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Determine the design axial load on the column section ISMB 350 given that height of column 3m and that is pin ended assume following fy=250mpa fe 410N/mm2 E=2×105N/M2
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ISMB 350

Area A=6671mm2 bd140 tf=142 tw=81

rxx=r28=142.9

ryy=28.4

length of column =3000mm

leff=k×L=1×3000=3000m

fy=250N/mm2

fa=410N/mm2

E=2×105N/mm2

Design strength of comp member

Pd=Ae×fcd

buckling class(Is pg 44)Is pg 35 table 7

hbf=350140=2.5

hbf=2.51.2

tf=1.42<40

huckling about zz axis class a

buckling about yy axis class b

α=0.21 class a

α=0.34 class b

buckling about z axis a

α=0.21

λz=fy×(klγzz)2)Π2E

=250×(2000/28.4)2Π2×2×105

λy=1.189

ϕ=0.5[1+0.34(1.1890.2)+(1.189)2]

ϕ=1.375

fcd=250/1.10.53+[0.5320.23622]0.5

fcd=225.2N/mm2

pd=Ae×fcd

6671×225.3

=1.502×106

=15.03×103N

=1503KN

buckling about class b

λ=0.34

λy=fy×(klrmm)2Π2E

=250×(2000/rmm)2Π2×2×105

λy=1.189

ϕ=0.5[1+0.34(1.1890.2)+(1.189)2]

ϕ=1.375

fcd=250/111.375+(1.37521.1892)0.5

fcd=110N/mm2

pd=110×6671

pd=733.7 KN

design load of column is 733.7 KN

buckling  z   Class a

α=0.21    class  a

klrxx    fcd

20      226

21    fcd

30    220

30203021=220223220fcd

fcd=225.3 N/mm2

buckling yy     class b

KLryy     fcd

100      118

106      fcd

110      104

fcd=110N/mm2

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