Horizontally and from Underside of Footing Vertically Summary Load Combinatons at Section a-a surcharge Vertical Reaction EQ, BR Live Load surcharge P H2 P V2 P1, M1 P2, M2 P3, M3 P4, M4
Trang 1Package: EX-10
Bridge name: Lach Tray Bridge
Abutment Name: A1, A2
1 MATERIALS
1.1 Material Unit Weights
• Unit Weight of Concrete = 2500kg/m3 → γc = 24.5 kN/m3
• Unit Weight of Soil = 1800kg/m3 → γs = 17.7 kN/m3
• Unit Weight of Saturated Soil = 1100kg/m4 → γsw = 11.0 kN/m4
1.2 Concrete
Compressive Strength of concrete at 28 days f'c = 35MPa
Reinforcing bar Area
2 LOADS FROM SUPERSTRUCTURE
2.1 Dead load Number of Girders ng = 8girders
Reaction due to dead load of super-T girder Span
Members
Live load Design Truck
Trang 2Dynamic Load Allowance
Deck Joint - All Limit States 75%
All Other ComponentsFatigue and Fracture Limit State 15%
Total span length Lst = 38.300 m
Calculation span leng Ls = 37.600 m
Notes: Reaction due to live load HL- 93 = max (Rtruck, Rtandem) + Rlane
Combine live load HL-93 and pedestrian for disadvantage
2.4 Temprature Load
Uniform temperature change ΔT = +/-20.0 oC
Coefficient of thermal expansion = 1.08E-5 /oC
Shear Modulus of Elastomer G = 1000KPa
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 = = - 20.2oC
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)
Trang 32.5.1 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 = 3.060m
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)
Vertical wind load
Vertical wind load Pv = 0.00045V2.Av= 643.2 KN 2.6 Earthquke effects (EQ)
Response Modification Factor
Elastic seismic response coefficient Csm = 0.292
Longitudinal Force due to Earthquake EQ = 3198.1 KN
Trang 43.1 Other input Data
• Skew angle (square bridge α = 90o
ABUTMENT DIMENSIONS (IN METRES)
Trang 53.2 Internal Forces at Bottom Footing
Loading Data:
H t = 10.000 m
K A = 0.297
kh = 0.175
kv = 0.070
θ = 0.19rad
δs = 0.35rad
K AE = 0.442
ABUTMENT LOADS
M Long
Moment
M Trans
Selfweight
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 - - - -
Description
Live Load Surcharge
+M +H
+V
Sign Convention
E
J
H t
(KA, KEA)γ s Ht
Vertical Reaction
EQ, BR
F H
B
D G
Live Load Surcharge
X
A1
H t /3
0.5H
δ
P H1
P V1
P 2
δ
P H2
P V2
I
Trang 6Total (EV) 7270.7 -12902.5 -
Buoyancy effect on Soil
Section F - - - -
Section G - - - -
Section H - - - -
Section K - - - -
Total (WA) - - -
PV1 = - 16.500 1481.4 7.500 -3.750 - -5555.1 - PV2 = - 16.500 180.7 7.500 -3.750 - -677.7 - PV1-EQ = - 16.500 2204.7 7.500 -3.750 - -8267.6 - ESv heq = 610 mm - 14.900 561.7 5.750 -2.000 - -1123.5 - Load from SuperStructure DC - - 4923.6 3.100 0.650 - 3200.3 - DW - - 552.7 3.100 0.650 - 359.3 - Live Load - - 1430.9 3.100 0.650 - 930.1 - WL - - 643.2 3.100 0.650 - 418.1 - Earth Pressure (EH) - 16.500 4070.0 - 3.333 - 13566.7 - Horizontal pressure due to Surcharge - 14.900 448.4 - 5.000 - 2242.0 - Braking force - - 105.6 - 7.850 - 829.2 - Longitudinal wind load on Structure - - 55.7 - 7.850 - 437.0 - Longitudinal wind load on Vehicle - - 14.4 - 7.850 - 112.7 - Temprature Load - - 65.6 - 7.850 - 514.9 - Earth Pressure due to EQ PH1-EQ - 16.500 6057.4 - 3.333 - 20191.2 - EQ from Superstructure - - 1599.1 - 7.850 - 12552.7 - EQ from Abutment Section A 6.25 16.500 738.5 - 1.250 - 923.1 - Section A1 3.75 16.500 443.1 - 1.250 - 553.9 - Section B 7.95 16.500 939.4 - 5.150 - 4837.8 - Section C 8.75 16.500 1033.9 - 1.250 - 1292.4 - Section D - - - -
Section E 1.10 16.500 130.0 - 8.900 - 1156.8 - Section F 15.93 1.600 182.5 - 4.775 - 871.3 - Section G - 1.600 - - 7.050 - - -
Trang 7Horizoltal Wind Load on Structure - - 222.7 - - 7.850 - 1748.0
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
SUMMARY LOADING AT FOOTING CENTER
Transverce Symbol
Longitudinal
Description
Trang 8+V
Trang 93.3 Section analysis
Internal Force at Section A-A
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
surcharge
Vertical Reaction
EQ, BR
Live Load surcharge
P H2
P V2
P1, M1 P2, M2 P3, M3 P4, M4
Trang 10Internal Force at Section B-b
Trang 11Horizoltal Wind Load on Structure - - 222.7 - - 5.350 - 1191.3
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
Trang 12Summary Load Combinatons at Section B-b
Internal Force at Section C-C
Note: • Self weight is included and
• Soil weight above pile cap is ignored
Extreme
Load Combinations
Description
Trang 13Internal Force at Section D-D
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
Wing Wall Calculation
Wing Wall is modeled and Calculated by ACES5.5 program:
Trang 14SERVICE – Element Moment X:
SERVICE – Element Moment Y:
Trang 15STRENGTH – Element Moment X:
STRENGTH – Element Moment Y:
Trang 16STRENGTH – Element Shear X:
STRENGTH – Element Shear Y:
Trang 17Resistance 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 = 28mm
Depth of the equivalent stress block: a = 22mm < 2d's
1.2 times the cracking moment 1.2Mcr = 2562kNm
1.33 times the factored moment 1.33Mu = 647kNm
Minimum Reinf. Min (1.2*M cr or 1.33*M u ) = 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 fsa, determined as:
y 3 / 1 c
)Ad(Z
Trang 18SECTION A-A CHECKShear Resistance
Regions requiring transverse reinforcement: Vu > 0.5 ϕVc
Vu = 487 < = 4937KN 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 β and θ:
Angle of inclination of transverse Reinf to longitudinal axis α = 900
Angle of inclination of diagonal compressive stresses θ = 24.30 (Supposition)
Shear stress on the concrete vu = Vu/(ϕbvdv)= 78KN/m2
Strain in the reinforcement on the flexural tension side of the member
0.00013
1000*εx = 0.128
Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) β = 3.2 [ β = F(v/f'c, 1000*εx)]
Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) θ = 24.30
Vc = 10972kN
Vs = 10519kN
Vn1 = 21492kN
Vn2 = 60459kN
Maximum spacing of transverse reinforcement
s
sin)gcotg(cotdfA
=
s s
u u v
u
x
AE2
gcotV5.0N5.0d
M
θ+
+
=ε
y
v c min
v
fsb'f083.0
Trang 19Resistance 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 = 72mm
Depth of the equivalent stress block: a = 58mm < 2d's
1.2 times the cracking moment 1.2Mcr = 23062kNm
1.33 times the factored moment 1.33Mu = 19507kNm
Minimum Reinf. Min (1.2*M cr or 1.33*M u ) = 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 fsa, determined as:
y 3 / 1 c
)Ad(Z
Trang 20SECTION B-B CHECKShear Resistance
Regions requiring transverse reinforcement: Vu > 0.5 ϕVc
Vu = 4210 < = 12710KN 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 β and θ:
Angle of inclination of transverse Reinf to longitudinal axis α = 900
Angle of inclination of diagonal compressive stresses θ = 31.90 (Supposition)
Shear stress on the concrete vu = Vu/(ϕbvdv)= 203KN/m2
Strain in the reinforcement on the flexural tension side of the member
0.00061
1000*εx = 0.611
Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) β = 2.5 [ β = F(v/f'c, 1000*εx)]
Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) θ = 31.90
Vc = 28244kN
Vs = 25477kN
Vn1 = 53721kN
Vn2 = 201576kN
Maximum spacing of transverse reinforcement
s
sin)gcotg(cotdfA
=
s s
u u v
u
x
AE2
gcotV5.0N5.0d
M
θ+
+
=ε
y
v c min
v
fsb'f083.0
Trang 21Resistance 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 = 72mm
Depth of the equivalent stress block: a = 58mm < 2d's
1.2 times the cracking moment 1.2Mcr = 64060kNm
1.33 times the factored moment 1.33Mu = 38979kNm
Minimum Reinf. Min (1.2*M cr or 1.33*M u ) = 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 fsa, determined as:
y 3 / 1 c
)Ad(Z
Trang 22SECTION C-C CHECKShear Resistance
Regions requiring transverse reinforcement: Vu > 0.5 ϕVc
Vu = 20499 < = 21404KN 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 β and θ:
Angle of inclination of transverse Reinf to longitudinal axis α = 900
Angle of inclination of diagonal compressive stresses θ = 31.90 (Supposition)
Shear stress on the concrete vu = Vu/(ϕbvdv)= 587KN/m2
Strain in the reinforcement on the flexural tension side of the member
0.00061
1000*εx = 0.611
Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) β = 2.5 [ β = F(v/f'c, 1000*εx)]
Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) θ = 31.90
Vc = 47565kN
Vs = 42905kN
Vn1 = 90470kN
Vn2 = 339455kN
Maximum spacing of transverse reinforcement
s
sin)gcotg(cotdfA
=
s s
u u v
u
x
AE2
gcotV5.0N5.0d
M
θ+
+
=ε
y
v c min
v
fsb'f083.0
Trang 23Resistance 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 = 55mm
Depth of the equivalent stress block: a = 44mm < 2d's
1.2 times the cracking moment 1.2Mcr = 64060kNm
1.33 times the factored moment 1.33Mu = 12670kNm
Minimum Reinf. Min (1.2*M cr or 1.33*M u ) = 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 fsa, determined as:
y 3 / 1 c
)Ad(Z
Trang 24SECTION D-D CHECKShear Resistance
Regions requiring transverse reinforcement: Vu > 0.5 ϕVc
Vu = 1697 < = 27279KN 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 β and θ:
Angle of inclination of transverse Reinf to longitudinal axis α = 900
Angle of inclination of diagonal compressive stresses θ = 24.30 (Supposition)
Shear stress on the concrete vu = Vu/(ϕbvdv)= 49KN/m2
Strain in the reinforcement on the flexural tension side of the member
0.00013
1000*εx = 0.127
Factor β taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) β = 3.2 [ β = F(v/f'c, 1000*εx)]
Factor θ taken from Table 5.8.3.4.2-1 (AASHTO LRFD 2004) θ = 24.30
Vc = 60620kN
Vs = 58124kN
Vn1 = 118744kN
Vn2 = 333816kN
Maximum spacing of transverse reinforcement
s
sin)gcotg(cotdfA
=
s s
u u v
u
x
AE2
gcotV5.0N5.0d
M
θ+
+
=ε
y
v c min
v
fsb'f083.0
Trang 25Factored Shear V u = 120 KN
Factored Moment M u = 180 KNm
Factored Axial Force N u = 0 KN
Service Moment M s = 115 KNm
h = 800mm
b = 1000mm
d1 = 71mm
d2 = 0mm
d3 = 659mm
d's = 70mm
de = ds = 729mm
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 = 55mm
Depth of the equivalent stress block: a = 44mm < 2d's
1.2 times the cracking moment 1.2Mcr = 398kNm
1.33 times the factored moment 1.33Mu = 239kNm
Minimum Reinf. Min (1.2*M cr or 1.33*M u ) = 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 fsa, determined as:
n = Es/Ec = 6.290 dc = 87mm
A = 21750mm
REINFORCEMENT
SECTION DIMENSIONS
SECTION E-E CHECK
0.85•f'c•a•b
As•fy
a
A's•fy
b
h ds
d2
d's
d3
d1
nv,Dv
ns, Ds n's, D's
y 3 / 1 c
) A d ( Z