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B66 S 0121(Orginal Sheet)

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  • 7/28/2019 B66 S 0121(Orginal Sheet)

    1/45

    JOB No. 0-3850-20 DOC No. C-667-1316-001

    SHEET 5 OF 45 .

    3- GEOMETRY & OUTLINE

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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

  • 7/28/2019 B66 S 0121(Orginal Sheet)

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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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    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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    JOB No. 0-3850-20 DOC No. C-667-1316

    SHEET 37

    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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    JOB No. 0-3850-20 DOC No. C-667-1316-0

    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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    SHEET 39 OF 45 .

    4.6- REINFORCEMENT

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    4.7- ATTACHMENT (MOMENT GRAPHS).

    4.7.1-

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    SHEET 42 OF 45 .

    4.7.2-

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    SHEET 43 OF 45 .

    4.7.3-

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    SHEET 44 OF 45 .

    4.7.4-

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    4.7.5-


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