Question

A built up section (as shown) has an effective length of 5.5 m is to be used as a compression member.

The sectional properties are as following
$A=6293m{m}^2,b_f=100mm$
$t_f=15.3mm,t_w=8.6mm$
$r_{xx}=154.8mm,g=60mm$
$C_{yy}=24.2mm$
$I_{xx}=15082.8\times {10}^{4}m{m}^4$
$I_{yy}=504.8\times {10}^{4}m{m}^4$
$\begin{array}{cc}\lambda & {\sigma }_{ac}\left(N/m{m}^2\right)\\ 30& 145\\ 40& 139\end{array}$
The maximum compressive load that the member can support is ____ kN.

• A. 2925.43

Answer 1

The correct option is A 2925.43

$\text{CG from top},\overline{y}=\frac{(2\times 6293\times 216+500\times 16\times 8)}{(2\times 6293+500\times 16)}$

$\Rightarrow \overline{y}=135.168mm$

$I_{xx}=2(15082.8\times {10}^4+6293(216-135.168{)}^2)+\frac{500\times {16}^3}{12}+500\times 16\times (135.168-8{)}^2$

$=51343.48\times {10}^4{mm}^4$

$I_{yy}=2\times 504.8\times {10}^4+2\times 6293\times (150+24.2{)}^2+16\times \frac{{500}^3}{12}$

$=55869.29\times {10}^{4}m{m}^4$

Minimum radius of gyration =

$r=\sqrt{\frac{I_{xx}}{A}}=\sqrt{\frac{51343.48\times {10}^4}{20586}}=157.93mm$

$\lambda =\frac{{\ell }_{eff}}{r}=\frac{5.5\times {10}^3}{157.92}=34.82$

$\text{By interpolation},{\sigma }_{ac}=145-\frac{(145-139)}{40-30}\left(4.82\right)=142.1N/m{m}^2$

$\text{Allowable safe load}=142.1\times 20586=2925.43kN$