DESIGN & INVESTIGATIVE PROJECTInstructed by : Dr. Nguyn Th Tuyt TrinhPrepared by : Ph Xun ThiClass: Advanced Program of Civil Engneering K49Student ID: 0912691

CONTENT:I INPUT DATA AND BASIC DIMENSIONS OF SPAN STRUCTURE1. Input data 2. Basis dimensions of cross section of truss bridge3. Dimensions of members in main truss4. Dimensions of members in floor beam system 5. Geometric properties of sectionII FLOOR BEAM DESIGN1. Stringer design2. Check stringer following strength limit state I3. Cross beam design design4. Check cross beam following strength limit state IIII MAIN TRUSS DESIGN 1. Determine dead loads acting on main truss 2. Internal force in truss members due to dead load3. Distribution factor of main truss4. Unfactored internal forces in truss member due to live load5. Load combination for live load 6. Load combination for main truss7. Geometric properties of members8. Check member in main trussIV BRACING SYSTEM DESIGN 1. Content of checking2. Lateral force acting on structure3. Internal forces in bracing member following strength limit state II4. Checking tensile member5. Checking compressive memberVI TRUSS JOINT DESIGN1. General principle2. Design procedure3. Design chord conection at jointsVII DRAWING.I INPUT DATA AND BASIC DIMENSIONS OF SUPER STRUCTURE1. Input data Span length : 80m Bridge width : 3 lanes Live load : HL 93 and pedestrian 3x103 kPa Material of truss : Steel M270, grade 345 Concrete grade of slab : 35 Mpa Connection : high strength bolt Thickness of slab: 20 cm Unit weight of concrete: 25 KN/m3 Thickness of wearing surface: fine grained asphalt layer: 7cm2. Basic dimensions of cross section of truss bridge Chosen span structure as in the below figure : GENERAL CROSS-SECTION OF BRIDGE

Bridge cross-section

Bridge longitudinal section

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 purposeIn 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 forcesIt is also wide enough for rational torsion rigidity of bridge. 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 = B1 + 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. b3 = 0.5 m.+ b4 : railing between side walk and main trusses. b4 is taken as 0.25 m.+ b5 : side walk width. We chose b5 = 2 m.+ b6 : the outside railing. Normally b6 = 0.2 0.5 m. We choose b6 = 0.5 m. Total bridge width : B = B1 + 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)Hd = 6 8 m.We choose d = 8 m.3. 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 :

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 L1 = B1 + 2b2 = 11 m+ The height h of section is chosen as simple beams, i.e. :h = (1/7-1/12)L1h = 733 440 mm.We choose h = 800 mm.+ The top and bottom flange width bt are the same and ranges from (1/3 2/3)h = 266 533 mm. We choose bt = 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)dh = 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 sectionSectionSectional areaA(m2)Moment of inertiaI(m4)Section modulusS(m3)

Cross beam0.0340.00340.0068

Stringer0.01920.00070.0033

II. FLOOR BEAM DESIGN1. 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 :

gmg1 = + For two or more lanes loaded :

gmg2 = Where: Eslab = 0.043xxfs0.5 = 3.18 x 104 MPa Eg = 2 x 105 MPa n = Es/Eg = 6.3 eg = (H ye)-hf/2 = 0.15 m Kg = 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 mm6000mm L = 8000 mm 73000 mmNb = 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 : gHL = 0.5yi , where yi is the ordinate of the ith axle on the influence line.+ For lane load : gLan = i, where i 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 = de = 400 mm.Range of applicability : 300 mm de 1700 mm. out of range of applicability, gmb2 = 0.666. Hence : gHL = max(gHL1,gmb2) gHL1 = 0.678gLan = max(gLan1,gmb2) gLan = 0.98 For shear For interior girder + For one lane loaded :

gvg1 = gvg1 = 0.56+ For two or more lanes loaded :

gvg2 = gvg2 = 0.62

check the range of application : 1100 mm S = 2200 mm 4900 mm.110 mm ts = 200 mm 300 mm6000mm L = 8000 mm 73000 mmNb = 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 = de = 400 mm inside the range of application (300 mm de 1700 mm ) gvb2 = 0.45Therefore : gvbHL = 0.678 gvblan = 0.98I establish the following table :

Girder no. For moment For shear

Truck loadLane loadDesign vehicleLane load

Exterior gider0.6780.980.6780.98

Interior girder0.460.460.620.62

b. Determine dead loads. Dead load stage I + Stringer self weight : DCdc = s.AA : cross sectional area. A = 0.0192 m2.s : Steel unit weight. s = 78.5 kN/m3 .DCdc = 1.51 kN/m.+ dead load of deck slab : for both interior and exterior girders :DCbm = S.ts. cS : distance between 2 adjacent girder. S = 1500 mmts : slab thickness. ts = 200 mm. c : concrete unit weight. c = 25 kN/m3 .DCbm = 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/mAssume that, dead load of railing is borne by only exterior girder ( girder 1 ).Girder 1 : DClc = 1.15 kN/mGirder 2 : DClc = 0 kN/m+ Dead load of wearing surface on one girder : water proofing layer : t1 = 0.4 cm , 1 = 18 kN/m3asphalt concrete layer : t2 = 7 cm.2 = 22 kN/m3. + Dead load of utility: DWuti = 5 KN/m DWas + DWuti = 36.5 KN/mFor interior girder : DCsur = (0.04x18 + 0.07x22)x1.5= 3.39 kN/mFor exterior girder : DCsur = (0.04x18 + 0.08x22)x2.0 = 4.52 kN/m Total dead load stage II : For exterior girder : DW = 3.39 kN/m For interior girder : DW = 5.67 kN/m.I establish the following table :Load typeInterior girderExterior girderUnit

DC9.019.01kN/m

DW3.395.67kN/m

Sum 12.414.68kN/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.

Internal forces due to design truck will be taken as the maximum value of two cases above. Bending moment + Moment due to design truck : Mtruck1 = 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 : Mxetk = max(Mtruck ; Mtandem ).+ Moment due to lane load : qlan= 9.3 kN/mMlan = 9.3.M Where M is the area of corresponding moment influence line. Shear force+ Shear force due to design truck : Vtruck = 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.V(+) .Where V(+) is the positive area of corresponding shear influence line.Internal forcesType of loadUnit

Design vehicleLane load

M137474.4kN.m

V0212.0637.2kN.

Load combination due to live load HL 93+ Moment combination (factored with distribution factor gm ).IM = 25 %For interior girder :MLLg = gmg ( 1 + IM ).Mxetk + gmg .Mlan .MLLg = 249.274 kN.mFor exterior girder :MLLb = gHL(1 + IM).Mxetk + glan.Mlan .MLLb = 389.88 kN.m+ Shear force combination (factored with distribution factor gv )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 = 216.18kN.d. Internal forces due to dead load. Dead load stage I .

Internal forcesDCInfluence line areaInterior girderExterior girderUnit

M1DC9.01872.872.8kN.m

V0DC9.01436.436.4kN

Dead load stage II.

Internal forcesDWInfluence line areaInterior girderExterior girderUnit

Gider 2Girder 1

M1DW3.395.67827.1245.36kN.m

V0DW3.395.67413.5622.68kN

e. Internal force combination at sections of exterior girder Strength limit state I : + Moment :MuCD1g = .(1.75MLLg + 1.25.MDCg + 1.5.MDWg ).Where : load modifier. = 1.0 for normal bridge. MuCD1g = 841.33 kN.m+ Shear :VuCD1g = .(1.75VLLg + 1.25.VDCg + 1.5.VDWg ). VuCD1g = 457.835 kNf. Internal force combination at sections of interior girder Strength limit state+ Moment : MuCD1b = .(1.75MLLb + 1.25.MDCb + 1.5.MDWb ).MuCD1b = 567.91 kN.m + Shear : VuCD1b = .(1.75VLLb + 1.25.VDCb + 1.5.VDWb ). VuCD1b = 393.81 kN

+Establish the following table.

LOAD COMBINATION RESULTSInternal forcesStrength limit state IUnit

Girder 2Girder 1

M1841.33567.91kN.m

V0457.835393.81kN

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.