pavement design spreadsheet - ccaa method

Upload: shoyeeb-ahmed

Post on 08-Mar-2016

150 views

Category:

Documents


11 download

DESCRIPTION

Transport

TRANSCRIPT

  • Prepared by :

    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

  • 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

  • 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=

  • 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

  • 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 =

  • 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 =

  • 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

  • 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 =

  • 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 =

  • 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

  • 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

  • Prepared by : Navaeed KhodabakusDesign ManagerKBR - Asia Pacific

  • 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