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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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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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Ha no i - Hai Phong Exp resswa y Projec t
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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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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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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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
f
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
f
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
f
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
f
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
f
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
f
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
Abutment-A1A2(OK).xls - CR&SH Date: 9/26/2011 Page: 1 of 2
8/15/2019 Package EX-10 Bridge Name Lach Tray Brid Abutment Calsulation
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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)