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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/m^{2}
. 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.

γ_{conc.} = 24kN/mm^{3}

__ 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:

γ_{fL} steel = 1.05

γ_{fL} concrete = 1.15

Live Load:

γ_{fL} 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 × 9.0^{2} / 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 σ_{y
}= 265 N/mm^{2}

Approximate modulus required:

= M × γ_{m} × γ_{f3} / σ_{y}

= 608 × 1.2 × 1.1 × 10^{6} / 265 = ** 3.03 × 10**^{6 }mm^{3}

Try a **610 × 229 × 125**kg/m UB (Z_{×} = 3.222 × 10^{6}, Z_{p} = 3.677 × 10^{6})

cl.9.3.7

Check for compact section:

cl.9.3.7.2

web:

web depth = 547; and m = 0.5

34t_{w}(355/σ_{yw})^{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:

b_{fo} = (229 - 11.9 - 2*12.7)/2 = 96

7t_{fo}(355/σ_{yf})^{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

l_{e} = k_{1} k_{2} k_{e}

k_{1} L = 1.0 (flange is free to rotate in plan)

k_{2} = 1.0 (load is not free to move laterally)

k_{e} = 1.0 (check later for initial value)

L= 9000mm

l_{e} = 1.0 × 1.0 × 1.0 × 9000 = 9000mm

cl.9.7

Slenderness:

cl.9.7.1

Half wavelength of buckling = l_{w} = L = 9000mm

M_{pe} = Z_{pe} × σ_{yc}

M_{pe} = 3.677 × 10^{6} ×
265 × 10^{-6} = 974kNm

cl.9.7.2

λ_{LT}
= l_{e} k_{4} ην / r_{y}

k_{4} = 0.9

η = 0.94 (From Fig. 9(b): M_{A}/M_{M} = M_{B}/M_{A} = 0)

λ_{F} = l_{e}/r_{y}(t_{f}/D)

λ_{F} = 9000/49.6 (19.6/611.9) = 5.81

i = I_{c}/(I_{c}+I_{t})

i = 0.5

ν = 0.78 (from Table 9)

λ_{LT} = 9000 × 0.9 × 0.94 × 0.78 / 49.6 = 120

cl.9.8.

Limiting moment of resistance: λ_{LT} ((σ_{yc}/355)(M_{ult}/M_{pe}))^{0.5}

Section is compact, hence M_{ult} = M_{pe} = 974kNm

λ_{LT} ((σ_{yc}/355)(M_{ult}/M_{pe}))^{0.5} = 120 × ((265/355)(974/974))^{0.5} = 104

l_{e}/ l_{w} = 1.0

From Fig.11_{(b)} : M_{R}/M_{ult} = 0.42

M_{R} = 0.42 × M_{ult} = 0.42 × 974 = 409 kN/m

cl.9.9.1.2

M_{D} = M_{R} /γ_{m} γ_{f3}

M_{D} = 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 l_{e}
= 0 (cl.9.6.4.2.1) which results in

M_{R} = M_{ult} = 974
kNm giving M_{D}
= 974 / (1.2 × 1.1) = 738 kNm
> 608 kNm}

Approx. Zpe required = 608/310 × 3.677×10^{6} = 7.21 × 10^{6} mm^{4}

Use a **762 × 267 × 197**kg/m UB

Z_{pe} = 7.167 × 10^{6} mm^{3}

M_{pe} = 7.167 × 10^{6} × 265 × 10^{-6}

M_{pe} = M_{ult} = 1899kNm

Repeating the procedure above will show :

Section is compact

λ_{F} = 5.2

ν= 0.81

λ_{LT} = 108

M_{R}/M_{ult} = 0.51

**M _{D} = 733 kNm** > 608 hence OK

Check effect of assuming ke = 1 (cl.9.6.4.2.1)

M_{D}
/ M_{ult} = 733 / 1899 = 0.39

From fig. 11(b) : λ_{LT}((σ_{yc}/355)(M_{ult}/M_{pe}))^{0.5}
= 110

giving λ_{LT}
= 127

λ_{LT}
= l_{e} k_{4} ην / r_{y}

approx le = (110×57.1) / (0.9×0.94×0.81) = 9166

Hence k_{e} maximum = 9166 / 9000 = 1.018

cl.9.6.2(a)

k_{e}^{2}
= 1/[1-(60Et_{f}βδ_{t}R_{v}/(W[L/r_{y}]^{3}ν^{4}
))]

R_{v}/W = 0.5 (load causing max moment in beam)

E = 205 000 (cl.6.6)

t_{f} = 25.4

β = 1.0

ν = 0.81

L/r_{y} = 9000 / 57.1

Hence max δ_{t} = 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 (λ_{LT
}= 0).

cl.9.14.2.1

Deflection of
cantilever :

δ_{t} = F a^{3} / 3 E I

0.000378 = 1×(744.2)^{3} / (3×20500× I )

I = 1.773 × 10^{6} mm^{4}

I of end stiffener :

I = 250×15.6^{3}/12 + t×250^{3}/12

t min = 1.3 mm

Use at least 10mm plate hence OK

Try
10mm end plate and check bearing stiffener :

I = 250×15.6^{3}/12 + 10×250^{3}/12 = 13.1×10^{6}
mm^{4}

δ_{t} = 1×(744.2)^{3} / (3×205000×13.1×10^{6})
= 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 (k_{1} = 1.0 in cl.9.6.2).

cl.9.12.5.2.2

F_{s1} = 0.005(M/(d_{f}{1-(σ_{fc/}σ_{ci})^{2}}))

d_{f} = 769.6 - 25.4 = 744mm

σ_{fc} = M / Zxc = 608×10^{6} / 6.234×10^{6} = 97.5 N/mm^{2}

σ_{ci} = π^{2}ES/λ_{LT}^{2}

S = Z_{pe}/Z_{xc} = 7.167 / 6.234 = 1.15

σ_{ci} = π^{2} × 205000 × 1.15 / 108^{2} = 199 N/mm^{2}

F_{s1} = 0.005(608×10^{6}/(744{1-(97.5/199)^{2}}))×10^{-3}

F_{s1} = 5.4 kN

cl.9.12.5.2.3

F_{s2} = β (Δ_{e1} + Δ_{e2})σ_{fc} / ((σ_{ci} - σ_{fc})Σδ)

Δ_{e1} = Δ_{e2} = D/200 = 769.6 / 200 = 3.848

β = 1

Σδ = 2δ_{t} = 2 × 51×10^{-6} = 0.102 × 10^{-3}

σ_{ci} is to be determined using l_{e} from 9.6.4.1.1.2b).

l_{e} = πk_{2}(EI_{c}(δ_{e1}+δ_{e2})/L)^{0.5}

I_{c} = 25.4 × 268^{3} / 12 = 40.7 × 10^{6}

l_{e} = π × 1.0(205000 × 40.7 × 10^{6} × 2 × 51 × 10^{-6} / 9000)^{0.5} = 967mm

λ_{F} = (l_{e}/r_{y})(t_{f}/D) = (967/57.1)(25.4/769.6) = 0.559

ν = 0.993 (From Table 9)

λ_{LT} = l_{e}k_{4}ην/r_{y} = 967 × 0.9 × 1.0 × 0.993 / 57.1 = 15.1

σ_{ci} = π^{2}ES/λ_{LT}^{2}

σ_{ci} = π^{2} × 205000 × 1.15 / 15.1^{2} = 10205 N/mm^{2}

F_{s2} = 3.848 × 97.5 / ((10205 - 97.5) × 0.102 × 10^{-3}) × 10^{-3}

F_{s2} = 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 :

F_{s3} = Rd_{L}(Δ/D+θ_{L}tanα)/D

α = 0

R = 60 × 9 / 2 = 270 kN

d_{L} = 810 say (allow 40mm for depth of bearing)

F_{s3} = 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 F_{s4} = 0

cl.9.12.5.2.1

F_{s} = 5.4 + 0.4 + 1.4 + 0 = 7.2 kN

Allowing additional effect for wind load ( which is generally small compared
to F_{s} ) then

say F_{s} = 9 kN

Moment at base of stiffener = 9 × ( D - 1.5 × t_{f })

M = 9 × ( 769.6 - 1.5 × 25.4 ) × 10^{-3} = 6.6 kNm

Bending capacity of stiffener = Z_{xc}
× σ_{yc
}/ (γ_{m
}× γ_{f3})

M_{D} = 6.6 × 10^{6 } = Z_{xc}× 265 / ( 1.2 × 1.1 )

Z_{xc} = 32.9 × 10^{3} mm^{3}

cl.9.14.2.1(c)

Portion of web plate = 16t_{w}= 16 × 15.6 = 250mm

Z_{xc} = t_{stiffener} × 250^{2}/6 + 250 × 15.6^{2}/6

Hence t_{stiffener } = 2.2mm < 10mm therefore 10mm plate is satisfactory.

Use **762 × 267 × 197** kg/m UB
with 10mm thick end bearing stiffener.

cl.9.9.2.2

web thickness = 15.6mm

d_{we} = 685.8mm

λ = (d_{we}/t_{w})*(σ_{yw}/355)^{0.5}

λ = (685.8/15.6)*(265/355)^{0.5} = 38

From Figures 11 to 17 τ_{l}/τ_{y} = 1

Note: if λ < 56 then τ_{l}/τ_{y} = 1

Transverse web stiffeners will not improve the shear strength of the web.

τ_{l} = τ_{y} = σ_{yw}/1.732 = 153 N/mm^{2}

V_{D} = (t_{w}(d_{w} - h_{h})/(γ_{m} γ_{f3}))τ_{l}

d_{w} = D = 769.6

h_{h} = 0

γ_{f3} = 1.1 (Clause 4.3.3)

γ_{m} = 1.05 (Table 2)

V_{D} = ((15.6 * 769.6)/(1.05 * 1.1)) * 153 * 10^{-3} kN

**V _{D} = 1590 kN** ⟩⟩ 270 kN Hence Section OK

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