Question

An unequal angle section $200\text{mm}\times 150\text{mm}\times 15\text{mm}$ is to be used in a truss as a strut of effective length 4.5 m The cross - sectional properties of the section are as follows: Area of cross - section $=5025m{m}^2$
$I_{xx}=2\times {10}^{7}m{m}^4,I_{yy}=9.7\times {10}^{6}m{m}^4$ and $I_{xy}=-8.3\times {10}^{6}m{m}^4$
By using the table of permissible compressive stress given below, the load on the member should be kN.
$\begin{array}{ccccccccc}\lambda & 100& 110& 120& 140& 150& 160& 170& 180\\ {\sigma }_{ac}\left(Mpa\right)& 80& 72& 64& 51& 45& 41& 37& 33\end{array}$

• A. 251.74

Answer 1

The correct option is A 251.74

The principal moment of inertia may be calculated by
$I_{uu}/I_{vv}=\frac{I_{xx}+I_{yy}}{2}\pm \sqrt{{\left(\frac{I_{xx}-I_{yy}}{2}\right)}^2+I_{xy}^2}$
$=\frac{2\times {10}^7+0.97\times {10}^7}{2}\pm \sqrt{{\left(\frac{2\times {10}^7-0.97\times {10}^7}{2}\right)}^2+(-0.83\times {10}^7{)}^2}$
$=1.485\times {10}^7\pm \sqrt{0.265225\times {10}^{14}+0.6889\times {10}^{14}}$
$=1.485\times {10}^7\pm 0.977\times {10}^7$
$I_{uu}=(1.485+0.977)\times {10}^7=2.462\times {10}^{7}m{m}^4$
$I_{vv}=(1.485-0.977)\times {10}^7=0.508\times {10}^{7}m{m}^4$
$\therefore I_{min}=I_{vv}=0.508\times {10}^{7}m{m}^4$
$\therefore r_{min}=\sqrt{\frac{I_{min}}{A}}=\sqrt{\frac{0.508\times {10}^7}{5025}}=31.8mm$
Effective length of the strut or $l_{eff}=4.5m$
$\therefore \lambda =\frac{l_{eff}}{r_{min}}=\frac{4.5\times {10}^3}{31.8}=141.51$
$\Rightarrow {\sigma }_{ac}=51+\frac{45-51}{150-140}\times (141.51-140)$
$=50.094\text{MPa}$
$\therefore$ Load on the member
$={\sigma }_{ac}A$
$=50.094\times 5025\times {10}^{-3}$
$=251.72kN$