STRUCTURAL STEEL ASSIGNMENT Lecturer: Student: Class: DESIGN OBJECTIVE Design a main girder, simple span on bridge way for vehicles with I-section, un- composite steel girder. Flange and girder wall connected by weld line in the factory and construction joint by high-pressure bolt. Designed lanes: 𝑛 = 2 làn 3. Main girder distance: 𝑆 =2.
Dead load on reinforced concrete deck: 𝑤 = 6. Dead load of wearing surface and utilities: 𝑤 = 6 kN/m 6. Designed car live load (with road level coefficient): HL – 93 7. Average Daily Traffic ADT: ADT = 1.5× 10 xe/ngày/lần 8.
Truck on lane ratio: k = 0. Lateral moment distribution coefficient: 𝑚𝑔 = 0. Lateral shear distribution coefficient: 𝑚𝑔 = 0. Lateral deflect distribution coefficient: 𝑚𝑔 = 0.
Lateral fatigue distribution coefficient: 𝑚𝑔 = 0. Road level coefficient: m = 0. Allowable deflection of live load: ∆ = L/800 15. Material: - Steel designed for girder: M270M Grade 345 - High-pressure bolt: ASTM A490M 16.
Design Standard: TCVN 11823-2017 II. CALCULATION AND DESIGN REQUIREMENTS A - Calculation 1. 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. Draw main section of girder and representative section.
Draw detail of deck slab, stiffener, joint 3. 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. 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 = ( ÷ )L (mm) Height of girder d should be chosen evenly to 5cm.
Select flange, girder web Flange’s width can be selected basically by this experience equation: 𝑏 = ÷ 𝑑 (𝑚𝑚 ) We have: 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 (A6.3), 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: = = 51 < 150 -> OK Flange ratio: Tensioned and compressed flange have this ratio: = = 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 ≥ 1.1 ≤ ≤ 1 -> Ok After choosing, girder section has this figure: 1.
Calculate signature geometry of girder section Ai A i hi Section hi (mm) I0i (mm4) Aiyi2 (mm4) Ii (mm4) (mm2) (mm3) Upper 7000 740 5180000 233333.33 932575000 932808333 flange Girder 9940 375 3727500 417562833 0 417562833 web Lower 7000 10 70000 233333.33 932575000 932808333 flange Total 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: ybot ytop ybotmid ytopmid Sbot Stop Sbotmid Stopmid Section (mm) (mm) (mm) (mm) (mm3) (mm3) (mm3) (mm3) Steel 375 375 365 365 6088478.3 girder 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. Calculate girder dead load Girder dead load on one meter long: WDC1 = A x γs = 23940*78. 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 = 1 m We numbered the beam section parts respectively: Moment influence line value: Influence line Mi Section xi (m) AMi (m2) (m) 1 1 0.5 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: 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.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. Table: Moment of strength limit state I MiStrengt M xi AMi Truck Tandem MiDC MiDW MiLL 𝑴𝑻𝒓𝒖𝒄𝒌 𝒊 𝑴𝑻𝒂𝒏𝒅𝒆𝒎 𝒊 hI C m m2 y1 y2 y3 y1 y2 kNm kNm kNm kNm kNm kNm 1 1 4.848 Table: Moment of service limit state II MiStrengt xi AMi Truck Tandem MiDC MiDW MiLL 𝑴𝑻𝒓𝒖𝒄𝒌 𝒊 𝑴𝑻𝒂𝒏𝒅𝒆𝒎 𝒊 hII MC m m2 y1 y2 y3 y1 y2 kNm kNm kNm kNm kNm kNm 1 1 4.357 Moment strength limit state I diagram 0 1 2 3 4 5 6 7 8 9 10 0 0 -255.42 Influence line values of shear force is calculated with the following table: Influence line Vi Section xi (m) AVi (m2) A1, vi (m2) (m) 0 0 1 5.250 Where: + xi: distance from bearing to numbers “i” section + Influence line Vi = orientation of bigger part of influence line Vi + Avi = total area of influence line Vi + A1,vi: area of influence line V1 (bigger area parts) Shear force at random section is calculated by: + With strength limit state I: + With service limit state: Where: mgv: lateral distribution for shear force (already calculated lane coefficient m) Av1: area of influence line Vi A1,vi: area of influence line Vi (bigger area parts) Arrange design trucks and tandems on influence line to find out orientation values off influence line correspond with each vehicle axle to each influence line. After arranging and calculating, we have: Calculating table for value V: Table of shear force value at strength limit state I M xi AVi A1,Vi Truck Tandem ViDC ViDW ViLL 𝑽𝑻𝒓𝒖𝒄𝒌 𝒊 𝑽𝑻𝒂𝒏𝒅𝒆𝒎 𝒊 ViStrengthI C m (m2) (m2) y1 y2 y3 y1 y2 kNm kNm kNm kNm kNm kNm 0 0 5.807 Table of shear force value at service limit state II M xi AVi A1,Vi Truck Tandem ViDC ViDW ViLL 𝑽𝑻𝒓𝒖𝒄𝒌 𝒊 𝑽𝑻𝒂𝒏𝒅𝒆𝒎 𝒊 ViStrengthI C m (m2) (m2) y1 y2 y3 y1 y2 kNm kNm kNm kNm kNm kNm 0 0 5.598 Shear force at strength limit state I diagram 290. CHECK GIRDER AT STRENGTH LIMIT STATE I 3.1: Check moment condition 3.1: Calculate stress in flange steel girder We arrange table of stresses in steel girders’ flange of middle section at strength limit state I: Section M Sbot Stop Sbotmid Stopmid fbot ftop fbotmid ftopmid (N.59 girder Where: Fbot= stress at under steel girder flange Ftop= stress on top of steel girder flange Fbotmid= stress in the middle point of lower steel girder web Ftopmid= stress in the middle point of upper steel girder web 3.2: Calculate yield moment of section Yield moment of un-composite section is defined by following equation My = FySNC Where: Fy: minimum yield strength by the requirement of steel Snc: bending resistance moment of un-composite section We have: Fy = 345 Mpa SNC = 6088478.3: Calculate plastic moment of section The height compressed plastic moment is defined as: (A6.2) With double symmetry section: Dcp= D/2= 355 mm Plastic moment of un-composite section Where: PW = FYWAW = 345x9940 = 3429300 N: Plastic force of web PC = FYCAC = 345x7000 = 2415000 N: Plastic force of upper compressed flange PT = FYTAT = 345x7000 = 2415000 N: Plastic force of lower compressed flange We have: Mp = 3429300*710/4 + 2415000*(710/2+20/2) + 2415000*(710/2+20/2) = 2371650750 N.
Check proportional of section Bended section I must be designed proportionally: IYC = inertia moment of compressed flange of steel section around axial through median point of web IYT = inertia moment of tensioned flange of steel section around axial through median point of web In this design, we choose double symmetric so IYC = IYT So = 1 ≤ 1 → 𝑂𝐾 3.5: Check modulus resistance (index A of TCVN 11823-6: 2017) These regulations only apply for straight girder’s section, which has square bearing or crosswise with angle smaller than 20 degree compare with square line of axle girder, has bulkhead or lateral beam in girder, placed in the middle of girder’s parallel lines. They musts meet these requirements: Minimum bending strength of flange and girder web do not allow to get over 485Mpa. We have Fy = 345 Mpa ≤ 485 Mpa -> OK Web must satisfy slenderness non air-free limit: 2𝐷 𝐸 < 5.7 𝑡 𝐹 With un-composite double symmetric section: 𝐷 = d/2 – tfc = 750/2 - 20 = 355 mm × = = 50.24 mm -> OK Webs must satisfy the ratio: = 1 ≥ 0.6: Bending resistance based on compression flange Check formula: 𝑀 + 𝑓𝑆 ≤ 𝜙 𝑀 𝜙f: bending resistance coefficient (𝜙 = 1.0) Mnc: defined bending resistance based on compression flange (Nmm) f1: caused by lateral bending without brace of flange (Mpa). f 1 must be chosen as the biggest stress value of lateral bending on unbrace length of flange.
