4. Bracing system design

4. Bracing system design4.1 Longitudinal bracing system4.1.1. DimensionNormally consist of trusses placed in the planes of top chords and bottom chords. Because the truss plane distance is much bigger than the distance betweem cross beam, to ensure the stability of structure and lateral resistance to wind, we use bracing type X with verticals. As experience, we choose:

DIMENSIONS OF LONGITUDINAL BRACING SYSTEMTypeINumbers38unitsHeight0.55mThickness of web0.03mFlange width0.4mFlange thickness0.035mArea of c-s0.0424m2Inertia moment0.002135913m3Dead load of 1 bar3.3284kN/m

4.2 Lateral forces in bracing member following strength limit state II4.2.1 Design wind velocity (TCN 3.8.1.1)V = VB.S =38x1.09 = 41.42 m/s.VB : basic 3 second gust wind velocity with 100 year return period approximate to the wind zone in which structure located. Assume that structure locates in wind zone I.VB = 38 m/s.S : correction factor for upwind terrain and deck height. Assume that the height of bridge deck above water level is H = 10 m. Then, S = 1.09.

4.2.2 Wind load on structure: WS (TCN 3.8.1.2)PD = 0.0006V2AtCd 1.8At (kN) V : design wind velocityAt : area of structure or elements for calculation of traverse wind load (m2)Cd : drag coefficient.

Lateral forces in bracing member following strength limit state II4.2.3 For main trussCdi : drag coefficient of the ith member, for angular steelAi : wind area of the ith member. A : area that limited by boundary lines of main trussResultant table:

For the second main truss: Cd2 = .Cd1 Cd1 : drag coefficient for first main truss. : adjustment factor

MemberNo.Ai (m2)CdiA (m2)CdHip vertical90.4x10=41.47200.29Diagonal80.4x12.8=5.121.4Chord180.5x10=51.2End post20.5x12.8=6.41.2Lateral forces in bracing member following strength limit state IIWind load on bridge

4.2.4 Wind load on vehicle: WL (TCN 3.8.1.3)When considering strength limit state III, the design wind load shall be applied to both structure and vehicle. Transverse wind load on vehicles shall be presented by a line of 1.5 kN/m acting horizontally, transverse to horizontal centerline of the structure at 1800 mm above the roadway

MemberAt (m2)Cd = Cd1 + Cd2PDtc (kN)1.8AtMain truss7200.566419.491296Floor beam561.5287.62100.8Railing161.5225.0328.8Sum532.141425.6Lateral forces in bracing member following strength limit state IIWind load distribution

qWS = 0.6* 1566/80 = 11.745 kN/m qWL = 0.6* 1.5 = 0.9 kN/m

4.3 Internal forces in bracing member following strength limit state IIAs requirement, we only consider the strength limit state II DC=DW=0 (in horizontal direction)

WS can be determined by placing load on influence line.

Internal forces in bracing member following strength limit state IILoad combination following strength limit state II

Member (45) shall be checked as tensile member and member (56) shall be checked as compressive member.

4.4 Check tensile member4.4.1 Tensile resistanceThe factored tensile resistance shall be taken as the lesser of two values below:Pr = yPny = yFyAg Pr = uPnu = uFuAnUWhere :Pny = tensile resistance for yielding gross section (N)Ag = gross cross sectional area of member (mm2)Pnu = nominal tensile resistance for fracture in net section (N) Fu = tensile strength. Fu = 450 N/mm2.y : resistance factor for yielding of tension member as specified in TCN 6.5.4.2.u : resistance factor for fracture of tension member as specified in TCN 6.5.4.2

Check tensile memberAn : net area of section. An= Ag tf.(d+2).n, where :n is the number of row of bolt and d is the nominal diameter of bolt. d = 20 mm. U : reduction factor, to account for shear lag specified in TCN 6.8.2.2, U = 0.85+ Pr is calculated as in table below :

Check members tensile resistance

No.F.SFyMPaAgmm2PnykNFuMPanAnmm2PnukNPrkN45Tension3459088313545030.00557 *10^633113605No.F.S Pu( kN )Pr( kN )Check1Tension713099OKCheck tensile memberChecking the results

No.F.SPukNPrkNPu/Pr(Mux/Mrx)FormulaLeftsideRightsideCheck1T53.826490.020.071(1)0.0811OK4.5 Check compressive member4.5.1 Compressive resistance Pr = cPn : (TCN 6.9.4.1)

Where:K : effective length factor, taken as in TCN 4.6.2.5, K = 0.75L : unbraced length of the columnrs : radius of gyration of steel section in plane bending. Fy : yield strength of steel. Fy = 345 MPaE : Youngs modulus of steel. E = 200000MpaAg : gross cross sectional area The results are shown in the table below :

MemberForce subjectedKLmrsmmFyN/mm2EN/mm2Agmm2PnkN2C0.7516.911473452x1053.159088443Check compressive memberThe factored compressive resistance :Pr = cPn, where c = 1 (TCN 6.5.4.2)

4.5.2 Check following combined compress sion and flexure conditions ( TCN 6.9.2.2)

MemberForce subjectedPnkNcPr (kN )Pu (kN )Check2Compression443144370.9OK

Check compressive memberCheck for vertical plates:

Check for the horizontal plates:

MemberF.SbmmtmmkEMPaFyMpab/tCheck2Compression150120.5620000034512.513.48OK

MemberF.SDmmtmmkEMPaFyMpaD/tCheck2Compression340121.4920000034512.535.87OK

Sheet1A(+) (m2)A(-) (m2)qw (kN/m)WS (kN)N (kN)Top element 4'510.38-5.8411.74553.32230.9570.918659Bottom element 56'8.07-12.611.745-53.204850.95-70.7624505