Package EX-10 Bridge Name Lach Tray Brid Abutment Calsulation

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    Package: EX-10

    Bridge name: Lach Tray Bridge

     Abutment Name:  A1, A2

    1. MATERIALS

    1.1. Material Unit Weights

      • Unit Weight of Concrete 2500= kg/m3 →   γc 24.5= kN/m

    3

      • Unit Weight of Soil 1800= kg/m3 →   γs 17.7= kN/m3

      • Unit Weight of Saturated Soil 1100= kg/m4 →   γsw 11.0= kN/m4

    1.2. Concrete

    Compressive Strength of concrete at 28 days f'c 35= MPa

    Modulus of Elasticity Ec 31799= MPa

    Stress Block Factor    β1 0.80=

    Modulus of Rupture f r  =0.63√f'c 3.73= MPa

    1.3. Reinforcement Reinf . Standart: 1

    Yield strength f y 420= MPa

    Modulus of elasticity Es 200000= MPa

    Reinforcing bar Area

    Diameter  10 13 16 19 22 25 29 32 36

    (mm2) 71 129 199 284 387 510 645 819 1006

    2. LOADS FROM SUPERSTRUCTURE

    2.1. Dead load Number of Girders ng 8= girders

    Width of Bridge W 16.50= m

    Volume Gravity Reaction

    (m3) (kN/m3) (kN) 1 - Main Girder  220.00 24.5 2697.8

     2 - Deck Slab 126.39 24.5 1549.9

     3 - Diaphargms 15.00 24.5 183.9

     4 - Precast plank 11.40 24.5 139.8

     5 - Curbs 28.73 24.5 352.2

     6 - Railing and other  10.0

     7 - Surface wearing 48.10 22.6 542.7

    DC 4923.6

    DW 552.7

    2.2. Live load

    Vehicular live loading name HL-93

    Number of lanes 4=

    Mutiple Presence Factors of Live Load 0.650=

    0.004448

    0.3048 P1 35 kN V1 4.3 m

    P2 145 kN V2 4.3 m

    P3 145 kN

    P4 110 kN V3 1.2 m

    P5 110 kN

    Wl 9.3 kN/m

     ABUTMENT CALCULATION

    Design Truck 

    Design Tandem

    Design Lane Load 

    Total

    Forces Wheel Spacing

    Reaction due to dead load of super-T girder Span

    Members

    Live load

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    V1 = 4.3m V2 = 4.3m - 9.0m

    Design Truck

    35kN 145kN 145kN

    Design Tandem

    1.200m

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    Dynamic Load Allowance

    Component IM

    Deck Joint - All Limit States 75%

     All Other Components

    Fatigue and Fracture Limit State 15%All Other Limit States 25%

    Total span length Lst 38.300= m

    Calculation span leng Ls 37.600= m

    Pedestrian load 0.0= kN/m2

    Width of 1 sidewalk Wsw 0.000= m

    Number of sidewalk nsw 0.000=

    Pedestrian load Wp 0.0= kN/m

    Pedestrian Live

    P1 P2 P3 Total P4 P5 Total Load LoadReaction 70.2 333.9 377.0 781.1 365.0 286.0 651.0 454.6 0 1430.9

    Notes: Reaction due to live load HL- 93 = max (Rtruck, Rtandem) + Rlane

    Combine live load HL-93 and pedestrian for disadvantage

    2.3. Braking force BR 211.3= KN

    2.4. Temprature Load

    Uniform temperature change   ΔT = +/-20.0 oCCoefficient of thermal expansion 1.08E-5= /

    oC

    Movement Δu 0.0081= m

    Shear Modulus of Elastomer G 1000= KPa

     Area of bearing  Ab 0.158= m2

    Height of bearing hrt 0.078= m

    Horizontal Force due to ΔT H = G.A.Δu/hrt 131.2= KN

    2.5. Creep and Shrinkage Load

    Convert to Uniform temperature change   ΔT = 20.2= - oCMovement Δu 0.0082= m

    Horizontal Force due to ΔT H = G.A.Δu/hrt 132.6= KN

    2.4. Wind Load

    The design wind velocity, V, shall be determined from: V = S.VB

    where: VB: Basic 3 second gust wind velocity with 100 years return period appropriate to the Wind Zone

    in which the bridge is located, as specified in table.

    VB

    (m/s)

    I 38 Wind zone: IVII 45 VB 59.0= m/s

    III 53

    IV 59

    S: Correction factor for up wind terrain and deck height S 1.140=

    V 67.3= m/s

    Wind zone according to

    TCVN 2737-1995

    Design Truck Design Tandem Lane

    Load

    Design Truck

    Design Lane Load

    R

    WL

    Ls

    LIVE LOAD APPLYING ON SUPERSTRUCTURE

    P1 P3P2

    R

    Design Tandem

    R

    P4 P5

    Design Lane load9.3 kN/m

    Design Tandem

    1.200m

    110kN 110kN

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    2.5.1. Wind load on structures

    Transverse wind load on structure

    Overall width of the bridge between outer faces of parapets b 15.700= m

    Depth of superstructure, including solid parapets d 3.060= m

    Ratio b/d 5.131=

    Dag coefficient   Cd = f(b/d) 1.4= Area of the structures for calculation of transverse wind load  At 58.599= m2

    Transverse wind load PD = max(0.0006V2.Cd. At,1.8At) = 222.7= KN

    Longitudinal wind load  FWSL  = 0.25PD = 111.3= KN

    2.5.2. Wind load on vehicles (WL)

    Transverse wind load on vehicle 28.7= KN

    Longitudinal wind load on vehicles 28.7= KN

    Vertical wind load 

     Area  Av 315.975= m2

    Vertical wind load   Pv = 0.00045V2.Av 643.2= KN

    2.6. Earthquke effects (EQ)

     Acceleration coefficient A 0.1168=

    Seismic Performance Zones 2

    Soil profile type IV

    Site Coefficients S 2=

    Response Modification Factor 

    For Stem wall R 1.5=

    For Foundation R 1.0=

    Elastic seismic response coefficient Csm 0.292=

    Longitudinal Force due to Earthquake EQ 3198.1= KN

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    3.2. Internal Forces at Bottom Footing

    Loading Data:

    Ht   10.000= m

    KA   0.297=

    kh 0.175=

    kv 0.070=

    θ 0.19= rad

    δs 0.35= rad

    KAE   0.442=

    ABUTMENT LOADS

    Area Length Force   X1 Arm 1 Arm 2

      Moment

    MLong

    Moment

    MTrans

    (m2)   (m) (kN) (m) (m) (m) (kN•m) (kN•m)

    Selfweight

      Section A 6.25 16.500 2529.1 1.250 2.500 - 6322.9 -

    Section A1 3.75 16.500 1517.5 3.250 0.500 - 758.7 -

    Section B 7.95 16.500 3217.1 3.250 0.500 - 1608.5 -

    Section C 8.75 16.500 3540.8 5.750 -2.000 - -7081.6 -

    Section D - - - - - - - -

    Section E 1.10 16.500 445.1 3.750 - - - -

    Section F 15.93 1.600 624.9 5.750 -2.000 - -1249.8 -

    Section G - - - - - - - -

    Section H 10.33 1.600 405.2 5.750 -2.000 - -810.3 -

    Section I 0.14 14.900 49.3 4.133 -0.383 - -18.9 -

    Section J 5.00 1.600 196.2 8.367 -4.617 - -905.8 -

    Bearing Seat 0.04 4.800 4.7 3.100 0.650 - 3.1 -

    Concrete Block 0.61 3.360 50.2 3.100 0.650 - 32.6 -

    Shield Wall 1.50 0.400 14.7 3.000 0.750 - 11.0 -

    Curb 0.75 6.000 110.4 6.500 -2.750 - -303.5 -

    Railing - - 6.0 6.500 -2.750 - -16.5 -Total (DC) 12525.2 -1649.5  

    Buoyancy effect on Abutment

      Section A - - - - - - - -

    Section A1 - - - - - - - -

    Section B - - - - - - - -

    Section C - - - - - - - -

    Section D - - - - - - - -

    Section F - - - - - - - -

    Section G - - - - - - - -

    Section H - - - - - - - -

    Section J - - - - - - - -

    Total (WA) - - -Soil weight

      Section F 15.93 14.900 4189.9 5.750 -2.000 - -8379.9 -

    Section G - - - - - - - -

    Section H 10.33 14.900 2716.6 5.750 -2.000 - -5433.1 -

    Section K 1.25 16.500 364.2 1.250 2.500 - 910.5 -

    Description

       V   E   R   T   I   C   A   L   L   O   A   D   S

    Live Load

    Surcharge

    +M

    +H

    +V

    Sign Convention

    EJ

    Ht

    (K A, KEA)γsHt

    Vertical

    Reaction

    EQ, BR

    F

    H

    B

    D

    G

     A C

    Live Load

    Surcharge

    X

     A1

    Ht/3

    0.5HKP1

    δ

    PH1

    PV1

    P2δ

    PH2

    PV2

    I

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Horizoltal Wind Load on Structure - - 222.7 - - 7.850 - 1748.0

    Horizontal Wind Load on Vehicle - - 28.7 - - 7.800 - 224.1

    EQ form Superstructure - - 479.7 - - 7.800 - 3741.8

    EQ from Abutment

      Section A 6.25 16.500 221.6 - - 1.250 - 276.9

      Section A1 3.75 16.500 132.9 - - 1.250 - 166.2

      Section B 7.95 16.500 281.8 - - 5.150 - 1451.3

      Section C 8.75 16.500 310.2 - - 1.250 - 387.7

      Section D - - - - - - - -

    Section E 1.10 16.500 39.0 - - 8.900 - 347.0

      Section F 15.93 1.600 54.7 - - 4.775 - 261.4

      Section G - 1.600 - - - 7.050 - -

    Section H 10.33 1.600 35.5 - - 8.525 - 302.6

      Section I 0.14 14.900 4.3 - - 9.250 - 40.0

      Section J 5.00 1.600 17.2 - - 9.060 - 155.7

      Bearing Seat 0.04 4.800 0.4 - - 7.850 - 3.2

      Concrete Block 0.61 3.360 4.4 - - 0.870 - 3.8

      Shield Wall 1.50 0.400 1.3 - - 8.550 - 11.0

      Curb 0.75 6.000 9.7 - - 10.600 - 102.5

      Railing - - 0.5 - - 11.400 - 6.0

    Total 1113.5 - 3515.4

      Notes: 1. Distance 'X' is measured horizontally from Toe of Abutment to C.G. of Section

    2. Moment 'Arm' is measured from Pile C.G. Horizontally and from Underside of Footing Vertically

    SUMMARY LOADING AT FOOTING CENTER

    Vertical

    V Hx My Hy Mx

    (kN) (kN) (kN•m) (kN) (kN•m)

    Seflweight of Abutment DC 12525 - -1650 - -

    DC of Superstructure DC 4924 - 3200 - -

    DW of Superstructure DW 553 - 359 - -

    Soil cover at toe EV 7271 - -12902 - -

    Earth Pressure EH 1481 4070 8012 - -

    Vertical pressure due to LL surcharge ESV 562 - -1123 - -

    Horizontal pressure due to LL surcharge ESL 181 448 1564

    Live Load from Superstructure LL 1431 - 930 - -

    Braking Force BR - 106 829 - -

    Longitudinal Wind Load on Superstructure WSL - 56 437 - -

    Longitudinal Wind Load on Vehicle WLL - 14 113 - -

    Horizontal Wind Load on Superstructure WST - - - 223 1748

    Horizontal Wind Load on Vehicle WLT - - - 29 224

    Vertical Wind LoadWS

    V 643 - 418 - -Temperature Load TU - 66 515 - -

    Earth pressure due to EQ EH-EQ 2205 6057 11924 - -

    EQ from Abutment EQ - 3712 11718 1113 3515

    EQ from Superstructure EQ - 1599 12553 480 3742

    Buoyancy effect on Abutment WA - - - - -

    Buoyancy effect on Soil WA - - - - -

    Transverce

    Symbol

    Longitudinal

       H   O   R   I   Z   O   N   T   A   L   L   O   A   D   S

    Description

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    LOAD FACTOR

    Service

    STR-IA STR-IB STR-IIIA STR-IIIB SER-I EXT-IA EXT-IB

    Seflweight of Abutment DC 1.25 0.90 1.25 0.90 1.00 1.25 0.90

    DC of Superstructure DC 1.25 0.90 1.25 0.90 1.00 1.25 0.90

    DW of Superstructure DW 1.50 0.65 1.50 0.65 1.00 1.50 0.65

    Soil cover at toe EV 1.35 0.90 1.35 0.90 1.00 1.35 0.90

    Earth Pressure EH 1.50 0.90 1.50 0.90 1.00 - -

    Vertical pressure due to LL surcharge ESV 1.75 1.75 1.35 1.35 1.00 0.50 0.50

    Horizontal pressure due to LL surcharge ESL 1.50 0.75 1.50 0.75 1.00 0.50 0.50

    Live Load from Superstructure LL 1.75 1.75 1.35 1.00 1.00 0.50 0.50

    Braking Force BR 1.75 1.75 1.35 1.00 1.00 0.50 0.50

    Longitudinal Wind Load on Superstructure WSL - - 0.40 0.40 0.30 - -

    Longitudinal Wind Load on Vehicle WLL - - 1.00 1.00 1.00 - -

    Horizontal Wind Load on Superstructure WST - - - - - - -

    Horizontal Wind Load on Vehicle WLT - - - - - - -

    Vertical Wind Load WSV - - - - - - -

    Temperature Load TU 0.50 0.50 0.50 0.50 0.50 - -

    Earth pressure due to EQ EH-EQ - - - - - 1.00 1.00

    EQ from Abutment EQ - - - - - 1.00 1.00

    EQ from Superstructure EQ - - - - - 1.00 1.00

    Buoyancy effect on Abutment WA 1.00 1.00 1.00 1.00 1.00 1.00 1.00

    Buoyancy effect on Soil WA 1.00 1.00 1.00 1.00 1.00 1.00 1.00

    LOAD COMBINATIONS

    Longitudinal Transverce

    N Hx My Hy Mx

    (kN) (kN) (kN.m) (kN) (kN.m)

    Description Symbol

    Strength IA STR-IA   38436 6995 793 0 0

    Strength IB STR-IB   27563 4217 -229 0 0

    Strength IIIA STR-IIIA   37639 6990 826 0 0

    Strength IIIB STR-IIIB   26265 4174 -812 0 0

    Service I SER-I   28927 4688 -279 0 0

    Extreme IA EXT-IA   35747 11645 22353 1593 7257

    Extreme IB EXT-IB   25898 11645 27311 1593 7257

    Internal Force at Head Pile

    Com.   P1   P2   P3   P4   M1   M2   M3   M4

    (kN) (kN) (kN) (kN) (kN.m) (kN.m) (kN.m) (kN.m)

    1 STR-IA -23660 - - -14775 -9599 - - -9599

    2 STR-IB -16335 - - -11228 -5860 - - -5860

    3 STR-IIIA -23264 - - -14375 -9587 - - -9587

    4 STR-IIIB -15557 - - -10708 -5861 - - -5861

    5 SER-I -17298 - - -11629 -6517 - - -6517

    6 EXT-IA -28971 - - -6775 -13794 - - -137947 EXT-IB -24920 - - -978 -13279 - - -13279

    Note: the above pile internal forces are taken from a 3D pile analysis software

    SymbolDescription ExtremeStrength

    No.

    Load combinations

     M

     V

    P1, M1

     H

    Back fill sideFront side

    +M

    +H

    +V

    Quy − íc P2, M2 P3, M3 P4, M4

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    3.3. Section analysis

    Internal Force at Section A-A

    Area Length Force   X1 Arm Moment

    (m2)   (m) (kN) (m) (m) (kN•m)

    Selfweight of Abutment

      Section I 0.14 14.900 49.3 4.133 -0.383 -18.9

      Section E 1.10 16.500 445.1 3.750 - -

    Curb 0.75 1.000 18.4 3.750 - -

    Total (DC) 494.5 -18.9

    Earth Pressure (EH) - 16.500 197.0 - 0.880 173.4

    Horizontal Pressure due to Surcharge - 16.500 109.2 - 1.100 120.2

    Earth Pressure due to EQ - 16.500 293.2 - 0.880 258.0

    EQ from Abutment

      Section I 0.14 14.900 14.4 - 0.750 10.8

      Section E 1.10 16.500 130.0 - 1.100 143.0

      Curb 0.75 1.000 5.4 - 2.800 15.0

    Total 144.4 153.8  

      Notes: 1. Distance 'X' is measured horizontally from Toe of Abutment to C.G. of Section

    2. Moment 'Arm' is measured from Pile C.G. Horizontally and from Underside of Footing Vertically

    Summary Load Combinatons at Section a-a

    Longitudinal

    N Hx My(kN) (kN) (kN)

    Description Symbol

    Service SER   494 306 275

    Strength IA STR-IA   618 487 447

    Extreme IA EXT-IA   618 492 448

    Description

    Load Combinations

     A

    X

    B

    C

    D

    (K A, KEA)γsHtLive Load

    surcharge

    Vertical

    Reaction

    EQ, BR

    Live Load

    surcharge

    Ht

    Ht/3

    0.5H P1

    δ

    PH1

    PV1

    P2δ

    PH2

    PV2

    P1, M1 P2, M2 P3, M3 P4, M4

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Horizoltal Wind Load on Structure - - 222.7 - - 5.350 - 1191.3

    Horizontal Wind Load on Vehicle - - 28.7 - - 5.350 - 153.7

    EQ form Superstructure - - 479.7 - - 5.350 - 2566.5

    EQ from Abutment

      Section B 7.95 16.500 281.8 - - 2.650 - 746.8

      Section D - - - - - - - -

    Section E 1.10 16.500 39.0 - - 6.400 - 249.6

      Section F 15.93 1.600 54.7 - - 2.275 - 124.5

      Section G - 1.600 - - - - - -

    Section H 10.33 1.600 35.5 - - 6.025 - 213.8

      Section I 0.14 14.900 4.3 - - 6.750 - 29.2

      Section J 5.00 1.600 17.2 - - 6.560 - 112.7

      Bearing Seat 0.04 4.800 0.4 - - 5.350 - 2.2

      Concrete Block 0.61 14.900 19.5 - - -1.630 - -31.8

      Shield Wall 1.50 0.400 1.3 - - 6.050 - 7.8

      Curb 0.75 6.000 9.7 - - 8.100 - 78.3

      Railing - - - - 8.900 - -

    Total 463.4 - 1533.2  

      Notes: 1. Distance 'X' is measured horizontally from Toe of Abutment to C.G. of Section

    2. Moment 'Arm' is measured from Pile C.G. Horizontally and from Underside of Footing Vertically

    SUMMARY LOADING AT SECTION B-B

    Vertical

    V Hx My Hy Mx

    (kN) (kN) (kN•m) (kN) (kN•m)

    Selfweight of Abutment DC 3842 - -254 - -

    DC of Superstructure DC 4924 - 739 - -

    DW of Superstructure DW 553 - 83 - -

    Earth Pressure EH - 2289 6868 - -

    Horizontal pressure due to LL surcharge ESH - 372 1397 - -

    Live Load from Superstructure LL 1431 - 215 - -

    Braking Force BR - 106 565 - -

    Longitudinal Wind Load on Superstructure WSL - 56 298 - -

    Longitudinal Wind Load on Vehicle WLL - 14 77 - -

    Horizontal Wind Load on Superstructure WST - - - 223 1191

    Horizontal Wind Load on Vehicle WLT - - - 29 154

    Vertical Wind Load WSV 643 - 96 - -

    Temperature Load TU - 66 351 - -

    Earth pressure due to EQ EH-EQ - 3407 10222 - -

    EQ from Abutment EQ - 1496 5208 463 1533

    EQ from Superstructure EQ - 1599 8555 480 2567

    Buoyancy effect on Abutment WA - - - - -

    Longitudinal Transverse

    Description Symbol

       H   O   R   I   Z   O   N   T   A   L   L   O   A   D

       S

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    LOAD FACTOR

    Service

    STR-IA STR-IB STR-IIIA STR-IIIB SER-I EXT-IA EXT-IB

    Selfweight of Abutment DC 1.25 0.90 1.25 0.90 1.00 1.25 0.90

    DC of Superstructure DC 1.25 0.90 1.25 0.90 1.00 1.25 0.90

    DW of Superstructure DW 1.50 0.65 1.50 0.65 1.00 1.50 0.65

    Earth Pressure EH 1.50 0.90 1.50 0.90 1.00 - -

    Horizontal pressure due to LL surcharge LLSl 1.50 0.75 1.50 0.75 1.00 0.50 0.50

    Live Load from Superstructure LL 1.75 1.75 1.35 1.35 1.00 0.50 0.50

    Braking Force BR 1.75 1.75 1.35 1.35 1.00 0.50 0.50

    Longitudinal Wind Load on Superstructure WSL - - 0.40 0.40 0.30 0.50 0.50

    Longitudinal Wind Load on Vehicle WLL - - 1.00 1.00 1.00 - -

    Horizontal Wind Load on Superstructure WST - - - - - - -

    Horizontal Wind Load on Vehicle WLT - - - - - - -

    Vertical Wind Load WSV - - - - - - -

    Temperature Load UT 0.50 0.50 0.50 0.50 0.50 - -

    Earth pressure due to EQ EQW - - - - - 1.00 1.00

    EQ from Abutment EQL - - - - - 1.00 1.00

    EQ from Superstructure EQ - - - - - 1.00 1.00

    Buoyancy effect on Abutment WA 1.00 1.00 1.00 1.00 1.00 1.00 1.00

    Summary Load Combinatons at Section B-b

    Longitudinal Transverse

    N Hx My Hy Mx

    (kN) (kN) (kN) (kN) (kN.m)

    Description Symbol

    Strength IA STR-IA   14291 4210 14667 0 0

    Strength IB STR-IB   10753 2557 9259 0 0

    Strength IIIA STR-IIIA   13718 4205 14551 0 0

    Strength IIIB STR-IIIB   10180 2552 9143 0 0

    Service I SER-I   10750 2831 9953 0 0

    Extreme IA EXT-IA   12502 6769 25952 943 4100

    Extreme IB EXT-IB   8964 6769 25712 943 4100

    Internal Force at Section C-C

    Load Internal Force at head pile Sefl Shear Moment

    Comb.   P1   P2   M1   M2   weight (kN/m) (kN•m)

    STR-IA -23660 N/A -9599 N/A 3161 -20499 -29307

    STR-IB -16335 N/A -5860 N/A 2276 -14059 -19350

    STR-IIIA -23264 N/A -9587 N/A 3161 -20103 -28900

    STR-IIIB -15557 N/A -5861 N/A 2276 -13280 -18572SER-I -17298 N/A -6517 N/A 2529 -14769 -20653

    EXT-IA -28971 N/A -13794 N/A 3161 -25810 -38814

    EXT-IB -24920 N/A -13279 N/A 2276 -22644 -35354

      Note: • Self weight is included and

    • Soil weight above pile cap is ignored

    Extreme

    Load Combinations

    Symbol StrengthDescription

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Internal Force at Section D-D

    Area Length Force   X1 Arm Moment

    (m2)   (m) (kN) (m) (m) (kN•m)

    Self weight

      Section C 8.75 16.500 3540.8 5.750 -1.750 -6196.4

      Section F 15.93 1.600 624.9 5.750 -1.750 -1093.6

      Section G - - - - - -

    Section H 10.33 1.600 405.2 5.750 -1.750 -709.0

      Section I 0.14 14.900 49.3 4.133 -0.133 -6.6

      Section J 5.00 1.600 196.2 8.367 -4.367 -856.7

      Curb 0.75 6.000 110.4 6.500 -2.500 -275.9

      Railing - - 6.0 6.500 -2.500 -15.0

    Total 4932.7 -9153.2  

    Soil

      Section F 15.93 14.900 4189.9 5.750 -1.750 -7332.4

      Section G - - - - - -

    Section H 10.33 14.900 2716.6 5.750 -1.750 -4754.0

    Total 6906.5 -12086.3

      ESv heq  = 610 mm - 14.900 561.7 5.750 -1.750 -983.0

      Notes: 1. Distance 'X' is measured horizontally from Toe of Abutment to C.G. of Section

    2. Moment 'Arm' is measured from Pile C.G. Horizontally and from Underside of Footing Vertically

    3. Buoyancy is ignored

    Summary Internal Force at Section d-d

    Comb. Shear Moment

    P4   P3   P2   M4   M3   M2   N M (kN/m) (kN•m)

    STR-IA -14775 N/A N/A -9599 N/A N/A 16473 -29478 1697 -9526

    STR-IB -11228 N/A N/A -5860 N/A N/A 11638 -20836 410 -4240

    STR-IIIA -14375 N/A N/A -9587 N/A N/A 16473 -29478 2098 -10316

    STR-IIIB -10708 N/A N/A -5861 N/A N/A 11638 -20836 931 -5281

    SER-I -11629 N/A N/A -6517 N/A N/A 12822 -22960 1193 -6219

    Wing Wall Calculation

    Wing Wall is modeled and Calculated by ACES5.5 program:

    d4 1.5000= m tgφ 5.250=

    h1 3.5000= m Earth Pressure due to LL Surcharge 3.202= KN/m

    h2 4.0000= m

    b2 3.5000= m

    b3 3.5000= m

    b4 2.0000= m

    Internal Force at Head Pile

       V   E   R   T   I   C   A   L   L   O   A   D   S

    Description

    F

    E

    h1

    h2

    b4b3

    b2

    Ht

    K AγsHtLL Surcharge

    LL Surcharge

    0.4Ht

    0.5H

    d4

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    SERVICE – Element Moment X:

    SERVICE – Element Moment Y:

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    STRENGTH – Element Shear X:

    STRENGTH – Element Shear Y:

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    Factored Shear  Vu 487= KN

    Factored Moment  Mu 487= KNm

    Factored Axial Force Nu 618= KN

    Service Moment  Ms 275= KNm

    h 500= mm

    b 16500= mm

    d1 70= mm

    d2 0= mm

    d3 362= mm

    d's 68= mm

    de = ds 430= mm

    Tension Reinforcement Compresion Reinf. Transverse Reinf.

    Number (bars) ns 132= n's 66= nv 68=

    Diameter (mm) Ds 16= D's 16= Dv 16=

     Area (mm2)  As 26268=  A's 13134=  Av 13532=

    Spacing (mm) s 125= d 250= sv 500=

    Resistance factor for Flexure: ϕ 0.90=

    Resistance factor for Shear: ϕv 0.90=

    Flexural Resistance

    Distance from extreme compression fiber to the neutral axis: c 28= mm

    Depth of the equivalent stress block: a 22= mm < 2d's

    Factored Flexural Resistance Mr  4158= kN•m > 487 kN•m O.K.

    Maximum Reinf. c / d e 0.065 = < 0.42   O.K.

    1.2 times the cracking moment 1.2Mcr  2562= kNm

    1.33 times the factored moment 1.33Mu 647= kNm

    Minimum Reinf. Min (1.2*Mcr  or 1.33*Mu) 647= kN•m < 4158 kN•m O.K.

    Control of cracking by distribution of reinforcement

    Components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state

    does not exceed f sa, determined as:

    n = Es/Ec 6.290= dc 84= mm

     A 21000= mm

    0.023= Z 30000= N/mm

    f sa 248= Mpa

    0.194= 0.6f y 252= Mpa

     j = (1-k/3) 0.935=

    26= Mpa Check: OK

    REINFORCEMENT

    SECTION DIMENSIONS

    SECTION A-A CHECK

    0.85•f'c•a•b

     As•f y

    a A's•f y

    b

    h ds

    d2

    d's

    d3

    d1

    nv,Dv

    ns, Ds

    n's, D's

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    y3/1c

    sa f 6.0) Ad(

    Zf  ≤=

    s

    s

    bd

    nAm =

    m2mmk 2 ++−=

    ss

    ss

     jd A

    Mf  =

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    SECTION A-A CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 419= mm

    Effective Shear Width bv 16500= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 487= < 4937= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 24.3= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 78= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00013= ≤ 0.001 OK

    1000*εx 0.128=

    Ratio vu/f'c 0.002=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   3.2= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   24.3= 0

    Vc 10972= kN

    Vs 10519= kN

    Vn1 21492= kN

    Vn2 60459= kN

    Norminal Shear Resistance Vn 21492= kN

    Factored Shear Resistance Vr  19342= kN > 487 kN O.K.

    Minimum transverse reinforcement  9645= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.08= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 335= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    Factored Shear  Vu 4210= KN

    Factored Moment  Mu 14667= KNm

    Factored Axial Force Nu 14291= KN

    Service Moment  Ms 9953= KNm

    h 1500= mm

    b 16500= mm

    d1 75= mm

    d2 0= mm

    d3 1355= mm

    d's 70= mm

    de = ds 1425= mm

    Tension Reinforcement Compresion Reinf. Transverse Reinf.

    Number (bars) ns 132= n's 66= nv 68=

    Diameter (mm) Ds 25= D's 19= Dv 16=

     Area (mm2)  As 67320=  A's 18744=  Av 13532=

    Spacing (mm) s 125= d 250= sv 500=

    Resistance factor for Flexure: ϕ 0.90=

    Resistance factor for Shear: ϕv 0.90=

    Flexural Resistance

    Distance from extreme compression fiber to the neutral axis: c 72= mm

    Depth of the equivalent stress block: a 58= mm < 2d's

    Factored Flexural Resistance Mr  35529= kN•m > 14667 kN•m O.K.

    Maximum Reinf. c / d e 0.051= < 0.42   O.K.

    1.2 times the cracking moment 1.2Mcr  23062= kNm

    1.33 times the factored moment 1.33Mu 19507= kNm

    Minimum Reinf. Min (1.2*Mcr  or 1.33*Mu) 19507= kN•m < 35529 kN•m O.K.

    Control of cracking by distribution of reinforcement

    Components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state

    does not exceed f sa, determined as:

    n = Es/Ec 6.290= dc 89= mm

     A 22125= mm

    0.018= Z 30000= N/mm

    f sa 240= Mpa

    0.173= 0.6f y 252= Mpa

     j = (1-k/3) 0.942=

    110= Mpa Check: OK

    REINFORCEMENT

    SECTION DIMENSIONS

    SECTION B-B CHECK

    0.85•f'c•a•b

     As•f y

    a A's•f y

    b

    h ds

    d2

    d's

    d3

    d1

    nv,Dv

    ns, Ds

    n's, D's

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    y3/1c

    sa f 6.0) Ad(

    Zf  ≤=

    s

    s

    bd

    nAm =

    m2mmk 2 ++−=

    ss

    ss

     jd A

    Mf  =

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    SECTION B-B CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 1396= mm

    Effective Shear Width bv 16500= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 4210= < 12710= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 31.9= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 203= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00061= ≤ 0.001 OK

    1000*εx 0.611=

    Ratio vu/f'c 0.006=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   2.5= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   31.9= 0

    Vc 28244= kN

    Vs 25477= kN

    Vn1 53721= kN

    Vn2 201576= kN

    Norminal Shear Resistance Vn 53721= kN

    Factored Shear Resistance Vr  48349= kN > 4210 kN O.K.

    Minimum transverse reinforcement  9645= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.20= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 600= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    Factored Shear  Vu 20499= KN

    Factored Moment  Mu 29307= KNm

    Factored Axial Force Nu 0= KN

    Service Moment  Ms 20653= KNm

    h 2500= mm

    b 16500= mm

    d1 120= mm

    d2 0= mm

    d3 2260= mm

    d's 120= mm

    de = ds 2380= mm

    Tension Reinforcement Compresion Reinf. Transverse Reinf.

    Number (bars) ns 132= n's 132= nv 68=

    Diameter (mm) Ds 25= D's 22= Dv 16=

     Area (mm2)  As 67320=  A's 51084=  Av 13532=

    Spacing (mm) s 125= d 125= sv 500=

    Resistance factor for Flexure: ϕ 0.90=

    Resistance factor for Shear: ϕv 0.90=

    Flexural Resistance

    Distance from extreme compression fiber to the neutral axis: c 72= mm

    Depth of the equivalent stress block: a 58= mm < 2d's

    Factored Flexural Resistance Mr  59831= kN•m > 29307 kN•m O.K.

    Maximum Reinf. c / d e 0.030 = < 0.42   O.K.

    1.2 times the cracking moment 1.2Mcr  64060= kNm

    1.33 times the factored moment 1.33Mu 38979= kNm

    Minimum Reinf. Min (1.2*Mcr  or 1.33*Mu) 38979= kN•m < 59831 kN•m O.K.

    Control of cracking by distribution of reinforcement

    Components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state

    does not exceed f sa, determined as:

    n = Es/Ec 6.290= dc 129= mm

     A 32125= mm

    0.011= Z 30000= N/mm

    f sa 187= Mpa

    0.136= 0.6f y 252= Mpa

     j = (1-k/3) 0.955=

    135= Mpa Check: OK

    REINFORCEMENT

    SECTION DIMENSIONS

    SECTION C-C CHECK

    0.85•f'c•a•b

     As•f y

    a A's•f y

    b

    h ds

    d2

    d's

    d3

    d1

    nv,Dv

    ns, Ds

    n's, D's

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    y3/1c

    sa f 6.0) Ad(

    Zf  ≤=

    s

    s

    bd

    nAm =

    m2mmk 2 ++−=

    ss

    ss

     jd A

    Mf  =

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    SECTION C-C CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 2351= mm

    Effective Shear Width bv 16500= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 20499= < 21404= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 31.9= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 587= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00061= ≤ 0.001 OK

    1000*εx 0.611=

    Ratio vu/f'c 0.017=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   2.5= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   31.9= 0

    Vc 47565= kN

    Vs 42905= kN

    Vn1 90470= kN

    Vn2 339455= kN

    Norminal Shear Resistance Vn 90470= kN

    Factored Shear Resistance Vr  81423= kN > 20499 kN O.K.

    Minimum transverse reinforcement  9645= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.59= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 600= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    Factored Shear  Vu 1697= KN

    Factored Moment  Mu 9526= KNm

    Factored Axial Force Nu 0= KN

    Service Moment  Ms 6219= KNm

    h 2500= mm

    b 16500= mm

    d1 166= mm

    d2 0= mm

    d3 2234= mm

    d's 100= mm

    de = ds 2334= mm

    Tension Reinforcement Compresion Reinf. Transverse Reinf.

    Number (bars) ns 132= n's 132= nv 68=

    Diameter (mm) Ds 22= D's 25= Dv 16=

     Area (mm2)  As 51084=  A's 67320=  Av 13532=

    Spacing (mm) s 125= d 125= sv 500=

    Resistance factor for Flexure: ϕ 0.90=

    Resistance factor for Shear: ϕv 0.90=

    Flexural Resistance

    Distance from extreme compression fiber to the neutral axis: c 55= mm

    Depth of the equivalent stress block: a 44= mm < 2d's

    Factored Flexural Resistance Mr  44647= kN•m > 9526 kN•m O.K.

    Maximum Reinf. c / d e 0.023= < 0.42   O.K.

    1.2 times the cracking moment 1.2Mcr  64060= kNm

    1.33 times the factored moment 1.33Mu 12670= kNm

    Minimum Reinf. Min (1.2*Mcr  or 1.33*Mu) 12670= kN•m < 44647 kN•m O.K.

    Control of cracking by distribution of reinforcement

    Components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state

    does not exceed f sa, determined as:

    n = Es/Ec 6.290= dc 77= mm

     A 19250= mm

    0.008= Z 30000= N/mm

    f sa 263= Mpa

    0.121= 0.6f y 252= Mpa

     j = (1-k/3) 0.960=

    54= Mpa Check: not.OK

    REINFORCEMENT

    SECTION DIMENSIONS

    SECTION D-D CHECK

    0.85•f'c•a•b

     As•f y

    a A's•f y

    b

    h ds

    d2

    d's

    d3

    d1

    nv,Dv

    ns, Ds

    n's, D's

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    y3/1c

    sa f 6.0) Ad(

    Zf  ≤=

    s

    s

    bd

    nAm =

    m2mmk 2 ++−=

    ss

    ss

     jd A

    Mf  =

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    SECTION D-D CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 2312= mm

    Effective Shear Width bv 16500= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 1697= < 27279= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 24.3= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 49= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00013= ≤ 0.001 OK

    1000*εx 0.127=

    Ratio vu/f'c 0.001=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   3.2= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   24.3= 0

    Vc 60620= kN

    Vs 58124= kN

    Vn1 118744= kN

    Vn2 333816= kN

    Norminal Shear Resistance Vn 118744= kN

    Factored Shear Resistance Vr  106870= kN > 1697 kN O.K.

    Minimum transverse reinforcement  9645= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.05= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 600= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    Factored Shear  Vu 120= KN

    Factored Moment  Mu 180= KNm

    Factored Axial Force Nu 0= KN

    Service Moment  Ms 115= KNm

    h 800= mm

    b 1000= mm

    d1 71= mm

    d2 0= mm

    d3 659= mm

    d's 70= mm

    de = ds 729= mm

    Tension Reinforcement Compresion Reinf. Transverse Reinf.

    Number (bars) ns 8= n's 4= nv 4=

    Diameter (mm) Ds 22= D's 19= Dv 16=

     Area (mm2)  As 3096=  A's 1136=  Av 796=

    Spacing (mm) s 125= d 250= sv 500=

    Resistance factor for Flexure: ϕ 0.90=

    Resistance factor for Shear: ϕv 0.90=

    Flexural Resistance

    Distance from extreme compression fiber to the neutral axis: c 55= mm

    Depth of the equivalent stress block: a 44= mm < 2d's

    Factored Flexural Resistance Mr  828= kN•m > 180 kN•m O.K.

    Maximum Reinf. c / d e 0.075 = < 0.42   O.K.

    1.2 times the cracking moment 1.2Mcr  398= kNm

    1.33 times the factored moment 1.33Mu 239= kNm

    Minimum Reinf. Min (1.2*Mcr  or 1.33*Mu) 239= kN•m < 828 kN•m O.K.

    Control of cracking by distribution of reinforcement

    Components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state

    does not exceed f sa, determined as:

    n = Es/Ec 6.290= dc 87= mm

     A 21750= mm

    0.027= Z 30000= N/mm

    f sa 243= Mpa

    0.206= 0.6f y 252= Mpa

     j = (1-k/3) 0.931=

    55= Mpa Check: OK

    REINFORCEMENT

    SECTION DIMENSIONS

    SECTION E-E CHECK

    0.85•f'c•a•b

     As•f y

    a A's•f y

    b

    h ds

    d2

    d's

    d3

    d1

    nv,Dv

    ns, Ds

    n's, D's

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    y3/1c

    sa f 6.0) Ad(

    Zf  ≤=

    s

    s

    bd

    nAm =

    m2mmk 2 ++−=

    ss

    ss

     jd A

    Mf  =

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    SECTION E-E CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 707= mm

    Effective Shear Width bv 1000= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 120= < 470= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 26.1= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 189= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00022= ≤ 0.001 OK

    1000*εx 0.223=

    Ratio vu/f'c 0.005=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   3.0= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   26.1= 0

    Vc 1044= kN

    Vs 965= kN

    Vn1 2009= kN

    Vn2 6188= kN

    Norminal Shear Resistance Vn 2009= kN

    Factored Shear Resistance Vr  1808= kN > 120 kN O.K.

    Minimum transverse reinforcement  585= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.19= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 566= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    SECTION F-F CHECK

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Shear Resistance

    Effective Shear Depth dv 708= mm

    Effective Shear Width bv 1000= mm

    Regions requiring transverse reinforcement:   Vu > 0.5 ϕVcVu 260= < 442= KN No need 

    The norminal shear resistance, Vn, shall be determined as the lesser of: (Vn1 = Vc + Vs, Vn2 = 0.25f'cbvdv)

    for which:

    Determination of and θ

     Angle of inclination of transverse Reinf. to longitudinal axis α 90= 0

     Angle of inclination of diagonal compressive stresses θ 27.9= 0 (Supposition)

    Shear stress on the concrete vu = Vu/(ϕbvdv) 408= KN/m2

    Strain in the reinforcement on the flexural tension side of the member 

    0.00033= ≤ 0.001 OK

    1000*εx 0.333=

    Ratio vu/f'c 0.012=

    Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   2.8= [ β = F(v/f'c, 1000*εx)]

    Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004)   27.9= 0

    Vc 982= kN

    Vs 895= kN

    Vn1 1877= kN

    Vn2 6196= kN

    Norminal Shear Resistance Vn 1877= kN

    Factored Shear Resistance Vr  1689= kN > 260 kN O.K.

    Minimum transverse reinforcement  585= mm2

    O.K.

    Maximum spacing of transverse reinforcement 

    If vu < 0.125f'c, then: s ≤ 0.8dv ≤ 600mm

    If vu ≥ 0.125f'c, then: s ≤ 0.4dv ≤ 300mm

    vu 0.41= Mpa < 0.125f'c 4.38= Mpa

    sv < smax 567= mm O.K.

    vvcc db'f 083.0V β=

    s

    sin)gcotg(cotdf  AV

    vyv

    s

    α+θ=

    ss

    uuv

    u

    x AE2

    gcotV5.0N5.0d

    M

    θ++=ε

    y

    vcminv

    sb'f 083.0 A =

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    CREEP AND SHRINKAGE

    General input

    Length of girder    L 38.3= m

    Number of Strands   ntr  42= strandsJacking Force   P j   195 KN

    Elastic Shortening   L = 14.1 mm

     Average ambient relative humidity   H 85= %

    Specified compressive strength of concrete of girder at 28 days f'c 50= Mpa

    Compressive strength of concrete of girder at time of initial prestress f'ci 42.5= Mpa

    Modulus of elasticity Ec 38007 Mpa

    Eci 35041 Mpa

    Coefficient of thermal expansion for normal concrete   ε 1.08E-5=Creep coefficient

    The creep coefficient may be estimated as:

    Where:

    kf  = 62/(42+f'c) : Factor for the effect of concrete strength

    t : Maturity of concrete (day)

    ti : Age of concrete when load is initially applied (day)   ti 3= days

    kc : Factor for the effect of the volume-to-surface ratio of the component specified

    Figure 5.4.2.3.2-1 of AASHTO LRFD 2004 or taken as:

    10 150 10950

     Area of Section, A (m2) 0.635 0.635 0.993

    Perimeter of Section, P (m) 10.248 10.248 12.434

    Volume to Surface ratio, A/P (mm) 62 62 80

    kc 0.665 0.806 0.818

    kf  0.674 0.674 0.674

    Creep coefficient, ψ(t,ti) 0.292 0.969 1.424Strain due to shrinkage

    The strain due to shrinkage, esh, at time, t, mat be taken as:

    Where:

    t : Drying time (day)

    kh : Humidity Factor 

      kh = (140-H ) / 70 for H < 80%

      kh = 3(100-H )/ 70 for H ≥ 80% kh 0.643=ks : Size factor specified Figure 5.4.2.3.3-2 of AASHTO LRFD 2004 or taken as:

    10 150 10950

     Area of Section, A (m2) 0.635 0.635 0.993

    Perimeter of Section, P (m) 10.248 10.248 12.434

    Volume to Surface ratio, A/P (mm) 62 62 80kh 0.643 0.643 0.643

    ks 0.684 0.829 0.830

    Strain due to shrinkage, εsh -0.00005 -0.00022 -0.00027

    Item

    Item

    Maturity of concrete (day)

    Maturity of concrete (day)

    Ha no i - Hai Phong Exp resswa y Projec t 

     

    ( )  ( )

    ( )6.0

    i

    6.0

    i118.0

    if ci

    tt0.10

    ttt

    120

    H58.1kk5.3t,t

    −+

    −⎟ ⎠

     ⎞⎜⎝ 

    ⎛ −=ψ   −

    3

    hssh 1051.0t0.35

    tkk   −×⎟

     ⎠

     ⎞⎜⎝ 

    ⎛ +

    −=ε

    ⎥⎦

    ⎤⎢⎣

    ⎡   −

    ⎥⎥⎥⎥

    ⎢⎢⎢⎢

    +

    +=923

    )P/ A(70.31064

    t45

    tte26

    t

    k)P/ A(0142.0

    s

    ⎥⎦

    ⎤⎢⎣

    ⎡   +

    ⎥⎥⎥⎥

    ⎢⎢⎢⎢

    +

    +=−

    587.2

    e77.180.1

    t45

    tte26

    t

    k)P/ A(0213.0)P/ A(0142.0

    c

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    Ha no i - Hai Phong Exp resswa y Projec t 

     

    Movement due to Creep

    10 150 10950

    Shortening (mm) 4.1 13.7 20.1

    Movement due to Shrinkage

    10 150 10000

    Shortening (mm) 1.9 8.4 10.4

    Movement due to Creep and Shrinkage   L 8.4= mm

    Convert to Uniform Temperature Change   T -20= 0C

    Item

    Item

    Maturity of concrete (day)

    Maturity of concrete (day)