1. Trang chủ
  2. » Cao đẳng - Đại học

Bảng tính mố chữ U cao tốc Hải Phòng

30 945 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 30
Dung lượng 351,18 KB

Các công cụ chuyển đổi và chỉnh sửa cho tài liệu này

Nội dung

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 1

Package: 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 2

Dynamic 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 3

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 = 3.060m

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 4

3.1 Other input Data

• Skew angle (square bridge α = 90o

ABUTMENT DIMENSIONS (IN METRES)

Trang 5

3.2 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 6

Total (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 7

Horizoltal 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 9

3.3 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 10

Internal Force at Section B-b

Trang 11

Horizoltal 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 12

Summary 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 13

Internal 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 14

SERVICE – Element Moment X:

SERVICE – Element Moment Y:

Trang 15

STRENGTH – Element Moment X:

STRENGTH – Element Moment Y:

Trang 16

STRENGTH – Element Shear X:

STRENGTH – Element Shear Y:

Trang 17

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 = 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 18

SECTION 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+

+

y

v c min

v

fsb'f083.0

Trang 19

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 = 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 20

SECTION 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+

+

y

v c min

v

fsb'f083.0

Trang 21

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 = 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 22

SECTION 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+

+

y

v c min

v

fsb'f083.0

Trang 23

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 = 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 24

SECTION 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+

+

y

v c min

v

fsb'f083.0

Trang 25

Factored 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

Ngày đăng: 01/01/2017, 18:23

TỪ KHÓA LIÊN QUAN

🧩 Sản phẩm bạn có thể quan tâm

w