Kiểm toán cống hộp 4x4m bằng phần mềm midas civil
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Project :
Item
Box Culvert 4x4M
CALCULATION SHEET OF BOX CULVERT ACCORDING TO 22TCN272-05
B CALCULATION
I. LOADS
1 Input
Kind of box Culvert
Explanation
=
Value
Single
Pavement layer thickness dađ(m)
Thickness of filling layer above the top of the box culvert H (m)
=
=
0.80
0.83
Length of box culvert segment Ltt (m)
Inside with of box culvert b (m)
Inside height of box culvert h (m)
=
=
=
1.00
4.00
4.00
Thickness of top slab dbn(m)
=
0.35
Thickness of wall dw (m)
=
0.35
Thickness of bottom slab dbđ (m)
=
0.35
Heigth of water calculation h1 (m)
=
4.00
Density of pavement structure γađ (T/m3)
=
2.40
Density of backfill γs (T/m3)
=
1.80
Density of water γw (T/m3)
=
1.00
Density of concrete γc (T/m3)
=
2.50
Angle of internal friction of soil ϕf (degree)
Gravitational acceleretion g (m/s2)
=
=
=
=
30
9.81
2.00
1.00
Number of caculated lanes (count for 1/2 box)
Multiple presence factor m
Width of box culvert Bc (m)
Bc = 2 x b + 3 x dw
=
4.70
Height of Box culvert Hc (m)
Hc = h + dt + db
=
4.70
Caculated width of box culvert btt (m)
btt = 2 x b + 2 x dw
=
4.35
Calcualted heigth of box culvert htt (m)
Self weigth of box culvert Pb (T)
htt = h + dn / 2 + dñ / 2
Pb = γc x (h x 2 x dw + Bc x (dn + dñ))
=
=
4.35
15.23
k0 = 1 - sinϕf
=
0.50
At-rest pressure coefficent k0
2 Dead loads
Wañ = γañ x dañ
=
1.92
We = γs x Ltt x H
=
1.49
Max horizontal pressure PA (T/m2)
PA = K0 x [γañ x dañ + γs x (H + dbn/2 + htt)]
=
5.78
Min horizontal pressure pA (T/m2)
PA = K0 x [γañ x dañ + γs x (H + dbn/2) ]
=
1.86
Self weigth of top slab Wbn (T/m2)
Wbn = γc x dbn
=
0.88
Self weigth of bottom slab Wbđ (T/m2)
Wbñ = γc x dbñ
=
0.88
Ww = γw x h1 x Ltt
=
4.00
3.6.1.2.5
=
0.51
y1=2.28x10-3γ (1+IM/100).P (3.6.1.2.5-1)
=
0.29
IM(%)=33x(1-4.1x10 (H+dad)) (3.6.2.2-1)
P (N) (3.6.1.2.5-1)
X= 1.8+1.2+1.8+x1+2.(H+dad)
Y=y1+2.(H+dad)
p=P/(X.Y)
=
=
=
=
=
33%
220000
8.57
3.55
0.74
m.WL/3.5
=
0.27
Z=H+dad+dbn/2+htt
=
6.155
3.11.6.2
=
0.75
Pavement load Wađ (T/m2)
Backfill load We (T/m2)
3 Water loads
Prerssure of water Ww (T/m2)
4 Live loads ( Design vehicular and Design Lane )
Distributing tire pressure on the top slab of box culvert
Width of horizontal contact x1 (m)
Length of longitudinal contact y1 (m)
In which: IM dynamic load (%)
Axle vehicular load (N)
Contact width on top slab of box culvert X (m)
Contact length on top slab ob box culvert y (m)
Total load on a calculated box culver segment (T/m2)
Design lane load WL (T/m2)
5 Additional backfill load of vehicular
Depth of bottom slab from road Z (m)
Equivalent heigth of soil for the design truck heq (m)
-4
Load of additional live load: ∆p (T/m2)
6 Defining coefficient of subgrade reaction
Scale factor of foundation K (T/m3)
Distance of two point pring support Lg (m)
Coefficent of subgrade reaction Kz (T/m2)
∆p=ko.γ.heq
=
0.68
22 TCN 18-79
=
=
=
400
0.54
218
Kz=KxLg
Project :
Item
Trung Luong - My Thuan Expressway
Box Culvert 4x4M
CALCULATION SHEET OF BOX CULVERT ACCORDING TO 22TCN272-05
II. Load combinations
No Kind of loads
Units
Service I
Strength Ia
Strength Ib
Factor
value
Factor
Value
Factor
Value
Dead load
1
Pavement load Wañ
T/m2
1.00
1.92
1.50
2.88
1.50
2.88
2
Backfill load We
T/m2
1.00
1.49
1.50
2.24
1.50
2.24
3
Max horizontal pressure PA
T/m2
1.00
5.78
1.50
8.67
0.90
5.20
4
Min horizontal pressure pA
T/m2
1.00
1.86
1.50
2.80
0.90
1.68
T/m2
1.00
4.00
1.00
4.00
1.00
4.00
T/m2
1.00
0.74
1.75
1.29
1.75
1.29
T/m2
1.00
0.27
1.75
0.47
1.75
0.47
T/m2
1.00
0.68
1.75
1.18
1.75
1.18
Water load
5
Water pressure Ww
Live load
6
Tire pressure distribution on the top slab of box culvert
7 Design lane load WL
10 Load of additional live load ∆p
KẾT QUẢ PHÂN TÍCH (RESULT OF ANALYSIS)
1. SƠ ĐỒ TÍNH TOÁN (DESIGN SHEME)
2. MOMENT DIAGRAMS
2.1 Mxx
a. Strength Ia
(Mặt cắt duyệt nội lực – Checking position of moment)
(Biểu đồ moment Mxx– Moment diagram Mxx)
b. Strength Ib
(Biểu đồ moment Mxx– Moment diagram Mxx)
c. Service
2.1 Myy
a. Strength Ia
(Mặt cắt duyệt nội lực – Checking position of moment)
(Biểu đồ moment Myy– Moment diagram Myy)
b. Strength Ib
(Biểu đồ moment Myy– Moment diagram Myy)
c. Service
(Biểu đồ moment Myy– Moment diagram Myy)
3. SHEAR DIAGRAMS
3.1 Qxx (Strength Ib)
(Mặt cắt duyệt nội lực – Checking position of shear)
(Biểu đồ lực cắt Qxx– Shear diagram Qxx)
3.2 Qyy (Strength Ia)
(Mặt cắt duyệt nội lực – Checking position of shear)
(Biểu đồ lực cắt Qyy– Shear diagram Qyy)
4. AXIAL FORCE DIAGRAMS
4.1 Fxx (strength Ib)
(Mặt cắt duyệt nội lực – Checking position of axial force)
(Biểu đồ lực dọc Fxx– Axial force diagram Fxx)
4.2 Fyy (strength Ia)
(Mặt cắt duyệt nội lực – Checking position of axial force)
(Biểu đồ lực dọc Fyy– Axial force diagram Fyy)
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to X axis - cross section of box culvert )
CHECKING FLEXURAL & AXIAL RESISTANCE
The factored flexural resistance Mr shall be taken as: Mr=ϕMn
Mn=Asfy(ds-a/2)-A'sf'y(d's-a/2)
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 β 1
− β1 : Stress block factor specified in Article 5.7.2.2
0.84
- c : Distance from neutral axis to the extreme compression fiber = [Asfy-A'sf'y]/(0.85f'cβ 1b)
1- Checking load
Maximum axial force
N
Maximum moment
M
176.6 kN
93.2 kN.m
2- Checking cross section
h
350 mm
b
1,000 mm
d's
91.5 mm
d
66.5 mm
ds
284 mm
3- Materials
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
4 - Reinforcement
Tension reinforcement As
D
As-1bar
25.0 mm
490.87 mm2
Nos
8
As
3,927 mm2
Compression reinforcement A's
D
25.0 mm
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to X axis - cross section of box culvert )
A's-1bar
490.87 mm2
Nos
8
A's
3927 mm2
5 - Checking flexural resistance
c=
73.70923 mm
a=
61.60 mm
Mn
302 kN.m
- ϕ : Flexural resistance and compression factor of concrete
0.9
Mr=ϕMn
271 kN.m
Checking:
Mr =
271
>
ϕM =
93
Satisfactory
6. For cracking moment
Apllied formular :
ϕMn ≥ Min (1.2 Mcr, 1.33M)
For which :
Mcr = frIg/yt
- Modulus of rupture of concrete fr = 0.63f'c0.5
3.45 Mpa
3,572,916,667 mm4
- Moment of inertia of gross concrete section Ig :
- Distance from neutral axis to the extreme tension fiber yt:
Checking:
175 mm
ϕMn =
271
> 1.2Mcr =
85 Satisfactory
7. Checking for maximum reinforcement :
The maximum amount of nonprestressed reinforcement shall be such that :
c/de ≤ 0.42
Where:
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/de =
0.26
<
0.42
Satisfatory
8. Checking for minimum reinforcement :
The minimum amount of nonprestressed reinforcement shall be such that :
ρmin ≥ 0.03f'c/f'y
− ρmin : ratio of tension reinforcement and effective concrete area
ρmin=As/Ac
- Area of tension reinforcement
As =
3,927 mm2
- Effective area of concrete
Ac =
283,500 mm2
ρmin=As/Ac
0.0139
0.03f'c/f'y =
Checking:
ρmin =
0.01385
>
0.0023
0.03f'c/f'y =
0.00225 Satisfactory
9. Control cracking by distribution of reinforcement:
Checking load combination is Service load
M=
52.0 kN.m
Condition:
fs ≤ fsa = Z/(dcA)1/3 ≤ 0.6 f'y
+ Minimum reinforcement
ρ
0.01385
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to X axis - cross section of box culvert )
+ Es/Ec
n
6.79
+k
k
0.350
+j
j
0.883
fs =
52.87 Mpa
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
dc (actually)=
75.0 mm
dc (choosen) =
50.0 mm
8,312.5 mm2
A=
Z=
17.5 kN/mm
fs =
52.9 Mpa
fsa=
234 Mpa
0.6f'y=
240 Mpa
Checking:
fs =
52.9
<
min(fsa; 06f'y)=
234
Satisfactory
and
fsa =
234
<
06f'y=
240
Satisfactory
176,580
Satisfactory
10. Checking axial compression:
Axial resistance of components shall be taken as: Pr = ϕ Pn
For members with spiral reinforcement:
Pn = 0.85 [0.85f'c(Ag-Ast) + fyAst]
For members with tie reinforcement:
Pn = 0.8 [0.85f'c(Ag-Ast) + fyAst]
Type of reinforcement:
Tie 0.75 -
ϕ: axial compression factor
350,000.0 mm2
Ag: gross area of section
7,854.0 mm2
Ast: total area of longitudinal reinforcement
9,493,052.9 N
Pn: nominal axial resistance
7,119,789.67 N
Pr: calculation axial resistance
Checking:
Pr =
7,119,790
>
min(fsa; 06f'y)=
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to X axis - cross section of box culvert )
CHECKING SHEAR RESISTANCE
Materials
Yield strength of reinforcement
fy
Elastic modulus of Prestressing Steel
Ep
197,000 MPa
Elastic modulus of Reinforcement
Es
200,000 MPa
Elastic modulus of Concrete
Ec
26,630 MPa
Compress strength of concrete
f'c
Maximum moment
Mu
Maximum shear
Vu
Depth
h
Effective web width
bw
1,000 mm
Distance from extreme tension fiber to the centroid tensile reinforcement
ds
66.5 mm
400 MPa
30 MPa
Checking load
93,195 N.mm
162,846 N
Section properties
350 mm
Tension reinforcement
- Diameter
mm
25
- Quantity
bar
- Area of cross section of tensile reinforcement
Ast
3,927 mm2
Distance from extreme compression fiber to the centroid tensile reinforcement
dst
284 mm
Effective shear depth
dv
252 mm
8
Sức kháng cắt Vr = ϕ Vn
Vn = min {ϕ(Vc + Vs + Vp); 0.25f'cbvdv + Vp}
+ Vc : nominal shear resistance of concrete
0.083*b*(f'c)^0.5*bw*dv
+ Vs : nominal shear resistance of reinforcement
Ay*fy*dv*(cosq+cosa)*sina)/s
* 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
Xác định b & q
Resistance factor for shear
ϕ
Factor indicating ability of diagonally cracked concrete to transmit tension
β
Strain in the tensile reinforcement
εξ
0.9
3.8
0.000
Inclination angle of diagonal compressive stress
θ
Vc
Nominal shear resistance of concrete
Vc
439,595 N
Vp
Component of effective prestresed force in the direction of the applied shear
Vp
0 N
Checking region requiring transverse reinforcement: Vu và 0.5ϕ(Vc+Vp)
Minimum tranverse reinforcement within distance s s:
27.46 độ
Unncessary
Av
170.479 mm2
Data of transverse reinforcement
Group 1
Angle of inclination of transverse reinforcement to longitudinal axis
α1
Spacing of stirrups
s1
90 degrees
150 mm
Area of shear reinforcement within a distance s: Av1
12 mm
- Diameter
- Quantity
- Area of cross section of shear reinforcement
Nominal shear resistance of reinforcement
Av1
-
mm2
Vs1
-
N
Group 2
Angle of inclination of transverse reinforcement to longitudinal axis
α2
Spacing of stirrups
s2
45 degrees
500 mm
Area of shear reinforcement within a distance s: Av2
- Diameter
12 mm
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to X axis - cross section of box culvert )
-
- Quantity
- Area of cross section of shear reinforcement
Nominal shear resistance of reinforcement
Av2
-
mm
Vs2
-
N
-
N
Vs1 + Vs2
Vn
Nominal shear resistance
Calculation shear resistance
Checking:
Vr =
395,636
439,595 N
Vr
>
2
395,636 N
Vu =
162,846
Satisfactory
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to Y axis - longitudinal section of box culvert )
CHECKING FLEXURAL & AXIAL RESISTANCE
The factored flexural resistance Mr shall be taken as: Mr=ϕMn
Mn=Asfy(ds-a/2)-A'sf'y(d's-a/2)
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 β 1
− β1 : Stress block factor specified in Article 5.7.2.2
0.84
- c : Distance from neutral axis to the extreme compression fiber = [Asfy-A'sf'y]/(0.85f'cβ 1b)
1- Checking load
Maximum axial force
N
19.6 kN
Maximum moment
M
14.7 kN.m
2- Checking cross section
h
350 mm
b
1,000 mm
d's
85 mm
d
110 mm
ds
240 mm
3- Materials
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
4 - Reinforcement
Tension reinforcement As
D
As-1bar
Nos
As
12.0 mm
113.10 mm2
5
565 mm2
Compression reinforcement A's
D
12.0 mm
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to Y axis - longitudinal section of box culvert )
A's-1bar
113.10 mm2
Nos
5
A's
565 mm2
5 - Checking flexural resistance
c=
10.61413 mm
a=
8.87 mm
Mn
35 kN.m
- ϕ : Flexural resistance and compression factor of concrete
0.9
Mr=ϕMn
32 kN.m
Checking:
Mr =
32
>
ϕM =
15
Satisfactory
6. For cracking moment
Apllied formular :
ϕMn ≥ Min (1.2 Mcr, 1.33M)
For which :
Mcr = frIg/yt
- Modulus of rupture of concrete fr = 0.63f'c0.5
3.45 Mpa
3,572,916,667 mm4
- Moment of inertia of gross concrete section Ig :
- Distance from neutral axis to the extreme tension fiber yt:
Checking:
175 mm
ϕMn =
32
> 1.2Mcr =
20 Satisfactory
7. Checking for maximum reinforcement :
The maximum amount of nonprestressed reinforcement shall be such that :
c/de ≤ 0.42
Where:
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/de =
0.04
<
0.42
Satisfatory
8. Checking for minimum reinforcement :
The minimum amount of nonprestressed reinforcement shall be such that :
ρmin ≥ 0.03f'c/f'y
− ρmin : ratio of tension reinforcement and effective concrete area
ρmin=As/Ac
- Area of tension reinforcement
As =
565 mm2
- Effective area of concrete
Ac =
240,000 mm2
ρmin=As/Ac
0.0024
0.03f'c/f'y =
Checking:
ρmin =
0.00236
>
0.0023
0.03f'c/f'y =
0.00225 Satisfactory
9. Control cracking by distribution of reinforcement:
Checking load combination is Service load
M=
10.8 kN.m
Condition:
fs ≤ fsa = Z/(dcA)1/3 ≤ 0.6 f'y
+ Minimum reinforcement
ρ
0.00236
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to Y axis - longitudinal section of box culvert )
+ Es/Ec
n
6.79
+k
k
0.164
+j
j
0.945
fs =
84.10 Mpa
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
dc (actually)=
75.0 mm
dc (choosen) =
50.0 mm
22,000.0 mm2
A=
Z=
17.5 kN/mm
fs =
84.1 Mpa
fsa=
170 Mpa
0.6f'y=
240 Mpa
Checking:
fs =
84.1
<
min(fsa; 06f'y)=
170
Satisfactory
and
fsa =
170
<
06f'y=
240
Satisfactory
19,620
Satisfactory
10. Checking axial compression:
Axial resistance of components shall be taken as: Pr = ϕ Pn
For members with spiral reinforcement:
Pn = 0.85 [0.85f'c(Ag-Ast) + fyAst]
For members with tie reinforcement:
Pn = 0.8 [0.85f'c(Ag-Ast) + fyAst]
Type of reinforcement:
Tie 0.75 -
ϕ: axial compression factor
350,000.0 mm2
Ag: gross area of section
1,131.0 mm2
Ast: total area of longitudinal reinforcement
7,478,839.6 N
Pn: nominal axial resistance
5,609,129.71 N
Pr: calculation axial resistance
Checking:
Pr =
5,609,130
>
min(fsa; 06f'y)=
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to Y axis - longitudinal section of box culvert )
CHECKING SHEAR RESISTANCE
Materials
Yield strength of reinforcement
fy
Elastic modulus of Prestressing Steel
Ep
197,000 MPa
Elastic modulus of Reinforcement
Es
200,000 MPa
Elastic modulus of Concrete
Ec
26,630 MPa
Compress strength of concrete
f'c
Maximum moment
Mu
14,715 N.mm
Maximum shear
Vu
27,468 N
Depth
h
Effective web width
bw
1,000 mm
Distance from extreme tension fiber to the centroid tensile reinforcement
ds
110 mm
400 MPa
30 MPa
Checking load
Section properties
350 mm
Tension reinforcement
- Diameter
mm
12
- Quantity
bar
- Area of cross section of tensile reinforcement
Ast
565 mm2
Distance from extreme compression fiber to the centroid tensile reinforcement
dst
240 mm
Effective shear depth
dv
252 mm
5
Sức kháng cắt Vr = ϕ Vn
Vn = min {ϕ(Vc + Vs + Vp); 0.25f'cbvdv + Vp}
+ Vc : nominal shear resistance of concrete
0.083*b*(f'c)^0.5*bw*dv
+ Vs : nominal shear resistance of reinforcement
Ay*fy*dv*(cosq+cosa)*sina)/s
* 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
Resistance factor for shear
ϕ
0.9
Factor indicating ability of diagonally cracked concrete to transmit tension
β
3.7
Strain in the tensile reinforcement
εξ
0.000
Inclination angle of diagonal compressive stress
θ
27.79 độ
Vc
Nominal shear resistance of concrete
Vc
Vp
Component of effective prestresed force in the direction of the applied shear
Vp
Xác định b & q
Checking region requiring transverse reinforcement: Vu và 0.5ϕ(Vc+Vp)
Minimum tranverse reinforcement within distance s s:
426,796 N
0 N
Unncessary
Av
170.479 mm2
Data of transverse reinforcement
Group 1
Angle of inclination of transverse reinforcement to longitudinal axis
α1
Spacing of stirrups
s1
90 degrees
150 mm
Area of shear reinforcement within a distance s: Av1
12 mm
- Diameter
- Quantity
- Area of cross section of shear reinforcement
Nominal shear resistance of reinforcement
Av1
-
mm2
Vs1
-
N
Group 2
Angle of inclination of transverse reinforcement to longitudinal axis
α2
Spacing of stirrups
s2
45 degrees
500 mm
Area of shear reinforcement within a distance s: Av2
- Diameter
- Quantity
12 mm
-
Project:
Trung Luong - My Thuan Expressway
Item
Box Culvert 4x4M
Section:
Bottom Slab (Checking to Y axis - longitudinal section of box culvert )
Av2
-
mm
Vs2
-
N
-
N
Nominal shear resistance
Vn
426,796 N
Calculation shear resistance
Vr
384,117 N
- Area of cross section of shear reinforcement
Nominal shear resistance of reinforcement
Vs1 + Vs2
Checking:
Vr =
384,117
>
Vu =
27,468
2
Satisfactory
