Beam Design to BS 5400 Part 3

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 x 0.15 x 3.0 = 10.8 kN/m beam = say 2.0 kN/m Nominal Live Load : = 10 x 3.0 = 30 kN/m

Load factors for ultimate limit state from BS 5400 Part 2 (or BD 37/01) 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 x 10.8)+(1.05 x 2.0)+ (1.5 x 30) 12.42 + 2.1 + 45.0

= =

60 kN/m

60 x 9.02 / 8 608 kNm

Design ultimate moment = =

Design ultimate shear = 60 x 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 x γm x γf3 / σy 608 x 1.2 x 1.1 x 106 / 265

= =

3.03 x 106 mm3

Try a 610 x 229 x 125kg/m UB (Zx = 3.222 x 106, Zp = 3.677 x 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 x 11.9 x (355/265)0.5/0.5 = 937 547 < 937 therefore web OK cl.9.3.7.3.1

compression flange: bfo = (229 - 11.9 - 2*12.7)/2 = 96 7 x 19.6 x (355/265)0.5 = 159

7tfo(355/σyf)0.5 =

96 < 159 therefore 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 9000mm cl.9.7

Slenderness: cl.9.7.1

Half wavelength of buckling = lw = L = 9000mm Mpe = Zpe x σyc Mpe = 3.677 x 106 x 265 x 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 x 0.9 x 0.94 x 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 x ((265/355)(974/974))0.5 = 104 le/ lw = 1.0 From Fig.11(b) : MR/Mult = 0.42 MR = 0.42 x Mult = 0.42 x 974 = 409 kN/m

le = 1.0 x 1.0 x 1.0 x 9000 =

cl.9.9.1.2

MD = MR /γm γf3 MD = 409 / (1.2 x 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 x 1.1) = 738 kNm > 608 kNm} Approx. Zpe required = 608/310 x 3.677x106 = 7.21 x 106 mm4 Use a 762 x 267 x 197kg/m UB Zpe = 7.167 x 106 mm3 Mpe = 7.167 x 106 x 265 x 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 = (110x57.1) / (0.9x0.94x0.81) = 9166 Hence ke maximum = 9166 / 9000 = 1.018 cl.9.6.2(a)

ke2 = 1/[1-(60EtfβδtRv/(W[L/ry]3ν4 ))] 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).

Deflection of cantilever : δ t = F a3 / 3 E I 0.000378 = 1x(744.2)3 / (3x20500x I ) I = 1.773 x 106 mm4 I of end stiffener : I = 250x15.63/12 + tx2503/12 t min = 1.3 mm Use at least 10mm plate hence OK

Try 10mm end plate and check bearing stiffener : I = 250x15.63/12 + 10x2503/12 = 13.1x106 mm4 δt = 1x(744.2)3 / (3x205000x13.1x106) = 51x10-6 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

Fs1 = 0.005(M/(df{1-(σfc/σci)2})) df = 769.6 - 25.4 = 744mm σfc = M / Zxc = 608x106 / 6.234x106 = 97.5 N/mm2 σci = π2ES/λLT2 S = Zpe/Zxc = 7.167 / 6.234 = 1.15 σci = π2 x 205000 x 1.15 / 1082 = 199 N/mm2 Fs1 = 0.005(608x106/(744{1-(97.5/199)2}))x10-3 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 x 51x10-6 = 0.102 x 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 x 2683 / 12 = 40.7 x 106 le = π x 1.0(205000 x 40.7 x 106 x 2 x 51 x 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 x 0.9 x 1.0 x 0.993 / 57.1 = 15.1 σci = π2ES/λLT2 σci = π2 x 205000 x 1.15 / 15.12 = 10205 N/mm2 Fs2 = 3.848 x 97.5 / ((10205 - 97.5) x 0.102 x 10-3) x 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 x 9 / 2 = 270 kN dL = 810 say (allow 40mm for depth of bearing) Fs3 = 270 x 0.81 x ( 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 9 x ( D - 1.5 x tf ) 9 x ( 769.6 - 1.5 x 25.4 ) x 10-3 6.6 kNm Zxc x σyc / (γm x γf3) Zxcx 265 / ( 1.2 x 1.1 ) 32.9 x 103 mm3

cl.9.14.2.1(c)

16 x 15.6 = 250mm tstiffener x 2502/6 + 250 x 15.62/6 2.2mm < 10mm therefore 10mm plate is satisfactory Use 762 x 267 x 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)*(> 270 kN Hence Section OK

Bending Capacity of Steel Beams 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.

Move the mouse pointer over the diagrams for animation effects. 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 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.

Restraint at support. The design procedure for checking a beam section is :

•

From Clause 9.6 Determine the effective length (le) 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 Cl. 9.12.5. This ensures that the compression flange can be assumed to have at least a pinned end support so the maximum k1 (Cl.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.

• • • •

From Clause 9.7 Determine the slenderness (λ

LT

) using the geometrical parameters of the beam.

From Clause 9.8 Determine the limiting moment of resistance (MR). From Clause 9.9 Determine the bending capacity (MD) From Clause 9.12.5 Determine the strength of the end bearing stiffener to provide adequate support to the compression flange.

Beam Design Example | Shear in Steel Beams | Back to Tutorial Index Last Updated : 4/06/02 For more information : Email: [email protected]