pavement design spreadsheet - ccaa method
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Pavement Design ( Wheel Loading ) - Using CCAA Approach
A concrete pavement is to be designed to support loading from mining trucks with an axle load of 640 kN with a wheel spacing of 4.97 m. All areas of the pavement may be traversed by the mining truck. The pavement design shall be designed for an operating life of 20 years, and it has been estimated that an average of 25 daily load repetitions may occur.
Wheel Spacing = 5.00 mTruck Axel Load = 640 kNPavement life span = 20 YearsTruck repetition = 25 times/day * or equal to = 130000 times / 20 Years - work days
1. The Soil Profile
1. 1 Soil Layer i Comments
Sands/Gravels - Dense 1.0 CBR=8% BASE Based on Geotechnical Report0 m 1
0 m
0 m ROCK
0 m - - Note : *) Assumption(Table 1.1)
1. 2
The equation to determine the Young's Modulus of an equivalent uniform soil layer is as follows :
Where
n = total number of layers in layered profile
Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
Depth (Hi)
Uniform Modulus Young (E se)
Hi = thickness of layer i (m)Esi = Young's modulus of layer i (Mpa)Wfi = weighting factor for layer i
E se=i=1
n
W fi .H i
i=1 s
n
W fi.H i/E si
Not Scale
5.00 m320 KN
320 KN
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
Soil Surface Soil Surface
(figure 1.2.a Young Modulus of an equivalent uniform soil layer)
wheel loading : X=S (wheel Spacing (m))
post loading : X=f(x,y) (average post spacing (m))
distributed loading : X=(aisle or loading width (m))
( figure 1.2.b Weighting factor Wfi, for the estimation of Young's modulus for an equivalent layer - Scaled)
= 0.50 m = 0.10 0.90 1.0 m
using figure 1.2, Ese (Wheel Loading) :
Wf1 --> z1 applied to figure 1.2b, W f1 = H1 =
Layer i Hi Esi
Zi
H H Ese
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
X 5.00 m
= 0.00 m = 0.00 0.00 0 mX 5.00 m
= 1.00 m = 0.20 0.00 0 mx 5.00 m
= 1.00 m = 0.20 0.00 0 mx 5.00 m
( 0.90 x 1.0 ) + ( 0.00 x 0.0 ) + ( 0.00 x 0.00 ) + ( 0.00 x 0 ) = 0.9
1. 3
(figure 1.2.c Correlation between long-term (figure 1.2.d Correlation between short-term Young's
Young's Modulus Esi and CBR) Modulus Es, N from SPT and Plasticity index)
The relationship beetween short-term and long term values of Es can be expressed as :
Soil type Correlation factor, BetaGravels 0.9Sands 0.8Silts, silty clays 0.7Stiff clays 0.6Soft clays 0.4
Based on soil property, the Young's modulus for each layer :
Assume soil : sand/gravel, with Young Modulus Value 20-50 Mpa, Use : 25 Mpa Es = 25 MPa
Wf2 --> z2 applied to figure 1.2, W f2 = H2 =
Wf3 --> z3 applied to figure 1.2, W f3 = H3 =
Wf4 --> z4 applied to figure 1.2, W f4 = H4 =
Calculating Young's Modulus on each layer (Hi)
Ess (short term) = Esl (longterm) / Beta
Es1
i=1
n
W fi.H i=
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
SPT (N) = Es = ### MPa
SPT (N) = Es = ### MPa
SPT (N) = Es = ### MPa
( 0.90 x 1.0 ) + ( 0.00 x 0.0 ) + ( 0.00 x 0.00 ) + ( 0.00 x 0 ) = 0.036( 25.00 ) ( 0.00 ) ( 0.00 ) ( 0.00 )
So, = 25 Mpa
Equivalent Young's Modulus for the 1-m deep layer is 25 MPa
1. 4 Flexural Strenght of Concrete
The design Flextural strenght of the concrete is determined from equation :
Where :
0.9 based on table below
Loading typewheel 0.85-0.95Post 0.75-0.85Distributed 0.75-0.85
### years for a daily repetitin of ### is 130000over the design life.
0.6
Load RepetitionsUnlimited 0.50400000 0.51300000 0.52200000 0.54100000 0.5650000 0.59
Es2
Es3
Es4
fall = k1.k2.f'cf
k1 = Material factork2 = load repetition factorf'cf = characteristic flextural tensile strength of concrete (Mpa)
Assume the value k1 =
range of k1
In order to determine the k2 factor, the number of repetitious over
Using Table below the k2 factor is
Load Repetition factor, k2
i=1 s
n
W fi.H i/Esi=
E se=i=1
n
W fi .H i
i=1 s
n
W fi.H i/E si
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
30000 0.6010000 0.6420000 0.701000 0.73
( based on RTA - NSW 1993) assumed Fc' = ### Mpa
0.7 x ( 32 3.96 MPa
0.9 x 0.6 x 3.96 = 2.138 MPa
1. 4 Concrete Thickness Due to INTERIOR Loading
The base thickness may now be determined based on the interior and edge loading conditions.
for interior loading
Where :
= 1.2 for internal loading= 1.5 for edge loading
using table below
fc'### 0.03### 1.07### 1.12### 1.16### 1.20
for Fc' = 32MPa 1.12
Using Chart 1.4.1
The value of f'cf = 0.7 (fc')0.5
f'cf = )0.5 =
So, Fall =
F1 = Fall FE1 FH1 FS1 k3 k4
Fall = design tensile stregth of concrete (Mpa)FE1 = Factor for short-term Young's modulus, Ess (of equivalent uniform layer of soil)FH1 = factor for depth of equivalent uniform layer of soil, HFS1 = Factor of center to center spacing wheelk3 = calibration factor for geothechnical behavior
k4 = calibration factor for concrete strength
k4
k4 =
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
for Es = ### Mpa 1.19
Using Chart 1.4.2
for S = 5 m 1.20
Using Chart 1.4.3
(Chart 1.4.1 Correlation between short-term Youngs Ess and FE1)
FE1 =
(Chart 1.4.2 Correlation between Wheel spacing and FS1)
FS1 =
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
for H = 9 m 1.20
1.2 (for internal loading)
F1 = 2.138 x 1.19 x 1.2 x 1.2 x 1.2 x 1.12
F1 = 4.925
From chart 1.4.4
(Chart 1.4.4 Slab Thickness and F1)
375 say 375 mm
1. 5 Concrete Thickness Due to Edge Loading
The base thickness may now be determined based on the interior and edge loading conditions.
for edge loading
(Chart 1.4.3 Correlation between depth soil layer and FS1)
FH1 =
k3 =
so, F1 = Fall FE1 FH1 FS1 k3 k4
t1 =
F2 = Fall FE2 FH2 FS2 k3 k4
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
Where :
= 1.2 for internal loading= 1.5 for edge loading
using table below
fc'### 0.03### 1.07### 1.12### 1.16### 1.20
1.12
Using Chart 1.5.1
for Es = 25 Mpa 1.25
Using Chart 1.5.2
Fall = design tensile stregth of concrete (Mpa)FE2 = Factor for short-term Young's modulus, Ess (of equivalent uniform layer of soil)FH2 = factor for depth of equivalent uniform layer of soil, HFS2 = Factor of center to center spacing wheelk3 = calibration factor for geothechnical behavior
k4 = calibration factor for concrete strength
k4
k4 =
(Chart 1.5.1 Correlation between short-term Youngs Ess and FE2)
FE2 =
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
for S = 5 m 1.15
Using Chart 1.5.3
for H = 9 m 1.2
1.05 (for internal loading)
(Chart 1.5.2 Correlation between Wheel spacing and FS2)
FS2 =
(Chart 1.5.3 Correlation between depth soil layer and FS1)
FH2 =
k3 =
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
2.138 x 1.25 x 1.2 x 1.15 x 1.05 x 1.12
4.338
From chart 1.5.4.
600 mm
The interior and edge loading conditions indicate that a 375 and 600 thick base is requiredrespectively. Using Table 1.20 ( clause 3.4.13 ), the recommended distance, e, for edge base thickeningis 8t for a stiff soil support. Therefore, the edge of the base 600 mm thick and the thickness varies fora total distance of 8 x 375 mm = 3000 mm from the edge
1. 6 Reinforcement for internal Slab
A. Reinforcement are required for shrinkage
Assume control joint spacing, L = 5 m
(Eqn E2 - Page 83 CCAA)
t = 375 mmL = 5 m
500 Mpa335 MPa
101
so, F2 = Fall FE2 FH2 FS2 k3 k4
F2 =
F2 =
(Chart 1.5.4 Slab Thickness and F2)
t2 =
So, Asreq = 18 t . L / fs (mm2/m)
fsy =fs = 0.67 fsy =
Asreq = mm2/m
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
B. Minimum tensile strength base on As 3600-2001 cl.9.1.1
Ast = Rho * b.d
Rho= 0.002b = 1000 mmD = 375 mmd0 = 75 mm (Concrete Cover)Ds = 16 mm (Steel Diameter)d = 276 mm
Ast = 552 mm2/m
C. Reinforcement in secondary direction in unrestrained slabs(AS 3600 cl 9.4.3.3)
Minimum Reinforcement area Ast
Ast = 483 mm2
670EWEF
SUMMARY :
Ast = (1.75-2.5 Tcp) b.D X 10-3
Ast = (1.75-2.5 * 0) * 1000 * 501 * 10 -3
So, Use N16-300 --> As = mm2/m
1500 1500
N16 - 300
600
150
800
375
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Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific
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BasePlate_R0 Page 25 of 37 02/14/2016
Southern Regional Water Pipeline (SRWP) ProjectArea 54 - Chambers FlatAmenity Building
Date : 02-Aug-2007Rev. : 0Eng. : Andre Maulana (JEC)
1. Design action (maximum load case) for base plateBased on the design calculation for SHS column:For tension: Nt* = (0.9 x 3.23) + (-20) = -17.09 kNFor shear: V* = (2.36 kNm / 4 m) = 0.59 kN
2. Check bolt capacityN*t N*t Ok!
checking combined shear and tension per bolt:Nt*f = 17.09 / 2 = 8.545 kNV*f = 0.59 / 2 = 0.295 kN
therefore: (8.545/78.4) + (0.295/44.6) = 0.1 < 1.0 Ok!
3. Check the plate capacity for axial tension in the columnN*t dc
(See AISC connection manual [2] clause 4.12.4)
where: ti = 16 mm ; base plate thicknessfyi = 250 Mpa ; base plate yield strengthnb = 2 nos ; bolt(s) numerous
0.9therefore:
549.72 kN > N*t = 17.09 kN Ok!
4. Check weldsN*t