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BUILT UP STEEL BEAMSPrepared by Ismail Ali

In designing built up sections , the minimum area of the section must be determined in order to beto the required section modulus (Zreq = M/0.6fy

Thickness of the web (t):

!he thic"ness of the #eb can be determined empirically from the follo#ing equation $ t = % & 'h must be ')/*h+

#here t is the #eb thic"ness in mmh is section depth in meter ) is the ma- shear force+ design shear strength of steel

r can be ta"en from the follo#ing table $

h(m 0. .*1 .1 .%1 * *.1 ' 2 1

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t(mm 63 3 0 43 0 03 * *3 2 2 23 6 63 *0 **3*2

h/t 003''

003*1

*1320

*1320

*1326

2' 163%

663%

*00 *0 3**%

!he minimum thic"ness of the #eb is ta"en t = 6 mm , and is increased in steps of * mm

Depth of the section (h) :

h = . 15Zreq/t ( / 0 333 /*0 7 for deflection control#here 7 is the beam length

Area of flange :

8rea of one flange = ' Zreq/2h for symmetric sections (identical upper and lo#er flanges

8rea of the t#o flanges = 'Zreq/*h for unsymmetrical sections

!hen 8fl = btfl

9here 8fl is area of flange b is #idth of flangetfl is thic"ness of flange

and b: '0 tfl

tfl t of #eb : *.1 t

Checks :

M/Z : +y) ;/I t : +

9hereZ is the actual section modulusI actual section moment of inertia

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; = 8fl - & 8# h#/

EXAMPLE

equired ;ection modulus = Zreq = M/0.6fy = *4'20000/0.6 *200 = *0'%1 cm'

!E"

hmin = 7/ 0 = */ 0 = .*mtmin = % &' h = % & ' .* = 0.6 mmta"e t = * mmh = . 15Zreq/t = . 15*0'%1/ .* = 24. cm ? 10 cmchoose beam depth = .1 m and #eb thic"ness = * mm

#lange

8fl = ' Zreq/2 h = ' *0'%1/2 10 = 0 . cm*tfl : *.1 t = *.1 * = '0 mmta"e tfl = *0 mm b * = 0 . b = 10 cm = 100 mmchoose flange section 100 *0 mm

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section properties

8fl = 00 cm*I = t h' / * & * 8fl @*I = 2'*%00 cm2Z = 2'*%00/%1 = 4 0* cm'

Check

M/Z = *4'20000/ 4 0* = 0 *.26 "g/cm* A +y = 0.6 *200 = 220 "g/cm* o"

; = 8fl - & 8# h#/ = 00 %2 & 26 .* 26/ = 014%.2 cm'

) ;/I t =4% 00 014%.2/ 2'*%00 .*=60*. "g/cm* A + =0.2 *200=460 "g/cm* o"