91505103 box culvert at chainage 83 10 m xlsx depth 4 1 m
TRANSCRIPT
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Project:
Element:
Job No.: M 133 Date: 26-Mar-13 Page No.: 1
Made by: RN Checked by: HKM Approved by: HKM
Culvert Details Vertical Stress
4.1
27.93
5
5
Soil Parameters 99.97
18
20
9.81
30
5
10
0.33
27.93
99.97
25
4.1
23
74.95
5
5
500
500
500
BS 5400 PART 2:2006
2.5
10.5
(a) Calculation of HA UDL
69.5
120
0.685
19.0
32.88
BD 37 Chapter 4
45
450
112.5
Soil Reaction16.2
11.5
Motorways and
Trunk Roads
Nominal Load per axle (k N)=
Nominal Load per wheel (k N)=
Class of Road carried by structure=
Knife Edge Load(KEL) per notional lane(k N)=
(b) Calculation of HB Load
Number of Units for Hb load (k N)=
BS 8002 :1994 Clause
3.3.4.1
Clause 6.2 Type HA
Loading
Culvert width (m)=
Culvert depth(m)=
Thickness of wall (mm)=
Thickness of base Slab(mm)=
Loaded length of Box Culvert, L(m)=
BD 37/01 Part 14 - Clause 3.2.9.3.1 -Notional lane width (m)=
HA UDL per m of loaded length (k N/m)= 336 (1/L)^0.67 =
Note that the loaded length will be the width of the Box Culvert
Surcharge due to vehicular Traffic (k N/m2)=
Coefficient of active pressure, ka=
Lateral Pressure at top of wall (kN/m2)=
Clause 6.3 Type HBLoading
Verdun Trianon Link Road
Box Culvert at Chainage 83.10 m- Chainage 35.46 to 49.4 m
Road Level
5
5
Thickness of Upper Slab (m)=
Depth of soil retained by Upper Slab(m)=
Assuming 50 mm thick surfacing layer
Unit weight of surfacing layer (k N/m3)=
Load per m2
of soil on upper slab (k N/m2)=
Unit Weight of Soil unsaturated , (k N/m3)=
Internal Friction angle,() =
Lateral Pressure at bottom of wall (kN/m2)=
Unit weight of concrete (k N/m3)=
Height of water table(m)=
Unit Weight of Soil saturated , sat(k N/m3)=
Unit weight of water=
BD 37 Part 14 Table 14
-HA Lane Factors
First Lane load Factor, 1 = 0.274bl=
(For loaded length 0
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Project:
Element:
Job No.: M 133 Date: 26-Mar-13 Page No.: 2
Made by: RN Checked by: HKM Approved by: HKM
BS 5400 Part 2 2006
Clause 6.26
The effect of the Vertical load is calculated using Boussineq's equatiion:
112.5 kN
Point Load, Q (kN) = 112.5
r= 2.21
Depth of Soil,z(m)= 4.10
r/z 0.54
Influence factor, Ip = 0.252
Vertical Stress, z (kN/m) = 1.688851192
Assumption
For 4 wheels,z (kN/m) = 6.8
Joint Dispersal of wheel load on deck of Culvert, therefore multiply vertical stress by 4.
4.10
CULVERT
Verdun Trianon Link Road
Box Culvert at Chainage 83.10 m- Chainage 35.46 to 49.4 m
Volume 2 Section 2
Part 12BD 31/01 Pg
3/5
Dispersal of the single nominal wheel load at a spread to depth ratio of 1 horizontally
to 2 Vertically through asphalt and similar surfacing may be assumed ,where it is
considered that this may take place.
Volume 2 Section 2
Part 12BD 31/01 Pg
3/3
For Cover exceeding depth 0.6m, the HAUDL/KEL does not adequately model traffic
loading. In these circumstances the HA UDL/KEL combination shall be replaced by 30
Units HB Loading, dispersed through the fill. However, in this case Hb Load = 45 k N,
hence for analysis purposes Ha loading has been ignored.
Dispersal of Wheel loads
concrete Designer's
Manual Handbook
11th Edn. Pg 9
Section 2.4.9
Dis erssal of wheel
For the Hb vehicle, one unit of Hb corresponds to 2.5 k N per wheel, the side of the
square contact area becomes approximately 260mm for 30 units,290 mm for 37.5
Units and 320mm for 45 Units.Therefore use 320 mm as we are designing for
motorways.
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ELEMENT DESIGN to BS 8110:1997
SOLID SLABSOriginated from RCC11.xls on CD 1999 BCA for RCC
INPUT Location Deck Mid Span
Design moment, M 1350.0 kNm/m fcu 35 N/mm gc = 1.50
b 1.00 fy 460 N/mm gs = 1.05
span 5600 mm
Height, h 800 mm Section location SIMPLY SUPPORTED S
Bar 25 mm
cover 100 mm to this reinforcement
OUTPUT Deck Mid Span Compression steel = Nominal
d = 800 - 100 - 25/2 = 687.5 mm .
(3.4.4.4) K' = 0.156 > K = 0.082 ok .
(3.4.4.4) z = 687.5 [0.5 + (0.25 - 0.082 /0.9)^ = 618.2 > 0.95d = 653.1 mm
(3.4.4.1) As = 1350.00E6 /460 /618.2 x 1.05 = 4985 > min As = 1040 mm/m
PROVIDE T25 @ 100 = 4909 mm/m.
(Eqn 8) fs = 2/3 x 460 x 4985 /4909 /1.00 = 311.4 N/mm
(Eqn 7) Tens mod factor = 0.55 + (477 - 311.4) /120 /(0.9 + 2.856) = 0.917
(Equation 9) Comp mod factor = 1 + 0.13/(3 + 0.13) = 1.042
(3.4.6.3) Permissible L/d = 20.0 x 0.917 x 1.042 = 19.109 .
Actual L/d = 5600 /687.5 = 8.145 ok .
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ELEMENT DESIGN to BS 8110:1997
SOLID SLABSOriginated from RCC11.xls on CD 1999 BCA for RCC
INPUT Location Upper Slab Support
Design moment, M 305.0 kNm/m fcu 35 N/mm gc = 1.50
b 1.00 fy 460 N/mm gs = 1.05
span 5500 mm
Height, h 500 mm Section location SUPPORT
Bar 16 mm
cover 50 mm to this reinforcement
OUTPUT Upper Slab Support Compression steel = None
d = 500 - 50 - 16/2 = 442.0 mm .
(3.4.4.4) K' = 0.156 > K = 0.045 ok .
(3.4.4.4) z = 442.0 [0.5 + (0.25 - 0.045 /0.9)^ = 418.9 > 0.95d = 419.9 mm
(3.4.4.1) As = 305.00E6 /460 /418.9 x 1.05 = 1662 > min As = 650 mm/m
PROVIDE T16 @ 100 = 2011 mm/m.
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Project
Box Culvert Verdun Trianon Link Road
ClientBCEG Ltd Made by Date Page
Location Upper Slab RN 26-Mar-2013 1
Crack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No
Originated from RCC14.xls on CD 1999 BCA for RCC HKM M133
CRACK WIDTH CALCULATIONS - FLEXURE -
INPUT
fcu= 35 N/mm2
fy= 460 N/mm2
Area of reinforcement " As " = 2011 mm2
= 1000 mm
h = 500 mm
d = 442 mm
Minimum cover to tension reinforcement " CO " = 50 mm
Maxmum bar spacing " S " = 100 mm
Bar dia " DIA " = 16 mm" acr " =(((S/2)^2+(CO+DIA/2) 2)^(1/2)-DIA/2) as default or enter other value = 68.6 mm
"acr " is distance from the point considered to the surface of the nearest longitudinal bar
Applied service moment " Ms "= 131.0 KNm
CALCULATIONS
moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 13.5 KN/mm
moduli of elasticity of steel " Es " = 200.0 KN/mm2
Modular ratio "a " = (Es/Ec) = 14.81 r = s = 0.005
depth to neutral axis, "x" = (-a.r +((a.r) + 2.a.r).
.d = 135 mm
" Z " = d-(x/3) = 397
Reinforcement stress " fs " = Ms/(As*Z) = 164 N/mmoncrete stress c = s s . x = 4.88 mm
ra n a so o concre e eam s a e1 = s s -x -x = 0.000976Strain due to stiffening effect of concrete between cracks " e2 " =e2 = b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.1 mm n/ae2 = 0.000360
Average strain for calculation of crack width "em "= e1-e2 = 0.000616Calculated crack width, " w " = 3.acr.em/(1+2.(acr-c)/(h-x))
CALCULATED CRACK WIDTH, 'w' = 0.12 mm
REINFORCED CONCRETE COUNCIL
HKM
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ELEMENT DESIGN to BS 8110:1997
SOLID SLABSOriginated from RCC11.xls on CD 1999 BCA for RCC
INPUT Location Wall
Design moment, M 230.0 kNm/m fcu 35 N/mm gc = 1.50
b 1.00 fy 460 N/mm gs = 1.05
span 5500 mm
Height, h 500 mm Section location SIMPLY SUPPORTED S
Bar 16 mm
cover 50 mm to this reinforcement
OUTPUT Wall Compression steel = None
d = 500 - 50 - 16/2 = 442.0 mm .
(3.4.4.4) K' = 0.156 > K = 0.034 ok .
(3.4.4.4) z = 442.0 [0.5 + (0.25 - 0.034 /0.9)^ = 424.8 > 0.95d = 419.9 mm
(3.4.4.1) As = 230.00E6 /460 /419.9 x 1.05 = 1250 > min As = 650 mm/m
PROVIDE T16 @ 150 = 1340 mm/m.
(Eqn 8) fs = 2/3 x 460 x 1250 /1340 /1.00 = 286.0 N/mm
(Eqn 7) Tens mod factor = 0.55 + (477 - 286.0) /120 /(0.9 + 1.177) = 1.316
(3.4.6.3) Permissible L/d = 20.0 x 1.316 = 26.320
. Actual L/d = 5500 /442.0 = 12.443 ok .
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Project
Box Culvert Verdun Trianon Link Road
ClientBCEG Ltd Made by Date Page
Location WALL CRACK WIDTH RN 26-Mar-2013 1
Crack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No
Originated from RCC14.xls on CD 1999 BCA for RCC HKM M 133
CRACK WIDTH CALCULATIONS - FLEXURE -
INPUT
fcu= 35 N/mm2
fy= 460 N/mm2
Area of reinforcement " As " = 2011 mm2
= 1000 mm
h = 500 mm
d = 442 mm
Minimum cover to tension reinforcement " CO " = 50 mm
Maxmum bar spacing " S " = 100 mm
Bar dia " DIA " = 16 mm" acr " =(((S/2)^2+(CO+DIA/2) 2)^(1/2)-DIA/2) as default or enter other value = 68.6 mm
"acr " is distance from the point considered to the surface of the nearest longitudinal bar
Applied service moment " Ms "= 191.0 KNm
CALCULATIONS
moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 13.5 KN/mm
moduli of elasticity of steel " Es " = 200.0 KN/mm2
Modular ratio "a " = (Es/Ec) = 14.81 r = s = 0.005
depth to neutral axis, "x" = (-a.r +((a.r) + 2.a.r).
.d = 135 mm
" Z " = d-(x/3) = 397
Reinforcement stress " fs " = Ms/(As*Z) = 239 N/mmoncrete stress c = s s . x = 7.12 mm
ra n a so o concre e eam s a e1 = s s -x -x = 0.001423Strain due to stiffening effect of concrete between cracks " e2 " =e2 = b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.1 mm n/ae2 = 0.000360
Average strain for calculation of crack width "em "= e1-e2 = 0.001063Calculated crack width, " w " = 3.acr.em/(1+2.(acr-c)/(h-x))
CALCULATED CRACK WIDTH, 'w' = 0.20 mm
REINFORCED CONCRETE COUNCIL
HKM
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ELEMENT DESIGN to BS 8110:1997
SOLID SLABSOriginated from RCC11.xls on CD 1999 BCA for RCC
INPUT Location Base Slab
Design moment, M 31.0 kNm/m fcu 35 N/mm gc = 1.50
b 1.00 fy 460 N/mm gs = 1.05
span 5500 mm
Height, h 500 mm Section location SIMPLY SUPPORTED S
Bar 16 mm
cover 50 mm to this reinforcement
OUTPUT Base Slab Compression steel = None
d = 500 - 50 - 16/2 = 442.0 mm .
(3.4.4.4) K' = 0.156 > K = 0.005 ok .
(3.4.4.4) z = 442.0 [0.5 + (0.25 - 0.005 /0.9)^ = 439.8 > 0.95d = 419.9 mm
(3.4.4.1) As = 31.00E6 /460 /419.9 x 1.05 = 169 < min As = 650 mm/m
PROVIDE T16 @ 300 = 670 mm/m.
(Eqn 8) fs = 2/3 x 460 x 169 /670 /1.00 = 77.1 N/mm
(Eqn 7) Tens mod factor = 0.55 + (477 - 77.1) /120 /(0.9 + 0.159) = 2.000
(3.4.6.3) Permissible L/d = 20.0 x 2.000 = 40.000
. Actual L/d = 5500 /442.0 = 12.443 ok .
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Project
Box Culvert Verdun Trianon Link Road
ClientBCEG Ltd Made by Date Page
Location BASE CRACK WIDTH RN 26-Mar-2013 1
Crack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No
Originated from RCC14.xls on CD 1999 BCA for RCC HKM M 133
CRACK WIDTH CALCULATIONS - FLEXURE -
INPUT
fcu= 35 N/mm2
fy= 460 N/mm2
Area of reinforcement " As " = 2011 mm2
= 1000 mm
h = 500 mm
d = 452 mm
Minimum cover to tension reinforcement " CO " = 40 mm
Maxmum bar spacing " S " = 100 mm
Bar dia " DIA " = 16 mm" acr " =(((S/2)^2+(CO+DIA/2) 2)^(1/2)-DIA/2) as default or enter other value = 61.3 mm
"acr " is distance from the point considered to the surface of the nearest longitudinal bar
Applied service moment " Ms "= 28.0 KNm
CALCULATIONS
moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 13.5 KN/mm
moduli of elasticity of steel " Es " = 200.0 KN/mm2
Modular ratio "a " = (Es/Ec) = 14.81 r = s = 0.004
depth to neutral axis, "x" = (-a.r +((a.r) + 2.a.r).
.d = 137 mm
" Z " = d-(x/3) = 406
Reinforcement stress " fs " = Ms/(As*Z) = 34 N/mmoncrete stress c = s s . x = 1.01 mm
ra n a so o concre e eam s a e1 = s s -x -x = 0.000197Strain due to stiffening effect of concrete between cracks " e2 " =e2 = b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x) /(3.Es.As.(d-x)) for crack widths of 0.1 mm n/ae2 = 0.000347
Average strain for calculation of crack width "em "= e1-e2 = -0.000149Calculated crack width, " w " = 3.acr.em/(1+2.(acr-c)/(h-x))
CALCULATED CRACK WIDTH, 'w' = -0.02 mm
REINFORCED CONCRETE COUNCIL
HKM
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9 (C)
10 (C)11 (C)
12 (C)
13 (C)
14 (C)
15 (C)
16 (C)
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ULS Dead+ Surfacing+Ha UDL+HaKEL(midspan)+Ev+Eh 1 1.15 2 1.75
ULS Dead+ Surfacing+Ha UDL+HaKEL(support)+Ev+Eh 1 1.15 2 1.75ULS Dead+ Surfacing+Hb(midspan)+Ev+Eh 1 1.15 2 1.75
ULS Dead+ Surfacing+Hb(support)+Ev+Eh 1 1.15 2 1.75
SLS Dead+ Surfacing+Ha UDL+HaKEL(midspan)+Ev+Eh 1 1 2 1.2
SLS Dead+ Surfacing+Ha UDL+HaKEL(support)+Ev+Eh 1 1 2 1.2
SLS Dead+ Surfacing+Hb(midspan)+Ev+Eh 1 1 2 1.2
SLS Dead+ Surfacing+Hb(support)+Ev+Eh 1 1 2 1.2
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4 1.5 5 1.5
4 1.5 6 1.54 1.5 7 1.3
4 1.5 8 1.3
4 1 5 1.2
4 1 6 1.2
4 1 7 1.1
4 1 8 1.1