Detemine factor Udl on laterally unsupported beam ISLB 300 of span 3.25m s.s at the ends

Given ISLB 300

D=300 bw=150 tf=9.4

tw=6.7 $\gamma_{1}$=15

Ze=488.9$\times 10^{3}$

Iz=7332.9$\times 10^{4}mm^{4}$

Iy=376.2$\times 10^{4}mm^{4}$

Zp=554.32$\times 10^{3}mm^{4}$-- Is code

span length =3250mm

d=300-2$\times9.4-2\times 15$=251.2

1) sectional classification

$\frac{b}{tf}=\frac{bf/2}{tf}=7.98 \lt$ 9.4 $\epsilon$

$\frac{d}{tw}=\frac{D-2tf-2R_{1}}{6.7}=\frac{251.2}{6.7}=37.49\lt 84 \epsilon$

section is plastic

Required factor Udl=w kN/m

md=Bd$\ast$ Zp$\ast $fdb

1) IT=$\sum \frac{blti^{3}}{3}$

=2[$\frac{150\times 9.4^{3}}{3}]+[\frac{hf \times t^{3}}{3}] $hf= D-1 thick flange

=2[$\frac{150\times 9.4^{3}}{3}]+[\frac{29.06+6.7^{3}}{3}]$=300-9.4=290.6

It=112.19$\times10^{3}$

Iw=(1-Bf)Bf$\ast Iy \ast hr^{2}$

=(1-0.5)0.5$\ast$376.2$\times$10$^{4}\ast $290.6$^{2}$

Iw=7.94$\times 10^{10}$

Eg=2g(1+u)

2$\times10^{5}$=2g(1+0.3) $\mu$=0.3

G=76.92$\times 10^{3}$ mpa

LLT=3250mm eff length=3250mm

mcr=$\sqrt{\frac{\pi^{2}\times(2\times10^{5}\times376.2\times10^{4})}{(3250)^{2}}[76.92\times 10^{3}\times112.19\times10^{3}+\frac{\pi^{2}\times2\times10^{5}\times7.94\times10^{10}}{(3250)^{2}}}$

=$\sqrt{838.47+153192.47}$

mcr=128.45$\times10^{6}$N.mm

laterally buckling movement

$\lambda LT=\sqrt{Bb,Zp,fy/mcr}\leq \sqrt{1.2\ast Ze\ast fy/mcr}$

=$\sqrt{\frac{0.5\times(554.32\times10^{3})\times 250}{128.45\times10^{6}}} \leq \sqrt{\frac{1.2\ast 488.9\times10^{3}\times250}{128.45\times10^{6}}}$

$\lambda LT=1.038\leq 1.06$ safe

$\Phi=0.5[1+\alpha LT(\lambda LT-0.2)+\lambda LT^{2}]$

=0.5[1+0.21(1.038-0.2)+1.038$^{2}]$

=0.5[1+0.21(1.038-0.2)+1.038$^{2}$]

$\Phi_{Lt}$=1.126

X$_{LT}=\frac{1}{\{\Phi_{LT}+[\Phi_{LT^{2}-\lambda^{2}LT]^{0.5}}\}}\leq$1

=$\frac{1}{[1.126+[1.126^{2}-1.038^{2}]^{0.5}]}$

$x_{LT}$=0.64

fbd=$\frac{xLD\times fy}{\gamma mo}$

=$\frac{0.64\times 250}{1.1}$

fbd=145.46mpa

md=B.P .ZP. fbd

=1$\times554.32\times10^{3}\times145.46$

md=80.63$\times 10^{6}$

80.63$\times 10^{6}=\frac{w\times 3250 ^{2}}{8}$

w=61.07kN/m

alternative method

$\frac{kL}{\gamma min}=\frac{3250}{28 \ steel tale}$=116.19

kl/$\gamma \ min$ h/tf=31.9

110

116.19

120

$\frac{kl}{\gamma min} \frac{h}{tf}$=30 $ \ \ \ $ 31.91 $ \ \ \ \ $ 35

110 $ \ \ \ \ \ $ 232.1 $ \ \ \ \ $ x $ \ \ \ \ \ $ 219.3

120 $ \ \ \ \ \ $ 202.4 $ \ \ \ \ $ y $ \ \ \ \ \ $190.1

1) 30 $ \ \ \ \ $ 232.1

31.91 $ \ \ \ \ \ $ x

35 $ \ \ \ \ \ \ \ \ \ $ 219.3

x=227.236

30 $ \ \ \ \ \ \ \ \ \ $ 202.4

31.41 $ \ \ \ \ \ \ \ $ y

35 $ \ \ \ \ \ \ \ \ \ $ 190.1

y=197.26

110 $ \ \ \ \ \ \ \ \ \ $ 227.23

116.19 $ \ \ \ \ $ tcrb

120 $ \ \ \ \ \ \ \ \ \ $197.26

fcrb=208.68

fcrb $ \ \ \ \ \ \ \ \ \ $ fdb

200 $ \ \ \ \ \ \ \ \ $ 134.1

208.68 $ \ \ \ \ \ \ $ x

250$ \ \ \ \ \ \ \ \ \ \ \ $152.3

fbd=137.26

md=Bd$\times ZP\ast $ fdb

md=76.08$\times 10^{6}$