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STRUCTURAL STEEL ASSIGNMENT Lecturer: Student: Class: DESIGN OBJECTIVE Design a main girder, simple span on bridge way for vehicles with I-section, uncomposite steel girder Flange and girder wall connected by weld line in the factory and construction joint by high-pressure bolt I HYPOTHETICAL METRICS Girder length (calculated span) Designed lanes: Main girder distance: Dead load on reinforced concrete deck: Dead load of wearing surface and utilities: Designed car live load (with road level coefficient): Average Daily Traffic ADT: Truck on lane ratio: Lateral moment distribution coefficient: 10 Lateral shear distribution coefficient: 11 Lateral deflect distribution coefficient: 12 Lateral fatigue distribution coefficient: 13 Road level coefficient: 14 Allowable deflection of live load: 15 Material: - Steel designed for girder: - High-pressure bolt: 16 Design Standard: II 𝐿 =10m 𝑛 = 𝑆 =2.2 𝑤 = 6.0 kN/m 𝑤 = kN/m HL – 93 ADT = 1.5× 10 xe/ngày/lần k = 0.1 𝑚𝑔 = 0.6 𝑚𝑔 = 0.7 𝑚𝑔 = 0.5 𝑚𝑔 = 0.45 m = 0.5 ∆ = L/800 M270M Grade 345 ASTM A490M TCVN 11823-2017 CALCULATION AND DESIGN REQUIREMENTS A - Calculation Select girder section, calculate signature geometry Calculate and draw internal forces diagram by influence line method Check girder by strength limit state, service limit state and fatigue limit state Calculate and design stiffener Calculate and design joint construction B - Drawing Draw main section of girder and representative section Draw detail of deck slab, stiffener, joint Statistic materials Paper type A1 or A3 DESIGN STEPS I SELECT GIRDER SECTION Girder section is selected by trial and error method, it means we select girder section size respectively by experience and control regulation of design standard, and check again Process is repeated until it meets the requirement LATERAL GIRDER SECTION 1.1 Steel girder height d (mm) The height of main girder has big influence on construction’s price, so we have to consider it carefully when select this value To car bridge way, simple span, we can choose them basically by experience With simple bridge beam, I section of un-composite reinforced concrete deck: d≥ L(mm), we choose d = ( Height of girder d should be chosen evenly to 5cm We have: L =10 m Chose: (1/25)L = 0.4 m (1/20)L = 0.5 m (1/12)L = 0.833 m d = 750 mm ÷ )L (mm) 1.2 Select flange, girder web Flange’s width can be selected basically by this experience equation: 𝑏 = We have: 𝑑 (𝑚𝑚 ) ÷ d = 750 mm 1/3d = 250 mm 1/2d = 375 mm So we choose upper compressed flange’s width bfc = 350 mm Lower compressed flange’s width bft = 350 mm Flange’s width and girder web: Follow the requirement of regulation (A6.7.3), the minimum thickness of flange and girder web is 8m This minimum thickness is used for anti-corrosive and transporting requirements, collapsing in construction Requirements of ASTM A6M We choose: Upper compressed flange’s thickness: tfc = 20 mm Lower compressed flange’s thickness: t ft = 20 mm Girder web’s thickness: tw = 14 mm So height of girder web is: D =𝑑 − 𝑡 − 𝑡 = 750-20-20 = 710 mm Check section ratio: Marker 10.2 of TCVN 11823-6:2017 introduce these limit dimension of lateral section: Girder web ratio Girder web without axial stiffeners: = Flange ratio: Tensioned and compressed flange have this ratio: = 51 < 150 -> OK = × = 8.75 < 12 -> OK This is the limitation in realistic to make sure that flange will not get over-deformed when weld it to girder web bf = 350 mm ≥ D/6 = 710/6 = 118 mm -> OK tf = 25 mm ≥ 1.1tw = 1.1*14 = 15 mm -> OK 0.1 ≤ ≤ -> Ok After choosing, girder section has this figure: 1.4 Calculate signature geometry of girder section Section Upper flange Girder web Lower flange Total Ai (mm2) hi (mm) A i hi (mm3) I0i (mm4) Aiyi2 (mm4) Ii (mm4) 7000 740 5180000 233333.33 932575000 932808333 9940 375 3727500 417562833 417562833 7000 10 70000 233333.33 932575000 932808333 23940 375 8977500 418029500 1865150000 2283179500 Where: Ai = section area numbers “i” (mm2) hi = distance from median point of section “i” to bottom girder (mm) I0i= inertia moment of section numbers “i” with cross axis through its median point (mm4) = distance from median point of section numbers “i” yi = distance from the median point of section numbers “i” to median point of girder section (mm) Ii= inertia moment of section numbers “i” with lateral axis through median point of girder section (mm4) Ii = I01 + Aiyi (mm4) We have: Section Steel girder ybot ytop ybotmid ytopmid (mm) (mm) (mm) (mm) 375 375 365 365 Sbot (mm3) Stop (mm3) Sbotmid (mm3) Stopmid (mm3) 6088478.7 6088478.7 6255286.3 6255286.3 ybot = distance from median point of girder section to under lower flange steel girder (mm) ytop = distance from median point of girder section to top of upper flange steel girder (mm) ybotmid = distance from median point of girder section to median point of under flange steel girder (mm) ytopmid = distance from median point of girder section to median point on top of flange steel girder (mm) Sbot = Bending moment resistance of girder section with y bot (mm3) Stop = Bending moment resistance of girfer section with y top (mm3) Sbotmid = Bending moment resistance of girder section with ybotmid (mm3) Stopmid = Bending moment resistance of girder section with ytopmid (mm3) 1.5 Calculate girder dead load Girder dead load on one meter long: WDC1 = A x γs = 23940*78.5*10-6 = 1.88 kN/m II Calculate and draw internal forces diagram 2.1 Calculate M, V by influence line method: Split girder to equal parts Choose Nđ = 10 parts Length of each part: Lđ = m We numbered the beam section parts respectively: Moment influence line value: Section xi (m) 5 Influence line Mi (m) 0.9 1.6 2.1 2.4 2.5 Where: xi = distance from bearing to numbers “i” section Influence line Mi = Ordination of influence line Mi AMi = Area of influence line Mi We have moment influence line figure of girder sections: AMi (m2) 4.5 10.5 12 12.5 Adjusting load coefficient for limit strength state: We consider total loads: + Live load (HL-93) + Dead load of structural components and nonstructural attachments, reinforced concrete surface deck (DC) + Dead load of wearing surface and utilities (DW) Moment in random section is calculated by these equations: + Strength limit state I + Service limit state: Where: LLL = designed lane load = uniform load = 9,3 kN/m = Effects of designed truck at “i” section = Effects of designed tandem at “i” section mgm = Lateral distribution coefficient calculated for moment (and lane ratio m) WDC = dead load of girder on a unit length WDW = uniform loads by wearing surface and utilities IM: impact factor = 0.33 AMi = Area of influence line Mi k: road level coefficient or reduction factor live load for designed car 1,0; 1,25; 1,3; 1,5, 1.7: load coefficient by requirement of load combination TCVN 11823:2017 Arrange design trucks and tandems on influence line to find out orientation values influence line correspond with each car axle to each influence line We have Arrange trucks and tandems on moment influence line Rg   Ag Fyt  LS =1  Ag Fyt  SS Rh=1.0  f cf    f Fyf  Rg  Rh  = Fcf   = 219.15 Mpa 0.75 f Fyf Rg = 0.75x1x1x345x1 = 258.75 Mpa So Fcf = 258.75 Mpa An   b ft  4dh  t ft = (350-4*22)*20 = 5240 mm2  F Ae   u u  F  y yt   An =  x5240 = 5755.61 mm2 < Ag= 350*20 = 7000 mm2 Design joint force in tensioned flange Pfc  Fncf Ae = 258.78*5755.61*10-3= 1489.26 kN Tensioned force resistance of tensioned flange is smaller value: Pru  u FuUAn = 0.8*450*1*5240*10-3 = 1886.4 kN Pry   y Fy Ag = 0.95*345*7000*10-3 = 2294.25 kN So Pft = 1443.789 kN < min(Pru;Pry) = 1886.4 kN Design force value in tensioned joint slab of strength limit state I: Pfc = 1489.26 kN Floating slab and connection with uncontrol flange must have enough minimum resistance, which equal to design stress, Fncf multiply with effective area value smaller than Ae of uncontrol flange: Fncf  Rcf fcf Rh  0.75 f Fyf Rg Where: Rcf is absolute value ratio of Fcf and fcf for control flange Fncf is maximum bending stress caused by calculate load in middle point of uncontrol flange’s thickness at joint position together with fcf (Mpa) Rg: adjust resistance coefficient of flange Rcf = Fcf/fcf = 258.75/93.3 = 2.77 Fncf = 2.77*93.3/1=258.441 Mpa 0.75 f Fyf R g = 0.75*1*1*345*1 = 258.75 Mpa So Fncf = 258.75 Mpa Design joint force in compressed flange: Pfc  Fncf Ae = 258.75*5755.61*10-3 = 1489.26 kN Design force in compressed floating slab not get over calculated compression resistance Rr: Rr  c Fy As = 0.9*345*7000*10-3 = 2173.5 kN So Pfc = 1489.26 kN Design force value in compressed flange joint at strength limit state I: Pfc = 1489.26 kN 7.2.3 Calculate minimum design force in flange at service limit state II: Bolted connection of flange connection must be designed by friction joint type for design force of flange Minimum design force of approving flange, check slip of flange bolted joint, it must be design stress at total using load II, Fs multiply to smaller value between original flange area to two parts of joints: Fs  fs Rh Where: fs is maximum bending stress caused by total using load II at the middle point of thickness for flange in smaller section and in smaller section of joint position (Mpa) Rh: hybrid coefficient Fs  fs = Rh = 61.5 Mpa Axial force in joint flange at service limit state II: Pservice-II = FsAf = 61.5*7000*10-3= 430.5 kN Table: Minimum design stress in steel girder flange at strength limit state I Position Upper flange Lower flange A(mm2) P(kN) 7000 7000 1489.26 1489.26 Table: Minimum design stress in steel girder flange at service limit state II Position Upper flange Lower flange A(mm2) P(kN) 7000 7000 430.5 430.5 7.3 Design flange joint 7.3.1 Choose joint size Joints are designed by trial and error method, it means we choose joint size respectively, which are based on experience and control regulation of design standard, and check again If it don’t meet the requirements, we choose and check again Process is repeated until it meets the requirements We choose joint size basically: The thickness of floating slab of flange t ≥ tf/2 = 20/2 = 10 mm - Inside floating slab size: thickness x width x length = 14x350x430mm - Outside floating slab size: thickness x width x length = 14x155x430mm - High-pressure bolt diameters: d = 20m - Standard hole: h = 22 mm - Number of bolt in each part of joint: N = 12 bolts Arrange bolts to rows, bolts for each row: - Distance of bolt by axial girder: Sl = 65 mm - Distance of bolt by lateral girder: Sh = 65 mm We have: Calculating for lower flange, upper flange has the same method 7.3.2 Check distance of high-pressure bolts 7.3.2.1 Minimum distance Minimum distance from center slab to center of bolts: Smin = 3*20 = 60 mm Check distance between bolts by the formula: min(SI,Sh) ≥ Smin Where: SI = distance between bolts by axial girder SH = distance between bolts by lateral girder We have: min(SI,Sh) = min(65;65) = 65 mm > 60 mm -> OK Conclusion: OK 7.3.2.2 Maximum distance To make sure closely press the joint, avoid moist, maximum distance from center slab to center of bolt of contigous bolt row with free side of floating slab or steel shape must meet these requirements: S ≤ 100 + 4t ≤ 175 Where: t = smaller thickness of floating slab or steel shape (mm) We have: S = 80 mm < 100+4*14 = 156 mm -> OK Conclusion: OK 7.3.2.3: Distance to edge Minimum distance to center bolt to bar edge must meet the requirements, table A6.13.2.6.6-1 Minimum distance to center bolt to bar edge is not eight times bigger than thin floating slab’s thickness or 125mm Check distance to edge: Semin ≤ Se ≤ Semax Where: Semin = minimum distance to center bolt to bar edge (mm) Semax = maximum distance to center bolt to bar edge (mm) Se = distance from outermost center bolt to bar edge (mm) We have: 34 mm < 50 mm < 112 mm -> OK Conclusion: OK 7.3.3: Check shear resistance of high-pressure bolt Calculated shear resistance of high-pressure bolt at strength limit state: Rrs = 𝜑sRns Where: 𝜑s = resistance coefficient for sheared bolt (A490M) by regulation (A6.5.4.2)𝛷 Rrs = Shear resistance of high-pressure bolt by regulation, use bolt’s length with dented thread lies out of clipping plane: Rns = 0.48AbFubNs Where: Ab = bolt area by defined diameter (mm2), Ab = 314 mm2 Fub = minimum tensioned strength of bolt (Mpa) (A6.4.3), F ub = 830 Mpa Ns = number of section for each bolt, Ns = We have: Rns = 0.48AbFubNs = 0.48*314.159*830*2 = 250322 N => Rrs = 𝛷s*Rns = 0.8*250321.89*10-3= 200.26 kN Calculated shear resistance of high-pressure bolt at strength limit state must meet these requirements: Ru < Rrs Where: Pbot = minimum design force in lower flange at strength limit state I (N) We have: Pu = 1489.26 kN -> = 1489.26/12 = 124 kN < Rrs = 200.26 kN -> OK 7.3.4 Check pressure resistance of connection of high-pressure bolt Because the thickness of flange equals 16mm and total thickness is 28 mm > t f = 25 mm so we calculate pressure for strain bar is flange: Close edge hole: Lc = Le - = 45-22/2 = 34 mm < 2d = 2*20 = 40 mm Rn1 = 1.2LctFu = 1.2*34*20*450*10-3= 367.2 kN Other holes: Lc = s – dh = 65-22 = 43 mm > 2d=2*20 = 40 mm Rn2 = 2.4dtFu = 2.4*20*20*450*10-3= 432 kN Pressure resistance on gusset plate of all bolt holes:  R  0.8(4Rn1  8Rn2 ) =0.8*(4*367.2+8*432) = 3939.84 kN bb n  R  3939.84 kN > Pft = 1489.26 kN OK bb n 7.3.5 Check slip resistance of high-pressure bolt We supposed uniform shear force for bolts, so shear force on a bolt at service limit state II is defined as: SDII Pbot  430.5/12 = 35.875 kN Slip resistance of high-pressure bolt at service limit state must meet this requirement: R a ≤ Rr = R n Where: Pa = minimum shear force on flange at service limit state II (N) Rn = slip resistance of high-pressure bolt by regulation (A6.13.2.8): R n = K h K s N s Pt Where: Ns = number of friction section for each bolt, Ns = Pt = minimum rrequired tensile force in bolt by regulation (A6.12.2.8-1); Pt = 179000 N Kh= hole size coefficient by regulation (A6.12.2.8-2); K h = 1,0 Ks = surface condition coefficient by regulation (A6.12.2.8-3); We use surface type A, so Ks = 0.33 Rrslip  Rn = 1x0.33x2x179000x10-3= 118 kN SDII = 35.875 kN < Rrslip  118 kN  OK Pbot 7.4 Calculate and design girder web joint 7.4.1 Choose joint size Joints are designed by trial and error method, it means we choose joint size respectively, which are based on experience and check again If it don’t meet the requirements, we choose and check again Process is repeated until it meets the requirements We choose joint size basically: The thickness of floating web slab: t ≥ tw/2 = mm Floating web size: thickness x width x length: 10x300x610 mm - High-pressure bolt diameters: d = 20 mm - Standard hole: h = 22 mm - Number of bolt in each part of joint: N = 18 bolts Arrange bolts to rows, bolts for each row: - Distance of bolt by axial girder: Sl = 65mm - Distance of bolt by lateral girder: Sv = 65 mm We have: 7.4.2 Calculate stress in joint web 7.4.2.1 Stress in joint web at strength limit state I Marker 13.6.1.4.2 of TCVN 11823-6:2017: design shear force at strength limit state for joint web is defined as: Vuw  1.5Vu Vuw  Vu  vVn Nếu Vu  0.5vVn Nếu Vu  0.5vVn Where: 𝛷v :shear resistance coefficient Vu: shear force at strength limit state in joint position Vn: defined shear resistance of connected section Design shear force at strength limit state for joint web equals to shear force at strength limit state in joint position Design moment at strength limit state for joint web: M = Muw + Muv Where: Muw: moment caused by design shear force at joint position eccentricity with median point of a part bolt group from joint to center of joint 𝑀uv = 𝑉uw 𝑒 AASHTO LRFD 2012: these equations are put forward to define design moment, M uw and lateral design force resultant, Huw at position of web’s height to design connecting plate web and joints at strength limit state tw D ( Rh Fcf  Fcf f ncf ) 12 t D2  w ( Rh Fcf  Fcf f ncf ) 12 M uw  H uw Where: Fcf: design stress for control flange at connecting point as 5.191 formula: odd for tensioned, even for compressed (Mpa) Rcf: absolute value of ratio between Fcf and fcf for control flange Fncf: bending stress caused by load, calculated in the middle of uncontrol flange at connecting point happens to fcf, odd for tensioned, even for compressed (Mpa) Vu = 142.307 kN < 0.5𝛷vVn = 0.5x1x1988.99 = 994.495 kN -> Vuw = 1.5Vu = 1.5*142.307 = 213.4605 kN Muv = Vu.e = 142.307*72.5*103*10 -6= 10.317 kN.m 𝑀uw = (𝑅 𝐹 − 𝑅 𝑓 ) = × × 258.75 − 2.77 × (−93.3) × 10 = 304.169 kN.m M uCDI  M uw  M uv  304.169 + 10.317 = 314.486 kN.m Because girder section is double symmetric so Huw = 7.4.2.2 Stress in joint web at service limit state II VuSDII  91.038 kN M uSDII  VuSDII e  91.038*72.5*103*10-6 = 6.6 kN.m 𝑀SDII = 𝑡 𝐷 (𝑓 − 𝑓 ) 12 + fs caused by total using load II in the middle thickness of flange for smaller section at connecting point fs = 61.5 Mpa + fs caused by total using load II in the middle height of other flange at connecting point, happens together with fs in flange Fos = - 61.5 Mpa M uSDII w  tw D ( f s  f os ) = 14*710^2*(61.5+61.5)/12 = 72338350 Nmm = 72.34 kN.m 12 M SDII  M uvSDII  M uSDII w  72.34 + 6.6 = 78.94 kN.m 7.4.4 Check distance of high-pressure bolt 7.4.4.1 Minimum distance Minimum distance from center to center bolt must meet these requirements: Smin = 3d = 3*20 = 60 mm Check distance between bolt by the formula: min(S1,Sv) ≥ Smin Where: S1 = Distance of bolt by axial girder (mm) Sv = Distance of bolt by (mm) min(S1,Sv) = min(65;65) = 65 mm > Smin = 60 mm -> OK 7.4.4.2 Maximum distance To make sure closely press the joint, avoid moist, maximum distance from center to center of bolt of contigous bolt row with free side of floating slab or steel shape must meet these requirements: S ≤ (100 + 4t) ≤ 175 Where: t = smaller thickness of floating slab steel shape (mm) S = 75 < (100 + 4t) = 100+4*10 =140 < 175 -> OK 7.4.4.3: Distance to edge Minimum distance to center bolt to bar edge must meet these requirements, table A6.13.2.6.6-1 Minimum distance to center bolt to bar edge is not eight times bigger than thin floating slab’s thickness or 125mm Check distance to edge by the formula: Semin ≤ Se ≤ Semax Where: Semin = minimum distance from center bolt to edge (mm) Semax = maximum distance from center bolt to edge (mm) Se = distance from center bolt outermost to edge (mm) Semin = 38 mm < Se = 50 mm < Semax = 65 mm -> OK 7.4.5 Calculate shear force in one high-pressure bolt We calculate with high-pressure bolt in the farthest position to center of bolt group in each part of joint, the bolt has the biggest shear force Calculated shear force in bolt at the farthest position: 𝑅 Where: = 𝑉 𝑀𝑥 + 𝑁 ∑ 𝑥 +𝑦 + 𝐻 𝑀𝑦 + 𝑁 ∑ 𝑥 +𝑦 N = number of bolt from each side of joint V = design shear force (N), Vuw= 213.4605 kN ; 𝑉 M = design moment (N.mm); 𝑀 = 91.038 kN = 314.486 kN.m; 𝑀 = 78.94 kN.m H = lateral design force (N); HCD = N J = Total square distance of joint group nail to center of group nail (mm 2) J=∑(𝑥 + 𝑦 ) = 32.5 × 18 + 260 × + 195 × + 130 × + 65 × = 526012.5 mm Xmax = distance from the farthest nail by lateral to median point of each part of joint group nail xmax = 32.5 mm Ymax = distance from the farthest nail by lateral to median point of each part of joint group nail ymax= 260 mm PCDI = × + × × + + × × =158563.532 N Calculated shear force in the farthest bolt at strength limit state: PSDII = × + × × + + × × = 40263.816 N 7.4.6 Check shear resistance of high-pressure bolt Shear resistance of high-pressure bolt at strength limit state I: Rrs = 𝜑sRns Where: 𝜑s = resistance factor for bolt (A490M) sheared by regulation (A6.5.4.2) Rns = Shear resistance of high-pressure bolt by regulation, we use bolts which have length with dented thread lies out of clipping plane Rns = 0.48AbFubNs Where: 𝐴𝑏= area bolt by defined diameter (mm2), Ab = 314.2 mm2 𝐹𝑢𝑏= minimum tensioned strength of bolt (Mpa) (A6.4.3), F ub = 830Mpa Ns = number of section for each bolt, Ns = Rns = 0.48AbFubNs = 0.48*314.2*830*2 = 250354.56 N So Rr1 = 𝜑sRn1 = 0.8*250354.56 = 200283.648 N Calculated shear resistance of high-pressure bolt at strength limit state I must meet these requirements: Pstrength-I ≤ Rr1 Where: Pstrength-I = calculated shear force on bolt at the farthest position at strength limit state I Pstrength-I = 158563.532 N < Rr2 = 200283.648 N -> OK 7.4.7 Check pressure resistance of high-pressure bolt Calculated pressure resistance of high-pressure bolt at strength limit state: Rrbb = 𝜑bbRnbb Where: 𝜑bb = pressure resistance coefficient of bolt on material by regulation (A6.5.4.2), 𝜑bb=0.8 Rnbb = defined pressure resistance of high-pressure bolt by regulation Rn2 = 2.4dtFu Where: T: thickness of floating slab (mm), t = 14 mm Fu is tensioned strength of connecting material (Mpa), Fu = 450Mpa Rrbb = 𝜑bbRnbb = 0.8*2.4*20*14*450 = 241920 N Pressure resistance of high-pressure bolt at strength limit state must meet these requirements: Pstrength-I ≤ Rr2 Where: Pstrength-I = calculated shear force on bolt of the farthest position at strength limit state I Pstrength-I = 158563.532 N < Rr2 = 241920 N  OK 7.4.8 Check slip resistance of high-pressure bolt at service limit state: Rr = R n Where: Rn = defined slip resistance of high-pressure bolt by regulation (A6.13.2.8) Rn= KhKsNsPt Where: Ns = number of friction section for each bolt, Ns = Pt = minimum required tensile force in bolt by regulation (A6.13.2.8-1), P t = 179000 N Kh = Hole size coefficient by regulation (A6.13.2.8-3) We use standard hole so K h = 1.0 Ks = Surface coefficient by regulation (A6.13.2.8-3) We use surface type A, so K s =0.33 𝑅𝑛 = 𝑅𝑟 = KhKsNsPt = 1*0.33*2*179000=118140 N Calculated slip resistance of high-pressure bolt at service limit state must meet this requirement: Pservice-II ≤ Rr Where: Pstrength-II = calculated shear force on bolt of the farthest position at service limit state II Pservice-II = 40263.816 N ≤ Rr = 118140 N -> OK