Reference Calculation Output
Area of concreteArea of concrete in compressionArea of tension reinforcementMinimum area of tension reinforcementLength of that part of member traversed by shear failure plane
b With (breath) or effective width of sectionc Cover to outer diameterd Effective depth of section
Basic force used in defining compressive forcesBasic force used in defining tie forcesCharacteristic strength of concreteEstimated design service stress in the tension reinforcementCharacteristic strength of reinforcement
G Shear modulusH Maximum horizontal force
Horizontal force in x directionHorizontal force in y direction
h Overall depthKEL Knife edge loadL Critical perimeter
Dimension of element on x directionDimension of element on y directionDimension of element on z direction
M Design ultimate resistance momentMoment on x axisMoment on y axisMoment on z axis
q Surcharge loadr Internal radius of bendSLS Serviceability limit stateT Traction forcet Thickness of the elementULS Ultimate limit stateV Shear force due to design ultimate loads or design ultimate value of a
concentrated loadv Design shear stressvc Design shear stress in concretex Neutral axis depthx' Distance from Y axis to the centroid of an elementy' Distance from X axis to the centroid of an elementz Lever armz' Distance from X - Y plane to point where the considered resultant
force acting Coefficient, variously defined, as appropriateStrain in tension reinforcementNominal range of movementSoil friction angle, or diameterActive earth pressureUnit weight of soilPartial load factorPartial load factor
Doc. No. DESIGN UNIT Designed
Ac
Acc
As
As min
av
Fc
Ft
fcu
fs
fy
Hx
Hy
lxlylz
Mx
My
Mz
DEC
Date
β∈sδφσ aγγ fLγ f 3
EPC DIVISION Checked Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page
Reference Calculation Output
DEC
Doc. No. DESIGN UNIT Designed EPC DIVISION Checked Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page
Reference Calculation Output
DEC
Date
Doc. No. DESIGN UNIT Designed EPC DIVISION Checked Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page
Reference Calculation Output
DEC
Date
Doc. No. DESIGN UNIT Designed EPC DIVISION Checked Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page
Reference Calculation Output
DEC
Date
Doc. No. DESIGN UNIT Designed EPC DIVISION Checked Date CENTRAL ENGINEERING CONSULTANCY BUREAU (CECB) Job Code Page
DEC
Date
Reference Calculation Output
Design of Box Culvert
Figure 01Dimentional Properties
h = 1.2 ml = 1.5 m
Soil Cover , H = 7.2 mSafe Bearing Pressure = 150 kN/m2Section Thickness = 0.2 m ( hw , h = span/(10 ~15))
Main R/F = 12 mmCover to R/F = 45 mmGrade of Concrete = 25 N/mm2Properties of Soil
γc = 24 kN/m3γs = 20 kN/m3γw = 9.81 kN/m3Φ' = 25
1 - Permanent Loads1.1 Dead Loads
The nominal dead doad consist of the weight of the materials and the part of the structure
Structural Unit Weight of Concrete shall be taken as 24 kN/m3Engineering Becouse of the arching of soil, check whether the depth above culvert is Design in > 3 x width of culvert ( in which case limit depth to 3 x width ) preactice(Roger - Depth of cover (H) = 7.2 mwestbrook) 3 x width = 3 x 1.6(page-94) = 4.8 m
3 x width < = 7.2 m SoDepth limited to = 4.8 m
Surcharge on RoofSurcharge Presure (qr) = 4.8 x 20
qr = 96 kN/m2
Soil Engineering Casses of conduit installation consider as Ditch Conduit (Spangler & Ditch Conduit Handy) A ditch conduit is defined as one which is instaled in a relatively narrow
ditch dug in passive or undisturbed soil and wich is then covered with earth backfill.
Ceylon Electricity Board Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked Date
o
C E B
Date
Y
hs
hw
Ground Level
hs
hw
A B
D C
H
l
h
X
Civil Structure Maintanance Job Code Page 1Reference Calculation Output
Maximum load on ditch condition
Depth of cover = 7.2 m
Surcharge on RoofSurcharge Presure (qr) ,
(qr) =
Cd =
=
K =
- coedicient of friction between fill materialand side of ditch
K - Active Lateral earth pressure coeficient- Horizontal width of ditch at top of conduit
γ - Unit weight (wet density) of filling materialH - Height of fill above top of conduite
Cd - Load coeficient for ditch condition
So, K = Bd = 3.60 m, Consider 1m length of Roof slab
= 0.406== 0.466
2.K.µ'.(H/Bd) = 0.76Cd = 1.403
(qr) =(qr) = 101.0 kN/m2
Structural 1.2 Horizontal Earth Pressure
Engineering
Design in If the backfill properties are known,preactice If wall friction is to be ignored (δ = 0 )(Roger -westbrook) = 1-sin Φ' = 0.577(page-94) = ( 1-sin Φ' ) / ( 1+sin Φ' ) = 0.406
q max = γ.Ka.h= 20 x 0.41 x 9.1= 73.9 kN/m2
= 20 x 0.41 x 1.9= 15.42 kN/m2
q =q = 58.44 kN/m2
Ceylon Electricity Board Doc. No.Dam Safety Designed S.M.P 31.05.2010
C E B
Cd.γ.Bd2
1-e-2Kµ'(H/Bd)
2.K.µ'µ' tan φ'
1-sin φ1+sin φ
µ'
Bd
1-sin φ1+sin φ
µ' tan φ'
Cd.γ.Bd2
K0
Ka
qep
qmax - qep
C E B
Date
Environmental & Checked DateCivil Structure Maintanance Job Code Page 1
Reference Calculation Output
AASHTO 2 - Vertical Live Loads3.7.1
For Fill Depths H ≥ 8 feet (2400 mm) and Culvert Clear Span Length,The effect of live load is neglected in design when the depth of fill is more than 8 feet
3 - Hydrostatic Pressure (Internal)
= C.h= 9.81 x 1.7= 16.68 kN/m2
4 - AnalysisReinforced Concrete Constant K = h { hs } 3 = 1.21Designers lManual k1 = K+1 = 2.21(ref-5.1) k3 = K+3 = 4.21
k5 = 2K+3 = 5.43k7 = 2K+7 = 9.43k8 = 3K+8 = 11.64
4.1 Load Case -01 Testing Condition4.1.1 Hydrostatic Pressure-(Internal)
Reinforced = =Concrete 60.k1.k3Designers = 0.99 kN.m/mManual(ref-5.1) = = Ma. K8
k7= 1.217 kN.m/m
4.1.2 Flexure due to weight of wall
Wall weight ( G ) = hw.γ.h q1 = 2.G = 10.20 kN/m2= 8.2 kN/m l.hw
Reinforced Concrete = =Designers 12.k1.k3Manual = 0.22 kN.m/m(ref-5.1)
= = Ma. K5K
= -0.97 kN.m/m
4.1.3 Flexure due to weight of Roofq = hs.γc = 4.8 kN/m2
Doc. No.
C E B
q ip
hw
MA MB qip.h2.K.k7
MC MD
MA MB q1.l2.K
MC MD
A B
D C
qip
q = qipB.M.DPressures
A B
D Cq1
G G
B.M.DPressures
Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 2
Reference Calculation Output
= = =
=12.k1
= -0.35 kN.m/mAddition of moment for Load case 01
Position γf Walls Roof γf
A and B 0.99 1.4 1.38 0.22 -0.35 -0.14 1.4 -0.19 1.19
C and D 1.22 1.4 1.70 -0.97 -0.35 -1.32 1.4 -1.85 -0.15
0.99 1.4 1.38 0.22**
1.04 1.4 1.45 2.830.82
1.22 1.4 1.70** **
2.35 1.4 3.29 5.001.53 0.82
*1.4 -2.88 -0.38 -0.35 -0.73 1.4 -1.02 -3.90
-2.06Table - 01
Fixed end mement of the wall for Hydrostatic load
= W.L = W.L15 10
= 1.607 kN.m/m = 2.41 kN.m/m
Maximum (-ve) moment = W.L(Where x is 0.45L from C) 23.3
= -1.0 kN.m/m
* Calculation of moment at mid span of walls done by aproximatly by adding
moment transferred to mid span from FEM to the Maximum negative meoment occurred at 0.45L after moment distribution
** Moment at mid span of the wall is calculated by considering full bending
Calculation of midspan moment due to wall loadNiutral axis depth from A = 0.26 m
4.2 Load Case -02 Culvert empty and trench filledLateral soil pressurees giving rise to flexture in the structure
4.2.1
Reinforced Concrete = =Designers 60.k1.k3Manual = -0.91 kN.m/m(ref-5.1)
= =k7
= -1.13 kN.m/m
C E B
Date
MA MB MC MD
q.l2
Hydrost-atic
uls- Mb
Walls + Roof
uls-Mb
Total uls
Roof mid-Span
Base mid-Span
Walls middle
MA MC
"q"is the rectanguler pressure and "qep" is the triangular pressure
Trianguler Pressure,qep
MA MB qep.h2.K.k7
MC MD MA. K8
A B
D Cqepqep
B.M.DPressures
A B
D Cq = q1
B.M.DPressures
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 3
Reference Calculation Output
4.2.2 Surcharge on walls,q= = =
Reinforced =Concrete 12.k1Designers = -7.72 kN.m/mManual 4.2.3 Surcharge on Roof ,qr(ref-5.1) = = =
=12.k1
= -7.45 kN.m/mAddition of moment for Load Case 2
Posotion q γf
A and B -0.91 -7.72 -0.14 -7.45 -16.22 1.4 -22.70
C and D -1.13 -7.72 -1.32 -7.45 -17.62 1.4 -24.66
-0.91 -7.72 1.04 17.29 9.70 1.4 13.58
-1.13 -7.72 2.35 17.29 10.80 1.4 15.12
Walls middle* **
-0.73 -7.45 6.65 1.4 9.311.43 13.39
= W.L = W.L15 10
= 1.486 kN.m/m = 2.229 kN.m/m
Maximum (-ve) moment = W.L(Where x is 0.45L from C) 23.3
= -1.0 kN.m/m
4.2 Load Case -034.2.1 This is load case 02 + Hydrostatic load from Load case 01
Posotion
A and B -16.22 0.99 -15.23 -22.70 1.38 -21.32
C and D -17.62 1.22 -16.40 -24.66 1.70 -22.96
9.70 0.99 10.69 13.58 1.38 14.96
10.80 1.22 12.02 15.12 1.70 16.83
Walls middle 6.65 -2.06 4.59 9.31 -2.88 6.43
C E B
Date
MA MB MC MD
q.h2.K
MA MB MC MD
q.l2
qepWalls &
Roof(LC-1)Surcharg -e (Roof)
Total (Survice)
Total U.L.S.
Roof mid-Span
Base mid-Span
Fixed end mement of the wall due to qep
MA MC
L.C.02 (Service)
Hydrost. (Service)
Total (Service)
L.C.02 (U.L.S.)
Hydrost. (U.L.S.)
Total (U.L.S.)
Roof mid-Span
Base mid-Span
A B
D CB.M.DPressures
Pressures
A B
D C
B.M.D
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 4
Reference Calculation Output
5 - Check on ground safe bearing pressure5.1 Load Case -01
Hydrostatic Pressure = 16.68 kN/m2Weight of walls = 10.20 kN/m2Weight of Roof + Floor = 9.60 kN/m2Total Pressure = 36.48 kN/m2
Total Pressure < 150 kN/m2 hence ok
5.2 Load Case -02
Weight of walls = 10.20 kN/m2Weight of Roof + Floor = 9.60 kN/m2Surcharge on Roof = 96.00 kN/m2Total Pressure = 115.80 kN/m2
Total Pressure < 150 kN/m2 hence ok
5.3 Load Case -03
Weight of walls = 10.20 kN/m2Weight of Roof + Floor = 9.60 kN/m2Surcharge on Roof = 96.00 kN/m2Hydrostatic Pressure = 16.68 kN/m2Total Pressure = 122.28 kN/m2
Total Pressure < 150 kN/m2 hence ok
6 - U.L.S. of FlextureMaximum Moments kN.m/m
Member Hogging SaggingRoof -22.70 (L.C-01) 14.96 (L.C-03)Walls -24.66 (L.C-02) 9.31 (L.C-02)Base -24.66 (L.C-02) 16.83 (L.C-03)
i - SlabsMaximum Moment = 24.15 kN.m/m
C E B
Date
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 5
Reference Calculation Output
6 - Design Calculation for Box Culvert
6.1 U.L.S. of FlextureAnalysis was carried out for several load cases of various loading arrangements to find out the maximum effect on the Box culvert
Diameter of main reinforcement = 12 mmDiameter of secondary reinforcement = 12 mmSection Thickness = 200 mm
Maximum Bending Moment = 24.15 kN.m/m
Assume severe environment condition, for driving rainCover = 45 mm
Effective depth, d = 200 - 45 - 6 d = 149 mm= 149 mm
k = 2== 0.044 < 0.156
Hence no compression r/f is required
M = equation 1z = equation 5 from these two equations
z =z =
= 141.41 < 0.950 d
Take Z as 0.95dZ = 0.95 d
= 0.95 x 149 = 142 mm
6.1.1 Design of main reinforcement== == 426 426
Use T 12 @ 250 ( As = 452 =
452Minimum area of main rainforcement for slabs
= 100x452/(1000x149) = 0.30 ### 0.13 Main r/fT 12 @ 250
Hence o.k
6.2 Design for Shear Reinforcement
Check shear in U.L.S. on roof and floor slabs Take Load case 02Shear across support = ( 115.80 - Wt of Base x γf )
= 109.08 kN/m2
C E B
Date
M / (bd2fcu)
(24.15x106 /(1000x1492x25)
(0.87fy)Asz(1 - 1.1fyAs/ fcubd) d
d (0.5+(0.25-k/0.9)1/2 d [0.5+(0.25-0.044/0.9)1/2
As M / 0.87fyz24.15 x106 / 0.87x460x142 As req
mm2/m mm2/m
mm2/m As pro
mm2/m
100As / bad
Therefore shear in the support = 109.08 x 1.2 /2= 65.45 kN/m
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 6
Reference Calculation Output
Design shear force, V design = 65.45 kN/mEffective depth, d = 149 mmTension steel across shear plane = Y12 -250 c/c
100 As/bd = 100 x 4521000x149
= 0.30
BS 8110 Effective depth = 149 mmPart 01 =table 3.1 = 0.54
Design shear stress v = V/bd== 0.44
v < vc Hence o.k
6.3 Check in U.L.S. on the ability of the wall to trasmit the axial loads
Bs 8110 Treat as a column with bending at right angle to wall3.9.3.6.2 Check h/hw = 1.7 / 0.23.4.4.1 = 8.5 < 12
hence column is shortBS 8110 indicates that the effect of the axial load may be ignored if this force does
hence 0.1.fcu.(C.S.A) = 0.1 x 30 x 200= 600 kN/m
Ultimate Load /m/Wall = 1/2( 96.0 x 1.7 x 1.4+ 0.2 x 1.7 x 24x1.4 )
= 120 kN/m < 600 kN/mhence o.k.
The above calculation assumes that the wall is cosidered as reignfoced and not mass concrete vertical R/F provided = Y 12 @ 200 2 Layers
so Area = 1131.0 mm2
Percentage of Concrete area = 1131.0 x 1001000 x 149
= 0.759 % > 0.4 %This is > Minimum of 0.4% hence o.k.
C E B
Date
vc 0.79x{(100As/bd)1/3.(400/d)1/4/1.25
(65.45x103)/(1000x149)N/mm2
not exceed 0.1.fcu.(c.s.a.)
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 7
Reference Calculation Output
C E B
Date
Doc. No.Dam Safety Designed S.M.P 31.05.2010Environmental & Checked DateCivil Structure Maintanance Job Code Page 8
C E B
Date
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