DESIGN & INVESTIGATIVE PROJECT Instructed by : Dr. Nguyễn Thị Tuyết Trinh Prepared by : Phú Xuân Thái Class: Advanced Program of Civil Engneering K49 Student ID: 0912691
CONTENT:
I INPUT DATA AND BASIC DIMENSIONS OF SPAN STRUCTURE 1. 2. 3. 4. 5.
Input data Basis dimensions of cross section of truss bridge Dimensions of members in main truss Dimensions of members in floor beam system Geometric properties of section
II FLOOR BEAM DESIGN 1. 2. 3. 4.
Stringer design Check stringer following strength limit state I Cross beam design design Check cross beam following strength limit state I
III MAIN TRUSS DESIGN 1. 2. 3. 4. 5. 6. 7. 8.
Determine dead loads acting on main truss Internal force in truss members due to dead load load Distribution factor of main truss Unfactored internal forces in truss member due to live load Load combination for live load Load combination for main truss Geometric properties of members Check member in main truss
IV BRACING SYSTEM DESIGN
1. 2. 3. 4. 5.
Content of checking Lateral force acting on structure Internal forces in bracing member following strength limit state II Checking tensile member Checking compressive member
VI TRUSS JOINT DESIGN 1. General principle 2. Design procedure 3. Design chord conection at joints VII DRAWING. I
INPUT DATA AND BASIC DIMENSIONS OF SUPER STRUCTURE 1. Input data - Span length : 80m - Bridge width : 3 lanes 3 - Live load : HL 93 and pedestrian 3x10 kPa - Material of truss : Steel M270, grade 345 - Concrete grade of slab : 35 Mpa - Connection : high strength bolt - Thickness of slab: 20 cm 3 - Unit weight of concrete: 25 KN/m - Thickness of wearing surface: fine grained asphalt layer: 7cm 2. Basic dimensions of cross – section of truss bridge - Chosen span structure as in the below figure : GENERAL CROSS-SECTION OF BRIDGE
12
3
1
bottomchord
2
diagonal , vertical
3
top chord
4
cross beam
5
longitudinalgirder
6
bracket
7
deck
8
pedestrian
9
top connection frame system
9
2
10
8
7
2%
10 clearance
2%
11 bottomlongitudinal connection framesystem 12 top longitudinal connection frame system
5 6 4 1 11
Bridge cross-section
I
I
T1
T2
T3
T4
T5
T6
T7
T8
T9
D1
D2
D3
D4
D5
D6
D7
D8
D9
Do
Bridge longitudinal section
D10
c r o s s - s ec t io n i - i t o p l o n g i t u d in al c o n n ec t io n f r a me s y s t e m 3
c r o s s - s ec t io n i i - ii b o t t o m l o n g i t u d in al c o n n ec t io n f r a me s y s t e m 1
4
-
Main truss height : H The main truss height is chosen following the requirements : + Small main truss self – weight + Insurance of navigational clearance level + Small architecture height for deck truss bridge + Insurance of vertical stiffness of span structure + Aesthetics purpose In experience, for simple span bridge, the height of main truss is normally chosen as below : H=
( )
Where L is the span length of the bridge. L = 80 m. Hence : H = 8 ÷ 11.43 m. However, the height of main truss should satisfy the vertical clearance specified in TCN 2.3.3.2, i.e. 5300 mm. We choose H = 10.0 m .
-
Distance between 2 main trusses : B B depends on the clearance. This distance should be large enough to prevent bridge from overturning under the action of lateral forces such as
wind forces, laterally fluctuating forces…It is also wide enough for rational torsion rigidity of bridge.
-
-
-
-
3. -
The number of lane is 3 lanes. Basing on the Specification for Highway Design TCVN 4054, the width of one lane depends on the class of road. It ranges from 2.75m to 3.75m. Here we choose it as 3.5m (3500 mm). Furthermore, the distance of one curb to separate carriageway and main truss is 0.25 m(also for curb to separate carriageway lanes), the width of main truss member is chosen as 500mm, so that the distance from center to center of 2 main trusses is : B = 3x3.5 + 2x0.25+2x0.25 + 0.5 = 12 m. Total bridge width : B = B 1 + 2b2 + 2b3 + 2b4 + 2b5 + 2b6 . Where : + B1 : carriageway width. B1 = 3x3.5m. + b2 : railing between main trusses and carriageway. b2 = 0.25 m. + b3 : main truss width. b 3 = 0.5 m. + b4 : railing between side walk and main trusses. b 4 is taken as 0.25 m. + b5 : side walk width. We chose b5 = 2 m. + b6 : the outside railing. Normally b 6 = 0.2 – 0.5 m. We choose b 6 = 0.5 m. Total bridge width : B = B 1 + 2b2 + 2b3 + 2b4 + 2b5 + 2b6 . B = 17 m. Length of panel : d. d is the distance between 2 adjacent cross-beam. In experience, d is normally chosen as below : d = (0.6 ÷ 0.8)H d = 6 ÷ 8 m. We choose d = 8 m. Dimensions of members in main truss. Box section is applied for chords and end post. I section is applied for hip vertical and diagonals. Dimensions of sections of members are shown in the below figure :
end post and chor d sect ion
hip ver t ical and diagonal sect ion
4. Dimensions of members in floor beam system. - Floor beam system includes cross beams and stringers. It bears the load of deck slab as well as live load, then transfers it to the main trusses. Stringers lay on cross beams and are connected to them by high – strength bolts. - Dimensions of section of cross beam and stringer are chosen as below : - Cross beams : the effective length used in calculating cross beam is equal to distance from 2 gussets of main trusses. We denotes by L 1 = B1 + 2b2 = 11 m + The height h of section is chosen as simple beams, i.e. : h = (1/7-1/12)L1 h = 733 ÷ 440 mm. We choose h = 800 mm. + The top and bottom flange width b t are the same and ranges from (1/3 ÷ 2/3)h = 266 ÷ 533 mm. We choose b t = 350 mm. + The thickness tw web is: According to the standard ASTM, the minimum thickness of web is 12mm. I choose tw = 22 mm. + The thickness flange: 25 mm - Stringer : the effective length of stringer is distance between 2 adjacent cross beam or the length of panel d, where d = 8.0 m. Hence, the height of section h : H=(1/10-1/15)d h = 533 ÷ 800 mm. I choose h = 500 m. The thickness of flange and web are taken as 20 mm for all. The flange width is chosen to be 250 mm.
These dimensions are shown in the figure below.
5.Geometric properties of section
Section
Sectional area
Moment of inertia
2
A(m )
4
Section modulus 3
I(m )
S(m )
Cross beam
0.034
0.0034
0.0068
Stringer
0.0192
0.0007
0.0033
II. FLOOR BEAM DESIGN
1. Stringer Design . a. Determination of load distribution factors for stringer. For moment For interior girder : using tabular method ( TCN 4.6.2.2.2 ) + For one lane loaded : K g gmg1 = 0.06 4300 L L.t s S
0.4
S
0.3
3
0.1
+ For two or more lanes loaded : S S K g gmg2 = 0.075 2900 L L.t s 0.6
0.2
0.1
3
Where: 0.5 4 Eslab = 0.043xγxf s = 3.18 x 10 MPa 5 Eg = 2 x 10 MPa n = Es/Eg = 6.3 eg = (H – ye)-hf /2 = 0.15 m K g = n(Id+Aeg2) = 7.13 x 10-2 m4 = 7.13x109 mm4 gmg1 = 0.38 gmg2 = 0.46 Check the range of applicability : 1100 mm ≤ S = 1500 mm ≤ 4900 mm. 110 mm ≤ ts = 200 mm ≤ 300 mm 6000mm ≤ L = 8000 mm ≤ 73000 mm N b = 7 > 4 OK. Choose the maximum value to be the load distribution factor for interior girder : gmg = max (gmg1, gmg2 ) gmg = 0.46 The level rule method is used for exterior girder. + Basing on the influence line of reaction of each girder ( from 1 to 4 ), we can calculate the load distribution factors for moment and shear for each girder due to live load ( truck load and lane load ). + For truck load : g HL = 0.5∑yi , where yi is the ordinate of the i th axle on the influence line. + For lane load : gLan = ωi, where ωi is the area of corresponding influence line. + These factors should be factored with the multiple presence factor m. -
For exterior girder :
+ For one lane loaded : m = 1.2
For design truck load :
gHL1 =
gHL1 = 0.678
For lane load :
gLan1 =
gLan1 = 0.98 + For two or more lanes loaded : m = 1.
gmb2 = g mg 0.77
-
d e
de = 400 mm.
2800
Range of applicability : 300 mm ≤ de ≤ 1700 mm. out of range of applicability, g mb2 = 0.666. Hence : gHL = max(gHL1,gmb2) gHL1 = 0.678 gLan = max(gLan1,gmb2) gLan = 0.98 For shear For interior girder + For one lane loaded : gvg1 =
0.36
S 7600
gvg1 = 0.56 + For two or more lanes loaded : S gvg2 = 0.2 3600 10700 S
2.0
gvg2 = 0.62 check the range of application : 1100 mm ≤ S = 2200 mm ≤ 4900 mm. 110 mm ≤ ts = 200 mm ≤ 300 mm 6000mm ≤ L = 8000 mm ≤ 73000 mm N b = 7 > 4 Choose the maximum value as the distribution factor for shear for interior girder : gvg = max(gvg1 ; gvg2 ) gvg = 0.62 - For exterior girder . + For one lane loaded : using level rule method. As calculated above : gHL1 = 0.678
gLan1 = 98 + For two or more lane loaded :
gvb2 = g vg 0.6
d e
3000
de = 400 mm inside the range of application (300 mm ≤ de gvb2 = 0.45 Therefore :
≤ 1700 mm )
gvbHL = 0.678
gvblan = 0.98 I establish the following table :
Girder no.
For moment
For shear
Truck load
Lane load
Design vehicle
Lane load
Exterior gider
0.678
0.98
0.678
0.98
Interior girder
0.46
0.46
0.62
0.62
b. Determine dead loads. - Dead load stage I + Stringer self – weight : DC dc = γs.A 2 A : cross sectional area. A = 0.0192 m . γs : Steel unit weight. γs = 78.5 kN/m3 . DCdc = 1.51 kN/m. + dead load of deck slab : for both interior and exterior girders : DC bm = S.ts. γc S : distance between 2 adjacent girder. S = 1500 mm ts : slab thickness. ts = 200 mm. γc : concrete unit weight. γc = 25 kN/m3 . DC bm = 7.5 kN/m Total dead load stage I : DC = 9.01 kN/m. - Dead load stage II + Dead load of concrete element : 1 kN/m
+ Dead load of steel element : 0.15 kN/m Assume that, dead load of railing is borne by only exterior girder ( girder 1 ). Girder 1 : DC lc = 1.15 kN/m Girder 2 : DC lc = 0 kN/m + Dead load of wearing surface on one girder : water proofing layer : t1 = 0.4 cm , γ1 = 18 kN/m3 asphalt concrete layer : t2 = 7 cm. γ2 = 22 kN/m3. + Dead load of utility: DWuti = 5 KN/m DWas + DWuti = 36.5 KN/m For interior girder : DCsur = (0.04x18 + 0.07x22)x1.5= 3.39 kN/m For exterior girder : DCsur = (0.04x18 + 0.08x22)x2.0 = 4.52 kN/m For exterior girder : DW = 3.39 kN/m Total dead load stage II : For interior girder : DW = 5.67 kN/m. I establish the following table : Load type
Interior girder
Exterior girder
Unit
DC
9.01
9.01
kN/m
DW
3.39
5.67
kN/m
Sum
12.4
14.68
kN/m
c. Internal forces due to live load determination. The influence line of moment and shear for stringer: For strength limit state I, the most critical section for moment is midspan section and the most critical one for shear is girder end section.
110kN
1.2m
tandem
110kN
Case 1 145kN
145kN
35kN 4.3m
4.3m
truck lane load
0
4 . 1
2
0
Influence line at L/2 110kN
Case 2
145kN
1.2m
110kN tandem
145kN 4.3m
35kN 4.3m
truck lane load
0
7 . 2 1
7 . 1
0
Influence line at L/2
110kN
1.2m
110kN
145kN
tandem
145kN 4.3m
35kN 4.3m
truck lane load
0 0 , 1
5 8 . 0
5 2 6 4 . 0
0
Influence line at bearing shoe
Internal forces due to design truck will be taken as the maximum value of two cases above.
Bending moment
+ Moment due to design truck : M truck1 = 145.yM1 + 145.yM3 + 35.yM4 Mtruck2 = 145.yM1 + 145.yM3 + 35.yM5 Mtruck = max(Mtruck1 ; Mtruck2 ). + Moment due to design tandem :
Mtandem1 = 110.yM1 + 110. yM2.
Mtandem2 = 110. yM2 + 110. yM4 . Mtandem = max(Mtandem1 ; Mtandem2 ). + Moment due to design vehicle :
M xetk = max(Mtruck ; Mtandem ).
+ Moment due to lane load : q lan= 9.3 kN/m Mlan = 9.3.ωM Where ωM is the area of corresponding moment influence line.
Shear force + Shear force due to design truck : V truck = 145.yM1 + 145.yM3 + 35.yM4 . + Shear force due to tandem : Vtandem = 145.yM1 + 145.yM2 . + Shear force due to lane load : Vlan = 9.3.ωV(+) .
Where ωV(+) is the positive area of corresponding shear influence line. Internal forces
-
Type of load
Unit
M1
Design vehicle 374
Lane load 74.4
kN.m
V0
212.06
37.2
kN.
Load combination due to live load HL – 93 + Moment combination (factored with distribution factor g m ). IM = 25 % For interior girder : MLLg = gmg ( 1 + IM ).M xetk + gmg .Mlan . MLLg = 249.274 kN.m For exterior girder : MLLb = gHL(1 + IM).Mxetk + glan.Mlan . MLLb = 389.88 kN.m + Shear force combination (factored with distribution factor g v ) IM = 25 % For interior girder : VLLg = gvg(1 + IM).Vxetk + gvg.Vlan . VLLg = 187.41 kN. For exterior girder :
VLLb = gHL(1 + IM).Vxetk + glan.Vlan . VLLb = 216.18 kN. d. Internal forces due to dead load. Dead load stage I .
-
Internal forces M1DC V0DC
-
V0DW
-
-
9.01
Influence line area 8
Interior girder 72.8
Exterior girder 72.8
Unit kN.m
9.01
4
36.4
36.4
kN
Dead load stage II.
Internal forces M1DW
DC
DW Gider 2 Girder 1 3.39 5.67 3.39
Influence line area
Interior girder
Exterior girder
Unit
8
27.12
45.36
kN.m
4
13.56
22.68
kN
5.67
e. Internal force combination at sections of exterior girder Strength limit state I : + Moment : M uCD1g = η.(1.75MLLg + 1.25.MDCg + 1.5.MDWg ). Where η : load modifier. η = 1.0 for normal bridge. MuCD1g = 841.33 kN.m + Shear : VuCD1g = η.(1.75VLLg + 1.25.VDCg + 1.5.VDWg ). VuCD1g = 457.835 kN f. Internal force combination at sections of interior girder Strength limit state + Moment : M uCD1b = η.(1.75MLLb + 1.25.MDCb + 1.5.MDWb ). MuCD1b = 567.91 kN.m + Shear : VuCD1b = η.(1.75VLLb + 1.25.VDCb + 1.5.VDWb ). VuCD1b = 393.81 kN +Establish the following table.
LOAD COMBINATION RESULTS Internal forces Strength limit state I
Unit
M1
Girder 2 841.33
Girder 1 567.91
kN.m
V0
457.835
393.81
kN
On comparing, the internal forces in the exterior girders are greater than that in interior girder so that, the exterior will be chosen to check.
