Question

Design the compressive strength of the column section formed by connecting 2 ISMC 400 section and a flat plate section 500 x 16 as shown in the figure below. Effective length of the member is 5.5 m. The distance of the neutral axis from the top is ____.
$\overline{y}=135.168mm.$


The relevant properties of ISMC 400 @ 481.6 N/m are

$\begin{array}{ccc}A=6293m{m}^2& b_f=100mm& t_f=15.3mm\\ t_w=8.6mm& r_z=154.8mm& g=60mm\\ I_x=15082.8\times {10}^{4}m{m}^4& I_y=504.8\times {10}^{4}m{m}^4& C_{yy}=24.2mm\end{array}$

Use the following table for buckling curve C

$\begin{array}{ccccccc}kL/r& 10& 20& 30& 40& 50& 60\\ f_{cd}\text{(MPa)}& 227& 224& 211& 198& 183& 168\end{array}$

• A. 3156 kN
• B. 5219 kN
• C. 4214 kN
• D. 4892 kN

Answer 1

The correct option is C 4214 kN

$\text{Area of channel section}=2\times 6293m{m}^2=12586m{m}^2$

Area of plate section
$=(500\times 16)=8000m{m}^2$

Effective area
$A_e=12586+8000=20586m{m}^2$

The distance of neutral axis from top is
$\overline{y}=135.168$ mm

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

$=51342.82\times {10}^{4}m{m}^4$

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

$=55869.22\times {10}^{4}m{m}^4$

$\text{As}{I}_x\text{is less than}{I}_y\text{, the minimum radius of gyration will be,}$

$r_z=\sqrt{\frac{51342.82\times {10}^4}{20586}}=157.92mm$

Effective slenderness ratio,
$\frac{L_{eff}}{r_z}=\frac{5.5\times {10}^3}{157.92}=34.82$

$For\lambda =34.82$, from design compressive stress table, using linear interpolation,

$f_{cd}=211-\left(\frac{211-198}{40-30}\right)\times (34.82-30)=204.734MPa$

Hence, Design compressive strength
$P_d=A_e{f}_{cd}$

$=20586m{m}^2\times 204.734N/m{m}^2$

$P_d=4214.65kN$