• Author / Uploaded
• Samuel Antobam

• Categories
• Viga (Estructura)
• Pandeo
• Doblar
• Braguero
• Resistencia de materiales

Citation preview

Bending Capacity of Steel Beams to BS 5400 Pt.3 Beams in bending develop tension and compression in their flanges. The bending capacity of the beam is limited by how much force can be carried by each flange. [Hold the mouse pointer over the diagrams for animation effects. (Scripts must be allowed to run in Internet Explorer)]

The tension flange acts like the string in an archer's bow and the maximum force that can be developed is limited only by the yield stress of the flange material. The compression flange acts like a strut and is susceptible to buckling before yield stress can be developed. Unless the compression flange is fully restrained then the beam may fail by lateral torsional buckling.

Lateral torsional buckling.

Clause 9.6.1 says if the second moment of area of a cross-section about the axis of bending (axis Y-Y in diagram below) is smaller than that about an axis perpendicular to it (axis X-X), the crosssection as a whole is stable against overall lateral torsional buckling and its effective length be taken as zero.

e

may

Clause 9.6.1 requires all beams to be restrained at their supports. The strength and stiffness of the restraints are checked using clause 9.12.5. The restraint is required to hold the compression flange in place and is usually provided at the support by use of the bearing stiffeners and a suitable bearing and/or a bracing system.

Clause 9.12.5 Restraint at support. The restraining system is designed to resist the force FS (or FS + FL in the case of U-frames subjected to vertical loading on cross-beams) together with any coexistent forces such as wind or frictional forces. The strength of the end restraint to resist FS is considered in isolation to the main beam(s) and any distribution effects along the length of the beam(s) are ignored.

Where only the bearing stiffener provides the torsional restraint {case (a)} then the stiffener has to be designed in accordance with Clause 9.14 as a load bearing support stiffener as well as satisfying the stiffness criteria of Clause 9.12.5.3. In cases (b) and (c) a bearing stiffener may be provided independently or incorporated into the restraining system.

The design procedure for checking a beam section is :

i.

From Clause 9.6 Determine the effective length ( e) based on the support condition of the compression flange. The Code requires the compression flange to be supported laterally at the beams‘ supports in accordance with Clause 9.12.5. This ensures that the compression flange can be assumed to have at least a pinned end support so the maximum k1 (Clause 9.6.2) that may be assumed is 1.0 for non-cantilever beams. If end diaphragms are provided to prevent the compression flange from rotating in plan then a smaller value of k1can be used.

Rotational end restraint.

1. Intermediate restraints to the compression flange can be provided to further reduce the effective length of the beam. 2. These can take the form of: a.

Plan bracing

b.

Torsional bracing

c.

U-Frames

3. a) Plan Bracing 4. This is only generally used as temporary bracing as it obstructs the deck construction on top of the beams. If the bracing is located on the underside of the compression flange then it is difficult to paint, for maintenance purposes, once the deck is in place.

5. Clause 9.12.2a requires plan bracing to be able to resist force FR + any wind and other laterally applied forces.

1. If the plan bracing is fully effective {i.e. δR ≤ 9.6.4.1.1.1} then otherwise

e

e

=

R

3

/ (40 EIc) in accordance with Clause

R

has to be calculated using Clause 9.6.4.1.1.2.

2.

b) Torsional Bracing (Clause 9.12.2b) Intermediate torsional bracing does not prevent the compression flange from moving laterally, but provides additional stiffness to resist the torsional displacement. Some of the common types of bracing in use are shown below.

1. Beams are often braced together in pairs before lifting on to their supports; this makes them stable during contruction. If the deck slab is designed to restrain the compression flange, as in composite decks with shear connectors, then the intermediate bracing will be redundant when the deck is complete. The cost of removing the bracing and repairing the steelwork protection system is usually more expensive than the reclaim value of the steelwork and is therefore usually left in place. If the bracing is left in place then it will pick up load when the beams deflect under traffic loading and will therefore need to be checked for this loading condition. Torsional restraints need to be checked in accordance with Clause 9.12.2b to be capable of resisting two transverse forces FR applied to the flanges. The effective length

e

is then calculated using Clause 9.6.4.1.2.

2.

c) U-Frames U-frames are a characteristic of the 'Half Through' deck construction. The 'U' shape of the U-frame is formed by two vertical web stiffeners and the deck cross beam. The web stiffeners are attached to the compression flanges of the main edge girders and are able to resist the lateral buckling movement of the flange (see Clause 9.12.3.1). The stiffness of the U-frame is evaluated by adding the stiffness of its components due to a unit force acting at the level of the compression flange and is determined using Clause 9.6.4.1.3: δR = d13 / 3EI1 + uBd22 / EI2 + fd22 i.

The deflection of the web stiffener acting as a cantilever (δi = d13 / 3EI1).

ii.

The effect from the cross beam bending (δii = uBd22 / EI2).

iii.

The flexibility of the joint between the cross beam and the vertical stiffeners (δiii = fd22).

δR = δi + δii + δiii For end U-frames:

1. The effective length

e

δe = δi + δii + δiii

is then calculated using Clause 9.6.4.1.1.2 with

R

equal to the

intermediate U-frame spacing. An example of determining the effective length for beams with U-frame restraint is given in the workshop section for BD 56/10 (The Assessment of Steel Highway Bridges).

ii.

From Clause 9.7 Determine the slenderness (λLT) using the geometrical parameters of the beam.

iii.

From Clause 9.8 Determine the limiting moment of resistance (MR).

iv.

From Clause 9.9 Determine the bending capacity (MD)

v.

From Clause 9.12.5 Determine the strength of the end restraint to provide adequate support to the compression flange.

Steel Beam Design Example to British Standards Beam Design to BS 5400 Part 3 : 2000 This example will illustrate the procedures to design a steel beam to BS 5400 Part 3. The concrete deck and live loading are included to demonstrate the use of load factors only, they do not represent a solution for a deck design. Problem: Design a simply supported beam which carries a 150mm thick concrete slab together with a nominal live load of 10.0 kN/m2 . The span of the beam is 9.0m centre to centre of bearings and the beams are spaced at 3.0m intervals. The slab will be assumed to be laid on top of the beams with no positive connection to the compression flange.

γconc. = 24kN/mm3 Loading per beam (at 3.0m c/c) Nominal Dead Loads : slab = 24 × 0.15 × 3.0 = 10.8 kN/m beam = say 2.0 kN/m Nominal Live Load : = 10 × 3.0 = 30 kN/m Load factors for ultimate limit state from BS 5400 Part 2 Table 1: Dead Load: γfL steel = 1.05 γfL concrete = 1.15 Live Load: γfL HA = 1.50 Total load for ultimate limit state: = (1.15 × 10.8)+(1.05 × 2.0)+ (1.5 × 30) = 60 kN/m Design ultimate moment = 60 × 9.02 / 8 = 608 kNm Design ultimate shear = 60 × 9.0 / 2 = 270 kN BS 5400 Pt. 3 1) Design for Bending Use BS EN 10 025 steel grade S275, then nominal yield stress σy = 265 N/mm2 Approximate modulus required: = M × γm × γf3 / σy = 608 × 1.2 × 1.1 × 106 / 265 = 3.03 × 106 mm3

Try a 610 × 229 × 125kg/m UB (Z× = 3.222 × 106, Zp = 3.677 × 106) cl.9.3.7

Check for compact section: cl.9.3.7.2

web: web depth = 547; and m = 0.5 34tw(355/σyw)0.5/m = 34 × 11.9 × (355/265)0.5/0.5 = 937 547 < 937 ∴ web OK cl.9.3.7.3.1

compression flange: bfo = (229 - 11.9 - 2*12.7)/2 = 96 7tfo(355/σyf)0.5 = 7 × 19.6 × (355/265)0.5 = 159 96 < 159 ∴ flange OK Hence section is compact. cl.9.6

Determine Effective Length: cl.9.6.2

le = k1 k2 ke k1 L = 1.0 (flange is free to rotate in plan) k2 = 1.0 (load is not free to move laterally) ke = 1.0 (check later for initial value) L= 9000mm le = 1.0 × 1.0 × 1.0 × 9000 = 9000mm cl.9.7

Slenderness: cl.9.7.1

Half wavelength of buckling = lw = L = 9000mm Mpe = Zpe × σyc Mpe = 3.677 × 106 × 265 × 10-6 = 974kNm cl.9.7.2

λLT = le k4 ην / ry k4 = 0.9 η = 0.94 (From Fig. 9(b): MA/MM = MB/MA = 0) λF = le/ry(tf/D) λF = 9000/49.6 (19.6/611.9) = 5.81 i = Ic/(Ic+It) i = 0.5 ν = 0.78 (from Table 9) λLT = 9000 × 0.9 × 0.94 × 0.78 / 49.6 = 120 cl.9.8.

Limiting moment of resistance: λLT ((σyc/355)(Mult/Mpe))0.5 Section is compact, hence Mult = Mpe = 974kNm λLT ((σyc/355)(Mult/Mpe))0.5 = 120 × ((265/355)(974/974))0.5 = 104

le/ lw = 1.0 From Fig.11(b) : MR/Mult = 0.42 MR = 0.42 × Mult = 0.42 × 974 = 409 kN/m cl.9.9.1.2

MD = MR /γm γf3 MD = 409 / (1.2 × 1.1) = 310 kNm < 608 kNm hence section too small. {Note: If the compression flange is cast into the deck slab then le = 0 (cl.9.6.4.2.1) which results in MR = Mult = 974 kNm giving MD = 974 / (1.2 × 1.1) = 738 kNm > 608 kNm} Approx. Zpe required = 608/310 × 3.677×106 = 7.21 × 106 mm4 Use a 762 × 267 × 197kg/m UB Zpe = 7.167 × 106 mm3 Mpe = 7.167 × 106 × 265 × 10-6 Mpe = Mult = 1899kNm Repeating the procedure above will show : Section is compact λF = 5.2 ν= 0.81 λLT = 108 MR/Mult = 0.51 MD = 733 kNm > 608 hence OK Check effect of assuming ke = 1 (cl.9.6.4.2.1) MD / Mult = 733 / 1899 = 0.39 From fig. 11(b) : λLT((σyc/355)(Mult/Mpe))0.5 = 110 giving λLT = 127 λLT = le k4 ην / ry approx le = (110×57.1) / (0.9×0.94×0.81) = 9166 Hence ke maximum = 9166 / 9000 = 1.018 cl.9.6.2(a) 2

3 4

ke = 1/[1-(60EtfβδtRv/(W[L/ry] ν ))] Rv/W = 0.5 (load causing max moment in beam) E = 205 000 (cl.6.6) tf = 25.4 β = 1.0 ν = 0.81 L/ry = 9000 / 57.1 Hence max δt = 0.000378 mm cl.9.14.2.1

Effective section for bearing stiffener :

The ends of bearing stiffeners should be closely fitted or adequately connected to both flanges (cl.9.14.1). Hence the compressive edge of the bearing stiffener is fully restrained at the point of maximum bending. Therefore yield stress can be developed in both tension and compression edges of the bearing stiffener (λLT = 0). cl.9.14.2.1

Deflection of cantilever : δt = F a3 / 3 E I 0.000378 = 1×(744.2)3 / (3×20500× I ) I = 1.773 × 106 mm4 I of end stiffener : I = 250×15.63/12 + t×2503/12 t min = 1.3 mm Use at least 10mm plate hence OK Try 10mm end plate and check bearing stiffener : I = 250×15.63/12 + 10×2503/12 = 13.1×106 mm4 δt = 1×(744.2)3 / (3×205000×13.1×106) = 51×10-6 mm/N cl.9.12.5

End stiffeners have to be provided to support the compression flange for a pinned end condition (k1 = 1.0 in cl.9.6.2). cl.9.12.5.2.2 2

Fs1 = 0.005(M/(df{1-(σfc/σci) })) df = 769.6 - 25.4 = 744mm σfc = M / Zxc = 608×106 / 6.234×106 = 97.5 N/mm2 σci = π2ES/λLT2 S = Zpe/Zxc = 7.167 / 6.234 = 1.15 σci = π2 × 205000 × 1.15 / 1082 = 199 N/mm2 Fs1 = 0.005(608×106/(744{1-(97.5/199)2}))×10-3 Fs1 = 5.4 kN cl.9.12.5.2.3

Fs2 = β (Δe1 + Δe2)σfc / ((σci - σfc)Σδ) Δe1 = Δe2 = D/200 = 769.6 / 200 = 3.848 β=1

Σδ = 2δt = 2 × 51×10-6 = 0.102 × 10-3 σci is to be determined using le from 9.6.4.1.1.2b). le = πk2(EIc(δe1+δe2)/L)0.5 Ic = 25.4 × 2683 / 12 = 40.7 × 106 le = π × 1.0(205000 × 40.7 × 106 × 2 × 51 × 10-6 / 9000)0.5 = 967mm λF = (le/ry)(tf/D) = (967/57.1)(25.4/769.6) = 0.559 ν = 0.993 (From Table 9) λLT = lek4ην/ry = 967 × 0.9 × 1.0 × 0.993 / 57.1 = 15.1 σci = π2ES/λLT2 σci = π2 × 205000 × 1.15 / 15.12 = 10205 N/mm2 Fs2 = 3.848 × 97.5 / ((10205 - 97.5) × 0.102 × 10-3) × 10-3 Fs2 = 0.4 kN cl.9.12.5.2.4

Assume no camber is provided to the beams and the bearings are aligned square to the longitudinal axis of the beam, then : Fs3 = RdL(Δ/D+θLtanα)/D α=0 R = 60 × 9 / 2 = 270 kN dL = 810 say (allow 40mm for depth of bearing) Fs3 = 270 × 0.81 × ( 1/200 ) / 0.77 = 1.4 kN cl.9.12.5.2.5

Bearings are aligned square to the longitudinal axis of the beam : Hence Fs4 = 0 cl.9.12.5.2.1

Fs = 5.4 + 0.4 + 1.4 + 0 = 7.2 kN Allowing additional effect for wind load ( which is generally small compared to Fs ) then say Fs = 9 kN Moment at base of stiffener = 9 × ( D - 1.5 × tf ) M = 9 × ( 769.6 - 1.5 × 25.4 ) × 10-3 = 6.6 kNm Bending capacity of stiffener = Zxc × σyc / (γm × γf3) MD = 6.6 × 106 = Zxc× 265 / ( 1.2 × 1.1 ) Zxc = 32.9 × 103 mm3 cl.9.14.2.1(c)

Portion of web plate = 16tw= 16 × 15.6 = 250mm Zxc = tstiffener × 2502/6 + 250 × 15.62/6 Hence tstiffener = 2.2mm < 10mm therefore 10mm plate is satisfactory. Use 762 × 267 × 197 kg/m UB with 10mm thick end bearing stiffener. 2) Design for Shear cl.9.9.2.2

web thickness = 15.6mm dwe = 685.8mm λ = (dwe/tw)*(σyw/355)0.5 λ = (685.8/15.6)*(265/355)0.5 = 38 From Figures 11 to 17 τl/τy = 1 Note: if λ < 56 then τl/τy = 1

Transverse web stiffeners will not improve the shear strength of the web. τl = τy = σyw/1.732 = 153 N/mm2 VD = (tw(dw - hh)/(γm γf3))τl dw = D = 769.6 hh = 0 γf3 = 1.1 (Clause 4.3.3) γm = 1.05 (Table 2) VD = ((15.6 * 769.6)/(1.05 * 1.1)) * 153 * 10-3 kN VD = 1590 kN ⟩⟩ 270 kN Hence Section OK

Shear Capacity of Steel Beams to BS 5400 The Code allows beams to be designed to fail at Ultimate Limit State. Shear failure of a beam, with transverse web stiffeners, develops in three stages : [Hold the mouse pointer over the diagrams for animation effects. (Scripts must be allowed to run in Internet Explorer)]

Stage 1 Initially pure shear is developed within the unbuckled web.

Stage 2 A diagonal tension field is developed when the web has buckled.

Stage 3 Plastic hinges develop in the flanges which produce a mechanism for collapse to occur.

When the web buckles, at the limit of stage 1, the diagonal compressive stress in the web is assumed to be at at its maximum. The diagonal tensile stress is usually far from its maximum at this stage. If transverse web stiffeners are provided then the beam can carry additional load by truss action (stage 2).

The web forms the tie members of the truss (with its spare tensile capacity), whilst the transverse stiffeners form the strut members. The formation of this truss action is referred to in the Code as "tension field action".

The top and bottom flanges act as the top and bottom chords of the truss. Bending of these members is increased as the tension in the web increases. Failure occurs when plastic hinges form in the flanges to produce a mechanism (stage 3). Figures 11 to 17 in the Code enable the web to be designed to achieve an ultimate capacity utilising this mechaism failure. The term mfw in the graphs represents the plastic moment of resistance of the flanges. If Figure 11 (mfw = 0) is used for design then this will ensure that the plastic hinges are not developed in the flanges.

The design rules for determining the initial web section are :

•

From Clause 9.9.2.2 Use Figure 11 (mfw = 0); this ensures that the flanges are prevented from buckling by tension field action.

•

Use a transverse stiffener spacing to web depth ratio of 1.

•

Use a web thickness with reserve capacity.

BD 56/10 ASSESSMENT OF STEEL HIGHWAY BRIDGE STRUCTURES

Live Load rating of a plate girder with U-frame restraint.

Clause 9.6.4.1.3 Beams with U-frame restraints

Problem: Based on the moment capacity, what is the live load rating of a riveted plate girder with U-frame restraint. ?

Example: Carriageway = 6m wide Verges = 1m and 0.64m wide Construction Depth at Verges = 0.9m Construction Depth at Carriageway = 0.8m

Main Girders - Mid Span Section: Top and Bottom Flanges: 508 × 38mm plate Web: 2286 × 9 mm plate Main Girders - End Section: Top and Bottom Flanges: 508 × 25mm plate Web: 2286 × 11 mm plate Flange Angles: 102 × 102 × 14mm

Cross Girders : Top Flange: 254 × 19mm plate Flange: 356 × 12 mm plate Web: 508 × 9mm plate Flange Angles: 89 × 89 × 12mm

Web stiffeners are positioned at each cross girder and consist of 12mm gusset plates with 89 × 89 × 12mm angles alternating with 127 × 76 × 11mm T's. The main girders are bedded on 508 × 1118 × 12mm thick plates on large masonry pad stones. Records show that the bridge was built in 1925 so, from BD 21/01 Clause 4.3, the matal shall be assumed to be of steel with a characteristic yield strength of 230 N/mm2.

BD 21/01 Dead Load : Cl. 4.1. Table 4.1.

Main Girder self weight : Cross sectional area of plates and angles (from spreadsheet ″Sectionprop5610.xls″) = 69822mm2/m Allowing 10% for web stiffeners and gusset plates then: Self weight = 69822 × 1.1 × 7850 × 9.81 × 10-9 = 5.91kN/m

Cross Girder self weight : Cross sectional area of plates and angles (from spreadsheet ″Sectionprop5610.xls″) = 21638mm2/m Allowing 10% for fittings then: Self weight = 21638 × 1.1 × 7850 × 9.81 × 10-9 = 1.83kN/m

Infill concrete :

Radius of jack arch = (6352 + 4052)/(2 × 405) = 700mm ψ = 2Sin-1(635/700) = 130° Area of arch = 0.5 × r2(

ψ° / 180° - Sinψ°) = 0.368m2

Area of infill concrete = (1.626 - 0.009)(0.508 + 0.019) - 0.368 = 0.484m2 Weight of infill concrete = 0.484 × 2300 × 9.81 × 10-3 = 10.92kN/m

Depth of Road construction above jack arch = (800 - 12 - 508 - 19) = say 260mm From Note to Table 3.1: surfacing depth = 100mm Therefore fill depth = 260 - 100 = 160mm Weight of fill = 0.16 × 2200(miscellaneous fill) × 9.81 × 10-3 = 3.45kN/m2

Weight of surfacing = 0.1 × 2300(Asphalt) × 9.81 × 10-3 = 2.26 kN/m2

Weight of verges (above fill) = 0.2 × 2300(Asphalt) × 9.81 × 10-3 = 4.51 kN/m2 Cl. 3.7. Table 3.1

Applying Partial Load Factors γfL we get: Main girder self weight = 1.05 × 5.91 = 6.21 kN/m Cross girder self weight = 1.05 × 1.83 = 1.92 kN/m Concrete = 1.15 × 10.92 = 12.56 kN/m Fill = 1.2 × 3.45 = 4.14 kN/m2 Surfacing = 1.75 × 2.26 = 3.96 kN/m2

Verges = 1.2 × 4.51 = 5.41 kN/m2

The load path for distributed loads on the deck to the main girders is via the cross girders. These loads could be represented more accurately by point load reactions at the end of the cross girders, however it will be sufficiently accurate to represent them as uniformly distributed loads to simplify the calculation.

Reactions from the loads applied as a unifomly distributed load to each main girder: Self Weight = 6.21 kN/m Cross Girders = 1.92 × 8.148 / (2 × 1.626) = 4.81 kN/m Concrete = 12.56 × 8.148 / (2 × 1.626) = 31.47 kN/m Fill = 4.14 × 8.148 / 2 = 16.85 kN/m Sub-Total = 59.34 kN/m

Uniformly distributed reactions from the surfacing and verge loads to each main girder: North Main Girder: Surfacing = 3.96 × 6.0 × 3.894 / 8.148 = 11.36 kN/m North Verge = 5.41 × 1.0 × 7.394 / 8.148 = 4.91 kN/m South Verge = 5.41 × 0.64 × 0.574 / 8.148 = 0.24 kN/m Sub-Total = 16.51 kN/m South Main Girder: Surfacing = 3.96 × 6.0 × 4.254 / 8.148 = 12.40 kN/m North Verge = 5.41 × 1.0 × 0.754 / 8.148 = 0.50 kN/m South Verge = 5.41 × 0.64 × 7.574 / 8.148 = 3.22 kN/m Sub-Total = 16.12 kN/m Total Assessment Dead Load on each girder: North Main Girder = 59.34 + 16.51 = 75.85 kN/m South Main Girder = 59.34 + 16.12 = 75.46 kN/m

BD 56/10 Cl. 16A Fig. 16.1A

The Effective Span (not to be confused with the effective length

e)

of the main girders, for

calculating the load effects only , may be determined by calculating the position of the 'Centroid of Pressure' in accordance with the pressure diagram shown in Fig. 16.1A of BD 56/10. Effective Span (from spreadsheet ″EffectiveSpan.xls″) = 17.725m

Maximum Mid span Dead Load Moment = 75.85 × 17.7252 / 8 = 2980 kNm Cl. 9.6.4.1.3

Restraint to the compression flange is only provided by U-frames. Assume only web stiffeners with gusset plates, at 3.251m centres, are effective in restraining the compression flange. Note: The 'T' stiffeners will act as web stiffeners but are unlikely to be strong enough to resist U-frame forces FR and Fc (calculated below).

Effective width of web plate in stiffener = 16t = 16 × 9 = 144mm

I1 = IXX: Gusset = 12 × 5083 / 12 = 131.097×106mm4 Angles = 2(12 × 1873 + 77 × 333) / 12 = 13.540×106mm4 Remaining web = 2 × 49 × 93 / 12 = 0.006×106mm4 I1 = (131.097 + 13.540 + 0.006)×106 = 144.643×106mm4 I2 = second moment of area of cross girder (from spreadsheet ″EffectiveSpan.xls″) = Ixxnet = 1.056×109mm4 d1 = 38/2 + 2286 - (12 + 508 + 19) = 1766mm d2 = 1766mm + depth to centroid of cross girder (ycnet)(from spreadsheet ″EffectiveSpan.xls″) = 1766 + 251 = 2017mm u = 0.5 for an outer girder B = 8148mm f = 0.5×10-10 for riveted cleat joint E = 205000 N/mm2 (BD 21/01 Table 4.2) δR = (d13/3EI1) + (uBd22/EI2) + fd22 δR = [17663 / (3 × 205000 × 144.643 × 106)] + [0.5 × 8148 × 20172 / (205000 × 1.056 × 109)] + 0.5 × 10-10 × 20172 δR = 6.192×10-5 + 7.656×10-5 + 20.341×10-5 = 34.189×10-5 Cl. 9.6.4.1.1.2

From Clause 9.6.4.1.3: R

= 3251mm

δe = δe, max = δR = 34.189×10-5 (all U-frame stiffeners are the same) a) e

= k2k3k5

1

but not less than k3

R

nor greater than L

L = Distance between end U-frames = 19304 - (2 × 1118) = 17068mm k2 = 1.0 (load applied via cross girder to bottom flange) k3 = 1.0 (initial conservative assumption) Ic = 38 × 5083 / 12 = 415 × 106mm4 1

= (EIc RδR)0.25

1

= (205000 × 415 × 106 × 3251 × 34.189×10-5)0.25 = 3118mm

Χ=

1

3

/ (√2EIcδe, max)

Χ = 31183 / (√2 × 205000 × 415 × 106 × 34.189×10-5) = 0.737 k5 = 2.22 + 0.69 / (Χ + 0.5) = 2.22 + 0.69 / (0.737 + 0.5) = 2.78 e

= k2k3k5

e

= 1.0 × 1.0 × 2.78 × 3118 = 8668mm

Minimum

1

e

= k3

R

= 1.0 × 3251mm < 8668mm ∴

e

= 8668mm ( 2701 kNm Hence mid span section OK for bending effects providing that the U-frames comply with Clause 9.12.3. Note: If the moment capacity of the girder is insufficient for the dead + live load moment then a suitable K factor (Clause 5.25) will need to be applied to the live live moment. The live load moment of 2701 kNm already includes a K factor of 0.91 so the revised K factor needs to be applied to the full moment of 2701 / 0.91 = 2968 kNm.

BD 56/10 Cl. 9.12.3.1

The cross members of each U frame may be assumed to be restrained by the concrete infill against movement in a direction normal to their compression flange. Cl. 9.12.3.2

Strength: U-frames need to be checked for force FR (plus wind load if critical) calculated from Clause 9.12.2. Also a force Fc, from Clause 9.12.3.3, resulting from loads on the cross member, needs to be added to FR. Cl. 9.12.2

FR = [ σfc / ( σci - σfc )] ×

w

/ (667δR) but not greater than

[ σfc / ( σci - σfc )] × (n + 1)EIc / (16.7n w

R

2

)

= 17068 mm

δR = 34.189 × 10-5 Ic = 415 × 106 mm4 R

= 3251 mm

Zpe net = 58.815 ×106 mm3 Zxc net = 59.922 ×106 mm3 λLT = 58.34 n = 5 (number of restraints in the half wavelength of buckling) M = MDead + MLive = 2980 + 2701 = 5681 kNm

S = Zpe / Zxc = 58.815 / 59.922 = 0.982 σci =

2ES

/ λLT2 =

2

× 205000 × 0.982 / 58.342 = 584 N/mm2

σfc = M / Zxc = 5681 / 59.922 = 94.8 N/mm2 FR = [ σfc / ( σci - σfc )] ×

w

/ (667δR) = [ 94.8 / (584 - 94.8)] × 17068 × 10-3 / (667 × 34.189 ×

10-5) = 14.5 kN [ σfc / ( σci - σfc )] × (n + 1)EIc / (16.7n

R

2)

= [ 94.8 / (584 - 94.8)] × (5 + 1) × 205000 × 415

×106 × 10-3 / (16.7 × 5 × 32512) = 112 kN ∴ FR = 14.5 kN (< 112 kN) Cl. 9.12.3.3

Fc = θd2 / [ 1.5δR + {

R

3

/ (12EIc)} ]

θ is the difference in rotation between a cross member and the mean rotations of the cross members on either side of it. As the dead load and HA UDL will have a similar effect on the rotaion for all cross members then θ for these loads = 0. The KEL however will be applied to only one cross member so we need to determine θ for the KEL only (average θ of adjacent members = 0).

To calculate θ we can use the Moment Area theorem which states: The change in slope of the deflected shape of a beam between points A and B (A at end and B at mid-span) is equal to the

area under the M/EI diagram between those two points. The bending moment M at the centre of the cross girder with the two KEL in the lanes at W2 shown above: Reactionnorth = 112.2(5.144 + 2.144) / 8.148 = 100.4 kN Mid-span moment M = [100.4 × 4.074] - [112.2 × {1.25 + (4.074 - 3.004)}2 / (2.5 × 2)] = 288 kNm It will be sufficiently accurate to assume a parabolic moment diagram as produced by a UDL. θ = Area of M/EI diagram between mid-span and end = 2 × 4074 × 288 ×

6

/ (3 × 205000 ×

1.056 × 109) = 3.613 × 10-3 d2 = 2017 mm δR = 34.189 × 10-5 Ic = 415 × 106 mm4 R

= 3251 mm

Fc = 3.613 × 10-3 × 2017 × 10-3 / [ 1.5 × 34.189 × 10-5 + { 32513 / (12 × 205000 × 415 × 106)} ] = 13.3 kN Cl. 9.12.3.2

FR + Fc = 14.5 + 13.3 = 27.8 kN

Wind loading should be calculated in accordance with Clause 5.3 of BD 37/01 however a value of 6 kN/m2 may be used for general cases (see Clause 5.3.1 of BD 37/01). Wind load on face of girder (per U-frame) = 6 × 2.362 × 3.251 = 46.1 kN Note: Wind loading is Combination 2 loading which has a γfL = 1.25 for HA loading instead of 1.5 as used above, so FR and Fc may be reduced when combined with wind load. However continue with the calculated values for the example.

Moment in stiffener at top of cross girder = (27.8 × 1.766) + (46.1 × 0.604) = 77 kNm

Check bending capacity of stiffener:

IXX = 144.643×106mm4 IYY = 9 × 2883/12 + 24 × 1903/12 + 154 × 363/12 + 321 × 123/12 = 32.279 × 106 mm4 Area = 288 × 9 + 499 × 12 + 4(89 + 77) × 12 = 16548 mm2 Cl. 9.7.3.2

34tw /m × √(355/σyw) = [34 × 12 / 0.5] × [√(355/230)] = 1014 > 508 hence section is compact Cl. 9.6.4 Table 8 e

= 1.4L = 1.4 × 1766 = 2472mm Cl. 9.7.2

λLT =

e

k4η

/ ry

t f = 0 ∴ λF = 0 and

= 1.0

k4 = 1.0 Figure 10b)

MA = -M and MM = MB = 0: MA / MM = -∞ and MB / MA = 0 then η = 0.76 ry = √(Iy/A) = √ ( 32.279 × 106 / 16548) = 44 mm λLT = 2472 × 0.76 / 44 = 43 Cl. 9.8 e

/

w

=1

Mult = Mpe (section compact) Zpe = 2 × [ (12 × 2542 / 2) + (24 × 93.52 / 2) + (154 × 16.52 / 2) + (98 × 16.52 / 2) = 1.053 × 106 mm3 Mpe = Zpeσyc = 1.053 × 106 × 230 × 10-6 = 242 kNm λLT = 2472 × 0.76 / 44 = 43 λLT√[(σyc/355)(Mult/Mpe)] = 43 × √(230 / 355) = 35 Figure 11

MR / Mult = 0.94 Cl. 9.9.1.2

MD = MR / (γmγf3 = 0.94 × 242 / (1.05 × 1.1) = 197 kNm 197 > 77 ∴ the intermediate stiffeners are strong enough to restrain the main girder compression flange by U-frame action. Cl. 9.12.5.1

The main girders also need to be restrained at their supports. Note: Where the restraint provided is less than required to resist force FS then the slenderness parameter λLT must be modified as described in the addition to Clause 6.1 in BD 56/10. Cl. 9.12.5.2.1

Restraining force FS = FS1 + FS2 + FS3 + FS4 Cl. 9.12.5.2.2

FS1 = 0.005 M / [df {1 - ( σfc / σci )2 } ] M = 5681 kNm df = 2286 + 25 = 2311 mm (25mm thick flanges at support) σfc = 94.8 N/mm2 σci = 584 N/mm2 FS1 = 0.005 × 5681 × 103 / [2311 × {1 - ( 94.8 / 584 )2 }] = 12.7 kN Cl. 9.12.5.2.3

FS2 = β( Δe1 + Δe2 ) σfc /({σci - σfc} Σδ ) β = 1.0 Δe1 = Δe2 = D / 200 = (2286 + 50) / 200 = 11.7 mm Σδ = δt1 + δt2 δt = d13 / 3EI1 = 17663 / (3 × 205000 × 144.643×106) = 6.2 × 10-5 Σδ = 2 × 6.2 × 10-5 = 12.4 × 10-5 FS2 = 1.0 × 2 × 11.7 × 94.8 × 10-3/ ({584 - 94.8} × 12.4 × 10-5) = 36.6 kN Cl. 9.12.5.2.4

FS3 = 0 as the load is applied directly through the bottom flange and bearing plate Cl. 9.12.5.2.5

FS4 = 0 as there is no skew. Cl. 9.12.5.2.6

FL = d2θ / [2.5δR + (δe / 2) + {

R

3

/ (3EIc)}]

d2 = 2017 mm θ = 3.613 × 10-3 radians δR = δe = 34.189 × 10-5 mm R

= 3251mm

Ic = 415 × 106 mm4 FL = 2017 × 3.613 × 10-3 × 10-3 / [(2.5 × 34.189 × 10-5) + (34.189 × 10-5 / 2) + {32513 / (3 × 205000 × 415 × 106)}] = 6.3 kN FS + FL = 12.7 + 36.6 + 0 + 0 + 6.3 = 55.6 kN Add in wind effects as for the internal U-frame, this is over-estimated as the end U-frames are closer together. Moment in stiffener at top of cross girder = (55.6 × 1.766) + (46.1 × 0.604) = 126 kNm Capacity of end stiffener is the same as the internal stiffener = 197 kNm > 126 ∴ end U-frame is

adequate.

Having established that the U-frames are adequate then the main girder mid span section is able carry the full range of vehicles up to 40/44 tonnes gross weight as described in BD 21/01.

Note: All sections of the girder where plate sizes change (flange curtailment for example) or where corrosion has reduced the plate thickness, need to be checked for adequacy. Web plates also need to checked for shear effects; joints need to considered to ensure they can transfer loads to the relevant elements. The live load capacity is determined by the weakest element in the bridge and a comprehensive check of every element needs to be caried out.