BẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONG

BẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONGBẢNG TÍNH MỐ HỘP MỐ CHUI MÔ KHUNG TIENG ANH CAU DUONG

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Input】

SHEET NO : 1 / 2

1 INPUT DATA:

1.1 Dimensional of abutment

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Input】

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Load】

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2 LOADS

2.1 Superstructure load:

2.1.1 Dead load:

2.1.1.1 Dead load of strucural components and nonstructural attachments - DC1

2.1.1.2 Dead load of wearing surfaces and utilities - DW1

2 - Pedestrian lane and parapet

2.1.2 Live load (3.6.1.2 - 22TCN272-05)

2.1.2 1 Live load on structure:

In case of 1: Live load on superstructure

- Distance from center bearing to center of pile cap eg = 1.70 m

(KN)

3 lane x m

1 lane x m

In case of 2: Live load on superstructure and abutment

- Distance from center lane on abutment to center of pile cap el = -1.2 m

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SHEET NO : 2 / 4

2.1.2 2 Live load inside abutment

- Live load inside abutment inculding:

2.1.2.3 Dynamic load allowance - IM (3.6.2 - 22TCN272-05)

The static effect of design truck or tandem shall be increased by the following percentages:

My (KN.m)

IMLL

3 lanes on span and abutment

IM

Live load on superstructure

Pz (KN)

Mx (KN.m)

LL

Items

My (KN.m) Items

IM

1 lanes inside abutmment

PL

Pz (KN) Items

My (KN.m)

LL

1 8 0 0

3 0 0 0

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2.3 Dead load of abutment

2.3.1 Dead load of structure - DC2

A - A (bottom of pile cap)

No.

Items

Items Total

No.

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2.5 Vertical pressure of earth backfilling - EV

2.6 Wind load

2.6.1 Wind load on superstructure - WS (3.8.1.2; 22TCN-272-05)

Horizontal wind load:

Transverse wind load

PD= 0.0006 xV2 x At x Cd ³ 1.8 x At (kN) (3.8.1.2.1-1; 22TCN-272-05)

Where:

2.6.1.1 Transverse wind load

2.6.2 Longitudinal wind load on vehicle: (3.8.1.3; 22TCN-272-05)

Wind load on vehicles taken as

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01 Bo Ao_Cal Box Abutment_25.7m.xls【ES-LS】

H3: Calculated height, from top of pile slab upward

+ Active earth pressure: pa=ka.γ.H3

δ : friction angle between fill and wall (degree) 20

θ : Angle of back face of wall to the horizontal (degree) 90

+ Passive earth pressure:

2

) sin(

sin

) ( sin

2

δ θ θ

φ

θ

+ Γ

−

=

) ) sin(

).

sin(

) sin(

).

sin(

1 (

β θ δ

θ

β φ δ

φ

+ +

−

= Γ

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01 Bo Ao_Cal Box Abutment_25.7m.xls【ES-LS】

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3.2 Lateral pressure due to effect of earthquake - EAE :

Lateral pressure due to effect of earthquake taken as:

3.3 Live load surcharge - LS: (3.11.6; 22TCN-272-05)

- The increase in horizontal pressure due to live load surcharge taken as :

Where:

K : Coefficient of earth pressure

heq : Equivalent height of soil for vehicular load LS=Dp.H3.L

MLS =0.5.LS.H3

(m)

Dp(KN/m2)

LS (KN)

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Summary】

SHEET NO : 1 / 1

4 SUMMARY OF LOAD AT BOTTOM OF PILE CAP (Section A - A)

Pz Longitidunal Bridge Horizontal Bridge (KN) Hx (KN) My (KN.m) Hy (KN) Mx (KN.m)

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Comb-A】

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Load factorLoad

No

Total

Load factorLoad

No

Total

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Load factor

Load

Total

No

Load factorLoad

Load

No

Load factor

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Total

factorNo

Load

No

Total

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Cal pile foundation】

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SHEET NO : 2 / 3

6.3 The forces at bottom of pile cap

No Force X Force Y Force Z Moment X Moment Y Moment Z

6.4 The result of analysis

No Combine Direction Z

Direction X Direction Y

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6.5 Checking pile section

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01 Bo Ao_Cal Box Abutment_25.7m.xls【Analysis】

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7 ANALYSIS STRUCTURE OF ABUTMENT

7.1 Assumes

Some assume for analysis struture of abutment inculding:

- Design for 1m length of box abutment

- Live load inside box abutment is tandem with axel load 0.5x∑ P (3.6.1.2.2)

- The value live load for pedestrian inside box abutment is 3x10-3 MPa

7.2 Loads

7.2.1 Dead Load

Units Value

Dead load from span

Dead load from box abutment

7.2.2 Live Load

7.2.2.1 Live Load on top box abutment

The equivalent strip widths for live load on box abutment according : (4.6.2.3)

- For one lane:

E1lane = 250+0.42x(L1 x W1)^0.5 = 2.92 m (4.6.2.3-1)Where:

- NL: number of design lane for box abutment = 1.00

- For mutilane:

E1lane = 2100+0.12x(L1 x W1)^0.5 ≤ W/NL = 3.18 m (4.6.2.3-2)Where:

- NL: number of design lane for box abutment = 3.00

Live load on equivalent strip width

- For one lane:

- Force action for 1m of equivalent strip width by truck load + IM, P1 = 74.41 KN

- Force action for 1m of equivalent strip width by tendem + IM, P2 = 51.89 KN

- Force action for 1m of equivalent strip width by lane load, Wl1 = 3.82 KN/m

- For mutilane:

- Force action for 1m of equivalent strip width by truck load + IM, P'1 = 96.89 KN

- Force action for 1m of equivalent strip width by tendem + IM, P'2 = 73.51 KN

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SHEET NO : 2 / 4

- Force action for 1m of equivalent strip width by lane load, Wl2 = 2.49 KN/m

===> Choose the equivalent strip width for Mutilane

7.2.2.2 Live Load inside box abutment

- Number of design lane inside box abutment, Nins = 1

- Tải trọng do hoạt tải xe +IM trong lòng mố, P3 = 37.21 KN

7.2.2.3 Live Load from span

Live load from span put in the bearings position

Units Value Note

7.2.3 Earth Pressure

Earth pressure consider for a meter width of abutment

Units Value Note

6

5

9

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5

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01 Bo Ao_Checking Section M, N, Q_25.7m.xls【M, N(1-1)】

SHEET NO : 1 / 3

7.6.3.1.1 CHECKING FLEXURAL & AXIAL RESISTANCE

The factored flexural resistance Mr shall be taken as: M r = ϕM n

Where :

- Mn : Nominal resistance

- ϕ : Resistance factor as specified in Article 5.5.4.2

- As : Area of nonprestressed tension reinforcement

- A's : Area of compression reinforcement

- fy : Specified yield strength of tension reinforcement

- f'y : Specified yield strength of compression reinforcement

- ds : Distance from extreme compression fiber to the centiod of nonprestressed tension reinforcement

- d : Distance from extreme tension fiber to the centriod of nonprestressed tension reinforcement

- d's : Distance from extreme compression fiber to the centriod of compression reinforcement

- a : Depth of the equivalent stress block = c β1

- c : Distance from neutral axis to the extreme compression fiber = [Asfy-A'sf'y]/(0.85f'cβ1b)

Specified yield strength of tension reinforcement fy 400 MPa

Specified yield strength of compression reinforcement f'y 400 MPa

Specified compressive strength of concrete f'c 30 MPa

Modulus of elasticity of steel Es 200,000 MPa

Modulus of elasticity of concrete Ec 29,440 MPa

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Mcr = frIg/yt

- Modulus of rupture of concrete fr = 0.63f'c

- Moment of inertia of gross concrete section Ig : 10,416,666,667 mm4

- Distance from neutral axis to the extreme tension fiber yt: 250 mm

Checking: ϕM n = 257 > 1.2M cr = 173 Satisfactory

7 Checking for maximum reinforcement :

The maximum amount of nonprestressed reinforcement shall be such that :

c/de ≤ 0.42Where:

de = (As*fy*ds)/(As*fy)

- de : the corresponding effective depth from the extreme compression fiber to the centriod of the tensile force in the tensile reinforcement (mm)

Checking: c/d e = 0.1001 < 0.42 Satisfatory

8 Checking for minimum reinforcement :

The minimum amount of nonprestressed reinforcement shall be such that :

ρmin ≥ 0.03f'c/f'y

- Area of tension reinforcement As = 2,199 mm2

- Gross area of concrete Ac = 412,500 mm2

ρmin=As/Ac 0.0053

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dc: depth of concrete measured from extreme tension fiber to center of bar located closest hereto, for calculation purpose, the thickness

of clear cover used to compute dc shall not be taken to be greater than 50mm

A: Area of concrete surrounding tension reinforcement

Z: crack width parameter

+ 30 kN/mm for members in moderate exposure conditions

+ 23 kN/mm for members in severe exposure conditions

+ 17.5 kN/mm for buried structures

+ f'y : Specified yield strength of compression reinforcement 400 Mpa

Checking: f s = 114.0 < min(f sa ; 06f' y )= 240 Satisfactory

10 Checking axial compression:

Axial resistance of components shall be taken as: P r = ϕ P n

For members with spiral reinforcement:

For members with tie reinforcement: Pn = 0.8 [0.85f'c(Ag-Ast) + fyAst]

ϕ: axial compression factor 0.75

Ag: gross area of section 500,000.0 mm2

Ast: total area of longitudinal reinforcement 4,398.2 mm2

Pn: nominal axial resistance 11,517,709.6 N

Pr: calculation axial resistance 8,638,282.22 N

Checking: P r = 8,638,282 > min(f sa ; 06f' y )= 317,540 Satisfactory

Pn = 0.85 [0.85f'c(Ag-Ast) + fyAst]

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01 Bo Ao_Checking Section M, N, Q_25.7m.xls【Q(1-1)】

SHEET NO : 1 / 2

7.6.3.1.2 CHECKING SHEAR RESISTANCE

Materials

Distance from extreme tension fiber to the centroid tensile reinforcement ds 87.5 mmTension reinforcement

* α : angle of inclination of transverse reinforcement to longitudinal axis

* b : factor indicating ability of diagonally cracked concrete to transmit tension

* q : inclination angle of diagonal compressive stress + Vp : component of effective prestresed force in the direction of the applied shear

Determine b & q Factor indicating ability of diagonally cracked concrete to transmit tension β 3.1

V p Component of effective prestresed force in the direction of the applied shear Vp 0N

Checking region requiring transverse reinforcement: V và 0.5ϕ(V +V ) Unncessary

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Checking: V r = 449,249 > V u = 172,880 Satisfactory

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SHEET NO : 1 / 3

Where :

7.6.3.2 SECTION 2 - 2

Mn=Asfy(ds-a/2)-A'sf'y(d's-a/2)

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Mcr = frIg/yt

- Modulus of rupture of concrete fr = 0.63f'c

- Moment of inertia of gross concrete section Ig : 10,416,666,667 mm4

- Distance from neutral axis to the extreme tension fiber yt: 250 mm

Checking: ϕM n = 257 > 1.2M cr = 70 Satisfactory

7 Checking for maximum reinforcement :

The maximum amount of nonprestressed reinforcement shall be such that :

c/de ≤ 0.42Where:

de = (As*fy*ds)/(As*fy)

- de : the corresponding effective depth from the extreme compression fiber to the centriod of the tensile force in the tensile reinforcement (mm)

Checking: c/d e = 0.1001 < 0.42 Satisfatory

8 Checking for minimum reinforcement :

The minimum amount of nonprestressed reinforcement shall be such that :

ρmin ≥ 0.03f'c/f'y

- Area of tension reinforcement As = 2,199 mm2

- Gross area of concrete Ac = 412,500 mm2

ρmin=As/Ac 0.0053 0.03f'c/f'y = 0.0023

Checking: ρ min = 0.00533 > 0.03f' c /f' y = 0.00225 Satisfactory

9 Control cracking by distribution of reinforcement:

Checking load combination is Service load

Condition:

fs≤ fsa = Z/(dcA)1/3 ≤ 0.6 f'y

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dc: depth of concrete measured from extreme tension fiber to center of bar located closest hereto, for calculation purpose, the thickness

of clear cover used to compute dc shall not be taken to be greater than 50mm

A: Area of concrete surrounding tension reinforcement

Z: crack width parameter

+ 30 kN/mm for members in moderate exposure conditions

+ 23 kN/mm for members in severe exposure conditions

+ 17.5 kN/mm for buried structures

+ f'y : Specified yield strength of compression reinforcement 400 Mpa

Checking: f s = 36.8 < min(f sa ; 06f' y )= 240 Satisfactory

10 Checking axial compression:

Axial resistance of components shall be taken as: P r = ϕ P n

For members with spiral reinforcement:

For members with tie reinforcement: Pn = 0.8 [0.85f'c(Ag-Ast) + fyAst]

ϕ: axial compression factor 0.75

Ag: gross area of section 500,000.0 mm2

Ast: total area of longitudinal reinforcement 4,398.2 mm2

Pn: nominal axial resistance 11,517,709.6 N

Pr: calculation axial resistance 8,638,282.22 N

Checking: P r = 8,638,282 > min(f sa ; 06f' y )= 281,610 Satisfactory

Pn = 0.85 [0.85f'c(Ag-Ast) + fyAst]

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01 Bo Ao_Checking Section M, N, Q_25.7m.xls【Q(2-2)】

SHEET NO : 1 / 2

7.6.3.2.2 CHECKING SHEAR RESISTANCE

Materials

Distance from extreme tension fiber to the centroid tensile reinforcement ds 87.5 mmTension reinforcement

* α : angle of inclination of transverse reinforcement to longitudinal axis

* b : factor indicating ability of diagonally cracked concrete to transmit tension

* q : inclination angle of diagonal compressive stress + Vp : component of effective prestresed force in the direction of the applied shear

Determine b & q Factor indicating ability of diagonally cracked concrete to transmit tension β 4.7

V p Component of effective prestresed force in the direction of the applied shear Vp 0N

Checking region requiring transverse reinforcement: V u và 0.5ϕ(V c +V p ) Unncessary

Minimum tranverse reinforcement within distance s s: Av 227.305 mm2

Data of transverse reinforcement

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Checking: V r = 693,897 > V u = 10,490 Satisfactory

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SHEET NO : 1 / 3

Where :

Modulus of elasticity of concrete Ec 29,440 MPa

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