Question

Details of a double angle web member of a steel roof truss are given below. The member consists of 2-ISA 90 x 90 x 6 connected to opposite sides of 8 mm gussets using 20 mm bolts. Calculate the compressive strength of the member.

Effective length for buckling about XX axis = 1.7 m
Effective length for buckling about YY axis = 4.0 m

Yield strength of steel used = 250 MPa
Properties of single ISA 90 x 90 x 6 are
$Area=1047m{m}^2,\overline{y}=24.2mm$,
$I_{xx}=801000m{m}^4$

$\text{Allowable compressive stress}=\frac{0.60f_y{f}_{cc}}{(f_y^{1.5}+f_{cc}^{1.5}{)}^{1/1.5}}$
where
$f_{cc}=\text{Euler buckling stress}$

Answer 1

x-x axis:

$I_{xx}=2\times 801000m{m}^4$

$=1602000m{m}^4$

$r_{xx}=\sqrt{\frac{I_{xx}}{A}}$

$=\sqrt{\frac{1602000}{2\times 1047}}=27.66mm$

$\text{Let}\lambda \text{be the slenderness ratio}$

$\lambda =\frac{\text{effective length for buckling about xx-axis}}{r_{xx}}$
$=\frac{1.7\times {10}^3}{27.66}$
$=61.46$

y-y axis:

$I_{yy}=2\times 801000+2\times 1047\times {(24.2+\frac{8}{2})}^2$

$=3267232.56{mm}^4$

$\lambda =\frac{\text{Effective length for buckling about yy axis}}{r_{yy}}$

$=\frac{4\times {10}^3}{39.50}=101.26$

We will adopt maximum value of slenderness ratio

$\therefore \lambda =101.26$

$Now{f}_{cc}=\frac{{\pi }^{2}E}{{\lambda }^2}=192.51N/m{m}^2$

$\text{Allowable compressive stress}\left({\sigma }_{ac}\right)$

$=\frac{0.6f_y{f}_{cc}}{(f_y^{1.5}+f_{cc}^{1.5}{)}^{1/1.5}}$

$=\frac{0.6\times 250\times 192.51}{({250}^{1.5}+{192.51}^{1.5}{)}^{1/1.5}}$

$=81.87N/m{m}^2$

$\therefore$ Compressive strength of member

$=81.87\times 2\times 1047N$

$=171.43kN$