Page: 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/mm$^{2}$ E=2$\times10^{5}N/M^{2}$
0

ISMB 350

Area A=$6671mm^{2}$ bd140 tf=142 tw=81

rxx=r28=142.9

ryy=28.4

length of column =3000mm

leff=$k\times L=1\times 3000=3000m$

fy=250N/mm$^{2}$

fa=410N/mm$^{2}$

E=2$\times10^{5}N/mm^{2}$

Design strength of comp member

Pd=$Ae\times fcd$

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

$\frac{h}{bf}=\frac{350}{140}=2.5$

${h}{bf}=2.5\geq1.2$

tf=1.42$\lt 40$

huckling about zz axis class a

buckling about yy axis class b

$\alpha=0.21$ class a

$\alpha=0.34$ class b

$\alpha$=0.21

$\lambda_{z}=\sqrt{\frac{fy\times(\frac{kl}{\gamma zz)^{2}})}{\Pi^{2} E}}$

=$\sqrt{\frac{250\times(2000/28.4)^{2}}{\Pi^{2}\times2\times 10^{5}}}$

$\lambda y=1.189$

$\phi =0.5[1+0.34(1.189-0.2)+(1.189)^{2}]$

$\phi$=1.375

fcd=$\frac{250/1.1}{0.53+[0.53^{2}-0.2362^{2}]^{0.5}}$

fcd=225.2N/mm$^{2}$

pd=Ae$\times$fcd

6671$\times225.3$

=1.502$\times10^{6}$

=15.03$\times10^{3}N$

=1503KN

$\lambda=0.34$

$\lambda y=\sqrt{\frac{fy\times(\frac{kl}{rmm})^{2}}{\Pi^{2}E}}$

$=\sqrt{\frac{250\times(2000/rmm)^{2}}{\Pi^{2}\times2\times10^{5}}}$

$\lambda y$=1.189

$\phi=0.5[1+0.34(1.189-0.2)+(1.189)^{2}]$

$\phi$=1.375

fcd=$\frac{250/11}{1.375+(1.375^{2}-1.189^{2})^{0.5}}$

fcd=110N/mm$^{2}$

pd=110$\times$6671

pd=733.7 KN

design load of column is 733.7 KN

$buckling \ \ z \ \ \ Class \ a$

$\alpha=0.21 \ \ \ \ class \ \ a$

$\frac{kl}{rxx} \ \ \ \ fcd$

20 $\ \ \ \$ 226

21$\ \ \ \$fcd

30$\ \ \ \$220

$\frac{30-20}{30-21}=\frac{220-223}{220-fcd}$

fcd=225.3 N/mm$^{2}$

buckling yy $\ \ \$ class b

$\frac{KL}{ryy} \ \ \ \ \ fcd$

100 $\ \ \ \$ 118

106 $\ \ \ \$ fcd

110 $\ \ \ \$ 104

fcd=110N/mm$^{2}$

page ddss(76) • 96 views