only submerged density and ground water pressure.)
lx = 2.600 m
Soil
ly
qs qq
qs = Ka x gs x Hs = 22.9
q = Ka x Surcharge = 2.49
Wv Total Factored Triangular Load = (1.6 x qs) 1.6 x 22.9 = 36.7 KN/m2
Wu Total Factored Uniform Load = (1.6 x qq) 1.6 x 2.49 = 3.98
201
Surcharge
100.15
37
5.00
KN/m3
KN/m3
KN/m2
1910
10.00
28
1000
0.00333
300
75
217
420
95.15
100
1.92
KN/m2
0.02125
1 - sin q
tan2
( 45o
- f / 2)
KN/m2
0.00150
0.00250
KN/m2
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JOB No. 0-3850-20 DOC No. C-667-1316-
SHEET 36
For Triangular Loading:- ( Due to soil)
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graphCF = Shear coefficient taken from the wall with 3 sides fixed graph
Wv= Total factored trinagular load in KN
= 0.008 x 36.7 x 6.8 = 2.0 KN-m
= 0.005 x 36.7 x 6.8 = 1.2 KN-m
= 0.040 x 36.7 x 2.6 = 3.8 KN
= 0.046 x 36.7 x 6.8 = 11.4 KN-m
= 0.095 x 36.7 x 6.8 = 23.5 KN-m
= 0.385 x 36.7 x 2.6 = 36.7 KN
For Rectangular Loading :- ( Due to surcharge)
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graph
CF = Shear coefficient taken from the wall with 3 sides fixed graphWu= Total factored Rectangular load in KN
= 0.083 x 4.0 x 6.8 = 2.2 KN-m
= 0.042 x 4.0 x 6.8 = 1.1 KN-m
= 0.510 x 4.0 x 2.6 = 5.3 KN
= 0.058 x 4.0 x 6.8 = 1.6 KN-m
= 0.012 x 4.0 x 6.8 = 0.3 KN-m
= 0.400 x 4.0 x 2.6 = 4.1 KN
Total Net Moments:
= 2.0 + 2.23 = 4.2 KN-m
= 1.2 + 1.13 = 2.4 KN-m
= 3.8 + 5.27 = 9.1 KN= 11.4 + 1.56 = 13.0 KN-m
= 23.5 + 0.32 = 23.9 KN-m
= 36.7 + 4.14 = 40.8 KN
b- Test Condition ( Pit filled with water and no back fill)
lx = 2.600 m
ly
qw
qw = gwater x H = 50
Wv Total Factored Triangular Load = 1.6 x qw = 80 KN/m2
My1 (-ve)
My2 (+ve)
Qy1
KN/m2
Mx1 (-ve)
Mx2 (+ve)
Qx1
Qy1
Mx2 (+ve)
Qx1My1 (-ve)
My2 (+ve)
Q= CF x Wu x H
Mx1 (-ve)
M= CF x Wv x H2
Q= CF x Wv x H
M= CF x Wu x H2
Water
Qy1
Mx1 (-ve)
Mx2 (+ve)
Qx1
My2 (+ve)
My1 (-ve)
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For Triangular Loading:- ( Due to water )
Where
CF = Moment coefficient taken from the wall with 3 sides fixed graphCF = Shear coefficient taken from the wall with 3 sides fixed graph
Wv= Total factored triangular load in KN
= 0.008 x 80.0 x 6.8 = 4.3 KN-m
= 0.005 x 80.0 x 6.8 = 2.7 KN-m
= 0.040 x 80.0 x 2.6 = 8.3 KN
= 0.046 x 80.0 x 6.8 = 24.9 KN-m
= 0.095 x 80.0 x 6.8 = 51.4 KN-m
= 0.385 x 80.0 x 2.6 = 80.1 KN
Total Net Moments:
= 4.3 = 4.3 KN-m= 2.7 = 2.7 KN-m
= 9.1 = 9.1 KN
= 24.9 = 24.9 KN-m
= 51.4 = 51.4 KN-m
= 80.1 = 80.1 KN
Controlling Mx Moment = 4.3 KN-m = N-mm
Controlling My Moment 51.4 KN-m = N-mm
Controll ing shear = 80.1 KN = 80.1 KN
Consider 1m strip:-
r = 0.85 x fc'
fy 0.85 x 0.9 x fc' x b x d2
rx = < So use :-
ry = < So use :-
1.33rx =1.33ry
= rx b x d1 = 70.3 mm2/m
= rminx b x h = 750mm /m
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SHEET 38 O
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Bar Dia = No. of Bars = Pitch =
As (provided) = mm > mm OK
Therefore
Providing D16 @ 250 C/C In Horizontal Direction Both Faces
Providing D16 @ 250 C/C In Vertical Direction Both Faces
Reinforcement Used in drawing:-
D16 C/C In Horizontal Direct ion Both Faces
D16 C/C In Vert ical Direct ion Both Faces
Shear Check:Vu(max) = 80.1 KN
f Vc = 0.85 x (f c')1/2
x b x d
6
= KN > 80.1 KN O.K Satisfi ed
REINFORCEMENT PROVIDED IN X- DIRECTION= HORIZONTAL
REINFORCEMENT PROVIDED IN Y- DIRECTION =VERTICAL
16 4
804.25 750.00
16 4 250
250
162.67
804.25 670.00
@ 200 mm
@ 200 mm
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4.6- REINFORCEMENT
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4.7- ATTACHMENT (MOMENT GRAPHS).
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4.7.5-