## 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

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:

γfL steel = 1.05

γfL concrete = 1.15

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 = MRm 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)

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).

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

Fs1 = 0.005(M/(df{1-(σfc/σci)2}))
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 + Δe2fc / ((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(EIce1e2)/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 τly = 1

Note: if λ < 56 then τly = 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

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