A compression member $AB($HE $400B)$ is pinned at both ends and braced in the weak direction as shown in the figure. Material and geometrical properties are given below. For the given loading and ASD load combination $),$ determine whether the section is satisfactory or not.

$15\%$

Material:

$S355Fy=355$MPa,$Fu=510$MPa

Section properties $($HE $400B)$:

A: $19780m{m}^2$

d: $,400mm$

tr$.24mm$

b: $,300mm$

$E=200000$MPa

h: $,298mm$

$t{w}_w$:$,13\mathrm{.}5mm$

ix: $,170\mathrm{.}8mm$

iy: $,74mm$

solve this question with the solution methods applied in another question in the image below

COMrassion MEMARKs

Local buckling timit state.

Por nange, $($Table $5\mathrm{.}1.,$ Case $1)$

Hence, flange asd web are NoT, Iender.

$,$ Critical stress,

Effective leagth factor, $K,$ for pin$-$ended member, $K=1\mathrm{.}0.$ Therefore,

$\frac{L_i}{i_4}=\frac{K_4{L}_4}{{}_4}=\frac{(1\mathrm{.}0)(9000)}{191\mathrm{.}4}=47\mathrm{.}02$

Note that,

$L_{,\text{a}\text{n}\text{d}}{L}_{,}$ are determined depending $on$

the locations $of$ the lateral braces and

$\frac{L_0}{t_2}=\frac{K_i{L}_2}{b_h}=\frac{(1\mathrm{.}0)(4890)}{73\mathrm{.}3}=61\mathrm{.}39$

If $\frac{L_{}}{l_{,\text{w}\text{i}\text{l}\text{l}}}{P}_n=F_e{A}_f=(267\mathrm{.}32)(21800){10}^{-1}=5828kN{P}_r=\frac{P_1}{{}_i}=\frac{5828}{1\mathrm{.}67}=34906N{P}_1=B_0+B_0=350+2400=3250kN\frac{P_s}{P_r}=\frac{3250}{3490}=0\mathrm{.}931\mathrm{.}00\mathrm{.}93-1\mathrm{.}0\frac{L_{ca}}{l_q}$ then$\frac{L_{}}{l_{,\text{w}\text{i}\text{l}\text{l}}}$ control the strength.

CoMPREs$5$ION MEM Bexs

Thenominal axial compressive strength,

$P_n=F_e{A}_f=(267\mathrm{.}32)(21800){10}^{-1}=5828kN$

The allowable axial compressive strength

$P_r=\frac{P_1}{{}_i}=\frac{5828}{1\mathrm{.}67}=34906N$

The required strength,

$P_1=B_0+B_0=350+2400=3250kN$

Cheek $if$ the member has adequate strength.

$\frac{P_s}{P_r}=\frac{3250}{3490}=0\mathrm{.}931\mathrm{.}0$

$If0\mathrm{.}93-1\mathrm{.}0$ then, the member $is$ satisfactary