Question

ISA 100 x 100 x 10mm (Cross-sectional area = 1908 mm${}^2$) serves as tensile member. This angle is welded to a gusset plate along A and B appropriately as shown. Assuming the yield strength of the steel to be 260 N/mm${}^2$ the tensile strength of the member can be taken to be approximately _______.

• A. 225 kN
• B. 300 kN
• C. 375 kN
• D. 500 kN

Answer 1

The correct option is A 225 kN

Shear lag reduces the effectiveness of outstanding leg of tension member.
hence,
$A_{net}=A_1+k{A}_2$
$A_1\to \text{net area of connected leg.}$
$A_2\to \text{net area of outstanding leg.}$


$A_1=(100-\frac{10}{2})\times 10=950m{m}^2$

$A_2=(100-\frac{10}{2})\times 10=950m{m}^2$

$k=\frac{3A_1}{3A_1+A_2}=0.75$

$A_{net}=950+0.75\times 950$

$=1662.5m{m}^2$

Tensile strength of the member

$T_{dn}=A_{net}\times f_{permissible}$

$=1662.5\times 0.6\times 260$

= 259.35 kN

Note:
$Permissiblestress=0.6f_y$