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


Design ultimate moment =

60 x 9.02 / 8

=

608 kNm


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

7tfo(355/σyf)0.5 =

7 x 19.6 x (355/265)0.5 = 159

 

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

le = 1.0 x 1.0 x 1.0 x 9000 =

 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


cl.9.9.1.2

MD = MRm 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 :
δt = 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 + Δe2fc / ((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(EIce1e2)/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

Moment at base of stiffener =

 9 x ( D - 1.5 x tf )

=

 9 x ( 769.6 - 1.5 x 25.4 ) x 10-3

=

 6.6 kNm

Bending capacity of stiffener =

 Zxc x σyc / (γm x γf3)

MD = 6.6 x 106 =

 Zxcx 265 / ( 1.2 x 1.1 )

Zxc =

 32.9 x 103 mm3

cl.9.14.2.1(c)

Portion of web plate = 16tw=

 16 x 15.6 = 250mm

Zxc =

 tstiffener x 2502/6 + 250 x 15.62/6

Hence tstiffener =

 2.2mm < 10mm therefore 10mm plate is satisfactory

 

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


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