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 soluton 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.


gconc. = 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

gfL steel = 1.05

 

gfL concrete = 1.15

Live Load

gfL 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 sy = 265 N/mm2

Approximate modulus required =

M x gm x gf3 / sy

=

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/syw)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/syf)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 syc

Mpe = 3.677 x 106 x 265 x 10-6
       = 974kNm

cl.9.7.2

lLT = le k4 h n / ry

k4 = 0.9

h = 0.94 (From Fig. 9(b): MA/MM = MB/MA = 0)

lF = le/ry(tf/D)

lF = 9000/49.6 (19.6/611.9) = 5.81

i = Ic/(Ic+It)

i = 0.5

n = 0.78 (from Table 9)

lLT = 9000 x 0.9 x 0.94 x 0.78 / 49.6 = 120


cl.9.8.

Limiting moment of resistance: lLT ((syc/355)(Mult/Mpe))0.5

Section is compact, hence Mult = Mpe = 974kNm

lLT ((syc/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 = MR / gm gf3

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

lF = 5.2

n = 0.81

lLT = 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) : lLT((syc/355)(Mult/Mpe))0.5 = 110
giving lLT = 127

lLT = le k4 h n / 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-(60EtfbdtRv/(W[L/ry]3n4 ))]

Rv/W = 0.5 (load causing max moment in beam)

E = 205 000  (cl.6.6)

tf = 25.4

b = 1.0

n = 0.81

L/ry = 9000 / 57.1

Hence max dt = 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 (lLT = 0).


 

Deflection of cantilever :
dt = 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 
dt = 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-(sfc/sci)2}))

df = 769.6 - 25.4 = 744mm

sfc = M / Zxc = 608x106 / 6.234x106 = 97.5 N/mm2

sci = p2ES/lLT2

S = Zpe/Zxc = 7.167 / 6.234 = 1.15

sci = p2 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 = b(De1 + De2)sfc / ((sci - sfc)Sd)

De1 = De2 = D/200 = 769.6 / 200 = 3.848

b = 1

Sd = 2dt = 2 x 51x10-6 = 0.102 x 10-3

sci is to be determined using le from 9.6.4.1.1.2b).

le = pk2(EIc(de1+de2)/L)0.5

Ic = 25.4 x 2683 / 12 = 40.7 x 106

le = p x 1.0(205000 x 40.7 x 106 x 2 x 51 x 10-6 / 9000)0.5 = 967mm

lF = (le/ry)(tf/D) = (967/57.1)(25.4/769.6) = 0.559

n = 0.993 (From Table 9)

lLT = lek4hn/ry = 967 x 0.9 x 1.0 x 0.993 / 57.1 = 15.1

sci = p2ES/lLT2

sci = p2 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(D/D+ qLtana)/ D

a = 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 syc / (gm x gf3)

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

 

l = (dwe/tw)*(syw/355)0.5

 

l = (685.8/15.6)*(265/355)0.5 = 38

 

From Figures 11 to 17 tl/ty = 1

 

Note: if l < 56 then tl/ty = 1

 

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

 

tl = ty = syw/1.732 = 153 N/mm2

VD = (tw(dw - hh)/(gm gf3))tl    

dw = D = 769.6
hh = 0
gf3 = 1.1 (Clause 4.3.3)
gm = 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